# ATAR Notes: Forum

## VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: FlorianK on March 30, 2013, 08:10:52 pm

Post by: FlorianK on March 30, 2013, 08:10:52 pm

If you have general questions about the VCE Physics course or how to improve in certain areas, this is the place to ask!

Everyone is welcome to contribute; even if you're unsure of yourself, providing different perspectives is incredibly valuable.

Please don't be dissuaded by the fact that you haven't finished Year 12, or didn't score as highly as others, or your advice contradicts something else you've seen on this thread, or whatever; none of this disqualifies you from helping others. And if you're worried you do have some sort of misconception, put it out there and someone else can clarify and modify your understanding!

There'll be a whole bunch of other high-scoring students with their own wealths of wisdom to share with you, including TuteSmart tutors! So you may even get multiple answers from different people offering their insights - very cool.

To ask a question or make a post, you will first need an ATAR Notes account. You probably already have one, but if you don't, it takes about four seconds to sign up - and completely free!

OTHER PHYSICS RESOURCES

Original post.
Title: Re: VCE Physics Question Thread!
Post by: ~T on April 01, 2013, 12:08:12 am
Because I don't "immerse" myself or believe that "beauty" is an appropriate term regarding the subject, can I still post here? ;)

Seriously though, seconding a question that someone started as a thread, how much do we need to know about diodes? I'm confident with their applications and circuit questions, but what about the mechanics? No idea whether to spend time learning the workings of a diode or not.
Similarly, the Heinemann textbook is all like "Transistors aren't covered in this course" and now it's showing me a 'physics in action' segment that is telling me (very poorly) how a transistor works. I think I already understand them, but do I need to for the course?

And lastly, question 4 of chapter 4.1. I swear it can't actually happen. Unless I'm missing something... but it seems that when there is a grounded point on the circuit, there are two different currents at different points on the circuit. The answers say so too. Now as I'm typing I realise I'm not even sure about the point/function of a ground point. I know it's a reference point for the measurement of voltage, but does it affect the flow of current, or the circuit, in any way?

Many questions, many thanks in advance :)
Title: Re: VCE Physics Question Thread!
Post by: b^3 on April 01, 2013, 01:10:03 am
Currents will always flow towards the lowest potential (i.e. in the context of VCE physics, the lowest 'voltage'), which in most cases is the ground, since it's 0 V.  Here though we have a voltage that is lower than the ground, that is -3 V, as in this case our ground is just a reference. For a, there is 5 V dropped across the first resistor, so I=V/R=5/5K=1 mA. For the second resistor there is 3 V dropped over the 2K resistor, so I=3/2k=1.5 mA.

As to the two different currents part, the second current is greater than the first current, that is because there will be some current leaking from the ground, as again, the current will flow towards the lower voltage. So from 0 V to -3 V, i.e. that is where your extra current is coming from.

For b, if we wanted to find the voltage dropped over each resistor, we would use a voltage divider, but we want the current through each, which is the same through each. So 8 V is dropped over 7 K worth of resistance, that is I=8/7K =1.14 mA.

As for the direction, the will all be flowing in the direction P, from higher potential to lower potential.

Hope that helps.
Title: Re: VCE Physics Question Thread!
Post by: ~T on April 01, 2013, 11:40:49 am
Thank you both! I think I'm fine with diodes then, if that's all we need.

As for the whole earthing thing. If there is some current that is leaking from the ground in this case, why don't we have to take that into consideration for other circuits? Or do we? Like, for simple circuit analysis where there is a circuit with a battery, a resistor, and ground at some point, can we not calculate the current simply because there is the ground??? I think ground just confuses me, I thought its only purpose was as a reference point :/
Title: Re: VCE Physics Question Thread!
Post by: EspoirTron on April 01, 2013, 11:56:25 am
Can someone please explain to me how an LDR works and how to do calculations involving LDR's. Also, can someone explain it using a ratio method, instead of 'regular formula' way. Anyone's help would be greatly appreciated!  :)
Title: Re: VCE Physics Question Thread!
Post by: b^3 on April 01, 2013, 03:07:42 pm
Thank you both! I think I'm fine with diodes then, if that's all we need.

As for the whole earthing thing. If there is some current that is leaking from the ground in this case, why don't we have to take that into consideration for other circuits? Or do we? Like, for simple circuit analysis where there is a circuit with a battery, a resistor, and ground at some point, can we not calculate the current simply because there is the ground??? I think ground just confuses me, I thought its only purpose was as a reference point :/
Normally the ground is the lowest reference, which is why we can do what we normally do with other circuits. But since there is a potential that is lower than this reference in this case, we have to look at the situation differently, as we could find the two currents in the two resistors. But as the second was greater than the first, the current had to come from somewhere (again current flows from higher potential to lower potential, so that's why some current was drawn from there). In the normal case, where the lowest potential is the ground, the currents are flowing towards this direction, and so we can look at the problem normally.

It's a little confusing, VCE physics doesn't explain it well (there was a fair bit of further electronics that I didn't fully understand (but was able to do the problems) until I got to uni and did an electrical engineering unit). Hope that makes sense.
Title: Re: VCE Physics Question Thread!
Post by: Homer on April 08, 2013, 08:53:48 pm
could someone help me with the first two
Title: Re: VCE Physics Question Thread!
Post by: FlorianK on April 08, 2013, 09:34:24 pm
could someone help me with the first two
let IB1 be the voltage drawn from the battery 1 and IB2 be the voltage drawn from battery 2

IR1 = IB1
IR2 = IR3 = 0.5 * IB1
IR4 = 2/3 * IB2
IR5 = IR6 = 1/3 * IB2

Ok, since both circuits will have different resistances IB1=/= IB2

The total resistance of circuit 1 is: [Rx is the resistance of any resistor in the circuit]

Rtotal = Rx + 1/2 * Rx     [When 2 resistors of the same value are in parallel the resitance of this parallel circuit is half the resitance of each of them]

--> Rtotal = 1.5 * Rx

Ok the other one is a parallel circuit. The equation you should use for a parallel circuit, with 2 components is:
$\frac{R1*R2}{R1+R2}$

So here R1 is R4 so Rx and R2 is R5+R6 so 2*Rx
--> R1=Rx R2=2*Rx
we get:
$\frac{Rx*2Rx}{Rx+2Rx}$ = 2/3 * Rx

So now to the current

IB1=V/(3/2*Rx) = 2/3 V/Rx
IB2=V/(2/3*Rx) = 3/2 V/Rx

So now putting 'values' into the equations from the top

IR1 = IB1 = 2/3 * V/Rx
IR2 = IR3 = 0.5 * IB1 = 1/3 V/Rx
IR4 = 2/3 * IB2 = V/Rx
IR5 = IR6 = 1/3 * IB2 = 1/2 V/Rx

So we see we have the highest Current in R4 and the lowest current in R5 and R6.

Hope I didn't make some stupid mistake :p
Title: Re: VCE Physics Question Thread!
Post by: Homer on April 08, 2013, 09:48:51 pm
oooo, i see! thanks a bunch FlorianK
Title: Re: VCE Physics Question Thread!
Post by: LazyZombie on April 10, 2013, 09:42:47 pm
as a rollercoaster goes through a dip, which of the two cars, front or back, will be the fastest or will they have the same speed? Explain your answer.
Title: Re: VCE Physics Question Thread!
Post by: Homer on April 10, 2013, 09:55:04 pm
im just guessing the second one since the first has to pull the second down, slowing itself down a lil?
Title: Re: VCE Physics Question Thread!
Post by: LazyZombie on April 11, 2013, 09:22:10 pm
Another:
Explain what happens to your body on corners using at least one of Newtons Laws.
What points would I have to cover?
I wrote: There is no force acting on the body initially so it conforms to newtons first law, continuing to move in the original forward motion so there is a feeling of being 'pushed' outwards from the corner.
Plus a little birds eye diagram of a roller coaster rounding a corner and an arrow pointing tangent to the curve labelled velocity. (right thing to label?)
Title: Re: VCE Physics Question Thread!
Post by: Homer on April 16, 2013, 07:28:29 pm
in a force distance graph for satellite, is the kinetic energy greater at a greater distance or is the kinetic energy greater at a smaller distance
Title: Re: VCE Physics Question Thread!
Post by: LazyZombie on April 16, 2013, 08:46:28 pm
in a force distance graph for satellite, is the kinetic energy greater at a greater distance or is the kinetic energy greater at a smaller distance
kinetic energy is greater at a smaller distance.
by finding the area of the force distance graph, you can find the work done and so find the kinetic energy
Title: Re: VCE Physics Question Thread!
Post by: Robert123 on April 17, 2013, 04:05:36 pm
Why can you have negative voltage (AC) ? Doesn't that imply we can have "negative" energy? Wouldn't it be best if we have negative and positive current for saying which way the current flow?
Also;
I lost a mark on a SAC with the question being about explainging how seatbelt and crumpling reduce the severity of the drivers injury, it was a 3 mark question. What type of exam "style" answers should I have. The teacher noted on it "how does the seat belt increase the time of collision?", how would I undergo answering that for an 'exam' quality answer?
Thanks :)
Title: Re: VCE Physics Question Thread!
Post by: LazyZombie on April 17, 2013, 08:46:40 pm
Why can you have negative voltage (AC) ? Doesn't that imply we can have "negative" energy? Wouldn't it be best if we have negative and positive current for saying which way the current flow?
Also;
I lost a mark on a SAC with the question being about explainging how seatbelt and crumpling reduce the severity of the drivers injury, it was a 3 mark question. What type of exam "style" answers should I have. The teacher noted on it "how does the seat belt increase the time of collision?", how would I undergo answering that for an 'exam' quality answer?
Thanks :)
I'm not sure about the voltage thing but since voltage is relative, I think it is possible. But I'm prettyy sure its not necessary in the course. (not in the study design) I might get back to this later :)

As for the crumple zone and seatbelt
$F=\frac{\Delta p}{t}$ therefore the crumble zone increases the time of the collision, thus decreasing the force.
Inertia causes the body to continue moving in the original motion of the car in a collision and cause injury or fatality, and seatbelts help to prevent this from happening.

Seatbelts don't increase the time of collision. I think that's what your teacher means. (?)
Title: Re: VCE Physics Question Thread!
Post by: availn on April 17, 2013, 08:53:02 pm
Put simply, if voltage is negative, then the current is also negative. As P = VI, the negative cancels out.

And yeah, seat belts aren't crumple zones, they don't increase the time of collision, they just make sure you don't shoot out of your car.
Title: Re: VCE Physics Question Thread!
Post by: Lasercookie on April 17, 2013, 10:45:43 pm
Ok, thanks for that. I will ask my teacher what her opinion of what the answer "should" be (even though I find people on this forum more knowledgable and reliable with answer).
Also, another question (this is more a general education question).
Does the only thing that matters with SACs is your ranking? Please only answer if you are 100% sure due to reading a report from VCAA or talking to an examiner?
If so, then does that means there is no different between averaging  90% and 100% on SACs if you go to a low scoring school?
Split this post and moved it to the Technical Score Discussion boards
Title: Re: VCE Physics Question Thread!
Post by: [paradox]+ on April 25, 2013, 04:41:12 pm
Jasper has a transistor radio with ﬂat batteries. Suggest a reason to explain why the sound is distorted when he tunes in to a strong radio signal.

Title: Re: VCE Physics Question Thread!
Post by: ~T on April 26, 2013, 05:16:06 pm
Jasper has a transistor radio with ﬂat batteries. Suggest a reason to explain why the sound is distorted when he tunes in to a strong radio signal.

I'm not entirely sure on this one but I'll give you my thoughts...

Basically, any amplification system works of a small varying AC signal (the radio signal here) and a larger DC input (the battery here) that amplifies the input signal to create a much larger output (the sound here).

So, if the batteries are running flat, then they will not supply enough power to the transistor(s) in order to amplify the larger voltages, and the linear amplification region may be smaller. The input voltages will still be the same, but the extreme values will no longer be amplified linearly, distorting the signal.

This only half makes sense in my head. But pretty much, there is no reason that the input signal would be larger (resulting in clipping outside of the region) so it must be that the linear region itself becomes smaller.
Title: Re: VCE Physics Question Thread!
Post by: Alwin on April 26, 2013, 09:11:33 pm
Jasper has a transistor radio with ﬂat batteries. Suggest a reason to explain why the sound is distorted when he tunes in to a strong radio signal.

For this one, I decided to add an image! just because.. I can haha

(http://www.electronics-tutorials.ws/amplifier/amp13.gif)

Quote
The input signal may be too large, causing the amplifier to be limited by the supply voltage.

This is true in the case of Jasper and his poor flat batteries. The DC power supply is the two parallel lines, where the voltage supplied is +Vcc and grounded at 0V. I won't go into the specifics of the transistor, since it is no longer on the study design, but simply the "in" signal is the current that enters the middle of the transistor and allows the larger current to flow through (top to bottom) from the DC power source.
Hence, if the DC source is not large enough then it cannont properly amplify the "in" signal because it cannot amplify a strong signal, ie it cannot amplify the strong +ve and strong -ve signals, ie smaller clipping range

Hope that makes sense!
Title: Re: VCE Physics Question Thread!
Post by: Robert123 on April 27, 2013, 03:04:38 pm
What does a negative gain amplifier actually achieve?
I know it inverts the Vout graph but how does that have a meaningful purpose? What would be the difference in a microphone that is hooked up to an amplifier with a negative gain rather than a positive one?
Thanks :)
Title: Re: VCE Physics Question Thread!
Post by: Alwin on April 27, 2013, 03:19:04 pm
What does a negative gain amplifier actually achieve?
I know it inverts the Vout graph but how does that have a meaningful purpose? What would be the difference in a microphone that is hooked up to an amplifier with a negative gain rather than a positive one?
Thanks :)

There actually is no "purpose" for wanting negative gain. It is just a side effect of using a simple single transistor amplifier, the npn transistor inverts the input signal. If you want a non-inverting amplifier, you need a two-stage amplifier (two transistors). That's the only reason why a lot of amplifiers in physics questions have negative gain since we deal with simple amplifiers.

As for the difference, well there is none. Remember when a microphone or speaker works, it is a diaphragm that vibrates back and forth creating a sound wave. If the initial current is positive, then the diaphragm moves forward, then back, then forward etc. But, if the signal had been inverted, initially the current flows in the other direction and so the diaphragm moves back first then forwards then back again if you get what I mean. So, the frequency of sound produced / recorded it the exact same.

Hope that clears things up for you!
Title: Re: VCE Physics Question Thread!
Post by: Robert123 on April 27, 2013, 03:42:51 pm
Thank you for that,  clears things up a lot :)
Title: Re: VCE Physics Question Thread!
Post by: Homer on May 02, 2013, 07:08:52 pm
Hey guys how would you calculate the value of R?

Thanks ;D

ANS: 309.1ohms
Title: Re: VCE Physics Question Thread!
Post by: sin0001 on May 02, 2013, 07:48:45 pm
For part b of the attached question, I get that the resistor, parallel to the LED, will have the same potential difference as the LED-2.5 V, so doesn't that mean the Resistor should have the same current flowing through it as the LED-11 mA; so can't we then use these values for current & voltage, to work out the resistance of Resistor-'R'?
But in the worked solutions they haven't assumed that current across 'R' will be same as I(LED), instead this is what how they solved resistance:

Vtherm. = 10 – 2.5 = 7.5 V, Itherm. = 7.5/500 = 0.015 A
since ILED = 0.011, IR = 0.004 A and RR = 2.5/0.004 = 625 Ω
Title: Re: VCE Physics Question Thread!
Post by: Alwin on May 02, 2013, 08:04:22 pm
Hey guys how would you calculate the value of R?

Thanks ;D

ANS: 309.1ohms

Well, first you look at the diodes:
Diode 1:
Since the current is 20 mA, V = 1. Hence, R = 1/0.020 = 50 ohm. Thus, you have a 50ohm 'resistor' and a 500 ohm resistor in parallel
$\text{effective resistance of vertical branches} = 45.45 \Omega$

Diode 2:
This diode is in reverse bias, hence no current flows. There are only the 2 100ohm resistors in parallel
$\text{effective resistance of horizontal branches} = 50 \Omega$

Total resistance:
$R_T=45.45+50+50+R$

Total current:
$I_T = 0.022 \text{A} \text{ (through parallel circuits ratios)}$

From Ohm's law, R = V/I
$45.45+50+50+R = \frac{10}{0.022}$
$R=309.1 \Omega$

Hope this makes sense!
Title: Re: VCE Physics Question Thread!
Post by: Alwin on May 02, 2013, 08:08:37 pm
For part b of the attached question, I get that the resistor, parallel to the LED, will have the same potential difference as the LED-2.5 V, so doesn't that mean the Resistor should have the same current flowing through it as the LED-11 mA; so can't we then use these values for current & voltage, to work out the resistance of Resistor-'R'?
But in the worked solutions they haven't assumed that current across 'R' will be same as I(LED), instead this is what how they solved resistance:

Vtherm. = 10 – 2.5 = 7.5 V, Itherm. = 7.5/500 = 0.015 A
since ILED = 0.011, IR = 0.004 A and RR = 2.5/0.004 = 625 Ω

I think you got a bit confused between parallel and series circuits. In parallel circuits, the potential difference across both branches is the same, BUT the current in each branches is different. The sum of the currents in each of the branches is equal to the total current, in this case 0.015A. This is different from series circuits.

Hopefully you can see how the solutions got the answer from here!
Title: Re: VCE Physics Question Thread!
Post by: sin0001 on May 02, 2013, 08:16:53 pm
I think you got a bit confused between parallel and series circuits. In parallel circuits, the potential difference across both branches is the same, BUT the current in each branches is different. The sum of the currents in each of the branches is equal to the total current, in this case 0.015A. This is different from series circuits.

Hopefully you can see how the solutions got the answer from here!
Ahhh yeah I had a mental blank lol
Also, how would you work out the resistance of these diodes, would you just plug it into the formula: R = V/I?
Thanks!
Title: Re: VCE Physics Question Thread!
Post by: Alwin on May 02, 2013, 08:53:31 pm
Ahhh yeah I had a mental blank lol
Also, how would you work out the resistance of these diodes, would you just plug it into the formula: R = V/I?
Thanks!

It's okay! just try not to have a mental blank in the exam ;)

Yeah, just R = V/I. I have an example in a previous post:
Title: Re: VCE Physics Question Thread!
Post by: Homer on May 02, 2013, 09:36:09 pm
Thanks :)
Title: Re: VCE Physics Question Thread!
Post by: jono88 on May 03, 2013, 05:47:40 pm
What is the difference between tensile strength and yield strength?
Title: Re: VCE Physics Question Thread!
Post by: Phy124 on May 03, 2013, 07:31:31 pm
Ultimate tensile strength is the maximum stress a material can withstand before failing under tension.

Yield strength is the maximum stress which a material can withstand before it begins to deform plastically, rendering it unable to return to its original state.
Title: Re: VCE Physics Question Thread!
Post by: Homer on May 13, 2013, 06:56:42 pm
How to?

EDIT: I actually figured out a way to do this now, but not sure if right or not

$\frac{2}{m^2}\times 16mm^2$

$\frac{2}{(1000mm)^2}\times 16mm^2$

$\frac{32mm^2}{(1000mm)^2}$

$\frac{32mm^2}{1000000mm^2}$

$=\frac{32}{1000000}$

$32\mu W$
Title: Re: VCE Physics Question Thread!
Post by: jono88 on May 13, 2013, 08:06:24 pm
I was wondering if anyone has had there EPI yet for physics. What topics are involved?
Title: Re: VCE Physics Question Thread!
Post by: Professor Polonsky on May 15, 2013, 08:17:33 pm
Can anyone explain why the net force (if there is no side friction) for a banked then is $mg\times\tan\theta$?
Title: Re: VCE Physics Question Thread!
Post by: [paradox]+ on May 15, 2013, 10:50:24 pm
Anyone mind helping me out with my circuit question (topic link below)

circuit question
Title: Re: VCE Physics Question Thread!
Post by: Robert123 on May 22, 2013, 04:45:08 pm
Question about the 'gain' in amplifiers
First off, can you have a negative gain? Does that occur when the gradient is negative?
Secondly, my interpretation of the gain is the factor in which the input voltage is multiplied to get the output voltage. However, in Qs 3a in Test 2 on electronic & photonics part in the atar notes book, it doesn't work because the gradient is negative.
Also, my original understanding with circuits, if you have two resistors in a circuit, the resistor with the highest resistance will have the highest power output. But this doesn't seem the case with resistors in parallel. Could someplease please confirm this and try to explain it a bit.
Thanks
Title: Re: VCE Physics Question Thread!
Post by: EspoirTron on May 22, 2013, 09:50:53 pm
Question about the 'gain' in amplifiers
First off, can you have a negative gain? Does that occur when the gradient is negative?
Secondly, my interpretation of the gain is the factor in which the input voltage is multiplied to get the output voltage. However, in Qs 3a in Test 2 on electronic & photonics part in the atar notes book, it doesn't work because the gradient is negative.
Also, my original understanding with circuits, if you have two resistors in a circuit, the resistor with the highest resistance will have the highest power output. But this doesn't seem the case with resistors in parallel. Could someplease please confirm this and try to explain it a bit.
Thanks

Remember that the formula for gain is constituted by 'change in' vout and vin. The reason you may get a 'negative' gain is that the amplifier is inverting and since the gradient of the graph is negative, you will get a negative answer. To my knowledge gain is only concerned with a magnitude; therefore, while using it to calculate vout it you simply multiply by the gain value.
To address your second question. The reason why in series that a resistor with a higher resistance will output more power is because that it will have a higher potential across it, if you recall in a series circuit each component has the same amount of current flowing through it; hence, to maintain the ratio of v2/v1 = r2/r1 the resistor with the high resistance must have a higher potential across it and consequently a higher power output. The reason why this isn't the case for resistors in parallel is that in parallel the resistors will have the same potential but varying current based on their resistance. Therefore, I believe the lower the resistance the higher the current; thus, the resistor with the lowest resistance will have the highest power output (in the parallel component).
I hope that helped you out!  :)
Title: Re: VCE Physics Question Thread!
Post by: sin0001 on May 22, 2013, 11:25:07 pm
Just a made-up example for help:
E.g. Suppose there's a circuit involving 2 resistors (2 and 6 ohms) in parallel, the battery supplies, let's say, 8 V. Current would be 1.5 A, the 2 ohm resistor would receive 3/4 of the total current, so 9/8 A. Whereas the 6 ohm resistor would get 1/4 of the total current, so 3/8 A.
P(2 ohm)= I^2 x R = (9/8)^2 x 2 = 2.53 W
P(6 ohm) = (3/8)^2 x 6 = 0.84 W
So it can be seen that the resistor with the lowest resistance, in a parallel circuit, would have the highest power output.
Hope this helps!
Title: Re: VCE Physics Question Thread!
Post by: ~T on May 27, 2013, 04:51:31 pm
^^ just getting resistance and current mixed up...

$\frac{1}{R_{T}} = \frac{1}{2} + \frac{1}{6} = \frac{2}{3}$

therefore $R_{T} = \frac{3}{2}$, not $I$

Current is thus given  $I_{T} = \frac{V}{R_{T}} = \frac{8}{1.5} = 5.33A$

The point still stands of course, but none of that calculation is even required.
The smaller resistor will receive $I_{2} = \frac{V_{T}}{R} = \frac{8}{2} = 4A$
The larger resistor will receive $I_{2} = \frac{V_{T}}{R} = \frac{8}{6} = 1.33A$

As $V$ is the same for both resistors (parallel circuit), the current will be greater for less resistance as $I=\frac{V}{R}$ and then the power will be greater because $P=VI$
Title: Re: VCE Physics Question Thread!
Post by: Homer on May 27, 2013, 07:54:53 pm
with question 6 in the vcaa 2011 exam 1, how do we know that the switch it attached across the resistor or thermistor?
Title: Re: VCE Physics Question Thread!
Post by: FlorianK on May 27, 2013, 08:38:44 pm
When the temperatur increases the resistance goes down and the voltage across the thermistor goes down, hence the voltage across the resistor goes up. If we would have a 500 Ohm resistor and the 1500Ohm Thermistor (at 20°) and the circuit switch across the thermistor then the circuit would switch when the temperature falls below 20° and not when it goes above 20°
Title: Re: VCE Physics Question Thread!
Post by: tote.moore on May 28, 2013, 06:24:46 pm
from memory the question was...

If a certain amount of Power is transmitted through power lines into a house, why does the voltage of an appliance go up when another appliance is turned off?
Title: Re: VCE Physics Question Thread!
Post by: Robert123 on May 29, 2013, 03:05:22 pm
Just a few question related to electronics & photonics.
What is RMS? How is it calculated? How much do we have to know about it for electronics & photonics?
Why do photodiodes have to be in reverse bias?
If a photodiode is in series with a resisitor, what would happen if the light intensity is doubled?
What are the 'reasons' for limits in an amplifier?
Any help will be greatly appreciated :)
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on May 29, 2013, 04:17:40 pm
1. RMS means root-mean-square. It really should be called "square-mean-root" as you square your function, average it and then square root it. Why do we need it? AC voltage is sinusoidally varying; its average value over a cycle is zero. That's not very useful. However, in power calculations, where P=V^2/R, an average of V^2 is just as good. This time, if we square the voltage, it's always positive and now an average makes sense. I won't go into the calculation details, but the RMS voltage is the peak voltage divided by sqrt 2, for a sinusoidal wave only (which is VCE physics). Just know how to calculate RMS voltages, currents, what they mean and the fact that P=I^2 R or V^2/R refers to RMS currents and voltages.

2. If a photodiode is forward-biased, it is a regular diode. Only when reverse-biased does its behaviour depend on light like in the course. Then, the photocurrent is directly proportional to the intensity of light striking the diode. If the light intensity doubles, the current doubles.

3. I'm not great with transistors, sorry. My electronics knowledge is strictly limited to what's in the course.
Title: Re: VCE Physics Question Thread!
Post by: Robert123 on May 29, 2013, 07:13:22 pm

Thanks for that, I had questions on those topics for a practice SAC and our class haven't cover much of this stuff in detail if at all.
Another question relating to springs.
Pinball game has a ball with mass 0.15kg which is launch by a compressed spring at 4.2m/s that has been compressed by 20 cm. If the spring is "ideal" what is the spring constant k in N/m?
Since k is N/m, would you use constant acceleration formulas to find the acceleration, multiply it by mass (since F=ma) and divide by distance (k=F/x)? Would you have to worry about the constant acceleration question or are you meant to substitute the final speed for acceleration?
Title: Re: VCE Physics Question Thread!
Post by: availn on May 29, 2013, 08:02:17 pm
Thanks for that, I had questions on those topics for a practice SAC and our class haven't cover much of this stuff in detail if at all.
Another question relating to springs.
Pinball game has a ball with mass 0.15kg which is launch by a compressed spring at 4.2m/s that has been compressed by 20 cm. If the spring is "ideal" what is the spring constant k in N/m?
Since k is N/m, would you use constant acceleration formulas to find the acceleration, multiply it by mass (since F=ma) and divide by distance (k=F/x)? Would you have to worry about the constant acceleration question or are you meant to substitute the final speed for acceleration?

You cannot do this easily with kinematics, because a spring does not accelerate a mass at a constant rate. You should do this with energy, spring potential energy before equals kinetic energy after.

0.5kx2 = 0.5mv2
k · 0.22 = 0.15 · 4.22
∴ k = 66.15Nm-1
Title: Re: VCE Physics Question Thread!
Post by: random_person on May 29, 2013, 08:04:22 pm
Thanks for that, I had questions on those topics for a practice SAC and our class haven't cover much of this stuff in detail if at all.
Another question relating to springs.
Pinball game has a ball with mass 0.15kg which is launch by a compressed spring at 4.2m/s that has been compressed by 20 cm. If the spring is "ideal" what is the spring constant k in N/m?
Since k is N/m, would you use constant acceleration formulas to find the acceleration, multiply it by mass (since F=ma) and divide by distance (k=F/x)? Would you have to worry about the constant acceleration question or are you meant to substitute the final speed for acceleration?

We have m=0.15kg, v=4.2ms-1, s=20cm=0.2m

EK=1/2 mv2
=1/2 0.15 x 4.22
=1.323 J
Us=1/2 ks2
therefore
1.323=1/2 k x 0.22
k=66.15Nm-1

Hope this helps
Title: Re: VCE Physics Question Thread!
Post by: Sentar on June 03, 2013, 04:01:23 pm
Found an exam question but can't find the answer anywhere, don't have the actual answers, seeing if anyone can help.

Explain how it is possible to transmit the information contained in an electrical signal which has both a positive and negative component, when it is impossible to produce a beam of light with negative intensity?

Cheers
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on June 03, 2013, 04:29:53 pm
You just need a light signal whose amplitude varies just like your electrical signal/has the same shape. E.g. move the electrical signal up so it's always positive.
Title: Re: VCE Physics Question Thread!
Post by: ~T on June 10, 2013, 11:19:35 pm
Just looking at the 2012 Exam 2 paper and confused with question 1. Basically, there is a point P and it asks for you to draw an arrow at P indicating the direction of the magnetic field. There is a solenoid which will produce a field (comparable to Earth's) going right, and there is an arrow pointing upwards that says "magnetic north."

The answers say that the arrow drawn should be "to the right and up the page at an angle of approximately 45°." However, in saying that "magnetic north" is upwards, does that not mean that the field lines due to the Earth should be going downwards? If magnetic north is up, magnetic south is down, and field lines go from north to south. I would have drawn an arrow "to the right and *DOWN* the page at an angle of approximately 45°."

Basically, in saying that magnetic north is up, does that not mean that geographical north is downwards?
Title: Re: VCE Physics Question Thread!
Post by: availn on June 11, 2013, 02:18:50 pm
Just looking at the 2012 Exam 2 paper and confused with question 1. Basically, there is a point P and it asks for you to draw an arrow at P indicating the direction of the magnetic field. There is a solenoid which will produce a field (comparable to Earth's) going right, and there is an arrow pointing upwards that says "magnetic north."

The answers say that the arrow drawn should be "to the right and up the page at an angle of approximately 45°." However, in saying that "magnetic north" is upwards, does that not mean that the field lines due to the Earth should be going downwards? If magnetic north is up, magnetic south is down, and field lines go from north to south. I would have drawn an arrow "to the right and *DOWN* the page at an angle of approximately 45°."

Basically, in saying that magnetic north is up, does that not mean that geographical north is downwards?

Nope. My friend got hit by this too.

Magnetic North and North Magnetic Pole is at the earth's north.
Magnetic North Pole is at the earth's south.
Title: Re: VCE Physics Question Thread!
Post by: ~T on June 12, 2013, 06:19:45 pm
Oh really? So magnetic north is actually a south pole (in dipole terms)? No wonder >50% got that wrong.
Title: Re: VCE Physics Question Thread!
Post by: jono88 on June 24, 2013, 07:43:46 pm
VCAA 2007 exam 2 question 1, the question asks you to sketch ﬁve magnetic ﬁeld lines around the magnet, yet in the assessors report they have 6 lines? wut?
Title: Re: VCE Physics Question Thread!
Post by: Homer on June 24, 2013, 07:55:32 pm
how would  i find the tension in each cable?
Title: Re: VCE Physics Question Thread!
Post by: Phy124 on June 24, 2013, 08:18:47 pm
how would  i find the tension in each cable?
As the structure is in rotational equilibrium you can sum the moments (I think you refer to it as torque in yr 12 physics?) about the point of rotation and equate them to zero.

Hint: The weight force of the bridge and the two vertical components of the tension forces will be the forces that cause rotation. (the horizontal component of tension does not cause rotation as it acts "through" the point of rotation)

Spoiler
$\sum M = -(5)(700)(9.8) + (2)(10)(T\sin(30^{\circ})) = 0$

$\Rightarrow T=3430 N$
Title: Re: VCE Physics Question Thread!
Post by: Homer on June 29, 2013, 11:54:24 am
Could someone walk me through this question :)

Thanks

ANS: a) fh=390N, fv=680N
b) 330N
Title: Re: VCE Physics Question Thread!
Post by: SocialRhubarb on June 29, 2013, 01:08:34 pm
Part A is quite simple:

The horizontal component is equal to $800\cos(60)\ N= 400\ N$

The vertical component is equal to $800\sin(60)\ N= 693\ N$

Umm, unless I've done something stupid I think your answers are slightly off.

Part B is a bit beyond the scope of VCE.

Essentially what you're doing is equating torques since the object isn't spinning off in one direction.

$\tau = r \times F$

The net torque on the fence post is 0, so the torque from one source must be equal to the torque from the other.

$0.85\ m\times 400\ N = 1.0\ m\times Ft$   Technically there should be a negative somewhere, but whatever.

$Ft = 340\ N$

Alternatively, using the answer of 390N,

$0.85\ m \times 390\ N = 1.0\ m\times Ft$

$Ft = 331.5\ N$
Title: Re: VCE Physics Question Thread!
Post by: Homer on June 29, 2013, 02:14:18 pm
yeah im not sure why theyve got different component values?
anyways

A 6m ladder of mass 7.5 kg leans against a wall at 60 degrees to the horizontal. What is the torque exerted on the ladder by the weight of the ladder itself?

I know it should be 7.5g x 3 x sin(30), but i dont understand why they take theta as 30 instead of 60?
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on June 29, 2013, 03:13:09 pm
It depends on how the angle is defined. The formula $\tau = rF sin{\theta}$ is only used when the angle is measured between the heads of the vectors r and F (it's a vector equation). In this case, 60 degrees is the angle between the ladder and the horizontal, while here theta is the angle between r and mg, in this case the ladder and the VERTICAL.

This is why I like thinking that torque = moment arm * force where the moment arm is the component of the radius that is perpendicular to the force. You'll never get it wrong this way.
Title: Re: VCE Physics Question Thread!
Post by: Homer on July 02, 2013, 11:41:55 am
The young's modulus for a rope of length 30m, cross-sectional area 30cm2, which stretches 2m under a load of 1000N is? ANS: 5 x 10^6 Nm-2

Title: Re: VCE Physics Question Thread!
Post by: sin0001 on July 02, 2013, 11:57:53 am
It depends on how the angle is defined. The formula $\tau = rF sin{\theta}$ is only used when the angle is measured between the heads of the vectors r and F (it's a vector equation). In this case, 60 degrees is the angle between the ladder and the horizontal, while here theta is the angle between r and mg, in this case the ladder and the VERTICAL.
Is the torque formula examinable material for our core topics, or is it only required for a particular detailed study because I haven't seen it in the Heinemann textbook :/
Also, are you guys gonna reinforce the Unit 3 knowledge-through some mid-year exams- or continue on with the course, as revision?
Title: Re: VCE Physics Question Thread!
Post by: Phy124 on July 02, 2013, 04:59:53 pm
The young's modulus for a rope of length 30m, cross-sectional area 30cm2, which stretches 2m under a load of 1000N is? ANS: 5 x 10^6 Nm-2

$E = \frac{\sigma}{\varepsilon }}$

$\sigma = \frac{F}{A}$

$\varepsilon = \frac{\Delta L}{L}$

$\Rightarrow E = \frac{FL}{A\Delta L} = \frac{(1000)(30)}{(0.003)(2)} = 5 \times 10^6 \ Nm^{-2}$

Is the torque formula examinable material for our core topics, or is it only required for a particular detailed study because I haven't seen it in the Heinemann textbook :/
Also, are you guys gonna reinforce the Unit 3 knowledge-through some mid-year exams- or continue on with the course, as revision?

I would think that torque is only relevant to the "Materials and their use in structures" detailed study.

Although your question wasn't directed at me, I would advise you to do some mid-year practice exams during these holidays to both reinforce key concepts and get a feel for VCAA physics exams.

Title: Re: VCE Physics Question Thread!
Post by: Homer on July 03, 2013, 10:11:24 am

$\Rightarrow E = \frac{FL}{A\Delta L} = \frac{(1000)(30)}{(0.003)(2)} = 5 \times 10^6 \ Nm^{-2}$

hey just a question why is it 0.003 wouldnt it be $(\frac{30}{100})^2$ so  $\frac{(1000)(30)}{(0.09)(2)}$ ?

Title: Re: VCE Physics Question Thread!
Post by: Homer on July 03, 2013, 11:45:34 am
Also, What would be the tension in the cables of the cranes? Thankyou
Title: Re: VCE Physics Question Thread!
Post by: Alwin on July 03, 2013, 12:58:15 pm
Also, What would be the tension in the cables of the cranes? Thankyou

Let the tensions of the cables be T1 and T2, corresponding the the crane numbers Crane 1 and Crane 2.
Note I have included units in my working, just a small personal habit.

Using Crane 2 as the pivot,
$\Rightarrow 25m \times T_1 = 20m \times 300,000 kg \times 10 m/s^{2}$
$\therefore T_1 = 2,400,000 N$

You can now either say,
$T_1 + T_2 = 300,000 kg \times 10 m/s^{2}$

or use Crane 1 as a pivot,
$25m \times T_2 = 5m \times 300,000 kg \times 10 m/s^{2}$

Either way, it works out that
$T_2 = 600,000 N$

So, then tensions on cables of the cranes 1 and 2 are 2,400,000N an 600,000N in that order

$\Rightarrow E = \frac{FL}{A\Delta L} = \frac{(1000)(30)}{(0.003)(2)} = 5 \times 10^6 \ Nm^{-2}$

hey just a question why is it 0.003 wouldnt it be $(\frac{30}{100})^2$ so  $\frac{(1000)(30)}{(0.09)(2)}$ ?

30cm2 is not the same as 30cm x 30cm = (0.3m)^2 = 0.09 m2

30cm2 is the same as 30cm x 1cm = (0.3m) x (0.01m) = 0.003 m2

The general rule is that 1cm2 = 0.0001m2

EDIT: answered both of homer's questions, rather than double post
Title: Re: VCE Physics Question Thread!
Post by: Homer on July 05, 2013, 02:05:26 pm
Having trouble with 3. I dont know what they are asking for and how to work it out. :(
Title: Re: VCE Physics Question Thread!
Post by: SocialRhubarb on July 05, 2013, 03:00:17 pm
They've given you a compass on the side, and the earth's magnetic field generally speaking runs south to north. Near the poles it's actually more into the ground, but usually it's mostly running south to north. Using right hand rule, if magnetic field is running from south to north, and current is going into the page, the field must be towards the right, or towards B.
Title: Re: VCE Physics Question Thread!
Post by: Nato on July 06, 2013, 11:16:12 pm
Hey guys,

i am having a little trouble getting my head around Newton's third law. so it's the whole equal and opposite reaction thing.
so if an object exerts a force on the another, that object while exert the same force of same magnitude in the opposite direct. what i don't understand is how object are still able to move (i understand how the forces are acting on different and can't be cancelled out).

for example if you kick a ball with 100N, the ball exerts 100N back on you. Where does that *extra* force to make the ball go flying??

thank you guys
Title: Re: VCE Physics Question Thread!
Post by: sin0001 on July 07, 2013, 12:09:18 am
Hey guys,

i am having a little trouble getting my head around Newton's third law. so it's the whole equal and opposite reaction thing.
so if an object exerts a force on the another, that object while exert the same force of same magnitude in the opposite direct. what i don't understand is how object are still able to move (i understand how the forces are acting on different and can't be cancelled out).

for example if you kick a ball with 100N, the ball exerts 100N back on you. Where does that *extra* force to make the ball go flying??

thank you guys
Okay so the forces are identical, but what is the difference here? Something has to be different between you and the ball, because clearly one is moving while the other is stable. The difference is mass; you weigh a lot more than the ball you've kicked and according to the formula: a = F/m, you are going to experience negligible acceleration while the ball is going to experience much greater acceleration, because if 'm' increases, then acceleration of an object will decrease. Therefore, you are going to 'absorb' a reaction force of 100 N by only experiencing negligible acceleration, but this force will cause the ball to go 'flying'.
Title: Re: VCE Physics Question Thread!
Post by: SocialRhubarb on July 07, 2013, 05:30:34 pm
You exert a force on the ball of 100N. The 'reaction force' which Newton's third law talks about is the ball's force applied to you. That force doesn't affect the ball's acceleration at all, because it doesn't act on the ball. It acts on you. So the original 100N of force which you apply isn't cancelled out by the reaction force from the ball because they act on different objects.
Title: Re: VCE Physics Question Thread!
Post by: Homer on July 08, 2013, 02:45:18 pm
What would be the direction of the magnetic force and how do we work it out?
Title: Re: VCE Physics Question Thread!
Post by: jssantucci on July 08, 2013, 03:58:21 pm
Could someone please explain why when the secondary coil in a transformer is connected to a circuit with an open switch (i.e. one with no load) the energy used by the primary coil is zero? Wouldn't the power being dissipated by the primary coil always just be IxV regardless of what's going on in the secondary circuit? This is in relation to question 9 of chapter 10.6 in Heinemann.

Thanks
Title: Re: VCE Physics Question Thread!
Post by: SocialRhubarb on July 08, 2013, 05:07:45 pm
Unfortunately I don't really understand the physics behind this but at the very least I can show you with the formulas.

Let the primary current be $I_P$, the secondary current be$I_S$, the turns on the primary coil be $N_P$ and the turns on the secondary coil be $N_S$.

We know that $I_S=0$, because the resistance across an open circuit is infinite, and $I=\frac{V}{R}$.

We also know that $\frac{I_S}{I_P}=\frac{N_P}{N_S}$.

Therefore, $I_P=\frac{N_S}{N_P}I_S=0$.

Thus, since $I_P=0$ and $P=VI$, the power dissipated in the primary circuit must be 0.

Like I said before, I don't actually understand the physics behind this, but somehow the secondary circuit must somehow influence the primary circuit? Maybe someone else can help with that.
Title: Re: VCE Physics Question Thread!
Post by: jssantucci on July 08, 2013, 05:21:27 pm
@SocialRhubarb this is what my text book says in the chapter about transformers...

When there is no load connected to the secondary coil no current flows in the secondary and so all the flux is being generated by the primary coil. However, this flux will induce a voltage in the primary coil that will oppose the change of flux. That is, as the current increases, this ‘back EMF’ will tend to reduce the current. In a good transformer, this process is very effective and very little current will flow in the primary if none is flowing in the secondary.

Not really sure how this works but it seems to be the answer to the question. Could someone clarify the above idea about 'back EMF' for me?

Thanks
Title: Re: VCE Physics Question Thread!
Post by: SocialRhubarb on July 08, 2013, 05:34:44 pm
Hmm. I'm familiar with the concept of back EMF but I haven't seen it being induced in a solenoid by its own flux.

Basically, in a simple DC motor with one loop, when you run a current through the loop, it runs through the magnetic field and generates a force, and your motor starts turning. But when your motor is turning, its flux is constantly changing. We know that a change in flux through a coil induces a voltage in that coil, so the changing flux actually induces an EMF which counteracts the voltage we applied to make the motor spin. This EMF we call a 'back EMF'.
Title: Re: VCE Physics Question Thread!
Post by: Guest on July 08, 2013, 08:17:52 pm
I originally posted this question in my unit 1&2 Physics thread but people suggest that it should be of more relevance here on this thread.
Question:
1) Sound travels with a speed of 343m/s in air. Find the speed of sound in steel where K=200 GPa and p= 7,870 kg/m^3, and in water where E=2,200 GPa and p=1000 kg/m^3. Compare to the speed in air.

Normally I can solve these questions by simply plugging the values into the equations but in this case the value of K is provided for a solid object and the value of E is given for the liquid. Shouldn't it be the other way around? Would some conversion be necessary? How would you "compare to the speed in air"?

The answer states that : the speed is 5,040 m/s in steel and 4.7*10^4 m/s in water. Could someone please explain how one might reach this conclusion?

Title: Re: VCE Physics Question Thread!
Post by: Robert123 on July 08, 2013, 09:21:34 pm
What would be the direction of the magnetic force and how do we work it out?

You missed an important piece of information there, the direction of the current (read 4 lines down from that picture in 4a & 4b).
From there, you use the right hand slap rule, fingers pointing to the South Pole and thumb down for 4a and up for 4b.
Clear?
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on July 08, 2013, 09:46:41 pm
i dont get the question from Hinemann Physics 12 Chapter 11 Exercise 11.2, question 4c.

the question is attached.

Lots of love. ahha.

sorry *i didnt notice the questions thread*

edit: damn sorry guys i just understood it. tell me if you want me to explain it.

Moderator edit [2/cos(c)]: Merged double post
Title: Re: VCE Physics Question Thread!
Post by: Nato on July 09, 2013, 12:06:54 pm
impulse/momentum question here:

as we know in cars, there are crumple zones which increases the amount of time the change in velocity occurs so from forumla
$F_{net}=\frac{\Delta p}{\Delta t}$. what i need help with understanding is for cars more time has resulted in less impulse, but why when talking about baseball/golf and following through more time results in greater impulse.

thanks guys
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on July 09, 2013, 01:07:49 pm
impulse/momentum question here:

as we know in cars, there are crumple zones which increases the amount of time the change in velocity occurs so from forumla
$F_{net}=\frac{\Delta p}{\Delta t}$. what i need help with understanding is for cars more time has resulted in less impulse, but why when talking about baseball/golf and following through more time results in greater impulse.

thanks guys

you have it wrong my friend. By increasing time in the car case you increase your impulse:
the equation is ΣF= ∆P/∆T. So if we increase the time the Net force decreases.

say ∆P=4 and ∆T=1 then the net force is 4N. but if we increase the change in time to, lets say, 2, then the Net force is 2N.

similarly, say we have a force of 5N and ∆T=1 then the impulse will be 5. however if we increase the change in time to, lets say, 2, then the impulse is 10.

You understand broda?
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on July 09, 2013, 02:07:38 pm
you have it wrong my friend. By increasing time in the car case you increase your impulse:
the equation is ΣF= ∆P/∆T. So if we increase the time the Net force decreases.

say ∆P=4 and ∆T=1 then the net force is 4N. but if we increase the change in time to, lets say, 2, then the Net force is 2N.

similarly, say we have a force of 5N and ∆T=1 then the impulse will be 5. however if we increase the change in time to, lets say, 2, then the impulse is 10.

You understand broda?

Actually your interpretation is slightly off as well. Impulse is just change in momentum. For a given collision, their initial and final velocities are going to be the same, so regardless of the time taken for the collision, the change in momentum is the same. The time, however, affects the force and acceleration that will be felt by the object as mentioned.

impulse/momentum question here:

as we know in cars, there are crumple zones which increases the amount of time the change in velocity occurs so from forumla
$F_{net}=\frac{\Delta p}{\Delta t}$. what i need help with understanding is for cars more time has resulted in less impulse, but why when talking about baseball/golf and following through more time results in greater impulse.

thanks guys

More time does not mean less impulse. It does, however, meant less force. Force and impulse are not the same quantity.
Title: Re: VCE Physics Question Thread!
Post by: Alwin on July 09, 2013, 05:15:48 pm
More time does not mean less impulse. It does, however, meant less force. Force and impulse are not the same quantity.

Since you're always soo nit-picky :P, ill add to this too: same impulse over the longer period of time means smaller average force (usually meaning smaller maximum force too). sorry, just had to go there :P

From a few posts ago:
Could someone please explain why when the secondary coil in a transformer is connected to a circuit with an open switch (i.e. one with no load) the energy used by the primary coil is zero? Wouldn't the power being dissipated by the primary coil always just be IxV regardless of what's going on in the secondary circuit? This is in relation to question 9 of chapter 10.6 in Heinemann.

Thanks

Since it's rather a lot to explain, I found a link that may interest you:

http://sound.westhost.com/xfmr.htm

Also, just something interesting from the Heinamann textbook:
Quote
To create an induced current we do not always need loops and magnets. Whenever a changing magnetic flux encounters a conducting material an induced current will occur. These currents are often called eddy currents and may result in lost energy in electrical machinery. The eddy currents produced in a moving conductor will themselves be subject to the IlB force.
p369 of the Heinemann textbook
Eddy currents are present in the core of transformers too.

I quite like how it includes topics outside the 3/4 course (I first came across this stuff in Singapore coursework so not sure what level it is in australia) but you might find it interesting too! :D
Title: Re: VCE Physics Question Thread!
Post by: jssantucci on July 09, 2013, 11:18:50 pm
http://www.physics.usyd.edu.au/~khachan/PTF/Transformer%20explanation.pdf

for anyone interested, this link explains the answer to my question quite nicely.
Title: Re: VCE Physics Question Thread!
Post by: Homer on July 12, 2013, 10:20:55 am
i know the top of the beam experiences compression and the bottom tension, but what about the centre? is it a little of both?
Title: Re: VCE Physics Question Thread!
Post by: Phy124 on July 12, 2013, 06:34:45 pm
i know the top of the beam experiences compression and the bottom tension, but what about the centre? is it a little of both?
In theory, the beam experiences no bending stresses at its centre. Although I believe in most cases there will be a very small amount, which is negligible.

These diagrams indicate the bending stress (RHS) and strain (LHS) patterns a beam which is symmetrical about its neutral axis experiences.

(http://www.fgg.uni-lj.si/kmk/esdep/media/wg07/f0820004.jpg)

As you can see when the beam is in its elastic state the bending stresses have a linear relationship with distance from the neutral axis. However, when the beam is plastically deforming the bending stress is constant throughout.

*Note: As far as I know for VCE Physics purposes you only need to know about the elastic state for bending stresses.
Title: Re: VCE Physics Question Thread!
Post by: sin0001 on July 24, 2013, 06:16:28 pm
I know that the force produced by the presence of current in a magnetic field is a result of the interactions of the conductor & magnet's magnetic fields, but can someone explain why this force is at MAX. value when the external magnetic field is perpendicular to the conductor?

Also, is 'Forces on moving charges' part of the course?
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on July 24, 2013, 06:24:17 pm
Because VCE physics is inadequate and doesn't give you the full picture. Period.

Jks.
The force is actually given by a cross product: Force = iL x B
where L is the current length vector; it has magnitude equal to the length of the wire and has the same direction as the wire
and B is the magnetic field vector which has a magnitude and a direction.

The magnitude of this force is given by i*|L||B| |sin theta| where theta is the angle between L and B. As you've seen from VCE physics, if the current wire is parallel to the field, no force results.
Likewise, if it's perpendicular, the sine function returns the maximum value of 1. Hence the force is maximum when the external magnetic field is perpendicular to the conductor.

Read up on wiki if you're not sure about cross products in general.
Title: Re: VCE Physics Question Thread!
Post by: sin0001 on July 24, 2013, 06:57:08 pm
Because VCE physics is inadequate and doesn't give you the full picture. Period.

Jks.
The force is actually given by a cross product: Force = iL x B
where L is the current length vector; it has magnitude equal to the length of the wire and has the same direction as the wire
and B is the magnetic field vector which has a magnitude and a direction.

The magnitude of this force is given by i*|L||B| |sin theta| where theta is the angle between L and B. As you've seen from VCE physics, if the current wire is parallel to the field, no force results.
Likewise, if it's perpendicular, the sine function returns the maximum value of 1. Hence the force is maximum when the external magnetic field is perpendicular to the conductor.

Read up on wiki if you're not sure about cross products in general.
Oh how I can relate to that! VCE physics is especially vague in explaining Electronics & Photonics.
Didn't the formula for cross product contain a 'cos', where did the 'sin' come from? O.o
Thanks, sorta makes sense
Title: Re: VCE Physics Question Thread!
Post by: Conic on July 24, 2013, 07:00:33 pm
Oh how I can relate to that! VCE physics is especially vague in explaining Electronics & Photonics.
Didn't the formula for cross product contain a 'cos', where did the 'sin' come from? O.o
Thanks, sorta makes sense
Basically, dot/scalar product is cos and cross product is sin. You use a few of each in VCE physics torque, force (in magnetism), work etc.
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on July 24, 2013, 07:56:27 pm
Basically, dot/scalar product is cos and cross product is sin. You use a few of each in VCE physics torque, force (in magnetism), work etc.

I think you mean "you're meant to use a few of each in VCE physics torque, force, work etc".
The dot product is "too difficult" for VCE physics. Funny. Maths was made primarily for physics, so why are we omitting it like this. Hmph.
Title: Re: VCE Physics Question Thread!
Post by: SocialRhubarb on July 24, 2013, 08:25:12 pm
To be fair, most of my classmates are struggling as is.
Title: Re: VCE Physics Question Thread!
Post by: Conic on July 24, 2013, 08:41:58 pm
I think you mean "you're meant to use a few of each in VCE physics torque, force, work etc".
The dot product is "too difficult" for VCE physics. Funny. Maths was made primarily for physics, so why are we omitting it like this. Hmph.
They use them without knowing that they are using them :P
Title: Re: VCE Physics Question Thread!
Post by: sin0001 on July 25, 2013, 11:08:14 pm
Hey, can someone please help me with question 2a, I thought the max. EMF was induced after the coil is turned 90 (i.e. 1/4 of a turn), but the worked solutions contain a '2pi'
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on July 26, 2013, 04:24:27 pm
They're looking for the max instantaneous emf induced in the coil, not the average. You'll need a derivative here.
Frequency of 100 Hz => 0.01 s = period
Let's denote the rotation speed by w rad/s. Then the angle made between the magnetic field and the coil changes, so the flux is going to be B*A*cos(wt)
As it's a uniform rotation rate, the coil rotates by 2pi radians in one period. Therefore the rotation rate is 2pi/T = 2pi rad/(0.01s) = 200pi rad/s = w
So now we can differentiate the flux with respect to time. This yields -w*BAsin(wt). The maximum of this is simply wBA, which by Faraday's law is the maximum emf induced at any time.
But from before, w = 200 pi rad/s, so plugging in all the values yields emf = 200 pi rad/s * 8*10^-4 T * 40*10^-4 m^2 = 64000pi * 10^-8 Wb/s = 2.01^-3 V

Someone check my working.
Title: Re: VCE Physics Question Thread!
Post by: sin0001 on July 26, 2013, 05:06:34 pm
They're looking for the max instantaneous emf induced in the coil, not the average. You'll need a derivative here.
Frequency of 100 Hz => 0.01 s = period
Let's denote the rotation speed by w rad/s. Then the angle made between the magnetic field and the coil changes, so the flux is going to be B*A*cos(wt)
As it's a uniform rotation rate, the coil rotates by 2pi radians in one period. Therefore the rotation rate is 2pi/T = 2pi rad/(0.01s) = 200pi rad/s = w
So now we can differentiate the flux with respect to time. This yields -w*BAsin(wt). The maximum of this is simply wBA, which by Faraday's law is the maximum emf induced at any time.
But from before, w = 200 pi rad/s, so plugging in all the values yields emf = 200 pi rad/s * 8*10^-4 T * 40*10^-4 m^2 = 64000pi * 10^-8 Wb/s = 2.01^-3 V

Someone check my working.
Thanks :)
It mathematically makes sense but are we allowed to solve it using differentiation? Otherwise, is there some alternative way of solving this (i.e. how would one answer it graphically?)
Also, can you please explain the 'cos(wt)' part of the flux formula?
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on July 26, 2013, 05:50:31 pm
Technically, you can't use differentiation in VCE physics as it's "too difficult". Somehow they decided that setting the mathematics level to year 9 was appropriate for a VCE level subject that is meant to have deep connections to mathematics.
Graphically? It's like asking someone what the maximum rate of change of a sine curve is. Graphically? Good luck with that. I'd be stuck for an answer.
As for the cosine term...I'm not surprised you ask. VCE physics's definition of flux has a serious problem. It only considers the one special case where the magnetic field is perfectly perpendicular to the surface/parallel to the surface normal. Now for a general magnetic field, if you resolve it into components parallel and perpendicular to the surface, the component parallel to the surface will not pass through the surface as they are parallel, so no flux there. Only the perpendicular component contributes to the flux, so if theta is the angle between the perpendicular component of the magnetic field and the field itself, the perpendicular component has magnitude B cos theta. Therefore the flux is BA cos theta.
Now initially, theta = 0 in the diagram; the field lines are perfectly perpendicular to the surface. As it rotates with constant angular velocity w, the angle theta at any time would be given by wt + c where c is a constant. But when t = 0, theta = 0, so c=0. This is where the cos(wt) term comes from.
Title: Re: VCE Physics Question Thread!
Post by: sin0001 on July 26, 2013, 06:47:57 pm
I think the only other approach would be to memorize the formula... typical VCE Physics...
I was surprised this question came up in the Heinemann book, coz I dont think the exercises showed us how to find the max. EMF; is it just me or is Electric Power a hard topic (especially since the Heinemann book doesn't help)
Thanks again!
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on July 27, 2013, 03:07:53 pm
I think the only other approach would be to memorize the formula... typical VCE Physics...
I was surprised this question came up in the Heinemann book, coz I dont think the exercises showed us how to find the max. EMF; is it just me or is Electric Power a hard topic (especially since the Heinemann book doesn't help)
Thanks again!

Dont even start with the Hinemann book. It is the worst. Im using an old Jacaranda one to accompany my understanding and a few youtube videos.
Title: Re: VCE Physics Question Thread!
Post by: Shyam995 on July 31, 2013, 01:06:36 pm
Hey Guys

Just been wondering, as i love maths and physics, i thought this question can bring out some curiosity.

As i have been researching, i was actually fascinated about the whole concept of a "second".

As i have been told and have studied, the result:

"the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom."

I have a possible rebuttal to this "too accurate" theory.

If this is a second, does that mean that, at  the big bang, how the different phases of the explosion measure in seconds, is this deemed to be inaccurate?

- sorry for the defective expression, as all of you know already, physics requires a substantial amount of communication and reasoning, as it can be hard to express, especially in writing.
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on August 02, 2013, 08:37:10 pm
question...

A double slit arrangement is illuminated by light that consists of two wavelengths, one of which is known to be 600nm. the interference pattern on a screen shows that the fourth dark fringe for the known wavelength coincides with the fifth bright fringe for the unknown wavelength. What is the unknown wavelength?

Title: Re: VCE Physics Question Thread!
Post by: lzxnl on August 02, 2013, 09:21:32 pm
Formula for dark fringes is d sin theta = (n+1/2) wavelength where the first fringe is at n = 0
So the fourth dark fringe for the known wavelength has a path difference of 9/2*600 nm
This corresponds to the fifth dark fringe, which has a path difference of (11/2)*w
9/2*600 = 11/2 * w
w = 5400/11 nm
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on August 02, 2013, 09:26:48 pm
Formula for dark fringes is d sin theta = (n+1/2) wavelength where the first fringe is at n = 0
So the fourth dark fringe for the known wavelength has a path difference of 9/2*600 nm
This corresponds to the fifth dark fringe, which has a path difference of (11/2)*w
9/2*600 = 11/2 * w
w = 5400/11 nm

are these sort of q's in vce?
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on August 02, 2013, 09:32:06 pm
I haven't bothered with many VCAA questions on interference yet, although I can tell you that UMEP physics has a LOT of questions like these.
Title: Re: VCE Physics Question Thread!
Post by: Robert123 on August 07, 2013, 02:07:09 pm
Having some difficulty understanding transformers:
So by the way I understand it is they change the current and voltage while (ideally) keeping the same power overall. A step up transformer increase voltage and decrease current while a step down transformer decrease voltage but increase current.
The difficulty I have is to relate this to Ohm's law (V=IR) which state that if you have an increase or decrease in voltage or current, you will get a proportional increase or decrease in the other. That is, you can't decrease one while increasing the other.

So, do transformers change the resistance? Could someone please explain to me how transformers work while taking into consideration ohms law.

Thank you
Title: Re: VCE Physics Question Thread!
Post by: SocialRhubarb on August 07, 2013, 05:24:02 pm
A transformer works based on electromagnetic induction - the changing magnetic flux induced by one coil induces a voltage in the other coil. Furthermore, any resistors attached to the coils of the transformers are still subject to Ohm's laws. Changes in the secondary coil also affect the primary coil. This is probably best illustrated with an example.

Let's take a 5V RMS AC source, connected to a step-up transformer with a 1:10 ratio. A 100 Ohm resistor is connected to the secondary coil. Let's try to work out the current being drawn.

The voltage across the secondary coil is pretty straight forward to work out:

$\frac{V_2}{V_1}=\frac{N_2}{N_1}\ \ \ \ \ \ \ \ \ \ \ \ \ V_2=\frac{10}{1}\times 5V=50V$

Applying Ohm's law:

$I_2=\frac{V_2}{R}=\frac{50V}{100\Omega}=0.5A$

Then we can find the current across the primary:

$\frac{I_2}{I_1}=\frac{N_1}{N_2}\ \ \ \ \ \ \ \ \ \ \ \ I_2=\frac{1}{10}\times 0.5A=0.05A$

Notice that if we were to attach a 50 Ohm instead of a 100 Ohm resistor, we would double the current through the secondary coil and also double the current through the primary coil. This is probably one of the concepts students find hardest to grasp - changing the resistance across the secondary coil changes the current in the primary coil as well as the secondary coil. Why? It's a bit difficult to explain, but it's to do with the primary coil inducing a back EMF on itself. For this reason, you also can't treat the primary coil as a conventional 'circuit' - the coil induces a back EMF which almost totally opposes the voltage drawn, even if there are no physical resistors in its path. Also, the current produced by a power source is not fixed - it can vary and will vary depending on the external circuit, such as what kind of transformers and resistors it is attached to.

So Ohm's law still applies in circuits with transformers, but we also have to take into account the effect of electromagnetic induction, and remember that generally speaking the current generated by a power source is not constant, but varies depending on the circuit it's attached to.
Title: Re: VCE Physics Question Thread!
Post by: Robert123 on August 07, 2013, 05:56:35 pm
Ok, continuing on from your example, say we have a 10:1 to ratio (step down transformer). Now the voltage would 0.5V. Using ohms law on the resistance to find the current (V=IR)
I= 0.5/100= 5*10^-3A
P=VI
P= 0.5 * 5*10^-3= 2.5*10^-3 W
For the step up transformer in the example you provided, the power is 25W which is greatly differently to the result above (transformers are meant to have the same power). This is due to not taking into account the increase in current but that means ohms law is wrong. Correct?
So then, how can ohms law and transformers agree?
Title: Re: VCE Physics Question Thread!
Post by: SocialRhubarb on August 07, 2013, 06:10:29 pm
The power on the coils of a transformer is the same. That is, the power you put into the primary of a transformer is equal to the power coming out of the secondary of a transformer. The rule is not that all transformers all have the same power, because we have different transformers in different circuits, but the power put into a transformer is equal to the power coming out of a transformer.
Title: Re: VCE Physics Question Thread!
Post by: ~T on August 11, 2013, 02:18:28 pm
Just a diffraction query (wary clarification I guess) as the Heinemann textbook is slightly unclear...

Sometimes it talks about the extent of diffraction being most significant when the wavelength approaches the size of the gap, and other times it suggests/states the proportionality: extent of diffraction is proportional to wavelength/gap.

I presume the latter is most correct and - as such - wavelengths that are larger than the gap size will produce even more diffraction than wavelengths equal to the gap size?

Thanks
Title: Re: VCE Physics Question Thread!
Post by: xenon2013 on August 11, 2013, 03:00:09 pm
can't figure out how to do this:

A sprinter reaches his maximum speed vmax in 2.5 seconds from rest with constant
acceleration. He then maintains that speed and finishes the 100 meters in the overall
time of 10.40 seconds. Determine his maximum speed vmax.

Thanks
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on August 11, 2013, 04:21:37 pm
So...his maximum speed is 2.5a
Maintains this for a time of 10.4 - 2.5 = 7.9 seconds
Distance travelled = 7.9 * 2.5a + 1/2*at^2 = 1/2*a*6.25 + 7.9*2.5a
22.875a = 100
a = 4.37 m/s^2
Max speed = 2.5a = 10.9 m/s
Title: Re: VCE Physics Question Thread!
Post by: Homer on August 11, 2013, 04:50:19 pm
an alternative:

$v(7.9)+\frac{1}{2}(2.5v)=100$

$7.9v + 1.25v =100$

$9.15v =100$

$v=10.9 ms^{-1}$

Title: Re: VCE Physics Question Thread!
Post by: xenon2013 on August 11, 2013, 04:57:17 pm
So...his maximum speed is 2.5a
Maintains this for a time of 10.4 - 2.5 = 7.9 seconds
Distance travelled = 7.9 * 2.5a + 1/2*at^2 = 1/2*a*6.25 + 7.9*2.5a
22.875a = 100
a = 4.37 m/s^2
Max speed = 2.5a = 10.9 m/s

im a bit confused because, if he's maintaing the vmax, shouldnt the acceleration be zero instead of 4.37? and within that 7.9 seconds, he travelled a distance less than 100m because he took 10.4s to finish, so i dont understand why t=7.9 and dist=100 were subbed together into one eqn, to me this implies that he ran 100m in the 7.9s

sorry if this is confusing but im having trouble understanding it all
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on August 11, 2013, 05:15:05 pm
im a bit confused because, if he's maintaing the vmax, shouldnt the acceleration be zero instead of 4.37? and within that 7.9 seconds, he travelled a distance less than 100m because he took 10.4s to finish, so i dont understand why t=7.9 and dist=100 were subbed together into one eqn, to me this implies that he ran 100m in the 7.9s

sorry if this is confusing but im having trouble understanding it all

OK. I'll clarify my working.
a is the acceleration in the period where he is accelerating. The acceleration is only equal to a in that time. When he is travelling at max speed, the acceleration is zero.

He accelerated for 2.5 seconds. He ran for 10.4 seconds in total. Therefore the time in which he maintained the max speed is 7.9 seconds. What I subbed in was 1/2*at^2, the distance travelled during the acceleration, and 7.9*vmax, or 7.9*2.5a, which is the distance travelled at constant speed.
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on August 15, 2013, 06:47:40 pm
i gotta question:

Why is the EMF induced in a DC generator a maximum when the coil is parallel to the magnetic field, and a minimum when perpendicular to the field?

Title: Re: VCE Physics Question Thread!
Post by: lzxnl on August 15, 2013, 07:11:02 pm
i gotta question:

Why is the EMF induced in a DC generator a maximum when the coil is parallel to the magnetic field, and a minimum when perpendicular to the field?

Assuming that the coil is rotating at a constant rate:

The magnetic flux is actually given by BA*sin(theta) where theta is the angle between the magnetic field lines and the coil. As the rotation is at a constant rate, theta = wt where w is the angular velocity and t is time. I am assuming that at t = 0 there is no flux for simplicity.

Then, flux = BA*sin(wt)
By Faraday's law, emf = -rate of change of flux = -d(flux)/dt = -wBA*cos(wt)
So, for the magnitude to be a maximum, we want wt = some integer multiple of pi. This corresponds to the field being parallel to the coil.
When the field and coil are perpendicular, cos(wt) = 0, i.e. minimum magnitude of the emf.
Title: Re: VCE Physics Question Thread!
Post by: sydneyboy on August 18, 2013, 07:32:20 pm
We are doing a prac at school with a circular motion kit, I've stumbled on a task that i have no clue how to go about:

"Given Fnet = mv^2/r

Determine a relationship between F and m so everything else must be kept constant. Think about how best to keep v^2/r constant while collecting your f vs m data.

Clue: what is another way of expressing v^2/r and how can this be kept constant even when radius is changing. "

Not sure how to star, any pointers would be appreciated.
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on August 18, 2013, 08:54:00 pm
well v^2/r = acceleration. and if you keep the magnitude of the velocity constant, than the acceleration should be constant.

also when radius increases, increase the velocity so the acceleration remains same. and vice versa when radius decreases.

hope it helps.
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on August 18, 2013, 11:10:19 pm
We are doing a prac at school with a circular motion kit, I've stumbled on a task that i have no clue how to go about:

"Given Fnet = mv^2/r

Determine a relationship between F and m so everything else must be kept constant. Think about how best to keep v^2/r constant while collecting your f vs m data.

Clue: what is another way of expressing v^2/r and how can this be kept constant even when radius is changing. "

Not sure how to star, any pointers would be appreciated.

v = radius * angular velocity
= rw

v^2/r = (rw)^2/r = w^2*r
Or v^2/r = (2pi*r/t)^2/r = 4pi*r/t^2

The second form is probably more useful, so if you keep the same distance and period you'll be fine. I don't know about the practicalities of that though.
Title: Re: VCE Physics Question Thread!
Post by: P0ppinfr3sh on August 21, 2013, 06:30:28 pm
This question has popped up in the textbook a couple of times an I can't seem to get the right answer:
How would you calculate the maximum induced voltage? I know that the maximum rate of change of magnetic flux occurs when the plane of the coil is parallel to the magnetic field lines but  I can't seem to translate this into Faraday's Law.

Thanks :)

P.S. (I've got a SAC on this tomorrow and this is the only thing I'm unsure about so please answer quickly  ;)).
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on August 21, 2013, 06:51:58 pm
This question has popped up in the textbook a couple of times an I can't seem to get the right answer:
How would you calculate the maximum induced voltage? I know that the maximum rate of change of magnetic flux occurs when the plane of the coil is parallel to the magnetic field lines but  I can't seem to translate this into Faraday's Law.

Thanks :)

P.S. (I've got a SAC on this tomorrow and this is the only thing I'm unsure about so please answer quickly  ;)).

can you tell me which question it is? is it from hinemann?
Title: Re: VCE Physics Question Thread!
Post by: SocialRhubarb on August 21, 2013, 06:55:35 pm
Find the average RMS voltage and multiply by $\sqrt{2}$ to get the peak voltage.
Title: Re: VCE Physics Question Thread!
Post by: P0ppinfr3sh on August 21, 2013, 06:58:31 pm
can you tell me which question it is? is it from hinemann?
Well question 5 and questions 8 of 10.4 involve peak induced voltage and magnetic flux; and yes, it's from Heinemann.

Find the average RMS voltage and multiply by $\sqrt{2}$ to get the peak voltage.

How would you go about finding the average RMS voltage of a generator? I can find the average induced voltage but I don't think this is equivalent to the average RMS voltage.

Also, I might as well post the related question(s) from the textbook:
5. 100 turns, area=20cm^2, B=5.0mT, coil rotates at a rate of 15degrees per millisecond. What is the peak value of the induced EMF for this coil?.
ANS: 0.263V.

8. Peak voltage=8.0kV, N=1000 turns, each coil has a radius of 10cm, magnetic field strength=B, frequency=50Hz. Calculate the strength of the magnetic field required to produce a peak voltage of 8.0kV.
ANS: B=0.81T.

Thanks.

Title: Re: VCE Physics Question Thread!
Post by: 09Ti08 on August 21, 2013, 07:07:12 pm
This question has popped up in the textbook a couple of times an I can't seem to get the right answer:
How would you calculate the maximum induced voltage? I know that the maximum rate of change of magnetic flux occurs when the plane of the coil is parallel to the magnetic field lines but  I can't seem to translate this into Faraday's Law.

Thanks :)

P.S. (I've got a SAC on this tomorrow and this is the only thing I'm unsure about so please answer quickly  ;)).

If you draw the graph, it should look like a cosine graph (fyi flux is the dot product of area and magnetic field, that's why we have the cos function). After drawing the graph, i think it should be clear that the gradient of the graph is greatest/lowest when magnetic flux=0 (if you know calculus, this is easy to prove). Greatest/lowest gradient (which is basically the biggest/smallest value of -d(phi)/dt) gives you the maximum value for the magnitude of emf.
Title: Re: VCE Physics Question Thread!
Post by: P0ppinfr3sh on August 21, 2013, 07:37:59 pm
If you draw the graph, it should look like a cosine graph (fyi flux is the dot product of area and magnetic field, that's why we have the cos function). After drawing the graph, i think it should be clear that the gradient of the graph is greatest/lowest when magnetic flux=0 (if you know calculus, this is easy to prove). Greatest/lowest gradient (which is basically the biggest/smallest value of -d(phi)/dt) gives you the maximum value for the magnitude of emf.

Yeah, that seems to make sense, but I don't really know how to apply it to the questions I've mentioned above.  :-\
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on August 21, 2013, 08:08:04 pm
This question has popped up in the textbook a couple of times an I can't seem to get the right answer:
How would you calculate the maximum induced voltage? I know that the maximum rate of change of magnetic flux occurs when the plane of the coil is parallel to the magnetic field lines but  I can't seem to translate this into Faraday's Law.

Thanks :)

P.S. (I've got a SAC on this tomorrow and this is the only thing I'm unsure about so please answer quickly  ;)).

Find the average RMS voltage and multiply by $\sqrt{2}$ to get the peak voltage.

Erm, RMS voltage and multiply by square root 2 only lets you find the RMS voltage of a sinusoidally voltage. Something whacky like a square wave in which the voltage is equal to V for half a period and -V for the other half would have an RMS value of V. Just an example.

Besides RMS has nothing to do with this question.

If you draw the graph, it should look like a cosine graph (fyi flux is the dot product of area and magnetic field, that's why we have the cos function). After drawing the graph, i think it should be clear that the gradient of the graph is greatest/lowest when magnetic flux=0 (if you know calculus, this is easy to prove). Greatest/lowest gradient (which is basically the biggest/smallest value of -d(phi)/dt) gives you the maximum value for the magnitude of emf.

Indeed. Another instance of where VCE physics just fails. You have flux = BA*cos(theta)
but as theta is changing at a constant rate (constant rotation), theta = wt
so flux = BA*cos(wt)
differentiate: emf = BAw*sin(wt) with the sign depending on how the voltage is measured

So peak emf is BAw
Title: Re: VCE Physics Question Thread!
Post by: Alwin on August 21, 2013, 08:26:56 pm
Yeah, that seems to make sense, but I don't really know how to apply it to the questions I've mentioned above.  :-\
How would you go about finding the average RMS voltage of a generator? I can find the average induced voltage but I don't think this is equivalent to the average RMS voltage.

Also, I might as well post the related question(s) from the textbook:
5. 100 turns, area=20cm^2, B=5.0mT, coil rotates at a rate of 15degrees per millisecond. What is the peak value of the induced EMF for this coil?.
ANS: 0.263V.

8. Peak voltage=8.0kV, N=1000 turns, each coil has a radius of 10cm, magnetic field strength=B, frequency=50Hz. Calculate the strength of the magnetic field required to produce a peak voltage of 8.0kV.
ANS: B=0.81T.

5. This is what happens when you TRY to calculate peak voltage with RMS...
\begin{alignedat}{1} EMF_{ave}&=-\frac{\Delta\phi}{\Delta t} \end{alignedat}
It rotates 15 degrees per millisecond. So, in 6 milliseconds it turns 90 degrees. Say if it start parallel to the field, then after 6 milliseconds it had turned to perpendicular to the field.
\begin{alignedat}{1} \Delta \phi &= \phi _{perpendicular} - \phi_{parallel}
\\ &= 0.0020 \times 100 \times 0.005
\\ &= 0.001 T m^{-2}
\\ \therefore EMF_{ave}&=-\frac{\Delta\phi}{\Delta t}=-\frac{0.001}{0.006}
\\ &=-0.167 V
\\ \text{since } EMF_{peak} &= \sqrt{2} EMF_{ave},
\\ EMF_{peak} &= \sqrt{2} \times 0.167
\\ &=0.236V
\end{alignedat}

^ close but no cigar. You should note that from the textbook, a rate of 15° per millisecond (i.e. a frequency of 42 Hz) is not actually correct since not accurate enough.. it is actually 15.12 degrees per millisecond.. but that's still a bit off the correct answer

What you should actually do is use the formula:
$NBA 2 \pi f = 100 \times 5 \times 10^{–3} \times 20 \times 10^{–4} \times 2\pi \times 42 = 0.263 V$

8. Pretty much same thing,
The maximum EMF is equal to NBA2πf, so:
$8 \times 10^{3} = 1000 \times B \times \pi \times 0.1^{2} \times 2 \pi \times 50 \text{ and so }B = 0.81 T.$

EDIT: didn't see nliu's post since was latexing. If you're wondering w is angular velocity, ω=2 pi f

Hope it helps / makes sense :)
Title: Re: VCE Physics Question Thread!
Post by: P0ppinfr3sh on August 21, 2013, 08:38:27 pm
Thanks alot guys for your help.  :D

You're all awesome!

(I can now sleep peacefully)
Title: Re: VCE Physics Question Thread!
Post by: SocialRhubarb on August 21, 2013, 09:57:19 pm
I guess sometimes you just have to use calculus, huh.
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on August 21, 2013, 10:13:58 pm
I guess sometimes you just have to use calculus, huh.

But that's too difficult for the delicate minds of VCE physics students.
Title: Re: VCE Physics Question Thread!
Post by: BasicAcid on August 21, 2013, 10:47:52 pm
But that's too difficult for the delicate minds of VCE physics students.

Well you seem to be one of those guys who has no respect for the VCE design (for all subjects) even though you'll be one of the people who simply ace VCE.

Also is it wrong that I never fully understood motion in physics (which we was the first thing we did in term 1) until I began kinematics in spesh a few weeks ago?
And do you know if we're allowed to use I and J systems in physics?
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on August 21, 2013, 11:03:59 pm
Well you seem to be one of those guys who has no respect for the VCE design (for all subjects) even though you'll be one of the people who simply ace VCE.

Also is it wrong that I never fully understood motion in physics (which we was the first thing we did in term 1) until I began kinematics in spesh a few weeks ago?
And do you know if we're allowed to use I and J systems in physics?

Why? VCE physics is an absolute joke. As you've seen yourself, maths teaches physics better than physics does. I'm not surprised; I actually learnt forces, energy and kinematics firstly from spesh in year ten before doing physics, and it worked perfectly fine.

Why is physics a joke?
1. They cut down the maths. In a subject which is built on maths. It's like teaching a guy to build a house but you only give him sand. He needs to find the bricks somewhere, but if you're giving them to him...

2. You have frigging notes for the exam. Come on. You can literally summarise EVERYTHING you need for the exam in four sides of paper as long as you write small. I mean come on. That's really lazy. I can understand giving the formulas, but letting us bringing in a cheat sheet? It's a wonder 90% is still an A+ on the exams.

3. Why the frick are we repeating motion after doing it in year 11? Why the frick are we going over basic circuits when that should have been covered in year 8, at most year 10?

4. Why the bloody heck are we cutting a course down when we're increasing the study time for the course by removing the mid-year exam study period and making it available for class time? Hint. English and Maths all have three hours worth of exams in total, but they didn't cut anything down.

5. Actually test us on physical knowledge instead of boring subbing into equations. I don't know how many I used E = hc/wavelength for revision for my last SAC.

That rant enough for you?

If you use vectors in physics, it'll work, but there's no need to use something so advanced for VCE physics. I'm not kidding.
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on August 21, 2013, 11:28:21 pm
nliu1995, I am very intrigued by you at the moment.

Never seen anyone who seems to actually be capable of virtually destroying VCE with minimal effort.

But what? Spesh in year 10? In year 10 I was still learning what a parabola was lol...

I'm just a more outspoken member of the "dissatisfied with VCE" group. There are heaps of people like me. They just don't want to be so vocal for fear of offending people.

You should read pi's VCE Physics 3/4 review that's stickied. It's even better.

lol well, I spent too much on maths in earlier life. Parabolas was year five for me. I still remember fumbling around to understand what this f(-b/2a) meant in primary school.
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on August 21, 2013, 11:42:23 pm
I am simply speechless...
Man what the hell, although deep down know you're telling the truth, I simply cannot believe what you are saying.

Year 5... Pokemon Emerald/FireRed dominated my life throughout that year?

I only learnt what that f(-b/2a) thing meant last year lol.

You see, I am where I am now because I sacrificed gaming time for study. I looked like a dork then. I was even bashed one day for having a year eleven maths book in year six. But now, I'm laughing at those who laughed at me. It's well worth it :D
I think my decision to pre-learn (at least read upon) all the uni courses that I'll be doing next year in the long break makes sense now.
Title: Re: VCE Physics Question Thread!
Post by: sydneyboy on August 25, 2013, 05:48:55 pm
v = radius * angular velocity
= rw

v^2/r = (rw)^2/r = w^2*r
Or v^2/r = (2pi*r/t)^2/r = 4pi*r/t^2

The second form is probably more useful, so if you keep the same distance and period you'll be fine. I don't know about the practicalities of that though.

Thanks, so i will select a time to make the period for every test i do and also raidus.. however do you know how i would go about keeping period the same w/o increasing speed (because it increases my r)
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on August 25, 2013, 06:04:17 pm
Thanks, so i will select a time to make the period for every test i do and also raidus.. however do you know how i would go about keeping period the same w/o increasing speed (because it increases my r)

t = 2pi*r/v. You can only keep one of the three variables constant.
Title: Re: VCE Physics Question Thread!
Post by: Alwin on August 25, 2013, 06:09:41 pm
I'm just a more outspoken member of the "dissatisfied with VCE" group. There are heaps of people like me. They just don't want to be so vocal for fear of offending people.
...
lol well, I spent too much on maths in earlier life. Parabolas was year five for me. I still remember fumbling around to understand what this f(-b/2a) meant in primary school.

with him on this^
I just cbbs ranting here + its a question thread not a bagging thread ;) (I just +1 him everytime ;) )
I'll just add for me parabolas in grade 3.  as for the  f(-b/2a) i 'proved' it by taking the mean of the two intercepts that can be found by the quadratic formula. Then proved it a couple years later with calculus, which is SO hard on physics... esp since on some exams they don't even bother to draw graphs correctly since we can count the squares XD

anyways:
Determine a relationship between F and m so everything else must be kept constant. Think about how best to keep v^2/r constant while collecting your f vs m data.
Thanks, so i will select a time to make the period for every test i do and also raidus.. however do you know how i would go about keeping period the same w/o increasing speed (because it increases my r)

you want it to increase your r..  say k=v^2/r, so when v increase you need r to increase (at a rate proportional to the square of the increase of v). Otherwise, v increases r stays same  -> k increases (ie not constant = fail).

BUT you are testing the relationship between F and m. So, you want to keep everything else (r, T, v etc) constant.

you change the mass...
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on August 25, 2013, 06:36:13 pm
To be honest, I had no idea what he wanted, as simply investigating the relationship between F and m is...trivial. F = ma arises from a definition of force in Newtonian mechanics and with a constant mass. So why would anyone measure the relationship between F and m? It's just measuring your experimental capabilities.
Title: Re: VCE Physics Question Thread!
Post by: Alwin on August 25, 2013, 06:46:45 pm
To be honest, I had no idea what he wanted, as simply investigating the relationship between F and m is...trivial. F = ma arises from a definition of force in Newtonian mechanics and with a constant mass. So why would anyone measure the relationship between F and m? It's just measuring your experimental capabilities.

its not like vce physics would test maths skills :P

I've given it some thought, sydneyboy, and i think i have an idea of what your prac is testing: try different masses and plot a graph of F vs m?? It should be straight line since when  v^2/r constant F ∝ m. If it's not straight, then you talk about errors like not keeping v^2/r constant...
Is this interpretation right?

Otherwise just post the prac sheet and we (esp nliu since he <3 vce physics :P ) will do our best to try help!! Or, ask your teacher for clarification.

EDIT (in spoiler):
SECRET TO SUCCEEDING IN ANY PRAC
Just copy what the smart people in your class are doing :P  Or worse comes to worse make up/tweak your data results after the prac then write some cohesive BS about errors and stuff :D
Or just blame the teacher hahahaha. that always goes down well..
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on August 25, 2013, 08:22:11 pm
question: how do i link the generation of AC EMF to its use in a DC motor.
Title: Re: VCE Physics Question Thread!
Post by: sydneyboy on August 25, 2013, 08:49:06 pm
hey guys thinks for all the replies, there was a part A which explored relationship between Tension and V^2 - this was worth 80% of the marks. This next section is worth 20% and is the 'challenge' part of the sac so all we were given is

"Given Fnet = mv^2/r
Determine a relationship between F and m so everything else must be kept constant. Think about how best to keep v^2/r constant while collecting your f vs m data.
Clue: what is another way of expressing v^2/r and how can this be kept constant even when radius is changing." on  a piece of paper.

@Alwin im sure your interpretation is right. I have 2 periods left to record data and plot the graph. I think it will be a tedious task to make everything constant.

Thanks for the help guys I'll have a crack tomorrow i have a much better idea of what to do now.
Title: Re: VCE Physics Question Thread!
Post by: SocialRhubarb on August 25, 2013, 09:34:56 pm
This is late but:

It might help to know what equipment you've been given.

But honestly, I have no idea how to approach this. Obviously you want to keep v^2/r constant, but I suppose the question is how can you best do it, so as to minimise the error in your results. Especially since it's telling you to keep it constant "even when radius is changing".

My first thought was to swing it in a vertical arc, and then use gravity to kind of 'calibrate' your v^2/r value, but that really presents more problems than solutions, especially if you're swinging your mass by hand.

I suppose I'll have to go with what everyone else is saying and just say "keep the period and radius constant" even though it's not really answering the question of how to keep v^2/r constant with a changing radius.
Title: Re: VCE Physics Question Thread!
Post by: sydneyboy on August 25, 2013, 09:43:13 pm
circular motion kit. http://shop.omegascientific.com.au/image/cache/data/products/80-161-500x500.jpg
Title: Re: VCE Physics Question Thread!
Post by: sydneyboy on August 25, 2013, 09:44:28 pm
There is a speed setting on it
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on August 27, 2013, 11:53:27 pm
in circuit construction questions how do you know where the switching circuit goes?
Title: Re: VCE Physics Question Thread!
Post by: Lasercookie on August 28, 2013, 04:10:34 pm
EDIT (in spoiler):
SECRET TO SUCCEEDING IN ANY PRAC
Just copy what the smart people in your class are doing :P  Or worse comes to worse make up/tweak your data results after the prac then write some cohesive BS about errors and stuff :D
Or just blame the teacher hahahaha. that always goes down well..
I'm guessing you meant it tongue in cheek, but anyway :P it depends if your teacher is expecting you to get the correct results from the experiment though, a well designed practical sac would allow you to still do pretty well if you're able to explain how your data turned out to be messed up.
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on September 02, 2013, 08:00:09 pm
question :

A proton and electron are accelerated from rest through the same potential difference. Given that their masses are in a ratio of 1836:1 , find the ratio of their de Broglie wavelengths.
Title: Re: VCE Physics Question Thread!
Post by: SocialRhubarb on September 02, 2013, 08:14:50 pm
Okay. Firstly, the magnitude of charge on the electron is the same as the magnitude of the charge on the proton. This means that accelerating them through the same potential difference gives them the same energy, since potential difference is energy per charge. Let's use the subscript 'p' for the proton, and the subscript 'e' for the electron.

$E_p=E_e$

$m_p=1836m_e$

$\lambda=\frac{h}{p}$

Using the formulas for momentum and energy of a particle, $p=mv,\ E=\frac{1}{2}mv^2$,     we can easily derive a relationship between the two.

$p=\sqrt{2Em}$

$p_e=\sqrt{2E_em_e}$

$p_p=\sqrt{2E_pm_p}=\sqrt{2(E_e)(1836m_e)}$, substituting in our two pieces of information from before.

$p_p=\sqrt{1836}\times\sqrt{2E_em_e}=42.85\sqrt{2E_em_e}$

$p_p:p_e=42.85:1$
Title: Re: VCE Physics Question Thread!
Post by: saifh on September 02, 2013, 10:06:18 pm
A question I need answered :)

1. Describe the photoelectric effect's implications for the wave model of light and how it forced the reintroduction of a particle model or photon model for light

Title: Re: VCE Physics Question Thread!
Post by: Conic on September 02, 2013, 10:17:49 pm
If the wave model was correct, there would be a delay before electrons were emitted, as it would require time for the electron to build up energy. There would also be electrons emitted for all frequencies of light, but the energy would take longer to build up. It also predicts that higher intensities would produce more current [This is actually true. This supports the particle model as higher intensities would be predicted to decrease the delay before emission, but instead there is more current]. These have been shown to be false through experiment. The particle model can explain all of these so it is more appropriate than the wave model. Since the wave model couldn't explain it and the particle model can, the particle model was reintroduced.
Title: Re: VCE Physics Question Thread!
Post by: saifh on September 02, 2013, 10:34:15 pm
If the wave model was correct, there would be a delay before electrons were emitted, as it would require time for the electron to build up energy. There would also be electrons emitted for all frequencies of light, but the energy would take longer to build up. It also predicts that higher intensities would produce more current. These have been shown to be false through experiment. The particle model can explain all of these so it is more appropriate than the wave model. Since the wave model couldn't explain it and the particle model can, the particle model was reintroduced.

explains it, thanks!
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on September 02, 2013, 10:38:12 pm
If the wave model was correct, there would be a delay before electrons were emitted, as it would require time for the electron to build up energy. There would also be electrons emitted for all frequencies of light, but the energy would take longer to build up. It also predicts that higher intensities would produce more current. These have been shown to be false through experiment. The particle model can explain all of these so it is more appropriate than the wave model. Since the wave model couldn't explain it and the particle model can, the particle model was reintroduced.

Erm...higher intensities DO produce more current. Another piece of evidence was the fact that intensities had no effect on the energies of the emitted electrons.
Title: Re: VCE Physics Question Thread!
Post by: Conic on September 02, 2013, 10:39:27 pm
Erm...higher intensities DO produce more current. Another piece of evidence was the fact that intensities had no effect on the energies of the emitted electrons.
True. I'll edit the post.
Title: Re: VCE Physics Question Thread!
Post by: saifh on September 03, 2013, 12:18:12 pm
Question

Explain the role of a reverse bias voltage and an ammeter in the photoelectric effect
Title: Re: VCE Physics Question Thread!
Post by: Robert123 on September 03, 2013, 07:35:33 pm
Question

Explain the role of a reverse bias voltage and an ammeter in the photoelectric effect
A reverse bias voltage can be used to find the stopping voltage (ie. when current=0A). In doing so, the kinetic energy of the electrons (since voltage=energy/coulomb(1.609*10^19 electrons I believe).  From this, we can determine the work function of the metal (amount of energy to liberate
electrons=hf-KE)

Why is momentum measured in Ns for some questions? I know the units all add up but I thought that you just keep it as kgms^-1
Title: Re: VCE Physics Question Thread!
Post by: ~T on September 09, 2013, 02:06:39 pm
Why is momentum measured in Ns for some questions? I know the units all add up but I thought that you just keep it as kgms^-1
I presume it's from Newton's second Law, which in its true form is $F = \frac{dp}{dt}$ (I'm not sure if you do methods so I don't know if you are familiar with differential notation - force is the rate of change of momentum with respect to time)

Presuming a constant mass, and given $p=mv$
we get $F=m\frac{dv}{dt}=ma$
...which should hopefully seem familiar!

But if we leave it in its original form and presume a constant force, then we can write $F = \frac{\Delta p}{\Delta t}$
Rearranging yields $\Delta p = F\Delta t$
Thus, momentum is in Ns
Title: Re: VCE Physics Question Thread!
Post by: Chazef on September 15, 2013, 03:04:10 pm
Do we need to know about the thermal motion of electrons (I think that's what it's called)? A question came up about it in the second atarnotes exam from the study guide in regards to an incandescent globe
Title: Re: VCE Physics Question Thread!
Post by: BasicAcid on September 15, 2013, 04:47:05 pm
Do we need to know about the thermal motion of electrons (I think that's what it's called)? A question came up about it in the second atarnotes exam from the study guide in regards to an incandescent globe

I'm pretty sure not but the guys at ATARnotes seem reliable.

Mind posting/PMing me the question? Maybe you're misinterpreting the question or something. :P
Title: Re: VCE Physics Question Thread!
Post by: Robert123 on September 18, 2013, 04:43:07 pm
Question related to the detailed study photonics?
By having more modes in fibre optics, does that give any more advantages?
Title: Re: VCE Physics Question Thread!
Post by: joey7 on September 23, 2013, 01:44:24 pm
In the examiners report of last years second exam, in relation to the last question it stated "a common misconception was that the electrons moved around the orbit in a wave pattern"
If someone explain to me how this is wrong that would be great
Title: Re: VCE Physics Question Thread!
Post by: availn on September 23, 2013, 01:57:32 pm
In the examiners report of last years second exam, in relation to the last question it stated "a common misconception was that the electrons moved around the orbit in a wave pattern"
If someone explain to me how this is wrong that would be great

The standing waves you draw around the nucleus are not the paths the electron takes, i.e. they do not continually move towards and away from the centre of orbit. The electrons are a matter wave, so the standing waves show the possible circumferences the electrons can use to orbit, because with these circumferences, constructive interference occurs.
Title: Re: VCE Physics Question Thread!
Post by: SocialRhubarb on September 23, 2013, 02:33:05 pm
The electrons don't orbit in a 'wave' pattern - electrons are the wave. This is probably a bit tricky for most people to understand, since it's difficult to comprehend something being a particle and a wave at the same time.

The electron exhibits both particle-like properties and wave-like properties. They have a rest mass, like a particle, but it's easiest to explain the orbitals of electrons using a wave model for electrons. The electrons can only form orbits when the 'wavelength' of the electron fits exactly into the circumference of their orbit, and they form a 'standing wave' structure. The electron cannot take any other orbital, or else it will 'destructively interfere' with itself, and so it can only exist at certain orbits and at certain energy levels. If you do chemistry, this is what causes the shells and sub-shells.

So the electron's orbit is not a wave-pattern, but instead we find the orbits by treating the electron as a wave.
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on September 23, 2013, 03:57:48 pm
The electrons don't orbit in a 'wave' pattern - electrons are the wave. This is probably a bit tricky for most people to understand, since it's difficult to comprehend something being a particle and a wave at the same time.

The electron exhibits both particle-like properties and wave-like properties. They have a rest mass, like a particle, but it's easiest to explain the orbitals of electrons using a wave model for electrons. The electrons can only form orbits when the 'wavelength' of the electron fits exactly into the circumference of their orbit, and they form a 'standing wave' structure. The electron cannot take any other orbital, or else it will 'destructively interfere' with itself, and so it can only exist at certain orbits and at certain energy levels. If you do chemistry, this is what causes the shells and sub-shells.

So the electron's orbit is not a wave-pattern, but instead we find the orbits by treating the electron as a wave.

At the same time...electrons in multi-electron atoms do not form orbitals solely when the wavelength of the electron fits the circumference of the orbit. There are electron-electron repulsions which cannot be neglected. This is why the Bohr model of the atom fails completely for even helium.
The VCE course isn't entirely right when they use the fitting standing waves argument to atoms in general. It is only valid for species with one electron, such as H, He+, Li 2+ etc...even Au 78+.

Just keep this in the back of your mind. VCE only requires the answer that electrons form standing waves around the nucleus, although they generally don't.
Title: Re: VCE Physics Question Thread!
Post by: sin0001 on September 23, 2013, 04:27:28 pm
Need some help with VCAA's 2013 Sample exam (linked below):
http://www.vcaa.vic.edu.au/Documents/exams/physics/physics-specs-samp-w.pdf
-For 6b, isn't the Impulse meant to calculated using the resultant force, which is 17 N up (i.e. 17 N x 1.5s)? In Itute's solutions, the force of 22 N was used o.O
-For question 16a, shouldn't the graph of the induced EMF begin under the x-axis from 0 to 1 seconds and be drawn above the x-axis for 2 to 4 seconds? Well Itute, their solutions, have drawn it starting from above the x-axis; who's wrong?
-And finally, for question 18, why is the total resistance of the two wires 5 ohms? I thought the two wires would represent a parallel circuit so that the current splits up equally amongst them, so that the total resistance becomes 1.25 ohms (using: 1/R= (1/r + 1/r))

Thanks, would appreciate any help!
Title: Re: VCE Physics Question Thread!
Post by: SocialRhubarb on September 23, 2013, 04:35:30 pm
6b.) Calculate the impulse of the force provided by the burning gunpowder on the rocket.
It asks for the impulse generated by the 22N force alone. A separate impulse is provided by gravity and together they make up the resultant change in momentum of the rocket.

16a.) Doesn't specify a direction for the voltmeter. Both solutions should be correct.

18.) It's not a parallel circuit, it's a series circuit.
Title: Re: VCE Physics Question Thread!
Post by: Robert123 on September 27, 2013, 02:30:26 pm
Some Questions related to light and matter...
How does the medium (air vs water) influence the Fringe Spacing?

Electrons of known energy are fired into Mercury vapour. The energy of the scattered electrons is then measured. When electrons of energy from 0eV to 4.8eV are fired into the mercury vapour, the energy of the scattered electrons equal the energy of the incident electrons. At 4.8eV the energy of some of the scattered electrons falls to zero. Which of the following statements best explains this observation?
Inelastic collisions within the atoms can occur for electrons of energy 4.8eV, but not at lower energies.
Could someone please explain this to me.

For electronics...
If a diode with a turn on voltage of 0.7V is in parallel with a 6.0V battery as well as a resistor, what would be the voltage drop across the resistor? Would it matter whether the diode is in forward bias or reverse bias ?

And for electric power...
Also, are we meant to know how to work the EMF generated from a moving object in a constant magnetic field using EMF=BLv  as it was in the STAV 2013 practice exam?

Thanks

Title: Re: VCE Physics Question Thread!
Post by: lzxnl on September 27, 2013, 02:40:08 pm
Some Questions related to light and matter...
How does the medium (air vs water) influence the Fringe Spacing?

Electrons of known energy are fired into Mercury vapour. The energy of the scattered electrons is then measured. When electrons of energy from 0eV to 4.8eV are fired into the mercury vapour, the energy of the scattered electrons equal the energy of the incident electrons. At 4.8eV the energy of some of the scattered electrons falls to zero. Which of the following statements best explains this observation?
Inelastic collisions within the atoms can occur for electrons of energy 4.8eV, but not at lower energies.
Could someone please explain this to me.

For electronics...
If a diode with a turn on voltage of 0.7V is in parallel with a 6.0V battery as well as a resistor, what would be the voltage drop across the resistor? Would it matter whether the diode is in forward bias or reverse bias ?

And for electric power...
Also, are we meant to know how to work the EMF generated from a moving object in a constant magnetic field using EMF=BLv  as it was in the STAV 2013 practice exam?

Thanks

Question 1:
Medium affects the speed of the wave. Frequency cannot change => wavelength changes. Fringe spacings affected.

Question 2:
I don't quite like how it says "energy falls to zero". Kinetic plus potential? I think the point is that if electrons from 0 eV to 4.8 eV are fired, they do not have enough energy to knock out another electron, so they just rebound back with the same energy. At 4.8 eV, some of the mercury bound electrons are given enough energy to just escape, so those electrons will have zero energy.

Diode question:
If it's in parallel, diode in forward bias takes 0.7 volts. So does resistor, as the voltage drops are the same in parallel circuits.
If the diode is reverse biased, no current flows through the diode. All current flows through the resistor instead. The diode essentially has infinite resistance. Voltage across diode is 6.0V, and so is the drop across the resistor.

As for electric power...it's just a formula, another what, two lines on your cheat sheet? Can't be that bad.
Title: Re: VCE Physics Question Thread!
Post by: BasicAcid on September 27, 2013, 05:10:16 pm
As for electric power...it's just a formula, another what, two lines on your cheat sheet? Can't be that bad.

That's literally 7 characters (EMF=Blv)

The word commutator takes up more space than that lol
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on September 27, 2013, 06:42:21 pm
That's literally 7 characters (EMF=Blv)

The word commutator takes up more space than that lol

You'd need an explanation of what the formula is and what it means.

Oh wait, wrong subject.
Title: Re: VCE Physics Question Thread!
Post by: ~T on September 27, 2013, 07:53:27 pm
TSFX Exam 2 2010, Light and Matter question 3. Basically, photoelectric effect measured on a metal with blue light and with ultraviolet light. The graph of current vs. potential difference shows that blue light produces a higher current for the forward potentials.

Question 3 asks "which of the lights had a greater intensity?" with options "blue," "ultraviolet" or "unable to determine." The answers say "unable to determine" because the intensity of the light isn't the only determining factor in the current - higher kinetic energy will also make a higher current, and this depends on the frequency of the incident light.

I chose "blue", because the ultraviolet light will have higher energy photons - E=hf. It would need more photons to equal the photocurrent of the ultraviolet light, let alone produce a larger photocurrent. Right?
Title: Re: VCE Physics Question Thread!
Post by: ~T on September 27, 2013, 08:00:16 pm
Another question... we're doing the Synchrotron detailed study, and I was wondering: what dictates whether Thomson scattering, Compton scattering, or the photoelectric effect occurs when a photon interacts with an electron?
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on September 27, 2013, 10:26:00 pm
TSFX Exam 2 2010, Light and Matter question 3. Basically, photoelectric effect measured on a metal with blue light and with ultraviolet light. The graph of current vs. potential difference shows that blue light produces a higher current for the forward potentials.

Question 3 asks "which of the lights had a greater intensity?" with options "blue," "ultraviolet" or "unable to determine." The answers say "unable to determine" because the intensity of the light isn't the only determining factor in the current - higher kinetic energy will also make a higher current, and this depends on the frequency of the incident light.

I chose "blue", because the ultraviolet light will have higher energy photons - E=hf. It would need more photons to equal the photocurrent of the ultraviolet light, let alone produce a larger photocurrent. Right?

I can see the issues here. What we're used to dealing with is the MAXIMUM kinetic energy of the electrons, and that is given by hf-W. However, not all of the electrons are at this max value. At a certain voltage, all of the electrons do get to this max value, which is why we eventually see the graph smoothen out. Before this, however, varying proportions of electrons are not at this max value. We can only work out what the maximum energy is; we don't know the energy distribution of the electrons though.
Title: Re: VCE Physics Question Thread!
Post by: SocialRhubarb on September 27, 2013, 11:27:44 pm
I chose "blue", because the ultraviolet light will have higher energy photons - E=hf. It would need more photons to equal the photocurrent of the ultraviolet light, let alone produce a larger photocurrent. Right?

If you have a laser shooting, say, 10000 photons of blue light per second, and a laser which is shooting 10000 photons of ultraviolet light per second, both on the same area, which laser has the higher intensity?

Well, the ultraviolet laser, since both the lasers emit the same number of photons, but the photons of ultraviolet light carry more energy per photon.

And that's really the issue here. Yes, from the graph we can tell that there are more photons of blue light being shone onto the metal. But intensity isn't only affected by the number of photons you shine, it's affected by the energy carried by each of those photons too. So while it is true that there are less photons of ultraviolet light, since each photon carries more energy than photons of blue light, it may be of a higher intensity, or lower intensity as well. It is unable to be determined.
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on October 03, 2013, 10:37:18 pm
question: attached
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on October 03, 2013, 10:46:53 pm
So...horizontal components of forces must be equal => tensions are equal as the angles are the same.
So now resolve vertical components.
2T*sin 25 = mg
T=mg/(2 sin 25). Whatever that is.
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on October 03, 2013, 11:28:58 pm
but the question says in the wire. so wont you multiply the answer by 2.
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on October 03, 2013, 11:35:02 pm
Is the top V-shaped thing one piece of wire?
If so...the tension will not be doubled. Even so. It's like one of those pulley questions. You could put a pulley there and it wouldn't change the force analysis.
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on October 04, 2013, 12:02:08 am
whats the 2 for in the equation?
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on October 04, 2013, 01:30:18 am
The string is split into two halves. Vertically, each half of the string contributes T sin 25 to the net force.
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on October 04, 2013, 02:50:57 pm
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on October 04, 2013, 03:06:45 pm
Yep along with thousands of other physics students.
I've got a fair few friends who have physics as their last exam as well.
We're all planning on keeping our cool together after they say pens down in the physics exam (we don't want to be reported to VCAA or anything haha) then we're gonna go absolutely ballistic outside.

Planning on getting smashed on the footy oval, hopefully it's a nice day as well considering it'll be halfway through November!

for me its gonna be a full day of sleep and PS4 session.
Title: Re: VCE Physics Question Thread!
Post by: BasicAcid on October 04, 2013, 03:25:58 pm
for me its gonna be a full day of sleep and PS4 session.

Plenty of time for that after the day in my signature, I'm buying GtaV on the 14th lol
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on October 04, 2013, 06:56:28 pm
Lucky. For me, after the physics exam, all I can say is three down, three to go.
Title: Re: VCE Physics Question Thread!
Post by: barydos on October 04, 2013, 07:12:23 pm
Just got some electricity questions from the VCAA 2007 exam
Questions: http://imgur.com/a/UhAma

1) How come the graph needed to be inverted?
2) How come it clips at 3.0 V?

Thanks

Edit: Also, it's fine for explanation questions to be answered in dot points, right? Or are we required to write full sentences?
Title: Re: VCE Physics Question Thread!
Post by: BasicAcid on October 04, 2013, 09:05:44 pm
Just got some electricity questions from the VCAA 2007 exam
Questions: http://imgur.com/a/UhAma

1) How come the graph needed to be inverted?
2) How come it clips at 3.0 V?

Thanks

Edit: Also, it's fine for explanation questions to be answered in dot points, right? Or are we required to write full sentences?

Yeah dot points are fine and wtf, I haven't done any VCAA physics exams yet (except for the 2013 sample one... That seemed incredibly easy) but I actually have no idea why it's inverted o_o

Oh well, another thing to add to my cheat sheet when nliu answers haha

Lucky. For me, after the physics exam, all I can say is three down, three to go.

Aww English Language is the day after and you still have Uni exams?
But look at the bright side, you'll be able to say three 50s down (the way you demolish those spesh, chem and physics questions... far out lol), one 45+ to go and the uni stuff doesn't really even count haha.
Title: Re: VCE Physics Question Thread!
Post by: Alwin on October 04, 2013, 10:37:09 pm
Oh well, another thing to add to my cheat sheet when nliu answers haha

Aww English Language is the day after and you still have Uni exams?
But look at the bright side, you'll be able to say three 50s down (the way you demolish those spesh, chem and physics questions... far out lol), one 45+ to go and the uni stuff doesn't really even count haha.

Im sorry to disappoint for I'm not nliu (@ nliu you have no idea what autocorrect on my iPod suggested for your nick LOL)

1) Anonymiza youre actually doing a paper from an older study design when students where actually required to know how things worked D; D; D; IKR THE SHOCK

2) Most simply put, what you have us a single transistor amplifier. Because you have an NPN transistor, when a large input signal comes in through the base, a "smaller" output voltage signal I recorded. Note if it was a double transistor amplifier then it would be non inverting

3) I really don't want to go into capacitors specifically, only because the textbook brushes on n-p and p-n junctions (not in stuy design) but no capacitors or the saturation point or cutoff point of transistors.

It's suffices to say you WON'T get a question like this. it's nice to know how an amplifiers work so you can answer questions like this, but 2013 VCE physics just doesnt require this standard.

EDIT: Sorry if I sound a bit too cynical of the physics subject. I put it down to my old age and that I did quite a few older exams (2009 was when they made the cuts/change ) just to spice up my revision last year :P
Title: Re: VCE Physics Question Thread!
Post by: barydos on October 04, 2013, 11:08:38 pm
In sorry to disappoint for I'm not nliu (@ nliu you have no idea what autocorrect on my iPod suggested for your nick LOL)

1) Anonymiza youre actually doing a paper from an older study design when students where actually required to know how things worked D; D; D; IKR THE SHOCK

2) Most simply put, what you have us a single transistor amplifier. Because you have an NPN transistor, when a large input signal comes in through the base, a "smaller" output voltage signal I recorded. Note if it was a double transistor amplifier then it would be non inverting

3) I really don't want to go into capacitors specifically, only because the textbook brushes on n-p and p-n junctions (not in stuy design) but no capacitors or the saturation point or cutoff point of transistors.

it's suffices to say you WON'T get a question like this. it's nice to know how an amplifiers work so you can answer questions like this, but VCE physics just doesnt require this standard.

1) LOL
2,3) I can see why you got 49 last year! Thanks a lot, much appreciated :)
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on October 04, 2013, 11:28:23 pm
Aww English Language is the day after and you still have Uni exams?
But look at the bright side, you'll be able to say three 50s down (the way you demolish those spesh, chem and physics questions... far out lol), one 45+ to go and the uni stuff doesn't really even count haha.

Yep, English the day after. Which is going to be one hell of an exam.
And please don't give me that. At the rate I'm going at, I'll be surprised if I even get a 50 in anything.
Title: Re: VCE Physics Question Thread!
Post by: barydos on October 05, 2013, 11:59:49 am
When solving for the voltage gain of an amplifier, and it's an inverting one, are we allowed to write a negative answer? Or must it only be the magnitude of the gain? (are both acceptable?)
Title: Re: VCE Physics Question Thread!
Post by: ~T on October 05, 2013, 12:05:42 pm
I believe both are acceptable, I usually just write $(-)100$ or whatever the gain is if it is inverting, otherwise just the number.
Title: Re: VCE Physics Question Thread!
Post by: barydos on October 05, 2013, 12:15:23 pm
I believe both are acceptable, I usually just write $(-)100$ or whatever the gain is if it is inverting, otherwise just the number.

Thanks :)

Another question: for the magnetic field graph, is it better to have the corresponding induced EMF graph to be joined (with vertical lines) or not?
(see this image: http://imgur.com/a/Gt5E2)
Title: Re: VCE Physics Question Thread!
Post by: ~T on October 05, 2013, 12:17:43 pm
Again, I'm not strictly certain on this, but I usually do dotted vertical lines. I doubt they are pedantic enough to pick up on either of your two questions though :)
Title: Re: VCE Physics Question Thread!
Post by: barydos on October 05, 2013, 12:29:55 pm
Again, I'm not strictly certain on this, but I usually do dotted vertical lines. I doubt they are pedantic enough to pick up on either of your two questions though :)

That's good to hear haha thanks!
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on October 05, 2013, 12:39:41 pm
I hear that we're meant to have vertical lines, but from a mathematical perspective and indeed a physical perspective, it's not possible to have a emf graph with vertical lines in it. The emf is single-valued; it cannot be multi-valued at any one time, so although I see vertical lines drawn in answers, I don't agree with them. It's like saying for one instant, the emf is both 1.1V, 1.2V, 1.3V and 1.4V.
Title: Re: VCE Physics Question Thread!
Post by: ~T on October 05, 2013, 09:40:44 pm
I definitely agree nliu, which is why I compromise by sitting on the fence with dotted lines. I think they clearly convey both the "I know what you want me to do" and "I know what is actually going on here" messages.
Title: Re: VCE Physics Question Thread!
Post by: barydos on October 06, 2013, 12:57:31 pm
In the VCAA 2009 paper there was a 3-mark question:
Spoiler
At the time of Young's double-slit experiment there were two competing models of the nature of light. Explain how Young's experiment supported one of these models compared with the other."
Assessment Report Comment:
Spoiler
Young’s experiment demonstrated interference effects, so his work supported the wave model of light. The particle model did not explain the interference effect.
Is a response like that enough for the whole 3 marks??

Also in general, for some 2-mark questions such as "What is the power dissipated in R1?"
Obviously it's expected that there's some working out involved. But say you did the working out straight into your calculator, and got the right answer (no working out written down). Would you still obtain the full marks? (like for multiple choice questions)
Title: Re: VCE Physics Question Thread!
Post by: Aurelian on October 06, 2013, 01:12:50 pm
In the VCAA 2009 paper there was a 3-mark question:
Spoiler
At the time of Young's double-slit experiment there were two competing models of the nature of light. Explain how Young's experiment supported one of these models compared with the other."
Assessment Report Comment:
Spoiler
Young’s experiment demonstrated interference effects, so his work supported the wave model of light. The particle model did not explain the interference effect.
Is a response like that enough for the whole 3 marks??

Personally I would add what the particle model would predict for the experiment, and perhaps mention that nature of the interference pattern actually observed as well (i.e. alternating dark and light fringes). I do think that some examiners would only award you two marks for the response given there by the examiner's report...

Also in general, for some 2-mark questions such as "What is the power dissipated in R1?"
Obviously it's expected that there's some working out involved. But say you did the working out straight into your calculator, and got the right answer (no working out written down). Would you still obtain the full marks? (like for multiple choice questions)

Why not just write down the working to be safe? Usually you'll only need one or two very short lines =) You'll also minimize the chance of making a silly mistake, as well as enabling the examiner to give you one of the two marks if you get the answer wrong but have valid working.
Title: Re: VCE Physics Question Thread!
Post by: barydos on October 06, 2013, 01:27:32 pm
Personally I would add what the particle model would predict for the experiment, and perhaps mention that nature of the interference pattern actually observed as well (i.e. alternating dark and light fringes). I do think that some examiners would only award you two marks for the response given there by the examiner's report...

Why not just write down the working to be safe? Usually you'll only need one or two very short lines =) You'll also minimize the chance of making a silly mistake, as well as enabling the examiner to give you one of the two marks if you get the answer wrong but have valid working.

Alright, I'll keep that in mind, thanks.

And yeah aha I know I know, I was just curious about it more than anything else :P
Title: Re: VCE Physics Question Thread!
Post by: ~T on October 06, 2013, 01:29:55 pm
Just to clarify on that, we don't strictly require working do we? I always write my working down anyway, but there have been a few two mark questions for which you could easily just write the answer.

I guess my query is:

No working but correct answer = full marks regardless of the number of marks?
Title: Re: VCE Physics Question Thread!
Post by: Alwin on October 06, 2013, 01:35:09 pm
In the VCAA 2009 paper there was a 3-mark question:
Spoiler
At the time of Young's double-slit experiment there were two competing models of the nature of light. Explain how Young's experiment supported one of these models compared with the other."
Assessment Report Comment:
Spoiler
Young’s experiment demonstrated interference effects, so his work supported the wave model of light. The particle model did not explain the interference effect.
Is a response like that enough for the whole 3 marks??

Also in general, for some 2-mark questions such as "What is the power dissipated in R1?"
Obviously it's expected that there's some working out involved. But say you did the working out straight into your calculator, and got the right answer (no working out written down). Would you still obtain the full marks? (like for multiple choice questions)

If I remember correctly the marking scheme for such questions went like this:

 Explain the wave model of light and what it predicts (interference pattern of light and dark) 1 Mark Explain the particle model of light and what it predicts (no such pattern, just one light band) 1 Mark Young's DS Experiment supported the wave model of light because a fringe interference pattern was observed on the screen 1 Mark

You have to remember that most assessment reports are brief because they don't want students to memorise answers from the reports. So, they just touch on important points

 Correct formula with wrong values or wrong conversion substituted in 0 Marks Correct formula with right values substituted in but wrong answer 1 Mark Correct formula with right values substituted in and correct answer 2 Mark No/wrong formula and wrong answer 0 Marks (no duh) Wrong formula/working but somehow correct answer 1 Mark (you showed bad physics in your working) No formula and correct answer 1 Mark I'm pretty sure for 2013 because for short answer questions it says: "In questions worth more than 1 mark appropriate working should be shown."

The moral of the story? Just show some working :))

Even if it's a MC question where anything that gets you the right answer goes (even closing your eyes and guessing without righting a single formula) you should write some working. Why? It's helps when you go back and double check your answer so you can make sure you used the right method and substituted correctly!

:D
Title: Re: VCE Physics Question Thread!
Post by: BasicAcid on October 06, 2013, 02:42:46 pm
You have to remember that most assessment reports are brief because they don't want students to memorise answers from the reports. So, they just touch on important points

Or better yet copy down the assessor's report onto your A3 cheat sheet
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on October 06, 2013, 02:55:19 pm
"In questions worth more than 1 mark appropriate working should be shown."

where does it say that?
Title: Re: VCE Physics Question Thread!
Post by: Alwin on October 06, 2013, 03:06:34 pm
where does it say that?

in the 2013  trial paper is says:
(http://i.imgur.com/9KVgf3d.png)
Title: Re: VCE Physics Question Thread!
Post by: ~T on October 06, 2013, 03:48:07 pm
I had always thought the fact that is explicitly states "should" rather than the "must" of Methods and Spesh exams suggests that they want you to but they can't actually mark you down for it.
Title: Re: VCE Physics Question Thread!
Post by: Alwin on October 06, 2013, 03:54:14 pm
I had always thought the fact that is explicitly states "should" rather than the "must" of Methods and Spesh exams suggests that they want you to but they can't actually mark you down for it.
"You must clean you room"

Either way you get yelled at if you don't do it :P

But seriously, better safe than sorry. Plus, what if you did the question all on your calculator get the answer wrong. Don't even get any marks for method or working. It's just safer this way :))

I always showed full working last year, but if someone on AN wants to try it with no working at all this year be my guest and let us know how it went :) (no, this is not a personal challenge at you nliu :P)
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on October 06, 2013, 04:39:19 pm
I had always thought the fact that is explicitly states "should" rather than the "must" of Methods and Spesh exams suggests that they want you to but they can't actually mark you down for it.

If you're willing to take the risk, be my guest.

"You must clean you room"

Either way you get yelled at if you don't do it :P

But seriously, better safe than sorry. Plus, what if you did the question all on your calculator get the answer wrong. Don't even get any marks for method or working. It's just safer this way :))

I always showed full working last year, but if someone on AN wants to try it with no working at all this year be my guest and let us know how it went :) (no, this is not a personal challenge at you nliu :P)

Alwin, I'm not going to potentially jeopardise my physics score just to test the system :P
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on October 06, 2013, 10:15:52 pm
in the 2013  trial paper is says:
(http://i.imgur.com/9KVgf3d.png)

Dont be smartass alright. It was a genuine question.

i just got the physics marking scheme from my teacher. it says "where questions dont specify that the working must be shown, if the correct answer is in the box the student receives full marks". Maybe you should get some suffice information yourself before absolutely classifying other peoples question invalid and, from the sound of your tone, stupid.
Title: Re: VCE Physics Question Thread!
Post by: Phy124 on October 07, 2013, 12:15:03 am
(http://cdn.meme.li/i/oy2pt.jpg)
Title: Re: VCE Physics Question Thread!
Post by: barydos on October 07, 2013, 06:04:42 pm
As you wish.

Are we still required to know about incandescent (filament) light globes?
E.g. "explain, in terms of electron behaviour how light is produced in an incandescent (filament) light globe?"
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on October 07, 2013, 06:12:34 pm
I've seen those questions before in past exams. Given the cutting of the physics course down, it's possible that you're not asked to know this.

In case you do, something on the lines of the thermal vibrations of the electrons creates an electromagnetic wave, light, would probably suffice.
Title: Re: VCE Physics Question Thread!
Post by: barydos on October 07, 2013, 06:30:28 pm
I've seen those questions before in past exams. Given the cutting of the physics course down, it's possible that you're not asked to know this.

In case you do, something on the lines of the thermal vibrations of the electrons creates an electromagnetic wave, light, would probably suffice.

Thanks nliu :)
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on October 07, 2013, 09:00:34 pm
how do you do question 8 from electronics section.
http://www.vcaa.vic.edu.au/Documents/exams/physics/2009physics1-w.pdf

Title: Re: VCE Physics Question Thread!
Post by: lzxnl on October 07, 2013, 09:12:54 pm
OK. P is definitely an electric signal as that's what microphones do; convert a sound signal into an electric one. I would say P is B.

It looks to me that Q, the wave from the RF source to the modulator is the carrier wave, high frequency, so A is at Q. Then R appears to be the modulated wave signal C. S is the original electric signal B.
Title: Re: VCE Physics Question Thread!
Post by: SocialRhubarb on October 07, 2013, 09:16:50 pm
The process of modulation involves varying the properties of a wave in order to transmit information.

So in this case we're transferring the sound coming through the microphone on the back of a radio wave.

Into the question now:

at P, all that we have is the signal from the microphone. It's a sound wave which has been converted into an electrical signal, and the other waves which we're required to label are radio waves. In general, the frequencies of light, even in the form of radio waves, will be much higher than the frequencies of sound. So if we're trying to pick a waveform which will match the sound, we'll probably pick B, as its frequencies seems the lowest.

At Q, all the we're receiving is the 'carrier wave', the wave that will be altered to carry the sound wave. In this case it is a radio wave. The wave hasn't yet reached the modulator, so what we're looking for is a flat, uniform wave, which in this case will be A.

At R, the wave has been altered to carry the signal. The answer will be C, as you can see that the waveform has been altered to 'carry' the original sound sample on top of the radio waves being emitted. You can imagine that if you multiplied the waves in A and B together, you might get something that looks like C.

At S, the wave has been demodulated to form the original electrical signal from the microphone, so we're back to B again.
Title: Re: VCE Physics Question Thread!
Post by: Jaswinder on October 07, 2013, 11:48:28 pm
is there a difference between cut off potential and cut off voltage? can the cut of potential have eV units?
Title: Re: VCE Physics Question Thread!
Post by: Robert123 on October 10, 2013, 06:48:59 pm
Photonics (detailed study) question,
Could someone please explain the what "numerical aperture" and "acceptance angle". I can usually get MC questions related to them right by applying the formula but I have no genuine idea what the numbers even mean.
Cheers
Title: Re: VCE Physics Question Thread!
Post by: ~T on October 11, 2013, 07:11:57 pm
To nliu and Alwin and potentially others, I got my hands on the VCE Physics Exam marking principles today. No working, correct answer = full marks regardless of number of available marks.

Some other absolute gems:
"If the student changes the units in the answer box and his answer is correct, give full marks."
"Scientific notation – 2 E 9 is acceptable"
Title: Re: VCE Physics Question Thread!
Post by: BasicAcid on October 11, 2013, 07:22:21 pm
Some other absolute gems:
"If the student changes the units in the answer box and his answer is correct, give full marks."
"Scientific notation – 2 E 9 is acceptable"

Wow are you serious lol

Do you think you could link us/post the marking principles?
Title: Re: VCE Physics Question Thread!
Post by: sin0001 on October 11, 2013, 07:33:05 pm
To nliu and Alwin and potentially others, I got my hands on the VCE Physics Exam marking principles today. No working, correct answer = full marks regardless of number of available marks.

Some other absolute gems:
"If the student changes the units in the answer box and his answer is correct, give full marks."
"Scientific notation – 2 E 9 is acceptable"
So if there are 3-4 marks devoted to a question, we can full mark it simply by writing the answer? Source please.
Title: Re: VCE Physics Question Thread!
Post by: ~T on October 11, 2013, 07:55:33 pm
Definitely not me asking here... it's um... for a friend... um... can you upload documents here?  :P

http://www.2shared.com/document/0BRMUi2k/General_marking_principles.html

P.S. Enjoy <3

P.P.S. I believe this was from the chief examiner during a 2012 teachers conference
Title: Re: VCE Physics Question Thread!
Post by: Phy124 on October 11, 2013, 08:18:11 pm
So if there are 3-4 marks devoted to a question, we can full mark it simply by writing the answer? Source please.

Why would anyone put just an answer when there are multiple marks allocated?

If you get the answer wrong and have no working out, then you get zero, it's as simple as that.

However, if you have the incorrect answer but have working out is certainly possible to obtain marks.

In a subject where you hear numerous stories of people mistyping things into their calculators and obtaining an incorrect final answer, I really don't understand why you would simply write an answer and nothing else, especially considering it takes, what, an extra 5 seconds to write?  ???
Title: Re: VCE Physics Question Thread!
Post by: ~T on October 11, 2013, 08:29:10 pm
I certainly agree. I was merely clarifying as it was seemingly debated earlier without any resolution. I would always do working if (almost always) I have the time, but you must admit that you occasionally get a two mark question - say, finding the current in a simple circuit - and if you happen to have stuffed up something earlier, and spent too much time fixing it or blanking on a question and find yourself under the pump, it's nice to know that you will still get the full marks IF the answer is right. To do so in a 4 mark question would of course be ridiculous, but it's also nice to know that you don't have to show a huge amount of steps if you are confident. Again, not saying I would.
Title: Re: VCE Physics Question Thread!
Post by: sin0001 on October 11, 2013, 09:57:51 pm
Why would anyone put just an answer when there are multiple marks allocated?

If you get the answer wrong and have no working out, then you get zero, it's as simple as that.

However, if you have the incorrect answer but have working out is certainly possible to obtain marks.

In a subject where you hear numerous stories of people mistyping things into their calculators and obtaining an incorrect final answer, I really don't understand why you would simply write an answer and nothing else, especially considering it takes, what, an extra 5 seconds to write?  ???
I'm the type of person that will show working even if the questions is worth 1 mark, there's no way I'm ever gonna write an answer without the working shown. I was just curious about whether the examiners have concrete rules for allocating marks
Title: Re: VCE Physics Question Thread!
Post by: Jaswinder on October 12, 2013, 03:59:53 pm
1)what effect does changing the slit width have on the diffraction pattern?
2) What would be the range of answers expected for 2005 VCAA Exam 2 question 2 from Light and Matter Section
3) Are we expected to know about the mechanism by which light is produced in an incandescent light?
4) In a transformer, Joan increases the load on the secondary side of the transformer. Suddenly, it
stops working. She suspects that the fuse in the primary circuit has blown and intends to replace it.

In order to replace the fuse as safely as possible, which of the following is the best precaution for Joan to
take?
A. stand on a rubber mat
B. switch off the mains supply
C. disconnect the transformer from the mains supply
D. remove the load from the transformer

Thanks
Title: Re: VCE Physics Question Thread!
Post by: Stevensmay on October 12, 2013, 05:59:22 pm
In order to replace the fuse as safely as possible, which of the following is the best precaution for Joan to
take?
A. stand on a rubber mat
B. switch off the mains supply
C. disconnect the transformer from the mains supply
D. remove the load from the transformer

This is just my thought process.
Switch could fail and not actually disconnect power to the fuse.
Unplugging it guarantees there will be no power to the fuse, therefore no chance of electrocution.

I think this would be a case of, one answer is slightly better than the other.
Title: Re: VCE Physics Question Thread!
Post by: Robert123 on October 12, 2013, 06:36:01 pm
Photonics (detailed study) question,
Could someone please explain the what "numerical aperture" and "acceptance angle". I can usually get MC questions related to them right by applying the formula but I have no genuine idea what the numbers even mean.
Cheers
Bumping this

Energy level diagram question:
E(eV)
0         N= infinite
-1.38   N=5
X         N=4
-1.94   N=3
Y         N=2
-5.13   N=1
A photon of 3.61eV is absorb by the ion. Moments later, photons with energy 0.42eV and 1.5eV are ejected from the atom. Use the information provide to calculate the energy levels at n=2 and n=4.

For n=1 to n=4,
5.13-3.61= 1.52eV
This agree with the energy level for n=3 since
N=4 to n=3
1.94-1.52= 0.42eV
The issue I'm having is getting the right n=2 since the 1.5eV could come from n=3 (giving a value of -3.44eV) or from n=4 (giving a value of -3.02eV).  The answer gives -3.02eV but how is this more right than -3.44eV since both could be ejected.
So which one is 'more right'? If vcaa ask a question like this would they give both answers as correct or just one?
Title: Re: VCE Physics Question Thread!
Post by: SocialRhubarb on October 12, 2013, 11:23:38 pm
The key here is that you're dealing with a single atom absorbing a single photon.

If the jump from N=4 to N=2 gives us a 1.5 eV photon, it can no longer produce a photon with an energy of 0.42 eV, hence it cannot jump from N=4 to N=2 to produce a 1.5 eV photon.
Title: Re: VCE Physics Question Thread!
Post by: Jaswinder on October 13, 2013, 08:26:20 am
what effect does changing the slit width have on the diffraction pattern?
Title: Re: VCE Physics Question Thread!
Post by: Stevensmay on October 13, 2013, 11:51:53 am
Our equation for the extent of diffraction is $extent \propto \frac{\lambda}{w}$ where w is the width of the slit.
We can see that as we increase w, $\frac{\lambda}{w}$ becomes smaller, thus less diffraction occurs.
As we decrease w, more diffraction occurs as $\frac{\lambda}{w}$ is now bigger.

This also explains why we need to use very thin slits/apertures for this experiment, as if they are too big any diffraction is not evident.
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on October 13, 2013, 12:16:14 pm
Our equation for the extent of diffraction is $extent \propto \frac{\lambda}{w}$ where w is the width of the slit.
We can see that as we increase w, $\frac{\lambda}{w}$ becomes smaller, thus less diffraction occurs.
As we decrease w, more diffraction occurs as $\frac{\lambda}{w}$ is now bigger.

This also explains why we need to use very thin slits/apertures for this experiment, as if they are too big any diffraction is not evident.

what about the intensity of the pattern? and what effect does it have on the fringe spacing?
Title: Re: VCE Physics Question Thread!
Post by: BasicAcid on October 13, 2013, 03:46:58 pm
what about the intensity of the pattern? and what effect does it have on the fringe spacing?

No effect on the intensity and as the diffraction is larger, the diffraction patterns are also larger, i.e. the fringe spacing is larger
Title: Re: VCE Physics Question Thread!
Post by: barydos on October 15, 2013, 09:25:20 pm
Hello
This question came from exam 1 of the VCAA 2006 paper:
Spoiler
(http://i.imgur.com/MVC5OOU.png)
The assessment report's comment:
Spoiler
(http://i.imgur.com/iWEThyJ.png)

I'd just like to know what your responses might be, because I feel the provided comment is insufficient.
Title: Re: VCE Physics Question Thread!
Post by: Alwin on October 15, 2013, 09:42:23 pm
offtopic
Dont be smartass alright. It was a genuine question.

i just got the physics marking scheme from my teacher. it says "where questions dont specify that the working must be shown, if the correct answer is in the box the student receives full marks". Maybe you should get some suffice information yourself before absolutely classifying other peoples question invalid and, from the sound of your tone, stupid.
Sorry lolipopper if I made you feel like you'd asked a "stupid question" or anything, I didn't mean it in that way. I was just joking with you, about the "read the instructions first" part I put in the picture. I haven't been on this thread in a while, been rather stressed for Indo oral exam so again I apologise for being or seeming a bit short with you.
I was just showing the source of my "information". I can only talk about the pre-2012 exam marking scheme, and as Tim...blahhh has confirmed from more accurate sources Re: Physics [3/4] Question Thread! it turns out I was wrong. Not going to blame the practise exam, it's just that it turns out I was wrong and I'm sorry if my response was inconsiderate.

Good luck to everyone with physics :)
Title: Re: VCE Physics Question Thread!
Post by: sin0001 on October 15, 2013, 10:17:45 pm
With questions talking about the extent of diffraction: do we assume that max. diffraction occurs when wavelength/width ratio ~ 1 or when this ratio is a large value?
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on October 15, 2013, 11:10:43 pm
offtopic
Sorry lolipopper if I made you feel like you'd asked a "stupid question" or anything, I didn't mean it in that way. I was just joking with you, about the "read the instructions first" part I put in the picture. I haven't been on this thread in a while, been rather stressed for Indo oral exam so again I apologise for being or seeming a bit short with you.
I was just showing the source of my "information". I can only talk about the pre-2012 exam marking scheme, and as Tim...blahhh has confirmed from more accurate sources Re: Physics [3/4] Question Thread! it turns out I was wrong. Not going to blame the practise exam, it's just that it turns out I was wrong and I'm sorry if my response was inconsiderate.

Good luck to everyone with physics :)

im sorry too bro. all good. and good luck to you too.

With questions talking about the extent of diffraction: do we assume that max. diffraction occurs when wavelength/width ratio ~ 1 or when this ratio is a large value?

yes, the size of the aperture must be approximately the same as that of the wavelength for significant diffraction to occur.

Mod Edit - Merged double posts (Phy124)
Title: Re: VCE Physics Question Thread!
Post by: joey7 on October 16, 2013, 08:46:01 pm
Quick question
"The shape of racing cars is designed to minimise air resistance. To a reasonable approximation the force (F) or air resistance is proportional to the square of the speed (v). That is: F=kv^2 where k is the drag coefficient.:

Which if the following choice is a possible unit for K
A. kgm^-1
B. s^2m^2
C. N
D. Nm^2s^-1
E. S^2M^-2
Title: Re: VCE Physics Question Thread!
Post by: SocialRhubarb on October 16, 2013, 08:51:46 pm
Force is equal to mass times acceleration, so it has the units of mass, kg, times acceleration, ms-2, so force has the units of kg m s-2.

v2 has the units of m2 s-2.

k=F/v2=kg m s-2 m-2 s2=kg m-1

So the answer is A, kg m-1.
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on October 17, 2013, 08:35:25 pm
what is a good answer to the following:
-what is modulation and demodulation?
Title: Re: VCE Physics Question Thread!
Post by: Jaswinder on October 17, 2013, 09:53:53 pm
I understand Question 9 and question 14 in motion  :-\
http://www.vcaa.vic.edu.au/Documents/exams/physics/physics1w06.pdf

thanks
Title: Re: VCE Physics Question Thread!
Post by: barydos on October 17, 2013, 09:54:22 pm
If asked what RMS voltage was, what would you respond with?
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on October 17, 2013, 10:04:56 pm
The corresponding constant DC voltage that provides the same average power output.
Title: Re: VCE Physics Question Thread!
Post by: barydos on October 17, 2013, 10:08:56 pm
The corresponding constant DC voltage that provides the same average power output.

Thanks!
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on October 17, 2013, 10:14:00 pm
what is a good answer to the following:
-what is modulation and demodulation?

Modulation is the conversion of an electric signal into a light signal. In the VCE course, we only look at amplitude modulation. In this form of modulation, the shape of the electric field signal is embedded in the shape of the carrier wave light signal, which generally has a very high frequency. When the electric field is negative, the light signal's amplitude is just lower than its average amplitude, which corresponds to the zero of the electric field signal. The purpose of the modulation is to allow the electric signal to be literally carried by the carrier way across a distance to where it is demodulated, or converted back into an electric signal, for use.

As an example, let's take two circuits, one with an LED and one with a photodiode. In the circuit with the LED, the current through the LED determines brightness of the light emitted. This light contains information about the current of the LED circuit; it is modulated. When it hits the photodiode, the intensity of the light then determines the current of the photodiode circuit; the light becomes demodulated. That's the example I use to remember what modulation does.
Title: Re: VCE Physics Question Thread!
Post by: Robert123 on October 18, 2013, 09:26:58 pm
VCAA 2010 Q16:
Basically a moving trolley with a spring on its front collide into a stationary trolley. The spring compress then uncompressed in the collision which  is elastic. How would the total momentum of the system varies with time. The answers says that momentum is always conserved in a collision so it will be a straight line with no variation when the springs come into play. But, my intuition tells me that some of the momentum will sort of be temporarily stored within the spring and decrease, could someone please explain to me why I'm wrong. Cheers :)
Title: Re: VCE Physics Question Thread!
Post by: barydos on October 18, 2013, 10:36:28 pm
This was a VCAA 2006 question:
Spoiler
(http://i.imgur.com/TN8UFLV.png)
(http://i.imgur.com/m9c0257.png)
Spoiler
$3.0*10^{-6} V$

So for max induced voltage, we would use that -NBA/t formula. The difficult for me was finding 't'
If the square was moving at 4.0 cms^-1, I thought the max change in flux would have been when it reaches the middle of the face of the magnet (aka 6cm in?) so t would be 6/4 = 1.5 seconds. But the answer suggests that it just takes 0.5 seconds meaning the square only just goes completely in the field.
Could someone just clarify if I'm being silly or not, and whether or not the square requires to be in the middle or not?
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on October 18, 2013, 10:44:59 pm
The square doesn't need to be in the middle; it just needs to be in a position such that the entire square is in the magnetic field. This is because the magnetic field is uniform.
Remember, the formula flux = BA refers to the area that is exposed to the magnetic field.
Now, the maximum voltage induced would then be due to the change in flux from when the square first enters the field to when it is fully immersed in the field. The side length is 2 cm, and the speed of motion is 4 cm/s, so the time we want is 0.5 s.
Just one loop only.
The change in flux is 3.7*10^-3 T * (2 cm)^2 = 14.8 * 10^-7 T m^2 (note change in units)
So dividing by the time, our max voltage is 29.6* 10^-7 V, so 2.96 uV which is the unrounded form of the answer given.

The trick is identifying what the time means. The time is the period of time over which the change in flux occurs. Note that when the square is in the middle of the magnetic field, the entire square will have remained in the field for a period of time, so the flux isn't changing; the voltage is zero in that case.

Honestly mate, I don't fully understand half the crap we're learning in physics and I got 96% on that sample VCAA 2013 exam.

All you need to do is sub in values.

We know speed = distance/time so therefore time = distance/speed = 0.02/0.04 = 0.5, all values directly from the question subbed into our formulas from our A3 cheat sheet.

Then sub time = 0.5 into the EMF equation and wallah, you get the answer.

Another example of how physics is becoming ridiculous; you don't even need any understanding of the mechanics of the formulas to get the marks.
Title: Re: VCE Physics Question Thread!
Post by: SocialRhubarb on October 18, 2013, 10:49:44 pm
VCAA 2010 Q16:
Basically a moving trolley with a spring on its front collide into a stationary trolley.
Spoiler
I think this is quite a difficult question to grasp conceptually, and difficult to answer without simply parroting the line that momentum is always conserved. But essentially what is happening is that forces are acting on the spring in both directions, and the spring is exerting a force on both trolleys. Because momentum is a vector quantity, that is, it has direction, exerting equal forces in opposite directions results in no change in momentum. The problem is it is a little more complicated than that, as the spring would not exert equal forces on both the trolleys, but essentially what would happen is that it would exert a slightly greater force on the stationary trolley than the moving trolley behind it, and the 'unbalanced' force actually contributes to the change in momentum of the spring itself. I expect that VCAA might expect you to briefly touch on some of these points and simply mention that momentum is always conserved.

I've talked about both forces on the spring and forces on the trolley, and used these interchangeably as they'll be of the same magnitude as they're Newton third law pairs, but it can make it a bit confusing.
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on October 18, 2013, 11:01:33 pm
Spoiler
I think this is quite a difficult question to grasp conceptually, and difficult to answer without simply parroting the line that momentum is always conserved. But essentially what is happening is that forces are acting on the spring in both directions, and the spring is exerting a force on both trolleys. Because momentum is a vector quantity, that is, it has direction, exerting equal forces in opposite directions results in no change in momentum. The problem is it is a little more complicated than that, as the spring would not exert equal forces on both the trolleys, but essentially what would happen is that it would exert a slightly greater force on the stationary trolley than the moving trolley behind it, and the 'unbalanced' force actually contributes to the change in momentum of the spring itself. I expect that VCAA might expect you to briefly touch on some of these points and simply mention that momentum is always conserved.

I've talked about both forces on the spring and forces on the trolley, and used these interchangeably as they'll be of the same magnitude as they're Newton third law pairs, but it can make it a bit confusing.

Why don't we look at a different reference frame, that of the "stationary trolley", so that the situation becomes a moving trolley collides with a trolley with a spring on it?
Let the now moving trolley be trolley A and the stationary trolley with a spring be trolley B.
Now, as trolley A runs into the spring, A exerts a force on the spring, which compresses it. Similarly, the spring exerts a retarding force on A which slows it down. The spring, meanwhile, also exerts a force on trolley B, which reduces the compression of the spring, effectively transferring momentum from A to B. As the spring compresses, it slows A down more and more. When it can be compressed no further, it pushes A backwards and this pushes B forwards. Even when the spring isn't being compressed and is 'stationary', the situation becomes similar to having two blocks; here, A effectively pushes directly onto B. When the spring unwinds itself, it pushes A backwards and B forwards. Momentum is conserved, but there are a number of forces at play here.

Note; conservation of momentum can be seen to be a consequence of Newton's third law in classical mechanics, which is partly the argument I'm making here.
Title: Re: VCE Physics Question Thread!
Post by: SocialRhubarb on October 18, 2013, 11:12:05 pm
I think the spring's on the other trolley in the question, although obviously it's essentially the same scenario anyway. I think what Robert was interested in though was how the compression of the spring affected the momentum of the trolleys.

Also, how does the spring exerting a force on trolley B reduce the compression of the spring? Surely at that stage of your argument the spring is still being compressed.
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on October 18, 2013, 11:21:38 pm
The spring is on the other trolley to begin with, but I decided to just switch things around for simplicity in my own mind. You're right, it doesn't make a difference.

The point about the force on trolley B is that the compression of the spring is occurring because trolley A is pushing onto the spring. Trolley B is in the other direction of the applied force, so pushing on that, and trolley B's resultant motion, relaxes the spring. It's probably most easily demonstrated with a wall. If you ran a trolley with a spring on it against a wall and the wall doesn't move, the spring is compressed a lot, because although it pushes on the wall, the wall doesn't budge. In my scenario, the spring pushes on trolley A and makes it move, and as the trolley is attached to the spring, the spring is stretched in the opposite direction it is being compressed.
Title: Re: VCE Physics Question Thread!
Post by: barydos on October 18, 2013, 11:28:31 pm
The square doesn't need to be in the middle; it just needs to be in a position such that the entire square is in the magnetic field. This is because the magnetic field is uniform.
Remember, the formula flux = BA refers to the area that is exposed to the magnetic field.
Now, the maximum voltage induced would then be due to the change in flux from when the square first enters the field to when it is fully immersed in the field. The side length is 2 cm, and the speed of motion is 4 cm/s, so the time we want is 0.5 s.
Just one loop only.
The change in flux is 3.7*10^-3 T * (2 cm)^2 = 14.8 * 10^-7 T m^2 (note change in units)
So dividing by the time, our max voltage is 29.6* 10^-7 V, so 2.96 uV which is the unrounded form of the answer given.

The trick is identifying what the time means. The time is the period of time over which the change in flux occurs. Note that when the square is in the middle of the magnetic field, the entire square will have remained in the field for a period of time, so the flux isn't changing; the voltage is zero in that case.

Another example of how physics is becoming ridiculous; you don't even need any understanding of the mechanics of the formulas to get the marks.

Ahhh that makes so much sense. Thanks for the clarification!
Title: Re: VCE Physics Question Thread!
Post by: SocialRhubarb on October 18, 2013, 11:29:01 pm
Oh, I think I see what you're saying, but surely you should state that it's the acceleration of trolley B that's important, not the force.

In your example of the wall, the spring is still exerting a force. In fact, it's probably exerting an even greater force. But the force doesn't 'decrease the compression'. The force increases the compression of the spring. You can imagine that if you had no force you'd have no compression. Which is why it's strange to say that the "force on trolley B ... reduces the compression of the spring".

But of course, the force accelerates the trolley, which increases its velocity, reducing its speed relative to the spring, reducing its compression, which I think is what you're trying to say?
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on October 18, 2013, 11:37:08 pm
But of course, the force accelerates the trolley, which increases its velocity, reducing its speed relative to the spring, reducing its compression, which I think is what you're trying to say?

Yes, that was my point that I failed to articulate clearly. Thank you for clarifying it :D
Title: Re: VCE Physics Question Thread!
Post by: barydos on October 19, 2013, 11:07:01 am
How do you do this question?
Spoiler
(http://i.imgur.com/RXyjDio.png)

What happens between the two?
Title: Re: VCE Physics Question Thread!
Post by: Jaswinder on October 19, 2013, 11:49:31 am
The answer is A, but shouldnt it be the other way around? SO negative current at start then positive? (because of the EMF being (-) change in flux?) Thanks
Title: Re: VCE Physics Question Thread!
Post by: bonappler on October 19, 2013, 12:01:31 pm
The answer is A, but shouldnt it be the other way around? SO negative current at start then positive? (because of the EMF being (-) change in flux?) Thanks
I don't think it matters, remember the motion has to be relative.
How do you do this question?
Spoiler
(http://i.imgur.com/RXyjDio.png)

What happens between the two?
Remember when there are two South poles to draw a dotted line (similar to an asymptote in maths) between the two. Then draw your field lines going from the two south poles and behaving asymptotically with the line you have drawn. I am pretty sure you can also draw some conventional lines from the north pole to the south pole as well. It's pretty hard to describe so I am sorry if you do not understand. Maybe someone can upload a picture.
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on October 19, 2013, 12:26:38 pm
How do you do this question?
Spoiler
(http://i.imgur.com/RXyjDio.png)

What happens between the two?

its explained in the pic
Title: Re: VCE Physics Question Thread!
Post by: barydos on October 19, 2013, 01:17:15 pm
I don't think it matters, remember the motion has to be relative.Remember when there are two South poles to draw a dotted line (similar to an asymptote in maths) between the two. Then draw your field lines going from the two south poles and behaving asymptotically with the line you have drawn. I am pretty sure you can also draw some conventional lines from the north pole to the south pole as well. It's pretty hard to describe so I am sorry if you do not understand. Maybe someone can upload a picture.
its explained in the pic

Thanks both of you :D

Another question haha
How do you draw the graph of question 11: http://www.vcaa.vic.edu.au/Documents/exams/physics/2011physics2-w.pdf
I thought it would've contained some "rectangular graphs" because constant change in flux = constant voltage? but apparently the graph is a curve.
This leaves me wondering why the answer to electric power q8: http://www.vcaa.vic.edu.au/Documents/exams/physics/2008physics2-w.pdf  is option C.
I would have thought they'd produce the same graphs (looking like q8 option C)

Title: Re: VCE Physics Question Thread!
Post by: SocialRhubarb on October 19, 2013, 03:22:08 pm
The difference in the questions is that the question in the 2008 VCAA exam states that the field can be considered 0 outside the poles, while it cannot be assumed so in the 2011 paper. In terms of realism, the curved graph is a more accurate representation, but sometimes for simplicity's sake, we make assumptions like no field outside the poles.
Title: Re: VCE Physics Question Thread!
Post by: barydos on October 19, 2013, 09:37:02 pm
The difference in the questions is that the question in the 2008 VCAA exam states that the field can be considered 0 outside the poles, while it cannot be assumed so in the 2011 paper. In terms of realism, the curved graph is a more accurate representation, but sometimes for simplicity's sake, we make assumptions like no field outside the poles.

Didn't even notice that small detail, thanks :)
Title: Re: VCE Physics Question Thread!
Post by: SocialRhubarb on October 19, 2013, 10:44:39 pm
The net force, which is the centripetal force in this case, goes towards the centre of the circular path of rotation.

If you think about it, if the net force was towards the centre of the chain, the person in the device would accelerate upwards, towards the centre of the chain, but they clearly instead follow a circular path.

We find the magnitude of the tension by realising that the net force must be the centripetal force, and so the vertical component of the tension must negate the person's weight force, while the horizontal component of the tension provides the centripetal force. With these two figures, we can calculate the magnitude of the tension in the chain with Pythagoras' theorem.
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on October 19, 2013, 11:10:36 pm
Let's clarify something. The net force is directed radially inwards from the object to the axis of rotation. For a constant axis of rotation, this axis of rotation is perpendicular to the force vector. Think about that. You can see how this works in the diagram.

As for astronauts in outer space, there is no contradiction. The net force is directed inwards in a direction that is perpendicular to the imaginary axis they are revolving around.

You can see that the tension has a horizontal component, which is the net force mv^2/r, and a vertical component to cancel out the weight force. As these components are perpendicular, use Pythag.
Title: Re: VCE Physics Question Thread!
Post by: barydos on October 20, 2013, 11:20:31 am
When drawing a circuit, how do we know when the switch has to be opposite a thermistor or LDR rather than a variable resistor (or vice versa)?
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on October 20, 2013, 02:19:32 pm
When drawing a circuit, how do we know when the switch has to be opposite a thermistor or LDR rather than a variable resistor (or vice versa)?

you have to see when the heater must be turned on. If it is to be turned on when the voltage increases then against the element whose resistance increases (Voltage increases across it as well) and vice versa.
Title: Re: VCE Physics Question Thread!
Post by: barydos on October 20, 2013, 09:48:28 pm
you have to see when the heater must be turned on. If it is to be turned on when the voltage increases then against the element whose resistance increases (Voltage increases across it as well) and vice versa.

That makes sense, thank you!
Title: Re: VCE Physics Question Thread!
Post by: Robert123 on October 21, 2013, 08:07:44 pm
This satellite question is driving me crazy!!
Q8 of motion in the 2013 sample exam by VCAA
R=1.35*10^7m
G=6.67*10^-11
M (earth)=5.98*10^24kg
m (satellite)=525kg
What is the period?
Both Vicphysics and Itute answer say it is1.53*10^4s but whenever I do it I always get 1.56*10^4s. It is so close yet so far away, could someone please test this question, I know how to do it but it just isn't working.

Title: Re: VCE Physics Question Thread!
Post by: random_person on October 21, 2013, 08:57:37 pm
This satellite question is driving me crazy!!
Q8 of motion in the 2013 sample exam by VCAA
R=1.35*10^7m
G=6.67*10^-11
M (earth)=5.98*10^24kg
m (satellite)=525kg
What is the period?
Both Vicphysics and Itute answer say it is1.53*10^4s but whenever I do it I always get 1.56*10^4s. It is so close yet so far away, could someone please test this question, I know how to do it but it just isn't working.

You're right!!!!!!?????????
Title: Re: VCE Physics Question Thread!
Post by: Thu Thu Train on October 21, 2013, 09:57:36 pm
Strange I also get 1.56x104 but two places(who you should assume are correct) are reporting the same answer which implies they've either both made mistakes or we're missing something obvious.
Title: Re: VCE Physics Question Thread!
Post by: Stevensmay on October 21, 2013, 10:03:53 pm
The answer you have gotten is correct. One upside of physics is that there often aren't very many technicalities to a question, it's just a matter of plugging numbers in.
Title: Re: VCE Physics Question Thread!
Post by: Thu Thu Train on October 21, 2013, 10:15:05 pm
The answer you have gotten is correct. One upside of physics is that there often aren't very many technicalities to a question, it's just a matter of plugging numbers in.

(http://i.imgur.com/nUqLD8D.gif)
Title: Re: VCE Physics Question Thread!
Post by: Stevensmay on October 21, 2013, 10:23:15 pm
Lets be honest here, this is VCAA style physics.
The only thing that has some degree of 'better watch out for that' I find is light.
Motion is essentially 'I've got these letters, let me look at my cheatsheet to see which formula has them.'
Electronics is 'can I read a graph, a formula and use V=IR.'

Light actually takes some learning, but only because VCAA tend to ask more theory orientated questions.
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on October 21, 2013, 11:55:56 pm
Lets be honest here, this is VCAA style physics.
The only thing that has some degree of 'better watch out for that' I find is light.
Motion is essentially 'I've got these letters, let me look at my cheatsheet to see which formula has them.'
Electronics is 'can I read a graph, a formula and use V=IR.'

Light actually takes some learning, but only because VCAA tend to ask more theory orientated questions.

i havent dropped a single mark for any unit 4 exam that i have done for physics. However i tend to drop 1 or 2 marks on the tension and relative motion questions for Unit 3. This is why only a quarter of my summary sheet is light and matter and electricity.The rest is motion.
Title: Re: VCE Physics Question Thread!
Post by: Robert123 on October 22, 2013, 07:39:39 am
Thanks for the response :)
I knew that vicphysics solutions has some other errors in their answer but when I check itute and they both have the same answer, it threw me off completely
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on October 23, 2013, 09:35:05 pm
quick question:

is relative motion still in Unit 3?
and also are the specific properties of materials to be known in "Detailed study : Materials and Structure"
Title: Re: VCE Physics Question Thread!
Post by: Alwin on October 23, 2013, 09:52:43 pm
This satellite question is driving me crazy!!
Q8 of motion in the 2013 sample exam by VCAA
R=1.35*10^7m
G=6.67*10^-11
M (earth)=5.98*10^24kg
m (satellite)=525kg
What is the period?
Both Vicphysics and Itute answer say it is1.53*10^4s but whenever I do it I always get 1.56*10^4s. It is so close yet so far away, could someone please test this question, I know how to do it but it just isn't working.

Oh this question :) VCAA copy + pasted that q from the 2009 exam iirc, but changed the numbers. SO the solutions companies posted up were wrong too because they copied their solutions from 2009 too lol

here, from a set of 'model' solutions I made myself for the 2013 practice exam :) :
(http://i.imgur.com/LrNa17g.png)

@lollipopper: when I did physics last year it wasn't, I think it was in Special Relativity detail study for some reason (apparently too hard for normal physics students?). And I don't recall there being any specifics needed to be rote learned for Materials I think. If there are you don't even need to learn it, just chuck it on your cheat sheet if anyone knows what needs to be memorised this year :D (sorry, it's been over a year and a half for me since materials so forgotten heaps =\)
Title: Re: VCE Physics Question Thread!
Post by: Jaswinder on October 24, 2013, 07:41:00 pm
how to do 1b from 2012 Exam 2? About the orientation the magnet will take?
Title: Re: VCE Physics Question Thread!
Post by: sin0001 on October 25, 2013, 11:22:23 am
Does it matter where you place the diode in a series circuit?

For example, you have the positive end of supply connected to reverse biased diode then diode connected to resistor then resistor connected back to supply.
The entire voltage supply drops across the diode as no current flows through it hence no drop across resistor.

But how about if the position of the diode and the resistor are swapped? Would the same thing happen?
What happens to the current that flows across the resistor first?
Current would first pass through the resistor, resulting in a drop, and THEN it'll be stopped by the diode in reverse-biased.
Title: Re: VCE Physics Question Thread!
Post by: sin0001 on October 25, 2013, 11:49:12 am
If the current passes through the reverse biased diode first, all 6 volts of the supply would dropped across it, meaning there will be NO drop across the resistor
Title: Re: VCE Physics Question Thread!
Post by: Robert123 on October 25, 2013, 12:38:14 pm
But what I'm asking is what if the current flows through the resistor first, then the reverse biased diode stops the current flow?
For the current to flow, you would need a complete circuit (ie doesn't flow in one half then stops)
Assuming the reverse bias diode has no leakage current, it would probably just act like an incomplete circuit.
The best way to find out is just to actually test it in real life.
Title: Re: VCE Physics Question Thread!
Post by: SocialRhubarb on October 25, 2013, 12:54:40 pm
No current flows if you have a diode in reverse bias anywhere in a series circuit.

Robert123 is right - the diode in reverse bias would act just like a break in the circuit.

You can't have current flowing halfway through a circuit and then stopping.
Title: Re: VCE Physics Question Thread!
Post by: SocialRhubarb on October 25, 2013, 01:38:45 pm
Okay, I don't actually do structures and materials so there may be an easier way to do this, but here goes:

I think in this case it may actually be helpful to at least think of the force diagram. There are three forces involved here - the force from the rod, the force from the wire and the weight force of the lamp, and we want the lamp to have a net force of 0. Now, which ways are the forces acting?

Weight force is quite simple, obviously straight down. But the rod is being compressed, so will be exerting a force diagonally upwards on the lamp, while the wire is in tension, so will be exerting a force to the left.

If we resolve these forces, the vertical component of the rod's force must balance out the lamp's weight force, and the horizontal component of the rod's force must balance out the tension in the string.

Spoiler
$mg=F\cos(30^\circ)$

$F=230.94\text{ N}$

$T=F\sin(30^)=115.47\text{ N}\approx 115\text{ N}$
Title: Re: VCE Physics Question Thread!
Post by: Jaswinder on October 25, 2013, 04:03:33 pm
how to do 1b from 2012 Exam 2? About the orientation the magnet will take?

bumb :)
Title: Re: VCE Physics Question Thread!
Post by: Thu Thu Train on October 25, 2013, 04:25:12 pm
bumb :)

Figure out the where the 'north' and 'south' end of the solenoid is. Your smaller magnet will act as a compass needle which(without any other magnetic fields) points at magnetic north. Placing the magnet near point Q and allowing it freely rotate means that the north end of the magnet will rotate to face the south end of the solenoid.

Spoiler
Title: Re: VCE Physics Question Thread!
Post by: Jaswinder on October 25, 2013, 05:56:15 pm
Figure out the where the 'north' and 'south' end of the solenoid is. Your smaller magnet will act as a compass needle which(without any other magnetic fields) points at magnetic north. Placing the magnet near point Q and allowing it freely rotate means that the north end of the magnet will rotate to face the south end of the solenoid.

Spoiler

wouldn't the magnetic field of earth have any effect on it?
Title: Re: VCE Physics Question Thread!
Post by: Jaswinder on October 25, 2013, 07:26:23 pm
also just to be 100% sure, we can take in 2 double sided a4 sheets into the exam yeh? :)
Title: Re: VCE Physics Question Thread!
Post by: BasicAcid on October 25, 2013, 08:05:13 pm
also just to be 100% sure, we can take in 2 double sided a4 sheets into the exam yeh? :)

They need to be stuck together, but yes.
Title: Re: VCE Physics Question Thread!
Post by: Jaswinder on October 25, 2013, 08:31:35 pm
stuck together? how do i do that? you can't bound 2 A4 sheets can you? would stapling them together work?
Title: Re: VCE Physics Question Thread!
Post by: Alwin on October 25, 2013, 08:34:46 pm
stuck together? how do i do that? you can't bound 2 A4 sheets can you? would stapling them together work?

you can try photocopying double sided onto A3 if you want :)
Title: Re: VCE Physics Question Thread!
Post by: Jaswinder on October 25, 2013, 08:38:40 pm
thatd be the last option  :( what else can we do?
Title: Re: VCE Physics Question Thread!
Post by: Robert123 on October 29, 2013, 08:18:43 pm
Having difficulty with this lenzs law from VCAA 2007 exam 2 Q14 & 15
Their reasoning for their answer is...
"when the switch was closed a magnetic field built up to the left. To oppose this, the induced current must produce a magnetic field to the left."
Wouldn't the induced current produce a field to the right?
Title: Re: VCE Physics Question Thread!
Post by: angus_grant on October 30, 2013, 05:35:15 pm

"when the switch was closed a magnetic field built up to the left. To oppose this, the induced current must produce a magnetic field to the left."

Yeah must have been a typo, without looking at the question id say the flux increased in the left direction, and so to oppose this, a current was induced that had a flux in the right direction, then do the right hand rule to find current flow (guessing its a solenoid question or something similar)
hope this helps :)
Title: Re: VCE Physics Question Thread!
Post by: Robert123 on October 30, 2013, 07:23:23 pm
Yeah must have been a typo, without looking at the question id say the flux increased in the left direction, and so to oppose this, a current was induced that had a flux in the right direction, then do the right hand rule to find current flow (guessing its a solenoid question or something similar)
hope this helps :)

I'm not sure if it was a typo, since it was a MC choice question, the answer would be the complete opposite. IF it is not too much of a hassle could be please have a look at it at http://www.vcaa.vic.edu.au/Documents/exams/physics/2007physics2.pdf page 11.
Title: Re: VCE Physics Question Thread!
Post by: SocialRhubarb on October 30, 2013, 07:30:32 pm
Highly doubt there would be an error in VCAA solutions. If something's wrong they will revise it.

Okay, so when the switch is closed, the current is increasing, creating an increasing magnetic field. The field's direction points towards the left. At the second coil, the direction of the magnetic field is also pointing towards the left. The current induced in the coil will be such that it opposes the change, hence we want to induce a field towards the right. Creating a field  pointing towards the right requires a current to flow from Y to X.

The current only flows momentarily, as once the current maintains a steady value there will no longer be any change in flux and hence no induced current.

Title: Re: VCE Physics Question Thread!
Post by: Robert123 on October 30, 2013, 07:43:16 pm
Highly doubt there would be an error in VCAA solutions. If something's wrong they will revise it.

Okay, so when the switch is closed, the current is increasing, creating an increasing magnetic field. The field's direction points towards the left. At the second coil, the direction of the magnetic field is also pointing towards the left. The current induced in the coil will be such that it opposes the change, hence we want to induce a field towards the right. Creating a field  pointing towards the right requires a current to flow from Y to X.

The current only flows momentarily, as once the current maintains a steady value there will no longer be any change in flux and hence no induced current.

But wouldn't a current flowing from Y to X produce a magnetic field to the left inside the solenoid? Using the RH grip rule on the leftest loop, your fingers would wrap around pointing to the left on the solenoid side
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on October 30, 2013, 07:54:50 pm
Close switch=>magnetic field to left induced=>increasing magnetic flux to left=>induced magnetic field to right=>use right hand rule, current goes from X to Y. OUTSIDE the ammeter. Through the ammeter, the direction is from Y to X.

The answer B is perfectly fine.
Title: Re: VCE Physics Question Thread!
Post by: Alwin on October 30, 2013, 08:14:06 pm
Alternate way of thinking:

> Close switch
> Magnetic field increasing to the left
^ using right hand grip rule
> Same as moving towards a south pole
> So south pole induced at the end closest to first coil
^ using Lenz's law
> Current flows from Y to X
^ using right hand grip rule again

This alternate method works best when there is on magnet and coil moving towards each other / in circles / whatever vcaa wants lol
Title: Re: VCE Physics Question Thread!
Post by: Jaswinder on October 30, 2013, 08:55:20 pm
whats a good cheatsheet appropriate definition for modulation and demodulation? thanks :D
Title: Re: VCE Physics Question Thread!
Post by: Stevensmay on October 30, 2013, 09:40:07 pm
Modulation: Imprinting a lower frequency data signal onto a higher frequency carrier wave.

Demodulation is the removal of the carrier wave to leave only the data wave. Reverse the above image.
Title: Re: VCE Physics Question Thread!
Post by: Jaswinder on October 30, 2013, 10:24:31 pm
thanks man!

also this one, i get q to p but answers say p to q
Title: Re: VCE Physics Question Thread!
Post by: BasicAcid on October 30, 2013, 10:36:52 pm
Lenz's law, the current produced opposes that of the change in flux
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on October 30, 2013, 10:37:53 pm
North pole moves to left=>induced magnetic field points to the the right=>use right hand rule, current is from q to p in the EXTERNAL circuit, but p to q through the ammeter.
Title: Re: VCE Physics Question Thread!
Post by: Jaswinder on October 30, 2013, 10:39:16 pm
North pole moves to left=>induced magnetic field points to the the right=>use right hand rule, current is from q to p in the EXTERNAL circuit, but p to q through the ammeter.

whats the external circuit and how is it different to this one  :-\ :-[
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on October 30, 2013, 10:51:54 pm
The external circuit is the solenoid that the ammeter is connected to.
Title: Re: VCE Physics Question Thread!
Post by: Robert123 on October 31, 2013, 02:04:43 pm
Having some difficulties with lenzs law.
A loop is sitting in a magnetic field that is going out of the page, it is quickly pulled to the left. Which direction will the current flow (clockwise or anticlockwise)?
Looking at the left hand side of the loop, you can use the right hand slap rule to determine the direction of the current( fingers going out of the page, palm to the right to oppose the force to  the left, current will be clockwise). However, if you do the same analysis on the other side, the thumb would be pointing the same way so current will be anticlockwise. Which one is correct and why?
Title: Re: VCE Physics Question Thread!
Post by: Stevensmay on October 31, 2013, 09:25:47 pm
Having some difficulties with lenzs law.
A loop is sitting in a magnetic field that is going out of the page, it is quickly pulled to the left. Which direction will the current flow (clockwise or anticlockwise)?
Looking at the left hand side of the loop, you can use the right hand slap rule to determine the direction of the current( fingers going out of the page, palm to the right to oppose the force to  the left, current will be clockwise). However, if you do the same analysis on the other side, the thumb would be pointing the same way so current will be anticlockwise. Which one is correct and why?

As the loop is moved out of the field, the amount of flux it encloses decreases. Lenz law then tells us that there will be an induced current that will oppose this decrease in flux by increasing it.

To increase flux we want to generate a B-field that is in the same direction as the original B-field, out of the page. Using our right hand grip rule, we can see that to generate a B-field going out of the page within the loop, the current must be flowing anti clockwise.

Title: Re: VCE Physics Question Thread!
Post by: Robert123 on November 01, 2013, 07:43:58 am
Yes The answer agree with that. I just want to know how to do using the RH slap rule as it could provide a quick answer check.

Also, with that reasoning, does it matter if the coil is move to the LEFT or to the RIGHT?
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on November 02, 2013, 12:17:38 am
are questions 9-11 from 2007 VCAA U4 AOS2 still in course?

also what is the range of wavelength-gap ratio that we should assume that allows diffraction, because in the same exam Q5 i got the ratio to be 0.667 and thought this was much lower than 1, hence no diffraction.

http://www.vcaa.vic.edu.au/Documents/exams/physics/2007physics2.pdf
Title: Re: VCE Physics Question Thread!
Post by: SocialRhubarb on November 02, 2013, 01:48:53 am
Question 9 would be an application of knowledge of electron energy levels in atoms, and its relation to light. In the course, but clearly the difficult here is recognising what the question needs.

I think incandescent lightbulbs are off the course, but you should understand the principles behind a vapour lamp, and how they work.

Quote from the study design: explain the production of atomic absorption and emission spectra, including those from metal
vapour lamps.

We were told that anything above 0.1 will give you noticeable diffraction. A ratio of 0.667 would give you about 45 degrees of diffraction, definitely enough to be noticed.
Title: Re: VCE Physics Question Thread!
Post by: ~T on November 02, 2013, 10:40:21 am
On that note, what happens to diffraction when the ratio becomes larger than one? Ie. Wavelength > Slit size

Does the diffraction become even more significant, or does it gradually decrease again, or does it stop entirely? (perhaps because the wave is "too large"???)
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on November 02, 2013, 10:46:40 am
For diffraction through a single slit, the angular positions of the intensity minima are given by $d\sin{\theta}=m\lambda$
where m is a constant (not zero), d is the slit width and theta is the angle the line from the slit to the minima makes with a line parallel to the slit opening.
As you can see, if the wavelength approaches the slit width, the first angle approaches ninety degrees, which means the first intensity minima spreads out completely. If the wavelength exceeds the slit width, then you will not have any minima and the wave will spread completely.

Title: Re: VCE Physics Question Thread!
Post by: sin0001 on November 02, 2013, 10:57:39 am
For those generic questions asking how Young's double-slit experiment supports the wave model of light, I talk about how a fringe pattern is observed and hence light is shown to interfere, which is a wave phenomenon and thus supports this model. But is it also correct to mention how diffraction is observed, because theoretically electrons/other particles also have De Broglie wavelengths and can diffract through the slits?
Don't see many company solutions, and even VCAA, mentioning diffraction for these questions.
Title: Re: VCE Physics Question Thread!
Post by: SocialRhubarb on November 02, 2013, 12:01:03 pm
I guess?

The thing is that it's a double slit diffraction experiment that you're being asked about. They've given you two slits, why only talk about single slit diffraction? It's clear that the question is directing you to talk about interference patterns, and if it asks you specifically what Young's double-slit experiment shows, the most notable thing about it is the interference pattern it shows. If you wanted an experiment to show simple diffraction of light, you probably wouldn't use Young's double-slit experiment.

I'm not sure what point you're trying to make about electrons and de Broglie wavelengths. The fact that electrons can diffract, and also interfere, speaks about the wave nature of matter, not the nature of light.
Title: Re: VCE Physics Question Thread!
Post by: sin0001 on November 02, 2013, 12:06:57 pm
I guess?

The thing is that it's a double slit diffraction experiment that you're being asked about. They've given you two slits, why only talk about single slit diffraction? It's clear that the question is directing you to talk about interference patterns, and if it asks you specifically what Young's double-slit experiment shows, the most notable thing about it is the interference pattern it shows. If you wanted an experiment to show simple diffraction of light, you probably wouldn't use Young's double-slit experiment.

I'm not sure what point you're trying to make about electrons and de Broglie wavelengths. The fact that electrons can diffract, and also interfere, speaks about the wave nature of matter, not the nature of light.
Ah okay, what I was trying to ask was whether *only* mentioning interference would suffice, or is it better to also talk about diffraction.
Title: Re: VCE Physics Question Thread!
Post by: ECheong on November 02, 2013, 12:26:59 pm
I guess if you wanted to be very safe you could say that the interference patterns can only exist if diffraction happened, i.e. creating path differences between the light that passes through Slit A and from Slit B and therefore, the two waves from each respective slit interfering?

On the whole though, I'd personally focus on constructive and destructive interference as that's typically what they want from an answer (judging by examiners reports). As long as its good physics you can't be pinged :)
Title: Re: VCE Physics Question Thread!
Post by: BasicAcid on November 02, 2013, 12:28:44 pm
For diffraction through a single slit, the angular positions of the intensity minima are given by $d\sin{\theta}=m\lambda$
where m is a constant (not zero), d is the slit width and theta is the angle the line from the slit to the minima makes with a line parallel to the slit opening.
As you can see, if the wavelength approaches the slit width, the first angle approaches ninety degrees, which means the first intensity minima spreads out completely. If the wavelength exceeds the slit width, then you will not have any minima and the wave will spread completely.

So actually answering Tim...Blah's question, there will still be diffraction if wavelength > slit size but just to a lesser extent?
Title: Re: VCE Physics Question Thread!
Post by: SocialRhubarb on November 02, 2013, 12:32:21 pm
Umm, no, there will be more diffraction if wavelength > slit.
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on November 02, 2013, 12:33:38 pm
mrrghaawd make up your mind guys !!!  :-*
Title: Re: VCE Physics Question Thread!
Post by: BasicAcid on November 02, 2013, 12:34:54 pm
Umm, no, there will be more diffraction if wavelength > slit.

I thought maximum diffraction occurred when wavelength/slit = 1?
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on November 02, 2013, 12:43:42 pm
Could someone verify this?

I'm a bit worried now after reading this

And I don't understand any of nlius explanation

nliu you sexy boy, explain it to us in simple terms.
Title: Re: VCE Physics Question Thread!
Post by: ECheong on November 02, 2013, 12:44:43 pm
well essentially, looking back at that dsin(theta)=lamda equation and applying it. If you have if you have lamda/d = 1 you get a theta of 90 degrees. Anything more than that, and it stays at a diffraction angle of 90 degrees.

If you imagine an aperture and a wave coming through it, it's not possible for the wave to diffract backwards at an angle >90degrees. This is because if you treat each single point on the edge of a wave as a point source (huygen's principle) there's no way for the point source to send a wave 'around the corner' of the aperture and start sending waves backwards.

Reading over this I've realised it's a pretty difficult concept to explain in words LOL, pictures would be better.

Tl;dr maximum diffraction at lamda/d =1 because you don't get any more diffraction if lamda/d>1
Title: Re: VCE Physics Question Thread!
Post by: SocialRhubarb on November 02, 2013, 12:46:24 pm
I haven't read anything outside of nliu1995's points, but they seem to make physical sense.

Basically, if you have a wavelength/slit ratio greater than 1, you don't get any minima and maxima forming, but instead just have a complete diffraction of the light source, which I'm assuming means an equal distribution of light intensity in all directions.

I should clarify that, obviously you can't have more than 90 degrees of diffraction, which occurs when wavelength approaches the slit width. If wavelength is greater than the slit width, you will get 90 degrees of diffraction as well. So if wavelength is greater than slit width, you will get the maximum possible amount of diffraction occurring.
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on November 02, 2013, 12:52:01 pm
I haven't read anything outside of nliu1995's points, but they seem to make physical sense.

Basically, if you have a wavelength/slit ratio greater than 1, you don't get any minima and maxima forming, but instead just have a complete diffraction of the light source, which I'm assuming means an equal distribution of light intensity in all directions.

I should clarify that, obviously you can't have more than 90 degrees of diffraction, which occurs when wavelength approaches the slit width. If wavelength is greater than the slit width, you will get 90 degrees of diffraction as well. So if wavelength is greater than slit width, you will get the maximum possible amount of diffraction occurring.

so the whole screen will be just bright, correct?
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on November 02, 2013, 01:01:38 pm
The point is, if the wavelength is larger than the slit width, the equation cannot be satisfied by any angle theta. Therefore, there IS no intensity minimum. That means the wave spreads out completely.
The equation I gave is a requirement for the waves in the slit to interfere with each other destructively. Failure to satisfy this equation means no complete destructive interference.
The reason why we say in VCE physics that "incomplete diffraction" occurs is that the width of the middle intensity maximum is small. After the first intensity minimum, the intensity drops off significantly.

Read up on diffraction on wiki for diagrams, but be warned, this subject can get maths-heavy. The equation I gave is a requirement for the waves in the slit to interfere with each other destructively. Failure to satisfy this equation means no complete destructive interference.

There's a neat diagram on wiki too; it'll make my point clearer.
Title: Re: VCE Physics Question Thread!
Post by: Alwin on November 02, 2013, 01:05:28 pm
For diffraction through a single slit, the angular positions of the intensity minima are given by $d\sin{\theta}=m\lambda$
where m is a constant (not zero), d is the slit width and theta is the angle the line from the slit to the minima makes with a line parallel to the slit opening.
As you can see, if the wavelength approaches the slit width, the first angle approaches ninety degrees, which means the first intensity minima spreads out completely. If the wavelength exceeds the slit width, then you will not have any minima and the wave will spread completely.

nliu, they kinda sorta mildly touch this in Synchrotron detail study, but I agree. Explained muchhhh more in depth in A level (eg singapore) or highschool equivalent in china and other asian countries.

Anyways, for those interested from page 496:
kinda tangential, only read it if you want to
A collimated beam of X-rays incident upon a layer of atoms will be scattered. Bragg’s law states that the beams will interfere constructively, producing maxima when the following relationship is satisfied:
2dsinθ = nλ where d is the distance between layers of atoms (m), θ is the angle the X-ray beam makes with the surface, λ is the wavelength of incident X-ray photons (m) and n is the number of the maxima occurring, 1, 2, 3, etc.

What nliu is trying to say is that as d (which can be considered as the slit width) and λ (wavelength) are approaching the same width, ie d/λ -> 1, the equation becomes:
2λsinθ = nλ -> sinθ= n/2.
n=1 is the centre band, so we look for n=2, the band next to it:
sinθ= n/2 -> sinθ= 2/2 -> θ= 90 degrees.
This implies that the wave has defracted so completely that the next band occurs at right angles to the path, ie parallel to the screen.

So, using a "similar" version of Bragg's Law (sadly barely touched upon in VCE) the max defraction occurs when the ratio of wavelength to slit with approaches one.

EDIT: beaten, but hope you get the gist
This is the page he is referring to: http://en.wikipedia.org/wiki/Diffraction#Single-slit_diffraction

From that page:
 Numerical approximation of diffraction pattern from a slit of width equal to wavelength of an incident plane wave in 3D spectrum visualization: Numerical approximation of diffraction pattern from a slit of width equal to five times the wavelength of an incident plane wave in 3D spectrum visualization Numerical approximation of diffraction pattern from a slit of width four wavelengths with an incident plane wave. The main central beam, nulls, and phase reversals are apparent. (http://upload.wikimedia.org/wikipedia/commons/thumb/4/46/Wavelength%3Dslitwidthspectrum.gif/220px-Wavelength%3Dslitwidthspectrum.gif) (http://upload.wikimedia.org/wikipedia/commons/thumb/0/0a/5wavelength%3Dslitwidthsprectrum.gif/220px-5wavelength%3Dslitwidthsprectrum.gif) (http://upload.wikimedia.org/wikipedia/commons/thumb/3/3c/Wave_Diffraction_4Lambda_Slit.png/220px-Wave_Diffraction_4Lambda_Slit.png)
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on November 02, 2013, 01:13:26 pm
nliu, they kinda sorta mildly touch this in Synchrotron detail study, but I agree. Explained muchhhh more in depth in A level (eg singapore) or highschool equivalent in china and other asian countries.

Anyways, for those interested from page 496:
A collimated beam of X-rays incident upon a layer of atoms will be scattered. Bragg’s law states that the beams will interfere constructively, producing maxima when the following relationship is satisfied:
2dsinθ = nλ where d is the distance between layers of atoms (m), θ is the angle the X-ray beam makes with the surface, λ is the wavelength of incident X-ray photons (m) and n is the number of the maxima occurring, 1, 2, 3, etc.

What nliu is trying to say is that as d (which can be considered as the slit width) and λ (wavelength) are approaching the same width, ie d/λ -> 1, the equation becomes:
2λsinθ = nλ -> sinθ= n/2.
n=1 is the centre band, so we look for n=2, the band next to it:
sinθ= n/2 -> sinθ= 2/2 -> θ= 90 degrees.
This implies that the wave has defracted so completely that the next band occurs at right angles to the path, ie parallel to the screen.

So, using a "similar" version of Bragg's Law (sadly barely touched upon in VCE) the max defraction occurs when the ratio of wavelength to slit with approaches one.

Ah Bragg's law. It's similar, although in that case you've got to remember where the two comes from; the path difference is 2 d sin theta, and that's because there are two layers in the crystal lattice and the 2 comes from a return journey from the first layer to the second layer. Look it up on wiki for a diagram.
Alwin, you'll confuse people xP

In double slit interference, the condition is d sin theta = m lambda for CONSTRUCTIVE interference because d sin theta is the path difference.

For single slit diffraction, the argument works differently. Again, look at some diagrams; they'll be better than what I can draw.
But essentially, if we split up the slit into half, to form two minislits of size d/2, then we have a double slit scenario with path difference d/2 * sin theta. This is equal to half a wavelength for minimum, so d/2*sin theta = w/2. d sin theta = w
If we split the slit into quarters, then quarter 1 can interfere destructively with quarter 2 and then 3 with 4. The slit width is now d/4, so the condition is d/4 * sin theta = w/2. d sin theta = 2w. Etc if we split the slit into 6, 8, 10, 12.
That's sort of where it comes from.

Define "max diffraction" though.
Title: Re: VCE Physics Question Thread!
Post by: ECheong on November 02, 2013, 01:21:35 pm
So essentially for this exam (which is all I care about):

- lamda/width < 1 = not a lot of diffraction
- lamda/width = 1 = max diffraction
- lamda/width > 1 = max diffraction

Correct?

At the core of it, pretty much haha :)
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on November 02, 2013, 01:23:47 pm
See, in the diagram with w=d, there is a beam that spreads across the entire wall, whereas with the third one, the most intense part of the wave is quite limited in range.
Title: Re: VCE Physics Question Thread!
Post by: Alwin on November 02, 2013, 01:31:00 pm
So essentially for this exam (which is all I care about):

- lamda/width < 1 = not a lot of diffraction
- lamda/width = 1 = max diffraction
- lamda/width > 1 = max diffraction

Correct?

pretty much.

*back to discussion with nliu* :P

For anyone who's interested in what we're talking about, this is the easiest explanation I can find:
scan of a textbook
(http://i.imgur.com/Lx5mhbq.jpg)
(http://i.imgur.com/hr0FPyH.jpg)

See, in the diagram with w=d, there is a beam that spreads across the entire wall, whereas with the third one, the most intense part of the wave is quite limited in range.
You referring to the ones in my post?
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on November 02, 2013, 03:27:23 pm
Oh yeah. That thing. I need to start making mine :P
Title: Re: VCE Physics Question Thread!
Post by: joey7 on November 02, 2013, 04:06:23 pm
What happens to magnetic field (B) when the current increases, and which equation tells us this
Title: Re: VCE Physics Question Thread!
Post by: Stevensmay on November 02, 2013, 04:47:36 pm
What happens to magnetic field (B) when the current increases, and which equation tells us this

$I \propto \frac{\Delta \phi_B}{\Delta t}$

So we can see that as the current increases, the magnetic field needs to increase or the time needs to be reduced.
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on November 02, 2013, 04:52:00 pm
Magnetic field strength is directly proportional to the current.
What equation tells you that?
Biot-Savart law: $d\vec{B} = \frac{\mu_0}{4\pi}\frac{i\vec{ds}\times\vec{r}}{|\vec{r}|^3}$

Oh wait, not part of course. My bad.
Title: Re: VCE Physics Question Thread!
Post by: Alwin on November 02, 2013, 05:16:20 pm
Magnetic field strength is directly proportional to the current.
What equation tells you that?
Biot-Savart law: $d\vec{B} = \frac{\mu_0}{4\pi}\frac{i\vec{ds}\times\vec{r}}{|\vec{r}|^3}$

Oh wait, not part of course. My bad.

really? :OOOOOOOOOOOOO dang, they must have really cut down on the course. I swear Biot-Savart's Law was in it last year :o

JOKES, no one panic please, just nliu deciding to demonstrate his knowledge again =P
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on November 02, 2013, 07:36:29 pm
$I \propto \frac{\Delta \phi_B}{\Delta t}$

So we can see that as the current increases, the magnetic field needs to increase or the time needs to be reduced.

This is only for an induced current, not for currents ttat create magnetic fields
Title: Re: VCE Physics Question Thread!
Post by: joey7 on November 02, 2013, 07:50:09 pm
Wait, so if a question talks about doubling the current how am I meant to know the effect on the field?
Also, I came across a question involving two parallel current carrying wires, where the distance between them was reduced by a factor of 2, In the answers it said field varies inversely with distance squared. Again how am I meant to know this?
Title: Re: VCE Physics Question Thread!
Post by: Stevensmay on November 02, 2013, 08:04:49 pm
This is only for an induced current, not for currents ttat create magnetic fields

Skipped over part of the question, thanks.
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on November 02, 2013, 11:18:00 pm
Wait, so if a question talks about doubling the current how am I meant to know the effect on the field?
Also, I came across a question involving two parallel current carrying wires, where the distance between them was reduced by a factor of 2, In the answers it said field varies inversely with distance squared. Again how am I meant to know this?

The awkward bit is...your answers are wrong. For parallel current carrying wires, the field strength is inversely proportional to the distance from the wire. I don't want to prove this here as 1. I would have to do some messy integration, or 2. I would have to introduce another physical law which would make no sense here.

You don't needed to know anything about magnetic field strengths in VCE, although maybe you need to know that increasing the distance generally decreases field strength (but not how).

That logic. We're taught that currents create a magnetic field, that increasing currents create larger magnetic fields and how magnetic forces operate, but we're not taught how to calculate magnetic field strengths. VCAA logic.
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on November 02, 2013, 11:48:54 pm
i'm not sure if i've asked this but are questions like Q2 of 2007 U3 motion still in course? i saw some on the older exams as well but didn't know how to do them well.
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on November 02, 2013, 11:56:44 pm
It's a parabola; the particle's horizontal velocity component is constant but the vertical component increases in magnitude. Just draw something like that and I think you'll be fine.
Title: Re: VCE Physics Question Thread!
Post by: lolipopper on November 03, 2013, 12:49:56 pm
It's a parabola; the particle's horizontal velocity component is constant but the vertical component increases in magnitude. Just draw something like that and I think you'll be fine.

thanks but are these types in the course, because it similar to relative motion using frames of references and stuff?
Title: Re: VCE Physics Question Thread!
Post by: Robert123 on November 03, 2013, 02:04:24 pm
thanks but are these types in the course, because it similar to relative motion using frames of references and stuff?
I think it is sort of on the course since you have to be aware that for projectile motion, the horizontal component of the velocity is constant
Title: Re: VCE Physics Question Thread!
Post by: sin0001 on November 03, 2013, 06:32:50 pm
Do we have to substitute values in the formulas we use, for 2 marks in the exam? I recently checked a company's solutions and they allocated 1 mark for 'substituting values into correct formulas' and 1 mark for the correct answer, I thought that a mark would've been allocated just for writing down the correct formula...
Title: Re: VCE Physics Question Thread!
Post by: Robert123 on November 03, 2013, 09:24:59 pm
Sin0001
Check out this link, it gives all the details regarding marking
http://www.vicphysics.org/documents/events/conf2009/Generalmarkingprinciples.doc
Title: Re: VCE Physics Question Thread!
Post by: Henreezy on November 09, 2013, 03:31:00 pm
Question 4 b) VCAA 2012 Exam 1:
Is the tension force referred to as the 'gravitational force - upwards' due to the fact that the reaction is due to gravity? I don't know and that confuses me.

Action: Gravity 'pulling' it downwards
Reaction: Gravity 'pushing' it upwards (results in 'tension' from string?)

I'm so confused.
Title: Re: VCE Physics Question Thread!
Post by: Robert123 on November 09, 2013, 03:35:44 pm
Henreezy, the reaction force rule is given by
F(a on b)= -F (b on a) that is, newton's 3rd law
From this, we can sort of sub in and solve so we get
F(earth's gravity on object)= -F(object's gravity on earth)
So what actually occur is the gravitation force from the ball actually pulls the earth up but since the earth is so massive, nothing is observed on the earth's behalf.

For young's double slit experiment, does it matter if the light source is coherent or not to achieve interference? I did an exam practice by ITUTE which said that using torch light would not allow interference to occur but in this video by veritasium, he achieved interference with sunlight
Title: Re: VCE Physics Question Thread!
Post by: Stevensmay on November 09, 2013, 03:36:25 pm
Question 4 b) VCAA 2012 Exam 1:
Is the tension force referred to as the 'gravitational force - upwards' due to the fact that the reaction is due to gravity? I don't know and that confuses me.

Action: Gravity 'pulling' it downwards
Reaction: Gravity 'pushing' it upwards (results in 'tension' from string?)

I'm so confused.

So we agree that the action force is the gravitational attraction exerted on the spheres by the earth.

To find our reactionary force, we simply swap the two objects around.

The reaction force is the gravitational attraction exerted on the earth by the spheres. This is effectively pulling the earth closer to the spheres, thus the direction is up.
Title: Re: VCE Physics Question Thread!
Post by: Stevensmay on November 09, 2013, 03:39:50 pm
For young's double slit experiment, does it matter if the light source is coherent or not to achieve interference? I did an exam practice by ITUTE which said that using torch light would not allow interference to occur but in this video by veritasium, he achieved interference with sunlight

Someone had the exact same question as you.

http://physics.stackexchange.com/questions/76692/is-coherent-light-required-for-interference-in-youngs-double-slit-experiment
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on November 09, 2013, 04:56:52 pm
Henreezy, the reaction force rule is given by
F(a on b)= -F (b on a) that is, newton's 3rd law
From this, we can sort of sub in and solve so we get
F(earth's gravity on object)= -F(object's gravity on earth)
So what actually occur is the gravitation force from the ball actually pulls the earth up but since the earth is so massive, nothing is observed on the earth's behalf.

For young's double slit experiment, does it matter if the light source is coherent or not to achieve interference? I did an exam practice by ITUTE which said that using torch light would not allow interference to occur but in this video by veritasium, he achieved interference with sunlight

Torch light does allow interference to occur, as you have waves at each slit interfering with each other, but you don't get a particularly convenient interference pattern as you can see in the video.
Question 4 b) VCAA 2012 Exam 1:
Is the tension force referred to as the 'gravitational force - upwards' due to the fact that the reaction is due to gravity? I don't know and that confuses me.

Action: Gravity 'pulling' it downwards
Reaction: Gravity 'pushing' it upwards (results in 'tension' from string?)

I'm so confused.

The tension is merely a force that exists to hold the object there. It doesn't have to exist if the object is falling.
However, as long as there is a gravitational force, there is a reaction to that force, and that force is always the same as the gravitational force in magnitude, but opposite in direction by Newton 3. That's the difference. In this case, as people before me have pointed out, the reaction is the attraction of the Earth by the object.
Title: Re: VCE Physics Question Thread!
Post by: Jaswinder on November 10, 2013, 10:39:11 am
01 - I thought the frequency doesn't change (hence A?) but the answers say C?  :-\

02 -For 14 I thought A, as the left of the battery equipped circuit would become north it would create flux going right to left in the other circuit and to oppose that you would need a current going from X to Y?

Thanks
Title: Re: VCE Physics Question Thread!
Post by: eddybaha on November 10, 2013, 10:56:01 am
1. the frequency is changing because the alternator is spinning faster and faster.
2. you are looking for the direction of current through the AMMETER, yes a bit tricky. so when the switch closes, it causes flux to thread the secondary coil to the left. Using lenz's law you need a magnetic field threading the coil to the right to oppose this change in flux. Then using grip rule you get your answer.
Title: Re: VCE Physics Question Thread!
Post by: This-is-not-me on November 10, 2013, 01:03:34 pm
Can someone help me with the concept in part of b of this question?
Title: Re: VCE Physics Question Thread!
Post by: eddybaha on November 10, 2013, 01:41:33 pm
because an electron has a debroglie wavelength it forms standing waves in orbit. the circumference of the orbit can only be an integer number of wavelengths and thus only these energy levels are stable.
Title: Re: VCE Physics Question Thread!
Post by: Alwin on November 10, 2013, 01:44:41 pm
Can someone help me with the concept in part of b of this question?

One (vce level reply) would be to say that electrons need to form stationary waves to be stable so the orbit circumference has to be a whole number multiple of the wavelength of the electron's de broglie wavelength.
If an electron orbited at radius such that the circumference was not a whole number multiple, destructive interference would occur and the electron would 'drop' to a lower stable orbit.
Thus, only certain energy levels, orbits, are possible for the electrons of the hydrogen atom... or something along those lines

inb4 nliu comments how they're not levels so much, but bands but that's beyond vce
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on November 10, 2013, 02:19:11 pm
One (vce level reply) would be to say that electrons need to form stationary waves to be stable so the orbit circumference has to be a whole number multiple of the wavelength of the electron's de broglie wavelength.
If an electron orbited at radius such that the circumference was not a whole number multiple, destructive interference would occur and the electron would 'drop' to a lower stable orbit.
Thus, only certain energy levels, orbits, are possible for the electrons of the hydrogen atom... or something along those lines

inb4 nliu comments how they're not levels so much, but bands but that's beyond vce

Actually, I'm happy with calling them energy levels; I'm not happy with the standing wave explanation in general :P
It only works for any atom with one electron. AKA H, He+, Li 2+ etc

Alwin, I think you might have say that as only specific wavelengths are permitted, only specific energies are allowed. This is VCE. We can't make any jumps in our thought processes apparently.
Title: Re: VCE Physics Question Thread!
Post by: Robert123 on November 11, 2013, 09:56:30 pm
Is diffraction stronger when wavelength equals slit width or when wavelength is greater than slit width?
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on November 11, 2013, 10:00:04 pm
Technically speaking, it's stronger when the wavelength is bigger than the slit width as the intensity of the diffracted wave then decreases more slowly as you get further from the intensity maximum at the middle, but for VCE purposes I'm not sure how relevant that is.
Title: Re: VCE Physics Question Thread!
Post by: BasicAcid on November 11, 2013, 10:55:52 pm
Is diffraction stronger when wavelength equals slit width or when wavelength is greater than slit width?

For VCE purposes, they are the same strength.
Title: Re: VCE Physics Question Thread!
Post by: Infinity Plus on November 12, 2013, 04:22:56 pm
Hey guys. With regards to diffraction I have read many sources that say different things so I would like to seek the opinions of my fellow trustworthy ANers. How large does the ratio of lambda/width have to be for there to be significant diffraction? MOst sources say ,1 others say a bit less than 1 is still considered a lot, and few say more than one.
Title: Re: VCE Physics Question Thread!
Post by: eddybaha on November 12, 2013, 04:31:31 pm
definitely noticeable diffraction at 1 but vcaa will either give you something obviously small like 0.000001 if theres no diffraction
Title: Re: VCE Physics Question Thread!
Post by: saifh on November 12, 2013, 04:48:02 pm
Can I stick stuff on my cheat sheet using tape?
Title: Re: VCE Physics Question Thread!
Post by: Infinity Plus on November 12, 2013, 04:55:49 pm
definitely noticeable diffraction at 1 but vcaa will either give you something obviously small like 0.000001 if theres no diffraction

Thank you man!
Title: Re: VCE Physics Question Thread!
Post by: Robert123 on November 12, 2013, 05:02:20 pm
For light and matter, do we need to know the required voltage to accelerate an electron? I have been given heaps of question related to it in private companies exams (particularly atar notes) but not VCAA
Title: Re: VCE Physics Question Thread!
Post by: Alwin on November 12, 2013, 05:37:02 pm
For light and matter, do we need to know the required voltage to accelerate an electron? I have been given heaps of question related to it in private companies exams (particularly atar notes) but not VCAA

It's not hard... but I'm pretty sure that that sounds like a detail study (synchrotron irrc) question not a core studies question...

Pretty sure it's not something you need to know.
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on November 12, 2013, 05:42:28 pm
For light and matter, do we need to know the required voltage to accelerate an electron? I have been given heaps of question related to it in private companies exams (particularly atar notes) but not VCAA

The definition of voltage is the potential energy difference per charge. So, a 1 V potential energy difference would give a 1 C charge 1 J of energy. Use E=Vq if you ever come across it.

It's not hard... but I'm pretty sure that that sounds like a detail study (synchrotron irrc) question not a core studies question...

Pretty sure it's not something you need to know.

Technically, in the photoelectric effect, if you have electron energy=hf-W, then the stopping voltage would be 1/q*(hf-W) which explains the gradient of the curve. It's not entirely out of the course.
Title: Re: VCE Physics Question Thread!
Post by: Alwin on November 12, 2013, 05:44:48 pm
The definition of voltage is the potential energy difference per charge. So, a 1 V potential energy difference would give a 1 C charge 1 J of energy. Use E=Vq if you ever come across it.

Technically, in the photoelectric effect, if you have electron energy=hf-W, then the stopping voltage would be 1/q*(hf-W) which explains the gradient of the curve. It's not entirely out of the course.

k

coz last year when I asked about this (forces on charged particles) my teacher said it wasn't strictly on the course. Who knows, maybe this could be your lucky year guys XD
Title: Re: VCE Physics Question Thread!
Post by: jono88 on November 12, 2013, 06:26:48 pm
-Identify rms voltage as an AC voltage which produces the same power in a resistive component as a DC voltage of the same magnitude
-explain the production of atomic absorption and emission spectra, including those from metal vapours?

I don't understand these 2 dot points, could someone elaborate further?
Title: Re: VCE Physics Question Thread!
Post by: jono88 on November 12, 2013, 06:31:16 pm
Spectra
The electromagnetic radiation that is emitted from a free atom only occurs at a number of discrete or separate wavelengths or frequencies.  This is usually represented as a set of lines.
Visible spectrum of hydrogen spectrum

Interpretation of Spectra
After Einstein, frequency is now linked to energy by E = hf.  Each line in the spectra now corresponds to a transition of an electron from one energy level to another level.

In fact, the discrete lines in the spectra must mean that electrons in atoms can only exist at specific energy levels, and the spectral lines represent transitions between these levels.

When an atom is excited electrons are sent to a higher energy level.  From there, the electron drops back to the ground state by any of the possible paths, each drop emitting a photon of specific energy and frequency.

That right or completely wrong?
Title: Re: VCE Physics Question Thread!
Post by: Robert123 on November 12, 2013, 06:31:49 pm
The definition of voltage is the potential energy difference per charge. So, a 1 V potential energy difference would give a 1 C charge 1 J of energy.
Thanks nilu, I understand that part it is just the random formulas out there for accelerating electrons from close to rest to form a diffraction pattern similar to x-rays, then work out what voltage is required to accelerate them. Just want to know if they are on the course?

I have a fairly good idea in deriving the formulas but if I do have to know it, it would be easier to just plug and play VCE style
Title: Re: VCE Physics Question Thread!
Post by: ECheong on November 12, 2013, 06:48:35 pm
Thanks nilu, I understand that part it is just the random formulas out there for accelerating electrons from close to rest to form a diffraction pattern similar to x-rays, then work out what voltage is required to accelerate them. Just want to know if they are on the course?

I have a fairly good idea in deriving the formulas but if I do have to know it, it would be easier to just plug and play VCE style

If you consider the electronvolt (eV), 1 eV is the amount of energy you get when you accelerate an electron through a potential difference of 1 volt. So, I guess what you can do is find the energy the electron has in eV, and then you can find the potential difference it was accelerated through.

Re: X-rays, you'd just need to find the velocity at which the electrons have a debroglie wavelength the same as xrays and then use E=0.5mv^2 to find energy, convert to eV and you'll know the potential that electron was accelerated through :)
Title: Re: VCE Physics Question Thread!
Post by: Alwin on November 12, 2013, 07:26:02 pm
-Identify rms voltage as an AC voltage which produces the same power in a resistive component as a DC voltage of the same magnitude
-explain the production of atomic absorption and emission spectra, including those from metal vapours?

I don't understand these 2 dot points, could someone elaborate further?

RMS: Vp indicates the voltage peak of an AC voltage vs power graph. Basically:
$V_{RMS}=\frac{V_{P}}{\sqrt{2}}$
It works on the prinicple the power output should be the same, but that's based on a perfect sine graph. Here's a generic proof if you were interested :P
http://masteringelectronicsdesign.com/how-to-derive-the-rms-value-of-a-sine-wave-with-a-dc-offset/

Absorption and emission in physics is pretty straight forward, specific/discrete/quantum (if you want to be fancy) amount of energy needed to excite electrons to a higher state, and they release this energy when they fall back to a lower state.
Most calculations in physics involve finding the energy in eV, wavelength of photon emitted and whether a photon of another different energy will have any impact on the atom in question.

Note that electron absorption is not on the course... but fun to read about :D
Title: Re: VCE Physics Question Thread!
Post by: iitwinz on November 12, 2013, 08:22:12 pm
hey guys, did we have to know the skin effect and optical fibres and how they work? I don't remember learning about it at all and haven't seen any on the exam.
Title: Re: VCE Physics Question Thread!
Post by: Alwin on November 12, 2013, 08:29:40 pm
hey guys, did we have to know the skin effect and optical fibres and how they work? I don't remember learning about it at all and haven't seen any on the exam.

It's been put into one of the detailed studies :) not in the core section of the exam
Title: Re: VCE Physics Question Thread!
Post by: iitwinz on November 12, 2013, 09:30:34 pm
It's been put into one of the detailed studies :) not in the core section of the exam

cool thanks man :P
Title: Re: VCE Physics Question Thread!
Post by: Lachjames on November 12, 2013, 09:45:31 pm
Hi guys, on the 2011 Exam 2 Electric Power Q.11, I've been talking with my friend from school and we can't figure the graph out. We looked at the VCAA solution and we get what they're saying but we were thinking something else - I put our train of thought in the attachment below.

Pretty much it's that since the flux will reduce once the magnet is actually inside the loop shouldn't that cause some emf generation, spiking in the opposite direction? The attachment will make what I'm saying clearer hopefully (even if it is nonsense lol)

Any advice would be great, thanks :)
Title: Re: VCE Physics Question Thread!
Post by: eddybaha on November 12, 2013, 09:59:20 pm
flux should be at a maximum inside the loop i believe
Title: Re: VCE Physics Question Thread!
Post by: ~T on November 12, 2013, 09:59:34 pm
Hi guys, on the 2011 Exam 2 Electric Power Q.11, I've been talking with my friend from school and we can't figure the graph out. We looked at the VCAA solution and we get what they're saying but we were thinking something else - I put our train of thought in the attachment below.

Pretty much it's that since the flux will reduce once the magnet is actually inside the loop shouldn't that cause some emf generation, spiking in the opposite direction? The attachment will make what I'm saying clearer hopefully (even if it is nonsense lol)

Any advice would be great, thanks :)
I'm a little confused but I think I see where you're coming from. The only thing is that the flux doesn't reduce when the magnet is inside the loop. Flux still "runs through" the inside of a magnet, and the field lines are still there when this magnet is in the centre of the loop.

At every point with the magnet approaching, going through, and leaving the loop, there is a leftwards flux, yes? It never changes direction. Now, given this constant direction, all that matters is whether this flux is increasing or decreasing. The field lines are closest together when near the magnet, so the flux will be greatest when the magnet is inside the loop. It will increase up until this point, and then decrease away as the magnet leaves.

I hope I've clarified, but I'm not sure it was super clear sorry :-\
Title: Re: VCE Physics Question Thread!
Post by: Lachjames on November 12, 2013, 10:05:32 pm
Thanks for that, I think I get it now :)
Title: Re: VCE Physics Question Thread!
Post by: papertowns on November 13, 2013, 01:29:12 am
A beam of red light of frequency 4 × 10^14 Hz is found to deliver 7.54 × 10^19 photons per second.

What would be the power of the beam?
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on November 13, 2013, 06:50:21 am
If anyone is here :)

I'm not sure, sorry, hopefully somebody catches this before the exam :(

I tried rotating it both ways in my head and both times got current from X to Y initially, not sure if it's a bad question or if I just stuffed up the logic somewhere.

A beam of red light of frequency 4 × 10^14 Hz is found to deliver 7.54 × 10^19 photons per second.

What would be the power of the beam?

P=E/t

E = number of photons x single photon energy
= 7.54 x 10^19 x hf
= 7.54 x 10^19 x 6.63 x 10^-34 x 4 x 10^14

so P=(that)/1=that

EDIT: I used the Js version of h because we want energy in Joules so that we can have Power in Watts (Joules/second), if we had have used the eVs value of h (the 4.14x10^-15 one) we would have got energy in eV meaning power in eV/s which isn't a standard unit for power, afaik.

Hopefully that makes sense, the key idea is that the energy of a beam is the number of photons times the energy of one photon :)

Good luck today everyone!
Title: Re: VCE Physics Question Thread!
Post by: Robert123 on November 13, 2013, 07:31:23 am
If anyone is here :)

Ok, to solve this we use lenz's law and the right hand slap rule. Let start off by analysining what would happen if we rotate it clockwise.
Firstly side XY would go up and due to lenzs law we know it would create a force in the opposite direction (that is down).
So using our right hand slap rule, palm  down and fingers point to the right, we can see the current will flow X to Y. Therefore, we can deduce that it must rotate the opposite way (anticlockwise).
You can do the same analysis for anticlockwise to confirm it.
Title: Re: VCE Physics Question Thread!
Post by: Rod on December 17, 2013, 09:43:51 pm
Can the year 11's moving into year 12 use this thread next year? Or do we have to make our own, new thread?
Title: Re: VCE Physics Question Thread!
Post by: Stevensmay on December 17, 2013, 09:44:48 pm
Can the year 11's moving into year 12 use this thread next year? Or do we have to make our own, new thread?

Feel free to post whatever.
Title: Re: VCE Physics Question Thread!
Post by: Einstein on March 01, 2014, 11:44:44 am
Can someone explain fissile and fissionable?

Title: Re: VCE Physics Question Thread!
Post by: katie101 on March 01, 2014, 05:31:53 pm
If a dead battery and a new battery were used in series, would the well used battery have any effect on the output voltage of the new battery? Why/why not?
Title: Re: VCE Physics Question Thread!
Post by: Thorium on March 05, 2014, 09:23:45 pm
A question says: "A spring has stiffness 20 N/m and is hang vertically. When a mass is attached, it stretches by 0.2 m. Assuming that the spring has no mass, what is the value of the mass."

What I did was that:
change(Elastic E)=Change(Gravitational E)
1/2kx^2=mgh
1/2(20)(0.2)^2=m(10)(0.2)
m=0.2 kg

But the ans says to use kx=mg, and the ans is 0.4 kg.

I am thinking that since the F due to gravity on the mass is constant, the 1/2kx^2 should not be used. Is that the reason why my method gave the wrong ans?
Title: Re: VCE Physics Question Thread!
Post by: RKTR on March 05, 2014, 09:31:18 pm
that day someone in my class asked a similar question. my teacher said maybe there is some energy lost. so it is better to use mg=kx if possible
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on March 05, 2014, 09:43:41 pm
A question says: "A spring has stiffness 20 N/m and is hang vertically. When a mass is attached, it stretches by 0.2 m. Assuming that the spring has no mass, what is the value of the mass."

What I did was that:
change(Elastic E)=Change(Gravitational E)
1/2kx^2=mgh
1/2(20)(0.2)^2=m(10)(0.2)
m=0.2 kg

But the ans says to use kx=mg, and the ans is 0.4 kg.

I am thinking that since the F due to gravity on the mass is constant, the 1/2kx^2 should not be used. Is that the reason why my method gave the wrong ans?

Let's think about this reasonably. The only thing you know about the spring when it is still is that the net force is zero, so you use forces.

Now, let's see what happens when you attach a mass to the relaxed spring and let go. Obviously, the spring is going to start moving. Initially, the spring only has gravitational potential energy. By equating mgx with 1/2 kx^2, all you are doing is finding when all of the gravitational potential energy goes completely into spring energy. By conservation of energy, this means the kinetic energy at this point (where all of the potential energy is spring) is still the same at the st art, where the kinetic energy is zero. This is where the object is momentarily stationary. HOWEVER, the forces are clearly not balanced as the spring is still headed upwards.

The point is, if you just let a spring go, it will oscillate forever assuming no resistive drag forces. Note how when you attach a spring, the forces are unbalanced. In order to balance the spring, the person holding the mass must physically move it with the spring to the point of stable equilibrium. This action drains energy from the spring and explains why you cannot just equate the energies.
Title: Re: VCE Physics Question Thread!
Post by: Rishi97 on March 09, 2014, 01:31:37 pm
Can I pls have help with the following questions?
1) An astronaut standing on the moon experiences a gravitational force of attraction of 160N. He moves away from the surface of the moon to an altitude where the gravitational force is 40N.
a) How far from the centre of the moon is this new location in terms of the radius of the moon?
b) The astronaut now travels to another location at a height of three moon radii above the surface. Calculate the gravitational force at this altitude

Cheers ;)
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on March 09, 2014, 01:48:53 pm
Can I pls have help with the following questions?
1) An astronaut standing on the moon experiences a gravitational force of attraction of 160N. He moves away from the surface of the moon to an altitude where the gravitational force is 40N.
a) How far from the centre of the moon is this new location in terms of the radius of the moon?
b) The astronaut now travels to another location at a height of three moon radii above the surface. Calculate the gravitational force at this altitude

Cheers ;)

So...F = k/r^2 where k is some constant.
If the force quarters, the distance r must double, so this new location is now two moon radii from the centre of the moon.

Now, three moon radii above the surface = four moon radii from centre of moon (r is ALWAYS measured from the centre of the moon), so we've quadrupled the distance r with regards to the initial distance, so the force drops to 1/16, or 10 N.
Title: Re: VCE Physics Question Thread!
Post by: Capristo on March 11, 2014, 10:27:50 pm
I have a uniform circular motion question.
Just a mass attached to a string moving in a horizontal circular motion. (Diagram attached). At the bottom of the vertical line of string there is some weights of mass M. At the end of the diagonal section of string that moves in the circular path, there is a rubber stopper of mass m.

The tension, F, is constant. Show the relationship between L and T (period) by completing the following.
Write a proportionality expression for L
Show an equation with constant, K. (L=K.....)
Write an expression for K in terms of pi, m, M and g (gravity).

Thanks!

Title: Re: VCE Physics Question Thread!
Post by: clıppy on March 22, 2014, 07:53:24 pm
A car of mass 600kg is being towed upward along an inclined plane at 5 degrees to the horizontal. Assume the frictional force on the car is constant throughout it's motion with a magnitude of 400N. The car is accelerating from rest at a rate of 0.2ms^-2
a) Find the magnitude of the tension in the towing cable
b) After traveling a distance of 100m the cable breaks. Determine the distance required for the car to come to a complete stop from the moment the cable breaks

I've found part a as 1043N which is correct but I have no idea how to find the answer to part b (15m)
Title: Re: VCE Physics Question Thread!
Post by: RKTR on March 29, 2014, 02:57:45 pm
a student is twirling a weight on the end of a string 50cm long in a vertical circle
speed of weight is 4.0ms^-1 at the top

to find speed at the bottom ,i let change in ke = change in gpe . is this right?
Title: Re: VCE Physics Question Thread!
Post by: rhinwarr on March 29, 2014, 03:17:16 pm
Yes, KE final = KE initial + change in GPE, then solve for speed.
Title: Re: VCE Physics Question Thread!
Post by: RKTR on March 29, 2014, 03:46:57 pm
Yes, KE final = KE initial + change in GPE, then solve for speed.

erm if the question says the person does not add any further energy into the system as he twirls it, is the method still correct? tension at top is found . will tension at bottom in this case be the same ?
Title: Re: VCE Physics Question Thread!
Post by: rhinwarr on March 29, 2014, 04:03:15 pm
That only works if the person does not add further energy. If it is in the vertical plane, tension at the bottom will be greater than the tension at the top because of the weight force acting against the centripetal force. Drawing force diagrams with the directions of the force will help you understand it.
At the top: centripetal force = tension + weight
At the bottom: centripetal force = tension - weight
Title: Re: VCE Physics Question Thread!
Post by: RKTR on March 29, 2014, 04:14:02 pm
That only works if the person does not add further energy. If it is in the vertical plane, tension at the bottom will be greater than the tension at the top because of the weight force acting against the centripetal force. Drawing force diagrams with the directions of the force will help you understand it.
At the top: centripetal force = tension + weight
At the bottom: centripetal force = tension - weight

thanks . i was confused because my friend says tension will be the same since the person doesnt add any energy.
Title: Re: VCE Physics Question Thread!
Post by: RKTR on March 31, 2014, 08:24:07 pm
http://a.imageshack.us/img831/1715/2shipprojectile.jpg

had this as one of my sac questions today. 1. which ship will get hit 1st? 2.explain by referring to relevant physics principle. can anyone explain? i keep thinking there's not enough info.
Title: Re: VCE Physics Question Thread!
Post by: rhinwarr on March 31, 2014, 09:11:53 pm
I think they would both get hit at the same time.
Vertical acceleration is the same for both so if the horizontal distances were the same (which they are not), ship B would get hit first as it reaches half the height of ship A's projectile.
However, since we need to take into account horizontal velocity (which remains constant for both), then the projectile for ship B actually has to travel twice as far so should land at the same time as ship A's projectile.
I'm not really sure so better check with someone else.
Title: Re: VCE Physics Question Thread!
Post by: RKTR on March 31, 2014, 10:34:55 pm
found the explanation online. vertical displacement =0
using x=ut+1/2 at^2
0= vsin(theta)t -5t^2
t=vsin(theta)/5

v for both are the same . when theta is larger, sin theta is also larger , time is also increased. so B will get hit first.  ohno already lost 3 marks out of 40
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on March 31, 2014, 10:54:03 pm
There's not enough information. You haven't been told about the speeds and angles of each missile.
Title: Re: VCE Physics Question Thread!
Post by: RKTR on March 31, 2014, 10:56:43 pm
There's not enough information. You haven't been told about the speeds and angles of each missile.

really? that's what i thought. but today the question on the sac was which ship will get hit 1st? not multiple choice. and how about the explanation i heard from other people and found online in my previous post?
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on March 31, 2014, 11:51:22 pm
I'm going to have a go at this and see what happens
Let the speed of the first projectile be v1 and the angle be s1
Then, the vertical displacement is 0 = v1*sin s1 t - 1/2 gt^2 => t = 2v1 sin s1/g
Hmm. The only way the two times are the same is if v1 sin s1 = v2 sin s2 and that's not always true. They haven't given you any numbers, so you really can't make a judgement.
Title: Re: VCE Physics Question Thread!
Post by: 09Ti08 on April 01, 2014, 11:57:43 am
*for the speed: the key information in the problem is the shells are fired "simultaneously", so I think it's reasonable to conclude that they fired at the same speed.

*for the angle: it is obvious that ship A is closer to the battleship, so the angle must be bigger for the shell to hit it.
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on April 02, 2014, 08:41:41 pm
Were the shells fired from the same cannon though? I know I'm being pedantic but I don't think I noticed that. Besides if they were, and the back shell was fired too hard, it could easily transfer some of its momentum to the shell in front. Imagine two bouncing basketballs on top of each other.
Title: Re: VCE Physics Question Thread!
Post by: 09Ti08 on April 02, 2014, 09:15:39 pm
No, I think they were not fired from the same cannon. Why? Let's assume that they were, I don't think that the trajectories can be that much different, even though we were not given information about the distances between the ships.

I think the force exerted by a canon is quite large, but we don't know the distance of the two shells from the mouth of the cannon and their masses and sizes... These numbers do matter, as I need them for calculations. I can't really do anything if I assume too much...

So I think they were fired from two identical canon balls from the same position. That would make sense.
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on April 02, 2014, 10:54:05 pm
If they weren't fired from the same cannon, they may well be fired at different speeds.

In any case, we can't conclusively say that the two balls will hit each boat at the same time.
Title: Re: VCE Physics Question Thread!
Post by: 09Ti08 on April 02, 2014, 11:04:04 pm
I totally agree. I was just trying to explore all the possibilities.
Anyway, the problem is getting unsolvable. However, because it's a question on a sac, we have to come up with an answer by making simplifications and/or assumptions.
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on April 02, 2014, 11:08:28 pm
Honestly, if I were in that situation, I'd just argue why there's no clear-cut answer.
Title: Re: VCE Physics Question Thread!
Post by: rhinwarr on April 06, 2014, 02:47:15 pm
Can someone help with this question please? I'm not sure how to rearrange the circuit. The answer is supposed to be 83.3 ohms but from the circuit I drew I get 116.7 ohms.
Title: Re: VCE Physics Question Thread!
Post by: alchemy on April 07, 2014, 11:37:43 am
Can someone help with this question please? I'm not sure how to rearrange the circuit. The answer is supposed to be 83.3 ohms but from the circuit I drew I get 116.7 ohms.

I don't do Physics, so I hope this is correct. I think you found the effective resistance for the entire circuit, whereas the question asks to find it between P and R. Check the attached image for my drawing of the circuit, which is pretty much similar to yours. The path marked in red is the area the question concerns. The effective resistance of the first resistor (in parallel) is 1/[(1/100)+(1/100)]=50. The effective resistance of the other resistors from Q to R and the one below them is 1/[(1/100)+(1/100)+(1/100)]=100/3=33.3. Thus, the total effective resistance is 50+33.3=83.3.
Title: Re: VCE Physics Question Thread!
Post by: Yacoubb on April 22, 2014, 09:38:26 pm
I have a few questions:

1. Describe the origin of the centripetal force that causes an object to follow a circular path.

2. I want to know whether I've approached this correctly.

Saturn has at least eighteen natural satellites, two of which are Titan and Tethys.

mass of Titan = 1.35 x 10^23 kg
radius of Titan = 2.6 x 10^6 m
period of Titan's orbit = 1.38 x 10^6 s
radius of Titan's orbit = 1.22 x 10^9 m
mass of Tethys = 7.4 x 10^20 kg
radius of Tethys orbit = 2.9 x 10^8 m

Calculate the gravitational field strength of the surface of titan.

g = GM/r^2.
G = 6.67 x 10^-11
M = 1.35 x 10^23 kg
r = 2.6 x 10^6 m

Is that value of r correct? Which radius do I use? How do I know which one to use in a case like this?

Help would be much appreciated. Thank you :)
Title: Re: VCE Physics Question Thread!
Post by: Bronzebottom64 on April 23, 2014, 04:52:37 pm

Is that value of r correct? Which radius do I use? How do I know which one to use in a case like this?

Help would be much appreciated. Thank you :)

Yes that is the correct value for the radius, the gravitational field strength is calculated purely on the object in question. Any gravitational field on the surface of a planet in the solar system (Earth) will also have external gravitational fields acting upon them as is the nature of matter in the known universe. The question is a bit ambiguous, however I believe what it's asking is what you have done.  :)
Title: Re: VCE Physics Question Thread!
Post by: EspoirTron on April 23, 2014, 05:34:17 pm
I have a few questions:

1. Describe the origin of the centripetal force that causes an object to follow a circular path.

2. I want to know whether I've approached this correctly.

Saturn has at least eighteen natural satellites, two of which are Titan and Tethys.

mass of Titan = 1.35 x 10^23 kg
radius of Titan = 2.6 x 10^6 m
period of Titan's orbit = 1.38 x 10^6 s
radius of Titan's orbit = 1.22 x 10^9 m
mass of Tethys = 7.4 x 10^20 kg
radius of Tethys orbit = 2.9 x 10^8 m

Calculate the gravitational field strength of the surface of titan.

g = GM/r^2.
G = 6.67 x 10^-11
M = 1.35 x 10^23 kg
r = 2.6 x 10^6 m

Is that value of r correct? Which radius do I use? How do I know which one to use in a case like this?

Help would be much appreciated. Thank you :)

For the first part to your question: the centripetal force is always supplied by a real force. This could include the tension in a string or the frictional force between the tyres of a car and the road.
Let's first take the example of the tension in the string scenario. Say if I have a bucket of water attached to a string, as I begin to swing it in the circular path the tension will act towards the center of the circle, which is by consequence the centripetal force.
The second example is one that has popped up a few times on exams. So consider a 'normal' road; that is, one that hasn't been rained on or has much due so it's pretty dry. If a car wanted to go in a circular path around this dry road it would be quite possible. This is because there is a frictional force between the surface of the road and the car tyres which acts towards the center the circular path, and this too is the centripetal force. Often in calculations on horizontal surfaces you will denote this as the net force.
Caution: One of the reasons why you can't maintain such motion on an icy track is because there is little friction between the car tyres and the surface of the road. So as you speed up it becomes difficult to control the car and the car will leave the circular path at a tangent to the path.
Title: Re: VCE Physics Question Thread!
Post by: Yacoubb on April 23, 2014, 06:45:54 pm
For the first part to your question: the centripetal force is always supplied by a real force. This could include the tension in a string or the frictional force between the tyres of a car and the road.
Let's first take the example of the tension in the string scenario. Say if I have a bucket of water attached to a string, as I begin to swing it in the circular path the tension will act towards the center of the circle, which is by consequence the centripetal force.
The second example is one that has popped up a few times on exams. So consider a 'normal' road; that is, one that hasn't been rained on or has much due so it's pretty dry. If a car wanted to go in a circular path around this dry road it would be quite possible. This is because there is a frictional force between the surface of the road and the car tyres which acts towards the center the circular path, and this too is the centripetal force. Often in calculations on horizontal surfaces you will denote this as the net force.
Caution: One of the reasons why you can't maintain such motion on an icy track is because there is little friction between the car tyres and the surface of the road. So as you speed up it becomes difficult to control the car and the car will leave the circular path at a tangent to the path.
Yes that is the correct value for the radius, the gravitational field strength is calculated purely on the object in question. Any gravitational field on the surface of a planet in the solar system (Earth) will also have external gravitational fields acting upon them as is the nature of matter in the known universe. The question is a bit ambiguous, however I believe what it's asking is what you have done.  :)

Thanks to you both!
Title: Re: VCE Physics Question Thread!
Post by: Rishi97 on April 27, 2014, 12:41:39 pm
Does anyone have any good ideas for EPI on electronics? I want to do somthing awesome/interesting :)
Title: Re: VCE Physics Question Thread!
Post by: Rod on April 27, 2014, 12:56:32 pm
Does anyone have any good ideas for EPI on electronics? I want to do somthing awesome/interesting :)
Maybe you could investigate factors that affect a modulated signal. Like for example my friend was telling me that the radio in his car always gets distorted when a motorcycle is near his car, maybe because of the fact those 'drumming' noises are interfering with the modulated signal being formed by the carrier and input signal.

If we were to do another EPI I would base it on that, looks so fun.

Good luck Rishi :)
Title: Re: VCE Physics Question Thread!
Post by: Rishi97 on April 27, 2014, 01:10:12 pm
Maybe you could investigate factors that affect a modulated signal. Like for example my friend was telling me that the radio in his car always gets distorted when a motorcycle is near his car, maybe because of the fact those 'drumming' noises are interfering with the modulated signal being formed by the carrier and input signal.

If we were to do another EPI I would base it on that, looks so fun.

Good luck Rishi :)

Yeah that does sound fun Rod. But we haven't learnt anything about modulated signals at school. So would I still be able to do it?
Title: Re: VCE Physics Question Thread!
Post by: Rod on April 27, 2014, 01:23:33 pm
Yeah that does sound fun Rod. But we haven't learnt anything about modulated signals at school. So would I still be able to do it?
Yeah you should be able to, it's unit 3 area of study 2 stuff, but confirm with your teacher just in case.

All my teacher when through was 'a low frequency input signal in increased by a carrier signal, and then we get our modulated signal.....'. And showed us one diagram. So yeah lol he didn't go through it with us a single bit but we still understand it.

It's just a suggestion, I would probably do it.

Best of luck!

Title: Re: VCE Physics Question Thread!
Post by: Rishi97 on April 27, 2014, 01:41:15 pm
Yeah you should be able to, it's unit 3 area of study 2 stuff, but confirm with your teacher just in case.

All my teacher when through was 'a low frequency input signal in increased by a carrier signal, and then we get our modulated signal.....'. And showed us one diagram. So yeah lol he didn't go through it with us a single bit but we still understand it.

It's just a suggestion, I would probably do it.

Best of luck!

I'll see some other options and ask my teacher as well. Thanks heaps Rod :) :) :)
Title: Re: VCE Physics Question Thread!
Post by: knightrider on April 27, 2014, 09:01:10 pm
how would you do these questions

Calculate the energy in electronvolts of
a)an alpha particle with 8.5*10^-12 J of energy
b)a beta particle with 6.4*10^-11 J of energy
c)a gamma ray with 4.7*10^-11 J of energy
Title: Re: VCE Physics Question Thread!
Post by: Alwin on April 30, 2014, 02:26:04 pm
how would you do these questions

Calculate the energy in electronvolts of
a)an alpha particle with 8.5*10^-12 J of energy
b)a beta particle with 6.4*10^-11 J of energy
c)a gamma ray with 4.7*10^-11 J of energy

Hint: 1 eV = 1.6×10−19 J

Eg for (a) we have:  8.5*10^-12 / (1.6×10−19) = 53 052 829.4 eV (or equivalent scientific notation)

You can try the rest yourself :)
Title: Re: VCE Physics Question Thread!
Post by: Rishi97 on April 30, 2014, 04:37:57 pm
ok so I got my physics EPI project info today and I found out that I have to design a circuit using either resistors, led or diodes. We had to buy this short circuit kit from jaycar and our EPI can only be made using the components which are spring connectors, battery holders, and all the basic electronic components.
I have no idea what to investigate. Any ideas would be greatly greatly appreciated :)
A massive thanks  :D Pls help me
Title: Re: VCE Physics Question Thread!
Post by: Stevensmay on April 30, 2014, 06:06:05 pm
ok so I got my physics EPI project info today and I found out that I have to design a circuit using either resistors, led or diodes. We had to buy this short circuit kit from jaycar and our EPI can only be made using the components which are spring connectors, battery holders, and all the basic electronic components.
I have no idea what to investigate. Any ideas would be greatly greatly appreciated :)
A massive thanks  :D Pls help me
What does 'all the basic electronic components' actually entail?
Title: Re: VCE Physics Question Thread!
Post by: Rishi97 on April 30, 2014, 06:11:05 pm
Diodes, resistors, LEDS, globes (it has more things but we are allowed to only use these)
Title: Re: VCE Physics Question Thread!
Post by: Stevensmay on April 30, 2014, 06:25:56 pm
Diodes, resistors, LEDS, globes (it has more things but we are allowed to only use these)

I asked for clarification just because it didn't seem like much sorry :).

Maybe a row of LED's with different brightness? Honestly I can't think of much you can do with that (at this level).
Title: Re: VCE Physics Question Thread!
Post by: Rishi97 on April 30, 2014, 06:51:39 pm
I asked for clarification just because it didn't seem like much sorry :).

Maybe a row of LED's with different brightness? Honestly I can't think of much you can do with that (at this level).

Yeah..,that's why it's so hard to choose an investigation. So when you said have a row of LED's, what exactly would I be measuring?
Title: Re: VCE Physics Question Thread!
Post by: Stevensmay on April 30, 2014, 07:11:51 pm
The effect of doubling resistance on brightness or something? Or different configurations of resistors eg series/parellel
Title: Re: VCE Physics Question Thread!
Post by: Rishi97 on April 30, 2014, 07:16:49 pm
The effect of doubling resistance on brightness or something? Or different configurations of resistors eg series/parellel

sounds great but how would I prove that the brightness has improved? Don't I need like some sort of measuring device?
Title: Re: VCE Physics Question Thread!
Post by: Orb on April 30, 2014, 08:08:38 pm
Hey guys,

bit stuck on this question:

In a laboratory class at school, Lee is given a spring with a stiffness of 15.4 N m–1 and unstretched length of 0.40 m. He hangs it vertically, and attaches a mass to it, so that the new length of the spring is 1.10 m when the mass is stationary. Assuming the spring has no mass, what was the value of the mass he attached?

Thanks!
Title: Re: VCE Physics Question Thread!
Post by: Rishi97 on April 30, 2014, 08:22:26 pm
Hey guys,

bit stuck on this question:

In a laboratory class at school, Lee is given a spring with a stiffness of 15.4 N m–1 and unstretched length of 0.40 m. He hangs it vertically, and attaches a mass to it, so that the new length of the spring is 1.10 m when the mass is stationary. Assuming the spring has no mass, what was the value of the mass he attached?

Thanks!

I got 0.18kg. Do you have the answers?
Title: Re: VCE Physics Question Thread!
Post by: Orb on April 30, 2014, 08:25:18 pm
I got 0.18kg. Do you have the answers?

it's 1.08.. :/
Title: Re: VCE Physics Question Thread!
Post by: Rishi97 on April 30, 2014, 08:40:15 pm
it's 1.08.. :/

ohhh. I'm waay off then. This is what I did....can someone please tell me where I'm going wrong
Us= 1/2kx2
=1/2 x 15.4 x 0.72
= 3.78 J
Since Us is 3.78 J, gpe must be 3.78
3.78 = mgh
And then I found mass by using the height as 1.1.
Title: Re: VCE Physics Question Thread!
Post by: Stevensmay on April 30, 2014, 08:58:51 pm

The mass will pull the spring down with a force equal to its mass * gravity.
Therefore the force exerted on the spring is $F_{spring} = mg$

We also know that $F_{spring} = kx$

Therefore $kx = mg$

x will be equal to the change in the springs length, so 1.1-0.4 = 0.7
k = 15.4
Substituting in and rearranging gives 1.08.
(I've left out all the negative signs so this isn't strictly true.)
Title: Re: VCE Physics Question Thread!
Post by: Zealous on April 30, 2014, 09:03:25 pm
I'm getting a different answer of 1.72...

The mass will pull the spring down with a force equal to its mass * gravity.
Therefore the force exerted on the spring is $F_{spring} = mg$

We also know that $F_{spring} = kx$

Therefore $kx = mg$

Substituting in and rearranging gives 1.72 which is wrong.
(I've left out all the negative signs so this isn't strictly true.)
Looks alright?

$kx=mg$

$m=\frac{kx}{g}=\frac{15.4 \times 0.7}{10}=1.08kg$

If we define downwards as being positive, our change in x (of the spring) will be a positive value as well as our acceleration of gravity so I don't think we'll encounter any negative answers.
Title: Re: VCE Physics Question Thread!
Post by: Stevensmay on April 30, 2014, 09:09:15 pm
Thanks Zealous, just realized I typed in 1.7 and not 0.7 by mistake.
Title: Re: VCE Physics Question Thread!
Post by: Orb on April 30, 2014, 09:09:57 pm
Looks alright?

$kx=mg$

$m=\frac{kx}{g}=\frac{15.4 \times 0.7}{10}=1.08kg$

If we define downwards as being positive, our change in x (of the spring) will be a positive value as well as our acceleration of gravity so I don't think we'll encounter any negative answers.

Thanks guys!
Title: Re: VCE Physics Question Thread!
Post by: Alwin on April 30, 2014, 10:22:59 pm
Diodes, resistors, LEDS, globes (it has more things but we are allowed to only use these)

Are you allowed to use a multimeter (or voltmeter/ammeter)??

This would considerably help with designing an experiment because you could measure the voltage and current (and hence calculate power) across components rather than buying some fancy device to measure light intensity of an LED.

If you were allowed these measuring devices, a simple (but hard depending if you're allowed a variable resistor or not) is you could vary the current in the circuit by adding resistors and measure the voltage across the diode (with an LED to dissipate excess power). Then plot voltage vs current and find the switch on voltage
(note that this could be VERY hard if you only have large-sized resistors, so google the diode before hand to see what the Vs is)

Otherwise, as others have suggested, you can create a variety of circuits and analyse. It all depends on what your teacher wants :)
(Perhaps even find someone who did Physics last year from your school and ask what they did, but don't do the exact some thing :P)

GLHF :D
Title: Re: VCE Physics Question Thread!
Post by: Rishi97 on May 01, 2014, 03:58:41 pm
Thanks for all your help guys...
Great news!!! I found what I want to do my EPI on.
I'm going to measure the effects of varying reistance has on the power outage of a light bulb. I'm also measuring current as the voltage will remain the same
It's simple but I didn't want to do anything over the required. Like I said, I'm not that good at physics :-\
Title: Re: VCE Physics Question Thread!
Post by: speedy on May 16, 2014, 10:10:32 pm
Ok I have no idea how to do this:

a) V1 decreases due to the negative gradient of the graph and I thought that V2 would stay the same due to it being in photocurrent mode but the answer says it increases.

b) This is where I really had no idea. The units for the graph is mW/m2 but we're given 3 W/m2 so I don't understand how we can use that graph. We can determine the current to be 15 microAmps from the other graph. But yeah that's all I get.

Question:
Spoiler
(http://i.imgur.com/hihb0Ye.png)

Graphs:
V1
Spoiler
(http://i.imgur.com/ML0BTk3.png)
V2
Spoiler
(http://i.imgur.com/oVvopFg.png)
Title: Re: VCE Physics Question Thread!
Post by: RKTR on May 16, 2014, 11:49:07 pm
3Wm^-2 =3000mWm^-2
Title: Re: VCE Physics Question Thread!
Post by: speedy on May 17, 2014, 12:21:52 am
3Wm^-2 =3000mWm^-2

Oh my god I am so stupid. For some reason I was thinking the units for the graph were mega watt / m^2... Thanks for the help!
Title: Re: VCE Physics Question Thread!
Post by: Einstein on May 17, 2014, 12:34:15 pm
Does anyone know any good youtube channels to watch vce physics 1/2 videos?
Title: Re: VCE Physics Question Thread!
Post by: coolrezaee on May 19, 2014, 10:21:35 pm
Does anyone know any good youtube channels to watch vce physics 1/2 videos?

Title: Re: VCE Physics Question Thread!
Post by: Rishi97 on May 20, 2014, 06:53:35 pm
Is anyone going to the TSFX Physics lecture this weekend???
Title: Re: VCE Physics Question Thread!
Post by: Rod on May 20, 2014, 08:27:35 pm
Is anyone going to the TSFX Physics lecture this weekend???
nah, are you?
Title: Re: VCE Physics Question Thread!
Post by: Rishi97 on May 20, 2014, 09:38:16 pm
nah, are you?

Yeppp :)
Title: Re: VCE Physics Question Thread!
Post by: speedy on May 21, 2014, 09:05:52 pm
Quick question:
Spoiler
(http://i.imgur.com/usMS3pS.png)

Generally with these graphs, to find gain must you find the gradient (rise/run) or can you just go Vout/Vin for any value? Because I was taught the latter but for this question that is wrong (must use gradient).
Title: Re: VCE Physics Question Thread!
Post by: rhinwarr on May 22, 2014, 04:30:31 pm
I think you should always use the gradient because I THINK with graphs like these if it doesn't go through zero, the Vout/Vin doesn't work.
Title: Re: VCE Physics Question Thread!
Post by: Nato on May 22, 2014, 04:50:09 pm
Quick question:
Spoiler
(http://i.imgur.com/usMS3pS.png)

Generally with these graphs, to find gain must you find the gradient (rise/run) or can you just go Vout/Vin for any value? Because I was taught the latter but for this question that is wrong (must use gradient).

In this situation, you can't just do Vout/Vin.
That works for when the line passes through the origin.
but in this case, the graph has been shifted - so when you simply do Vout/Vin, you are not taking the shifting of the graph into consideration.
Title: Re: VCE Physics Question Thread!
Post by: speedy on May 23, 2014, 10:33:59 am
Alright, thanks for the help guys.

Another question:
I've seen the formula V1/V2 = R1/R2 on some notes and stuff but I don't understand how to use it, and when it would be used as opposed to the regular voltage divider formula.
Title: Re: VCE Physics Question Thread!
Post by: speedy on May 31, 2014, 09:11:09 am
I really don't get this. the question asks to sketch the demodulated signal wave:

This is the answer, the top being the modulated wave... So how come there are "two" signal waves? Would it be wrong to just draw a regular sinusoidal wave?
(http://i.imgur.com/9mdRc8d.png)
Title: Re: VCE Physics Question Thread!
Post by: PB on May 31, 2014, 06:11:10 pm

Ok first things first, just do me a favour and completely ignore the answers because I think it is wrong. Either that, or I just don't understand it.
Now, this is an example of amplitude modulation, not frequency modulation, I think we gathered that much.
Secondly, I think the carrier wave is a monoamplitude sinusoidal wave with the same frequency as the mondulated wave.
Thus, the signal wave would have to be the other component of the modulated wave. And the only other wave that can summate with the carrier wave I just described to give you the modulated wave shown would be the positive sin looking graph in the answers (not the negative sin graph).

Why? because only a maximum on the signal wave would boost the positive and negative amplitudes of the carrier wave to give you that first section of that modulated wave.  and only a negative minimum on the signal wave would squeeze the positive and negative amplitudes of the carrier wave to give you that section at 2ms of the modulated wave (that part which looks like a noose had been wrung tightly over the wave)
Title: Re: VCE Physics Question Thread!
Post by: PB on May 31, 2014, 06:15:40 pm

So how come there are "two" signal waves?

It would be illogical to have the two signal waves looking like what the answer has shown you because they would simply cancel each other out upon wave summation. Thus, the 'modulated wave' and the carrier wave would look exactly the same. Which is why I don't understand what they are trying to say there...
Title: Re: VCE Physics Question Thread!
Post by: Yacoubb on June 01, 2014, 09:33:46 am
Silly mistakes will be the death of me in this subject. Argh.

Anyone else doing Materials and Structures for their detailed study? It actually looks like a decent detailed study. We then have to do data analysis on materials and structures for this SAC.

Is unit 4 Phys more difficult or easier than U3?
Title: Re: VCE Physics Question Thread!
Post by: Rishi97 on June 01, 2014, 10:21:39 am
Silly mistakes will be the death of me in this subject. Argh.

Anyone else doing Materials and Structures for their detailed study? It actually looks like a decent detailed study. We then have to do data analysis on materials and structures for this SAC.

Is unit 4 Phys more difficult or easier than U3?

We're doing materials and structures for our detailed study as well.. although, I would much prefer to do sound. It 'sounds' more interesting. hehe lol
Title: Re: VCE Physics Question Thread!
Post by: Thorium on June 01, 2014, 10:40:23 am
We're doing materials and structures for our detailed study as well.. although, I would much prefer to do sound. It 'sounds' more interesting. hehe lol

I love to do materials and structures as it hss to do with my dream job, civil engineering. But in vain, the teacher told i have to do sound coz it is related to the wave model of light.
Title: Re: VCE Physics Question Thread!
Post by: speedy on June 01, 2014, 12:32:59 pm
Ok first things first, just do me a favour and completely ignore the answers because I think it is wrong. Either that, or I just don't understand it.
Now, this is an example of amplitude modulation, not frequency modulation, I think we gathered that much.
Secondly, I think the carrier wave is a monoamplitude sinusoidal wave with the same frequency as the mondulated wave.
Thus, the signal wave would have to be the other component of the modulated wave. And the only other wave that can summate with the carrier wave I just described to give you the modulated wave shown would be the positive sin looking graph in the answers (not the negative sin graph).

Why? because only a maximum on the signal wave would boost the positive and negative amplitudes of the carrier wave to give you that first section of that modulated wave.  and only a negative minimum on the signal wave would squeeze the positive and negative amplitudes of the carrier wave to give you that section at 2ms of the modulated wave (that part which looks like a noose had been wrung tightly over the wave)

Alright thanks heaps. My answer was just the positive sin wave so yeah that's good. However, generally I still don't understand why the modulated wave has both a crest and a trough at the same time.
Title: Re: VCE Physics Question Thread!
Post by: PB on June 01, 2014, 03:17:04 pm
I understand where you are coming from and to be honest, I didn't really understand myself only that it is that way. However, I have never encountered something this in depth with modulation before and it is unlikely that you will get anything like this in the exam or SACs.
Title: Re: VCE Physics Question Thread!
Post by: hyunah on June 02, 2014, 08:59:56 pm
can someone help me with this question 886?
Title: Re: VCE Physics Question Thread!
Post by: Alwin on June 02, 2014, 09:23:46 pm
can someone help me with this question 886?

hi :)  (assuming you are doing sound as your detail study and know standing waves etc)

So the pitch is most important in this question. Now, we know that the pitch is essentially the frequency and that v = λ f

First of all consider the piano:
The piano creates music by the means of a hammer hitting a string, which would create a standing wave along the string of constant λ. Since we know that the speed of sound is greater, this implies that f is also 10% larger. Hence, the piano will sound off-pitch (ie need to be retuned)

Now consider the flute:
The flute creates music by means of a standing wave in the air column, kinda like those questions about standing waves in open pipes. Now, since the pipe has a constant length, this means that λ is also constant for say the fundamental frequency. But since the speed of sound is 10% greater, the flute will also sound too high pitched since the frequency of sound produced is 10% greater.

So, I would choose [A] ...although it has been a while since I touched sound :P
Title: Re: VCE Physics Question Thread!
Post by: hyunah on June 02, 2014, 09:40:43 pm
yup thanks alwin,

i am also wondering why at short wavelengths the sound is more likely to travel in straight line?

Title: Re: VCE Physics Question Thread!
Post by: lzxnl on June 02, 2014, 10:46:39 pm
Using VCE terminology:

Smaller wavelengths => smaller wavelength/gap size ratio => less 'diffraction' occurring

Really, a smaller wavelength/gap size ratio just means the wave spreads out less
Title: Re: VCE Physics Question Thread!
Post by: Einstein on June 04, 2014, 03:37:23 pm
for units 1/2 this year, i scored 92% on my radioactivity sac, 100% on Flight  (dual highest) and just recently electricity 98% (highest). Despite this, i have chosen to drop physics, im not enjoying it but i am getting really good marks, if you were in my position what would you do? do you think ive made the right decision to drop out of it for Accounting.
Title: Re: VCE Physics Question Thread!
Post by: Thorium on June 04, 2014, 04:05:21 pm
for units 1/2 this year, i scored 92% on my radioactivity sac, 100% on Flight  (dual highest) and just recently electricity 98% (highest). Despite this, i have chosen to drop physics, im not enjoying it but i am getting really good marks, if you were in my position what would you do? do you think ive made the right decision to drop out of it for Accounting.

If you get similar or better marks in accounting, then go for it
Title: Re: VCE Physics Question Thread!
Post by: Einstein on June 04, 2014, 06:13:30 pm
how and what is the answer to this question, from electricity sac.

Thanks
Title: Re: VCE Physics Question Thread!
Post by: b^3 on June 04, 2014, 06:20:43 pm
how and what is the answer to this question, from electricity sac.

Thanks
They need to be a multiple of the elementary charge, which is approximately $e=1.6\times 10^{-19}\:C$ (I can't remember how many sig figs you do it to in physics but it doesn't matter here).
So if you divide each of the charges by the elementarycharge, you should get an integer.
Doing this gives
A. 20
B. 5
C. 37.5
D. 90

So option C is the only one that isn't an integer multiple of the elementary charge, and so that charge could not exist. i.e. Option C.
Title: Re: VCE Physics Question Thread!
Post by: Einstein on June 04, 2014, 07:23:57 pm
how is this derived?
Title: Re: VCE Physics Question Thread!
Post by: RKTR on June 04, 2014, 11:11:06 pm
how is this derived?
the symbol before the Sv and Gy is micro which is 10^-6 . To determine the most damaging, you have to convert all them to dose equivalent. dose equivalent = absorbed dose x quality factor.
A 20 x 10^-6
B 30 x 10^-6 x (1) =30 x 10^-6
C 30 x 10^-6
D 20 x 10^-6  x( quality factor of 10-20) =(200  to 400) x 10^-6

dose equivalent of D is highest so it is the most damaging.
Title: Re: VCE Physics Question Thread!
Post by: Stew_822 on June 14, 2014, 12:33:28 pm
Hi guys,

I'm really stuck with a voltage divider question, I thought I had them down but this one is confusing me.

It's question 32 and 33 of A+ Notes for physics.

I realise the photo isn't clear - Change in Vout = 6V and change in Vin = 0.4V

Answer to Q30 is 0.6V and 4V - no worries.
Answer to Q31 is 15V - wtf? Isn't it negative 15V because the amplifier is inverting?
Answer to Q32 is 2.5V - That's reading from the graph but when you do Vout = Vgain * Vin = -15 * 0.7 = 10.5V, it doesn't agree with the graph. Can anyone explain why this is the case?
Answer to Q33 is 5.5V - Same problem as Q32.

Thanks,
Stewart
Title: Re: VCE Physics Question Thread!
Post by: speedy on June 14, 2014, 05:30:43 pm
Hi guys,

I'm really stuck with a voltage divider question, I thought I had them down but this one is confusing me.

It's question 32 and 33 of A+ Notes for physics.

I realise the photo isn't clear - Change in Vout = 6V and change in Vin = 0.4V

Answer to Q30 is 0.6V and 4V - no worries.
Answer to Q31 is 15V - wtf? Isn't it negative 15V because the amplifier is inverting?
Answer to Q32 is 2.5V - That's reading from the graph but when you do Vout = Vgain * Vin = -15 * 0.7 = 10.5V, it doesn't agree with the graph. Can anyone explain why this is the case?
Answer to Q33 is 5.5V - Same problem as Q32.

Thanks,
Stewart

Gain is an absolute value and is usually given as positive regardless of it inverting or not. However the assessors accept either positive or negative values.

What you must realise is that this amplifier is biased - its transfer characteristics graph doesn't pass through the origin. When you simply multiply Vin by the gain, you are treating the amplifier as if it is unbiased. To take the biasing into account, the graph must be used.
Title: Re: VCE Physics Question Thread!
Post by: Stew_822 on June 14, 2014, 09:55:46 pm
Ah, I see - I didn't realise there was a difference (or more to the point didn't realise it made a difference). The Heinemann textbook only seems to contain non-biased amplifiers, lucky then that I got these questions now rather than in an exam.

Is that the only way you can find V_out, by reading it off the graph? Or is there a mathematical method also that takes into account the biased nature of the amplifier. Final question, if there is a mathematical method do I need to know it or will the graph be given in the exam?

Stewart
Title: Re: VCE Physics Question Thread!
Post by: Rishi97 on June 21, 2014, 12:01:24 pm
Power lines carry an electric current in the Earth's magnetic field. Which would experience the greater magnetic force: a north-south power line or an east-west power line? Explain
Many thanks :)
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on June 21, 2014, 06:52:22 pm
Earth's magnetic field runs south to north (at least along the Earth's surface), so any current parallel or antiparallel to this would experience no force. Hence your power line needs to be east-west for there to be a force.
Title: Re: VCE Physics Question Thread!
Post by: Orb on June 23, 2014, 01:35:28 pm
When we're calculating the co-efficient of friction, formula Fr= m x N
where m is the co-efficient of friction,

What unit do we attribute to N, the normal reaction force? Do we use the mass provided or do we convert it into newtons?
Title: Re: VCE Physics Question Thread!
Post by: Jacyan on June 23, 2014, 02:01:23 pm
When we're calculating the co-efficient of friction, formula Fr= m x N
where m is the co-efficient of friction,

What unit do we attribute to N, the normal reaction force? Do we use the mass provided or do we convert it into newtons?

Fr is in newtons, N is in newtons, and the co-efficient of friction has no units.
Title: Re: VCE Physics Question Thread!
Post by: Orb on June 23, 2014, 02:07:12 pm
So how should I approach this question...

I get a result that's 10x more than the actual answer :/
Title: Re: VCE Physics Question Thread!
Post by: Ancora_Imparo on June 23, 2014, 07:17:31 pm
So how should I approach this question...

I get a result that's 10x more than the actual answer :/

As we are told the system is moving left, let left be positive. Let the tension in the left rope be $T1$ and the tension in the right rope be $T2$.

Draw the forces the acting on each block.
For M1: mg is down and T1 is up
For M2: mg is down, N is up, T1 is left, T2 is right, friction is right (as we are told block is moving to the left)
For M3: mg is down, T2 is up

Since acceleration of whole system is the same, we can apply Newton's second law to each block individually.
For M1:
$(M1)g - T1 = (M1)a$
$(1.0)g - T1 = (1.0)a$ --- Eq. 1

For M2: $T1 - T2 - Fr = (M2)a$
You also know that $Fr = \mu(N) = \mu(M2)g$
Thus: $T1 - T2 - \mu(M2)g = (M2)a$
$T1 - T2 - (0.50)(5.4)g = (5.4)a$
$T1 - T2 - (2.7)g = (5.4)a$ --- Eq. 2

For M3:
$T2 - (M3)g = (M3)a$
$T2 - (4.3)g = (4.3)a$ --- Eq. 3

Adding Eq. 1 and Eq 3. gives:
$-T1 + T2 - (3.3)g = (5.3)a$ --- Eq. 4

Adding Eq. 2 and Eq 4. gives:
$(-6.0)g = (10.7)a$
[EDIT: Whoops, got the fraction the wrong way around the first time]
$a = \frac{-6.0g}{10.7} m s^{-2}$ (negative sign shows that it's decelerating)

Hope that helps.
Title: Re: VCE Physics Question Thread!
Post by: Orb on June 24, 2014, 10:32:47 am
As we are told the system is moving left, let left be positive. Let the tension in the left rope be $T1$ and the tension in the right rope be $T2$.

Draw the forces the acting on each block.
For M1: mg is down and T1 is up
For M2: mg is down, N is up, T1 is left, T2 is right, friction is right (as we are told block is moving to the left)
For M3: mg is down, T2 is up

Since acceleration of whole system is the same, we can apply Newton's second law to each block individually.
For M1:
$(M1)g - T1 = (M1)a$
$(1.0)g - T1 = (1.0)a$ --- Eq. 1

For M2: $T1 - T2 - Fr = (M2)a$
You also know that $Fr = \mu(N) = \mu(M2)g$
Thus: $T1 - T2 - \mu(M2)g = (M2)a$
$T1 - T2 - (0.50)(5.4)g = (5.4)a$
$T1 - T2 - (2.7)g = (5.4)a$ --- Eq. 2

For M3:
$T2 - (M3)g = (M3)a$
$T2 - (4.3)g = (4.3)a$ --- Eq. 3

Adding Eq. 1 and Eq 3. gives:
$-T1 + T2 - (3.3)g = (5.3)a$ --- Eq. 4

Adding Eq. 2 and Eq 4. gives:
$(-6.0)g = (10.7)a$
$a = \frac{-10.7}{6.0g} m s^{-2}$ (negative sign shows that it's decelerating)

Hope that helps.

I'm sorry, i'm not sure what you did wrong but apparently that wasn't what the answer is...

Solution for the question was 5.6 ms ^-2

Title: Re: VCE Physics Question Thread!
Post by: Zealous on June 24, 2014, 11:01:53 am
I'm sorry, i'm not sure what you did wrong but apparently that wasn't what the answer is...

Solution for the question was 5.6 ms ^-2
I'll give this a shot. Hopefully my explanation makes sense.

First let's look at the forces due to gravity. We have a force due to gravity of M3 which is $F_{M3}=mg=4.3\times 10=43N$ and the force due to gravity on M1 which is $F_{M1}=mg=1\times 10=10N$.

Now let's consider the frictional force of M2:
$F_{M2}=mg=54N$

$F_{Friction}=\mu mg=0.5 \times 54=27N$

Friction acts against a motion, so since M2 was initially moving left, the force of friction will be acting in the right/clockwise direction.

If we consider moving right/clockwise to be positive, the sum of the forces will be:
$43N + 27N - 10N=60N$
M3 and M2 are causing the system to move right so they are positive, and M1 is working against both of them so is negative.

Hence:
$F=ma \implies 60=(4.3+5.4+1)a \implies a=\frac{60}{10.7}\approx 5.1 \ (\frac{m}{s^2})$

Title: Re: VCE Physics Question Thread!
Post by: Ancora_Imparo on June 24, 2014, 12:03:31 pm
Whoops, just got the last line wrong.
$-6.0g=(10.7)a$
$a=\frac{-6.0g}{10.7}=-5.5 m s^{-2}$ if $g=9.8 m s^{-2}$

Zealous, in the post above, also got the same, but just used $g=10 m s^{-2}$, which is why I believe the answer is slightly different.
Title: Re: VCE Physics Question Thread!
Post by: knightrider on June 28, 2014, 09:37:29 pm
How would you do the following question
Title: Re: VCE Physics Question Thread!
Post by: Homer on June 28, 2014, 09:49:58 pm
a)50+30=80km
b)50-30=20km north
Title: Re: VCE Physics Question Thread!
Post by: Stew_822 on June 28, 2014, 09:55:06 pm
Distance is a scaler value so you simply add the two together. Since displacement is a vector, you have to do vector addition, so if we define north as positive, south would be in the opposite direction and hence negative.
So 50 + (-30) = +20km and since we have said positive is north, the full answer is 20km north.
Title: Re: VCE Physics Question Thread!
Post by: knightrider on June 28, 2014, 10:40:20 pm

but in my book it says displacement=final position-initial position

in this case it would be 30-50=-20 which would be south

Can anyone explain how this works
Title: Re: VCE Physics Question Thread!
Post by: Yacoubb on June 29, 2014, 01:19:36 am

but in my book it says displacement=final position-initial position

in this case it would be 30-50=-20 which would be south

Can anyone explain how this works

So total distance would be 50km + 30km = 80km (no direction needed as this is a scalar quantity).

Now, let's make south positive, and thus north negative.

-50km + 30km = -20km
So, we know that the displacement is 20km NORTH.

Let's try making south negative, and thus north positive.

-30km + 50km = 20km
So, displacement = 20km north (given that this value is positive, denoting north).

Hope this helps (:
Title: Re: VCE Physics Question Thread!
Post by: Stew_822 on June 29, 2014, 10:02:01 am
It still works using that definition, so again we will say north is positive. Let's also say his initial position is 0. I'll do a little diagram for you:

Code: [Select]
- - - - - >     <- - -So he moves 50km right (north) then 30km south. His initial position is 0, then +50, then he moves back 30km so his final position is +20km.

Using that formula, displacement = final position - initial = +20 - 0 = +20km and since positive then its north.

An easier way to do these questions where they list all of the distance is to say that displacement = vector sum of all movements, ie. how its been calculated in the last few posts.
Title: Re: VCE Physics Question Thread!
Post by: knightrider on June 30, 2014, 11:08:45 pm
Thanks so much guys and girls it makes so much more sense now love you guys  :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) ::)
Title: Re: VCE Physics Question Thread!
Post by: knightrider on July 09, 2014, 08:16:41 pm
Can anyone help with this question

Title: Re: VCE Physics Question Thread!
Post by: PB on July 09, 2014, 10:56:41 pm
Can anyone help with this question
Well the instantaneous velocity would require you to find the gradient at point t=35s.  So that requires differentiation - which is basically impossible because you have no formula to differentiate!
In this case, I reckon you should try draw a tangent from the point t=35s and find the gradient of that line :P
A bit of a shoddy technique but its VCE Physics. Besides, VCAA is unlikely to give this kind of question because there would be too much variation in the answers obtained. Hope that helps
Title: Re: VCE Physics Question Thread!
Post by: knightrider on July 09, 2014, 11:58:11 pm
Thanks
how would you do this question
Title: Re: VCE Physics Question Thread!
Post by: Rishi97 on July 10, 2014, 11:29:35 am
A lamp purchased in the USA is designed to operate at optimum efficiency when it is connected to an AC supply that has a peak voltage of 170 V and a frequency of 60 Hz. The lamp has an operating resistance of 100 ohms

What is the peak power produced by this lamp?
I know the peak power = I peak x V peak but this isn't working for me

Thanks :)
Title: Re: VCE Physics Question Thread!
Post by: PB on July 10, 2014, 12:02:30 pm
Thanks
how would you do this question
seeing as you are going from v/t to pos/t graph this time. ANTIdifferentiation is required (area under the graph=displacement).  You can figure out the v-t graph formulas this time as they are linear so it is possible to antidiff.
Title: Re: VCE Physics Question Thread!
Post by: PB on July 10, 2014, 12:14:40 pm
A lamp purchased in the USA is designed to operate at optimum efficiency when it is connected to an AC supply that has a peak voltage of 170 V and a frequency of 60 Hz. The lamp has an operating resistance of 100 ohms

What is the peak power produced by this lamp?
I know the peak power = I peak x V peak but this isn't working for me

Thanks :)
Just wondering what the answer was?
Title: Re: VCE Physics Question Thread!
Post by: Rishi97 on July 10, 2014, 12:28:25 pm
Just wondering what the answer was?

289 W
Title: Re: VCE Physics Question Thread!
Post by: PB on July 10, 2014, 01:08:27 pm
289 W
Yep, I am not sure what you did wrong but your formula was right. Pp=VpIp
we have Vp so we need to find Ip.   Ip= Vp/R =170/100 =1.7

Pp=1.7*170 = 289W

Title: Re: VCE Physics Question Thread!
Post by: Rishi97 on July 10, 2014, 01:37:08 pm
Yep, I am not sure what you did wrong but your formula was right. Pp=VpIp
we have Vp so we need to find Ip.   Ip= Vp/R =170/100 =1.7

Pp=1.7*170 = 289W

Thanks heaps for that PB :)
Title: Re: VCE Physics Question Thread!
Post by: knightrider on July 10, 2014, 02:21:09 pm
seeing as you are going from v/t to pos/t graph this time. ANTIdifferentiation is required (area under the graph=displacement).  You can figure out the v-t graph formulas this time as they are linear so it is possible to antidiff.

Thanks for your help but i still dont understand what will be my position points for my position/time grapgh and how do i work them out
Title: Re: VCE Physics Question Thread!
Post by: PB on July 10, 2014, 02:56:14 pm
Thanks for your help but i still dont understand what will be my position points for my position/time grapgh and how do i work them out
Ok, I just realised that you don't even need to antidiff any formulas. You can just count the squares - its the same thing...area under the graph and all.
So at t=1 the area under the graph is 1m/s  x 1 s = 1m (seconds cancel out to give you metres - the position!).  t=2   ---->  1m/s x 2s = 2m.  t=3   ---->  3 and a half squares so 3.5m  So on and so forth.
Just find the area under the graph at each second and plot those values on the Pos/time graph. Then connect the dots.  Do you understand how that works?
Title: Re: VCE Physics Question Thread!
Post by: knightrider on July 10, 2014, 03:40:25 pm
Ok, I just realised that you don't even need to antidiff any formulas. You can just count the squares - its the same thing...area under the graph and all.
So at t=1 the area under the graph is 1m/s  x 1 s = 1m (seconds cancel out to give you metres - the position!).  t=2   ---->  1m/s x 2s = 2m.  t=3   ---->  3 and a half squares so 3.5m  So on and so forth.
Just find the area under the graph at each second and plot those values on the Pos/time graph. Then connect the dots.  Do you understand how that works?

yes i do thanks so much :)
Title: Re: VCE Physics Question Thread!
Post by: PB on July 10, 2014, 04:34:38 pm

yes i do thanks so much :)

Cool beans :) BTW, I have a physics lecture on next Tuesday  which teaches you how to deal with exactly these kinds of stuff and includes shortcuts for some questions and much much more. Would you be interested in coming?
I have held this lecture before and it seems to have been very beneficial for those who attended! Alll the details can be found in the link in my signature below. Please do check it out! It will be the last time I will be holding it, so don't miss out!
Title: Re: VCE Physics Question Thread!
Post by: Rishi97 on July 10, 2014, 07:17:18 pm
What is the difference between a slip-ring and a split ring communtator?
Which is better and why?
Title: Re: VCE Physics Question Thread!
Post by: knightrider on July 10, 2014, 08:04:31 pm
how would you do these questions

What is the acceleration of the train 10 s after starting?

What is the acceleration of the train 40 s after starting?

Title: Re: VCE Physics Question Thread!
Post by: Stew_822 on July 11, 2014, 11:06:00 am
The acceleration is given by the gradient, I think. So, if you can find the gradient of the line (generally found by drawing a straight line at the position you want, ie. 10s and for the second question 40s) then you can use the rise over run formula to find the gradient which gives you the acceleration.

eg. the acceleration at 10s I would say is 20/20 = 1m/s^2.

This is described here (scroll down to finding the gradient of a curve): http://www.mathsrevision.net/gcse-maths-revision/algebra/gradients-and-graphs

Does that make sense? Hope it helps,
Stewart
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on July 11, 2014, 11:56:33 am
What is the difference between a slip-ring and a split ring communtator?
Which is better and why?

They do completely different things. Slip rings don't change current direction => AC
Split rings do reverse current direction => DC
Hopefully you learn how split rings reverse the current direction
Title: Re: VCE Physics Question Thread!
Post by: knightrider on July 11, 2014, 01:01:02 pm
The acceleration is given by the gradient, I think. So, if you can find the gradient of the line (generally found by drawing a straight line at the position you want, ie. 10s and for the second question 40s) then you can use the rise over run formula to find the gradient which gives you the acceleration.

eg. the acceleration at 10s I would say is 20/20 = 1m/s^2.

This is described here (scroll down to finding the gradient of a curve): http://www.mathsrevision.net/gcse-maths-revision/algebra/gradients-and-graphs

Does that make sense? Hope it helps,Stewart

i still dont get how you got 20/20 = 1m/s^2. because it is a curve so you cant do rise on run and if you did that it would be 20/10=2m/s^2  the answers say something else to both the answers  can anyone help
Title: Re: VCE Physics Question Thread!
Post by: Stew_822 on July 11, 2014, 01:45:00 pm
Hi there,

Check these links for finding the gradient of a non-linear graph, you have to draw a line and find the gradient of that line as the other link explains. This is the acceleration.
http://cstl.syr.edu/fipse/graphb/unit8/unit8a.html
http://www.columbia.edu/itc/sipa/math/slope_nonlinear.html
Title: Re: VCE Physics Question Thread!
Post by: PB on July 11, 2014, 02:18:17 pm
as I mentioned before knightrider. This type of question will never be tested by VCAA - there will be too much variation in the answers. My advice is to not spend too much time on these types of questions and focus on other more important stuff.
Title: Re: VCE Physics Question Thread!
Post by: Rishi97 on July 11, 2014, 04:13:45 pm
A sphere of mass 12kg is allowed to roll down a plane inclined at 30 degrees to the horizontal. There is a constant frictional force of 20N acting on the sphere.
WHat is the acceleration?
Using the a=gsin(angle) formula, I am getting the answer of 5m/s/s but the answer says 3.3m/s/s

What am I doing wrong? Why doesn't the formula work?

Title: Re: VCE Physics Question Thread!
Post by: rhinwarr on July 11, 2014, 04:49:57 pm
You need to take away the acceleration in the opposite direction caused by the frictional force.
a = F/m = 20/12 = 1.7m/s/s
So the 'net' acceleration would be 5-1.7=3.3m/s/s
Title: Re: VCE Physics Question Thread!
Post by: PB on July 11, 2014, 05:16:44 pm
yep, that formula only works when there is no friction.
Title: Re: VCE Physics Question Thread!
Post by: PB on July 11, 2014, 06:28:12 pm
Oh and Rishi, for your split/slip ring question. The full definitions can be found in my cheat sheet under the section Electric Power. And neither is "better" per se, rather they both have different functions.
SLIP rings are used to "collect" the AC current generated from generators while SPLIT rings facilitate the constant rotation of motors.
You should try to understand how these rings achieve those feats though, and not just memorise the answers :)
Title: Re: VCE Physics Question Thread!
Post by: Yacoubb on July 11, 2014, 07:49:13 pm
how would you do these questions

What is the acceleration of the train 10 s after starting?

What is the acceleration of the train 40 s after starting?

With velocity-time graphs, acceleration can be calculated from the gradient.

So:

(a) 20m/s / 10s = 2m/s/s

(b) 40m/s / 40s = 1m/s/s

We can validate this because we can see that the graph eventually plateus; an indication of the fact that the object is travelling at constant speed (at which gradient = 0).
Title: Re: VCE Physics Question Thread!
Post by: Rishi97 on July 11, 2014, 10:25:02 pm
Oh and Rishi, for your split/slip ring question. The full definitions can be found in my cheat sheet under the section Electric Power. And neither is "better" per se, rather they both have different functions.
SLIP rings are used to "collect" the AC current generated from generators while SPLIT rings facilitate the constant rotation of motors.
You should try to understand how these rings achieve those feats though, and not just memorise the answers :)

Thanks PB
Yeah I'll check your cheat sheet on this info.  :)
Title: Re: VCE Physics Question Thread!
Post by: knightrider on July 11, 2014, 11:24:05 pm
thanks guys :) the answers say ~1.3 m s-2 ,  ~0.5 m s-2

Title: Re: VCE Physics Question Thread!
Post by: knightrider on July 11, 2014, 11:25:09 pm

how would you draw an acceleration–time graph for the bus
Title: Re: VCE Physics Question Thread!
Post by: Stew_822 on July 12, 2014, 11:25:38 am
With velocity-time graphs, acceleration can be calculated from the gradient.

So:

(a) 20m/s / 10s = 2m/s/s

(b) 40m/s / 40s = 1m/s/s

We can validate this because we can see that the graph eventually plateus; an indication of the fact that the object is travelling at constant speed (at which gradient = 0).
Hello there,
The acceleration you've calculated in (a) is the average acceleration for the first 10 seconds. The question is asking the accerelation at ten seconds after starting, meaning that you need to draw a line at a tangent to the gradient at that point and calculate the gradient from that line, as the acceleration isn't linear since the graph isn't linear. You can see that there is a lot of initial acceleration, but it dies off later. So although the average acceleration for the first 10s is 2m/s/s, the acceleration at say 2s is greater than this and the acceleration at say 8s is less than this (and obviously then the acceleration at 10s is less).

I hope this makes sense and helps. When I calculated the gradient, I didn't bother drawing a line - I basically guestimated. I can draw you guys a pic if you like.
Cheers,
Stewart
Title: Re: VCE Physics Question Thread!
Post by: Stew_822 on July 12, 2014, 11:28:02 am
how would you draw an acceleration–time graph for the bus

You find the accelation of each linear portion of the graph (you can find it using rise over run as this v-t graph is linear) and graph the acceleration against time.
Title: Re: VCE Physics Question Thread!
Post by: knightrider on July 12, 2014, 01:09:30 pm

You find the accelation of each linear portion of the graph (you can find it using rise over run as this v-t graph is linear) and graph the acceleration against time.

Yea thanks but how would you find the acceleration for 6 seconds or 8 seconds for the bus
Title: Re: VCE Physics Question Thread!
Post by: Stew_822 on July 12, 2014, 03:40:42 pm
You can see the acceleration is the same for 6s or 8s as the gradient is the same. So you do rise over run for that section of the graph and that is your acceleration for that section of the graph. Try googling "finding gradient of a graph" if you are having trouble doing this, there would be plenty of tutorials out there.
Title: Re: VCE Physics Question Thread!
Post by: knightrider on July 14, 2014, 04:18:28 pm
How would you do the following questions

Title: Re: VCE Physics Question Thread!
Post by: knightrider on July 14, 2014, 10:47:23 pm
Q1 Hint: What does the gradient of a velocity-time graph tell you?

Spoiler
Acceleration

Q2,3,4 Hint: What does the area under a velocity-time graph tell you?

Spoiler
Displacement

If you still need help just post again

Thanks for tips but i still need help
Title: Re: VCE Physics Question Thread!
Post by: knightrider on July 14, 2014, 11:43:03 pm
how would you do the following questions
Title: Re: VCE Physics Question Thread!
Post by: Stew_822 on July 15, 2014, 10:25:56 am
Hello knightrider,

When asking questions and for help, it is better if you let us know what you are struggling with specifically, rather than just saying you need help. For example, you could tell us how you attempted the problem and that way we would know where a mistake may have been made and point you in the right direction.

I remember that question with the bus and Anna from last year, I had to ask my teacher.

You can usually find the worked solutions by googling "heinemann physics 11 worked solutions chapter X" where X is obviously the chaper you're up to.

I hope this helps.

Cheers,
Stewart
Title: Re: VCE Physics Question Thread!
Post by: hyunah on July 15, 2014, 07:04:21 pm
An electron moving north enters a magnetic field
that is directed vertically upwards.
If the electron’s motion was inclined upwards
at an angle, as well as travelling north, what
would be the path of the electron?
Title: Re: VCE Physics Question Thread!
Post by: rlenora on July 15, 2014, 09:52:30 pm
An electron moving north enters a magnetic field
that is directed vertically upwards.
If the electron’s motion was inclined upwards
at an angle, as well as travelling north, what
would be the path of the electron?

Use right hand palm rule (slap rule :P)

Direction of magnetic field: up (use fingers of right hand to point up)

Direction of current, i.e. direction of positive charge: south (opposite to direction of electron) (point thumb to point south on right hand)

Direction of force on electron: West (given by the direction of the palm of right hand)

Hence electron will initially move West, and since force applied on the electron will always be perpendicular to its motion (like it is in circular motion), the electron will travel in a circular path.

We do not have to worry about the component of the electron's motion upwards (i.e. parallel to the magnetic field), since components of current  (velocity of the electron in this case) parallel to the magnetic field doesn't result in a force being applied on the electron.

Hope this helps :)

Let me know if you'd like clarification on part!
Title: Re: VCE Physics Question Thread!
Post by: rlenora on July 15, 2014, 10:27:41 pm
Hello knightrider,

When asking questions and for help, it is better if you let us know what you are struggling with specifically, rather than just saying you need help. For example, you could tell us how you attempted the problem and that way we would know where a mistake may have been made and point you in the right direction.

I remember that question with the bus and Anna from last year, I had to ask my teacher.

You can usually find the worked solutions by googling "heinemann physics 11 worked solutions chapter X" where X is obviously the chaper you're up to.

I hope this helps.

Cheers,
Stewart

Exactly!

Title: Re: VCE Physics Question Thread!
Post by: hyunah on July 15, 2014, 11:18:02 pm
hey Lenora,
the part which I didn't get is the inclines angle bit and yet as well as travelling north, does that when my hand is like slanted north - west or north east?
Title: Re: VCE Physics Question Thread!
Post by: rlenora on July 15, 2014, 11:51:44 pm
You don't have to worry about the inclined bit of the motion of the electron, since it is parallel to the direction of the magnetic field (which is also upwards), it doesn't result in a force being applied on the electron.

So all you have to worry about is that the electron is moving north.
Which is the same as if a positive charge was moving south.

and since the thumb points in the direction of the conventional current (i.e. direction of movement of positive charge), thumb will need to point south.

So, fingers up. thumb south. palm faces west. Force applied on electron is to the west :)
Title: Re: VCE Physics Question Thread!
Post by: knightrider on July 18, 2014, 11:27:47 pm
hi guys,
For these 2 equations how do you know when to use the one with the plus 1/2 and the one with -1/2

What does the plus and minus mean and how would you know to use which formula
Title: Re: VCE Physics Question Thread!
Post by: Thorium on July 19, 2014, 02:37:32 am
hi guys,
For these 2 equations how do you know when to use the one with the plus 1/2 and the one with -1/2

What does the plus and minus mean and how would you know to use which formula

Apparently the second one is not officially recognised (maybe only by my school), but it is quite useful in some situations.

They are both used under one main condition, that is, when acceleration is constant. You can notice that one involves u (initial velocity) and the other involves v (final velocity). So if you have one quantity missing out of x, u, t, a, then you use the equation x=ut+1/2at^2. If you have one quantity missing out of x, v, t, a, then you use the equation x=vt-1/2at^2.

Hope that helps :)
Title: Re: VCE Physics Question Thread!
Post by: knightrider on July 19, 2014, 09:07:57 am
Apparently the second one is not officially recognised (maybe only by my school), but it is quite useful in some situations.

They are both used under one main condition, that is, when acceleration is constant. You can notice that one involves u (initial velocity) and the other involves v (final velocity). So if you have one quantity missing out of x, u, t, a, then you use the equation x=ut+1/2at^2. If you have one quantity missing out of x, v, t, a, then you use the equation x=vt-1/2at^2.

Hope that helps :)

Didn't realize the equations were different i thought they bith had u's that were negative and positive thankyou  for your help  :)
Title: Re: VCE Physics Question Thread!
Post by: knightrider on July 19, 2014, 05:29:13 pm
Hi how would you do the following question

I have worked out the initial speed to be 20m/s and the maximum height to be 20m

how would you find the speed of the cork as it returned to its
starting point?
Title: Re: VCE Physics Question Thread!
Post by: Butterscotch on July 20, 2014, 04:09:17 pm
Hi how would you do the following question

I have worked out the initial speed to be 20m/s and the maximum height to be 20m

how would you find the speed of the cork as it returned to its
starting point?

An easy way to go about solving such types of questions is to ignore the rise of the cap. Let's just look at the fall of the cork from the maximum point.
u = 0 ms^-1
a = 9.8 ms^-2
t = 2 s (remember we're only looking at the return journey of the cork)
v = ?

You'd plug these into the formula: v = u + at

EDIT: Orrr an alternative (also one which saves time) is one that PB has said below :)
Title: Re: VCE Physics Question Thread!
Post by: PB on July 21, 2014, 11:58:40 am
Well, technically you don't even have to do any working out... i mean, the cork decelerates on the way up and accelerates by the same amount on the way down.  It is bound to have the same speed at the end of the flight as it did at the beginning.

edit: speed, not velocity
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on July 21, 2014, 02:46:12 pm
Well, technically you don't even have to do any working out... i mean, the cork decelerates on the way up and accelerates by the same amount on the way down.  It is bound to have the same velocity at the end of the flight as it did at the beginning.

You've invoked the conservative nature of the gravitational force, and a rigorous proof of that isn't going to be given before second year university :P

Technically, it should reach the starting height slower than its initial speed because of the non-conservative nature of air resistance. In addition, PB I think you've made a typo there that I'll highlight. I trust you can see what is wrong with it :P
Title: Re: VCE Physics Question Thread!
Post by: yang_dong on July 21, 2014, 09:11:35 pm
why no force is produced when the wire is parallel to the magnetic field?

thank you
Title: Re: VCE Physics Question Thread!
Post by: PB on July 22, 2014, 01:04:09 am
You've invoked the conservative nature of the gravitational force, and a rigorous proof of that isn't going to be given before second year university :P

Technically, it should reach the starting height slower than its initial speed because of the non-conservative nature of air resistance. In addition, PB I think you've made a typo there that I'll highlight. I trust you can see what is wrong with it :P

Well yeah of course, but I didn't want to confuse anyone else given that this is VCE Physics we are talking about and not real life physics.
But your second point is valid. My apologies people, it should really say speed instead of velocity because 'same velocity' would imply that the cork has the same direction of travel too..which is of course false - it is the opposite on the way down.
Title: Re: VCE Physics Question Thread!
Post by: PB on July 22, 2014, 01:19:14 am
why no force is produced when the wire is parallel to the magnetic field?

thank you

Interesting phenomenon and sadly way outside the course content of VCE Physics :P
If it were to be asked though, just prove it using the formula for the force acting on  current carrying wire "F=nILBsin(theta)".  When M-field and current is parallel, theta is 0 which makes the whole left side of the equation equal 0. Thus, force is zilch when wire is parallel to M-field.
Always good to link it back to a formula to prove a point.
Title: Re: VCE Physics Question Thread!
Post by: yang_dong on July 22, 2014, 11:28:00 am
thank you PB,

thank you
Title: Re: VCE Physics Question Thread!
Post by: Saikyo on July 27, 2014, 05:30:11 pm
My hatred for the jacaranda physics textbook is at my boiling point here...

In an elastic collision between two objects of mass m1 and m2, show that the speed of approach (u2-u1) is equal to the speed of seperation (v2+v1). The symbols u1, u2, v1 and v2 each represent speeds, not velocities.

there is NO WAY that this is a vcaa type question right guys??

[sorry for the bad drawing i made from the photo for the diagram in the question!]

Title: Re: VCE Physics Question Thread!
Post by: lzxnl on July 27, 2014, 08:30:08 pm
This is just proving the conservation of momentum (if initial total masses are equal to final total masses, then total initial speeds are equal to total final speeds)

VCAA won't give anything this vague, it will always have actual figures for you to plug into equations

It's not even proving the conservation of momentum; that is a principle you'd have to use to prove it

I wish VCAA gave things without figures.
Title: Re: VCE Physics Question Thread!
Post by: Brunette15 on July 28, 2014, 07:25:23 pm
Hi everyone 8) ,
Estimate the size of the smallest object that can be clearly imaged by a microscope that uses visible light. Explain this limitation.
Title: Re: VCE Physics Question Thread!
Post by: PB on July 29, 2014, 12:01:08 am
Hi everyone 8) ,
Estimate the size of the smallest object that can be clearly imaged by a microscope that uses visible light. Explain this limitation.
A microscope works by shining light through a slide and the specimen absorbs some frequencies of light, producing an image that enters the lens and into your eye (or something like that).
Now, we all know that light whose wavelength is greater than the width of a gap will diffract significantly. The same thing goes if the wavelength is greater than the width of an object! Fortunately, this doesn't affect our day to day lives because every-day objects are definitely not smaller than the wavelengths of the visible light spectrum. If not, there will be crazy diffractions going everywhere, and nobody will be able to see anything.
However when we are dealing with extremely tiny objects, whose width is actually smaller than the wavelength of visible light, significant diffraction starts to occur, and (as you can imagine) starts to create blurry images.
So there you go. Now go and find the smallest wavelength of visible light, and the object of interest should be no smaller than that in order to be clearly imaged by a light microscope.
Title: Re: VCE Physics Question Thread!
Post by: Brunette15 on July 29, 2014, 12:51:47 pm
A microscope works by shining light through a slide and the specimen absorbs some frequencies of light, producing an image that enters the lens and into your eye (or something like that).
Now, we all know that light whose wavelength is greater than the width of a gap will diffract significantly. The same thing goes if the wavelength is greater than the width of an object! Fortunately, this doesn't affect our day to day lives because every-day objects are definitely not smaller than the wavelengths of the visible light spectrum. If not, there will be crazy diffractions going everywhere, and nobody will be able to see anything.
However when we are dealing with extremely tiny objects, whose width is actually smaller than the wavelength of visible light, significant diffraction starts to occur, and (as you can imagine) starts to create blurry images.
So there you go. Now go and find the smallest wavelength of visible light, and the object of interest should be no smaller than that in order to be clearly imaged by a light microscope.

Thanks so much!  :D
Title: Re: VCE Physics Question Thread!
Post by: hyunah on July 30, 2014, 10:27:03 pm
a ball is thrown vertically upward from a platform 16m above the ground and has a initial velocity of 24.5m/s. Find:
a) max. height the ball reaches
b) time taken to reach max height
c) time taken to reach ground from max height
d) total time the ball is in the air
e) speed of ball when it returns to the platform
f) speed of ball when it hits the ground

Title: Re: VCE Physics Question Thread!
Post by: RKTR on July 31, 2014, 01:00:40 am
a ball is thrown vertically upward from a platform 16m above the ground and has a initial velocity of 24.5m/s. Find:
a) max. height the ball reaches
b) time taken to reach max height
c) time taken to reach ground from max height
d) total time the ball is in the air
e) speed of ball when it returns to the platform
f) speed of ball when it hits the ground

(a) v^2=u^2+2ax
at max height,v=0
0=24.5^2+2(-10)x
x=30m
but this is the displacement from the platform therefore max height=30+16=46m
(b) v=u+at
0=24.5-10t
t=2.45s
(c) x=ut+1/2 a t^2
from max height to ground x=-46
-46=1/2(-10)(t^2)
t=3.03s
(d) total time = ans of (b) + ans of (c)
=2.45+3.03
=5.48s
(e)v^2=u^2+2ax
v^2=2(-10)(-30)
v^2=600
speed=24.5 m/s
(f)v^2=u^2+2ax
v^2=2(10)(46)
v^2=920
speed=30.3 m/s
Title: Re: VCE Physics Question Thread!
Post by: hyunah on July 31, 2014, 11:23:43 am
thank you RKTR

A bus travels 60 metres in 10 seconds and the next 60 metres in 15 seconds. If the acceleration is constant, find:
i) how much further it will travel before ocming to rest
ii) how many more seconds it takes before coming to rest

Title: Re: VCE Physics Question Thread!
Post by: speedy on August 03, 2014, 01:05:42 pm
When using the RHG rule for a solenoid, the side that interacts as north is at the tip of the thumb, and the side that interacts as south is at the base of the thumb...? Right? My teacher says the opposite, which is true for inside the solenoid, but when trying to determine the direction of current flow using lenz's law, for example a bar magnet approaching a solenoid, wouldn't you want to focus on the interactions between the poles outside of the solenoid.

Take this image -
Spoiler
(http://www.geocities.ws/motorac2002/field_dir_poles_e.gif)
- my teacher would say that the poles should be switched, and we use those to determine how solenoids will interact.
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on August 03, 2014, 02:16:24 pm
Ask your teacher why he/she thinks that. The magnetic field lines are leaving at the marked N terminal, so I see nothing wrong with the diagram given.
Title: Re: VCE Physics Question Thread!
Post by: speedy on August 03, 2014, 06:04:49 pm
Ask your teacher why he/she thinks that. The magnetic field lines are leaving at the marked N terminal, so I see nothing wrong with the diagram given.

What he says is that any point is north relative to a point further down the line (in the direction of the field), which is a south. Basically that labelling a north and south is relative, so INSIDE the solenoid, the field starts at a north and goes to a south, thus doing the opposite to the diagram. He argues that you should never label the poles of a solenoid as they are relative (he's a VCAA assesor, and took a mark off my friend in a test for labelling the poles). However when I asked him a question on lenz's law, with a bar magnet going into a loop, he said to determine the direction of current, you need a north end produced by the coil that opposes the north coming towards it - to do this he used the north and south relative to inside the solenoid. Which, when looking over some questions, I found to be wrong.
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on August 03, 2014, 07:17:20 pm
What he says is that any point is north relative to a point further down the line (in the direction of the field), which is a south. Basically that labelling a north and south is relative, so INSIDE the solenoid, the field starts at a north and goes to a south, thus doing the opposite to the diagram. He argues that you should never label the poles of a solenoid as they are relative (he's a VCAA assesor, and took a mark off my friend in a test for labelling the poles). However when I asked him a question on lenz's law, with a bar magnet going into a loop, he said to determine the direction of current, you need a north end produced by the coil that opposes the north coming towards it - to do this he used the north and south relative to inside the solenoid. Which, when looking over some questions, I found to be wrong.

The problem is, you can't say that north and south are relative terms. They're not. Magnetic field lines form closed loops, so by his logic you can keep going more and more and more north, which is ludicrous.
Inside the solenoid, you go from a south pole to a north pole, not the other way around, exactly like a bar magnet. I don't know what he's thinking, but honestly I don't follow his logic.

As for Lenz's law and a bar magnet going into the loop, if it's the north pole of the bar magnet facing the loop, you have a stronger magnetic field going into the loop so the resulting current should produce a magnetic field which opposes the original magnetic field. And vice versa.
Title: Re: VCE Physics Question Thread!
Post by: speedy on August 03, 2014, 07:33:04 pm
The problem is, you can't say that north and south are relative terms. They're not. Magnetic field lines form closed loops, so by his logic you can keep going more and more and more north, which is ludicrous.
Inside the solenoid, you go from a south pole to a north pole, not the other way around, exactly like a bar magnet. I don't know what he's thinking, but honestly I don't follow his logic.

As for Lenz's law and a bar magnet going into the loop, if it's the north pole of the bar magnet facing the loop, you have a stronger magnetic field going into the loop so the resulting current should produce a magnetic field which opposes the original magnetic field. And vice versa.

Yeah I completely agree. This is really annoying because he's a very good teacher and does know his stuff. I'll bring it up with him tomorrow, tell him what you said, show him the textbook and a few questions. Thank you :)

Also with that, for a magnet approaching a loop from the left, would you say that the field is increasing to the left, thus the induced current must produce a field increasing to the right, which would be coming out of the loop in the same direction the magnet was moving?

Also, just another question I came across, how do you find the average voltage produced by a coil that has been turned one complete cycle? You use faraday's law with the time for one quarter of the cycle, right?
Title: Re: VCE Physics Question Thread!
Post by: Brunette15 on August 04, 2014, 08:02:50 pm
What accelerating potential difference would be required to give an alpha particle a de Broglie wavelength of 2.0nm?

mass of alpha is 6.67x10^-27kg

I attached the solution and am just unsure why they doubled the coulomb of charge. The way i tried to work it out is find the velocity using de broglie's wavelength formula which gave me 49.7m/s. Then use this to find the kinetic energy which was 8.24x10^-24J. Once i did this i used the formula W=qV and subbed in 8.24x10^-24=(1.6x10^-19)V which gave me the answer of 5.1x10^-5V required.
Title: Re: VCE Physics Question Thread!
Post by: hyunah on August 04, 2014, 08:17:21 pm

what is the importance of frequency in determining the non-ideal behaviour of transformers. are there any other parameters?

thank you
Title: Re: VCE Physics Question Thread!
Post by: Bestie on August 05, 2014, 10:29:50 pm
question: in a DC motor why does the armature continue to spin even when at one point on force (the forces cancel out) are acting on it?
I know it has something to do with momentum but I don't quite get it....
Title: Re: VCE Physics Question Thread!
Post by: knightrider on August 16, 2014, 11:38:14 pm
How would you do this question

A man of mass 70 kg steps forward out of a boat and
onto the nearby river bank with a velocity, when he
leaves the boat, of 2.5 m s−1 relative to the ground.
The boat has a mass of 400 kg and was initially at
rest. With what velocity relative to the ground does
the boat begin to move?
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on August 17, 2014, 10:56:07 am
How would you do this question

A man of mass 70 kg steps forward out of a boat and
onto the nearby river bank with a velocity, when he
leaves the boat, of 2.5 m s−1 relative to the ground.
The boat has a mass of 400 kg and was initially at
rest. With what velocity relative to the ground does
the boat begin to move?

Conservation of momentum. Initially both the man and the boat are at rest. Then, the man pushes on the boat to start moving. Have a think about it.
Title: Re: VCE Physics Question Thread!
Post by: knightrider on August 17, 2014, 12:00:57 pm
Conservation of momentum. Initially both the man and the boat are at rest. Then, the man pushes on the boat to start moving. Have a think about it.

Thanks would this be right :)

using m1u1+m2u2=m1v1+m2v2

(70*0)+(400*0)=(70*2.5)+(400*v2)
0=175+400v2
v2=-0.44metres per second what does the negative mean in the answer

Title: Re: VCE Physics Question Thread!
Post by: Zealous on August 17, 2014, 04:54:10 pm
Thanks would this be right :)

using m1u1+m2u2=m1v1+m2v2

(70*0)+(400*0)=(70*2.5)+(400*v2)
0=175+400v2
v2=-0.44metres per second what does the negative mean in the answer

The negative simply means the boat is moving in the opposite direction to the man. If you define the man's velocity to be positive 2.5m/s, the boat's velocity will be 0.44m/s in the other direction - this is because the man's movement pushes the two apart, they are not in the same direction.
Title: Re: VCE Physics Question Thread!
Post by: Alwin on August 17, 2014, 05:42:21 pm
What accelerating potential difference would be required to give an alpha particle a de Broglie wavelength of 2.0nm?

mass of alpha is 6.67x10^-27kg

I attached the solution and am just unsure why they doubled the coulomb of charge. The way i tried to work it out is find the velocity using de broglie's wavelength formula which gave me 49.7m/s. Then use this to find the kinetic energy which was 8.24x10^-24J. Once i did this i used the formula W=qV and subbed in 8.24x10^-24=(1.6x10^-19)V which gave me the answer of 5.1x10^-5V required.

Think about the charge on an alpha particle ;) It's a bit of a trick question because you have to had remembered what the charge is from year 11, or do it the way you did it with year 12 methods

what is the importance of frequency in determining the non-ideal behaviour of transformers. are there any other parameters?

Hmm interesting question but usually not examinable. A quick run down of things that can make transformers less ideal:
Eddy currents
Non ideal joins (side note is that Wilson Transformers actually makes all their transforms by hand and its an amazing process to get maximum efficiency)
Heat in the coil (usually transformers are oil-convection cooled)
Whether or not the core is laminated (this goes back to the eddy currents issue)
And the last one I can think of is flux leakage, but tbh I don't really get this either

But basically in essence the most important one as you've pointed how is the frequency as this will have an effect on eddy currents (imagine it as a sort of like a residue current in the coil)

question: in a DC motor why does the armature continue to spin even when at one point on force (the forces cancel out) are acting on it?
I know it has something to do with momentum but I don't quite get it....

I assume you mean something like this :)
(http://www.tpub.com/neets/book5/32NE0420.GIF)

I'll start with B first, because it's the most obvious: Force on side is going up and the force on the other is going down, so the coil spins, seems simple enough :)
But then when it's in points A and C, the force on the top is going upwards and the force on the bottom is acting downwards so I get there can be some confusion as how the coil keeps on turning. In fact, it's because of the commutator. At that instant the coil briefly loses connection with the batter (the 'split' part of the split ring commutator) so there is no current flowing and thus no force acting on the coil. Now, the coil was originally spinning (see point B) so it will continue to spin as there is no braking or frictional force acting on the coil. The terminals of the coil then touch the other side of the split ring commutator (reversing the current) and now the forces are reversed as well meaning the coil keeps on spinning.

So in essence, the coil's spinning momentum just before the perpendicular position is what carries to through the vertical position and allows the direction of the current to be changed via the commutator
Title: Re: VCE Physics Question Thread!
Post by: Stick on August 18, 2014, 12:08:26 pm
Lol, I know I'm not doing VCE anymore but I'm doing a VCE equivalent (kinda) Physics subject at university and I'm getting confused on the question attached. My answer for the second situation doesn't make sense so I'm assuming I'm doing it wrong. Thanks for the help. :)
Title: Re: VCE Physics Question Thread!
Post by: Zealous on August 18, 2014, 05:08:56 pm
Lol, I know I'm not doing VCE anymore but I'm doing a VCE equivalent (kinda) Physics subject at university and I'm getting confused on the question attached. My answer for the second situation doesn't make sense so I'm assuming I'm doing it wrong. Thanks for the help. :)

Hey Stick,

From what I've calculated, Rachel would be 6m from the centre pivot point, which is not actually on the see-saw. Yeah - doesn't make sense to me also.
Title: Re: VCE Physics Question Thread!
Post by: Stick on August 18, 2014, 05:11:51 pm
I got the same answer, so it must be a dodgy question lol. Thanks for checking that. :)
Title: Re: VCE Physics Question Thread!
Post by: knightrider on August 18, 2014, 09:05:34 pm
The negative simply means the boat is moving in the opposite direction to the man. If you define the man's velocity to be positive 2.5m/s, the boat's velocity will be 0.44m/s in the other direction - this is because the man's movement pushes the two apart, they are not in the same direction.

Thanks Zealous :)
Can you help with these questions?

A small research rocket of mass 250 kg is launched
vertically as part of a weather study. It sends out
50 kg of burnt fuel and exhaust gases with a velocity
of 180 m s−1 in a 2 s initial acceleration period.

a)What is the velocity of the rocket after this initial
acceleration?
b)What upward force does this apply to the rocket?
c)What is the net upward acceleration acting on the
rocket?
Title: Re: VCE Physics Question Thread!
Post by: Alwin on August 19, 2014, 12:46:36 pm
Thanks Zealous :)
Can you help with these questions?

A small research rocket of mass 250 kg is launched
vertically as part of a weather study. It sends out
50 kg of burnt fuel and exhaust gases with a velocity
of 180 m s−1 in a 2 s initial acceleration period.

a)What is the velocity of the rocket after this initial
acceleration?
b)What upward force does this apply to the rocket?
c)What is the net upward acceleration acting on the
rocket?

Sorry, I'm not Zealous but I hope you'll accept my help too :P

To me, the trick to this question is that the mass of the rocket changes from 250 to 200kg (50 kg of fuel is sent out)
So, I would use the 'average' mass of 225kg in my calculations. (there are more scientific methods to find the average mass than just choosing the middle value, but they're too bothersome and not required for VCE)

Part a
Conservation of momentum tells us that the momentum before the launch is equal to that after the launch. In maths form,
$p_{\text{ rocket}+\text{fuel,initial}}=p_{\text{ rocket,final}} + p_{\text{ fuel,final}}$

Now initially we know that the rocket was stationary and we're given all the information in the question (just don't forget to use the average mass!). Note that I take up as positive
$\qquad \quad 0 = v_{\text{rocket}} \times 225 + (-180) \times 50
\\ \therefore v_{\text{rocket}} = 40\,\text{ms}^{-1}$

Part b
To find the upwards force, we can use the impulse as we know the momentum and we are given the time
$\Delta p_{\text{ rocket}} = I = F\times t
\\ 225 \times 40 = F \times 2
\\ \,\qquad \therefore F = 4500\,\text{N}$

Part c
To find the net acceleration we just use Newton's 2nd law remembering to use the average mass again
$\:\: F_{\text{net}} = m \times a
\\ 4500 = 225 \times a
\\ \therefore a = 20 \,\text{ms}^{-1}$

Title: Re: VCE Physics Question Thread!
Post by: knightrider on August 19, 2014, 06:13:39 pm
Sorry, I'm not Zealous but I hope you'll accept my help too :P

To me, the trick to this question is that the mass of the rocket changes from 250 to 200kg (50 kg of fuel is sent out)
So, I would use the 'average' mass of 225kg in my calculations. (there are more scientific methods to find the average mass than just choosing the middle value, but they're too bothersome and not required for VCE)

Part a
Conservation of momentum tells us that the momentum before the launch is equal to that after the launch. In maths form,
$p_{\text{ rocket}+\text{fuel,initial}}=p_{\text{ rocket,final}} + p_{\text{ fuel,final}}$

Now initially we know that the rocket was stationary and we're given all the information in the question (just don't forget to use the average mass!). Note that I take up as positive
$\qquad \quad 0 = v_{\text{rocket}} \times 225 + (-180) \times 50
\\ \therefore v_{\text{rocket}} = 40\,\text{ms}^{-1}$

Part b
To find the upwards force, we can use the impulse as we know the momentum and we are given the time
$\Delta p_{\text{ rocket}} = I = F\times t
\\ 225 \times 40 = F \times 2
\\ \,\qquad \therefore F = 4500\,\text{N}$

Part c
To find the net acceleration we just use Newton's 2nd law remembering to use the average mass again
$\:\: F_{\text{net}} = m \times a
\\ 4500 = 225 \times a
\\ \therefore a = 20 \,\text{ms}^{-1}$

Thank you so much Alwin i really appreciate your help, you are very nice :)
Title: Re: VCE Physics Question Thread!
Post by: allstar on August 20, 2014, 10:44:45 pm
how would I show acceleration in the opposite direction on a velocity time graph?
thanks
Title: Re: VCE Physics Question Thread!
Post by: Conic on August 21, 2014, 11:38:07 pm
Acceleration is the slope of a velocity time graph, as $a=\frac{\Delta v}{\Delta t}$, so if the velocity is positive, you need a negative gradient for the acceleration to be in the opposite direction of the velocity. If the velocity is negative, then the gradient must be positive.
Title: Re: VCE Physics Question Thread!
Post by: chemdeath on August 22, 2014, 02:28:18 am
Cold-blooded animals (ectotherms) such as reptiles are able to sustain a higher population density than similarly sized warm-blooded animals (endotherms) on the same terrain. Can you
explain this observation in terms of energy and heat flow? (
Title: Re: VCE Physics Question Thread!
Post by: Bestie on August 22, 2014, 12:00:17 pm
mass of 10kg is place on a plane inclined at 45 degrees to the horizontal and allowed to slide down the plane. WHat is the accleration of mass if the frictional force opposing motion is 3.2 N?

Title: Re: VCE Physics Question Thread!
Post by: allstar on August 22, 2014, 12:04:40 pm
thank you conic :)

but what happens if it was initially constant velocity and then accelerate in the opposite direction, is the gradient of acceleration positive or negative in this case?
Title: Re: VCE Physics Question Thread!
Post by: Conic on August 22, 2014, 03:42:19 pm
It's the same situation, but the gradient will start off at 0 then change when the object starts accelerating.

(http://i.imgur.com/SZGEgH8.png)
Title: Re: VCE Physics Question Thread!
Post by: knightrider on August 25, 2014, 08:59:57 pm
how would you do this question?

Each has a mass of 10 kg and a height of 30 cm. The
tray of the truck is 1.5 m above the ground and the
removalist is placing each box on top of the previous
one.

What is the total work done on the boxes in lifting
all the boxes onto the truck as described?
Title: Re: VCE Physics Question Thread!
Post by: JHardwickVCE on August 25, 2014, 09:11:07 pm
Work (W) = Force (F) * Displacement (x)

As we are lifting boxes, the force is made up by the weight force = mg
Hence, each box, when lifted, has a force of (10)(10)=100N pulling it down.

For the first box:
x=1.5m
W=(100)(1.5) = 150N

You will have to lift each box an extra 0.3m

Total work done = (100)(1.5+1.8+2.1+2.4+2.7) = (100)(10.5) = 1050N = 1.1*10^3 N
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on August 25, 2014, 11:20:09 pm
Work (W) = Force (F) * Displacement (x)

As we are lifting boxes, the force is made up by the weight force = mg
Hence, each box, when lifted, has a force of (10)(10)=100N pulling it down.

For the first box:
x=1.5m
W=(100)(1.5) = 150N

You will have to lift each box an extra 0.3m

Total work done = (100)(1.5+1.8+2.1+2.4+2.7) = (100)(10.5) = 1050N = 1.1*10^3 N

For instance, if you were to actually try to exert a force equal to the weight force on the box, the net force on the box would be zero. Problem. In reality, we pull up with a greater force than the weight force to get the box moving, and then we reduce the force when we want to slow it down, so the force isn't constant. W = Fx only works for constant forces.

What you should actually do is note that the kinetic energy at the beginning and the end is zero, so the work done is the change in gravitational potential energy. The working out is the same, the reasoning isn't.
Title: Re: VCE Physics Question Thread!
Post by: knightrider on August 25, 2014, 11:40:45 pm
Work (W) = Force (F) * Displacement (x)

As we are lifting boxes, the force is made up by the weight force = mg
Hence, each box, when lifted, has a force of (10)(10)=100N pulling it down.

For the first box:
x=1.5m
W=(100)(1.5) = 150N

You will have to lift each box an extra 0.3m

Total work done = (100)(1.5+1.8+2.1+2.4+2.7) = (100)(10.5) = 1050N = 1.1*10^3 N

For instance, if you were to actually try to exert a force equal to the weight force on the box, the net force on the box would be zero. Problem. In reality, we pull up with a greater force than the weight force to get the box moving, and then we reduce the force when we want to slow it down, so the force isn't constant. W = Fx only works for constant forces.

What you should actually do is note that the kinetic energy at the beginning and the end is zero, so the work done is the change in gravitational potential energy. The working out is the same, the reasoning isn't.

Thx lzxnl and jhardwickvce :)

But remember jhardwickvce that work is measured in joules or newton metres
in your working out you wrote newtons at the end :)

By the way jhardwickvce and lzxnl how different is physics 1/2 to 3/4 physics and how are you guys finding it/found it
Title: Re: VCE Physics Question Thread!
Post by: Phenomenol on August 26, 2014, 12:13:30 am
Thx lzxnl and jhardwickvce :)

But remember jhardwickvce that work is measured in joules or newton metres
in your working out you wrote newtons at the end :)

By the way jhardwickvce and lzxnl how different is physics 1/2 to 3/4 physics and how are you guys finding it/found it

I know I'm not jhardwickvce nor lzxnl but I hope you don't mind me answering this question.

Motion in 3/4 builds upon what was learnt in 1/2 very well. Some new concepts involve circular motion, more intricate block/ramp systems and newtons's law of gravity.

Electricity involved familiar stuff too. Some additions included diodes, thermistors, photodiodes, light-dependent resistors in circuits. A common question is to figure out where to place a cooling/heating element in a circuit.

Light is a different story entirely (at least from what I learned in 1/2). In 1/2 we did a lot of ray tracing and calculations based on focal lengths of lenses/mirrors, but in 3/4 the content shifted towards wave/particle duality of light and the experiments that proved such light properties (photoelectric effect, interference patterns). I found this topic the driest and most repetitive.

Electromagnetism was not something we were taught in 1/2 so concepts like flux and field took a bit of time to get comfortable with, but otherwise it's fine.
Title: Re: VCE Physics Question Thread!
Post by: knightrider on August 26, 2014, 06:29:57 pm
I know I'm not jhardwickvce nor lzxnl but I hope you don't mind me answering this question.

Motion in 3/4 builds upon what was learnt in 1/2 very well. Some new concepts involve circular motion, more intricate block/ramp systems and newtons's law of gravity.

Electricity involved familiar stuff too. Some additions included diodes, thermistors, photodiodes, light-dependent resistors in circuits. A common question is to figure out where to place a cooling/heating element in a circuit.

Light is a different story entirely (at least from what I learned in 1/2). In 1/2 we did a lot of ray tracing and calculations based on focal lengths of lenses/mirrors, but in 3/4 the content shifted towards wave/particle duality of light and the experiments that proved such light properties (photoelectric effect, interference patterns). I found this topic the driest and most repetitive.

Electromagnetism was not something we were taught in 1/2 so concepts like flux and field took a bit of time to get comfortable with, but otherwise it's fine.

Thankyou Phenomenol :)
Title: Re: VCE Physics Question Thread!
Post by: knightrider on August 28, 2014, 04:06:30 pm
How would you do this question?

A rope that is at 35° to the horizontal is used to pull a
10.0 kg crate across a rough floor. The crate is initially
at rest and is dragged for a distance of 4.00 m. The
tension in the rope is 60.0 N and the frictional force
opposing the motion is 10.0 N.

Determine the energy lost from the system as heat
and sound due to the frictional force.
Title: Re: VCE Physics Question Thread!
Post by: allstar on August 30, 2014, 01:33:30 pm
The maximum rate at which a bus can accelerate or decelerate is 2 m/s2. It has a
maximum speed of 60 km/h. Find the shortest time the bus can take to travel between
two bus stops 1 km apart on a straight stretch of road.

Title: Re: VCE Physics Question Thread!
Post by: Brunette15 on August 30, 2014, 02:29:36 pm

A winch uses a steel cable to lift a large piece of machinery
of the steel cable increases to 1.001 times its original
value. What is the tensile strain on the cable when
percentage 8)
Title: Re: VCE Physics Question Thread!
Post by: Rishi97 on August 30, 2014, 05:14:55 pm

A winch uses a steel cable to lift a large piece of machinery
of the steel cable increases to 1.001 times its original
value. What is the tensile strain on the cable when
percentage 8)

Hey
Ok when I did this question, I just made up random numbers to get an idea of what the question is trying to say.
Lets say that the original length is 5m
When the cable increases to 1.001 times its original length, then 5x1.001 = 5.005
That means that the change in length is 0.005
Sub into the strain=change in length/ original length and you should get an answer of 0.001
Since the answer says to give as a percentage, 0.001 x 100 = 0.1%
That should be right :)
Title: Re: VCE Physics Question Thread!
Post by: qwerty04 on August 30, 2014, 09:48:53 pm
2 quick questions  (i have a very basic understanding of the photoelectric effect so far)

Why is a vacuum needed in a photocell?

Why do photoelectric cells need to be stored in a dark space?

Title: Re: VCE Physics Question Thread!
Post by: Brunette15 on September 02, 2014, 06:33:19 pm
Is the yield strength the same as the elastic limit of a material? If not what is the difference?  8) :-\
Title: Re: VCE Physics Question Thread!
Post by: Phenomenol on September 02, 2014, 09:34:51 pm
Is the yield strength the same as the elastic limit of a material? If not what is the difference?  8) :-\

I believe they mean the same thing.
Title: Re: VCE Physics Question Thread!
Post by: speedy on September 04, 2014, 04:58:59 pm
We don't have to know about incandescent light for L&M, right?
Title: Re: VCE Physics Question Thread!
Post by: Rishi97 on September 08, 2014, 08:53:02 pm
Hey everyone

In the physics exam, can we use formulas that are not in the study design? Because my tutor said we can only use formulas if we know how to derive them :/

Thanks
Title: Re: VCE Physics Question Thread!
Post by: Phenomenol on September 08, 2014, 09:00:43 pm
Hey everyone

In the physics exam, can we use formulas that are not in the study design? Because my tutor said we can only use formulas if we know how to derive them :/

Thanks

Don't use obscure ones - use ones that the examiner can clearly see are rearrangements of a formula on the formula sheet, or when very simple substitutions are made. For example, e = hc/lambda is completely acceptable. (substituting f = c/lambda into the photon energy equation)
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on September 08, 2014, 09:09:28 pm
Hey everyone

In the physics exam, can we use formulas that are not in the study design? Because my tutor said we can only use formulas if we know how to derive them :/

Thanks

Which formulas would you even want to use? I never felt the need to use extra formulas explicitly.
Title: Re: VCE Physics Question Thread!
Post by: Rishi97 on September 08, 2014, 09:32:57 pm
Which formulas would you even want to use? I never felt the need to use extra formulas explicitly.

just ones for projectile motion such as quicker formulas to find range and time
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on September 08, 2014, 10:13:22 pm
You sure those aren't on the formula sheet? Maybe my memory isn't working properly for me here but I swear the range is at least.
In any case, deriving it wouldn't take you more than half a minute.
Title: Re: VCE Physics Question Thread!
Post by: speedy on September 08, 2014, 10:17:33 pm
You sure those aren't on the formula sheet? Maybe my memory isn't working properly for me here but I swear the range is at least.
In any case, deriving it wouldn't take you more than half a minute.

It's not - that's all the motion ones:
Spoiler
(http://i.imgur.com/1yiTWX9.png)

I always thought that if you used a different formula, and got the answer wrong, you wouldn't get any marks for working. May be wrong though, but I would say always derive it first.
Title: Re: VCE Physics Question Thread!
Post by: Rishi97 on September 08, 2014, 10:32:17 pm
You sure those aren't on the formula sheet? Maybe my memory isn't working properly for me here but I swear the range is at least.
In any case, deriving it wouldn't take you more than half a minute.

Nope not on there. Could you by any chance give me tips on how to derive? And from what formula?
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on September 08, 2014, 10:38:10 pm
That is really weird.

Anyway, I'm assuming you mean the formula for no change in height projectile motion.

In the vertical direction, y=0. So ut + 1/2 at^2 = 0
We don't want t=0 = 2u + at = 0
Letting up be positive, a = -g
t = 2u/g. But u = v sin theta => t = 2v sin theta / g

Distance travelled = v cos theta * t = 2v sin theta cos theta v /g = v^2 sin(2 theta)/g by trig identity
Title: Re: VCE Physics Question Thread!
Post by: allstar on September 17, 2014, 10:39:05 pm
any help is greatly appreciated :)

an object, projected vertically upwards with a speed U, returns with a speed V. Assuming constant gravity and air resistance proportional to the square of the speed, find the total time taken.
Title: Re: VCE Physics Question Thread!
Post by: Yacoubb on September 17, 2014, 10:50:52 pm
any help is greatly appreciated :)

an object, projected vertically upwards with a speed U, returns with a speed V. Assuming constant gravity and air resistance proportional to the square of the speed, find the total time taken.

V= u -10t
t = -(v-u)/10
t = u-v/10 *2
t= u-v/5

Not sure if this is your answer but that's my jab! :-)
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on September 17, 2014, 10:59:12 pm
any help is greatly appreciated :)

an object, projected vertically upwards with a speed U, returns with a speed V. Assuming constant gravity and air resistance proportional to the square of the speed, find the total time taken.

What's the proportionality constant?
You'd have to break this question up into two parts as the air resistance changes direction, and even then it's a question involving calculus which wouldn't be asked for a VCE physics student.

V= u -10t
t = -(v-u)/10
t = u-v/10 *2
t= u-v/5

Not sure if this is your answer but that's my jab! :-)

Not constant acceleration; there's air resistance here
Title: Re: VCE Physics Question Thread!
Post by: Yacoubb on September 17, 2014, 11:04:50 pm

Not constant acceleration; there's air resistance here

Would this be right if we were told to ignore air resistance?
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on September 17, 2014, 11:47:11 pm
Your working missed a nuance in the question. The question says 'speed v'. But it's returning downwards, so its final velocity is actually -v, not v.
Title: Re: VCE Physics Question Thread!
Post by: Yacoubb on September 17, 2014, 11:55:58 pm
Your working missed a nuance in the question. The question says 'speed v'. But it's returning downwards, so its final velocity is actually -v, not v.

Ah yes, I see. Thanks lzxnl!
Title: Re: VCE Physics Question Thread!
Post by: maurlock on September 21, 2014, 01:01:16 pm
Why is the cosine angle used in this question instead of sine?
Thanks!
Title: Re: VCE Physics Question Thread!
Post by: Conic on September 21, 2014, 02:01:40 pm
The angle used in the formula is the angle between the force and the lever. In this case, as the angle with the ground is 65, the angle with the lever is 25, so the torque is given by

$\tau = F r \sin(\theta) = (7.5\times10\:\mathrm{N})(3.0\:\mathrm{m})(\sin(25^{\circ}))\approx 95\:\mathrm{Nm},$

which means that E is correct. Because $\sin(90-\theta)=\cos(\theta)$ it appears that the cosine of the angle is used, but you need to make sure you use the right angle. The cosine of the angle with the ground happens to be equal to the sine of the angle between the weight force and the lever, which is what is actually used in the calculation.
Title: Re: VCE Physics Question Thread!
Post by: maurlock on September 22, 2014, 05:50:39 pm
The angle used in the formula is the angle between the force and the lever. In this case, as the angle with the ground is 65, the angle with the lever is 25, so the torque is given by

$\tau = F r \sin(\theta) = (7.5\times10\:\mathrm{N})(3.0\:\mathrm{m})(\sin(25^{\circ}))\approx 95\:\mathrm{Nm},$

which means that E is correct. Because $\sin(90-\theta)=\cos(\theta)$ it appears that the cosine of the angle is used, but you need to make sure you use the right angle. The cosine of the angle with the ground happens to be equal to the sine of the angle between the weight force and the lever, which is what is actually used in the calculation.
Oooh I see, thank you so much!
Title: Re: VCE Physics Question Thread!
Post by: speedy on September 23, 2014, 05:01:06 pm
Why is Fbox on trailer > Ftrailer on box?
Spoiler
(http://i.imgur.com/syISyRb.png)
Title: Re: VCE Physics Question Thread!
Post by: Yacoubb on September 23, 2014, 06:15:52 pm
Why is Fbox on trailer > Ftrailer on box?
Spoiler
(http://i.imgur.com/syISyRb.png)

I could be wrong, but I think you're not looking at the diagram correctly. The arrow from the trailer upwards is actually from the bottom of the box (where the trailer makes contact with the box), and the arrow downwards is from the bottom of the trailer downwards, so in reality the proportion of upwards to downward arrow size is the same. It would have to be according to Newton's 3rd Law of Motion...
Title: Re: VCE Physics Question Thread!
Post by: speedy on September 23, 2014, 06:42:25 pm
I could be wrong, but I think you're not looking at the diagram correctly. The arrow from the trailer upwards is actually from the bottom of the box (where the trailer makes contact with the box), and the arrow downwards is from the bottom of the trailer downwards, so in reality the proportion of upwards to downward arrow size is the same. It would have to be according to Newton's 3rd Law of Motion...

Well I just drew that on so idk. I guess it comes down to interpretation of the image in the question:
Spoiler
(http://i.imgur.com/JcbEd2w.png)

If we say that the box is within the trailer, then I agree with you. If we say that it just sits on top then clearly the arrow downwards is larger. I don't know why the answer doesn't show both the trailer and box - makes the origin very ambiguous.

When I did the question I took it to be sitting on top of the trailer (see diagram), but I upon looking at it again, it does appear to go deeper.
Title: Re: VCE Physics Question Thread!
Post by: allstar on September 23, 2014, 06:55:02 pm
for the gain of an amplifier, (even if it is an inverting amplifier), should it always be a positive value?
Title: Re: VCE Physics Question Thread!
Post by: Zealous on September 23, 2014, 07:02:12 pm
for the gain of an amplifier, (even if it is an inverting amplifier), should it always be a positive value?
I quite sure that as long as you take the gradient of the Vin/Vout graph, whether negative or positive VCAA will accept it (from the assessment reports I've seen). I usually just take the magnitude anyway though - negative gain just sounds awkward.
Title: Re: VCE Physics Question Thread!
Post by: Yacoubb on September 23, 2014, 07:28:24 pm
I quite sure that as long as you take the gradient of the Vin/Vout graph, whether negative or positive VCAA will accept it (from the assessment reports I've seen). I usually just take the magnitude anyway though - negative gain just sounds awkward.

Yeah I always just find the absolute value. Negative gain...it does sound awkward lol.
Title: Re: VCE Physics Question Thread!
Post by: allstar on September 23, 2014, 07:31:00 pm
thank you guys!

just another one:
Can a light dependant resistor be used as a demodulator?
i know LED are used for the modulator
Title: Re: VCE Physics Question Thread!
Post by: Yacoubb on September 23, 2014, 07:31:57 pm
thank you guys!

just another one:
Can a light dependant resistor be used as a demodulator?
i know LED are used for the modulator

Yep! :-)
Title: Re: VCE Physics Question Thread!
Post by: allstar on September 23, 2014, 07:38:55 pm
but the ans says i can't? why?
Title: Re: VCE Physics Question Thread!
Post by: Yacoubb on September 23, 2014, 07:43:03 pm
but the ans says i can't? why?

Does it say photodiode?
Title: Re: VCE Physics Question Thread!
Post by: allstar on September 23, 2014, 07:49:17 pm
it just says no?

i dunno probably just another dodgy question

what is the purpose of slip rings in AC generator?
Title: Re: VCE Physics Question Thread!
Post by: Zealous on September 24, 2014, 05:15:51 pm

what is the purpose of slip rings in AC generator?
Slip Rings maintain contact with the rotating coil in an AC generator as a way to transfer the current out from a coil into a circuit for use.

I've got a quick Electric Power question.

How come the magnetic field lines diverge in this image? Can't they all be parallel? (VCAA 2008)

(http://i1282.photobucket.com/albums/a531/Ovazealous/screenshot305_zps56e7fd60.png)

Current flows in an anti-clockwise direction, by the way.
Title: Re: VCE Physics Question Thread!
Post by: Makutar on September 24, 2014, 07:43:39 pm
The lines diverge because the magnetic field (B) gets smaller the further away from the loop you go.
Title: Re: VCE Physics Question Thread!
Post by: Zealous on September 24, 2014, 08:20:26 pm
The lines diverge because the magnetic field (B) gets smaller the further away from the loop you go.
Thanks for the response.

Just on that, wouldn't the magnetic field strength have no effect on the direction?
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on September 24, 2014, 09:34:30 pm
Just on that, wouldn't the magnetic field strength have no effect on the direction?

field strength is represented by field line density (closer lines equals stronger field) so if they stayed parallel, you'd be saying that the strength of field is the same no matter how far away from the loop you go

this isn't the case, and the only way for the strength to dissipate in a field-line drawing is if the lines spread out. the direction IS consistent with these lines and does spread out

something else that might help you is this; i assume you used a right hand rule to determine that an anticlockwise current (as viewed from the side the arrows point in this case) causes the field to point the way it does. this actually comes from the 'grip rule' for a single current-carrying wire, and happens to also work for a loop.

if you imagine gripping the wire anywhere around the loop, with your right thumb pointing in the direction of the conventional current (anticlockwise) you'll see that inside the loop, the current around the wire points in that direction, (this is the same everywhere in the loop).
the circular field around a wire wrapped in a loop gives the net result of having a field which is stronger inside the loop because all points around the loop contribute to that direction. but their contributions are only parallel in the plane of the loop, and everywhere else (including outside) they are weaker and not parallel.

As per this image, just showing the lines in one plane;
Spoiler
(https://online.science.psu.edu/sites/default/files/phys010/W5electron/500px-VFPt_dipole_magnetic3.svg_.png)

it's not very important to draw the lines that do loop back around and it's pretty hard to draw clearly on an exam, but they are there
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on September 24, 2014, 09:47:36 pm
Can a light dependant resistor be used as a demodulator?
i know LED are used for the modulator

LDR's have too slow of a response rate to be at all effective in this application. you need a component that is fast enough to respond to the rapid changes in signal, but an LDR's response rate is never much faster than milliseconds, and is usually order 0.1s

For this reason, they're more useful as light sensors as in those circuits the response time is acceptable
In actual fact, sometimes they're used in audio compression for softening audio signals (utilising their delayed response)
Title: Re: VCE Physics Question Thread!
Post by: Brunette15 on September 26, 2014, 05:32:45 pm
Can someone please provide a worked solution to this qu from the 2007 exam? The answer is 1.01m...
Title: Re: VCE Physics Question Thread!
Post by: kinslayer on September 26, 2014, 05:38:18 pm
The light begins from rest and then travels under constant acceleration of g m/s^2.

The equation of motion to use here is s = ut + 1/2gt^2 where u = 0 and t = 0.45s.

s = 1/2*(9.8 m/s^2)(0.45 s)^2 = 0.99 m

The tail-light was 0.99 m above the ground before it fell off.

edit: that is with g accurate to two significant figures... if you use g = 10 m/s^2 you get 1.01 m. I never did 3/4 physics, is it common to use g = 10 m/s^2?  ???
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on September 27, 2014, 12:30:44 am
edit: that is with g accurate to two significant figures... if you use g = 10 m/s^2 you get 1.01 m. I never did 3/4 physics, is it common to use g = 10 m/s^2?  ???

VCE Physics exams have g as 10m/s^2 on the formula sheet and in the study design, probably to save on calculations and make it more about the physics. The mechanics section of Specialist Maths uses 9.8m/s^2, and VCE physics questions aren't marked wrong if you use 9.8 instead of 10 (nor in fact any error in sigfigs unless you give like 7 decimal places or something silly) afaik.
Title: Re: VCE Physics Question Thread!
Post by: Brunette15 on September 27, 2014, 01:45:34 pm
Can someone please help me with question 12 for structures...I have looked at the worked solutions on itute and i don't understand how you can neglect the torque of one of the beams when answering this question :/

The ans is B...
Title: Re: VCE Physics Question Thread!
Post by: Zealous on September 27, 2014, 02:37:56 pm
Can someone please help me with question 12 for structures...I have looked at the worked solutions on itute and i don't understand how you can neglect the torque of one of the beams when answering this question :/

The ans is B...
Torque is calculated as $\tau=Fx$ - the force multiplied by the distance from a "pivot point".

When we solve this question, we look at it almost like a seesaw, where Y is the centre and we take torque about the point Y. So the torque created by the beam Y is equal to $\tau=F_{Beam Y}\times (0)=0$. So we can neglect the force of the beam Y as we are calculating torque about this point, so the distance is 0 and the force isn't working to create any torque. Then we can just look at the force from the mass at Z and the force from the mass of the beam to solve Q12.
Title: Re: VCE Physics Question Thread!
Post by: Brunette15 on September 27, 2014, 04:23:08 pm
Torque is calculated as $\tau=Fx$ - the force multiplied by the distance from a "pivot point".

When we solve this question, we look at it almost like a seesaw, where Y is the centre and we take torque about the point Y. So the torque created by the beam Y is equal to $\tau=F_{Beam Y}\times (0)=0$. So we can neglect the force of the beam Y as we are calculating torque about this point, so the distance is 0 and the force isn't working to create any torque. Then we can just look at the force from the mass at Z and the force from the mass of the beam to solve Q12.

Ah yes...thankyou!  :D
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on October 03, 2014, 10:14:34 pm
Torque is calculated as $\tau=Fx$ - the force multiplied by the distance from a "pivot point".

When we solve this question, we look at it almost like a seesaw, where Y is the centre and we take torque about the point Y. So the torque created by the beam Y is equal to $\tau=F_{Beam Y}\times (0)=0$. So we can neglect the force of the beam Y as we are calculating torque about this point, so the distance is 0 and the force isn't working to create any torque. Then we can just look at the force from the mass at Z and the force from the mass of the beam to solve Q12.

Not quite. The torque is a product of a force and a distance, but this distance isn't just the distance from the pivot point to the force. That only holds if the force acts perpendicular to the vector from the pivot point to the point of application of the force.

Think about a see-saw. Logically, it would move the most if you pushed straight down on the seesaw, not if you pushed diagonally down on it. The angle matters too.
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on October 03, 2014, 10:47:47 pm
I always thought of it as the distance between the force and the pivot and the perpendicular component of the force, since there's also a component parallel which causes no torque if your force is at an angle. I know taking the angle into account on the force or the distance is equivalent. Anyway, in this case, if the distance is 0, which it is here, it doesnt matter what the force is or what the angle of the force is, the torque will still be 0 if d is 0 :D
Title: Re: VCE Physics Question Thread!
Post by: yang_dong on October 07, 2014, 11:00:44 am
Can someone please clarify this concept?
The forward driving force is provided by the friction exerted by the ground on the rear wheel. It is the action – reaction pair to the friction force of the wheel pushing backwards on the ground
Retarding forces acting on the bicycle include air resistance (drag) and friction on the wheels (especially when the brakes are applied).
So what does friction do? How is part of the driving force and the retarding force? So if the bicycle is travelling where there is no air resistance, it won’t ever stop? Because the retarding force is trying to stop it, but instead it actually helps it to move, but as it moves the friction on the wheels try to stop it, and then the cycle repeats itself?

Title: Re: VCE Physics Question Thread!
Post by: Bestie on October 07, 2014, 03:23:06 pm
Why is it that no work is done when the component is in a perpendicular direction to the objects movement?

Title: Re: VCE Physics Question Thread!
Post by: Rishi97 on October 07, 2014, 05:28:57 pm
Why is it that no work is done when the component is in a perpendicular direction to the objects movement?

I'm not too sure of the "real" reason but I'm a more mathematical person so I think of it using the formula
W=Fxcos(angle)
If the angle is 90 which as you said is perpendicular, the work will equal 0
Title: Re: VCE Physics Question Thread!
Post by: myanacondadont on October 07, 2014, 06:01:20 pm
Heyyy; Just curious as to where NEAP are in terms of difficulty of papers. We had the 2014 trial exam at school. I got ~85% however it's a bit disheartening after doing VCAA ones and getting high-mid 90's. Should I be worried or are they just tough?
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on October 07, 2014, 06:49:26 pm
Can someone please clarify this concept?
The forward driving force is provided by the friction exerted by the ground on the rear wheel. It is the action – reaction pair to the friction force of the wheel pushing backwards on the ground
Retarding forces acting on the bicycle include air resistance (drag) and friction on the wheels (especially when the brakes are applied).
So what does friction do? How is part of the driving force and the retarding force? So if the bicycle is travelling where there is no air resistance, it won’t ever stop? Because the retarding force is trying to stop it, but instead it actually helps it to move, but as it moves the friction on the wheels try to stop it, and then the cycle repeats itself?
As I understand it depends on what the wheels are trying to do;

If you're pedalling and trying to turn the wheels, consider the contact point of each wheel with the ground. It's trying to push the ground back because that's the way the wheel is turning when you try to pedal, and the reaction force pushes the entire bike/rider system forward. In addition to this there's a little bit of 'rolling resistance' which is a force opposing the propulsion due to the surface but for most surfaces and certainly most VCE physics questions this can be ignored.

If you're braking it's a slightly different story. Brakes apply a torque to the wheels to stop them from rotating rather than a torque to make them rotate which is the case above. As the wheels are slowing down, the situation is reversed. The wheel pushes against the ground the other way at the point of contact with the surface, and the surface pushes back and you slow down.

At all times, there's going to be air drag which is proportional to your velocity.

Hope that helps!
Title: Re: VCE Physics Question Thread!
Post by: PB on October 07, 2014, 09:26:18 pm
Why is it that no work is done when the component is in a perpendicular direction to the objects movement?

Work is what I like to call - "useful energy". The consumed energy that actually contributed towards an object's propagation in a certain direction.
Lets say that there are a billion forces acting on an object which all add up to a net force accelerating in a northerly direction. Now lets pick one of these forces which happens to be acting in a NW direction. We can say that this force has done some work because it has a north bound force component which has contributed some energy to the object's propagation in the northerly direction.

However, I would call a westerly-bound force a lazy, if not counter productive, force. All it is doing is expending all its energy trying to drag the object to the west instead of to the desired north. Simply because it doesn't have a north bound force component to contribute some work in the northerly direction.

Hence, no work is done by a force acting in a direction perpendicular to the object's movement.
Title: Re: VCE Physics Question Thread!
Post by: qwerty04 on October 07, 2014, 11:07:21 pm
Hey everyone
I have a some questions about a dodgy momentum prac report.

The aim of this experiment was to investigate whether momentum is conserved in elastic and/or inelastic two dimensional collisions.
An air table provides a surface with minimal friction for the pucks to move across.
Multi-image photography of the motion of the pucks on the air table can be obtained by illuminating the air table with a stroboscopic lamp.

(For the elastic collision)

Puck A was launched at Puck B (initially stationary)
A program was used to obtain the motion of the pucks at different time intervals in coordinates scaled to metres.
Strobe was set a 6 flashes per second, so the time interval between each image (coordinate points) was 1/6 of a second.

Now the annoying part......
the analysis wants a graph of momentum against time of each puck on the same axes
Puck A has a nice slope, momentum is increasing and then hits Puck B and starts decreasing cause momentum must be conserved.
it looks weird cause the puck B is initially stationary, so for the first 3 time intervals it is 0 and then goes up.
So now we state a relationship between momenta before and momenta after judging by the graph shapes.

Anyways, the next step is to adjust quantities of the graph to somehow show that momentum is conserved (of course some error will be present) so somehow get a relationship or two things for before and after, which when compared will be very close and hence momentum conserved.

Oh and so far i have tried a bunch of things but this one seemed the least wrong m*delta(s) vs time

Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on October 08, 2014, 12:17:55 am
i think you should be graphing total momentum vs time and showing that it's constant.

total momentum will be the sum of the momenta of the two pucks, i.e. m1 v1 + m2 v2 which should remain the same across the collision

sounds like a pretty epic prac though, stroboscopes!
Title: Re: VCE Physics Question Thread!
Post by: Rod on October 08, 2014, 07:49:24 am
Hi everyone

Just did 2006 VCAA, and I found motion section so hard, much harder than any neap, insight etc commercial exam motion section. How is everyone else feeling about motion 06? Will the motion section be of that difficulty in 14? Does the difficulty stay the same from 06-13?

thanks
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on October 08, 2014, 09:23:35 am
Hi everyone

Just did 2006 VCAA, and I found motion section so hard, much harder than any neap, insight etc commercial exam motion section. How is everyone else feeling about motion 06? Will the motion section be of that difficulty in 14? Does the difficulty stay the same from 06-13?

thanks

based on grade distributions that paper was the hardest one evar, A+ cutoff was 79% for unit 3 2006 and 0 people got full marks

all unit 3 and 4 exams except that from 2004-2012 have A+ cutoffs between 85% and 96% with the years towards 85% being considered the harder papers (they try to go for ~90%) (with the exception of unit 3 2004 which was 81%) (I havent added 2013 because this is my stats from last year)

should be fine!
Title: Re: VCE Physics Question Thread!
Post by: Rod on October 08, 2014, 04:09:19 pm
based on grade distributions that paper was the hardest one evar, A+ cutoff was 79% for unit 3 2006 and 0 people got full marks

all unit 3 and 4 exams except that from 2004-2012 have A+ cutoffs between 85% and 96% with the years towards 85% being considered the harder papers (they try to go for ~90%) (with the exception of unit 3 2004 which was 81%) (I havent added 2013 because this is my stats from last year)

should be fine!
Thank god.

I did so many commercial exams, and thinking I would now be able to ace VCAA, I did 2006 and got so freaking depressed haha. Would you reccomend me going over all those questions? Can they come up?

Thanks
Title: Re: VCE Physics Question Thread!
Post by: knightrider on October 08, 2014, 09:09:34 pm
based on grade distributions that paper was the hardest one evar, A+ cutoff was 79% for unit 3 2006 and 0 people got full marks

all unit 3 and 4 exams except that from 2004-2012 have A+ cutoffs between 85% and 96% with the years towards 85% being considered the harder papers (they try to go for ~90%) (with the exception of unit 3 2004 which was 81%) (I havent added 2013 because this is my stats from last year)

should be fine!

How can you tell how many people got full marks for particular exams
Title: Re: VCE Physics Question Thread!
Post by: Rod on October 08, 2014, 09:28:45 pm
How can you tell how many people got full marks for particular exams
Usually it says in the assessors report. For the 2013 one it didnt, but for years like 2009 where 100s and 100s people got 100% they did
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on October 08, 2014, 10:57:51 pm
How can you tell how many people got full marks for particular exams

Yes some older physics/specialist reports seemed to have this information, but I'm going based off the spreadsheet that my physics teacher gave me towards the end of last year which was his info on grade distributions and everything so idk he may have gotten the information from some other source related to being a VCE physics teacher. he had all sorts of info like that.
Title: Re: VCE Physics Question Thread!
Post by: allstar on October 08, 2014, 10:59:54 pm

I guess im stuck between those three big words: weight = mg
weightlessness? - whats that?
aparent weightlessness - when normal = 0?
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on October 08, 2014, 11:43:29 pm

I guess im stuck between those three big words: weight = mg correct, force due to gravity
weightlessness? - whats that? actually having weight = 0, i.e. m=0 or g=0, doesnt truly happen at any times here
aparent weightlessness - when normal = 0 yep

so 1.8g refers to the gravitational force the plane 'feels' even though the actual force on it is always based on g. it's to do with apparent weight. near the start and end of the maneuver the plane is in the bottom half of a vertical circular path and the force on the wings towards the centre of this circle is kinda the normal force from the air, providing this greater force.
same story with zero g, apparent weightlessness because there is no normal force as the plane is at the top of another vertical circle, its weight force is all that is needed to cause circular motion.
at all times the weight of the plane is unchanged and it is not truly weightless because weight is not zero.

dont think you'd need to talk about circles because that's not what it asked for, i just mentioned it to help explain!
Title: Re: VCE Physics Question Thread!
Post by: knightrider on October 08, 2014, 11:46:58 pm
Where would you find the A+ cutoffs and things like that on the vcaa website for physics
Title: Re: VCE Physics Question Thread!
Post by: Rod on October 08, 2014, 11:50:50 pm
Yes some older physics/specialist reports seemed to have this information, but I'm going based off the spreadsheet that my physics teacher gave me towards the end of last year which was his info on grade distributions and everything so idk he may have gotten the information from some other source related to being a VCE physics teacher. he had all sorts of info like that.
Hey silver sorry for the questions!

Would you reccomend me going through the qs for motion for 06? Or leave it?
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on October 09, 2014, 08:30:40 am
Where would you find the A+ cutoffs and things like that on the vcaa website for physics

go to VCAA website and search "grade distributions 2013" or whatever year you want, you should find a page with links to the distributions for each subject that year

Hey silver sorry for the questions!

Would you reccomend me going through the qs for motion for 06? Or leave it?

all the questions except 9* are still relevant, but it's an overall long section with some difficult ones. i think going through it with that mindset (that it's a source of challenging questions) could be beneficial! only if practice at harder questions is what you're after though.
it's all valid and in the course, it's just that motion exams usually don't contain so many difficult questions, there are normally only a few spread between a lot of more basic ones.

*question 9 is about reference frames, interesting physics but no longer in the course
Title: Re: VCE Physics Question Thread!
Post by: allstar on October 09, 2014, 11:18:41 am
thank you! silverpixeli

but at the top of the flight, wouldn't it still experience centripetal acceleration/therefore centripetal force downwards in that situation? in the same direction as its weight force? but its weight force is zero? unknow? im confused....
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on October 09, 2014, 11:43:24 am
thank you! silverpixeli

but at the top of the flight, wouldn't it still experience centripetal acceleration/therefore centripetal force downwards in that situation? in the same direction as its weight force? but its weight force is zero? unknow? im confused....

its weight force is always mg and so never zero (like unless you have no gravity or no mass, so realistically a plane would always have weight)

at the top of the flight, it does experience centripetal motion and therefore there must be a centripetal force, which is the weight force in this situation (a centripetal force has to be a real force, it's not so much a type of force like friction or gravitational attraction as a label for a real force that causes circular motion. an actual force (gravity, friction, tension) is called centripetal if it causes circular motion)

because at the top of the curve the weight force is all that is needed to cause circular motion (weight force is acting as centripetal) there is no need for a normal reaction force on the wings, meaning Normal force = 0 (this is called apparent weightlessness)
Title: Re: VCE Physics Question Thread!
Post by: allstar on October 09, 2014, 10:01:50 pm
so for that question I would ans it like this?

the plane has a weight force because it has mass and experiences 9.8N (g) on the Earth.

weightlessness can never happen, cause for this to happen it requires g=0, which is like never? so it does not apply to any situation?

apparent weightlessness occurs bcause there is no normal reaction force acting on the plane because only the weight force is needed? but how do we know this?
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on October 09, 2014, 11:10:23 pm
so for that question I would ans it like this?

the plane has a weight force because it has mass and experiences 9.8N (g) on the Earth.

weightlessness can never happen, cause for this to happen it requires g=0, which is like never? so it does not apply to any situation?

apparent weightlessness occurs bcause there is no normal reaction force acting on the plane because only the weight force is needed? but how do we know this?

yeah that's pretty much right, i just looked back at the question and noticed it wants answers directly about the passengers not the aircraft itself but it's all the same

• the astronauts have a weight at all times, because of the gravitational field of the earth
• weightlessness doesn't apply, the astronauts have weight, though they do 'feel' weightless at the top of the arc
• this is apparent weightlessness, as there is no normal reaction force acting at this point (the only force acting is weight, which accounts fully for the plane's circular motion)

as for how we know only the weight force is needed, the diagram says 'zero-g' for the top of the arc. when we talk about something's 'g-force' we mean how much apparent weight they're feeling compared to actual weight. one-g is normal conditions, 1.8-g is feeling heavier than normal, and zero-g is feeling weightless which means Normal = 0
Title: Re: VCE Physics Question Thread!
Post by: myanacondadont on October 11, 2014, 09:53:36 am
Hi all, I was doing the 2008 VCAA exam 1. In the electronics and photonics section it shows a npn transistor and stuff like that. That's no longer in our course right? Just want to confirm. I don't even know how to go about the question.
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on October 11, 2014, 10:15:03 am
Hi all, I was doing the 2008 VCAA exam 1. In the electronics and photonics section it shows a npn transistor and stuff like that. That's no longer in our course right? Just want to confirm. I don't even know how to go about the question.

yeah transistors were taken out of the study design and circuits involving them are no longer examinable
Title: Re: VCE Physics Question Thread!
Post by: Brunette15 on October 11, 2014, 03:32:11 pm
I am unsure how they found the change in time. I presume they converted the 4m/s using a distance but I can't interpret what the distance would be. :-[
Title: Re: VCE Physics Question Thread!
Post by: Rishi97 on October 11, 2014, 03:54:41 pm
I am unsure how they found the change in time. I presume they converted the 4m/s using a distance but I can't interpret what the distance would be. :-[

Hey Brunette
Speed=Distance/time
the speed they used is obviously 4cm/s but the distance is 2cm. They used the length of the loop.
so 2/4 = 0.5s

let me know if this makes sense or not
Title: Re: VCE Physics Question Thread!
Post by: qwerty04 on October 13, 2014, 06:09:37 pm
hey guys, another quick momentum question

when the slope of a momentum vs time graph is equal to 0, this shows that momentum in conserved right.

So for explaining this:

Do i say, because the slope of p vs t graph is force, if it is equal to 0, there is no net external force acting on the system and hence it is isolated and hence momentum is conserved??

Btw what i have is a graph of p vs t, with two points before collision and two after, the teacher said to put in two lines of best fit separately for before and after collsion and then compare them....
The slope values of both lines of best fit are very close to 0 but i am just having trouble explaining/comparing them to come to a conclusion that momentum is conserved.

Title: Re: VCE Physics Question Thread!
Post by: myanacondadont on October 13, 2014, 07:21:08 pm
^ Sorry not sure how to help. Did like 1 prac all year with our teacher and even though he said we had to like 4 we never did. Odd huh?

Anyway; does anyone have any 2013/2014 practice exams they could spare? My teacher gave us a total of 6 commercial exams, and I've gone and completed all the VCAA ones (some of which I'm now onto doing twice) soooooo yeah. If anyone has any I'd greatly appreciate it
Title: Re: VCE Physics Question Thread!
Post by: maurlock on October 20, 2014, 08:10:20 pm
Hey guys, quick question
How do you know whether to use the formula net force=Fn+Fg, or net force=Fn-Fg?
Thanks!
Title: Re: VCE Physics Question Thread!
Post by: myanacondadont on October 20, 2014, 08:33:58 pm
Have a look at my attachment Maurlock. I hope it is clear, if not I can help explain further.

edit: Should just add that we know it is Fn-Fg since net centripetal force is always directed to the center.
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on October 20, 2014, 10:10:42 pm
Hey guys, quick question
How do you know whether to use the formula net force=Fn+Fg, or net force=Fn-Fg?
Thanks!

Don't remember formulas. Think about the direction of the net force and where the normal, weight forces are pointed.
For a car on top of the hill, the net force must be directed down towards the centre of the hill. It therefore makes sense to define down as positive (you can choose whatever direction you want to be positive). Therefore, the normal force, pushing up on the car, is going to be negative. You'll have net force = weight force - normal force as the normal force is opposing the net force
Title: Re: VCE Physics Question Thread!
Post by: maurlock on October 20, 2014, 10:31:43 pm
Thanks so much both of you, this makes perfect sense to me now!
Title: Re: VCE Physics Question Thread!
Post by: Brunette15 on October 22, 2014, 07:20:48 pm
I understand the method to work it out but i always get confused when the resistance illustrated is the total resistance or you have to add it? How do i know what the total resistance is for these types of questions?
Title: Re: VCE Physics Question Thread!
Post by: Zealous on October 22, 2014, 08:30:42 pm
I understand the method to work it out but i always get confused when the resistance illustrated is the total resistance or you have to add it? How do i know what the total resistance is for these types of questions?

Maybe it will help to imagine the resistance of the wires as constant resistors in a circuit. In this case, we can imagine the resistance of the cables to be 2x2 ohm resistors for a total of 4 ohms resistance in the circuit. As V=IR, the voltage drop across the resistance of the wire will be V=0.5x4=2. The remaining 10V (12-2) will be across the voltmeter V1.
Title: Re: VCE Physics Question Thread!
Post by: knightrider on October 22, 2014, 09:15:39 pm
Title: Re: VCE Physics Question Thread!
Post by: myanacondadont on October 23, 2014, 09:22:08 am
hey I was wondering whether pn junctions and those type of things are in our study design. I remember my teaching doing something brief on them mid-year but every practice exam I've done I've not witnessed anything to do with them. Theyre also in my textbook - n and p type semiconductors and all that stuff.
Title: Re: VCE Physics Question Thread!
Post by: faredcarsking123 on October 23, 2014, 02:46:11 pm
hey I was wondering whether pn junctions and those type of things are in our study design. I remember my teaching doing something brief on them mid-year but every practice exam I've done I've not witnessed anything to do with them. Theyre also in my textbook - n and p type semiconductors and all that stuff.

No they aren't.
Title: Re: VCE Physics Question Thread!
Post by: bts on October 24, 2014, 08:49:02 am
In the textbook under the section about banked curves what does it mean when they say: at speed below the design car speed (speed that allows the car to travel without friction? Huh?) the car will experience a force point down the bank and at speeds above the design speed the car will experience a force pointing up the bank?
Title: Re: VCE Physics Question Thread!
Post by: allstar on October 24, 2014, 11:35:50 am
Whats the difference between elastic and isolated systems?
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on October 24, 2014, 08:19:09 pm
Whats the difference between elastic and isolated systems?

Isolated = no external forces
Elastic systems? I haven't heard that term used. If you mean elastic as in collisions, you can have an isolated system in which collisions are not elastic, namely that of light and matter, or a collision that involves heat transfer. Friction can still act in an isolated system.
Title: Re: VCE Physics Question Thread!
Post by: myanacondadont on October 24, 2014, 08:23:18 pm
Isolated = no external forces
Elastic systems? I haven't heard that term used. If you mean elastic as in collisions, you can have an isolated system in which collisions are not elastic, namely that of light and matter, or a collision that involves heat transfer. Friction can still act in an isolated system.
Yeah I think he means elastic collisions. Elastic collisions (in terms of physics 3&4) are collisions in which no kinetic energy is lost. (Ek(initial) = Ek(final)). And then inelastic is when kinetic energy is lost.
Title: Re: VCE Physics Question Thread!
Post by: speedy on October 29, 2014, 07:49:13 pm
This is a stupidly simple question, but I've got myself confused, when you find the 'direction' of a projectile, which angle to you take?
Title: Re: VCE Physics Question Thread!
Post by: Thorium on October 29, 2014, 07:56:39 pm
This is a stupidly simple question, but I've got myself confused, when you find the 'direction' of a projectile, which angle to you take?

Usually when the object is moving up or down, we take the acute angle that the direction has with the horizontal.

Hope that makes sense, and best of luck with the exams.
Title: Re: VCE Physics Question Thread!
Post by: speedy on October 29, 2014, 11:10:11 pm
Usually when the object is moving up or down, we take the acute angle that the direction has with the horizontal.

Hope that makes sense, and best of luck with the exams.

Alright thanks :)

Resistance here is ~19kOhm right?
Spoiler
(http://i.imgur.com/K6OR6yw.png)

Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on October 30, 2014, 01:04:30 pm
Resistance here is ~19kOhm right?

careful of the scale of the illumination axis, the question gives you $100Wm^{-2}$ but the graph is in $mWm^{-2}$

but yeah if it was the point you thought, you are reading the graph correctly and ~19 would have been correct
Title: Re: VCE Physics Question Thread!
Post by: speedy on October 30, 2014, 01:59:33 pm
careful of the scale of the illumination axis, the question gives you $100Wm^{-2}$ but the graph is in $mWm^{-2}$

but yeah if it was the point you thought, you are reading the graph correctly and ~19 would have been correct

Oh ffs lol, thanks a lot :)
Title: Re: VCE Physics Question Thread!
Post by: Bestie on October 31, 2014, 11:16:00 pm
For the vcaa 2013 exam question 1a) the assessors report claims I could use x=ut+1/2at^2 and they got 1.74? I got 1.75 using that method and the report says 1.75 is the wrong ans?

And also...
Q6c) shouldn't there be kinetic energy at points q and p? If not why?

Q15b)was I supposed to give it to two sig fig cause I used 18v which is two sig fig? The assessors gave it to 3 sig fig?

Q15c) how do I know whether to use the RMS voltage or the voltage peak?
Title: Re: VCE Physics Question Thread!
Post by: speedy on October 31, 2014, 11:41:12 pm
Is this seriously an answer that would give full marks?

Spoiler
(http://i.imgur.com/PtIPW2h.png)
(http://i.imgur.com/rNvoTSd.png)
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on October 31, 2014, 11:44:11 pm
Is this seriously an answer that would give full marks?

Spoiler
(http://i.imgur.com/PtIPW2h.png)
(http://i.imgur.com/rNvoTSd.png)

It seems REALLY brief, but it actually does summarise essentially why Young's model supported the wave theory of light. You should, however, explain why it suggests interference.
Title: Re: VCE Physics Question Thread!
Post by: Zealous on October 31, 2014, 11:48:57 pm
Is this seriously an answer that would give full marks?

Spoiler
(http://i.imgur.com/PtIPW2h.png)
(http://i.imgur.com/rNvoTSd.png)

I guess you could also talk about diffraction as a wave property?

Quick question: electrons can only make transitions upward (higher energy levels) from ground state right? So if there's n=3 at 4eV and n=4 at 7eV, an electron won't be able to absorb 3eV and take that transition in energy (or it can't go from n=2 straight to ionisation)? Is this because we expect the electron to stabilise by dropping back to ground state first before it can accept any more energy?

Just need some clarification thanks. :D
Title: Re: VCE Physics Question Thread!
Post by: speedy on October 31, 2014, 11:54:52 pm
I guess you could also talk about diffraction as a wave property?

Quick question: electrons can only make transitions upward (higher energy levels) from ground state right? So if there's n=3 at 4eV and n=4 at 7eV, an electron won't be able to absorb 3eV and take that transition in energy?

Eh, I wouldn't, as VCAA count Young's slit experiment as producing an 'ideal' interference pattern - not a true one. Ie. it doesn't take into account diffraction. (This is from my teacher, who is an assesor). You wouldn't lose marks for it though because it is 'good physics'.

It just seemed crazily simple for 3 marks. I also said what particle model would predict and that bright/dark bands were constructive/destructive interference.

Yeah funny you say that, because the exam I just did (2009 Exam 2 - Q11) assumes an electron is in the 1st excited state for the answer, which then is promoted higher.

Truthfully, it can happen, but it is very unlikely, at this level I have been told to assume it is impossible. But as I just mentioned, this question kinda throws it out.
Title: Re: VCE Physics Question Thread!
Post by: speedy on October 31, 2014, 11:57:03 pm
It seems REALLY brief, but it actually does summarise essentially why Young's model supported the wave theory of light. You should, however, explain why it suggests interference.

Yeah that's what I thought, I definitely wouldn't be this blunt on an exam lol.
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on November 01, 2014, 01:41:13 am
I guess you could also talk about diffraction as a wave property?

Quick question: electrons can only make transitions upward (higher energy levels) from ground state right? So if there's n=3 at 4eV and n=4 at 7eV, an electron won't be able to absorb 3eV and take that transition in energy (or it can't go from n=2 straight to ionisation)? Is this because we expect the electron to stabilise by dropping back to ground state first before it can accept any more energy?

Just need some clarification thanks. :D

Sigh. This gets complicated :P you have all these issues about which energy is it more favourable for it to relax to, or will it preferentially absorb energy...don't worry about that xP

Fluorescence is essentially when excited electrons go via intermediate energy levels back down to the ground state. So to be honest, I don't exactly know how particles determine which energy level to go to. Need to ask a quantum physicist there :P
Title: Re: VCE Physics Question Thread!
Post by: Yacoubb on November 01, 2014, 08:04:16 am
Is this seriously an answer that would give full marks?

Spoiler
(http://i.imgur.com/PtIPW2h.png)
(http://i.imgur.com/rNvoTSd.png)

For these questions, I mentioned that interference is a typical property of waves. Young's double-slit experiment demonstrated that light interferes constructively and destructively to produce light and dark bands respectively. Thus, Young's experiment supports the wave-model.
Title: Re: VCE Physics Question Thread!
Post by: speedy on November 01, 2014, 08:59:12 am

For these questions, I mentioned that interference is a typical property of waves. Young's double-slit experiment demonstrated that light interferes constructively and destructively to produce light and dark bands respectively. Thus, Young's experiment supports the wave-model.

You should also explain why it doesn't support the particle model - for any of these questions, you have to explain why it 'disproves' the other too. (You might have, but just going off what you wrote ^^ :) )
Title: Re: VCE Physics Question Thread!
Post by: Yacoubb on November 01, 2014, 09:01:25 am
You should also explain why it doesn't support the particle model - for any of these questions, you have to explain why it 'disproves' the other too. (You might have, but just going off what you wrote ^^ :) )

Yeah no you're right. On my cheat sheet I've got how X supports the wave-model but not particle, and vice versa. :) Definitely important in fully answering the question!
Title: Re: VCE Physics Question Thread!
Post by: Yacoubb on November 01, 2014, 10:48:17 am
Not sure if this has been mentioned yet, but VCAA don't provide super detailed solutions (enough to get the full 3 marks) to worded questions because everyone would just copy it down onto their cheat sheet.

So for a 3 mark question, would we be expected to say for instance:
1 mark - whether it supports the wave or particle model
1 mark - why it supports the particle model (or vice versa)
1 mark - why it doesn't support the wave model (or vice versa)

??
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on November 01, 2014, 10:55:20 am
Not sure if this has been mentioned yet, but VCAA don't provide super detailed solutions (enough to get the full 3 marks) to worded questions because everyone would just copy it down onto their cheat sheet.

Easiest fix ever: remove the bloody cheat sheet and make people actually remember stuff. Like, what, 99% of the other subjects?
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on November 01, 2014, 10:59:52 am
Quick question: electrons can only make transitions upward (higher energy levels) from ground state right? So if there's n=3 at 4eV and n=4 at 7eV, an electron won't be able to absorb 3eV and take that transition in energy (or it can't go from n=2 straight to ionisation)? Is this because we expect the electron to stabilise by dropping back to ground state first before it can accept any more energy?

Just need some clarification thanks. :D

Sigh. This gets complicated :P you have all these issues about which energy is it more favourable for it to relax to, or will it preferentially absorb energy...don't worry about that xP

Fluorescence is essentially when excited electrons go via intermediate energy levels back down to the ground state. So to be honest, I don't exactly know how particles determine which energy level to go to. Need to ask a quantum physicist there :P

This is a question I had all of last year! My physics teacher explained it by saying that the electrons are not excited for long enough for there to be any significant chance for a collision with another photon (unless you have a loooot more photons coming in) and then my first year lecturer mentioned that they're only out of ground state for a few nanoseconds.

So my understanding is that an already-excited electron COULD accept another photon with the right energy to promote it higher (or ionise) but the chance of that collision occurring is just really really slim.

As for which levels they fall back down to, as lzxnl said, that's beyond the scope of the course and you just need to know that it can fall down to anything below it on its way back to ground state!

EDIT: I actually JUST learned how to determine whether a particular path downwards towards n=1 ground state is allowed! Including a simple-english explanation for anyone interested.
Spoiler
It's to do with electron orbitals, which, at a particular energy level, basically 'point' different ways, and you can only move to one which points in a similar way to you. At ground state, there's only a single orbital and it points in direction '0' so if you're at (n=6) and pointing in the maximum direction, '5', you can't reach ground straight away you'd need to go to (n=5) where there's a '4' direction, etc. But if you're in (n=6) and you're only pointing in direction '1', you can jump straight down into (n=1)'s direction '0'.

If you want to learn more, look up orbitals after exams are done and once you understand those, look up the selection rule for emission and absorption!
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on November 01, 2014, 11:10:21 am
Easiest fix ever: remove the bloody cheat sheet and make people actually remember stuff. Like, what, 99% of the other subjects?

While probably more fair, I think this would have a negative effect (certainly in the year the change occurred) because people would spend time memorising formulas and answers by rote instead and that would take away from solving physics problems. I do agree that the provided formula sheet should be all you need in a physics exam, and if the change was made then they could improve the current one. But it's not really an issue, I mean you're going to do better if you have some intuition behind a concept than if you 'just have it on your cheat sheet' for obvious reasons.

So for a 3 mark question, would we be expected to say for instance:
1 mark - whether it supports the wave or particle model
1 mark - why it supports the particle model (or vice versa)
1 mark - why it doesn't support the wave model (or vice versa)

??

Looks good to me, I'd say;

The experiment supports the wave model for light, as the pattern that is observed on the screen can be explained by diffraction and interference, both wave phenomena. It contradicts the particle model, which does not account for diffraction or interference, instead predicting two bright bands of light directly behind the slits.

Since the question seems to be geared towards a comparison, I'd chuck in a quick diagram of the two expectations if I had time! That's probably more than needed for 3/3, but it can't hurt to give more than required (as long as it's all correct)
Title: Re: VCE Physics Question Thread!
Post by: Yacoubb on November 01, 2014, 12:18:13 pm
Perfect. :)

Thank you :)

Quote
Looks good to me, I'd say;

The experiment supports the wave model for light, as the pattern that is observed on the screen can be explained by diffraction and interference, both wave phenomena. It contradicts the particle model, which does not account for diffraction or interference, instead predicting two bright bands of light directly behind the slits.

Since the question seems to be geared towards a comparison, I'd chuck in a quick diagram of the two expectations if I had time! That's probably more than needed for 3/3, but it can't hurt to give more than required (as long as it's all correct)

Yeah so would that be sufficient to mention how the wave-model is supported? Mentioning that observations of diffraction/interference are consistent with waves, supporting the wave-model, and then describing how the particle-model would predict only two bright bands behind the slits?

I just want to be sure how much to put in, not omitting anything important :)

Also, are incandescent globes on the course? I've been doing pre-2006 exams and in the sections for Light and Matter, it's got questions on incandescent lights and I'm not sure whether or not they're still on the course. :/ thanks!
Title: Re: VCE Physics Question Thread!
Post by: speedy on November 01, 2014, 04:40:43 pm
Not sure if this has been mentioned yet, but VCAA don't provide super detailed solutions (enough to get the full 3 marks) to worded questions because everyone would just copy it down onto their cheat sheet.

Lol I never knew that aha... pretty funny...

Easiest fix ever: remove the bloody cheat sheet and make people actually remember stuff. Like, what, 99% of the other subjects?

I agree with you, but from what I've seen (in my class), if people don't actually understand the content, they do badly, regardless of the fact that they have a full cheat sheet with size 6 font lol.

Also, are incandescent globes on the course? I've been doing pre-2006 exams and in the sections for Light and Matter, it's got questions on incandescent lights and I'm not sure whether or not they're still on the course. :/ thanks!

Production of incoherent light was moved to the photonics detailed study.
Title: Re: VCE Physics Question Thread!
Post by: speedy on November 01, 2014, 11:52:38 pm
When answering Lenz's law questions, do we have to go through ALL of the subsequent effects:

eg. "thus induced EMF will produce a current that flows in a direction which produces a field that opposes this change in flux"

or

"thus a field will be produced that opposes this change in flux"

(This might seem stupid but I usually write the former, I want to know what others write :) )
Title: Re: VCE Physics Question Thread!
Post by: PB on November 02, 2014, 01:15:59 am
I usually write the former too. I think its actually better as it follows a logical progression of explanations that makes more sense to me rather than just "a field will be produced that opposes this change in flux".
Also, the former would probably impress the examiner more too as it shows that you know WHY an opposing field is created, not just that it happens because of lenz's law.
Title: Re: VCE Physics Question Thread!
Post by: yang_dong on November 02, 2014, 05:27:50 pm
what's the difference between dynamic loudspeakers and velocity microphones? Cause velocity microphones aren't on the Jacaranda textbook that my school studies off? :(
Title: Re: VCE Physics Question Thread!
Post by: magneto on November 02, 2014, 05:50:49 pm
what mass should i use 33, 44 or 77??
N = mg - ma?
Title: Re: VCE Physics Question Thread!
Post by: bts on November 02, 2014, 06:06:29 pm
Explain why adding a soft iron core increases the strength of an electromagnet.

Title: Re: VCE Physics Question Thread!
Post by: Zealous on November 02, 2014, 06:34:35 pm
what mass should i use 33, 44 or 77??
N = mg - ma?
Depends on how you want to approach the question.

Simplest way is to use ma=mg-N with the 33kg mass where N is the force exerted by B on A (working against the gravitational force of A).

So $33 \times 0.7=33 \times 10 -N \implies N=306.9$

Title: Re: VCE Physics Question Thread!
Post by: lzxnl on November 02, 2014, 07:19:34 pm
Explain why adding a soft iron core increases the strength of an electromagnet.

To put it simply, the iron itself is magnetic, so you're adding another magnet on top of an existing magnet.
Title: Re: VCE Physics Question Thread!
Post by: Yacoubb on November 02, 2014, 07:31:32 pm
Explain why adding a soft iron core increases the strength of an electromagnet.

I believe that soft iron has high magnetic permeability, and so by adding another magnet, you're increasing the strength of this electromagnetic.
Title: Re: VCE Physics Question Thread!
Post by: Bestie on November 02, 2014, 08:46:38 pm
STAV 2013
can someone please help me with question 3,4 and 14 in the attachment? all multiple choice questions
thank you
Title: Re: VCE Physics Question Thread!
Post by: speedy on November 02, 2014, 09:13:16 pm
STAV 2013
can someone please help me with question 3,4 and 14 in the attachment? all multiple choice questions
thank you

These questions are weird as lol...

3) Don't have the image?

4) Bombardment of electrons is off the study design now... This doesn't seem to be a question aligned with the current course, or maybe my knowledge is lacking lol. My guess is A though.

14) Is it D? Although it wouldn't exactly half the current, it is the best way.
Title: Re: VCE Physics Question Thread!
Post by: Bestie on November 03, 2014, 08:36:12 pm
hey speedy! :)

question 3 is above question 4 in the same attachment.

ans to question 4 is B?

Title: Re: VCE Physics Question Thread!
Post by: speedy on November 03, 2014, 10:05:28 pm
hey speedy! :)

question 3 is above question 4 in the same attachment.

ans to question 4 is B?

As in there is no diagram.

Yeah, maybe someone else could help, but bombardment with electrons has been removed from the study design so it's not something you're meant to know.
Title: Re: VCE Physics Question Thread!
Post by: rui97 on November 05, 2014, 11:05:02 pm
hey guys
for significant diffraction to occur, does lambda/slit width have to be greater than or equal to one or does it need to have the same order of magnitude? I havent been able to get a definite answer, what do you guys think?
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on November 06, 2014, 09:17:59 am
hey guys
for significant diffraction to occur, does lambda/slit width have to be greater than or equal to one or does it need to have the same order of magnitude? I havent been able to get a definite answer, what do you guys think?

For noticeable diffraction, they need to be of the same order of magnitude. For complete diffraction, you need your slit to be smaller than the wavelength.
Title: Re: VCE Physics Question Thread!
Post by: rui97 on November 06, 2014, 08:22:44 pm
For complete diffraction, you need your slit to be smaller than the wavelength.

Is complete diffraction the same as significant diffraction?
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on November 06, 2014, 11:28:44 pm
No. Complete diffraction is when your light wave completely spreads out around the slit with no intensity minimum. Significant diffraction is when the most significant part of the light wave is wide, but it does allow for intensity minima. Look at http://en.wikipedia.org/wiki/Diffraction and the pictures for this.
Title: Re: VCE Physics Question Thread!
Post by: Brunette15 on November 07, 2014, 02:31:15 pm
Are there any main difference's between a split ring commutator and slit ring commutator that we should note? So far on my cheat sheet I have only put that split rings are for DC and produce a DC output, and slip rings are for AC and produce an AC output...  :-\
Title: Re: VCE Physics Question Thread!
Post by: RKTR on November 07, 2014, 02:35:51 pm
Are there any main difference's between a split ring commutator and slit ring commutator that we should note? So far on my cheat sheet I have only put that split rings are for DC and produce a DC output, and slip rings are for AC and produce an AC output...  :-\
Split ring reverse direction of current every half turn but slip ring maintains it.
Can anyone confirm?
Title: Re: VCE Physics Question Thread!
Post by: Thorium on November 07, 2014, 02:46:04 pm
Split ring reverse direction of current every half turn but slip ring maintains it.
Can anyone confirm?

Yep that is right.

In addition, in generators, slip rings are used to produce AC EMF. This type of generator is also called alternator. And split ring is used to reverse current every half cycle, so that the produced EMF remains in the same direction. (So DC)

The Voltage vs time graph of an alternator looks like a sinusoidal graph.
If a split ring is used, the negative values of the sinusoidal graph are flipped, i.e. they become positive.

Hope that helps :)
Title: Re: VCE Physics Question Thread!
Post by: speedy on November 07, 2014, 03:11:32 pm
For the cheat sheet, do we have to stick the two A4 pieces together?
Title: Re: VCE Physics Question Thread!
Post by: myanacondadont on November 07, 2014, 03:14:10 pm
For the cheat sheet, do we have to stick the two A4 pieces together?

Afaik they have to be bound together by tape. Otherwise you can use a single piece of A3 doublesided.
Title: Re: VCE Physics Question Thread!
Post by: myanacondadont on November 07, 2014, 04:27:49 pm
Got a few questions:

What should I chuck on my cheat sheet for modulation? I don't think we touched on it during the year but the exams seem to look to ask questions on it. I was thinking of just putting what an information signal/unmodulated signal/modulated signal/carrier wave looks like.

If anyone is doing structures & materials; What's the difference between strain energy and toughness? Are they both the same thing or is one multiplied by volume? Also, when reading a f-x graph do we calculate the area under the graph (similar to a stress-strain graph) for strain energy?
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on November 07, 2014, 05:59:37 pm
Got a few questions:

What should I chuck on my cheat sheet for modulation? I don't think we touched on it during the year but the exams seem to look to ask questions on it. I was thinking of just putting what an information signal/unmodulated signal/modulated signal/carrier wave looks like.

If anyone is doing structures & materials; What's the difference between strain energy and toughness? Are they both the same thing or is one multiplied by volume? Also, when reading a f-x graph do we calculate the area under the graph (similar to a stress-strain graph) for strain energy?

modulation: yeah just a diagram is fine, make sure you know the few key points they look for in worded questions like
• the reason for modulating signals for transmission
• the process, from modulating to sending to demodulating and what waves are involved (signal wave, carrier wave, modulated carrier wave, etc)

strain energy: the energy of an object under load due to its strain which is given by the area under stress-strain graph, then multiplied by volume. the area in question might not be the whole graph, all the way to failure.

toughness: the maximum strain energy density of a material, a property of the material that describes how much energy a unit of volume or that material can take to failure. because of this, it's always calculated as the full area under the stress strain graph, and because it's not object-specific, it's material specific, you don't multiply by volume.

hope that helps!
Title: Re: VCE Physics Question Thread!
Post by: speedy on November 07, 2014, 06:26:29 pm
Saw this in the 2011 Exam 2 assesors report...
Spoiler
(http://i.imgur.com/3A5PKeJ.png)

I usually just used the first and last points that are given to estimate Planck's constant -> would you lose marks for this (let's say they weren't exactly on the line of best fit)?

Edit: Yup, not marks was confirmed later in the report.
Spoiler
(http://i.imgur.com/kD4UL22.png)

I'm still kinda unsure though, because my points were on the line of best fit, but I'm sure my answer was the "accepted value"
Title: Re: VCE Physics Question Thread!
Post by: jumcakes on November 07, 2014, 11:30:18 pm
does anybody have a suitable definition for single slit interference? I have it such that when light passes through a slit it diffracts and the waves from each side interfere with each other creating bright/dark bands but am still a bit unclear.

furthermore would the purpose of modulation be to transmit information more effectively since higher frequency waves diffract less (hence why we impose the signal wave onto the carrier wave)
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on November 08, 2014, 09:13:47 am
does anybody have a suitable definition for single slit interference? I have it such that when light passes through a slit it diffracts and the waves from each side interfere with each other creating bright/dark bands but am still a bit unclear.

furthermore would the purpose of modulation be to transmit information more effectively since higher frequency waves diffract less (hence why we impose the signal wave onto the carrier wave)

It's not like you really need a definition for this and I haven't tried to define it myself before so I'll see what I come up with here.
Huygen's wave model of light assumes that in a light wave, every point on the edge of the light wave acts as a new site of propagation of the light wave. In other words,you can think of light as originated from every single point on a light wave. Therefore, in single slit diffraction, you can think of light as radiating from every single point in the slit. Considering light being emitted from different parts of the slit and considering their path differences allows for analysis of the single slit diffraction phenomenon.

Yes, modulation transmits information more effectively, but I would have thought simply that it would have been because the shape of the original signal is more readily maintained in a high frequency wave than in a low frequency wave. Just what I always thought last year.
Title: Re: VCE Physics Question Thread!
Post by: speedy on November 08, 2014, 09:35:31 am
It's not like you really need a definition for this and I haven't tried to define it myself before so I'll see what I come up with here.
Huygen's wave model of light assumes that in a light wave, every point on the edge of the light wave acts as a new site of propagation of the light wave. In other words,you can think of light as originated from every single point on a light wave. Therefore, in single slit diffraction, you can think of light as radiating from every single point in the slit. Considering light being emitted from different parts of the slit and considering their path differences allows for analysis of the single slit diffraction phenomenon.

Do we need to know about Huygen's wavelet model? It hasn't been assessed before has it?
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on November 08, 2014, 09:39:37 am
Heck no. Hence why I said you don't need a definition of this.
Title: Re: VCE Physics Question Thread!
Post by: speedy on November 08, 2014, 10:15:43 am
Heck no. Hence why I said you don't need a definition of this.

Ahk, yeah I just recall my teacher mentioning it and wanted to clarify :)
Title: Re: VCE Physics Question Thread!
Post by: Rod on November 08, 2014, 12:51:38 pm
Hey guys, need help with question 22.c

http://www.vcaa.vic.edu.au/Documents/exams/physics/2013/2013physics-cpr-w.pdf

Here is my working out:

Let / = wavelength (because I don't know how to do it on this LOL)

So since it is the second brightest band, path different must be 1/ because central maxima (the first brightest band) is 0/.

Therefore 1/ = PD

1/ = 1.4X10^3X10^-9

/ = 1.4X10^3 nm

Therefore, since we have the wavelength, we can find the path difference of the first dark band

(n-1/2)/ = PD

1-1/2/ = PD

1/2 x 1.4x10^3x10^-9 = PD

PD = 7x10-7 m

-----------------------------------------

Okay so I got 2 out of the possible 3 marks for this. Where have I gone wrong? According to the assesors report my path difference was wrong for the bright band, it should have been 2/. Why 2? THanks
Title: Re: VCE Physics Question Thread!
Post by: Rod on November 08, 2014, 12:58:28 pm
And just another quick q sorry;

23.B

How does momentum influence the wavelength of an electon? ANd is 'fringe spacing' just referring to the diffraction pattern? Thanks
Title: Re: VCE Physics Question Thread!
Post by: Zealous on November 08, 2014, 01:10:21 pm
Hey guys, need help with question 22.c

http://www.vcaa.vic.edu.au/Documents/exams/physics/2013/2013physics-cpr-w.pdf

Here is my working out:
....

-----------------------------------------

Okay so I got 2 out of the possible 3 marks for this. Where have I gone wrong? According to the assesors report my path difference was wrong for the bright band, it should have been 2/. Why 2? THanks

The path difference is 2 whole wavelengths. The path difference to the first bright band is 1 whole wavelength and the path difference to the second bright band is 2 whole wavelengths. Recall the formula $PD=n \lambda$, we are looking at the 2nd bright band so n=2, and we get: $1400 \times 10^{-9}=2 \lambda$.

And just another quick q sorry;
23.B
How does momentum influence the wavelength of an electon? ANd is 'fringe spacing' just referring to the diffraction pattern? Thanks

The formula for de Broglie matter wavelength is $\lambda =\frac{h}{p}$. So the wavelength of an electron (matter) is inversely proportional to the momentum of the electron. Basically, the faster an electron is travelling (greater momentum), the smaller it's de Broglie wavelength and conversely, the slower an electron (and lower momentum), the larger it's de Broglie wavelength.

You're right with "fringe spacing" - the bands in the circular diffraction pattern (or rings) are refered to as fringes, and their spacing depends on the wavelength of electrons/light. It's a bit like taking Young's Double Slit interference pattern and converting it into a circle so you've got bright and dark sections from the inside to outside.
Title: Re: VCE Physics Question Thread!
Post by: Rod on November 08, 2014, 01:58:38 pm
The path difference is 2 whole wavelengths. The path difference to the first bright band is 1 whole wavelength and the path difference to the second bright band is 2 whole wavelengths. Recall the formula $PD=n \lambda$, we are looking at the 2nd bright band so n=2, and we get: $1400 \times 10^{-9}=2 \lambda$.

The formula for de Broglie matter wavelength is $\lambda =\frac{h}{p}$. So the wavelength of an electron (matter) is inversely proportional to the momentum of the electron. Basically, the faster an electron is travelling (greater momentum), the smaller it's de Broglie wavelength and conversely, the slower an electron (and lower momentum), the larger it's de Broglie wavelength.

You're right with "fringe spacing" - the bands in the circular diffraction pattern (or rings) are refered to as fringes, and their spacing depends on the wavelength of electrons/light. It's a bit like taking Young's Double Slit interference pattern and converting it into a circle so you've got bright and dark sections from the inside to outside.
Oh I see! So the central maxima does not count as a bright band? Thanks!

And I get 23.b now. Thanks! Awesome explanations ....
Title: Re: VCE Physics Question Thread!
Post by: Zealous on November 08, 2014, 02:13:54 pm
Oh I see! So the central maxima does not count as a bright band? Thanks!
Well, the central maximum is still a bright band, but you don't use it in the PD formula because there isn't any path difference. The central band is exactly half way between the two slits, so waves from both slits would've travelled the exact same distance, there's no difference in the paths they've taken so they'll interfere constructively.

It becomes a central maximum (brighter than the other bands) because the intensity of light is inversely proportional to the square of the distance ($I \propto \frac{1}{d^2}$), so the brightest band would be in the center (where the waves have travelled the least) and then the pattern will get dimmer as you go outwards (and the light travels further).
Title: Re: VCE Physics Question Thread!
Post by: Rod on November 08, 2014, 02:26:21 pm
Well, the central maximum is still a bright band, but you don't use it in the PD formula because there isn't any path difference. The central band is exactly half way between the two slits, so waves from both slits would've travelled the exact same distance, there's no difference in the paths they've taken so they'll interfere constructively.

It becomes a central maximum (brighter than the other bands) because the intensity of light is inversely proportional to the square of the distance ($I \propto \frac{1}{d^2}$), so the brightest band would be in the center (where the waves have travelled the least) and then the pattern will get dimmer as you go outwards (and the light travels further).
Thanks again

Sorry for bothering but would it be okay if you please explaind two more

2b and 6c

http://www.vcaa.vic.edu.au/Documents/exams/physics/2013/physics_examrep13.pdf

Don't get 6c AT ALL. I get how they have assumed the spring potential to be 0 N as incorrect, but how do I prove that it is not 0?
Title: Re: VCE Physics Question Thread!
Post by: Zealous on November 08, 2014, 02:45:42 pm
2b and 6c

http://www.vcaa.vic.edu.au/Documents/exams/physics/2013/physics_examrep13.pdf

Don't get 6c AT ALL. I get how they have assumed the spring potential to be 0 N as incorrect, but how do I prove that it is not 0?
2b:
Look at the two components of the system separately.

So for m1, we've got the gravitational force working in one direction and tension in the string working in the other:

$m_1 a=m_1 g-T \implies 2a=2g-T$

For m2, the only force acting on m2 is the tension in the string which is pulling it to the left:

$m_2 a=T$

The tension in a string is always the same, so we can sub in m2a=T into the first equation to get:

$2a=2g-6a \implies a=2.5$   Now we can sub a back into the second equation:

$6 \times 2.5=T \implies T=15$

So this is the simultaneous equation method, where you look at the forces acting on each block - alternatively you can look at the whole entire system as the whole, imagine both blocks are attached together and there's only a force of gravity acting on it, find the acceleration then look at individual blocks (the method in the examiner report).

6c:
So they've assumed that when the mass is at position Q, that this will be a point of 0 spring potential which is incorrect. Because the spring has already been stretched by 0.5m, it is incorrect to say there's no energy there. So to prove that it's not 0, do a calculation and show that $E=\frac{1}{2}kx^2 \implies E=\frac{1}{2}10 (0.5)^2=1.25$ - so there is in fact 1.25J stored in the spring, not 0. Furthermore you could show that to properly calculate the spring potential, you'd need to do $E=\frac{1}{2}k x_{final}^2-\frac{1}{2}k x_{initial}^2=\frac{1}{2}(10) (1.5^2)-\frac{1}{2}(10) (0.5^2)=10$.
Title: Re: VCE Physics Question Thread!
Post by: Yacoubb on November 08, 2014, 02:49:17 pm
Thanks again

Sorry for bothering but would it be okay if you please explaind two more

2b and 6c

http://www.vcaa.vic.edu.au/Documents/exams/physics/2013/physics_examrep13.pdf

Don't get 6c AT ALL. I get how they have assumed the spring potential to be 0 N as incorrect, but how do I prove that it is not 0?

2b. To calculate tension, we firstly need to calculate the acceleration of m2. So:

W = mg = 2*10 = 20N.
Fnet = ma.
20 = (6+2)a
a = 2.5 m/s^2

T = ma = 2.5 * 6 = 15N.

______________________________________________________________________________________________

6c. The spring potential is not 0 because the spring is being stretched 1m if you can see in the second diagram (so it has been stretched by 0.5m). Thus the strain potential energy is not equal to 0, and this is the mistake made in the collated data.

Edit: just realised Zealous beat me to the punch :)
Title: Re: VCE Physics Question Thread!
Post by: Rishi97 on November 08, 2014, 03:59:10 pm
What is an oscilloscope?
Thanks  :)
Title: Re: VCE Physics Question Thread!
Post by: myanacondadont on November 08, 2014, 05:48:52 pm
What is an oscilloscope?
Thanks  :)

When used in VCE physics it refers to a voltmeter/ammeter thingo in the sense it reads the voltage/current. But it produces a graph of the voltage/current aswell on a screen. (That's all I know). Someone can probably offer a better explanation.

Is this correct to say about transformers?: The current in the primary coil creates a magnetic field which is strengthened by the iron core, therefore threading the secondary coil and inducing a current proportional to the amount of turns in the coil.
Title: Re: VCE Physics Question Thread!
Post by: Zealous on November 08, 2014, 05:52:55 pm
Is this correct to say about transformers?: The current in the primary coil creates a magnetic field which is strengthened by the iron core, therefore threading the secondary coil and inducing a current proportional to the amount of turns in the coil.
I don't even think you need to mention an iron core unless they've given it in the question. I'd probably say "The AC current through the primary coil creates a rapidly changing flux through the secondary coil. This results in an EMF induced in the secondary coil, as dictated by Faraday's Law. The magnitude of the EMF induced is directly proportional to the amount of turns in the secondary coil." or something like that. But there's nothing wrong with your definition, just saying it in different ways.
Title: Re: VCE Physics Question Thread!
Post by: myanacondadont on November 08, 2014, 05:58:16 pm
I don't even think you need to mention an iron core unless they've given it in the question. I'd probably say "The AC current through the primary coil creates a rapidly changing flux through the secondary coil. This results in an EMF induced in the secondary coil, as dictated by Faraday's Law. The magnitude of the EMF induced is directly proportional to the amount of turns in the secondary coil." or something like that. But there's nothing wrong with your definition, just saying it in different ways.

Thankyou! Another question too :D

When a coil is placed in an external magnetic field and connected with slip rings, does the coil oscillate 90 degrees back and forth? Give or take an extra bit because of the momentum of the coil.

And when it's connected with slip rings to a DC power source, does it just align itself with the external field?
Title: Re: VCE Physics Question Thread!
Post by: Zealous on November 08, 2014, 06:04:19 pm
When a coil is placed in an external magnetic field and connected with slip rings, does the coil oscillate 90 degrees back and forth? Give or take an extra bit because of the momentum of the coil.

I'm not sure if you're talking about generators or motors. If you're talking about AC power generation, the coil will still rotate in the same direction (not going back and forth), but the current induced will create a sinosidual curve (sine curve) because the change in flux is positive and negative at different parts in the rotation. The momentum of the coil in rotation will completely depend on what you've got rotating the coil to generate the power.

And when it's connected with slip rings to a DC power source, does it just align itself with the external field?
Connecting a coil to a DC power source through slip rings is just like connecting it directly to the circuit. Once current goes through it, it will rotate to a position of maximum flux and get stuck at that position.
Title: Re: VCE Physics Question Thread!
Post by: myanacondadont on November 09, 2014, 11:53:23 am
I'm not sure if you're talking about generators or motors. If you're talking about AC power generation, the coil will still rotate in the same direction (not going back and forth), but the current induced will create a sinosidual curve (sine curve) because the change in flux is positive and negative at different parts in the rotation. The momentum of the coil in rotation will completely depend on what you've got rotating the coil to generate the power.
Connecting a coil to a DC power source through slip rings is just like connecting it directly to the circuit. Once current goes through it, it will rotate to a position of maximum flux and get stuck at that position.

Thankyou - I think I have some understanding wrong. What's the scenario when a coil will oscillate 90 degrees back and forth?
Title: Re: VCE Physics Question Thread!
Post by: Rod on November 09, 2014, 02:11:57 pm
Thanks Zeal and yacoub :)

http://www.vcaa.vic.edu.au/Documents/exams/physics/2012/physics_assessrep_12.pdf

Having trouble understanding 2d.

Agree:

Old -> 2\
New -> 1\

Don't understand:

How can we conclude that the path differences are the same?

Thanks
Title: Re: VCE Physics Question Thread!
Post by: Zealous on November 09, 2014, 03:44:08 pm
How can we conclude that the path differences are the same?
You can make that conclusion because the position Y is in the exact same spot with both patterns.  The light from both slits has still travelled the exact same distance to reach point Y, so the difference in the path they take will still be the same.

(http://i1282.photobucket.com/albums/a531/Ovazealous/ROF_zps0a8084a5.png)

If you see the diagram, there has been no change to the path of the light to reach point Y, the only thing that has changed is the spacing of the pattern because of the new wavelength of light.
Title: Re: VCE Physics Question Thread!
Post by: Plitzer on November 09, 2014, 05:04:15 pm
Could someone explain slip rings in reference to AC generators for me please? I'm having trouble understanding textbook definitions, thanks  :)
Title: Re: VCE Physics Question Thread!
Post by: Rishi97 on November 09, 2014, 05:19:08 pm
Could someone explain slip rings in reference to AC generators for me please? I'm having trouble understanding textbook definitions, thanks  :)

Basically, slip rings allow the AC current induced in the loops to be transferred to the LOAD. Generators that are fitted with slip rings produce AC
Title: Re: VCE Physics Question Thread!
Post by: Mujteba on November 09, 2014, 05:43:11 pm
Could someone explain slip rings in reference to AC generators for me please? I'm having trouble understanding textbook definitions, thanks  :)

Also compared to the commutator, slip-rings remain stationary and do not rotate with the coil. The only place they are seen as far as the physics course is concerned is on the AC generator. One problem with slip rings though is because of the metal to metal contact between the coil and the rings they are very prone to wear and also cannot handle vibration too well as the coil can temporarily lose contact with the slip rings and cause sparks. This is why in many motor vehicles an alternator is used which rotates the magnet rather than the coil and hence eliminates the need for slip rings.

Went a bit off track but hope that covered all bases  ;D
Title: Re: VCE Physics Question Thread!
Post by: Plitzer on November 09, 2014, 06:06:37 pm
Also compared to the commutator, slip-rings remain stationary and do not rotate with the coil. The only place they are seen as far as the physics course is concerned is on the AC generator. One problem with slip rings though is because of the metal to metal contact between the coil and the rings they are very prone to wear and also cannot handle vibration too well as the coil can temporarily lose contact with the slip rings and cause sparks. This is why in many motor vehicles an alternator is used which rotates the magnet rather than the coil and hence eliminates the need for slip rings.

Went a bit off track but hope that covered all bases  ;D
How is the current directed in the coil, though? Like a split-ring commutator reverses the current every 180degrees but slip rings maintain an AC current within the coil? How does this keep it rotating? Maybe I should have been more specific as this is really what I want to know. That gave me heaps of useful information though - so thank you!
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on November 09, 2014, 06:20:33 pm
They're just a circular electrical contact so that as the wires spin, they're always in contact with these rings which then connect the spinning wire back to the rest of the circuit.
As for the current in the coil, if you have an alternator, the current is generated by the motion of the coil through principles you're hopefully aware of by now. Its direction is determined by whether the flux is increasing or decreasing.
Title: Re: VCE Physics Question Thread!
Post by: speedy on November 09, 2014, 06:36:24 pm
For the exam, are you not allowed to write anywhere in the border? Or is it only on the side closest to the centre?
Title: Re: VCE Physics Question Thread!
Post by: Yacoubb on November 10, 2014, 08:56:28 am
For the exam, are you not allowed to write anywhere in the border? Or is it only on the side closest to the centre?

You can't write outside the border. So only within the vicinity of that box. :)
Title: Re: VCE Physics Question Thread!
Post by: davomac on November 10, 2014, 09:11:59 am
Would somebody be able to explain modulation and demodulation? Thanks  :)
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on November 10, 2014, 10:46:25 am
Would somebody be able to explain modulation and demodulation? Thanks  :)

You have some information you want to transmit, and it's in the form of a low-frequency wave. We'll call this the signal.

You know that a low frequency signal wont make it very far if you try to transmit it through a wire or the air, or something, so you want to encode your information onto a high frequency wave that we'll call the carrier. This might be a light wave, for example. The modulated wave wont suffer as much of a loss when transmitted over a distance, so your information goes further.

Modulation is the process of encoding the information into the carrier wave, and in VCE physics we talk about intensity/amplitude modulation which means we encode the changes in signal into changes in amplitude of the high-frequency carrier. (there's also frequency modulation where changes are mapped onto changes in frequency).

Demodulation is the reverse process, where the information is extracted from the modulated wave (after transmission), so basically you read the changes in amplitude of the incoming wave and that's your signal.
Title: Re: VCE Physics Question Thread!
Post by: Yacoubb on November 10, 2014, 10:53:44 am
Modulation is the process of encoding the information into the carrier wave, and in VCE physics we talk about intensity/amplitude modulation which means we encode the changes in signal into changes in amplitude of the high-frequency carrier. (there's also frequency modulation where changes are mapped onto changes in frequency).

So if we were using a light beam, would modulation be that we encode the changes in the signal into changes in the light brightness?
Title: Re: VCE Physics Question Thread!
Post by: myanacondadont on November 10, 2014, 11:40:27 am
Is it necessary to put a negative in front of voltage gain if the amplifier is inverting? Do you lose marks if you don't?
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on November 10, 2014, 11:44:07 am
So if we were using a light beam, would modulation be that we encode the changes in the signal into changes in the light brightness?

yeah, intensity (brightness) is proportional to amplitude squared, so essentially you're making it brighter/dimmer in accordance with the information signal you're trying to send.

Is it necessary to put a negative in front of voltage gain if the amplifier is inverting? Do you lose marks if you don't?

Do gain as the absolute value of the gradient (no minus sign either way), and then if necessary mention that it's an inverting amplifier when they ask. Putting the negative there is fine too, afaik.
Title: Re: VCE Physics Question Thread!
Post by: myanacondadont on November 10, 2014, 11:48:35 am
Also, Question 17b in VCAA exam 2013.

It says that there is zero emf when there is no change in flux (i.e gradient of the graph is 0) and list the possible times as 0.5,1,1.5. Wouldnt 0, 2 and 2.5 be included?

Nevermind I'm an idiot. It specifies a time restriction, gotta watch out for those aye.
Title: Re: VCE Physics Question Thread!
Post by: jumcakes on November 10, 2014, 04:03:51 pm
How should we deal with the negative sign in Faraday's law when computing the average emf induced?

Quite clearly we have emf = -N(change in flux/change in time), which is likely to come out as a negative value - how should we 'convert' this to a positive average emf induced, in terms of mathematically sound working?
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on November 10, 2014, 04:09:50 pm
How should we deal with the negative sign in Faraday's law when computing the average emf induced?

Quite clearly we have emf = -N(change in flux/change in time), which is likely to come out as a negative value - how should we 'convert' this to a positive average emf induced, in terms of mathematically sound working?

If you have a negative change in flux, the negatives will cancel. Otherwise you get a negative number of volts, but it doesn't matter because in VCE you figure out the direction of current analytically with right hand rules and you wont be tested on the specifics of which way is 'positive voltage' because that's not the focus of these questions.
If you do find yourself going through the working and get a negative answer at the end, don't sweat, just leave it or say the magnitude of the voltage is |your answer| = your answer without the negative.
Title: Re: VCE Physics Question Thread!
Post by: myanacondadont on November 10, 2014, 06:38:45 pm
Is there solutions to the sample exam VCAA provided last year? http://www.vcaa.vic.edu.au/Documents/exams/physics/physics-specs-samp-w.pdf

Or am I missing something?
Title: Re: VCE Physics Question Thread!
Post by: Zealous on November 10, 2014, 06:41:12 pm
Is there solutions to the sample exam VCAA provided last year? http://www.vcaa.vic.edu.au/Documents/exams/physics/physics-specs-samp-w.pdf

Or am I missing something?

VCAA didn't put out any solutions - interestingly the sample exam is just a compilation of old VCAA questions.

iTute has some solutions here: http://www.itute.com/wp-content/uploads/2013-2016-vcaa-physics-sample-exam-solutions.pdf
Title: Re: VCE Physics Question Thread!
Post by: theshunpo on November 10, 2014, 08:15:35 pm
Is there solutions to the sample exam VCAA provided last year? http://www.vcaa.vic.edu.au/Documents/exams/physics/physics-specs-samp-w.pdf

Or am I missing something?

In addition to Zealous's link i found this while looking through VicPhysics once, hope it helps.
Title: Re: VCE Physics Question Thread!
Post by: rui97 on November 10, 2014, 10:01:09 pm
Could someone explain why I cannot find the answer to Q14 as kx=mg?

The answer to Q13 is 19.6 btw
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on November 10, 2014, 10:16:43 pm
kx=mg applies to when the guy was hanging in equilibrium (net F = 0 = kx up - mg down --> kx = mg), and we don't know what the spring's equilibrium length is (natural length is x=0 and the point where he comes to rest isnt equilibrium because he's oscillating)
so in short, it's a valid formula but we aren't given what x is in that situation

this means you need to use conservation of energy here!
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on November 10, 2014, 10:33:54 pm
Quick check for what to use for springs:
If you're told energies or speeds, use conservation of energy. If you're told something isn't moving, using forces (net force = 0)
Title: Re: VCE Physics Question Thread!
Post by: Bestie on November 11, 2014, 01:03:48 pm
hello can somebody please explain the difference and hoe they work: dynamic and velocity micriphones?

whats the difference between dynamic microphones and dynamic loudspeaker loudspeakers? VCAA separated the two in the study design???
Title: Re: VCE Physics Question Thread!
Post by: chickenfries on November 11, 2014, 02:40:36 pm
can someone pwease explain the highlighted bit? what can't it be squarish shape?
VCAA 2011 Exam 2 q11 electric power
http://www.vcaa.vic.edu.au/Pages/vce/studies/physics/exams.aspx

thank you
Title: Re: VCE Physics Question Thread!
Post by: myanacondadont on November 11, 2014, 03:48:09 pm
It's a bit late but oh well. How do I tackle these questions? I found the acceleration and then consider about the forces in question.

In part c I considered the forces acting on box C. I found net force from F=ma and then subtracted friction force and since thats the 'driving' force on box C then Force of B on C must be equal to it?

However I always get 'em wrong :(
Title: Re: VCE Physics Question Thread!
Post by: Plitzer on November 11, 2014, 04:16:04 pm
It's a bit late but oh well. How do I tackle these questions? I found the acceleration and then consider about the forces in question.

In part c I considered the forces acting on box C. I found net force from F=ma and then subtracted friction force and since thats the 'driving' force on box C then Force of B on C must be equal to it?

However I always get 'em wrong :(
The net force on Box A is equal to the 140N driving force minus the frictional force of Box A.
Frictional force is 0.2 x ( 10 x 9.8 ) = 19.6N. Therefore the net force on Box A is 120.4N to the right, which is now exerted on to Box B.
Net force on box B = 120.4 - (0.2 x (9.8 x 7)) = 106.68 to the right, which is exerted on to Box C.
i.e. the Force of Box B on Box C is 106.68N to the right, is this correct?
Title: Re: VCE Physics Question Thread!
Post by: myanacondadont on November 11, 2014, 04:24:03 pm
The net force on Box A is equal to the 140N driving force minus the frictional force of Box A.
Frictional force is 0.2 x ( 10 x 9.8 ) = 19.6N. Therefore the net force on Box A is 120.4N to the right, which is now exerted on to Box B.
Net force on box B = 120.4 - (0.2 x (9.8 x 7)) = 106.68 to the right, which is exerted on to Box C.
i.e. the Force of Box B on Box C is 106.68N to the right, is this correct?

No :( Sorry. The answer for part c is 26.7N and the answer for part d is 73.4N
Title: Re: VCE Physics Question Thread!
Post by: myanacondadont on November 11, 2014, 04:37:29 pm
And one more sorry - For Q7 they times the torque supplied by the pole by 1.5. I thought you times it by 3 since its 3m from the pivot point?

Why is that?
Title: Re: VCE Physics Question Thread!
Post by: Camo15 on November 11, 2014, 04:41:58 pm
What are some devices that can act as modulators and demodulators, and what can act as a carrier wave?

EDIT: Also, how do you go about finding the rebound Force of a wall that is holding a cantilever as well as the angle of said force?

And one more sorry - For Q7 they times the torque supplied by the pole by 1.5. I thought you times it by 3 since its 3m from the pivot point?

Why is that?

The overall torque clockwise due to the pole is 1.5*100 because that's its horizontal length from the pivot, I believe. It's the same reason why the torque due to the 20kg mass is 3*200, even though the mass is at the end of the 6m rod
Title: Re: VCE Physics Question Thread!
Post by: Zealous on November 11, 2014, 05:21:20 pm
In part c I considered the forces acting on box C. I found net force from F=ma and then subtracted friction force and since thats the 'driving' force on box C then Force of B on C must be equal to it?

However I always get 'em wrong :(

Look at all three blocks as a whole to find the acceleration of the system: m=21kg, F(applied)=140N, friction=42N, acceleration=14/3

Force of B on C, set up an equation for block C individually:

$ma=F_{BonC}-Friction \implies 4(14/3)=F-8 \implies F=26.7$

Force of B on A, look at Box A by itself, you've got the push force, force of B on A and the frictional forces:

$ma=F_{Applied}-Friction-F_{BonA} \implies 10(14/3)=140-20-F_{BonA}$$\implies F_{BonA}=73.3$

And one more sorry - For Q7 they times the torque supplied by the pole by 1.5. I thought you times it by 3 since its 3m from the pivot point?

Why is that?

Torque is calculated as distance multiplied by perpendicular component of force, the 100N of weight is not perpendicular to the pole PQ, so we multiply the 100N of weight by the component of the pole which is perpendicular to the weight force which is 1.5m (using a bit of trigonometry).

What are some devices that can act as modulators and demodulators, and what can act as a carrier wave?
Modulators: you're mainly looking for transducers (electro-optic) which convert electrical signals to some physical signal, where their output can be adjusted based on an input signal. So things like an LED (can't think of any others right now).

Demodulators: take the opposite, so opto-electric devices such as LDR's and photodiodes. They will basically take the variations in the signal and convert it back into an electrical signal.

Carrier Waves: VCAA usually uses light waves in their scenarios, or even the brightness of an LED.

EDIT: Also, how do you go about finding the rebound Force of a wall that is holding a cantilever as well as the angle of said force?
Well, if you've got a rope or something at an angle holding up the cantilever, the force from the wall will just ensure the whole system is equilibrium. So find the value of tension in a cable/rope using torque calculations (to oppose the vertical weight force of the cantilever), then look at translational equilibrium - if the system isn't in translational equilibrium then the wall is likely applying a force to the cantilever in some direction.

(http://questions.transtutors.com/Transtutors001/Images/Transtutors001_e5ae4d3b-0bf7-4ba8-9555-a895c99d2e5b.PNG)
So in this example, we'd find the tension in the cable such that it opposes the weight of the beam. But the tension in the cable would be made up of a vertical component which opposes the weight of the beam, and a horizontal component (because the tension is at an angle). So the wall (point A) will provide a horizontal force to work against the tension so the whole thing stays in equilibrium.
Title: Re: VCE Physics Question Thread!
Post by: Camo15 on November 11, 2014, 05:44:12 pm

Well, if you've got a rope or something at an angle holding up the cantilever, the force from the wall will just ensure the whole system is equilibrium. So find the value of tension in a cable/rope using torque calculations (to oppose the vertical weight force of the cantilever), then look at translational equilibrium - if the system isn't in translational equilibrium then the wall is likely applying a force to the cantilever in some direction.

(http://questions.transtutors.com/Transtutors001/Images/Transtutors001_e5ae4d3b-0bf7-4ba8-9555-a895c99d2e5b.PNG)
So in this example, we'd find the tension in the cable such that it opposes the weight of the beam. But the tension in the cable would be made up of a vertical component which opposes the weight of the beam, and a horizontal component (because the tension is at an angle). So the wall (point A) will provide a horizontal force to work against the tension so the whole thing stays in equilibrium.

Ok, so the rebound force is working against the horizontal component of the tension?

Say that there's a cable with a tension of 5N that's got an upward component of 4N and a horizontal component of 3N, what would I use to figure out the rebound force?

If anything I've said is completely wrong let me know haha
Title: Re: VCE Physics Question Thread!
Post by: Marrogi12 on November 11, 2014, 07:58:42 pm
Hey guys,  I've got like 1 quarter of a page empty on my summary sheet , and I was wonder if I should put examples or find other theory to put in there , I don't really find examples helpful as I only use my summary sheet when I forget a formula or constant , but you never know in an exam ahaha
Title: Re: VCE Physics Question Thread!
Post by: S61778 on November 11, 2014, 08:16:14 pm
Hi,

Could someone please explain how standing waves support the existence of discrete energy levels in the atom?
Title: Re: VCE Physics Question Thread!
Post by: Camo15 on November 11, 2014, 09:15:44 pm
Hi,

Could someone please explain how standing waves support the existence of discrete energy levels in the atom?

Standing waves mean that if orbit is equal to a whole number of wavelengths, then a standing wave is formed. This means that electrons of those particular wavelengths can exist in the atom. Since different wavelengths correspond to different energies, only those energy states are possible for electron orbit, meaning that discrete energy levels exist.
Title: Re: VCE Physics Question Thread!
Post by: allstar on November 11, 2014, 09:36:49 pm
hello,
can someone please explain this question?
why can't i use"
mgh = (1/2)kx^2
1 x 10 x 0.3 = (1/2) k (0.3)^2

thank you
Title: Re: VCE Physics Question Thread!
Post by: Camo15 on November 11, 2014, 09:46:14 pm
hello,
can someone please explain this question?
why can't i use"
mgh = (1/2)kx^2
1 x 10 x 0.3 = (1/2) k (0.3)^2

thank you

I think it has to do with the fact that you can't take ugh from a certain point.
Title: Re: VCE Physics Question Thread!
Post by: allstar on November 11, 2014, 09:47:47 pm
ans says k =33?

how would i use their formula f=kx?
Title: Re: VCE Physics Question Thread!
Post by: theshunpo on November 11, 2014, 09:55:29 pm
ans says k =33?

how would i use their formula f=kx?

When it is stationary, the change in x is 30cm (0.3m). At this point The netforce =0, which means the force kx-mg=0. using x=0.3 you should be able to work out k=33
Title: Re: VCE Physics Question Thread!
Post by: speedy on November 11, 2014, 09:55:36 pm
hello,
can someone please explain this question?
why can't i use"
mgh = (1/2)kx^2
1 x 10 x 0.3 = (1/2) k (0.3)^2

thank you

Just do -> F = kx -> mg = kx -> 10 = k(.03) -> k = 33.33
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on November 11, 2014, 09:59:28 pm
Standing waves mean that if orbit is equal to a whole number of wavelengths, then a standing wave is formed. This means that electrons of those particular wavelengths can exist in the atom. Since different wavelengths correspond to different energies, only those energy states are possible for electron orbit, meaning that discrete energy levels exist.

Your answer is fine for VCE purposes, but I'm going to say again: this model is heavily oversimplified. The Bohr model of the atom only works for hydrogen atoms, where there is only one electron. Even for helium, the fact that you have two electrons and repulsions between electrons means you'll have problems. Indeed, Bohr was utterly unable to do anything about atoms with more than one electron.

hello,
can someone please explain this question?
why can't i use"
mgh = (1/2)kx^2
1 x 10 x 0.3 = (1/2) k (0.3)^2

thank you

Let me ask you this: what scenario does the question ask for? It's NOT an energy question because the masses are stationary at both times. This means the net force = 0. An extra extension of 30 cm counters the addition of 1 kg of mass. So k*0.3m = 1kg * g = 9.8 N
k = 9.8/0.3 N/m or around 33 N/m

I'll explain why you can't use energy. When you use mgh = 1/2 kx^2, you are saying the spring energy change equals the gravitational energy change. As mechanical energy is conserved and as the total potential energy seems to have remained constant, the kinetic energy must also have remained constant. Initially, before adding a mass, the spring wasn't moving. Therefore the kinetic energy initially is zero. Solving mgh = 1/2 kx^2 thus solves for the next position where the spring isn't moving either. This is all assuming NO EXTERNAL INFLUENCE! Conservation of energy only applies to a closed system.

Let me ask you this. If you got a spring, attached a mass to it and let go, what would happen? It'd oscillate up and down like a slinky. At its lowest point where the kinetic energy is momentarily zero, the net force isn't zero. It was moving down before, it's now not moving. Clearly the net force is going straight up. But in the diagram your mass is remaining stationary, so the net force is zero. So evidently, the stationary point you solve for with mgh = 1/2 kx^2 is a different point to the point where you can just hang the mass indefinitely.

The solution to this dilemma is: once you put the new mass on, you yourself have to then gradually pull the spring down until it stretches a bit more and the extra stretch of the spring balances the extra weight force. This is when you let go of the spring and nothing happens. See how you had to actively hold the spring? The system is no longer closed. Conservation of energy no longer applies.

In summary, use net force because the spring isn't moving and is staying still. That's not an energy question.
Title: Re: VCE Physics Question Thread!
Post by: Marrogi12 on November 11, 2014, 10:02:18 pm
Your answer is fine for VCE purposes, but I'm going to say again: this model is heavily oversimplified. The Bohr model of the atom only works for hydrogen atoms, where there is only one electron. Even for helium, the fact that you have two electrons and repulsions between electrons means you'll have problems. Indeed, Bohr was utterly unable to do anything about atoms with more than one electron.

Let me ask you this: what scenario does the question ask for? It's NOT an energy question because the masses are stationary at both times. This means the net force = 0. An extra extension of 30 cm counters the addition of 1 kg of mass. So k*0.3m = 1kg * g = 9.8 N
k = 9.8/0.3 N/m or around 33 N/m

I'll explain why you can't use energy. When you use mgh = 1/2 kx^2, you are saying the spring energy change equals the gravitational energy change. As mechanical energy is conserved and as the total potential energy seems to have remained constant, the kinetic energy must also have remained constant. Initially, before adding a mass, the spring wasn't moving. Therefore the kinetic energy initially is zero. Solving mgh = 1/2 kx^2 thus solves for the next position where the spring isn't moving either. This is all assuming NO EXTERNAL INFLUENCE! Conservation of energy only applies to a closed system.

Let me ask you this. If you got a spring, attached a mass to it and let go, what would happen? It'd oscillate up and down like a slinky. At its lowest point where the kinetic energy is momentarily zero, the net force isn't zero. It was moving down before, it's now not moving. Clearly the net force is going straight up. But in the diagram your mass is remaining stationary, so the net force is zero. So evidently, the stationary point you solve for with mgh = 1/2 kx^2 is a different point to the point where you can just hang the mass indefinitely.

The solution to this dilemma is: once you put the new mass on, you yourself have to then gradually pull the spring down until it stretches a bit more and the extra stretch of the spring balances the extra weight force. This is when you let go of the spring and nothing happens. See how you had to actively hold the spring? The system is no longer closed. Conservation of energy no longer applies.

In summary, use net force because the spring isn't moving and is staying still. That's not an energy question.
Gg man gg
Title: Re: VCE Physics Question Thread!
Post by: Yacoubb on November 12, 2014, 08:16:21 pm
Did SO bad in Physics lol. Easily lost ~30 marks. Oh well, it's my bottom subject!
Title: Re: VCE Physics Question Thread!
Post by: lolaishappy on December 19, 2014, 08:00:25 pm
Hellooo, I need clarification on tension in these types of problems (like below) and how to solve this one... part (ii)

(a) A car of mass 1400kg tows a trailer of mass 600kg due north along a level road at constant speed. The forces resisting the motion of the car and trailer are 400N and 100N respectively.

(b)If the car and trailer in part (a) with the same resistance forces, have a northerly acceleration of 2.0ms^2, what is:
(ii) the magnitude of the tension in the bar between the car and trailer?
Title: Re: VCE Physics Question Thread!
Post by: Zealous on December 20, 2014, 12:27:26 am
Hellooo, I need clarification on tension in these types of problems (like below) and how to solve this one... part (ii)

(a) A car of mass 1400kg tows a trailer of mass 600kg due north along a level road at constant speed. The forces resisting the motion of the car and trailer are 400N and 100N respectively.

(b)If the car and trailer in part (a) with the same resistance forces, have a northerly acceleration of 2.0ms^2, what is:
(ii) the magnitude of the tension in the bar between the car and trailer?
Let's resolve the forces acting on both the mass and trailer - so let's treat them as one object and ignore the trailer for now.

The car will need to apply a force to accelerate the object at 2ms-2 as well as overcoming friction so the required force is:
$F_{applied}=ma+Friction=(1400+600)\times 2+400+100=4500$

Now to find the tension in the bar and the trailer, we can resolve forces on the individual items - the car and the trailer so lets look at just the car. Have a look at the diagram below.

(http://i1282.photobucket.com/albums/a531/Ovazealous/diagram_zps1564eb19.png)

There's 3 forces acting on the car: the applied force of 4500N, the friction force of 400N and some tension force.  We also know the car by itself will also accelerate at 2ms-2. Therefore the NET force acting on just the car is F=ma=1400x2=2800.

Now we can setup an equation of forces:

F(NET)=F(Applied)-F(Friction)-F(Tension) which means F(Tension)=F(Applied)-F(Friction)-F(NET)=4500-400-2800=1300. Therefore the tension force is 1300N.

You can also check this result using the trailer - the forces acting on the trailer would be the tension force and the friction on the trailer. So F(NET)=1300-100=1200, a=1200/600=2m/s2. So we can confirm that the trailer will accelerate correctly if the tension force is 1300N.

So in summary, when solving tension questions:
1) Resolve the forces acting on the system as a whole (without considering the tension forces).
2) Resolve the forces of an individual item in the system to find the tension.

Hope this helps!
Title: Re: VCE Physics Question Thread!
Post by: lolaishappy on December 20, 2014, 09:02:20 am
Let's resolve the forces acting on both the mass and trailer - so let's treat them as one object and ignore the trailer for now.

The car will need to apply a force to accelerate the object at 2ms-2 as well as overcoming friction so the required force is:
$F_{applied}=ma+Friction=(1400+600)\times 2+400+100=4500$

Now to find the tension in the bar and the trailer, we can resolve forces on the individual items - the car and the trailer so lets look at just the car. Have a look at the diagram below.

(http://i1282.photobucket.com/albums/a531/Ovazealous/diagram_zps1564eb19.png)

There's 3 forces acting on the car: the applied force of 4500N, the friction force of 400N and some tension force.  We also know the car by itself will also accelerate at 2ms-2. Therefore the NET force acting on just the car is F=ma=1400x2=2800.

Now we can setup an equation of forces:

F(NET)=F(Applied)-F(Friction)-F(Tension) which means F(Tension)=F(Applied)-F(Friction)-F(NET)=4500-400-2800=1300. Therefore the tension force is 1300N.

You can also check this result using the trailer - the forces acting on the trailer would be the tension force and the friction on the trailer. So F(NET)=1300-100=1200, a=1200/600=2m/s2. So we can confirm that the trailer will accelerate correctly if the tension force is 1300N.

So in summary, when solving tension questions:
1) Resolve the forces acting on the system as a whole (without considering the tension forces).
2) Resolve the forces of an individual item in the system to find the tension.

Hope this helps!

Thanks Zealous! Btw are applied force and Thrust the same thing? ???
Title: Re: VCE Physics Question Thread!
Post by: FarAwaySS2 on December 31, 2014, 08:18:57 pm
I know this question is meant to be quite easy but I can't seem to get my head around it:

A speeding motorbike travels past a stationary police car at a speed of 35 m/s. The police car starts accelerating immediately at 4 m/s/s, and keeps accelerating at this rate until it has passed the bike.

How far does the police car travel before it overtakes the motorbike?

Thanks!
Title: Re: VCE Physics Question Thread!
Post by: dankfrank420 on December 31, 2014, 08:56:36 pm
I know this question is meant to be quite easy but I can't seem to get my head around it:

A speeding motorbike travels past a stationary police car at a speed of 35 m/s. The police car starts accelerating immediately at 4 m/s/s, and keeps accelerating at this rate until it has passed the bike.

How far does the police car travel before it overtakes the motorbike?

Thanks!

That's the thing with physics, it appears easy on the surface but is actually a bit challenging underneath.

Is the answer 612.5m? Here's how I worked it out:

d (motorbike) = 35t
d (car) = (0)t + 1/2(4)t^2

Equate the two formulas, and t = 0 or t = 17.5

sub in 17.5 into the initial motorbike formula, and you get 612.5m.
Title: Re: VCE Physics Question Thread!
Post by: AirLandBus on December 31, 2014, 09:16:24 pm
This thread seems so dead. We need some more activity for us physics kiddies.
Title: Re: VCE Physics Question Thread!
Post by: FarAwaySS2 on January 01, 2015, 01:31:05 pm
Yup. That's the answer! Thank youuuuu!  :D
That's the thing with physics, it appears easy on the surface but is actually a bit challenging underneath.

Is the answer 612.5m? Here's how I worked it out:

d (motorbike) = 35t
d (car) = (0)t + 1/2(4)t^2

Equate the two formulas, and t = 0 or t = 17.5

sub in 17.5 into the initial motorbike formula, and you get 612.5m.
Title: Re: VCE Physics Question Thread!
Post by: bts on January 05, 2015, 05:03:02 pm
Did they first assume light was the particle model or the wave model? who?

Thank you
Title: Re: VCE Physics Question Thread!
Post by: Kel9901 on January 06, 2015, 08:56:59 am
Did they first assume light was the particle model or the wave model? who?

Thank you

Newton proposed the particle (corpuscular) theory for light, but Young's double slit experiment's results could only be explained by the wave model, so the wave model became the 'only' model for light as there was not yet any evidence supporting the particle model only. The experiment that lead to wave-particle duality was the photoelectric effect, whose results could be explained only by the particle model
Title: Re: VCE Physics Question Thread!
Post by: RedCapsicum on January 13, 2015, 11:23:05 pm
Hi guys can some one help me with this question.

So a 2.5kg mass is rotated in a conical pendulum where the length of the string is 0.68 metres and thr angle between the string and the vertical is 35 degrees.
Find
a. The tension in the string
b. The speed of the mass

If the pendulum is now spun faster so that it's period is now 1.2 seconds find
A. The tension in the string
b. The angel the string makes with the vertical
Title: Re: VCE Physics Question Thread!
Post by: JackSonSmith on January 14, 2015, 02:56:16 pm
How do I express direction in terms of degrees. eg. 53.1 degrees counterclockwise from the ground (is there a more "scientific" way to express this?)
Title: Re: VCE Physics Question Thread!
Post by: Kel9901 on January 14, 2015, 03:03:04 pm
Hi guys can some one help me with this question.

So a 2.5kg mass is rotated in a conical pendulum where the length of the string is 0.68 metres and thr angle between the string and the vertical is 35 degrees.
Find
a. The tension in the string
b. The speed of the mass

If the pendulum is now spun faster so that it's period is now 1.2 seconds find
A. The tension in the string
b. The angel the string makes with the vertical

q1:
a. net force vertically on the mass is 0 newtons, therefore downward force due to weight=upward force due to tension.
T cos(35)=mg=2.5 * 10=25 therefore T=30.5 N

b. net force horizontally is equal to T sin(35)=17.5 N.
a=F/m=7 ms^-2
r=0.68 sin(35)=0.39 m
v^2=ar=2.73 --> v=1.65 ms^-1

q2: remember mass is still 2.5 kg and length of string is still 0.68 m
a. r=0.68 sin(θ), v=2πr/T=3.56 sin(θ)
a=v^2/r=18.6 sin(θ)
F=ma=46.6 sin(θ)
horizontal component of tension force=T sin(θ) --> T=46.6 N

b. vertical component of tension force=T cos(θ)=mg
46.6 cos(θ)=25
cos(θ)=0.54
θ=57.6 degrees

Hope this helped!

edit:
How do I express direction in terms of degrees. eg. 53.1 degrees counterclockwise from the ground (is there a more "scientific" way to express this?)
53.1 degrees from the horizontal is probably best. I definitely wouldn't use bearings.
Title: Re: VCE Physics Question Thread!
Post by: RedCapsicum on January 14, 2015, 04:48:06 pm
Thank-you so much for your help!
Title: Re: VCE Physics Question Thread!
Post by: JackSonSmith on January 14, 2015, 05:08:05 pm
A person pushes a lawnmower of mass 20kg at a constant speed with a force of 100N directed along the handle, which is at 35 degrees to the horizontal.

What force must the person exert on the lawnmower in order to give it a velocity of 2.0 m/s in 2.5 seconds?
Title: Re: VCE Physics Question Thread!
Post by: AirLandBus on January 14, 2015, 05:54:11 pm
A person pushes a lawnmower of mass 20kg at a constant speed with a force of 100N directed along the handle, which is at 35 degrees to the horizontal.

What force must the person exert on the lawnmower in order to give it a velocity of 2.0 m/s in 2.5 seconds?

I have a answer, whether it is right or not is another issue. Whats the answer in the book say? Cross check with mine.
Title: Re: VCE Physics Question Thread!
Post by: Kel9901 on January 14, 2015, 07:26:50 pm
A person pushes a lawnmower of mass 20kg at a constant speed with a force of 100N directed along the handle, which is at 35 degrees to the horizontal.

What force must the person exert on the lawnmower in order to give it a velocity of 2.0 m/s in 2.5 seconds?

Lawnmover moving at constant speed therefore acc. to Newton's first law forces are balanced therefore frictional forces=100 cos(35)=81.9 N.

assimung it accelerates from rest:
u=0, v=2, t=2.5, a=?
v=u+at
2=0+2.5a
a=0.8 ms^-2
Net force acting on lawnmower=ma=16 N=driving force (in the direction of motion)-frictional forces=F cos(35)-81.9
F cos(35)=97.9 N
F=120 N

Person must exert a force of 120 N.
Title: Re: VCE Physics Question Thread!
Post by: AirLandBus on January 14, 2015, 07:50:44 pm
Lawnmover moving at constant speed therefore acc. to Newton's first law forces are balanced therefore frictional forces=100 cos(35)=81.9 N.

assimung it accelerates from rest:
u=0, v=2, t=2.5, a=?
v=u+at
2=0+2.5a
a=0.8 ms^-2
Net force acting on lawnmower=ma=16 N=driving force (in the direction of motion)-frictional forces=F cos(35)-81.9
F cos(35)=97.9 N
F=120 N

Person must exert a force of 120 N.

Can confirm, also got 120 N.
Title: Re: VCE Physics Question Thread!
Post by: JackSonSmith on January 19, 2015, 04:54:31 pm
A ball of mass 0.15kg hits the floor at 12 m/s and rebounds at 9 m/s. The ball was in contact with the floor for 0.12s.

a) What is the maximum height the ball will rise to on the rebound?  Ans = 4.05m

b) What is the average size of the reaction force from the floor on the ball? Ans = 26.25N

Could someone please give me a clue as to how to work this out?
Title: Re: VCE Physics Question Thread!
Post by: Conic on January 19, 2015, 05:26:46 pm
When the ball rebounds it is moving upwards at a speed of 9 m/s. If we define upwards to be positive, the velocity is 9 m/s, the acceleration is -10 m/s/s. When the ball reaches its maximum height it has a velocity of 0 m/s. Now we use one of the constant acceleration formulas.

$v^2=u^2+2ax \implies (0)^2=(9)^2+2(-10)x \implies x=\frac{9^2}{20}=4.05,$

so the ball reaches a height of 4.05 m. Now the second question asks for the average force on the ball. We can work out the change of momentum, and use this to find the average force. Firstly, the change in momentum is given by

$\Delta p = p_{\mathrm{final}}-p_{\mathrm{initial}} = 0.15\times9-0.15\times(-12) = 3.15,$

so the change in momentum is 3.15 kgm/s, and we are given that the contact time is 0.12 s. Now we use this in the impulse equation.

$\Sigma F_{\mathrm{avg}}\Delta t = \Delta p\implies \Sigma F_{\mathrm{avg}}\times0.12 =3.15 \implies \Sigma F_{\mathrm{avg}}=\frac{3.15}{0.12}=26.25,$

so the average reaction force is 26.25 N.
Title: Re: VCE Physics Question Thread!
Post by: JackSonSmith on January 19, 2015, 06:02:23 pm
Thanks Conic.

A box of mass 15kg is pulled 5m along a rough surface by a constant force of 80N.

The coefficient of sliding friction for the box over this surface is 0.4.

a) What is the magnitude of the friction force acting on the box? Ans = 60N

b) What is the change in kinetic energy of the box over its 5 m journey? Ans = 100J
Title: Re: VCE Physics Question Thread!
Post by: Conic on January 19, 2015, 06:16:40 pm
I'm pretty sure you aren't expected to know about the coefficient of friction in VCE Physics, but it does appear in Spesh.  The coefficient of friction is 0.4, and the normal force is 150 N (to balance the weight force mg = 150 N). Now the friction is given by

$Fr = \mu N = 0.4\times150 = 60,$

so the frictional force is 60 N in the direction opposing motion. Therefore there is a net force of 20 N acting in he direction of motion. Now, we want to find the change in kinetic energy. This will be equal to the net work on the system, as no energy is being lost or converted to potential energy. The net work done is

$\Sigma W = F\Delta x = 20\times5 = 100,$

so the box gains 100 J of kinetic energy over its journey.
Title: Re: VCE Physics Question Thread!
Post by: FarAwaySS2 on January 19, 2015, 06:20:49 pm
I keep getting this question wrong.

A bicycle accelerates from rest, covering 16m in 4 s. The total mass of the bicycle and its rider is 90 kg. What is its average acceleration?

Thanks!
Title: Re: VCE Physics Question Thread!
Post by: Kel9901 on January 20, 2015, 08:31:02 am
I keep getting this question wrong.

A bicycle accelerates from rest, covering 16m in 4 s. The total mass of the bicycle and its rider is 90 kg. What is its average acceleration?

Thanks!

u=0, t=4, s=16, a=?
s=ut+1/2at^2
16=0+1/2*a*16=8a
a=2 ms^-2
Title: Re: VCE Physics Question Thread!
Post by: AirLandBus on January 20, 2015, 12:53:30 pm
u=0, t=4, s=16, a=?
s=ut+1/2at^2
16=0+1/2*a*16=8a
a=2 ms^-2

Just out of curiosity, do you have to use the motion equations?
Why couldnt you do something along the lines of:
V = 16/4 = 4m/s
Acceleration = DeltaV/Time
= 4-0/4
=1m/s^2
Why doesnt that work?
Title: Re: VCE Physics Question Thread!
Post by: Conic on January 20, 2015, 01:23:50 pm
The problem is that you are assuming that the speed at the end of the 4 seconds (i.e., v) is equal to the average speed of the journey. This will not be true if there is non-zero constant acceleration.
Title: Re: VCE Physics Question Thread!
Post by: Kel9901 on January 20, 2015, 04:54:14 pm
Just out of curiosity, do you have to use the motion equations?
Why couldnt you do something along the lines of:
V = 16/4 = 4m/s
Acceleration = DeltaV/Time
= 4-0/4
=1m/s^2
Why doesnt that work?

Because the change in velocity is not equal to the average velocity (which you calculated to be 4 m/s).

Yes, you have to use the constant acceleration equations, as work-energy can't make use of the time figure (it needs the final velocity rather than time) and momentum-impulse can't make use of the change in displacement figure (it needs the final velocity rather than change in displacement)
Title: Re: VCE Physics Question Thread!
Post by: alchemy on January 22, 2015, 03:26:54 pm
A softball of mass 250 g is thrown with an initial velocity of 16 m s–1 at an angle θ to the horizontal. When the ball reaches its maximum height, its kinetic energy is 16 J.
a What is the maximum height achieved by the ball from its point of release?
b Calculate the initial vertical velocity of the ball.
c What is the value of θ?
d Whatisthespeedoftheballafter1.0s?
e What is the displacement of the ball after 1.0 s?
f How long after the ball is thrown will it return to
the ground?
g Calculate the horizontal distance that the ball will
travel during its flight.

Need help with part b and part e specifically.
Thanks.
Title: Re: VCE Physics Question Thread!
Post by: Kel9901 on January 22, 2015, 04:05:51 pm
A softball of mass 250 g is thrown with an initial velocity of 16 m s–1 at an angle θ to the horizontal. When the ball reaches its maximum height, its kinetic energy is 16 J.
a What is the maximum height achieved by the ball from its point of release?
b Calculate the initial vertical velocity of the ball.
c What is the value of θ?
d Whatisthespeedoftheballafter1.0s?
e What is the displacement of the ball after 1.0 s?
f How long after the ball is thrown will it return to
the ground?
g Calculate the horizontal distance that the ball will
travel during its flight.

Need help with part b and part e specifically.
Thanks.

I'll just answer the whole Q systematically because answers to previous parts are probably needed for the later parts

a) When it was thrown KE=1/2 mv^2=32 J, at the top KE=16 J therefore GPE at the top=16 J=mgh
h=6.4 m

b) vertically:
v=0, a=-10, s=6.4, u=?
v^2=u^2+2as
0=u^2-128
u=11.3 ms^-1

c) sin(θ)=11.3/16=0.71
θ=45°

d) horizontal velocity=16 cos(45)=11.3
vertically:
u=11.3, t=1, a=-10, v=?
v=u+at=11.3-10=1.3
Using Pythagoras, speed=11.4 ms^-1

e) After 1 second:
horizontal: s=ut=11.3 m
vertically:
u=11.3, t=1, a=-10, s=?
s=ut+1/2 at^2=11.3-5=6.3 m
11.3 m to the right of (assuming it was launched to the right) and 6.3 m above the launch point
OR 13.0 m (Pythagoras) from the launch point at an angle of tan⁻¹(6.3/11.3)=29° from the horizontal

f) vertically:
u=11.3, a=-10, s=0 (as ball returns to the ground at the same height as where it was launched), t=?
s=ut+1/2 at^2
0=11.3t-5 t^2
0=t(11.3-5t)
t=0, 11.3-5t=0 --> t=2.26
t>0 therefore t=2.26 s

g) horizontal:
u=11.3, t=2.26, s=?
s=ut=25.6 m
Title: Re: VCE Physics Question Thread!
Post by: RedCapsicum on January 23, 2015, 12:24:51 am
Need help with this question:

A 0.50kg puck rests on a level air table and is connected by a light thread passing through a hole in the table to support a hanging mass of 3.0 kg. A stable orbit is achieved when the puck is sent into a circular path of radius 0.15 metres around the hole.
(A) neglecting friction at the edge of the hole, calculate the period of revolution of the puck in its orbit.

(B) Suppose that the speed of the puck in its orbit is now doubled, while the radius remains fixed. what central mass will be needed to achieve a stable orbit.
Title: Re: VCE Physics Question Thread!
Post by: Kel9901 on January 23, 2015, 09:13:07 pm
Need help with this question:

A 0.50kg puck rests on a level air table and is connected by a light thread passing through a hole in the table to support a hanging mass of 3.0 kg. A stable orbit is achieved when the puck is sent into a circular path of radius 0.15 metres around the hole.
(A) neglecting friction at the edge of the hole, calculate the period of revolution of the puck in its orbit.

(B) Suppose that the speed of the puck in its orbit is now doubled, while the radius remains fixed. what central mass will be needed to achieve a stable orbit.

I took a while to understand what this question was asking and the directions of the forces. The tension in the string counteracts the weight of the 3.0 kg and provides the centripetal force for the 0.50 kg

a) On 3.0 kg: Fnet=0=Tension-mg=Tension-30
Tension=30
On 0.50 kg: Fnet=Tension=30=ma
a=60=v^2/r=v^2/0.15
v^2=9
v=3=2πr/T
T=0.31 s

b) Let the new velocity, acceleration and net force on the 0.50 kg be v', a' and Fnet'
0.50 kg:
v'=2*3=6
a'=v'^2/r=240
Fnet'=ma'=120=T'
Other mass:
120=mg
m=12 kg
Title: Re: VCE Physics Question Thread!
Post by: Cosec on January 27, 2015, 07:24:53 pm
Can someone answer these couple of questions. Ive answered them and my questions different ot the BOB. Checked over them and they appear right (probs wrong).

1) A child rolls a 50 gram marble up a playground slide that is inclined at 15 degrees to the horizontal. The slide is 3.5m long and the marble is launched with a speed of 4.8m/s.
How fast is the marble travelling when it is halfway up the slide?

2) A car travelling with a constant speed of 80km/hr passes a stationary motorcycle policeman. The cop sets off in pursuit, acclerating uniformly to 80km/hr in 10secs, and 100km/hr after a further 5 secs. he stays at 100km/hr for the rest of the journey. At what time will the policeman catch up with the car?
I know the distance of both need to be the same and you solve for time, but i keep getting a different answer.

3) Cassie starts from rest at the top of a 3.2m long smooth playground slide and slides to thwe bottom with a constant acceleration. If she takes 2.4 secs to reach the bottom. Calculate her average acceleration?

Last bit. If your asked to find the speed of a golf ball which has fallen 2 meters and find the speed when it has rebounded. How is this done. I got the right answer but im trying to think why?
For speed would it be just pick a point that is jsut above the point of impact (ie. 0.5 meters) assuming that down is engative? So youve picked a point on its upwards journey.
And for velocity, the displacement would be say 1.5m from the top?
Im just kind of making a educated guess here ^.
Title: Re: VCE Physics Question Thread!
Post by: odeaa on January 27, 2015, 07:59:38 pm
I tried the first one but I'm pretty sure i was wrong ahaha
Our teacher has forced us to do the second type of question to death though so I'll give it a crack

A car travelling with a constant speed of 80km/hr passes a stationary motorcycle policeman. The cop sets off in pursuit, acclerating uniformly to 80km/hr in 10secs, and 100km/hr after a further 5 secs. he stays at 100km/hr for the rest of the journey. At what time will the policeman catch up with the car?

First of all I would convert to m/s
Car=22.2m/s
Cop=27.8m/s
It takes the cop 15 seconds to accelerate to 27.8m/s, during which time the car has travelled 333m
In the first 10 seconds the cop travels $22.2*10/2=111m$
In the following 5 seconds the cop travels $22.2*5+27.8*5/2=180.5$ bringing him to a total of 291.5m
The cop is travelling 5.5m/s faster than the car, and has to catch up 41.5m, which will take him 7.5 seconds, bringing the time up to 22.5 seconds
Usually it isnt that messy and you have a nice 20m/s and 30m/s values to work with (classic VCE Physics)

Cassie starts from rest at the top of a 3.2m long smooth playground slide and slides to the bottom with a constant acceleration. If she takes 2.4 secs to reach the bottom. Calculate her average acceleration?

$s=ut+\dfrac{1}{2}at^2$
$3.2=ut+\dfrac{1}{2}a2.4^2$
$3.2=2.88a$

$a=1.11ms^-2$

I hope this is right ahaha
Title: Re: VCE Physics Question Thread!
Post by: Cosec on January 27, 2015, 09:24:48 pm
I tried the first one but I'm pretty sure i was wrong ahaha
Our teacher has forced us to do the second type of question to death though so I'll give it a crack

A car travelling with a constant speed of 80km/hr passes a stationary motorcycle policeman. The cop sets off in pursuit, acclerating uniformly to 80km/hr in 10secs, and 100km/hr after a further 5 secs. he stays at 100km/hr for the rest of the journey. At what time will the policeman catch up with the car?

First of all I would convert to m/s
Car=22.2m/s
Cop=27.8m/s
It takes the cop 15 seconds to accelerate to 27.8m/s, during which time the car has travelled 333m
In the first 10 seconds the cop travels $22.2*10/2=111m$
In the following 5 seconds the cop travels $22.2*5+27.8*5/2=180.5$ bringing him to a total of 291.5m
The cop is travelling 5.5m/s faster than the car, and has to catch up 41.5m, which will take him 7.5 seconds, bringing the time up to 22.5 seconds
Usually it isnt that messy and you have a nice 20m/s and 30m/s values to work with (classic VCE Physics)

Cassie starts from rest at the top of a 3.2m long smooth playground slide and slides to the bottom with a constant acceleration. If she takes 2.4 secs to reach the bottom. Calculate her average acceleration?

$s=ut+\dfrac{1}{2}at^2$
$3.2=ut+\dfrac{1}{2}a2.4^2$
$3.2=2.88a$

$a=1.11ms^-2$

I hope this is right ahaha

Ive done the car one to death to. But this ones weird. When i added the values for the distance initially covered by the copy im getting a different value to you.
I think its where you calculated
$22.2*5+27.8*5/2=180.5$

Shouldnt it be 22.2*5 + (27.8-22.2 *5/2)?
Take a look at the graph attached.
Imgur: http://imgur.com/wmdtr5g

And thats what i got for the second equation. Book must be wrong.
thanks buddy.

Edit: The way I got taught is a bit differnt for the first question
Where told to make two equations for each vehicle, find the distance both will cover in terms of time and then let them equal each other and solve for time.
Not sure.
Title: Re: VCE Physics Question Thread!
Post by: odeaa on January 27, 2015, 09:47:09 pm
Yeah my bad, I ahould have drawn a graph but I was too lazy
I'm gonna have a crack at the first one but no promises
Title: Re: VCE Physics Question Thread!
Post by: Cosec on January 27, 2015, 09:55:27 pm
Yeah my bad, I ahould have drawn a graph but I was too lazy
I'm gonna have a crack at the first one but no promises

Haha no worries. The problem is when i try to solve for t im getting a negative value. Which isnt right. My dads a engineer and cant work it out either.  :o :)
Title: Re: VCE Physics Question Thread!
Post by: Cosec on January 27, 2015, 10:24:12 pm
All good dude. Found my error with some help. The distance covered in the first 15 seconds travelled i calculated was right, when i was solving for time i forgot it should be (for the cop car) t-15 and not just t. Cause the distance added to the first 15 seconds is the time after 15 seconds (hence the t-15 bit. All good. Thanks buddy.
Title: Re: VCE Physics Question Thread!
Post by: odeaa on January 27, 2015, 10:27:46 pm
All good dude. Found my error with some help. The distance covered in the first 15 seconds travelled i calculated was right, when i was solving for time i forgot it should be (for the cop car) t-15 and not just t. Cause the distance added to the first 15 seconds is the time after 15 seconds (hence the t-15 bit. All good. Thanks buddy.
(http://i.imgur.com/aUQgBHx.jpg)
Gave this one a go, have no idea if it's right but it's feasible
Title: Re: VCE Physics Question Thread!
Post by: dankfrank420 on January 27, 2015, 10:35:27 pm
Edit: fucked it up
Title: Re: VCE Physics Question Thread!
Post by: Kel9901 on January 28, 2015, 09:45:29 am
Can someone answer these couple of questions. Ive answered them and my questions different ot the BOB. Checked over them and they appear right (probs wrong).

1) A child rolls a 50 gram marble up a playground slide that is inclined at 15 degrees to the horizontal. The slide is 3.5m long and the marble is launched with a speed of 4.8m/s.
How fast is the marble travelling when it is halfway up the slide?

2) A car travelling with a constant speed of 80km/hr passes a stationary motorcycle policeman. The cop sets off in pursuit, acclerating uniformly to 80km/hr in 10secs, and 100km/hr after a further 5 secs. he stays at 100km/hr for the rest of the journey. At what time will the policeman catch up with the car?
I know the distance of both need to be the same and you solve for time, but i keep getting a different answer.

3) Cassie starts from rest at the top of a 3.2m long smooth playground slide and slides to thwe bottom with a constant acceleration. If she takes 2.4 secs to reach the bottom. Calculate her average acceleration?

Last bit. If your asked to find the speed of a golf ball which has fallen 2 meters and find the speed when it has rebounded. How is this done. I got the right answer but im trying to think why?
For speed would it be just pick a point that is jsut above the point of impact (ie. 0.5 meters) assuming that down is engative? So youve picked a point on its upwards journey.
And for velocity, the displacement would be say 1.5m from the top?
Im just kind of making a educated guess here ^.

1. u=4.8, a=10sin(15)=-2.6, s=1.75, v=?
v^2=u^2+2as=23.04-9.06=13.98
v=3.74 ms^-1

You can also use energy conservation like this:
Bottom: GPE=0, KE=1/2 mv^2=0.576
1/2 way up slide: GPE=mgh=mg(1.75sin(15))=0.23
KE=0.576-0.23=0.35=1/2 mv^2
v^2=14
v=3.74 ms^-1

2. 80 km/hr=22.2 ms^-1
100 km/hr=27.8 ms^-1
In the first 15 seconds:
Car travels 22.2*15=333.3 m
Policeman travels 1/2*10*22.2 + 1/2*5*(22.2+27.8)=236.1 m

After exactly 15 seconds the policeman is behind by 97.2 m, and the difference in velocity is 27.8-22.2=5.6 ms^-1

The policeman takes a further 97.2/5.6=17.5 s to catch up, so total time taken is 15+17.5=32.5 s

3. u=0, t=2.4, s=3.2, a=?
s=ut+1/2 at^2
3.2=2.88a
a=1.11 ms^-2

For the last question, you can find out the KE just before the ball hits the ground (=GPE at top=mgh=20m). Assuming an elastic collision (I think that's what you're supposed to assume), the KE just after the ball bounces is also 20m. This means that in both cases, the speed can be calculated via:
20m=KE=1/2 mv^2
v^2=40
v=6.32 ms^-1
Title: Re: VCE Physics Question Thread!
Post by: Kel9901 on January 29, 2015, 10:18:29 am
(http://i.imgur.com/aUQgBHx.jpg)
Gave this one a go, have no idea if it's right but it's feasible

You can't just add initial speed in; remember that energy is proportional to speed squared, so you have to calculate kinetic energy initially, gravitational potential energy at the end and subtract and calculate
Title: Re: VCE Physics Question Thread!
Post by: odeaa on January 29, 2015, 10:37:53 am
You can't just add initial speed in; remember that energy is proportional to speed squared, so you have to calculate kinetic energy initially, gravitational potential energy at the end and subtract and calculate
I get it now, thanks so much! I asked the class genius on facebook how to do that, I didn't even doubt his answer for one second ahaha I can't believe he was wrong
Title: Re: VCE Physics Question Thread!
Post by: JackSonSmith on February 04, 2015, 05:32:14 pm
A speeding motorbike travels past a stationary police car at a speed of 35 m/s. The police car starts accelerating immediately at 4 m/s^2, and keeps accelerating at this rate until it has passed the bike.

a) How far does the police car travel before it overtakes the motorbike?

b) At what time does the police car overtake the motorbike?
Title: Re: VCE Physics Question Thread!
Post by: odeaa on February 04, 2015, 05:33:20 pm
A speeding motorbike travels past a stationary police car at a speed of 35 m/s. The police car starts accelerating immediately at 4 m/s^2, and keeps accelerating at this rate until it has passed the bike.

a) How far does the police car travel before it overtakes the motorbike?

b) At what time does the police car overtake the motorbike?
Almost this exact same question got answered a few days ago if you scroll back, different values but same concepts
Title: Re: VCE Physics Question Thread!
Post by: JackSonSmith on February 04, 2015, 05:35:27 pm
I am having trouble working this one out as I can not seem to find time taken. I had read the previous question before asking.
Title: Re: VCE Physics Question Thread!
Post by: odeaa on February 04, 2015, 05:37:42 pm
Sorry if I came across rude
The time taken if you arrange d=vt is t=d/v
If you sub in 35m/s from the constant speed of the bike you will get an answer
I'm assuming you got the distance?
Title: Re: VCE Physics Question Thread!
Post by: Kel9901 on February 04, 2015, 07:49:32 pm
A speeding motorbike travels past a stationary police car at a speed of 35 m/s. The police car starts accelerating immediately at 4 m/s^2, and keeps accelerating at this rate until it has passed the bike.

a) How far does the police car travel before it overtakes the motorbike?

b) At what time does the police car overtake the motorbike?

a) , b)
Motorbike: s=ut=35t
Car: u=0, a=4, s=?, t=?
s=ut+1/2 at^2
s=2t^2

So the equations are s=35t and s=2t^2
35t=2t^2
t=35/2 (as t>0)=17.5 s
s=35t(=2t^2)=612.5 m
Title: Re: VCE Physics Question Thread!
Post by: Cosec on February 04, 2015, 07:56:44 pm
I am having trouble working this one out as I can not seem to find time taken. I had read the previous question before asking.

Here:
Imgur: http://imgur.com/qCeuwvq

Basically, when the cop pass the car? When their displacement is the same.
Therefore, its easiest to draw a simple graph and set up two equations like shown and solve for time.
Title: Re: VCE Physics Question Thread!
Post by: Fusuy on February 06, 2015, 05:25:03 pm
Hey Guys, is anyone selling the Nelson or Jacaranda 3/4 Physics book?
Title: Re: VCE Physics Question Thread!
Post by: paper-back on February 08, 2015, 12:57:37 pm
In momentum questions which require the use of the formula m1u1+m2u2=m1v1+m2v2, do we assign directions?
E.g. Question
A cart with mass 2kg moving at 2m/s in westerly direction collides with cart moving at 3m/s in easterly direction. Upon collision, carts connect and move of at 1m/s in westerly direction
What's the mass of second cart?

Would we go;
(2)(2)-(3)m2=(2+m2)(1)?
Title: Re: VCE Physics Question Thread!
Post by: Maths Forever on February 08, 2015, 01:11:56 pm
In momentum questions which require the use of the formula m1u1+m2u2=m1v1+m2v2, do we assign directions?
E.g. Question
A cart with mass 2kg moving at 2m/s in westerly direction collides with cart moving at 3m/s in easterly direction. Upon collision, carts connect and move of at 1m/s in westerly direction
What's the mass of second cart?

Would we go;
(2)(2)-(3)m2=(2+m2)(1)?

Always choose a direction to be positive! Let the easterly direction be positive.

p (Before) = m1v1 + m2v2 = 2 x -2 + m2 x 3

p (After) = (m1 + m2) v = (2 + m2) x -1

p (Before) = p (After) ......conservation of momentum

Therefore, -4 + 3 x m2 = -2 - m2

-2 = -4 m2

Hence, m2 = 0.5 kg

But your way (letting westerly direction to be positive) would also work. Hope this helps!

Title: Re: VCE Physics Question Thread!
Post by: Cosec on February 08, 2015, 01:16:04 pm
As far as im aware the general rule of thumb is, if its a vector, assign a direction. Which should be first nature now.
Title: Re: VCE Physics Question Thread!
Post by: keltingmeith on February 08, 2015, 01:18:27 pm
As far as im aware the general rule of thumb is, if its a vector, assign a direction. Which should be first nature now.

That's not even the rule of thumb - if you don't assign a direction, the maths breaks, lol.
Title: Re: VCE Physics Question Thread!
Post by: Cosec on February 08, 2015, 01:33:39 pm
That's not even the rule of thumb - if you don't assign a direction, the maths breaks, lol.

Hahha, gez cant get away from you. But yeah, thats what i was implying.
Title: Re: VCE Physics Question Thread!
Post by: keltingmeith on February 08, 2015, 01:35:14 pm
Hahha, gez cant get away from you. But yeah, thats what i was implying.

(http://i.imgur.com/CKi9aMl.jpg)
Title: Re: VCE Physics Question Thread!
Post by: stockstamp on February 09, 2015, 06:30:37 pm
Hi,

Can someone please fully explain the following question for me? I'm confused about the way gravity and the normal force are involved for these sort of questions:

A rubber ball of mass 80 g bounces vertically on a concrete floor. The ball strikes the floor at 10 m/s and rebounds at 8.0 m/s

The time of contact between the ball and the floor during the bounce was 0.050 s.

a) Calculate the average net force acting on the ball during its contact with the floor.
b) Calculate the average force that the floor exerts on the ball.
c) Calculate the average force that the ball exerts on the floor.

thanks
Title: Re: VCE Physics Question Thread!
Post by: Maths Forever on February 09, 2015, 07:13:47 pm
Hi,

Can someone please fully explain the following question for me? I'm confused about the way gravity and the normal force are involved for these sort of questions:

A rubber ball of mass 80 g bounces vertically on a concrete floor. The ball strikes the floor at 10 m/s and rebounds at 8.0 m/s

The time of contact between the ball and the floor during the bounce was 0.050 s.

a) Calculate the average net force acting on the ball during its contact with the floor.
b) Calculate the average force that the floor exerts on the ball.
c) Calculate the average force that the ball exerts on the floor.

thanks

I = Δp = F(average) Δt = m Δv

Always assign a positive direction! Let Up be positive.

a) Sub in values: Δt = 0.050 s, m = 0.080 kg, Δv = Final - Initial = (8.0) - (- 10.0) = 18.0 m/s

F(average) x 0.050 = 0.080 x 18.0

Therefore F (average) = 1.44 / 0.050 = 28.8 N

Therefore average net force is 28.8 N in the upwards direction.

b) F (av) (floor on ball) - W (weight force) = F (av) (net)

F (av) (floor on ball) = N (normal force)

Therefore, N - mg = 28.8 (from part a)

N = 28.8 + (0.080 x 10) = 28.8 + 0.8 = 29.6 N

Therefore F (av) (floor on ball) = 29.6 N in the upwards direction.

c) F (av) (floor on ball) and F (av) (ball on floor) are a Newton's 3rd Law action reaction pair.

Hence F (av) (ball on floor) = - 29.6 N

Therefore F (av) (ball on floor) = 29.6 N in the downwards direction.

This should be correct! Hope it helps!
Title: Re: VCE Physics Question Thread!
Post by: Cosec on February 09, 2015, 07:38:17 pm
This question relates to impulse.

Since I = Δp = F(average) Δt
and I = Δp = m Δv

Always assign a positive direction! Let Up be positive.

Therefore F(average) Δt = m Δv

a) Sub in values: Δt = 0.050 s, m = 0.080 kg, Δv = Final - Initial = (8.0) - (- 10.0) = 18.0 m/s

F(average) x 0.050 = 0.080 x 18.0

Therefore F (average) = 1.44 / 0.050 = 28.8 N

Therefore average net force is 28.8 N in the upwards direction.

b) I (floor on ball) = Δp (ball) = p (final) - p (initial) = m ( v (final) - v (initial) ) = 0.080 x (8.0 - (-10.0) )
= 0.080 x 18.0 = 1.44 N s (standard unit for impulse: Newton second)

Therefore I (floor on ball) = 1.44 N s in the upwards direction.

c) I (floor on ball) and I (ball on floor) are a Newton's 3rd law action reaction pair.

Hence I (ball on floor) = - 1.44 N s

Therefore I (ball on floor) = 1.44 N s in the downwards direction.

I think this is all correct! Hope it helps!

Looks good to me! Instead though i did part a with the constant acceleration motion formulae. Still got 28.8. Got a bit stumped though when i hit b and c, then i re read it and noticed "average", gets me every time. Haahha.
Title: Re: VCE Physics Question Thread!
Post by: stockstamp on February 09, 2015, 07:47:42 pm
This question relates to impulse.

Since I = Δp = F(average) Δt
and I = Δp = m Δv

Always assign a positive direction! Let Up be positive.

Therefore F(average) Δt = m Δv

a) Sub in values: Δt = 0.050 s, m = 0.080 kg, Δv = Final - Initial = (8.0) - (- 10.0) = 18.0 m/s

F(average) x 0.050 = 0.080 x 18.0

Therefore F (average) = 1.44 / 0.050 = 28.8 N

Therefore average net force is 28.8 N in the upwards direction.

b) I (floor on ball) = Δp (ball) = p (final) - p (initial) = m ( v (final) - v (initial) ) = 0.080 x (8.0 - (-10.0) )
= 0.080 x 18.0 = 1.44 N s (standard unit for impulse: Newton second)

Therefore I (floor on ball) = 1.44 N s in the upwards direction.

c) I (floor on ball) and I (ball on floor) are a Newton's 3rd law action reaction pair.

Hence I (ball on floor) = - 1.44 N s

Therefore I (ball on floor) = 1.44 N s in the downwards direction.

I think this is all correct! Hope it helps!

Thanks for the reply but I'm still slightly confused.
If the question asks for the 'average force' how can the answer be in N s? Should it not just be N?

Part a) you gave an answer in Newtons, but in part b) and c) you gave an answer in N s, even though it was asking for force? Can you please explain this to me because it's really got me confused.

Also
it might be worth mentioning that this question was taken from the god-awful Heinemann textbook - below is their version of the fully-worked solution for parts b) and c).

b)   The forces acting are gravity Fg downwards and normal reaction force FN upwards.
Fg = mg = 0.78 N
totalF = Fg + FN
This gives FN = 29 N up.
c)   As described by Newton’s third law, this is equal and opposite to the force that the floor exerts on the ball, so F = 29 N down.

Rendering me totally confused

thanks and sorry for my ignorance in advance
Title: Re: VCE Physics Question Thread!
Post by: Kel9901 on February 09, 2015, 07:54:10 pm
Hi,

Can someone please fully explain the following question for me? I'm confused about the way gravity and the normal force are involved for these sort of questions:

A rubber ball of mass 80 g bounces vertically on a concrete floor. The ball strikes the floor at 10 m/s and rebounds at 8.0 m/s

The time of contact between the ball and the floor during the bounce was 0.050 s.

a) Calculate the average net force acting on the ball during its contact with the floor.
b) Calculate the average force that the floor exerts on the ball.
c) Calculate the average force that the ball exerts on the floor.

thanks

regarding nature of forces: net force=both normal reaction and weight forces, forces in parts b and c are just normal reaction forces.

a) FnetΔt=mΔv
Fnet*0.050=0.08*18
Fnet=28.8 N (UP)

b) Fnet=N-W=28.8
N-0.08*10=28.8
N=29.6 N

c) According to Newton's 3rd law, this is also 29.6 N

edit: this question looks perfectly fine imo no need to hate on heinemann
Title: Re: VCE Physics Question Thread!
Post by: Maths Forever on February 09, 2015, 08:33:26 pm
Thanks for the reply but I'm still slightly confused.
If the question asks for the 'average force' how can the answer be in N s? Should it not just be N?

Part a) you gave an answer in Newtons, but in part b) and c) you gave an answer in N s, even though it was asking for force? Can you please explain this to me because it's really got me confused.

Also
it might be worth mentioning that this question was taken from the god-awful Heinemann textbook - below is their version of the fully-worked solution for parts b) and c).

b)   The forces acting are gravity Fg downwards and normal reaction force FN upwards.
Fg = mg = 0.78 N
totalF = Fg + FN
This gives FN = 29 N up.
c)   As described by Newton’s third law, this is equal and opposite to the force that the floor exerts on the ball, so F = 29 N down.

Rendering me totally confused

thanks and sorry for my ignorance in advance

Sorry, I took parts b and c to mean impulse. A silly misinterpretation! Please see my edited version above, and thanks to Kel9901!
Title: Re: VCE Physics Question Thread!
Post by: Cosec on February 09, 2015, 09:52:11 pm
Im confused now, haha. It says average, when i think of average i think of how the value changes over a period of time (im thinking calculus) and as a result thought that it would how the force changes over that time interval of impact.
Additionally, when your looking at a object that the only two forces acting on it are the weight and normal reaction forces and then you form the equation
Fw = -Fn  cause you have to take into account the different directions in which they act.
And therefore the Sum of forces = Fw + Fn.

Why above is it (Sum of forces) expressed as F (av) (floor on ball) - W (weight force) = F (av) (net)

Can somebody explain just to clarify for me.
Title: Re: VCE Physics Question Thread!
Post by: Maths Forever on February 09, 2015, 10:12:20 pm
Im confused now, haha. It says average, when i think of average i think of how the value changes over a period of time (im thinking calculus) and as a result thought that it would how the force changes over that time interval of impact.
Additionally, when your looking at a object that the only two forces acting on it are the weight and normal reaction forces and then you form the equation
Fw = -Fn  cause you have to take into account the different directions in which they act.
And therefore the Sum of forces = Fw + Fn.

Why above is it (Sum of forces) expressed as F (av) (floor on ball) - W (weight force) = F (av) (net)

Can somebody explain just to clarify for me.

Sorry, the notation you are confused about is only because of my lack of mathematical symbols.

The average was asked for in part (a). So therefore, we also take the average net force as the sum of all forces in parts (b) and (c).

We are calculating the average reaction force in part (b), or the average force that the floor exerts on the ball?

Does this help? Sorry for any confusion.
Title: Re: VCE Physics Question Thread!
Post by: Adequace on February 10, 2015, 03:55:40 pm
Hey, I'm just wondering if electricity and motion the only topics that are taught in 1/2 and then expanded on in 3/4?
Title: Re: VCE Physics Question Thread!
Post by: Maths Forever on February 10, 2015, 04:15:14 pm
Hey, I'm just wondering if electricity and motion the only topics that are taught in 1/2 and then expanded on in 3/4?

In units 1 and 2 the topics in my school were:

- Electricity
- Light
- Motion
- 2 detailed studies

In units 3 and 4 all schools cover:

- Motion in one and two dimensions
- Electronics and Photonics
- Electromagnetism
- Light and Matter
- 1 detailed study (11 multiple choice questions on the end of year exam)

I hope this helps!
Title: Re: VCE Physics Question Thread!
Post by: Kel9901 on February 10, 2015, 06:09:05 pm
In units 1 and 2 the topics in my school were:

- Electricity
- Light
- Motion
- 2 detailed studies

In units 3 and 4 all schools cover:

- Motion in one and two dimensions
- Electronics and Photonics
- Electromagnetism
- Light and Matter
- 1 detailed study (11 multiple choice questions on the end of year exam)

I hope this helps!

In other words, motion+electricity are very relevant to 3/4, light is only slightly relevant (only part that is relevant is a bit of wave theory like v=fλ, radiation has 0 relevance, and the detailed studies probably aren't relevant
Title: Re: VCE Physics Question Thread!
Post by: bobisnotmyname on February 11, 2015, 05:24:34 pm
hey guys im having a little trouble with rounding. Well as we know in physics the numbers arn't always exact so its left to do some rounding, however how do we know how much to round by and how many decimal places our answers should have. thanks :)
Title: Re: VCE Physics Question Thread!
Post by: Maths Forever on February 11, 2015, 05:54:02 pm
hey guys im having a little trouble with rounding. Well as we know in physics the numbers arn't always exact so its left to do some rounding, however how do we know how much to round by and how many decimal places our answers should have. thanks :)

The answer should have the same number of significant figures (sig figs) as the least accurate piece of information in the question.

E.g. Calculate the net force as a cart of mass 200 kg (3 sig figs) increases its speed from rest to 30 m/s (2 sig figs) in a time of 3.60 seconds (3 sig figs) with constant acceleration.

Now, using the constant acceleration formula, v = u + at:

30 = 0 + a (3.60)
a = 30 / 3.60 = 8.3 m/s2 ....the answer should be rounded to 2 significant figures, since 30 m/s was the least accurate piece of data.

F (net) = ma = 3.60 x 200 = 7.2 x 10^2 N ....leave this as 2 significant figures, as we still take into account that 30 m/s was the least accurate piece of data in the question.

Scientific notation is a good method to use if rounding to significant figures. But really, as long as you don't leave your answer to an absurd amount of decimal places, the examiner shouldn't be too harsh. Two or three significant figures are usually fine.
Title: Re: VCE Physics Question Thread!
Post by: Kel9901 on February 11, 2015, 10:48:01 pm
The answer should have the same number of significant figures (sig figs) as the least accurate piece of information in the question.

E.g. Calculate the net force as a cart of mass 200 kg (3 sig figs) increases its speed from rest to 30 m/s (2 sig figs) in a time of 3.60 seconds (3 sig figs) with constant acceleration.

Now, using the constant acceleration formula, v = u + at:

30 = 0 + a (3.60)
a = 30 / 3.60 = 8.3 m/s2 ....the answer should be rounded to 2 significant figures, since 30 m/s was the least accurate piece of data.

F (net) = ma = 3.60 x 200 = 7.2 x 10^2 N ....leave this as 2 significant figures, as we still take into account that 30 m/s was the least accurate piece of data in the question.

Scientific notation is a good method to use if rounding to significant figures. But really, as long as you don't leave your answer to an absurd amount of decimal places, the examiner shouldn't be too harsh. Two or three significant figures are usually fine.

Nope, that's for chem.

In physics, you put in all numbers before the decimal point as accurately as possible (ie use unrounded figures), and after the decimal point, do what's reasonable in your opinion.

Of course, scientific notation can be used when very large numbers are present

To show what I'm saying:

Q1: calculate the acceleration of a car travelling around a (circular) roundabout with a radius of 9m at a speed of 15 ms^-1
a=v^2/r=225/9=25 ms^-2 (not 20 or 30)

Q2 a person accelerates from rest to 10ms^-1 in 3 seconds. Calculate its acceleration
u=0, v=10, t=3, a=?
v=u+at
10=3a
a=3.333333333
You should probably say a=3.3 ms^-2, or perhaps 3.33 ms^-2 as they are appropriate

Q3 People on the Earth's surface experience an acceleration due to gravity of 10 ms^-2. The radius of the earth is 6.37*10^6 m. Calculate the mass of the earth

g=GM/R^2
10=1.64*10^-24 M
M=6.08*10^24 kg
Title: Re: VCE Physics Question Thread!
Post by: Maths Forever on February 12, 2015, 08:02:13 am
Nope, that's for chem.

In physics, you put in all numbers before the decimal point as accurately as possible (ie use unrounded figures), and after the decimal point, do what's reasonable in your opinion.

Of course, scientific notation can be used when very large numbers are present

To show what I'm saying:

Q1: calculate the acceleration of a car travelling around a (circular) roundabout with a radius of 9m at a speed of 15 ms^-1
a=v^2/r=225/9=25 ms^-2 (not 20 or 30)

Q2 a person accelerates from rest to 10ms^-1 in 3 seconds. Calculate its acceleration
u=0, v=10, t=3, a=?
v=u+at
10=3a
a=3.333333333
You should probably say a=3.3 ms^-2, or perhaps 3.33 ms^-2 as they are appropriate

Q3 People on the Earth's surface experience an acceleration due to gravity of 10 ms^-2. The radius of the earth is 6.37*10^6 m. Calculate the mass of the earth

g=GM/R^2
10=1.64*10^-24 M
M=6.08*10^24 kg

Like I said, 2 to 3 significant figures should be fine. But it is important that the answer is not left with more figures than what the information provides. If the answer does not exceed three significant figures (e.g. 450 m), just leave it as it is.

I.e. With your question 2, 3.3 m/s or 3.33 m/s would be appropriate. Examiners are not too strict on that. Like Kel9901 said, this is more of a concern for Chemistry. Don't worry too much for Physics!

Title: Re: VCE Physics Question Thread!
Post by: JackSonSmith on February 12, 2015, 06:31:27 pm
Is work done force x displacement or force x distance?
Title: Re: VCE Physics Question Thread!
Post by: Cosec on February 12, 2015, 07:56:41 pm
Is work done force x displacement or force x distance?

Force x distance.
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on February 12, 2015, 10:37:31 pm
Is work done force x displacement or force x distance?

Technically it's a dot product between the force and displacement vectors; think of it as the force times the component of the displacement in the direction of the force.
Title: Re: VCE Physics Question Thread!
Post by: Cosec on February 14, 2015, 02:48:20 pm
Can someone explain this to me.

Say if we have to forces acting
The weight and the normal force.
We can setup a equation like so (from my knowledge) - taking down as negative
-Fw = Fn
Or
Fw=-Fn
Net force = Fn + Fw
And if given a net force we can solve for the normal force acting.

Like wise with friction and the pulling motion of a subject (the horizontal component) - take left as positive
Fd=-Ff
-Fd=Ff

So can some explain why this same approach isnt set out in the Vtextbook video for circular motion?
Have they just not taken into account direction when doing the forces.

Or have i totally just screwed up with the theory?

Title: Re: VCE Physics Question Thread!
Post by: Alwin on February 14, 2015, 06:25:04 pm
Can someone explain this to me.

Say if we have to forces acting
The weight and the normal force.
We can setup a equation like so (from my knowledge) - taking down as negative
-Fw = Fn
Or
Fw=-Fn
Net force = Fn + Fw
And if given a net force we can solve for the normal force acting.

Like wise with friction and the pulling motion of a subject (the horizontal component) - take left as positive
Fd=-Ff
-Fd=Ff

So can some explain why this same approach isnt set out in the Vtextbook video for circular motion?
Have they just not taken into account direction when doing the forces.

Or have i totally just screwed up with the theory?

Hi Cosec :) Nice to see someone using VT :P

For your first equation, let's assume that we have a box resting on a table. That way we have a zero net force, and there only two forces that act on the box are the weight force, FW and FN.
As forces are vectors, we should put tilde, use arrows or use boldface to show that (see above). Now, if we don't assign a direction as being positive, then yes of course we can write:
$\overrightarrow{\Sigma F} = \overrightarrow{F_\text{N}} + \overrightarrow{F_\text{W}}$
Since we know that the net force is zero, we can let the left hand side be zero and move FN to the other side:
$\overrightarrow{0} &= \overrightarrow{F_\text{N}} + \overrightarrow{F_\text{W}}$
$\Leftrightarrow -\overrightarrow{F_\text{N}} &= \overrightarrow{F_\text{W}}$
^ which is the set of equations you got.

Now, the alternative method is to take into account the directions first, and then work with just magnitudes (which is what's done throughout the VT vids, as working with vector notation usually confuses the hell out of ppl not doing spesh).
1) Let up be positive for the box example
2) Now, 'convert' the vectors
FW equals to a force with magnitude of FW (note no more boldface) downwards. ie: $\overrightarrow{F_\text{W}} = -F_\text{W}$ (note how the arrow disappears)
FN equals to a force with magnitude of FN upwards. ie: $\overrightarrow{F_\text{N}} = +F_\text{N}$
3) So our equation for the net force becomes, using only magnitudes since we have already taken into account the directions
$\Sigma F = (+F_\text{N}) + (- F_\text{W})$
$\Sigma F = F_\text{N} - F_\text{W}$
When this is rearranged, and we sub in zero for the net force, we would get:
$F_\text{N} = F_\text{W}$ ... notice the difference in sign because we've already taken into account the direction, whereas in the first example we were still working with vectors!

As for the circular motion in VT vid, let's look at the diagram on the left:
1) Choose on which direction is positive
Take up as positive
2) Look at the forces and add the sign by looking at the direction
Normal force acts up, so it is negative: - N
Weight force acts down, so it is positive: + W
Net force acts down, so it is positive: + Fnet
3) Write the equation of motion
Fnet = W - N
This is what it would look like if we used vectors
$\overrightarrow{\Sigma F} = \overrightarrow{F_\text{N}} + \overrightarrow{F_\text{W}}$

Apologies if this wasn't clear on the video, it's been quite a while haha. Hope it makes sense now :)
Title: Re: VCE Physics Question Thread!
Post by: JackSonSmith on February 15, 2015, 11:14:58 am
A bicycle accelerates from rest, covering 16m in 4s. The total mass of the bicycle and its rider is 90kg. What is its average acceleration?
Title: Re: VCE Physics Question Thread!
Post by: Gentoo on February 15, 2015, 11:52:19 am
A bicycle accelerates from rest, covering 16m in 4s. The total mass of the bicycle and its rider is 90kg. What is its average acceleration?

s= 16   t=4    u=0    a=?

Sub into s=ut+1/2at^2  to find a.
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on February 15, 2015, 12:40:22 pm
A bicycle accelerates from rest, covering 16m in 4s. The total mass of the bicycle and its rider is 90kg. What is its average acceleration?

The important bit is not to use the right formula, but to know WHY you need to use that formula.
Here you're given the distance, time and initial velocity (0) and you want the acceleration. The only irrelevant piece of information is the final velocity, so you use the one constant acceleration equation that doesn't involve the final velocity. That's why you use x = 1/2 at^2 + ut
Title: Re: VCE Physics Question Thread!
Post by: Floatzel98 on February 15, 2015, 08:21:29 pm
Hey guys. I just started gravity and i'm already stuck on this question 3a. What does in terms of the radius on the moon mean. Am i supposed to know the radius of the moon already from my textbook?
The force is 4x weaker at the new radius so the radius is 4x greater? Does this mean its a variation question. Am i supposed to introduce a constant to solve it?

Spoiler
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on February 15, 2015, 08:57:44 pm
Hey guys. I just started gravity and i'm already stuck on this question 3a. What does in terms of the radius on the moon mean. Am i supposed to know the radius of the moon already from my textbook?
The force is 4x weaker at the new radius so the radius is 4x greater? Does this mean its a variation question. Am i supposed to introduce a constant to solve it?

Spoiler

The gravitational force depends on 1/r^2, so if the force has quartered, the radius has doubled
Title: Re: VCE Physics Question Thread!
Post by: Floatzel98 on February 15, 2015, 09:48:20 pm
So because I know how much the force has decreased by, I can then just solve it like so:

$\frac{1}{4}=\frac{1}{R^2}$

$R^2=4$

$R=2$

? And also how do you do 3b then?
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on February 15, 2015, 10:20:31 pm
More like

F = k/r^2
F1/F2 = (r2/r1)^2
Force ratio is 4 as the second force F2 is a quarter of the original one, so 4 = (r2/r1)^2
r2/r1 = 2, r2  = 2 r1

As for the next one, three Moon radii above the surface = four Moon radii from the centre of the Moon => quadruple the radius => force shrinks to 1/16 of original
Title: Re: VCE Physics Question Thread!
Post by: orgekas on February 17, 2015, 04:37:43 pm
Hi guys,

I've got the following investigation(see photo) coming up and I have no idea how to do it.
Any ideas?
The aim is to design an experiment that will prove the formula F= mv^2/r for a body moving in a uniform circle.
The teacher said that we should use graphs.

Thank you.
Title: Re: VCE Physics Question Thread!
Post by: odeaa on February 17, 2015, 09:10:19 pm
Need help with this question, the 'worked' solutions are useless.
I also don't understand the whole dilation of gravity thing
Thanks

(http://i.imgur.com/gl96p42.jpg)
Title: Re: VCE Physics Question Thread!
Post by: Stevensmay on February 18, 2015, 12:13:08 am
Need help with this question, the 'worked' solutions are useless.
I also don't understand the whole dilation of gravity thing
Thanks

(http://i.imgur.com/gl96p42.jpg)

Torque is defined as a force multiplied by the perpendicular distant from that point to the force. (More technically it's the cross product of radius and force, this is why it's perpendicular). In the real world it's effectively a force that makes something want to rotate. In the picture the 20kg force wants to make the beam rotate around point P.

In our case the force from the 20kg mass is 20*9.8N (mg).

(http://i.imgur.com/HJT3vBT.png)

So we've decided that a torque is just a measure of how much something wants to rotate. Higher torque means something is wanting to rotate more. In the image above, would the left hand side 20kg mass produce a torque smaller than, equal to or greater than the torque on the right hand side?

Spoiler
Equal to, because only the perpendicular distance matters and it is the same in each diagram. Similarly, if we put the 20kg force above point P it would have a torque of 0, as it is not trying to rotate the beam.

So we now what our perpendicular distance is, it's just a matter of going T = r*F to get an answer of (hopefully, if I read the question right) 588Nm.

If this doesn't make any sense my apologies.

With the second part, I can't remember if you do summation of torques and forces.
Title: Re: VCE Physics Question Thread!
Post by: Floatzel98 on February 20, 2015, 04:00:50 pm
I found 7a, but don't know what to do for 7b or 8.

Spoiler
(https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-xpf1/v/t34.0-12/11009026_648111405317277_42696173_n.jpg?oh=f18f9743dda9710cc67b8e13a3037e4e&oe=54E8B477&__gda__=1424530282_c625470941f4d77c7d98f714267bf350)
Title: Re: VCE Physics Question Thread!
Post by: Kel9901 on February 20, 2015, 06:15:07 pm
I found 7a, but don't know what to do for 7b or 8.

Spoiler
(https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-xpf1/v/t34.0-12/11009026_648111405317277_42696173_n.jpg?oh=f18f9743dda9710cc67b8e13a3037e4e&oe=54E8B477&__gda__=1424530282_c625470941f4d77c7d98f714267bf350)

Let the distance that must be calculated be r
m is the mass of the asteroid

F(asteroid, Alpha)=GMm/r^2
F(asteroid, Beta)=G(0.01M)m)/(R-r)^2

F(asteroid, Alpha)/F(asteroid, Beta)=100(R-r)^2/r^2=8100
(R-r)^2/r^2=81
(R-r)/r=9 (remember that r and R are both positive, and that R>r)
R-r=9r
10r=R
r=0.1R
Title: Re: VCE Physics Question Thread!
Post by: Floatzel98 on February 20, 2015, 07:21:57 pm
Thank you. Makes a lot more sense now
Title: Re: VCE Physics Question Thread!
Post by: Kel9901 on February 20, 2015, 08:54:13 pm
For question 8:
GMm/r^2=G(0.01M)m/(R-r)^2
1/r^2=0.01/(R-r)^2
100/r^2=1/(R-r)^2
r^2/100=(R-r)^2
r/10=R-r
1.1r=R
r=0.91R
Title: Re: VCE Physics Question Thread!
Post by: lolaishappy on February 21, 2015, 02:53:06 pm
Can anyone confirm answers to questions on the 2014 vcaa exam for Q2 about the spring?
Title: Re: VCE Physics Question Thread!
Post by: Kel9901 on February 22, 2015, 08:25:18 am
Can anyone confirm answers to questions on the 2014 vcaa exam for Q2 about the spring?
Yep, I did it last year and got it right (acc. to statement of marks)

a) k=F/x=mg/x pretty simple show that

b) 80 cm

c) Kinetic energy not included

d) 2 ms^-1
Title: Re: VCE Physics Question Thread!
Post by: lolaishappy on February 22, 2015, 08:32:50 am
Yep, I did it last year and got it right (acc. to statement of marks)

a) k=F/x=mg/x pretty simple show that

b) 80 cm

c) Kinetic energy not included

d) 2 ms^-1
Thank you :)
Title: Re: VCE Physics Question Thread!
Post by: paper-back on February 25, 2015, 08:11:34 pm
For questions such as the following:
A person pushes a 14.5KG mower at a constant speed with a force of 80N directed along the handle, which is at an angle of 45 degrees to the horizontal
(A) Find the normal force

Do we need to go; .
9.8x14.5+sin(45)x80?
Title: Re: VCE Physics Question Thread!
Post by: Cosec on February 25, 2015, 08:40:06 pm
For questions such as the following:
A person pushes a 14.5KG mower at a constant speed with a force of 80N directed along the handle, which is at an angle of 45 degrees to the horizontal
(A) Find the normal force

Do we need to go; .
9.8x14.5+sin(45)x80?

Normal force always acts perpendicular to the direction of motion. In this case, if its traveling along flat ground, the net force vertically is 0 as the mower isnt moving up or down, but rather left and right. Thus, using f=mg, f=14.5x10 (take gravity as 10m/s^2) gives us 145N (force of gravity acting downwards). Thus the reaction (normal) force is 145N in the opposite direction (acting upwards). In this case, the ground is pushing the mower with a force of 145N. Check the answers because i might of interpreted the question wrong (next time post a photo of the actual whole question with diagram if there is, makes it a bit easier).
Title: Re: VCE Physics Question Thread!
Post by: paper-back on February 25, 2015, 08:57:04 pm
Normal force always acts perpendicular to the direction of motion. In this case, if its traveling along flat ground, the net force vertically is 0 as the mower isnt moving up or down, but rather left and right. Thus, using f=mg, f=14.5x10 (take gravity as 10m/s^2) gives us 145N (force of gravity acting downwards). Thus the reaction (normal) force is 145N in the opposite direction (acting upwards). In this case, the ground is pushing the mower with a force of 145N. Check the answers because i might of interpreted the question wrong (next time post a photo of the actual whole question with diagram if there is, makes it a bit easier).
Thanks Cosec.
That's what I thought, however I believe the answer for the question adds in the verticle component of the force being exerted on the mower for the verticle force as it's close to 200
Title: Re: VCE Physics Question Thread!
Post by: Cosec on February 25, 2015, 09:21:49 pm
Thanks Cosec.
That's what I thought, however I believe the answer for the question adds in the verticle component of the force being exerted on the mower for the verticle force as it's close to 200

Hmm, i thought of that too, but i dont see why it would be added and ive never encountered a question that takes it into consideration. But if thats the answer, i guess thats the way they expect you to work it out in which case you just find the vertical component and add it to the weight force. Someone else with a bit more experience might have to chime in.
Title: Re: VCE Physics Question Thread!
Post by: Conic on February 25, 2015, 09:46:52 pm
You do have to consider the component of the pushing force on the lawnmower. If the normal force simply cancelled out the force of gravity, the net force in the vertical direction would be the downward component of the pushing force. This would mean the lawnmower accelerates into the ground, but this is clearly not the case.

What do we know in this situation? We know the gravitational force, and we can work out downward component of the pushing force. We need to relate the normal force to the things we already know. Using Newton's Second Law, we know that the net vertical force is 0, as there is no acceleration in this direction. The normal force is in the opposite direction to the weight force and the pushing force, so

$F_{\mathrm{Normal}}=F_{\mathrm{Push}}+F_{\mathrm{Weight}} .$

This gives

$F_{\mathrm{Normal}} = 80\mathrm{N} \times \sin(45^{\circ})+10\mathrm{ms}^{-2}\times15.5\mathrm{kg} \approx 202\mathrm{N}.$
Title: Re: VCE Physics Question Thread!
Post by: lolaishappy on February 26, 2015, 05:22:11 pm
The average speed for the uphill part of the journey is 4ms^1 while the average speed for travelling the same distance for downhill is 8ms^1. What is the average speed of the rider for whole journey. You may ignore time taken to turn around. Theres an algebraic method, in letting distances equal right, but from there i get confused
Title: Re: VCE Physics Question Thread!
Post by: lolaishappy on February 26, 2015, 05:22:31 pm
Acceleration is 0
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on February 26, 2015, 05:42:55 pm
The average speed for the uphill part of the journey is 4ms^1 while the average speed for travelling the same distance for downhill is 8ms^1. What is the average speed of the rider for whole journey. You may ignore time taken to turn around. Theres an algebraic method, in letting distances equal right, but from there i get confused

Let the distance be x. Work out the time travelled going up and down hill in terms of x, find the total distance and total time to work out the average speed.
Title: Re: VCE Physics Question Thread!
Post by: lolaishappy on February 26, 2015, 08:17:02 pm
Let the distance be x. Work out the time travelled going up and down hill in terms of x, find the total distance and total time to work out the average speed.
I tried to find time as 4t=x and 8t=x however i get t=0?
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on February 26, 2015, 10:44:31 pm
The two times clearly aren't the same as the speeds aren't the same. Hence you need x = 4t1 = 8t2
Total time = t1 + t2, total distance = 2x
So average speed should be 2x / (t1 + t2)
Simplify with what you have now.
Title: Re: VCE Physics Question Thread!
Post by: Floatzel98 on March 02, 2015, 08:21:06 pm
Can someone help me with question 5. This is from the detailed study materials and structures, Young's Modulus.
Title: Re: VCE Physics Question Thread!
Post by: odeaa on March 02, 2015, 08:38:48 pm
Can someone help me with question 5. This is from the detailed study materials and structures, Young's Modulus.
A strain percentage of 0.075% is equal to a strain of 0.00075. Also, note that while I don't convert the cm measurements to meters, the output is also in cm.
$\epsilon=\frac{\Delta l}{l}$

$0.00075=\frac{\Delta l}{8cm}$

$\Delta l=8*0.00075$

$\Delta l=0.006cm$

$\therefore length=8.006cm$

I hope that was correct ahah!
Title: Re: VCE Physics Question Thread!
Post by: Floatzel98 on March 02, 2015, 09:22:32 pm
Thank you :) you were correct haha. Also what detailed study do most schools do? My teacher says everyone does materials and structures because it is the easiest.
Title: Re: VCE Physics Question Thread!
Post by: odeaa on March 02, 2015, 09:26:41 pm
Thank you :) you were correct haha. Also what detailed study do most schools do? My teacher says everyone does materials and structures because it is the easiest.
I think the examiners report tells you
But yeah definately structures, it's so straightforward.
Title: Re: VCE Physics Question Thread!
Post by: Kel9901 on March 03, 2015, 09:22:23 am
I think the examiners report tells you
But yeah definately structures, it's so straightforward.

yeah either structure and sound (what i did) are the easiest afaik
Title: Re: VCE Physics Question Thread!
Post by: StressedAlready on March 03, 2015, 05:53:55 pm
Could someone please explain to me how the equation for resistance total for parallel circuits proves the validity of Kirchoff's Law?
Title: Re: VCE Physics Question Thread!
Post by: paper-back on March 03, 2015, 09:15:36 pm
When finding the kinetic energy of a projectile in motion do we use the horizontal component, verticle component or the resultant vector component?
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on March 03, 2015, 09:49:44 pm
Could someone please explain to me how the equation for resistance total for parallel circuits proves the validity of Kirchoff's Law?

I'm not sure if the resistance equation establishes Kirchhoff's law or the other way around...but anyway

If you have two resistors in parallel, the total resistance satisfies 1/Reff = 1/R1 + 1/R2 where R1, R2 are the resistances of each parallel part.
The total current is therefore V/Reff = V/R1 + V/R2 = I1 + I2
This shows that it's valid to say that the current after the junction is the sum of the currents before the junction. Also, I've used the same letter V for both resistors; it's clearly also valid to say that both parts of the circuit are at the same voltage.

When finding the kinetic energy of a projectile in motion do we use the horizontal component, verticle component or the resultant vector component?

KE = 1/2*mass*(speed)^2. Find the speed. It's not a vector.
Title: Re: VCE Physics Question Thread!
Post by: paper-back on March 03, 2015, 10:27:16 pm
So is the speed the resultant component?
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on March 03, 2015, 11:01:10 pm
Speed = size of velocity
Title: Re: VCE Physics Question Thread!
Post by: orgekas on March 04, 2015, 12:55:44 pm
Hi guys,

So we did this prac in physics for circular motion. The aim was to derive the formula F=mv^2/r. The apparatus we used contained paper clip, tube, fishing line, rubber stoper and slotted masses. So what we did with my partner was  measure the time and record it for the rubber stopper to make 10 revolutions and then keep radius constant change mass and calculate the velocity. Then, we repeated the same thing instead we changed the radius and kept the mass constant.
And now we have to use graphs to show that F=mv^2/r. We know that Fc=ma=F by the hanging masses but we dont know what graphs to draw in order to prove the formula.

Thanks.
Title: Re: VCE Physics Question Thread!
Post by: Stevensmay on March 04, 2015, 01:29:23 pm
Hi guys,

So we did this prac in physics for circular motion. The aim was to derive the formula F=mv^2/r. The apparatus we used contained paper clip, tube, fishing line, rubber stoper and slotted masses. So what we did with my partner was  measure the time and record it for the rubber stopper to make 10 revolutions and then keep radius constant change mass and calculate the velocity. Then, we repeated the same thing instead we changed the radius and kept the mass constant.
And now we have to use graphs to show that F=mv^2/r. We know that Fc=ma=F by the hanging masses but we dont know what graphs to draw in order to prove the formula.

Thanks.

I'm not too sure how to prove it with a graph (I've never seen anything proved like that before).

I'm happy to try writing up a derivation but it involves angular velocity/acceleration and I can't remember whether this is in the course. And a bit of calculus (basic).
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on March 04, 2015, 02:49:30 pm
Hi guys,

So we did this prac in physics for circular motion. The aim was to derive the formula F=mv^2/r. The apparatus we used contained paper clip, tube, fishing line, rubber stoper and slotted masses. So what we did with my partner was  measure the time and record it for the rubber stopper to make 10 revolutions and then keep radius constant change mass and calculate the velocity. Then, we repeated the same thing instead we changed the radius and kept the mass constant.
And now we have to use graphs to show that F=mv^2/r. We know that Fc=ma=F by the hanging masses but we dont know what graphs to draw in order to prove the formula.

Thanks.

Graph F against v^2. You should get a straight line of slope m/r
Title: Re: VCE Physics Question Thread!
Post by: orgekas on March 04, 2015, 03:03:49 pm
Graph F against v^2. You should get a straight line of slope m/r

If you can see the attachment those are the results that I have. And the graph that Excel produces. Is that right? If yes then what?
Title: Re: VCE Physics Question Thread!
Post by: JackSonSmith on March 04, 2015, 04:08:44 pm
We did a prac about friction on inclined planes. We got the length of hypotenuse (1m), 3 angles (0,12.7 and 24.8 degrees) and 3 times (0.81,1.07 and 1.17) respectively.

A 600g trolley was pulled by a 400g weight.

My calculation for 0 degrees, was friction = 0.95N.

I used constant acceleration formula to calculate acceleration to be 3.05.
Then
Fnet = Fapplied - Fopposing
3.05 = 4 - friction.
Friction = .95N

I have had trouble finding the friction when the plane was inclined. Do I have to somehow factor in the force down the plane due to the component of weight down? Could someone help please?
Title: Re: VCE Physics Question Thread!
Post by: Kel9901 on March 04, 2015, 05:03:58 pm
We did a prac about friction on inclined planes. We got the length of hypotenuse (1m), 3 angles (0,12.7 and 24.8 degrees) and 3 times (0.81,1.07 and 1.17) respectively.

A 600g trolley was pulled by a 400g weight.

My calculation for 0 degrees, was friction = 0.95N.

I used constant acceleration formula to calculate acceleration to be 3.05.
Then
Fnet = Fapplied - Fopposing
3.05 = 4 - friction.
Friction = .95N

I have had trouble finding the friction when the plane was inclined. Do I have to somehow factor in the force down the plane due to the component of weight down? Could someone help please?

You need to take into account the force down the plane due to, as you said, the weight force, and that's the only difference. What you've done so far is correct. I won't do the calculations for you (since it's your prac), but expect lower friction forces
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on March 04, 2015, 09:22:06 pm
If you can see the attachment those are the results that I have. And the graph that Excel produces. Is that right? If yes then what?

Something like that. Get Excel to display the line equation to get the slope. Right click the trendline.
Title: Re: VCE Physics Question Thread!
Post by: chekside on March 06, 2015, 10:40:22 pm
I found a question that dealt with an IR emitter and an IR phototransistor receiver. The question asked if the voltage drop across the receiver would increase or decrease when the IR emitter was on. The answer stated that it would decrease, but didn't give an explanation, could someone please explain why the voltage drop would decrease when the IR began hitting the receiver?
Thanks
Title: Re: VCE Physics Question Thread!
Post by: JackSonSmith on March 07, 2015, 10:14:54 pm
A geosynchronous satellite is one with a period of 24h positioned exactly above the equator. It appears motionless viewed from the surface of Earth. Explain why it must be in an orbit above the equator.

Two questions. 1: How important is knowing this theory. 2: Could someone explain in layman's terms what is going on.

Thanks.
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on March 07, 2015, 11:47:40 pm
I found a question that dealt with an IR emitter and an IR phototransistor receiver. The question asked if the voltage drop across the receiver would increase or decrease when the IR emitter was on. The answer stated that it would decrease, but didn't give an explanation, could someone please explain why the voltage drop would decrease when the IR began hitting the receiver?
Thanks

Not part of the course I'm sure. Don't need to know about transistors

A geosynchronous satellite is one with a period of 24h positioned exactly above the equator. It appears motionless viewed from the surface of Earth. Explain why it must be in an orbit above the equator.

Two questions. 1: How important is knowing this theory. 2: Could someone explain in layman's terms what is going on.

Thanks.
1. I guess it's unlikely VCAA will test this
2. Any orbit must have the centre of the Earth as the centre of the orbit. If you're not over the equator, you will still orbit around the centre of the Earth so your orbit will be lopsided.
Title: Re: VCE Physics Question Thread!
Post by: stockstamp on March 08, 2015, 12:18:00 am
Hoping someone can clarify this:

If I drop an object why does it stop when it hits the ground? I'm not challenging whether or not it should, but If I was asked this question in a test I doubt I could answer it satisfactorily.

I know that every action creates and equal and opposite reaction, but I don't understand how the size of the normal force can change during the contact time, and how equal and opposite reactions can cause a decelleration as a result of that.

Not sure if I've expressed my self properly here, but I hope someone follows my train of thought...

Thanks
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on March 08, 2015, 10:52:12 am
Hoping someone can clarify this:

If I drop an object why does it stop when it hits the ground? I'm not challenging whether or not it should, but If I was asked this question in a test I doubt I could answer it satisfactorily.

I know that every action creates and equal and opposite reaction, but I don't understand how the size of the normal force can change during the contact time, and how equal and opposite reactions can cause a decelleration as a result of that.

Not sure if I've expressed my self properly here, but I hope someone follows my train of thought...

Thanks

Equal and opposite forces act on DIFFERENT OBJECTS. That's why a deceleration can result. If you had the equal but opposite forces acting on the same object, then of course no net force would result.

As for why objects stop when they hit the ground, it's because collisions with the ground are inelastic and the object tends to lose its kinetic energy thermally.
Title: Re: VCE Physics Question Thread!
Post by: Adequace on March 08, 2015, 09:30:43 pm
Not sure where to post this but, I'm undertaking physics 1/2 a year earlier this year and want to build a solid foundation for 3/4. I'm aware of that electricity and motion overlap in to 3/4 but I'm not sure what else I can do to prepare myself for 3/4.

I'm looking to get at least 40, or so. What else can I do to prepare better? What test percentages should I be aiming to get? What should I do right now since I have more time?

It might be worth nothing that my confidence for physics is kind of low...I struggled with nuclear physics and will be lucky to get at least 80% on my outcome test when we get our results back.
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on March 13, 2015, 10:56:26 am
Not sure where to post this but, I'm undertaking physics 1/2 a year earlier this year and want to build a solid foundation for 3/4. I'm aware of that electricity and motion overlap in to 3/4 but I'm not sure what else I can do to prepare myself for 3/4.

I'm looking to get at least 40, or so. What else can I do to prepare better? What test percentages should I be aiming to get? What should I do right now since I have more time?

It might be worth nothing that my confidence for physics is kind of low...I struggled with nuclear physics and will be lucky to get at least 80% on my outcome test when we get our results back.

luckily nuclear physics is entirely irrelevant to units 3 and 4 (except the occasional non-VCAA practice paper where they assume you know the charge on an alpha particle is +2e but yeah)

forming solid foundations for basic electrical and mechanical concepts should be your goal in year 11 because there's plenty of new stuff that you'll have to wait until year 12 to learn, both content and problem-solving skills.

to help with this, i highly recommend khanacademy's physics playlists, the first 5 of which are relevant to VCE motion and the second last (electricity and magnetism) is relevant to that. This is a resource I regularly go to for a clear and comprehensive explanation of a physics concept! it can't hurt to give it a try :)
[EDIT: I should add, I'm currently watching the thermodynamics playlist because second year thermal physics isn't taught in a very organised way at the university of melbourne, and salman khan has everything sorted. khanacademy is also great for maths and other sciences]

the other thing you can do is work at improving your algebra skills (for VCE physics, algebra is as far as you need to go, no need for calculus). I'm talking about notation, solving for x, fractions skills and mental arithmetic. Being good at maths makes physics problems waaaay more fun than if you're not a maths-y student and you kinda get the physics but you can't follow the working out. Being clear and formal in your working forces you to think methodically and improve your problem solving skills.

By improving your physics intuition and maths skills now you're forming a really really good starting point for year 12 studies!
i mean, you can work ahead if you like, but you'll have plenty of time in year 12 to learn the year 12 content. If you focus on the foundations now you'll be set when things get far busier next year.
also don't worry about test scores, seriously the only thing that matters at this point is your intuition :)
Title: Re: VCE Physics Question Thread!
Post by: knightrider on March 16, 2015, 01:31:27 am
How do you get from 16KWh to 5.76*10^7 J
Title: Re: VCE Physics Question Thread!
Post by: nerdgasm on March 16, 2015, 02:00:21 am
16KWh means "sixteen kilowatt-hours" or "sixteen thousand watt-hours". What precisely is a 'watt-hour'?

Consider that the watt (W) is a unit of power. One formula for power is Power = Energy/Time. Specifically, 1 Watt = 1 Joule/second. This unit is useful when we want to talk about the energy output of something (such as a lightbulb, you may have noticed things like "60W" or "40W" on lightbulb cartons). Now that we know what a Watt (pardon the pun) is, let's think about what a "watt-hour" is. A 'watt-hour' is the amount of energy that a 1W object would produce in an hour. In other words, this would be (1 Joule/sec) * (3600sec) = 3600 Joules. In short, a "Watt-hour" is a unit of energy, and it is equivalent to 3600J. If you like, you can think of this as the result of multiplying the units 'watt' and 'hour' together.

Therefore, 16KWh = 16000Watt-hours = 16000*3600J = 57600000J = 5.76*10^7 J .
Title: Re: VCE Physics Question Thread!
Post by: knightrider on March 16, 2015, 02:11:00 am
16KWh means "sixteen kilowatt-hours" or "sixteen thousand watt-hours". What precisely is a 'watt-hour'?

Consider that the watt (W) is a unit of power. One formula for power is Power = Energy/Time. Specifically, 1 Watt = 1 Joule/second. This unit is useful when we want to talk about the energy output of something (such as a lightbulb, you may have noticed things like "60W" or "40W" on lightbulb cartons). Now that we know what a Watt (pardon the pun) is, let's think about what a "watt-hour" is. A 'watt-hour' is the amount of energy that a 1W object would produce in an hour. In other words, this would be (1 Joule/sec) * (3600sec) = 3600 Joules. In short, a "Watt-hour" is a unit of energy, and it is equivalent to 3600J. If you like, you can think of this as the result of multiplying the units 'watt' and 'hour' together.

Therefore, 16KWh = 16000Watt-hours = 16000*3600J = 57600000J = 5.76*10^7 J .

Thanks so much nerdgasm  :)
Title: Re: VCE Physics Question Thread!
Post by: lolaishappy on March 17, 2015, 06:01:51 pm
When finding tension in rope or rod of a object going in circular motion, why do some questions consider the weight force in finding T when others don't? And what's the difference between vertical and horizontal motion?
Title: Re: VCE Physics Question Thread!
Post by: Zealous on March 17, 2015, 06:35:01 pm
When finding tension in rope or rod of a object going in circular motion, why do some questions consider the weight force in finding T when others don't? And what's the difference between vertical and horizontal motion?

This image below might help:

(http://www.ic.sunysb.edu/Class/phy141md/lib/exe/fetch.php?media=phy141:lectures:ballonstring.png)

When we're looking at an object in circular motion we've got to consider all the forces acting on it - there's going to be gravity always acting downwards and a tension force which needs to apply an inwards force to the object so that it has a centripetal acceleration. At the bottom of the circle, the tension in the rope has to be able to overcome the force of gravity on the object, and then have the required extra force for the centripetal acceleration, that's why we have $T=\frac{mv^2}{r}+mg$ at the bottom. At the top of the circle, gravity is working in the direction required for centripetal acceleration, so the tension is decreased as dictated by $T=\frac{mv^2}{r}-mg$. If you're looking at the tension when the object is on the edges, we don't need to consider the weight force as it is acting perpendicular to the tension in the rope so there's no component of the weight force working against the tension.

Horizontal circular motion questions usually look like this:

(http://www.a-levelmathstutor.com/images/kinetics/kin-conpend.jpg)

The main difference is that the forces just work in different directions, and there are sometimes angles involved, but you can resolve them using simple trigonometry.
Title: Re: VCE Physics Question Thread!
Post by: lolaishappy on March 17, 2015, 06:41:50 pm
This image below might help:

(http://www.ic.sunysb.edu/Class/phy141md/lib/exe/fetch.php?media=phy141:lectures:ballonstring.png)

When we're looking at an object in circular motion we've got to consider all the forces acting on it - there's going to be gravity always acting downwards and a tension force which needs to apply an inwards force to the object so that it has a centripetal acceleration. At the bottom of the circle, the tension in the rope has to be able to overcome the force of gravity on the object, and then have the required extra force for the centripetal acceleration, that's why we have $T=\frac{mv^2}{r}+mg$ at the bottom. At the top of the circle, gravity is working in the direction required for centripetal acceleration, so the tension is decreased as dictated by $T=\frac{mv^2}{r}-mg$. If you're looking at the tension when the object is on the edges, we don't need to consider the weight force as it is acting perpendicular to the tension in the rope so there's no component of the weight force working against the tension.

Horizontal circular motion questions usually look like this:

(http://www.a-levelmathstutor.com/images/kinetics/kin-conpend.jpg)

The main difference is that the forces just work in different directions, and there are sometimes angles involved, but you can resolve them using simple trigonometry.
Thanks zealous, what happens in between the points of the top diagram?
Title: Re: VCE Physics Question Thread!
Post by: Kel9901 on March 17, 2015, 08:01:51 pm
Thanks zealous, what happens in between the points of the top diagram?

From the study design
Quote
apply Newton's second law to circular motion in a vertical plane; consider forces at the highest and lowest positions only

Title: Re: VCE Physics Question Thread!
Post by: lolaishappy on March 17, 2015, 08:04:26 pm
From the study design
Thank you kel9901!
Title: Re: VCE Physics Question Thread!
Post by: lolaishappy on March 22, 2015, 09:29:51 pm
How do we distinguish between v and u, and which to put in a equation in projectile motion?
In general I'm struggling to know which equations to apply to different questions :^(
Title: Re: VCE Physics Question Thread!
Post by: Floatzel98 on March 22, 2015, 11:01:25 pm
How do we distinguish between v and u, and which to put in a equation in projectile motion?
In general I'm struggling to know which equations to apply to different questions :^(
The simplest thing i do to figure out which equation I should use is to write out all the values for u,v,x,t,a that have been given in the question and also find the one that i need to solve for. Match the equation on the information given. If you aren't already given a diagram, try to draw one and try to draw the values on the diagram to get a sense of where each part applies. You are always given the velocity at the max height and the acceleration (vertical component) so if you split the parabola that makes up the projectile motion into 2, you will always have a final velocity if you use the first half. Then depending on what the questions asks, you can solve from there. You can always solve for initial velocity  if a height or time is given from this point.

Sorry if i am not too clear or anything. It's my first time actually trying to help someone on here. Usually i just get help haha. I can try to explain it a bit more if you need me to or if no one else tries to help.
Title: Re: VCE Physics Question Thread!
Post by: Cosec on March 23, 2015, 06:41:23 pm
Can somebody explain why when connected in series, a resistor and a thermistor, with the output voltage being measured across the fixed resistor which is placed second after the thermistor, that when the temp increases, so does the resistance of the thermistor and thus the output voltage increases. Im kind of confused with it all.
Title: Re: VCE Physics Question Thread!
Post by: Kel9901 on March 23, 2015, 09:13:59 pm
Can somebody explain why when connected in series, a resistor and a thermistor, with the output voltage being measured across the fixed resistor which is placed second after the thermistor, that when the temp increases, so does the resistance of the thermistor and thus the output voltage increases. Im kind of confused with it all.

I'm pretty sure that when the temperature increases, the resistance of thermistors decrease. Let R be the resistance of the normal resistor and R(th) that of the thermistor. Vout=Vin*R/R(th), so when R(th) decreases, the denominator only decreases and hence the voltage drop across the normal resistor increases

edit: fuck latex
Title: Re: VCE Physics Question Thread!
Post by: lolaishappy on March 23, 2015, 10:23:31 pm
The simplest thing i do to figure out which equation I should use is to write out all the values for u,v,x,t,a that have been given in the question and also find the one that i need to solve for. Match the equation on the information given. If you aren't already given a diagram, try to draw one and try to draw the values on the diagram to get a sense of where each part applies. You are always given the velocity at the max height and the acceleration (vertical component) so if you split the parabola that makes up the projectile motion into 2, you will always have a final velocity if you use the first half. Then depending on what the questions asks, you can solve from there. You can always solve for initial velocity  if a height or time is given from this point.

Sorry if i am not too clear or anything. It's my first time actually trying to help someone on here. Usually i just get help haha. I can try to explain it a bit more if you need me to or if no one else tries to help.
Thanks!
Title: Re: VCE Physics Question Thread!
Post by: Ha_Nguyen on March 25, 2015, 08:50:21 pm
Hi, can someone help me with question 3????

if an object of mass m is moving uniformly (constant speed, v) in a circle of radius r:
1. state the equation that relates the centripetal force, F to m,v,r

F= mv^2/r

2. state the equation that relates the period for 1 revolution (T) to v,r
T= (2 pi r )/v

3. from the two previous equations, write an equation relating T to m&r?
Title: Re: VCE Physics Question Thread!
Post by: Kel9901 on March 27, 2015, 09:40:18 am
Hi, can someone help me with question 3????

if an object of mass m is moving uniformly (constant speed, v) in a circle of radius r:
1. state the equation that relates the centripetal force, F to m,v,r

F= mv^2/r

2. state the equation that relates the period for 1 revolution (T) to v,r
T= (2 pi r )/v

3. from the two previous equations, write an equation relating T to m&r?

I think you'll need F in the equation too... or does it mean T to a and r?

Title: Re: VCE Physics Question Thread!
Post by: RedCapsicum on April 02, 2015, 11:37:26 pm
Hi guys, can someone help me with this question:

The mass of Earth is 6.0*10^24 kg and the amass of the moon is 7.4*10^22 kg. The radius of the earth is 6.4*10^6 metres and the radius of the moon is 1.7*10^6 metres. The orbital radius of the moon around the Earth taken from the centre of the Moon and the centre of the Earth is 3.8*10^8 metres.

1) Calculate the gravitational force acting between the Earth and the moon
2) Calculate to two decimal places how long it takes for the moon to orbit the Earth in days
3) Calculate the orbital speed of the moon around the earth in km/hr
Title: Re: VCE Physics Question Thread!
Post by: JackSonSmith on April 03, 2015, 08:10:02 pm
For electricity: does VIt equal energy or work?
Title: Re: VCE Physics Question Thread!
Post by: Zealous on April 03, 2015, 08:58:36 pm
Hi guys, can someone help me with this question:

The mass of Earth is 6.0*10^24 kg and the amass of the moon is 7.4*10^22 kg. The radius of the earth is 6.4*10^6 metres and the radius of the moon is 1.7*10^6 metres. The orbital radius of the moon around the Earth taken from the centre of the Moon and the centre of the Earth is 3.8*10^8 metres.

1) Calculate the gravitational force acting between the Earth and the moon
2) Calculate to two decimal places how long it takes for the moon to orbit the Earth in days
3) Calculate the orbital speed of the moon around the earth in km/hr

1) Use $F = \frac{GMm}{R^2}$, and substitute in the given values.

2) $a=\frac{4 \pi^2R}{T^2}=\frac{GM}{R^2}$ which means that $T=\sqrt{\frac{4 \pi^2R^3}{GM}$. Now you can substitute in the radius of the moons orbit around earth, the gravitational constant and the mass of earth. Remember to use the mass of the body which is being orbited around, not the mass of the body in orbit.

3) $\frac{GM}{R^2}=\frac{V^2}{R} \implies v=\sqrt{\frac{GM}{R}$. Now you can substitute in the gravitational constant, the mass of the earth and the radius of the orbit.

So with these sorts of questions, firstly write down all the variables that you are given, and then write down the ones you are trying to find. Then take the circular motion formulas and rearrange them until you find the value you are looking for. Once you've got enough practice, these questions can be done quite quickly as long as you know what value you're looking for and what formula you need to find it.
Title: Re: VCE Physics Question Thread!
Post by: dankfrank420 on April 03, 2015, 10:36:45 pm
For electricity: does VIt equal energy or work?

Aren't they the same thing in this context?
Title: Re: VCE Physics Question Thread!
Post by: JackSonSmith on April 04, 2015, 03:12:41 pm
Aren't they the same thing in this context?

I was a little confused about the difference between energy and work
Title: Re: VCE Physics Question Thread!
Post by: Kel9901 on April 04, 2015, 05:32:24 pm
I was a little confused about the difference between energy and work

work is pretty much change in energy, in this case they are the same because the 'energy' means energy dissipated anyway.
Title: Re: VCE Physics Question Thread!
Post by: dankfrank420 on April 09, 2015, 05:45:28 pm

There is a battery of 12v, a resistor of 150 ohms and a forward biased diode in circuit. The I-V characteristic of the diode says that the diode needs 0.6 v for current to flow.

Therefore, there is 11.4 volts across the resistor.

When working out the current flowing through the diode, why do we use 11.4/150 instead of 12/150? Since they're in series, shouldn't the current be constant across both the diode and the resistor? Why do we take away the 0.6v before we apply ohms law to find the current?
Title: Re: VCE Physics Question Thread!
Post by: alchemy on April 09, 2015, 09:30:59 pm

There is a battery of 12v, a resistor of 150 ohms and a forward biased diode in circuit. The I-V characteristic of the diode says that the diode needs 0.6 v for current to flow.

Therefore, there is 11.4 volts across the resistor.

When working out the current flowing through the diode, why do we use 11.4/150 instead of 12/150? Since they're in series, shouldn't the current be constant across both the diode and the resistor? Why do we take away the 0.6v before we apply ohms law to find the current?

The diode consumes 0.6V so what's left is 11.4V. This 11.4V is for the resistor. Since the current is the same for each component, you work out the current for the resistor first by doing VR/R=11.4/150=0.076A. Hence, since this is a series circuit, that same amount of current (0.076A) passes through the diode. Hope that made sense!
Title: Re: VCE Physics Question Thread!
Post by: dankfrank420 on April 09, 2015, 11:07:21 pm
Yes I understood.

However, why isn't the 0.6 v included? Isn't the current Vtotal/Rtotal?

So what you're saying is that the diode literally takes 0.6v away from the circuit for current to flow through it? But why isn't that voltage counted in the calculation of the total current?
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on April 09, 2015, 11:58:56 pm
Yes I understood.

However, why isn't the 0.6 v included? Isn't the current Vtotal/Rtotal?

So what you're saying is that the diode literally takes 0.6v away from the circuit for current to flow through it? But why isn't that voltage counted in the calculation of the total current?

Vtot/Rtot only works if every circuit component you have in your circuit is ohmic, or V/I is constant for every device you have.
Ohm's law isn't V=IR. Ohm's law is actually an assertion that V/I is a constant and devices that satisfy this assertion are ohmic, like normal resistors.

As you can quite clearly, diodes are NOT ohmic. Hence you can't just use normal voltage divider techniques to solve this question. You have to consider the voltage drop across each component separately.

This is what I don't like about VCE physics. Lots of concepts are not taught very well IMO.
Title: Re: VCE Physics Question Thread!
Post by: dankfrank420 on April 10, 2015, 01:04:14 am
Vtot/Rtot only works if every circuit component you have in your circuit is ohmic, or V/I is constant for every device you have.
Ohm's law isn't V=IR. Ohm's law is actually an assertion that V/I is a constant and devices that satisfy this assertion are ohmic, like normal resistors.

As you can quite clearly, diodes are NOT ohmic. Hence you can't just use normal voltage divider techniques to solve this question. You have to consider the voltage drop across each component separately.

Thanks! Really cleared up some conceptual misconceptions (is that a phrase?) I had.

So moral of the story is don't treat a diode like another element in the circuit because it isn't ohmic.

Quote
This is what I don't like about VCE physics. Lots of concepts are not taught very well IMO.

I find this alot.

Most teachers simply teach you the way of getting the right answer instead of the concepts behind it. Most of the course can simply be ROTE learned or worst-still copied onto the cheat sheet.
Title: Re: VCE Physics Question Thread!
Post by: Rishi97 on April 11, 2015, 04:13:29 pm
Hii guys

Does anyone want some physics exams ? If yes, just inbox me your email and I'll send them !

Rishi
Title: Re: VCE Physics Question Thread!
Post by: Floatzel98 on April 11, 2015, 10:02:02 pm
For any of the detailed studies on the exam, is it possible to get 1 mark for a question if you showed workings? I'm asking because we were doing past detailed studies exams in class and our teacher and he was making us show all of our workings for the questions on the page and saying we may get method marks for the workings out if they are correct but we get an answer wrong. It kind of makes sense because they are all out of 2 but I'm still not sure and a bit confused. Even though it might just be good practice now to show workings, i don't want to waste time on the exam showing pointless steps.
Title: Re: VCE Physics Question Thread!
Post by: Zealous on April 11, 2015, 10:19:44 pm
Nope. Not at all. The detailed study consists of 13 multiple choice questions, each worth 2 marks. It's either get 0 marks or 2 marks, and you answer on a multiple choice sheet and shade bubbles, so there's no way the computer will pick out working out.

However, older study design exams from maybe around 2000-2006 used to have sections for a detailed study where the answers were short answer, so working out made a difference back then.
Title: Re: VCE Physics Question Thread!
Post by: Floatzel98 on April 12, 2015, 12:51:07 am
Nope. Not at all. The detailed study consists of 13 multiple choice questions, each worth 2 marks. It's either get 0 marks or 2 marks, and you answer on a multiple choice sheet and shade bubbles, so there's no way the computer will pick out working out.

However, older study design exams from maybe around 2000-2006 used to have sections for a detailed study where the answers were short answer, so working out made a difference back then.
Okay, thanks for the clarification :)
Title: Re: VCE Physics Question Thread!
Post by: RedCapsicum on April 12, 2015, 10:35:12 am
Hey can someone help me with this question:

a bullet of mass 'm' grams is fired with a velocity 'v' m/s into a wooden block of mass 'M' kg at rest on a friction less table of height 'H'. The bullet is embedded in the block which slides off the table landing at a distance of 'D'.

a) What is the velocity of the block immediately after the bullet is embedded in it?
b) How much time expires before the block lands on the floor?
Title: Re: VCE Physics Question Thread!
Post by: Kel9901 on April 13, 2015, 11:09:11 am
Hey can someone help me with this question:

a bullet of mass 'm' grams is fired with a velocity 'v' m/s into a wooden block of mass 'M' kg at rest on a friction less table of height 'H'. The bullet is embedded in the block which slides off the table landing at a distance of 'D'.

a) What is the velocity of the block immediately after the bullet is embedded in it?
b) How much time expires before the block lands on the floor?

I think it's reasonable to assume that the block is on the edge of the table.

a) Conservation of momentum
m1v1=m2v2
mv=(m+M)v2
v2=mv/(m+M) m/s

b) u=0, a=10, s=H, t=?
s=ut+1/2 at^2
H=5t^2
t^2=H/5
t= sqrt(H/5) (time is always positive)
Title: Re: VCE Physics Question Thread!
Post by: Floatzel98 on April 15, 2015, 09:55:39 pm
Well i have started Electricity now and I've actually forgot a lot from last year about basic circuit stuff and i need some help with some questions from the image below.

For question 1, before the switch is closed is the circuit working at all? Or does closing the switch turn it on? What exactly is the switch doing here? Why does the current increase when it is closed? I thought the current was constant in series.

For question 2, how do R4 and R5 have more current flowing through them than R2 and R3? In parallel current is  I(total) = I1 + 12.... isnt it? If the total current at a point before it hits the resistors is the same for both of them, then how can R4 and R5 and a greater current since  I(total) = I1 + 12... Do not the first two have the greatest and equal current going through them?

Most likely my basic understanding of this is just all wrong, but hopefully if someone can help explain this i can get back on track.

Thanks guys :)

Title: Re: VCE Physics Question Thread!
Post by: Floatzel98 on April 17, 2015, 04:31:24 pm
Well i have started Electricity now and I've actually forgot a lot from last year about basic circuit stuff and i need some help with some questions from the image below.

For question 1, before the switch is closed is the circuit working at all? Or does closing the switch turn it on? What exactly is the switch doing here? Why does the current increase when it is closed? I thought the current was constant in series.

For question 2, how do R4 and R5 have more current flowing through them than R2 and R3? In parallel current is  I(total) = I1 + 12.... isnt it? If the total current at a point before it hits the resistors is the same for both of them, then how can R4 and R5 and a greater current since  I(total) = I1 + 12... Do not the first two have the greatest and equal current going through them?

Most likely my basic understanding of this is just all wrong, but hopefully if someone can help explain this i can get back on track.

Thanks guys :)
Anyone?
Title: Re: VCE Physics Question Thread!
Post by: Zealous on April 17, 2015, 04:54:43 pm
Well i have started Electricity now and I've actually forgot a lot from last year about basic circuit stuff and i need some help with some questions from the image below.

For question 1, before the switch is closed is the circuit working at all? Or does closing the switch turn it on? What exactly is the switch doing here? Why does the current increase when it is closed? I thought the current was constant in series.

For question 2, how do R4 and R5 have more current flowing through them than R2 and R3? In parallel current is  I(total) = I1 + 12.... isnt it? If the total current at a point before it hits the resistors is the same for both of them, then how can R4 and R5 and a greater current since  I(total) = I1 + 12... Do not the first two have the greatest and equal current going through them?

Most likely my basic understanding of this is just all wrong, but hopefully if someone can help explain this i can get back on track.

Thanks guys :)

1) Yes the circuit will work fine without the switch closed. There is still wire connecting all the elements together. When the switch is open, current is forced to flow from A, to B then to C and encounter all of the resistance. So when you close the switch, you provide another path for current to flow, the current can flow from A to B, but then can split off which effectively decreases the resistance as more current can flow through the circuit.

2) Similar reasoning to the previous question. Resistors in parallel will have a lower overall resistance than resisters in series. This is because by putting element in parallel, we provide more pathways for current to flow in which actually decreases the resistance overall. So, let's just imagine the resistor value for all the resistors in question 2 was 4 ohms and the voltage source was 12V.

For circuit 1, the current flowing will be V/R which is 12/4=3A.
For circuit 2, we can imagine R2 and R3 as one resistor of 8 ohms, so the current will be 12/8 = 1.5A. So 1.5A of current will flow through R2 and R3.
For circuit 3, since the elements are in parallel, they will have the same voltage drop or potential of 12V. Using V=IR, I=V/R which for R4 would be 12/4=3A.

You can see that the current through each resistor in the third circuit is double the current in the second circuit because they are in parallel.
Title: Re: VCE Physics Question Thread!
Post by: Floatzel98 on April 19, 2015, 01:18:50 pm
1) Yes the circuit will work fine without the switch closed. There is still wire connecting all the elements together. When the switch is open, current is forced to flow from A, to B then to C and encounter all of the resistance. So when you close the switch, you provide another path for current to flow, the current can flow from A to B, but then can split off which effectively decreases the resistance as more current can flow through the circuit.

2) Similar reasoning to the previous question. Resistors in parallel will have a lower overall resistance than resisters in series. This is because by putting element in parallel, we provide more pathways for current to flow in which actually decreases the resistance overall. So, let's just imagine the resistor value for all the resistors in question 2 was 4 ohms and the voltage source was 12V.

For circuit 1, the current flowing will be V/R which is 12/4=3A.
For circuit 2, we can imagine R2 and R3 as one resistor of 8 ohms, so the current will be 12/8 = 1.5A. So 1.5A of current will flow through R2 and R3.
For circuit 3, since the elements are in parallel, they will have the same voltage drop or potential of 12V. Using V=IR, I=V/R which for R4 would be 12/4=3A.

You can see that the current through each resistor in the third circuit is double the current in the second circuit because they are in parallel.
I've run into some other questions where there is a switch like in question 1 of the above question. Because it isn't actually in parallel with another resistor does the current still split between them or does it all go through the switch circuit and turn off the light at C? Is it because there is no resistance in the other path and it is easier to pass through it?
Title: Re: VCE Physics Question Thread!
Post by: Adequace on April 19, 2015, 01:57:25 pm
I've run into some other questions where there is a switch like in question 1 of the above question. Because it isn't actually in parallel with another resistor does the current still split between them or does it all go through the switch circuit and turn off the light at C? Is it because there is no resistance in the other path and it is easier to pass through it?
For example, for 1c. I don't think Globe C lights up because when the switch is closed it creates a short circuit. I'm not sure if this is the sort of question you are referring to.
Title: Re: VCE Physics Question Thread!
Post by: Floatzel98 on April 19, 2015, 01:59:38 pm
For example, for 1c. I don't think Globe C lights up because when the switch is closed it creates a short circuit. I'm not sure if this is the sort of question you are referring to.
Yeah this is what I'm talking about. Why does that happen for?

EDIT: Another question. Is this a correct way of drawing the circuit in a more simplified way?

EDIT 2: Wait I'm pretty sure it's wrong because the 10 and the 30 should be in series with the 5 right? Not parallel. And the 5 and the second 30 should actually be in parallel i now  think. So would the 3rd picture be correct now?
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on April 19, 2015, 06:19:47 pm
Yeah this is what I'm talking about. Why does that happen for?

EDIT: Another question. Is this a correct way of drawing the circuit in a more simplified way?

EDIT 2: Wait I'm pretty sure it's wrong because the 10 and the 30 should be in series with the 5 right? Not parallel. And the 5 and the second 30 should actually be in parallel i now  think. So would the 3rd picture be correct now?

Yeah picture number 2 is wrong, number 3 looks right!

You could 'clean' it a bit by drawing the 5 resistor going vertically, and then joining it with the 30-branch. Then it will look more like the circuits you are used to. You can't see the numbers but this is the shape I'm talking about:
circuit
(http://i.imgur.com/2EQNIu8.jpg)

I'm not sure if it's a technical term but I call this 'linearising' a circuit. The easiest way to do it is to start at the battery and trace yourself around the circuit, adding components in series and then branching into parallel when the circuit splits. When the splits come back together, you close those parallel bits off. This is really hard to explain by text haha.

As for the short circuit question, yeah no current will flow through a resistor if there's an alternate path with zero resistance.
here's my way of justifying it: consider a circuit with just a resistor in parallel with nothing (one of the branches is just an empty branch with only wire), and maybe another resistor somewhere else in series. The effective resistance is 0 because of the resistor-less branch, meaning the voltage across the component is also 0. Having no voltage across the resistor-less branch isnt a problem, current will still flow. But if there is zero volts across the resistor in the other branch, no current is going to be pushed through it.
Title: Re: VCE Physics Question Thread!
Post by: Floatzel98 on April 19, 2015, 07:14:33 pm
Yeah picture number 2 is wrong, number 3 looks right!

You could 'clean' it a bit by drawing the 5 resistor going vertically, and then joining it with the 30-branch. Then it will look more like the circuits you are used to. You can't see the numbers but this is the shape I'm talking about:
circuit
(http://i.imgur.com/2EQNIu8.jpg)

I'm not sure if it's a technical term but I call this 'linearising' a circuit. The easiest way to do it is to start at the battery and trace yourself around the circuit, adding components in series and then branching into parallel when the circuit splits. When the splits come back together, you close those parallel bits off. This is really hard to explain by text haha.

As for the short circuit question, yeah no current will flow through a resistor if there's an alternate path with zero resistance.
here's my way of justifying it: consider a circuit with just a resistor in parallel with nothing (one of the branches is just an empty branch with only wire), and maybe another resistor somewhere else in series. The effective resistance is 0 because of the resistor-less branch, meaning the voltage across the component is also 0. Having no voltage across the resistor-less branch isnt a problem, current will still flow. But if there is zero volts across the resistor in the other branch, no current is going to be pushed through it.
Everything makes much more sense now. Thanks so much!
Title: Re: VCE Physics Question Thread!
Post by: knightrider on April 20, 2015, 08:55:17 pm
When they say initial velocity does this always mean that time=0secs?
When is it like or not like this?
Title: Re: VCE Physics Question Thread!
Post by: odeaa on April 20, 2015, 09:17:20 pm
When they say initial velocity does this always mean that time=0secs?
When is it like or not like this?
Yeah, when t=0, v=u
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on April 20, 2015, 09:29:50 pm
When they say initial velocity does this always mean that time=0secs?
When is it like or not like this?

Depends on the context. Usually, yeah, initial velocity refers to velocity at time t=0. But t=0 is kinda arbitrary, right? Otherwise t=0 might refer to the start of the universe or whatever absolute zero point you want to define. The idea is, we define a zero point that works best for us.

In many kinds of motion, we set t=0 to the start of that motion of interest, so 'initial velocity' relates directly to that starting instant.

If we're ever talking about a change in velocity, initial velocity will refer to the start of the change.

Sometimes, in projectile motion questions, we'll deal with different values of u, (different 'initial velocities') at different points in the question. For example if you want to find the time taken to fall from the peak of the projectile path to the landing point, you'd consider the start of this motion to be the peak of the path, and hence vertical u=0. In another part of the question, you might take u to be the launch speed because that's where you're measuring from (this is far more common).
Title: Re: VCE Physics Question Thread!
Post by: knightrider on April 21, 2015, 06:16:29 pm
Depends on the context. Usually, yeah, initial velocity refers to velocity at time t=0. But t=0 is kinda arbitrary, right? Otherwise t=0 might refer to the start of the universe or whatever absolute zero point you want to define. The idea is, we define a zero point that works best for us.

In many kinds of motion, we set t=0 to the start of that motion of interest, so 'initial velocity' relates directly to that starting instant.

If we're ever talking about a change in velocity, initial velocity will refer to the start of the change.

Sometimes, in projectile motion questions, we'll deal with different values of u, (different 'initial velocities') at different points in the question. For example if you want to find the time taken to fall from the peak of the projectile path to the landing point, you'd consider the start of this motion to be the peak of the path, and hence vertical u=0. In another part of the question, you might take u to be the launch speed because that's where you're measuring from (this is far more common).

Thanks silverpixeli  :)
Title: Re: VCE Physics Question Thread!
Post by: knightrider on April 23, 2015, 08:45:42 pm
How would you draw a velocity time graph for the following situation. A particle  starts with a positive velocity  and  undergoes constant acceleration until the  end of the time period?
Title: Re: VCE Physics Question Thread!
Post by: odeaa on April 23, 2015, 09:32:36 pm
It would just increase linearly because it has no friction
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on April 23, 2015, 10:00:26 pm
It would just increase linearly because it has no friction

The question hasn't actually said the acceleration is positive :P
Title: Re: VCE Physics Question Thread!
Post by: odeaa on April 23, 2015, 10:01:11 pm
The question hasn't actually said the acceleration is positive
Lzxnl 1 odeaa 0
Title: Re: VCE Physics Question Thread!
Post by: knightrider on April 24, 2015, 12:01:39 am
How would you draw a velocity time graph for the following situation. A particle  starts with a positive velocity  and  undergoes constant acceleration until the  end of the time period?

How would i draw this graph?
Title: Re: VCE Physics Question Thread!
Post by: Adequace on April 27, 2015, 08:06:31 pm
I'm interested to know what should I be doing in units 1/2 if I'm aiming for a 40SS for 3/4? I've been putting a significant amount of effort in to electricity and will be doing the some for motion but is there anything else?
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on April 27, 2015, 09:56:32 pm
I'm interested to know what should I be doing in units 1/2 if I'm aiming for a 40SS for 3/4? I've been putting a significant amount of effort in to electricity and will be doing the some for motion but is there anything else?

Pretty sure only motion and electricity carry forward to 3/4, sounds like you're setting yourself up with some solid foundations for these topics in year 12. Place emphasis on developing really solid problem solving skills! You can of course work on them in year 12 as well, but you should ideally be able to tackle any practice question you see without any trouble.
Title: Re: VCE Physics Question Thread!
Post by: Kel9901 on April 28, 2015, 01:58:31 pm
Pretty sure only motion and electricity carry forward to 3/4, sounds like you're setting yourself up with some solid foundations for these topics in year 12. Place emphasis on developing really solid problem solving skills! You can of course work on them in year 12 as well, but you should ideally be able to tackle any practice question you see without any trouble.

don't forget to still put work in to the other parts of the course though, even if they aren't that relevant, because it will help you develop a good work ethic.
Title: Re: VCE Physics Question Thread!
Post by: Floatzel98 on April 29, 2015, 08:14:57 pm
I have a couple questions about diodes:

Firstly, I  don't really understand the purpose of diodes. What are they used for exactly and where? How does forward and reverse biased work? Like why does it have almost no resistance and then infinite resistance? I don't understand. Why do they have a switch on voltage? Before the switch on voltage is reached, my book says that the diode does not conduct. Does this mean, if it were in series for example with another resistor, the circuit would never actually turn on/be complete? Is this phase exactly the same as what happens in reverse bias?

Secondly, in class we went over how they work, a lot of the chemistry behind them, but I really didn't understand most of it. Do we need to know this at all for any reason? Would someone be able to explain how they work anyway because I still want to understand it. I looked at a couple videos online but i still don't really understand.

Thanks :)
Title: Re: VCE Physics Question Thread!
Post by: Adequace on April 29, 2015, 09:15:11 pm
I started doing Checkpoints Physics 1/2's electric circuit chapter, I'm getting most of the questions wrong or don't understand the terminology...killed my self-esteem.
Title: Re: VCE Physics Question Thread!
Post by: alchemy on April 29, 2015, 09:30:13 pm
I started doing Checkpoints Physics 1/2's electric circuit chapter, I'm getting most of the questions wrong or don't understand the terminology...killed my self-esteem.

Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on April 29, 2015, 11:39:23 pm
I have a couple questions about diodes:

Firstly, I  don't really understand the purpose of diodes. What are they used for exactly and where? How does forward and reverse biased work? Like why does it have almost no resistance and then infinite resistance? I don't understand. Why do they have a switch on voltage? Before the switch on voltage is reached, my book says that the diode does not conduct. Does this mean, if it were in series for example with another resistor, the circuit would never actually turn on/be complete? Is this phase exactly the same as what happens in reverse bias?

Secondly, in class we went over how they work, a lot of the chemistry behind them, but I really didn't understand most of it. Do we need to know this at all for any reason? Would someone be able to explain how they work anyway because I still want to understand it. I looked at a couple videos online but i still don't really understand.

Thanks :)

Can answer some of your questions but can't explain how diodes actually work... it's not assessable so I didn't bother :)

forward or reverse bias?

first of all, forward bias refers to when conventional current in a circuit flows in the same direction that the triangle in the diode symbol points. this is opposite to the flow of electrons, but it's not like the physical diode is actually an arrow, it's just chemically a one-way thing and so we chose to line the triangle up with conventional current in our diagrams.

reverse bias is obviously the other way. if you see your conventional current hitting the point side of a triangle, your diode will not conduct because it's in reverse bias mode.

why wont it conduct below the switch on voltage?

a diode is not made of material that can inherently conduct electricity. however, because of the magical chemistry going on under the hood, the diode can ---  in one direction --- 'carry' electrons across. this is different to electrons being pushed through a resistor because of potential difference from one side to the other. electrons are unable to push themselves through a diode, but they can hitch a ride through and so the diode appears to be conducting, just like a resistor does.

if there isn't enough potential across the diode, however, this 'carrying' service wont actually start and the diode wont conduct. likewise, if the circuit is pushing the other way, the diode is gonna be a dead end because electrons can't push themselves through, and the diode can only carry them in the forward direction.

i'm thinking about this 'carrying' analogy now, and it seems appropriate because of the one-way property and it's close to how diodes actually work, but theres another property of diodes that doesnt really sit well with it. When you're an electron and you push your way through a resistor, you can spend as much of your energy as you like (and arrive at the other side with less potential, hence potential 'difference' across the resistor). but in a diode, assuming we're above the switch on voltage so that the diode conducts, we'll never spend more than the switch on voltage.

because of this, maybe you should think of diodes as a one-way toll gate. you can go through as fast as you like (indeed, diodes don't influence current in a circuit -- that's decided by the elements in the rest of the circuit) but you always have to pay the same amount (the voltage across the diode will always be the switch on voltage, or less, never more). if you don't have enough potential to pay the toll, you wont even be allowed through! (this is the case where the diode has a potential difference below the switch on voltage across it) and of course, if you try to go back, you won't be allowed, because this gate is on a one way street (no conduction in reverse bias mode)

diodes dont influence current?

one final thing about the properties of a diode is the current through them. I said that diodes allow any current just like a toll gate may allow things to move through at any speed - as long as they pay their toll of potential.

think of a circuit where the battery provides the electrons with 6V of potential at the start, and they have to pass through a 0.8V diode and a 520 ohm resistor, in series. the electrons have the job of deciding how to spend their potential across the circuit before they get back to the battery - where they must have zero. if the circuit were two identical resistors, they would decide to spend their voltage evenly.
But because there's now a 'toll gate', their first priority will be to pay the toll to move through the diode. but they wont spend any more than the toll. then, they each have 6-0.8=5.2V of potential left to spend on the rest of the circuit. of course, there's only the resistor left, so they will splurge their remaining 5.2V on the 520 ohm resistor and move through it as fast as they can: I=V/R=5.2/520=0.01A. Their current through this resistor must be their current everywhere (series circuit) so they must also speed through the diode at 0.1A as well.

The point i'm getting at here is that it's the REST of the circuit that determines the current, you can't tell anything about currents just from a diode because they dont have a well defined resistance, but you can use them to tell you about voltage drops elsewhere because they have a well defined cost in crossing them.

in your example, a diode in series with a resistor, assuming the electrons have enough potential to switch on the diode, they will spend that potential on the diode and spend the rest on getting through the resistor with whatever current they can afford.
but if they weren't given enough potential to even pay the diode toll (less than switch on voltage), then none of them could get around the circuit and it wont conduct, just like if the diode was n reverse bias mode which is effectively the same as cutting one of the wires - no electrons can push through because diodes are a one-way toll gate.

non ideal diodes

in reality, diodes aren't perfectly on or off components. they will begin to conduct a little before the switch on voltage and they will charge slightly more for high current electrons - the I V graph doesn't have infinite gradient at V=switch on voltage.

also, they aren't perfectly impossible to pass in reverse bias mode. if there's enough voltage across a reverse bias diode, the materials inside will stop blocking electrons from flowing and it will begin to conduct freely (well, not freely, in this case you'd be paying a huge voltage toll to keep the gates open the wrong way)

what are they used for?

the one-way thing is actually hugely advantageous. there are many applications of regular diodes in circuitry, for example you can use them and a clever circuit layout called a bridge rectifier to turn AC current into positive-only current. (not studied in core electricity).

in photonics you'll also look at light-emitting diodes (LEDs) which I am sure you are familiar with. probably a bunch of lights in your home, school or any building are LED these days, meaning instead of a hot, glowing globe filament used in old light bulbs they just fire current through a particular type of diode and you get that blue/white light due to the emission of photons. computers, torches, toys, LOTS of things have little LED lights because it's an efficient way to illuminate stuff.
aside from looking really cool, LEDs can be used in optical applications like flashing on and off really fast sending pulsating light signals down an optic fibre for speed-of-light, highly efficient signal transmission.

there are also 'photodiodes' which you'll look at in photonics as well. These again have slightly different chemistry behind them which i dont really understand, but they basically leak a small amount of current that depends on the amount of light that illuminates their surface. In this way, they can capture changing lighting conditions and translate them into changing electrical current - we call this an opto-electric transducer, or something (i forget the technical terminology), and it's useful for doing these sorts of signal conversions

there are probably heaps more examples of applications of diodes! I've probably only scratched the surface.
Title: Re: VCE Physics Question Thread!
Post by: Floatzel98 on April 30, 2015, 07:29:56 am
^Holy crap, thanks so much for all that information. That toll-gate analogy actually helps a lot :)
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on April 30, 2015, 09:52:39 am
^Holy crap, thanks so much for all that information. That toll-gate analogy actually helps a lot :)

I know, I'm very proud of it, I understand diodes a lot better now after writing that :P
Title: Re: VCE Physics Question Thread!
Post by: Adequace on April 30, 2015, 08:02:44 pm
Can someone help me with understanding questions? It seems like a broad question but I think that I understand the theory and concepts behind electronics but I can't seem to interpret the questions and understand what they want me to do.

It ranges from the different terminology used in questions and just sometimes just being confused..
Title: Re: VCE Physics Question Thread!
Post by: Floatzel98 on April 30, 2015, 08:21:24 pm
Can someone help me with understanding questions? It seems like a broad question but I think that I understand the theory and concepts behind electronics but I can't seem to interpret the questions and understand what they want me to do.

It ranges from the different terminology used in questions and just sometimes just being confused..
Keep doing the questions. If you get them wrong or don't understand what it was asking specifically, work it out by looking at the answers, asking somebody or even posting here. You said that you understand the theories and concepts, so even if you don't understand a certain question, once you have it explained you should try to relate it back to what you know.

Try to find patterns in questions. There is only a certain number of ways they can ask you things. Maybe it involves graphs of ohmic devices. Gather a whole heap of questions and attempt them. Find patterns in them and maybe even methods or steps of what to look for first when solving them.

Maybe you do actually have some gaps in your knowledge and you don't actually know what you think you should know. If this is the case, broaden your knowledge from more than just the textbook. Look at videos online about the concepts. Maybe they will fill in the unfamiliar terminology your textbook is using or will just give you a different perspective of the topic.

I think you should start by just posting one of the questions you are having trouble with and getting help from us with it.
Title: Re: VCE Physics Question Thread!
Post by: Adequace on April 30, 2015, 09:40:20 pm
Keep doing the questions. If you get them wrong or don't understand what it was asking specifically, work it out by looking at the answers, asking somebody or even posting here. You said that you understand the theories and concepts, so even if you don't understand a certain question, once you have it explained you should try to relate it back to what you know.

Try to find patterns in questions. There is only a certain number of ways they can ask you things. Maybe it involves graphs of ohmic devices. Gather a whole heap of questions and attempt them. Find patterns in them and maybe even methods or steps of what to look for first when solving them.

Maybe you do actually have some gaps in your knowledge and you don't actually know what you think you should know. If this is the case, broaden your knowledge from more than just the textbook. Look at videos online about the concepts. Maybe they will fill in the unfamiliar terminology your textbook is using or will just give you a different perspective of the topic.

I think you should start by just posting one of the questions you are having trouble with and getting help from us with it.
Cheers for the help, I agree I probably do have gaps in my knowledge without me knowing. I'll definitely post questions I have trouble with in the future.
Title: Re: VCE Physics Question Thread!
Post by: StressedAlready on April 30, 2015, 10:23:26 pm
A scientist has a 120g sample of the radioisotope polonium-218. The first three steps in the decay series of polonium-218 are an alpha emission followed by two beta particle emissions. These decays have half-lives of 3, 27 and 19 mins respectively. Calculate how many polonium-218 remains after 15 mins.

Realise this is year 11 physics. Should totally know but don't despite doing 3+4 this year. T.T

Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on May 01, 2015, 08:26:21 am
A scientist has a 120g sample of the radioisotope polonium-218. The first three steps in the decay series of polonium-218 are an alpha emission followed by two beta particle emissions. These decays have half-lives of 3, 27 and 19 mins respectively. Calculate how many polonium-218 remains after 15 mins.

Realise this is year 11 physics. Should totally know but don't despite doing 3+4 this year. T.T

there's some info we don't need here. The decay series it's talking about it this:

$^{218}Po \rightarrow ^{214}Pb \rightarrow ^{214}Bi \rightarrow ^{214}Po$

We see that after it decays for the first time, it's never going to be Polonium-218 again. So "Calculate how many polonium-218 remains after 15 mins" is just asking about the first decay, and the other halflives are meaningless to us.

The question becomes "how much of your 120g sample,half life 3min, is left after 15m?"

15/3=5 half lives pass in the 15 minutes, so the sample size is halved 5 times.
120/2/2/2/2/2 = 120/32 = 3.75g

thus 3.75 grams remain after 15 mins.
Title: Re: VCE Physics Question Thread!
Post by: lolaishappy on May 02, 2015, 05:56:37 pm
When it comes to photonics, what theory is a must in knowing? Like modulation, demodulation, how about attenuation? Particularly from the Jacaranda text book.
Also How do I do sketch modulation graphs? :-[
Title: Re: VCE Physics Question Thread!
Post by: Cosec on May 02, 2015, 06:17:44 pm
When it comes to photonics, what theory is a must in knowing? Like modulation, demodulation, how about attenuation? Particularly from the Jacaranda text book.
Also How do I do sketch modulation graphs? :-[

A modulated graph, which im guessing is your just refeering to the graph of the wave itself, is a wave in which the information signal, normally one of low frequency, is super imposed on the carrier wave, one which has a much higher frequenc. producing a wave that is used to carry the information over a certain medium. Like this: http://www.sfu.ca/sonic-studio/handbook/Graphics/Amplitude_Modulation2.gif

I just had my sac, and learnt all of them which you described above. When i completed several practice sacs, a majority include a question about the modulation/demodulation process. So i would strongly advise you learnt them.
Title: Re: VCE Physics Question Thread!
Post by: Adequace on May 02, 2015, 07:44:33 pm
We have our electricity SAC coming up in 2 weeks, we're allowed to bring a double sided cheat sheet. I've just finished mine but I have a lot of spare space - a whole page worth. I'm wondering if I should use that space to write questions I had trouble on or to use it for additional information about the less important things such as electric shocks and resistivity of your skin, I know it's going to be on the MCQ as well.

This is for unit 1/2 if you guys don't mind, thanks.
Title: Re: VCE Physics Question Thread!
Post by: Floatzel98 on May 02, 2015, 08:29:49 pm
We have our electricity SAC coming up in 2 weeks, we're allowed to bring a double sided cheat sheet. I've just finished mine but I have a lot of spare space - a whole page worth. I'm wondering if I should use that space to write questions I had trouble on or to use it for additional information about the less important things such as electric shocks and resistivity of your skin, I know it's going to be on the MCQ as well.

This is for unit 1/2 if you guys don't mind, thanks.
If you know what is coming up on the test for the most part and you know what you will struggle with, definitely try to fill it up with that. Specific examples of questions you struggled with. I think simple sentence definitions could suffice for the less important information or even just try to rote learn it.

I know that we even get a cheat sheet to use in the 3/4 exam, but i would also recommend trying not to rely on it too much. It might seem counter-intuitive but the less you have to rely on a cheat sheet means that you know more of what you need to know already. You should / you usually are given a formula sheet with the standard equations you should know, so if you can try to dedicate more space on your cheat sheet to specifics of what you struggled on and not things you already know. This is not necessarily for this SAC now, but in the future as well. I know sometimes i just fill up my cheat sheets with all the equations that are already on the cheat sheet, sometimes with a few derived equations for myself. I spent time on it when i could have just been studying or i actually could have filled it up with useful information and examples that would aid me.

I know I'm just rambling on now and that my first sentence probably answers your question, but hopefully the rest helps as well if you can apply it at all. Anyway goodluck on your upcoming SAC and post any other questions you have! :)
Title: Re: VCE Physics Question Thread!
Post by: kk.08 on May 02, 2015, 10:18:20 pm
Was just wondering, is it possible to find the elastic limit of a material without a stress-strain graph?

Thanks :)
Title: Re: VCE Physics Question Thread!
Post by: Floatzel98 on May 03, 2015, 05:41:24 pm
If a resistor or even another diode is placed in series in front of a reversed biased diode (the current flows through the resistor before it hits the reverse biased diode), why doesn't  current still flow through. For example why isn't any power dissipated? The current has to go through the resistors and such to get to the reverse biased diode?

What happens to the current in the diode when its in reverse biased. Does it send it back the other way? Does it just store it? Or does it get used 100% as heat energy?

For diode labeling conventions. Are we supposed to always change the polarity of the circuit or can we just draw the diode facing the other way? Does it matter at all?  Because sometimes I just don't look at which way the circuit is and assume a diode is in forward bias when it is really in reverse.

What exactly determines the switch on voltage of  a diode. I know the material but what are the differences in material. Is a diode technically still a resistor? Are there other non-ohmic resistor type things as well?

Just some question i thought of while doing my homework today. Thanks to anyone that can answer them :)
Title: Re: VCE Physics Question Thread!
Post by: Adequace on May 03, 2015, 06:02:05 pm
http://imgur.com/wPXnz5R

In this circuit, what is voltage at X? I got the answer but I'm still confused on the method of getting the answer since I just guessed. We found the total resistance in the question above.
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on May 03, 2015, 06:14:09 pm
Was just wondering, is it possible to find the elastic limit of a material without a stress-strain graph?

Thanks :)

I'm not sure if it's possible without SOME knowledge of the behaviour (a graph, a table of stress and strain values, a force/extension graph), like there's certainly no VCE level formula for calculating the elastic limit of a material based on other quantities. There may be some formula that involves a lot of other properties of the material (but I doubt it).

If a resistor or even another diode is placed in series in front of a reversed biased diode (the current flows through the resistor before it hits the reverse biased diode), why doesn't  current still flow through. For example why isn't any power dissipated? The current has to go through the resistors and such to get to the reverse biased diode?

Good question. Current in series has to be the same everywhere. Think of a road with a fixed amount of lanes and lots and lots of cars. The rate of the slowest moving cars determines the rate of all cars because otherwise fast moving cars would catch up and crash into slower cars. This analogy has plenty of holes but the basic idea is that current cant be high in some sections and low in others if those sections are in series.

So, even when current wants to leave the battery, the electrons can't physically go anywhere because there's a blockage all the way along the wires at the diode. (a reverse biased diode is just like a hole in the circuit - no current flows around then either)
Thus, no current flows through the wires, or the resistor, and that's why P=VI=V*0=0.

What happens to the current in the diode when its in reverse biased. Does it send it back the other way? Does it just store it? Or does it get used 100% as heat energy?

Same explanation as first question, there is no current in this circuit. Nothing happens with it. If a small amount of current were to flow (all real reverse bias diodes let a tiny leakage current through, i think), it would just go round the circuit like normal - the diode acting like a very high-resistance resistor.

For diode labeling conventions. Are we supposed to always change the polarity of the circuit or can we just draw the diode facing the other way? Does it matter at all?  Because sometimes I just don't look at which way the circuit is and assume a diode is in forward bias when it is really in reverse.

Conventional current flows out of the bigger terminal of the battery (the positive one) and if that direction around the circuit points the same way that the triangle points, the diode is forward biased.

Electrons actually flow the other way in a real circuit, but the convention still holds.

Always watch out for trick questions 'what is the current in this circuit' when they sneakily put it in reverse mode.

If you're asked to draw a circuit with the diode in the other mode, it's probably easiest to reverse the diode direction rather then the battery direction: it's a lot easier to tell the difference for the diode. An assessor might not see that you changed the battery.

What exactly determines the switch on voltage of  a diode. I know the material but what are the differences in material. Is a diode technically still a resistor? Are there other non-ohmic resistor type things as well?

This is to do with the magical chemistry that makes diodes function so I'll let someone else try to answer it.

BUT for the other bit, a diode can be thought of as a resistor but it has no well defined resistance because it is non ohmic (resistance is not the same for different voltages/currents, V-I graph not linear, etc)

You can calculate R=V/I for any given situation but your result is meaningless for the same diode in a different circuit. It's merely the 'effective resistance' of the diode, if you replaced it with a resistor of that same resistance you would get the same behaviour for that circuit.

Other non ohmic devices studied in VCE include temperature dependant resistors, light dependent resistors, and the other types of diodes (LED's and photodiodes)
Title: Re: VCE Physics Question Thread!
Post by: Floatzel98 on May 03, 2015, 06:19:53 pm
http://imgur.com/wPXnz5R

In this circuit, what is voltage at X? I got the answer but I'm still confused on the method of getting the answer since I just guessed. We found the total resistance in the question above.
Well first you can find the total resistance which should be $(\frac{1}{4 + 2} + \frac{1}{12}})^{-1} = 4 \Omega$

Then the total current should be $I = \frac{V}{R} = \frac{12}{4} = 3 A$

The current though 12 ohm resistor should be $\frac{12}{12} = 1$ since votlage is constant in parallel.

That leaves 2 amps to be found in the series part of the circuit and since current is constant in series they both have the 2 amps through them.

That leaves the voltage across the 2 ohm resistor to be $IR = 2*2 = 4V$

The voltage in the 4 ohm resistor is $IR = 2*4 = 8V$

You can verify this because voltage is constant in parallel and the 2 series resistors voltage add up to 12 (4 + 8 )

I'm pretty sure this is right. I'm notorious for forgetting at least something which stuffs up all the other calculations. It might be a long working but i hoped it highlighted everything for you and answered the question :)

Spoiler
I'm not sure if it's possible without SOME knowledge of the behaviour (a graph, a table of stress and strain values, a force/extension graph), like there's certainly no VCE level formula for calculating the elastic limit of a material based on other quantities. There may be some formula that involves a lot of other properties of the material (but I doubt it).

Good question. Current in series has to be the same everywhere. Think of a road with a fixed amount of lanes and lots and lots of cars. The rate of the slowest moving cars determines the rate of all cars because otherwise fast moving cars would catch up and crash into slower cars. This analogy has plenty of holes but the basic idea is that current cant be high in some sections and low in others if those sections are in series.

So, even when current wants to leave the battery, the electrons can't physically go anywhere because there's a blockage all the way along the wires at the diode. (a reverse biased diode is just like a hole in the circuit - no current flows around then either)
Thus, no current flows through the wires, or the resistor, and that's why P=VI=V*0=0.

Same explanation as first question, there is no current in this circuit. Nothing happens with it. If a small amount of current were to flow (all real reverse bias diodes let a tiny leakage current through, i think), it would just go round the circuit like normal - the diode acting like a very high-resistance resistor.

Conventional current flows out of the bigger terminal of the battery (the positive one) and if that direction around the circuit points the same way that the triangle points, the diode is forward biased.

Electrons actually flow the other way in a real circuit, but the convention still holds.

Always watch out for trick questions 'what is the current in this circuit' when they sneakily put it in reverse mode.

If you're asked to draw a circuit with the diode in the other mode, it's probably easiest to reverse the diode direction rather then the battery direction: it's a lot easier to tell the difference for the diode. An assessor might not see that you changed the battery.

This is to do with the magical chemistry that makes diodes function so I'll let someone else try to answer it.

BUT for the other bit, a diode can be thought of as a resistor but it has no well defined resistance because it is non ohmic (resistance is not the same for different voltages/currents, V-I graph not linear, etc)

You can calculate R=V/I for any given situation but your result is meaningless for the same diode in a different circuit. It's merely the 'effective resistance' of the diode, if you replaced it with a resistor of that same resistance you would get the same behaviour for that circuit.

Other non ohmic devices studied in VCE include temperature dependant resistors, light dependent resistors, and the other types of diodes (LED's and photodiodes)
EDIT: Thank you for clearing everything up for me again. I really appreciate it :) For the part about the current flowing into a reverse biased diode: I know the answer stays the same, but does that mean that for the instant when it is turned on, current will flow up past everything to the diode? Also when you turn off the power source to a circuit, do the electrons that were currently in the circuit all leave the wire/circuit back into the power supply or can they get 'trapped' in there? Thanks again!
Title: Re: VCE Physics Question Thread!
Post by: Adequace on May 03, 2015, 06:27:59 pm
Well first you can find the total resistance which should be $(\frac{1}{4 + 2} + \frac{1}{12}})^{-1} = 4 \Omega$

Then the total current should be $I = \frac{V}{R} = \frac{12}{4} = 3 A$

The current though 12 ohm resistor should be $\frac{12}{12} = 1$ since votlage is constant in parallel.

That leaves 2 amps to be found in the series part of the circuit and since current is constant in series they both have the 2 amps through them.

That leaves the voltage across the 2 ohm resistor to be $IR = 2*2 = 4V$

The voltage in the 4 ohm resistor is $IR = 2*4 = 8V$

You can verify this because voltage is constant in parallel and the 2 series resistors voltage add up to 12 (4 + 8 )

I'm pretty sure this is right. I'm notorious for forgetting at least something which stuffs up all the other calculations. It might be a long working but i hoped it highlighted everything for you and answered the question :)
Thanks for the fast reply. Additionally, could I find the current of the first arm which would be 2A then substitute to find the voltage at X? I did it like that, would that also be correct in terms of method?
Title: Re: VCE Physics Question Thread!
Post by: Floatzel98 on May 03, 2015, 06:43:45 pm
Thanks for the fast reply. Additionally, could I find the current of the first arm which would be 2A then substitute to find the voltage at X? I did it like that, would that also be correct in terms of method?
If you mean find the total voltage of the series part of the circuit, then you would get a different answer to the voltage at X. You would get 6*2 = 12 V, which is the total voltage across the series arm. From there you could go and find the voltage across the 2 individual resistors, which is exactly the same as doing what i showed anyway.  Otherwise as long as you get the correct value for current and you use it to find the actual voltage across X and not the entire series arm, then yes. The method would be correct.
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on May 04, 2015, 09:15:48 am
Thank you for clearing everything up for me again. I really appreciate it :) For the part about the current flowing into a reverse biased diode: I know the answer stays the same, but does that mean that for the instant when it is turned on, current will flow up past everything to the diode? Also when you turn off the power source to a circuit, do the electrons that were currently in the circuit all leave the wire/circuit back into the power supply or can they get 'trapped' in there? Thanks again!

Actually, this is getting into the electromagnetism side, but there are always electrons distributed at all points along the wire, (conduction electrons bounce around fairly randomly when there's no potential difference affecting them)
When you 'turn on' the circuit, you're providing a 'potential difference' between one point and another. We think of this as a difference in energy, but the electrons are actually feeling a force, called 'electromotive force', because the battery is actually providing an electric field. The randomly bouncing electrons start to gradually experience a net movement with the force.
Compare this with the force of gravity, a uniform field when you're near the surface of the Earth. But sometimes we talk about potential energy (mgh). A grain of sand will fall downwards in a hollow tube, because there is a gravitational force pushing it that way. You could equivalently think of it as falling to the bottom of the tube, because then it will be at a lower gravitational potential energy (there is a 'potential difference' between the bottom and the top of the tube).
This is what's happening to electrons in your circuit. But we don't normally talk about it in terms of force in a circuit, we talk about the potential energy difference between the positive terminal and the negative terminal. In the case of the sand, we're talking about gravitational potential energy. In the case of the electron, we're talking about electrical potential energy.

'Electricity' travels at the speed of light, but this isn't the motion of the electrons, it's the speed that the field travels out and starts affecting the electrons that are further away. The electrons actually move relatively slowly :)
Title: Re: VCE Physics Question Thread!
Post by: Floatzel98 on May 04, 2015, 03:47:44 pm
Thanks again Silverpixeli! But alas i have more questions again. This time about voltage amplifiers. I went over it in class today and we watched a few videos and i even watched the VTextbook video on it just then, but I'm still confused.

Firstly, i don't really understand how a transistor works in a circuit. From what i understood, it has a voltage input that goes through the collector and then one that comes through the base? If that's correct, are they actually 2 different power sources or do they come from the same power supply. If so, then how do you regulate how much voltage goes through it from the base input? From a picture i saw in class today, the base is apparently usually connected to a capacitor? I don't really know what a capacitor is either, but doesn't that store energy/voltage? Is that connected back to the same power supply of the circuit if that's the case. How do you regulate how much voltage goes through that anyway?

Next, from what i gathered, when no input voltage comes through the base, the resistance is infinite since it is basically 2 diodes facing each other and one of them will be in reverse bias? When there is a voltage input from the base that goes above the switch on voltage it will work and the resistance will basically be 0. How does the 'switch on voltage' allow the diode to work if it is approaching it in reverse bias? You would need like 50V to break it but it is still only about 0.7 V according to my book.

Even if that all makes sense to me, i still don't really know what it does. It amplifies the voltage but how? Is it because you are basically adding 2 voltages from 2 power sources together. Almost like as if you were adding ordinates of 2 sine graphs? I hardly understand what all the graphs are about either, but if someone can enlighten me on how all the above works, maybe i will be able to figure it out for myself.

Thanks :)
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on May 04, 2015, 10:26:42 pm
Thanks again Silverpixeli! But alas i have more questions again. This time about voltage amplifiers. I went over it in class today and we watched a few videos and i even watched the VTextbook video on it just then, but I'm still confused.

Firstly, i don't really understand how a transistor works in a circuit. From what i understood, it has a voltage input that goes through the collector and then one that comes through the base? If that's correct, are they actually 2 different power sources or do they come from the same power supply. If so, then how do you regulate how much voltage goes through it from the base input? From a picture i saw in class today, the base is apparently usually connected to a capacitor? I don't really know what a capacitor is either, but doesn't that store energy/voltage? Is that connected back to the same power supply of the circuit if that's the case. How do you regulate how much voltage goes through that anyway?

Next, from what i gathered, when no input voltage comes through the base, the resistance is infinite since it is basically 2 diodes facing each other and one of them will be in reverse bias? When there is a voltage input from the base that goes above the switch on voltage it will work and the resistance will basically be 0. How does the 'switch on voltage' allow the diode to work if it is approaching it in reverse bias? You would need like 50V to break it but it is still only about 0.7 V according to my book.

Even if that all makes sense to me, i still don't really know what it does. It amplifies the voltage but how? Is it because you are basically adding 2 voltages from 2 power sources together. Almost like as if you were adding ordinates of 2 sine graphs? I hardly understand what all the graphs are about either, but if someone can enlighten me on how all the above works, maybe i will be able to figure it out for myself.

Thanks :)

Transistor workings are also not really part of the course. Capacitors, also, are left for the detailed study 'further electronics'.
Voltage amplification questions range from interpreting graphs, talking about cutoff/saturation, voltage gain and converting from input signals to output signals and back.

A transistor definitely has the two source things going on, but I'm not sure what you're talking about when you mention two diodes and one in reverse bias mode.
Whatever voltage applies at the base is like the voltage across a diode, in that if it's lower than some switch on value, there will be no conduction. A transistor is different in that when the diode IS switched on, the majority of the current to the emitter doesn't come through the base, it comes through the collector. Basically, the voltage across the base regulates how much current can come through the collector. Typically, you have the collector hooked up to some more powerful supply that can provide as much current to the collector as the transistor will allow through it.

This is useful for signal amplification as follows: Typically you have a signal captured as small variations in voltage/current in a circuit. If you feed this varying voltage into the base of a transistor, it will allow proportionate amounts of current through from the high-power collector. The result will be a higher power version of your original signal coming out of the emitter!

This assumes that your input signal is always within the working range of the transistor. Too low, and you won't open the gate for current from the collector. We refer to this as 'cutoff'. Too high, and you wont be able to get any more current through the transistor, and all higher signals will output that same max. We call this saturation. Both are examples of 'clipping' which result in the extremes of a signal being 'chopped off' from signals that dont lie within this range.

The complicated circuits that actually make this work are not studied and I have no understanding of them personally other than the high-level sketch above. Sorry :P

As for the amplifier graphs, Vin - Vout graphs map input voltages to output voltages. If you put in an input signal (often represented as a Vin - time graph) you can use the amplifier's characteristic graph to map it to an output signal v. time graph. This process, and the reverse, are semi-common exam questions.
Title: Re: VCE Physics Question Thread!
Post by: TheAspiringDoc on May 06, 2015, 03:53:08 pm
Hi guys, just wondering, is it possible to make a black (light emitting) LED/Torch? For example, if you were in a brightly lit room (such as an art gallery) with lots of big blank white walls and you were to shine this torch, it would make a dark patch..?
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on May 06, 2015, 04:23:18 pm
Hi guys, just wondering, is it possible to make a black (light emitting) LED/Torch? For example, if you were in a brightly lit room (such as an art gallery) with lots of big blank white walls and you were to shine this torch, it would make a dark patch..?

Good question. afaik a 'black light' is impossible.

When we look at something black, it looks black because there are no photons (light particles) reaching our eyes from that place. If there were some red photons, it would look red. If there were some red+green+blue photons (or another combination of lots of colours) it might look white.
To make it look black, we'd have to stop it from reflecting photons towards our eyes. We can't do this with a torch.
I guess we could alternatively try to stop the photons from reaching our eyes once they have been reflected, but I don't see how this could be done either.

In short, black things look black because there is no light coming from there. There are no black photons that you could shine on something.
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on May 06, 2015, 08:52:31 pm
Hi guys, just wondering, is it possible to make a black (light emitting) LED/Torch? For example, if you were in a brightly lit room (such as an art gallery) with lots of big blank white walls and you were to shine this torch, it would make a dark patch..?

Hmmm....

I wonder if a molecule exists that absorbs essentially completely in the visible range and has such a low fluorescent quantum yield that all of the light absorbed is re-radiated as heat. In theory, if you could make a molecule like that, said molecule would block out all light hitting it, so you would have a dark trail in the air. But I can't imagine what on earth would have such a property.

But yeah, no black photons though.
Title: Re: VCE Physics Question Thread!
Post by: Fusuy on May 10, 2015, 02:10:32 pm
Hey Guys, I've got my Electronics and Photonics SAC this week and was wondering if anyone had some past Sacs or resources they'd be kind enough to share. Thanks :D
Title: Re: VCE Physics Question Thread!
Post by: Floatzel98 on May 10, 2015, 08:15:37 pm
I'm having a bit of trouble with the attached question. I have looked at the worked solutions for question 3 and i understand the steps but i don't understand how they thought the steps up and how they got to them. I find that a lot of these ratio questions stump me, especially when they were used in Gravity and Satellites. I can kind of see how it wants me to answer it but i can never actually go through with it and get the correct answer. Can someone show me how they went through the steps to solve this? Thanks :)

EDIT: Another question. There is a circuit with 3 identical resistors of 100 Ohms that are connected with one in series and then the other 2 in parallel and the max power from any one resistor is 25W.

The question was to find the max voltage the can be applied. I used $P = \frac{V^2}{R}$ and had already found the total resistance and went $V = \sqrt{150\Omega*75W}$ for the 3 resistors, but my answer seems to be wrong.
Title: Re: VCE Physics Question Thread!
Post by: odeaa on May 10, 2015, 08:34:09 pm
I'm having a bit of trouble with the attached question. I have looked at the worked solutions for question 3 and i understand the steps but i don't understand how they thought the steps up and how they got to them. I find that a lot of these ratio questions stump me, especially when they were used in Gravity and Satellites. I can kind of see how it wants me to answer it but i can never actually go through with it and get the correct answer. Can someone show me how they went through the steps to solve this? Thanks :)

Q3) Total resistance $= 3R + R =4R$
Using the voltage divider formula, $V_{R_{A}}=V_{in}\frac{R_{A}}{R_{total}}
=V_{in}*\frac{1}{4}$

Alternatively, we can just use ratios- because $R_{A}$ is only 1/3 the size of $R_{B}$, it will only have a third of the voltage drop across it (from here you can just figure it out using basic maths)
Q4) $V=IR$
$R=\frac{V}{I}
=\frac{12}{200*10^{-3}}
=60\Omega$

Then, remembering that  $R_{A}$ is 1/4 of the total and $R_{B}$ is the other 3/4, they will be 15 and 45 ohms respectively
b) $P_{B}=\frac{V^{2}}{R}$
$=\frac{81V}{45\Omega}$
$=1.8 W$
I hope thats right

edit: I suck at latex, I think its all good now
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on May 10, 2015, 08:54:55 pm
Another question. There is a circuit with 3 identical resistors of 100 Ohms that are connected with one in series and then the other 2 in parallel and the max power from any one resistor is 25W.

The question was to find the max voltage the can be applied. I used $P = \frac{V^2}{R}$ and had already found the total resistance and went $V = \sqrt{150\Omega + 75W}$ for the 3 resistors, but my answer seems to be wrong.

Why are you adding ohms and watts together? I assume typo?

Okay so just because the maximum power of one resistor is 25 Watts doesnt mean it's operating at that power. That just means that as you increase the voltage applied, you have to be careful not to take ANY of them over the max. In particular, With more resistance and twice the current, the resistor OUTSIDe of the parallel section will probably have the most power at any given time. So you should solve for the total voltage in the circuit when all you know is that the series resistor has 25W, 100Ohms.

Title: Re: VCE Physics Question Thread!
Post by: Floatzel98 on May 10, 2015, 09:02:43 pm
Thanks, using the voltage divider formula makes more sense :)

Plus i have another question now. If you have a voltage divider over a resistor that is in parallel, do we take voltage divider resistors parallel voltage or it's individual voltage? From the questions I've done it's apparently the total parallel resistance but i don't think that makes sense for some reason. Is it because the voltage in parallel is constant? That's my only reasoning. If i turn the parallel resistor into a single resistor ( and put it in series with another resistor) is that still the same thing.

While typing that i think i've answered my own question. But does that make sense still? If i have a voltage divider over one resistor in parallel, i can just simplify that into a single resistor and everything will still work the same?

Thanks :)
Title: Re: VCE Physics Question Thread!
Post by: Floatzel98 on May 10, 2015, 09:11:30 pm
Why are you adding ohms and watts together? I assume typo?

Okay so just because the maximum power of one resistor is 25 Watts doesnt mean it's operating at that power. That just means that as you increase the voltage applied, you have to be careful not to take ANY of them over the max. In particular, With more resistance and twice the current, the resistor OUTSIDe of the parallel section will probably have the most power at any given time. So you should solve for the total voltage in the circuit when all you know is that the series resistor has 25W, 100Ohms.
Yes, that was a typo. My original workings were $P = \frac{V^2}{R}$ and solve for V such that $V = \sqrt{P*R}$. The total resistance is 150 ohms and if each resistor can have a max output of 25 W, then the total power output could be 75 W, so $V = \sqrt{(150)*(75)}$. Is the total power across the parallel resistors not 50W? Because i'm pretty sure my answer is wrong because of my watts working out. If the total voltage is the same and the resistors are identical, the the current is exactly split across them, which would then give 12.5W + 12.5W each, changing the equation to  $V = \sqrt{(150)*(25 + 25)}$, but that also doesn't give the correct answer. The question was what is the total allowed voltage that can pass across them. I might still be interpreting it wrongly though. Should i try to find the voltage across the parallel and series separately and then  add them together?
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on May 10, 2015, 09:23:29 pm
Yes, that was a typo. My original workings were $P = \frac{V^2}{R}$ and solve for V such that $V = \sqrt{P*R}$. The total resistance is 150 ohms and if each resistor can have a max output of 25 W, then the total power output could be 75 W, so $V = \sqrt{(150)*(75)}$. Is the total power across the parallel resistors not 50W? Because i'm pretty sure my answer is wrong because of my watts working out. If the total voltage is the same and the resistors are identical, the the current is exactly split across them, which would then give 12.5W + 12.5W each, changing the equation to  $V = \sqrt{(150)*(25 + 25)}$, but that also doesn't give the correct answer. The question was what is the total allowed voltage that can pass across them. I might still be interpreting it wrongly though. Should i try to find the voltage across the parallel and series separately and then  add them together?

Actually because the resistance is down over the parallel branch, they will actually not have as much voltage as across the series resistor.
Title: Re: VCE Physics Question Thread!
Post by: odeaa on May 10, 2015, 09:45:25 pm
What does it mean to have a negative voltage? Is it only when you have a photodiode or something in reverse bias and hence a negative current?
Title: Re: VCE Physics Question Thread!
Post by: lzxnl on May 10, 2015, 10:04:31 pm
What does it mean to have a negative voltage? Is it only when you have a photodiode or something in reverse bias and hence a negative current?

Firstly, what is a voltage?
A voltage is a potential difference between two points. Well that doesn't help, does it?
The electric potential is defined as the electric potential energy per unit charge. So you know how things like to go from high to low potential energy? For a positive charge, this means go from high to low potential but for a negative charge, it means it'll want to go from low to high potential.

So a negative potential just means you've measured a positive potential in the other way. This has implications for devices that are direction-specific, like diodes.
Title: Re: VCE Physics Question Thread!
Post by: odeaa on May 10, 2015, 10:05:32 pm
Firstly, what is a voltage?
A voltage is a potential difference between two points. Well that doesn't help, does it?
The electric potential is defined as the electric potential energy per unit charge. So you know how things like to go from high to low potential energy? For a positive charge, this means go from high to low potential but for a negative charge, it means it'll want to go from low to high potential.

So a negative potential just means you've measured a positive potential in the other way. This has implications for devices that are direction-specific, like diodes.
Legend, thanks for clearing that up
Title: Re: VCE Physics Question Thread!
Post by: Floatzel98 on May 13, 2015, 09:19:50 pm
Hey guys. I have a couple questions today about Voltage RMS.

I have a physics SAC tomorrow on the first half of the Electronics and Photonics Area of Study and my teacher said today that there would be a question about Voltage RMS on it, even though we haven't actually come across it in the book yet. He explained it to us and basically told us that the question was just reading the peak AC current and finding the RMS Voltage, which is easy enough. I understand the formula and how to use it for the question we will get, but i still don't understand how they get it. In the Heinemann textbook it comes up in the further electronics AOS and quickly in Electric Power AOS, but I still wasn't able to understand it.

So basically my questions are:

Where/how do they derive it?
What is it? I gathered it was like the average voltage provided by an AC signal. I also read that it is the same the DC voltage for something? What is the relationship between them?
How does the current RMS work? Where do you ever find the peak current? Are there usually graphs of these or is it something you can solve after finding the voltage RMS?

Also i have some questions about AC and DC signals that i don't think i ever really understood, and because we are finally using them a lot more now i think my knowledge isn't really matching up with what i'm learning. And some other random questions if anybody can be bothered answering them

What actually happens in an AC supplied circuit. If it was connected to a light, does that constantly turn off and on?
What does a resistor actually do to the electrons in a circuit? Does it slow them down so less can pass through or..?
In a normal DC circuit for example, do the same amount of electrons keep running the circuit or do they just get replaced each time they finish it? Also, how do they get the electrons in the first place?
What are some different uses for AC and DC current? Like which one is usually used in our lights at home etc..?
How do you actually make an alternating current? Are magnets used to do this or something else?

Some of these might have simple answers that i probably already know. I feel like I've been trying to think to deep into some of these things and I keep confusing myself and throwing what i already know out the door. The first part is what i need help with now anyway, so if anybody can help with anything at all that would be much appreciated.

Thanks! :)

Title: Re: VCE Physics Question Thread!
Post by: Kel9901 on May 13, 2015, 10:27:12 pm
Hey guys. I have a couple questions today about Voltage RMS.

I have a physics SAC tomorrow on the first half of the Electronics and Photonics Area of Study and my teacher said today that there would be a question about Voltage RMS on it, even though we haven't actually come across it in the book yet. He explained it to us and basically told us that the question was just reading the peak AC current and finding the RMS Voltage, which is easy enough. I understand the formula and how to use it for the question we will get, but i still don't understand how they get it. In the Heinemann textbook it comes up in the further electronics AOS and quickly in Electric Power AOS, but I still wasn't able to understand it.

So basically my questions are:

Where/how do they derive it?
What is it? I gathered it was like the average voltage provided by an AC signal. I also read that it is the same the DC voltage for something? What is the relationship between them?
How does the current RMS work? Where do you ever find the peak current? Are there usually graphs of these or is it something you can solve after finding the voltage RMS?

Also i have some questions about AC and DC signals that i don't think i ever really understood, and because we are finally using them a lot more now i think my knowledge isn't really matching up with what i'm learning. And some other random questions if anybody can be bothered answering them

What actually happens in an AC supplied circuit. If it was connected to a light, does that constantly turn off and on?
What does a resistor actually do to the electrons in a circuit? Does it slow them down so less can pass through or..?
In a normal DC circuit for example, do the same amount of electrons keep running the circuit or do they just get replaced each time they finish it? Also, how do they get the electrons in the first place?
What are some different uses for AC and DC current? Like which one is usually used in our lights at home etc..?
How do you actually make an alternating current? Are magnets used to do this or something else?

Some of these might have simple answers that i probably already know. I feel like I've been trying to think to deep into some of these things and I keep confusing myself and throwing what i already know out the door. The first part is what i need help with now anyway, so if anybody can help with anything at all that would be much appreciated.

Thanks! :)

i'll just answer the first part.

Try to think about it this way.
In AC, both voltage and current have a sinusoidal shape. When multiplied together to give power, it will be a graph that looks something like (sin(x))^2, ie something like this http://imgur.com/lMxj3Hf. It's pretty visually obvious that the 'average' power is half the 'peak' (top) power.

Hence, P(RMS)=P(peak)/2
V(RMS)I(RMS)=V(peak)I(peak)/2=(V(peak)/sqrt(2))(I(peak)/sqrt(2)
V(RMS)=V(peak)/sqrt(2), and I(RMS)=I(peak)/sqrt(2).

The RMS voltage of an AC signal represents the DC voltage that would provide the same average power.

current is pretty much treated the same way a voltage.
Title: Re: VCE Physics Question Thread!
Post by: Floatzel98 on May 13, 2015, 10:44:49 pm
i'll just answer the first part.

Try to think about it this way.
In AC, both voltage and current have a sinusoidal shape. When multiplied together to give power, it will be a graph that looks something like (sin(x))^2, ie something like this http://imgur.com/lMxj3Hf. It's pretty visually obvious that the 'average' power is half the 'peak' (top) power.

Hence, P(RMS)=P(peak)/2
V(RMS)I(RMS)=V(peak)I(peak)/2=(V(peak)/sqrt(2))(I(peak)/sqrt(2)
V(RMS)=V(peak)/sqrt(2), and I(RMS)=I(peak)/sqrt(2).

The RMS voltage of an AC signal represents the DC voltage that would provide the same average power.

current is pretty much treated the same way a voltage.
Okay, the explanation makes a lot more sense, but I'm a bit confused as to where the root2 came from though. Maybe it's just the formatting that's confusing me. Can you explain those steps? Other wise thanks, it all makes a lot more sense now :)
Title: Re: VCE Physics Question Thread!
Post by: Kel9901 on May 14, 2015, 12:46:01 pm
Okay, the explanation makes a lot more sense, but I'm a bit confused as to where the root2 came from though. Maybe it's just the formatting that's confusing me. Can you explain those steps? Other wise thanks, it all makes a lot more sense now :)

If the RMS power is half the peak power, then the RMS voltage x the RMS current will be 1/2 the peak voltage x the peak current. This half is 'split' equally between the voltage and the current as 1/sqrt(2) each, because 1/sqrt(2) x 1/sqrt(2)=1/2
Title: Re: VCE Physics Question Thread!
Post by: Peanut Butter on May 17, 2015, 04:26:15 pm

Thanks :)
Title: Re: VCE Physics Question Thread!
Post by: Floatzel98 on May 17, 2015, 05:08:53 pm

Thanks :)
For the first one, all we need to do is go find the voltage across the 6000 ohm resistor done by V = IR. Voltage is constant in parallel and there are no other components in the circuit, so that equals the EMF of the battery.

Since voltage is the same in parallel, we can work out the current in the other resistor by going I = V/R. Use the voltage you just calculated and the resistance provided. If you wanted to check if this is right, you could find the total parallel resistance by going $(\frac{1}{4000} + \frac{1}{6000})^{-1}$ and divide the total voltage by that. The two currents should add up to that individually.

With the break in the wire, nothing will flow through there. Every should just go through the 6000 ohm resistor, basically putting it in series with the battery. You should be able to work out the current of a single resistor given the total voltage and the resistance.

For the last question (and i might be wrong on this) you can think of it as putting a voltmeter across the 6000 ohm resistor in the same series scenario. X1 is on before the resistor and x2 is after it. I don't think 4000 ohm resistor will change that, but it might. If we think of it as just being in series that is right, but if we are literally testing from point x1 to x2, you would still have the 4000 ohm resistor in the way. But no current is flowing through it anyway, so it shouldn't matter.
Title: Re: VCE Physics Question Thread!
Post by: Kel9901 on May 18, 2015, 10:37:18 am
What actually happens in an AC supplied circuit. If it was connected to a light, does that constantly turn off and on?
No, the frequency (how quickly the current goes up and down) is so high that humans don't notice it; it appears to stay at constant intensity
What does a resistor actually do to the electrons in a circuit? Does it slow them down so less can pass through or..?
Pretty much slows them down, yeah. The resistor pretty much makes the electrons give up some energy (potential) to get past it.
In a normal DC circuit for example, do the same amount of electrons keep running the circuit or do they just get replaced each time they finish it? Also, how do they get the electrons in the first place?
The same electrons run the circuit, and are 'recharged' or given back the potential they lost when going through the resistors when they reach the battery. The electrons come from the wires/resistors, as they are usually metal and hence have free moving electrons.
What are some different uses for AC and DC current? Like which one is usually used in our lights at home etc..?
I think pretty much everything uses AC. The main exception I can think of, aside from lab experiments (lol), is trains, which use DC power.
How do you actually make an alternating current? Are magnets used to do this or something else?
You'll learn this in unit 4, but yeah, magnets.
Some of these might have simple answers that i probably already know. I feel like I've been trying to think to deep into some of these things and I keep confusing myself and throwing what i already know out the door. The first part is what i need help with now anyway, so if anybody can help with anything at all that would be much appreciated.

Thanks! :)
Title: Re: VCE Physics Question Thread!
Post by: Floatzel98 on May 18, 2015, 03:26:26 pm

Thanks!
Title: Re: VCE Physics Question Thread!
Post by: Adequace on May 18, 2015, 06:02:59 pm
What's the best way to revise for a mid-year exam? Practice exams only?
Title: Re: VCE Physics Question Thread!
Post by: Cosec on May 18, 2015, 06:43:25 pm
What's the best way to revise for a mid-year exam? Practice exams only?

Yolo, whats revision for practice exams?! Haha,
Yeah, thats what ill be doing! Smashing dem out. If i can find them. Do you know any sources that has a bunch of them?
Title: Re: VCE Physics Question Thread!
Post by: odeaa on May 18, 2015, 07:34:29 pm
Yolo, whats revision for practice exams?! Haha,
Yeah, thats what ill be doing! Smashing dem out. If i can find them. Do you know any sources that has a bunch of them?
our teacher gave us a USB with all the VCAA exams back to 1997, ask your library I think they will have  a CD with all the past exams
Title: Re: VCE Physics Question Thread!
Post by: Floatzel98 on May 19, 2015, 04:07:36 pm
Can someone help explain photodiodes to me. I'm still kind of confused how they work and especially how the I-V characteristic graph works. I'm not really interpreting it properly. If they are placed in reverse bias, how do they even work? I think i'm a bit confused about the "dark current" and how it is negative. Also, what is the photoconductive mode and the photovoltaic mode?

Thanks!

EDIT: Whoops i had another question as well (now attached). I don't exactly understand how they got that equation for part b, the number of photons part. Is this something we should know as well. Because i have never really come across anything like it where we introduce something into an equation like that.
Title: Re: VCE Physics Question Thread!
Post by: silverpixeli on May 19, 2015, 04:17:51 pm
Can someone help explain photodiodes to me. I'm still kind of confused how they work and especially how the I-V characteristic graph works. I'm not really interpreting it properly. If they are placed in reverse bias, how do they even work? I think i'm a bit confused about the "dark current" and how it is negative. Also, what is the photoconductive mode and the photovoltaic mode?

Thanks!

quick and dirty answer because i gotta run, but basically you place the photodiode in reverse bias and when light lands on it it lets a little trickle of current through (negative w.r.t. photodiode, but since the diode is reversed this is positive w.r.t. circuit as in, it's in the same direction as the voltage drop across the photodiode and the way conventional current would flow if there wasnt a diode there blocking it). usefully, this means we can get current that depends on light. the dark current is just the trickle that leaks through the diode when there's no light falling on it.

The graphs for these are most interesting in the negative voltage region (whichever of photovoltaic/photoconductive this corresponds to, i cant remember) because of this light-dependent current property, and the graphs usually have multiple lines telling us what current leaks through for a certain intensity. since these lines are flat, higher reverse-bias voltages dont affect the leakage current - it is only dependent on light intensity (for an ideal photodiode anyway)

the other mode isnt as interesting and i cant remember but i think it just functions like a regular diode when in forward bias, at least that's what the graphs suggest for voltages > switch on voltage.
Title: Re: VCE Physics Question Thread!
Post by: Kel9901 on May 19, 2015, 05:48:40 pm
Can someone help explain photodiodes to me. I'm still kind of confused how they work and especially how the I-V characteristic graph works. I'm not really interpreting it properly. If they are placed in reverse bias, how do they even work? I think i'm a bit confused about the "dark current" and how it is negative. Also, what is the photoconductive mode and the photovoltaic mode?

Thanks!

EDIT: Whoops i had another question as well (now attached). I don't exactly understand how they got that equation for part b, the number of photons part. Is this something we should know as well. Because i have never really come across anything like it where we introduce something into an equation like that.

there is a lot of stuff in the electronics+photonics section of heinemann that is irrelevant. this is one of those things; phototransistors aren't on the course.

to answer your question, the formula for the energy of a single photon is E=hf, where h is Planck's constant and f is the frequency (which is also equal to speed of light/wavelength). you'll learn about it in unit 4
Title: Re: VCE Physics Question Thread!
Post by: Floatzel98 on May 19, 2015, 06:27:23 pm
quick and dirty answer because i gotta run, but basically you place the photodiode in reverse bias and when light lands on it it lets a little trickle of current through (negative w.r.t. photodiode, but since the diode is reversed this is positive w.r.t. circuit as in, it's in the same direction as the voltage drop across the photodiode and the way conventional current would flow if there wasnt a diode there blocking it). usefully, this means we can get current that depends on light. the dark current is just the trickle that leaks through the diode when there's no light falling on it.

The graphs for these are most interesting in the negative voltage region (whicheve