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March 28, 2024, 10:53:38 pm

Author Topic: VCE Physics Question Thread!  (Read 603200 times)  Share 

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Zealous

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Re: VCE Physics Question Thread!
« Reply #930 on: April 11, 2015, 10:19:44 pm »
+1
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.
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Floatzel98

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Re: VCE Physics Question Thread!
« Reply #931 on: April 12, 2015, 12:51:07 am »
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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 :)
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Re: VCE Physics Question Thread!
« Reply #932 on: April 12, 2015, 10:35:12 am »
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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?

Kel9901

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Re: VCE Physics Question Thread!
« Reply #933 on: April 13, 2015, 11:09:11 am »
+1
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)
s=change in displacement for physics
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Floatzel98

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Re: VCE Physics Question Thread!
« Reply #934 on: April 15, 2015, 09:55:39 pm »
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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 :)



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Floatzel98

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Re: VCE Physics Question Thread!
« Reply #935 on: April 17, 2015, 04:31:24 pm »
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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?
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Zealous

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Re: VCE Physics Question Thread!
« Reply #936 on: April 17, 2015, 04:54:43 pm »
+2
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.
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Floatzel98

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Re: VCE Physics Question Thread!
« Reply #937 on: April 19, 2015, 01:18:50 pm »
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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?
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Adequace

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Re: VCE Physics Question Thread!
« Reply #938 on: April 19, 2015, 01:57:25 pm »
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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.

Floatzel98

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Re: VCE Physics Question Thread!
« Reply #939 on: April 19, 2015, 01:59:38 pm »
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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?
« Last Edit: April 19, 2015, 02:29:44 pm by Floatzel98 »
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silverpixeli

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Re: VCE Physics Question Thread!
« Reply #940 on: April 19, 2015, 06:19:47 pm »
+1
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

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.
« Last Edit: April 19, 2015, 06:26:27 pm by silverpixeli »
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Floatzel98

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Re: VCE Physics Question Thread!
« Reply #941 on: April 19, 2015, 07:14:33 pm »
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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

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!
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Re: VCE Physics Question Thread!
« Reply #942 on: April 20, 2015, 08:55:17 pm »
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When they say initial velocity does this always mean that time=0secs?
When is it like or not like this?

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Re: VCE Physics Question Thread!
« Reply #943 on: April 20, 2015, 09:17:20 pm »
+1
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
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Re: VCE Physics Question Thread!
« Reply #944 on: April 20, 2015, 09:29:50 pm »
+1
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).
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