Cort's 3/4 Physics Thread

[Rod]

Thank you for that Izxnl. Since I'm nearly finished doing some scouting for photonics (before I dive into a real understanding), I just wanted to ask: What is the mentality that you have/use/adopt when you're imagining and thinking about frequency waves/signals? It's a concept that I really cannot apply into any situation as of yet, because it feels so paper thin. That kinda sucks because this makes it harder for me to grasp modulation.

Thanks,
Cort.

,
Think of you computer! Go to a speed check and check out your bandwith. Go to phone and see what kind of wire it is, is it copper, coaxial or optical Use a shitty phone to prank call someone in Ireland, how attenuated does your signal get?

Is that better?

Sorry for not replying sooner, I read this but I skimmed it any didn't know you were asking a question

Rod

[Cort]

Cheers, Roddo.

I've been attempting to answer some motion basics again. Frankly, it's quite an embarrassment. I even struggle with Checkpoints motion basic. This is not good at all.

A car travelling with a constant speed of 80kmh^-1 passes a stationary motorcycle policeman. The policeman sets off in pursuit, accelerating uniformly to 80kmh^-1 in 10.0s and reaching a constant speed of 100 kmh^-1 after a further 5s. At what time will the policeman catch up with the car?

This was from Heineman Physics 12.

On another note, I think the biggest problem for me is that, I have no idea what to do with the numbers/equations after I've been given. Even though I'm aware of the problem, and what's its asking me, I can't seem to make the connection from thinking outside the box.  I think an equation is an equation, and that's it. Plug and play. Perhaps I need to just 'practice more'? Any tips to break this restraint?

[lzxnl]

First of all, learn what all of the symbols in each equation mean. For instance, in v^2 = u^2 + 2ax, v is the final velocity, u is the initial velocity, a is the acceleration and x is the displacement. This equation is only valid for constant acceleration, so a is constant. Also, x is the difference in position of the particle from when the velocity is u to when to the velocity is v. Don't just rote learn and formula bash.

So for your equation, for instance, the policeman will catch up with the car when it travels the same distance as the car.
The question you gave is slightly complicated. Let us break it down.
After a time t hours, the car will travel 80 km/h * t hours = 80t km.
The policeman's movement falls into three stages. When accelerating to 80 km/h in 10s or 1/360 hours, we're given an initial speed (0), final speed (80 km/h) and time (10 s). To find the displacement, we use x=t(u+v)/2 = 1/360 hours*(0+80)/2 km/hr = 1/9 km.
Then, in the second stage, the policeman goes from 80 to 100 km/h in 5 s. Using the above formula again, we have x = t/2*(u+v) = 1/2*5/3600 hours*(80+100)/2 km/hr = 450/7200 km = 45/720 km = 9/144 km = 1/16 km. (as you can see I'm doing this as I go LOL)

Note how I've written in all of the units. You can cancel them like they were regular pronumerals and this shows you the importance of converting units; if I had 1 km/h * 1 second = 1 km s/h, I would have to use the fact that 1 second is 1/3600 of an hour to get 1/3600 km.

In the third stage, the policeman travels for a time t hours - 15 seconds = (t-15/3600) hours = (t-1/240) hours. As it moves at a constant 100 km/hr speed, the distance travelled in this third stage is speed*time = (t-1/240)*100 km. In total, for a time t, the policeman travels 1/16 + 1/9 + 100(t-1/240) km. This is equal to the distance travelled by the car, so set this equal to 80t. Solve for t and you'll have the time in hours.
I get 20t = 100/240 - 25/144 = 50/120 - 25/144 = 60/144 - 25/144 = 35/144
t = 7/576 hours. Someone please check my working. In an exam, please use your calculator

[Cort]

First of all, learn what all of the symbols in each equation mean. For instance, in v^2 = u^2 + 2ax, v is the final velocity, u is the initial velocity, a is the acceleration and x is the displacement. This equation is only valid for constant acceleration, so a is constant. Also, x is the difference in position of the particle from when the velocity is u to when to the velocity is v. Don't just rote learn and formula bash.

So for your equation, for instance, the policeman will catch up with the car when it travels the same distance as the car.
The question you gave is slightly complicated. Let us break it down.
After a time t hours, the car will travel 80 km/h * t hours = 80t km.
The policeman's movement falls into three stages. When accelerating to 80 km/h in 10s or 1/360 hours, we're given an initial speed (0), final speed (80 km/h) and time (10 s). To find the displacement, we use x=t(u+v)/2 = 1/360 hours*(0+80)/2 km/hr = 1/9 km.
Then, in the second stage, the policeman goes from 80 to 100 km/h in 5 s. Using the above formula again, we have x = t/2*(u+v) = 1/2*5/3600 hours*(80+100)/2 km/hr = 450/7200 km = 45/720 km = 9/144 km = 1/16 km. (as you can see I'm doing this as I go LOL)

Note how I've written in all of the units. You can cancel them like they were regular pronumerals and this shows you the importance of converting units; if I had 1 km/h * 1 second = 1 km s/h, I would have to use the fact that 1 second is 1/3600 of an hour to get 1/3600 km.

In the third stage, the policeman travels for a time t hours - 15 seconds = (t-15/3600) hours = (t-1/240) hours. As it moves at a constant 100 km/hr speed, the distance travelled in this third stage is speed*time = (t-1/240)*100 km. In total, for a time t, the policeman travels 1/16 + 1/9 + 100(t-1/240) km. This is equal to the distance travelled by the car, so set this equal to 80t. Solve for t and you'll have the time in hours.
I get 20t = 100/240 - 25/144 = 50/120 - 25/144 = 60/144 - 25/144 = 35/144
t = 7/576 hours. Someone please check my working. In an exam, please use your calculator

I appreciate the help, but I'm afraid you'll have to drop a view IQ points whilst I'm here. I can understand the first stage and how you made the situation -- but what boggles me is why we have to find the displacement, when it just asks at the time that it's catching up to the car? Is it because we have to find time, we normally find the distance..or? Also, I lobbed "180 *(5=3600) /2" and it instead gave me 1/8 instead of 1/16km. But I apologise, beyond that I cannot understand it further, (that's my vault on my end), especially when you're talking about the third stage.

[hongkyho]

Also, I lobbed "180 *(5=3600) /2" and it instead gave me 1/8 instead of 1/16km.

You are right

why we have to find the displacement

Think of this situation like you would for two simultaneous questions in maths. Common knowledge in maths says that for two simultaneous equation, we need to find two variables that are present in both equations; in this case, displacement and time.

Or, just look at the question. It states: At what time will the policeman catch up with the car? The policeman will only catch up to the car once it has traveled the same distance from the starting point. Therefore, the time element we are trying to look for is specific to this certain condition.

The answer you should get is 65/2 seconds or 13/1440 hours.

Hope this helps

[lzxnl]

I appreciate the help, but I'm afraid you'll have to drop a view IQ points whilst I'm here. I can understand the first stage and how you made the situation -- but what boggles me is why we have to find the displacement, when it just asks at the time that it's catching up to the car? Is it because we have to find time, we normally find the distance..or? Also, I lobbed "180 *(5=3600) /2" and it instead gave me 1/8 instead of 1/16km. But I apologise, beyond that I cannot understand it further, (that's my vault on my end), especially when you're talking about the third stage.

I plugged my working into a calculator and I still seem to get 1/16. Hmm.
Note that in my working, it's 1/2*5/3600 hours*(80+100)/2 km/hr. On the numerator we have 1*5*180. On the denominator we have 2*3600*2. Dividing these through seems to get 1/16 for me.

As hongkyho said, the question is fundamentally asking for when the policeman's displacement is equal to that of the motorist's.
This question works in three stages, as there are three stages of motion. The policeman travels for 10 and 5 seconds in the first and second stages respectively. The distance travelled by the cyclist is given by 80t, where t is measured from the beginning. At the end, the policeman travels at a constant 100 km/h, so his distance travelled is 100T, where T is the time he actually travels at 100 km/h.
As the time for the motorcyclist and the policeman move the same, the t for the cyclist is the same t for the policeman. However, the policeman has spent 15 seconds in the first two stages of motion, so he really only spends t-15 seconds in the third stage of constant speed. That's why I've put in 100(t-15) is the distance travelled in the third stage.

[hongkyho]

Note that in my working, it's 1/2*5/3600 hours*(80+100)/2 km/hr.

I think you are halving it twice.

[lzxnl]

Yeah I see now.

The flaws of working things out on a computer screen

[Cort]

Right now I'm doing past exam questions and some checkpoints. The thing is, I'm getting them wrong. I'm writing the reason why it is..but my concern is, is there a way that you suggest  to actually remember the mistakes you've did? Keeping some kind of error notebook? So that if you're faced with a similar encounter, you whip it out again? My memory has never been great, nor has my logical reasoning.

I normally understand what to do if the question is the same for at least 3/4 times...however this doesn't happen. So, suggestions?

[PB]

I think keeping an error book is only good for exam prep...because if you update it throughout the year while doing checkpoints it would just be too much ttime and effort due to the inevitable mountain of mistakes you will make.
What I find really useful is to note down a general warning in a box in your notes of that topic. Say for motion, if you keep forgetting to convert kmh-1 to ms-1. Put a WARNING sign in your motion notes and note down that you should remember to convert etc.  This is so that when you are reading through notes prior to SACs, these little notices helps you cover each weakness you might have for each topic.

[Rod]

There are 783 questions in the 2014 checkpoints D::::

Where are you up to Cort? I'm only up to Q 21, but I've been doing more chem and health checkpoints.

We should keep each other update to keep up motivated !!

PS - PB, Lxnl - did you freaks end up completing all 783 qs? When do you think me and Cort should complete them by?

Cheers

[lzxnl]

There are 783 questions in the 2014 checkpoints D::::

Where are you up to Cort? I'm only up to Q 21, but I've been doing more chem and health checkpoints.

We should keep each other update to keep up motivated !!

PS - PB, Lxnl - did you freaks end up completing all 783 qs? When do you think me and Cort should complete them by?

Cheers

Some of the questions in Checkpoints are for a different detailed study; I didn't do those.
But yes, I did all the other questions in the physics 2013 Checkpoints that were relevant.

I think I completed the questions in SWOTVAC LOL

[Cort]

Heya Rod. I'm about up to 40-something. But I'm being butt-blasted in every question it's not funny at all. And thanks again PB, great help, haha.

Now, I've got some questions concerning tension itself. I understand that it acts as the 'driving force' for the object that it is being pulled (eg. trailer). My general conceptual question is: How do I exactly know which equation do I exactly use? Or, simply, how should I be aware of what the question is actually asking me to find? I'll demonstrate. I'll compare the questions, what I did and what the answers was.

ex.1: A car of mass 1000kg is being towed on a level road by a van of mass 2000kg. There is a constant retarding force due to air resistance and friction, of 500N on the van, and 300N on the car. The vehicles are travelling at a constant speed.
   Q: What is the value of the tension, T, in the towbar?
My answer and reasoning: Since tension is being pulled BY the van, that means that tension is only being acted on the car. Hence:
T-Fr= Fnet. Since fnet=0 from constant speed, then
T-Fr=0.
T-500N (from the van)=0.
Thus T=500N. This question was wrong, because T=300N from the car itself, not that van that was towing it. Why is this so? I thought tension was dependent on the object that is pulling it..hence the FR from the van. An object that is being pulled (the car) provides no form of any tension.


ex2. A car of mass 1300kg has a caravan of mass 900kg attached to it. The car and caravan move off from rest. They have an initial acceleration of 1.25ms^2.
Q: What is the tension in the coupling between the car and the caravan as they start to accelerate?
Again, my reasoning was this: Since tension occurs from the car pulling from the car, the only orce acting on the trailer is tension. Thus to find the tension, I did this.
T-Fr=Fnet. Since Fr is not provided, I assumed it was zero.
T=2750N.
This is again wrong; as the answer states that Fd is being provided solely by the coupling for the trailer. Hence, T=m*a
T= 900kg*1.25ms^2.

So, besides wondering why I got those two questions wrong, my third question is from these two examples: How do I know which type of 'tension' am I meant to exactly find? I cannot seem and put my tongue on it, but evidently there's a fault in my logical thinking that I'll link to amend. Do you might know the reason why I am thinking like this?

Thanks,
Cort.

[Cort]

Some of the questions in Checkpoints are for a different detailed study; I didn't do those.
But yes, I did all the other questions in the physics 2013 Checkpoints that were relevant.

I think I completed the questions in SWOTVAC LOL

Yes, but given your scores, I think you're a freak of nature (No intent to offend). But really, you are.

[lzxnl]

Heya Rod. I'm about up to 40-something. But I'm being butt-blasted in every question it's not funny at all. And thanks again PB, great help, haha.

Now, I've got some questions concerning tension itself. I understand that it acts as the 'driving force' for the object that it is being pulled (eg. trailer). My general conceptual question is: How do I exactly know which equation do I exactly use? Or, simply, how should I be aware of what the question is actually asking me to find? I'll demonstrate. I'll compare the questions, what I did and what the answers was.

ex.1: A car of mass 1000kg is being towed on a level road by a van of mass 2000kg. There is a constant retarding force due to air resistance and friction, of 500N on the van, and 300N on the car. The vehicles are travelling at a constant speed.
   Q: What is the value of the tension, T, in the towbar?
My answer and reasoning: Since tension is being pulled BY the van, that means that tension is only being acted on the car. Hence:
T-Fr= Fnet. Since fnet=0 from constant speed, then
T-Fr=0.
T-500N (from the van)=0.
Thus T=500N. This question was wrong, because T=300N from the car itself, not that van that was towing it. Why is this so? I thought tension was dependent on the object that is pulling it..hence the FR from the van. An object that is being pulled (the car) provides no form of any tension.


ex2. A car of mass 1300kg has a caravan of mass 900kg attached to it. The car and caravan move off from rest. They have an initial acceleration of 1.25ms^2.
Q: What is the tension in the coupling between the car and the caravan as they start to accelerate?
Again, my reasoning was this: Since tension occurs from the car pulling from the car, the only orce acting on the trailer is tension. Thus to find the tension, I did this.
T-Fr=Fnet. Since Fr is not provided, I assumed it was zero.
T=2750N.
This is again wrong; as the answer states that Fd is being provided solely by the coupling for the trailer. Hence, T=m*a
T= 900kg*1.25ms^2.

So, besides wondering why I got those two questions wrong, my third question is from these two examples: How do I know which type of 'tension' am I meant to exactly find? I cannot seem and put my tongue on it, but evidently there's a fault in my logical thinking that I'll link to amend. Do you might know the reason why I am thinking like this?

Thanks,
Cort.

Let's look at the forces on the car in the first question. There are only two forces on that: tension, and the resistive force of 300 N. As the car is moving at constant speed (presumably in a straight line), the net force is zero. Thus, the tension must equal the resistive force. Here, the resistive force on the van only affects the driving force necessary to maintain a constant speed, not the tension force itself.
Just consider tension as a regular force with a direction. Then, consider objects separately.

If you apply this logic to the second question, you see that the tension is the only thing accelerating the caravan (the tension slows the car down; a larger driving force is thus required for the car to tow a caravan, but the driving force isn't mentioned in these questions; perhaps that is confusing you). Now, your mistake here was assuming that the mass was the mass of the combined car-caravan system.  If you just consider the caravan separately and replace the tension force with any other pulling force, you see that the pulling force is the only force responsible for the acceleration on the caravan. Also, as the car is now not considered (this pulling force could be any pulling force), the mass in question is the mass of the caravan only. Therefore, we have T=ma=900*1.25=1125 N.

In any question with systems of objects, consider each object separately and find out what forces are actually acting on each object. The car only provides tension to the caravan and trailer, nothing more.

Yes, but given your scores, I think you're a freak of nature (No intent to offend). But really, you are.

Me? Freak of nature? I'll take that as a compliment