Mr. Study's 3 and 4 Question.

[Bhootnike]

Oh! I see why you would do the net speed and NOT the individual components of that net speed.

Thanks for that Bhootnike.

Haha yeah, gotta be careful :p
no worries 

[Mr. Study]

A car is travelling on a straight highway, there is a constant resistive force of 2500N in total against the motion. Assume the motor drives the front wheels only.

a) Draw an arrow to represent the force of the road on the rear wheels due to friction.
     I represented it as the red arrow.
b) Draw an arrow to represent the force of the road on the front wheels, due to friction.
     I do not know, as I answer this as the same direction as the red arrow but on the centre of the affected wheels but that is always the wrong answer.

I don't want the questions answered, I would like to know why exactly, in the answers, the frictional force is usually diagonal?
(Incase, My diagram is completely wrong but that was how I answered it).

Also, Since the motor only affected the front wheels, I assumed that the only force acting on the rear tyres were the friction of the road, with the direction on the diagram, but thats wrong. :S

Thank you to anyone who can help me.

[yawho]

To keep the rear wheels turning clockwise against the resistive force, the force of the road on the rear wheels due to friction is shown by the red arrow.
To keep the the tanker moving forward against the resistive force, the motor drives the front wheels clockwise. The front wheels push the road backward due to friction, so the road pushes the front wheels forward.

[Mr. Study]

Thanks yawho.

Just this question from the itute 2008 exam



I'm trying to figure out the total force. I drew it to be pointing towards the centre of the circle but the answer is slightly below it. I tried splitting the total force into its components, which I think, should be the frictional force, going in the opposite direction of the boxs movement, and the driving force, as the platform speed is increasing which would mean acceleration is being applied to the box, through the platform.

Are my assumptions correct though?

[Lasercookie]

First off, what on earth is that squiggly line on that stick figure dude's head? Also, I think that this question might be a bit off the course with the coefficient of friction and other stuff.

Anyway, the platform is increasing in speed - the motion here is not uniform circular motion (so like what you have for the vertical circular motion, with the rollercoasters and speed humps etc). There's now more factors involved. The velocity of the parcel here is not only changing in direction, but also in magnitude. This means that acceleration is also affected. As you would know from uniform circular motion, the centripetal acceleration is due to the constantly changing direction of the velocity.

You would also know that the instantaneous velocity at any point is at a tangent to the circle. However, since this velocity is now changing in magnitude, there is an acceleration in this tangential direction. So this alters the acceleration a bit too (you now have to take the vector sum of the two components of acceleration here). This means that the net acceleration no longer points directly towards the centre of the circle and therefore also the same for the direction of the net force.

So yeah, the reasoning above should make it clear why it wouldn't be directly to the centre. On to actually figuring out what it should be: The direction of net force is given by the direction of acceleration, right? ( e.g. )

So we need to figure out the direction of the net acceleration - which will be the vector sum of the 'centripetal acceleration' and the 'tangential acceleration' (I think there was a better word to describe that second one, can't remember it). If you draw them, then you should get something facing slightly away from the centre of the circle.

[DisaFear]

I drew it to be pointing towards the centre of the circle but the answer is slightly below it.

You mean like this?

[Mr. Study]

Haha, Thank you so much for that laseredd.

I see the reasoning behind it now!

Disafear, the answer is actually slightly higher than the answer in your diagram. Maybe VCAA exams will be a little more... lenient? haha

This question was definitely raising my hair as I never dealt with circular motion as a result of a rotating 'table' or what ever. :O

Thank you laseredd, for the explanation, and Disafear, for the funny editing.

[Lasercookie]

Disafear, the answer is actually slightly higher than the answer in your diagram. Maybe VCAA exams will be a little more... lenient? haha

This question was definitely raising my hair as I never dealt with circular motion as a result of a rotating 'table' or what ever. :O

You won't see it on a VCAA exam, it's way off the course (not sure about in 2008 though). It is a nice question though.

But yeah, non-uniform circular motion in VCE Physics extends to this:

• apply Newton’s second law to circular motion in a vertical plane; consider forces at the highest and lowest positions only;

At the highest and lowest points of vertical circular motion, you have that moment where the speed is constant, so no tangential acceleration, leaving only centrepretal acceleration hence the net force being towards the centre.

Coefficient of friction is not on the course either.

I'm guessing you'll give a shot at the other questions on that exam (probably don't worry about them too much, but they are interesting), so looking at that set of questions on the iTute exam, it refers to reference frames too. I think at one stage this was under the motion section of VCE Physics, but now the only mention of reference frames is in the special relativity detailed study.

The iTute question referred to accelerating frames of reference (special relativity relates only to frames of references with constant speeds), this page here is probably the best for a quick explanation of reference frames:
http://www.phys.unsw.edu.au/einsteinlight/jw/module1_Inertial.htm

[Mr. Study]

Thank you so much for that laserred!

Very, very, informative post!

[Mr. Study]

Okay, This is more of a general question as the Insight 2012 solutions are being really inconsistent.

I got question 10 right but question 9 wrong. >.>

I got the answer, for Q 9, as 18 ms^-1 left and I was assigning right to be positive and left to be negative but the solutions say it doesn't matter. They justify this by saying that velocity is a scalar, and that is why it's 2 ms^-1.
 >.>

However, in solutions for Question 10, they don't use the same value for the change in velocity but it's now 18 ms^-1.

Could someone help me out?

Thank you very much.

[Bhootnike]

Okay, This is more of a general question as the Insight 2012 solutions are being really inconsistent.

I got question 10 right but question 9 wrong. >.>

I got the answer, for Q 9, as 18 ms^-1 left and I was assigning right to be positive and left to be negative but the solutions say it doesn't matter. They justify this by saying that velocity is a scalar, and that is why it's 2 ms^-1.
 >.>

However, in solutions for Question 10, they don't use the same value for the change in velocity but it's now 18 ms^-1.

Could someone help me out?

Thank you very much.

They dont say velocity is scalar - the question asked for speed, (which is indeed scalar) and so you don't have to work with vectors. thus 2m/s. change in speed involves no direction. in the worked solutions however, they seem to be mixing it up, as in, theyve used FINAL - INITIAL, which is not what you would do with scalars i think, i.e. , with scalars theres no direction so to find the change in speed, your gonna say, what was the speed at the start and what is it now. so 10 -8 = 2.
but maybe for the sake of using the formula, theyve used v - u,?

secondly, for q10, it asked for the change in magnitude of momentum. momentum is a vector quanitity, so you have to work with vectors. momentum = velocity x mass.
so when considering velocity, you have to take careful note of the vectors, as you've done in q9 to get -18.

[Lasercookie]

They dont say velocity is scalar - the question asked for speed, (which is indeed scalar) and so you don't have to work with vectors.
...
but maybe for the sake of using the formula, theyve used v - u,?

Yeah, basically what Bhootnike said, just because they used 'v' doesn't mean they are referring to velocity.

You'll find that 'v' is often used to represent speed as well. It's also a common convention to come across things like using 's' to represent distance, displacement and a few other variable choices that might seem a bit weird. It really doesn't matter what convention you choose, but you shouldn't be getting tripped up by them. Taking into context is important - e.g. with the whole 'I' can refer to current, imaginary number stuff, moment of inertia or sometimes also Impulse.

Another thing is:

they seem to be mixing it up, as in, theyve used FINAL - INITIAL, which is not what you would do with scalars i think, i.e. , with scalars theres no direction so to find the change in speed, your gonna say, what was the speed at the start and what is it now. so 10 -8 = 2

I would always use final - initial. Why? Well what if you wanted to take into account which way the change is (is it increasing or is it decreasing). In this case you have 8 m/s - 10 m/s = -2 m/s, which you can then interpret the negative to mean: A decrease of 2 m/s. This has taken into account the fact that speed is scalar, as we haven't taken into account the directions.

You'll also notice that the solutions do allow an answer of -2 m/s for the change in speed. It also allows the interpretation that it was asking for the magnitude of the change in speed ("how much the speed changed"), and hence ignore the negative and provide the answer of 2 m/s. They're both correct answers, it's a minor detail not worth worrying about and I don't believe VCAA would discriminate on this.

However, the importance of using 'final - initial' is significant. What if you went onto do more things with the value? If you had gotten the answer using the incorrect means of 'initial - final', and hence ignored if the values was a decrease or increase, it would affect subsequent calculations.

Consider if we had a man that was 1 metre to the right of a pole. And let's say that he moved 2 metres further right, ending up to be 3m away from the tree. How do we represent this displacement?

x = final - initial = 3 - 1 = 2m right.

Let's say for simplicity that he took 1 second to move, what was his velocity?
v = x/t = 2/1 = 2 m/s right.

If we had used 'initial - final', we would have ended up with x = -2 = 2 m left and v = 2 m/s left, an answer for velocity that is clearly incorrect (since we explicitly stated that the man moves to the right)

[Mr. Study]

Thank you very much for the help guys, much appreciated.

Just this quick question on momentum. The question, from a 2011 exam, asked Can blah blah object feel this after blah blah collided with it. I answered yes but the answer was '1 x 10-15 m is about the diameter of a proton. So the Earth moves an incredibly small distance in 1 second. We could not detect this movement'.

So my question is should I even bother with this answer? Or Would you guys like me to post up the question? I thought this answer was a 'wtf moment'.

Thanks!

EDIT: Just wanted to add this question below.

I could not work it out and hence, looked at the answers.  However, the solution was to do F x d + mgh = 200 x 15 + 80 x 10 x 2 = 11000J.

However, I thought it wanted to work out the work going up hill and NOT the part where ben is going down.

Could someone show me a way to solve this question? OR are the solutions wrong?!

Thank you so much again! 

Worked out why the answer is the way it is!

EDIT 2: My guaranteed last questions for today. If a question asks to explain your reasoning, Are calculations and a bit of an explanation on the calculations more than enough to gain marks?

Regarding the cheat sheet for the exam, It says 'up to two pages (one A4 sheet)'. Would that mean we can bring in two sheets of paper but NOT double sided or one sheet of paper that is double sided?

Thank you again! 

[Bhootnike]

So you're good with that question?

And, yeah calculations if necessary in the context of the question, and an explanation would be good. e.g. for the questions in momentum when they ask you, what is the benefit of a crash barrier/crumple zone? - you'd show your understand through the equation and knowledge of impulse and momentum.

I think its just a double sided a4. cause 2 pages implies you can double up on the 1 sheet, and cause it says in the brackets , 'one a4 sheet' i think that further supports what i just said :p

[Mr. Study]

Thanks for that Bhootnike.

I answered all my question already. o.o

Could anyone clarify my understanding of energy conservation.

Scenario 1: A guy on a high building jumps. This potential energy is converted to kinetic energy. So that would mean we could do mgh=0.5mv^2. However, if we could, why can we say they equal? Wouldn't intial energy = potential energy + kinetic? So mid way through the jump, the total energy in the system is the same but the proportions of kinetic/potential are different when compared to before the jump.

Does that make sense? :S

Scenerio 2: The same guy, on a high building, has a bungee cord attached to his leg. He runs then jumps off. Would we then say kinetic energy and potentional energy will add up to give elastic potential energy in the bungee cord?

Thank you very much! And sorry for the basic questions but I just need to clarify my assumptions. :