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# Edexcel M2/M3 June 6th/10th 2013 watch

1. (Original post by Anonymous1717)
Thanks so much. I actually understand it now.
Ahh its fine, i was actually so worried, thought noone would make sense of it!
2. For COM what is the difference between the angle it makes with the vertical and angle it makes with the horizontal?
3. Does anyone happen to have the official Jan 2013 mark scheme?
4. Right M2 question that keeps popping up about interpreting collisions equations. How do you know the direction/velocity/speed by looking at equations involving e and u. The example I can give is in the mark scheme of June 2007. However you solve the answer for the velocity is u/6(1-5e) but the mark scheme says you should interpret the 'speed' as u/6(5e-1), how do you see the difference and what does it mean in terms of the symbols i can't find a tutorial anywhere as to why the markscheme's done this
5. (Original post by harisd95)
Right M2 question that keeps popping up about interpreting collisions equations. How do you know the direction/velocity/speed by looking at equations involving e and u. The example I can give is in the mark scheme of June 2007. However you solve the answer for the velocity is u/6(1-5e) but the mark scheme says you should interpret the 'speed' as u/6(5e-1), how do you see the difference and what does it mean in terms of the symbols i can't find a tutorial anywhere as to why the markscheme's done this
You know that e is greater than a fifth therefore the velocity of the sphere is negative. The question asks for the speed of the sphere rather than the velocity. Since speed is a scalar it only has magnitude and no direction so it can't be negative.

Hope that helps

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6. This is similar to a previous poster's problem, but:

Two particles, P, of mass 2m, and Q, of mass m, are moving along the same straight line on a smooth horizontal plane. They are moving in opposite directions towards each other and collide. Immediately before the collision the speed of P is 2u and the speed of Q is u. The coefficient of restitution between the particles is e, where e < 1. Find, in terms of u and e,
(i) the speed of P immediately after the collision, (ii) the speed of Q immediately after the collision.

Why doesn't the following method work?

2m * 2u - m*u = 2mv1 + 2mv2

v2 = 3u - 2v1

v1 = ((3u - v2)/2 - v2) / 3u

e= (v1-v2)/(3u)

Substituting the value of v2 from above gives v1 = u(e+1) and v2 = -u(2e+1)

However, the answers are v1 =u(1−e) and v2 =u(1+2e)

Where have I made a mistake? And how can I avoid it in the future?

7. (Original post by Anonymous1717)
This is similar to a previous poster's problem, but:

Two particles, P, of mass 2m, and Q, of mass m, are moving along the same straight line on a smooth horizontal plane. They are moving in opposite directions towards each other and collide. Immediately before the collision the speed of P is 2u and the speed of Q is u. The coefficient of restitution between the particles is e, where e < 1. Find, in terms of u and e,
(i) the speed of P immediately after the collision, (ii) the speed of Q immediately after the collision.

Why doesn't the following method work?

2m * 2u - m*u = 2mv1 + 2mv2

v2 = 3u - 2v1

v1 = ((3u - v2)/2 - v2) / 3u

e= (v1-v2)/(3u)

Substituting the value of v2 from above gives v1 = u(e+1) and v2 = -u(2e+1)

However, the answers are v1 =u(1−e) and v2 =u(1+2e)

Where have I made a mistake? And how can I avoid it in the future?

I always think about the collision in terms of real life. If you think about it, you have a ball with a higher mass and speed crashing into another ball - that 2nd ball is bound to go faster! This means that in your equation for e, for the difference to be positive it would be v2 - v1, as v2 is the larger speed. Does that make sense?
8. (Original post by Zaphod77)
I always think about the collision in terms of real life. If you think about it, you have a ball with a higher mass and speed crashing into another ball - that 2nd ball is bound to go faster! This means that in your equation for e, for the difference to be positive it would be v2 - v1, as v2 is the larger speed. Does that make sense?
It does. Thanks. Is that the only way to figure it out though?
9. (Original post by Anonymous1717)
It does. Thanks. Is that the only way to figure it out though?
I'll be honest, I don't know. There may well be another way, but I find that the easiest way to think about it! I knock my knuckles together to picture it, it probably looks quite funny
10. Does any one have a easier method to to Centre of Mass Question i never understand them???
11. (Original post by Zaphod77)
I'll be honest, I don't know. There may well be another way, but I find that the easiest way to think about it! I knock my knuckles together to picture it, it probably looks quite funny
I use a similar approach for my impulse questions as I act it out with my arms to see if it would be feasible or not, I probably look quite weird when doing it but I don't tend to make many mistakes from it all so it doesn't bother me
12. What are the full solutions of some mechanics questions:
Q1: A light elastic string of natural length 0.2m has its ends attached to 2 fixed points A and B which are on the same horizontal level with AB=0.2m. A particle of mass 5kg is attached to the string at the point P where AP=0.15m. The system is released and P hangs in equilibrium below AB with [angle APB= 90 degrees].
a) If angle BAP = x, show that the ratio of the extension of AP and BP is [(4cosx - 3)/(4sinx - 1)]
b) Hence show that (cosx)(4cosx - 3)=(3sinx)(4sinx - 1)

Q2: A particle of mass 3kg is attached to one end of a light string, of natural length 1m and modulus of elasticity 14.7N. The other end of the string is attached to a fixed point. The particle is held in equilibrium by a horizontal force of magnitude 9.8N with the string inclined to the vertical at an angle x.
If the horizontal force is removed, find the magnitude of the least force that will keep the string inclined at the same angle. [ANS:9.3N]

Q3: The particle P of mass 4kg is attached to one end of a light inelastic string of length 1m. The other end of the string is fixed at point O. P is hanging in equilibrium below O when it is projected horizontally with speed u ms-1. When OP is horizontal it meets a small smooth peg at Q, where OQ=0.8m. Calculate the minimum value of u if P is to describe a complete circle about Q. [ANS: (2.6g)^(1/2)]

Thxxx
13. (Original post by Zaphod77)
I always think about the collision in terms of real life. If you think about it, you have a ball with a higher mass and speed crashing into another ball - that 2nd ball is bound to go faster! This means that in your equation for e, for the difference to be positive it would be v2 - v1, as v2 is the larger speed. Does that make sense?
Has anyone else been told to do it like this?
14. (Original post by Anonymous1717)
Has anyone else been told to do it like this?
Ahhh no i don't i can never make the assumption, like lets say a spehere of mass 2m with speed 2u, hits another sphere at rest of mass 3m, i could never know if the first sphere reverses or not, all dependant on the coefficent of restitiution
15. (Original post by JoshThomas)
Ahhh no i don't i can never make the assumption, like lets say a spehere of mass 2m with speed 2u, hits another sphere at rest of mass 3m, i could never know if the first sphere reverses or not, all dependant on the coefficent of restitiution
Exactly... So have you been told how to do this?
16. (Original post by Anonymous1717)
Exactly... So have you been told how to do this?
Ahhh no, just done a few questions and different scenarios and look at what they do in the markscheme

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17. Doing both- printed every paper for M2 and M3 the other day from 2008.
Bloody hell carrying them felt like a load of bricks haha .

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18. Hey guys can someone help me with question 7 in the jan M2 2003 paper.

Basically I am having trouble with part b.

Conservation of energy means energy b4 equals energy after.

So this particle B is clearly starting from some height which is 1.2 m. So, initially there will be GPE and KE after there will be just KE

However on the mark scheme it only took into account KE why did they miss out GPE ?

Thanks !
19. (Original post by Marblepumpkin)
Is this the right thread for m2 help? Can anyone please explain the work energy principle to me, it's driving me insane!
Okay so work done against friction =the loss of energy minus the gain in either kinetic or potential?

What about total work done? Is that the work done against friction plus any change in potential? And then this also equals the change in kinetic?

What also then confuses me is how kinetic energy has to be positive! When working out a question I never know whether to the the final minus the initial or vice versa, it seems to change!

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Hi. I sympathise about CofE/WEP; the textbooks don't do it very well. (The current Heinemann book isn't quite as bad, but the previous one was horrendous - it did almost every example a different way, and if you used any of the methods on the wrong example, you got the answer wrong!)

It's actually quite easy if you forget about all this "change of KE" , "change of GPE" , "change of EPE" nonsense and just think about the total energy of the system.

The final total energy (of all sorts - so KE and GPE, and EPE if you are doing M3) is equal to the initial total energy plus any gains minus any losses. In M2 and M3 questions, gains and losses only come from work done by external forces. If a force is helping the motion (i.e. is in the direction of motion or within 90 degrees of it), it is doing positive work on the system and causes a gain in energy. If it is opposing the motion (i.e. opposite to the direction of motion, to within 90 degrees ) it is doing negative work and causes a loss of energy.

One last thing - work done by or against weights don't count; we have already taken account of gravity by putting in GPE.
20. There is always a question on centre of mass in every m2 paper, framework came up in the Jan 2013 paper, but lamina usually comes up

Which one do you personally think would come up?????

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