Edexcel M2/M3 June 6th/10th 2013

Thanks so much. I actually understand it now.

M2

http://www.thestudentroom.co.uk/showthread.php?t=1618497

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

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?

Thanks in advance.

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?

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 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?

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?

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!
please help, I can't understand any of the Internet sites! :'( thank you!


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