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# M2 collisions watch

1. Hey everyone!!!

Particle A of mass 2m is moving with speed 2u on a smooth horizontal table. Particle collides directly with particle B of mass 4m moving with speed u in the same direction as A. Co-efficient of restitution between A and B is 0.5u

a) show that the speed B aftyer the collision is 1.5u
b) find the speed of A after the collision

B then collides with particle C of mass m, which is at rest on the table. The co-efficient of restitution between B and C is e.

c) Given that there are no further collisions, fid the possible values for e

Thanks everyone!!!
2. 0 < e < 2/3
3. You're offline now but I'll quickly explain my working:
Call velocity of B after the collision w, velocity of C v
Law of restitution gives: e = (v - w)/(3u/2)
v - w = 3ue/2 (1)
Conservation of momentum: 6mu = 4mw + mv
6u = 4w + v (2)
(2) - (1): 5w = 6u - 3ue/2
For B not to collide with A, it must continue moving in the same direction with speed greater than or equal u (that's what I worked speed of A out to be)
so w > u
5w > 5u
6u - 3ue/2 > 5u
u > 3ue/2
so e < 2/3
4. Thanks!!! I understand it now!!!

Also, could you use the speed of C is greater than B, cos i tried that and it didn't work so i don't know why. To be honest, if there are 3 particles, i don't know which two particles to use, like in this question.

Also one last thing, if you have to use speeds of paricles to find the co-efficient of restitution and were told that the particle either changed direction or continued in the same direction, would you put the final speed of the particle less than or more than 0 depending on the direction that the particle is moving in - i.e. if you assume it is moving in the positive direction after the collision, you put the speed more than 0?

Sorry if that is confusing!!!
5. As you're talking about the collision of B and C, the velocity of C is always going to be greater than or equal to the velocity of B (assuming C starts on the positive side of B)
Taking right as positive, before collision of B and C:
A --> u, B--> 3u/2, C at rest
After collision:
A --> u, B--> w, C--> v (arrows are meant to be direction of travel)
w obviously can't be greater than v or B would have overtaken C! (I always define everything in the positive direction even if it can't actually go that way because then it's easier with signs) So there isn't a useful inequality looking at B and C, so you look at A and B, the only way they won't collide is if B moves to the right at a speed greater than or equal to A's speed ie w > u

Your way is how you do the question if it talks about change in direction or staying in the same direction, yes
6. (Original post by Bezza)
As you're talking about the collision of B and C, the velocity of C is always going to be greater than or equal to the velocity of B (assuming C starts on the positive side of B)
Taking right as positive, before collision of B and C:
A --> u, B--> 3u/2, C at rest
After collision:
A --> u, B--> w, C--> v (arrows are meant to be direction of travel)
w obviously can't be greater than v or B would have overtaken C! (I always define everything in the positive direction even if it can't actually go that way because then it's easier with signs) So there isn't a useful inequality looking at B and C, so you look at A and B, the only way they won't collide is if B moves to the right at a speed greater than or equal to A's speed ie w > u

Your way is how you do the question if it talks about change in direction or staying in the same direction, yes
Thank you SOOO much!!! You're a life saver!!!

I added to your rep cos you have been so nice and kind!!!
7. (Original post by Silly Sally)
Thank you SOOO much!!! You're a life saver!!!

I added to your rep cos you have been so nice and kind!!!
That's ok - any time. Well, not any time as I have to sleep, but you know what I mean.

Thanks for the rep!

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