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    Particles A, B and C of masses 4m, 3m and m respectively, lie at rest in a straight horizontal plane with B between A and C. Particles A and B are projected towards each other with speeds u ms^{-1} and v ms^{-1} respectively, and collide directly.

    As a result of the collision, A is brought to rest and B is rebounded with speed kv ms^{-1}. The coefficient of restitution between A and B is \frac{3}{4}.

    a) Show that u = 3v. (6)

    b) Find the value of k. (2)

    Immediately after the collision between A and B, particle C is projected with speed 2v ms^{-1} towards B so that B and C collide directly.

    c) Show that there is no further collision between A and B. (4)
    My workings so far (Diagrams are attatched):

    a)
    Conservation of momentum:

    4mu - 3mv = 3mkv

    Newton's Law of Restitution:

    e = \frac{separation speed}{approach speed}
    \frac{3}{4} = \frac{3mkv}{4mu + 3mv}

    Now I am unsure as to what I can do next??

    b)
     u = 3v

    12mv -3mv = 3mkv

    k = 3

    c)
    Conservation of momentum:

    9mv - 2mv = 3mv^{B} + mv^{C}

    Again I am unsure as to what I do next?? It's obvious that B won't collide with A again as B will keep moving in the same direction after hitting C but I can't see how to show that??
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    (Original post by JamesyB)
    Newton's Law of Restitution:

    e = \frac{separation speed}{approach speed}
    \frac{3}{4} = \frac{3mkv}{4mu + 3mv}

    Now I am unsure as to what I can do next??
    I think you are misunderstanding the law of restitution. It relates to speed (as you said above) not momentum (as you used below it).

    You can also divide your momentum equations through by m as it's the same value for all of them. This will leave you with some simultaneous equations which you should be able to solve.

    (Original post by JamesyB)
    c)
    Conservation of momentum:

    9mv - 2mv = 3mv^{B} + mv^{C}

    Again I am unsure as to what I do next?? It's obvious that B won't collide with A again as B will keep moving in the same direction after hitting C but I can't see how to show that??
    Don't know if it's the most elegant solution, but if you use restitution as in part a (using e), and solve simultaneously again, you should get a speed of B after the collision (in terms of v and e). The condition this velocity needs to satisfy is (taking away from a as positive) Velocity_B \geq 0. If you find it in terms of v and e, and use e as 0 \leq e \leq 1, it should be clear this is positive (you'll see what I mean if you get it)

    Sorry if that's not too clear. If you need a more explanation on part of it, ask.
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    (Original post by gcseeeman)
    I think you are misunderstanding the law of restitution. It relates to speed (as you said above) not momentum (as you used below it).

    You can also divide your momentum equations through by m as it's the same value for all of them. This will leave you with some simultaneous equations which you should be able to solve.



    Don't know if it's the most elegant solution, but if you use restitution as in part a (using e), and solve simultaneously again, you should get a speed of B after the collision (in terms of v and e). the condition this speed needs to satisfy is (taking away from a as positive) Velocity_B \geq 0. If you find it in terms of v and e, and use e as 0 \leq e \leq 1, it should be clear this is positive (you'll see what I mean if you get it)

    Sorry if that's not too clear. If you need a more explanation on part of it, ask.
    Ah cheers, mechanics is really not my strong point haha.

    Got both parts now
 
 
 
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