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    There's three balls on a smooth surface, A, B, and C, with masses m, 2m, and 3m respectively. Ball A is given a velocity u and collides elastically with B, which collides similarly with C.
    I need to find the fraction of u that ball C has immediately after its impact with B.

    I did the following:

    0.5 * u^2 * m = 0.5 * ku^2 * 2m
    1 = 2k
    k = 1/2

    Then:

    0.5 * u^2/4 * 2m = 0.5 * ku^2 *3m
    2(u^2)/4 = 3ku^2
    1/2 = 3k
    k = 1/6
    Which I'm told I need to input as a decimal to 3 s.f. I typed in: 0.167 (no units) but it's wrong? Is my method wrong?
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    (Original post by pineneedles)
    There's three balls on a smooth surface, A, B, and C, with masses m, 2m, and 3m respectively. Ball A is given a velocity u and collides elastically with B, which collides similarly with C.
    I need to find the fraction of u that ball C has immediately after its impact with B.

    I did the following:

    0.5 * u^2 * m = 0.5 * ku^2 * 2m
    1 = 2k
    k = 1/2

    Then:

    0.5 * u^2/4 * 2m = 0.5 * ku^2 *3m
    2(u^2)/4 = 3ku^2
    1/2 = 3k
    k = 1/6
    Which I'm told I need to input as a decimal to 3 s.f. I typed in: 0.167 (no units) but it's wrong? Is my method wrong?
    I'm not really sure what method you're using but I would use the conservation of momentum formula and say for the first one that (m x u) + (2m x 0) = (m x 0) + (2m x ku). This gives k=1/2 for the first one however it doesn't give 1/6 as the second answer.
    Also conscious of the fact that we're having to assume that A stops moving instantly after the collision as it doesn't actually say this :erm:

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    (Original post by smartalan73)
    I'm not really sure what method you're using but I would use the conservation of momentum formula and say for the first one that (m x u) + (2m x 0) = (m x 0) + (2m x ku). This gives k=1/2 for the first one however it doesn't give 1/6 as the second answer.
    Also conscious of the fact that we're having to assume that A stops moving instantly after the collision as it doesn't actually say this :erm:

    Posted from TSR Mobile
    I'm using the idea of kinetic energy being conserved to answer it, but your method seems fine.
    What did you get as your second answer?
    Also, that's a good point...
    [edit
    ] It's a smooth surface, so assuming air resistance is negligible and considering the balls are close together as in the diagram I have, it's not unreasonable to assume velocity on impact is equal to initial velocity.
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    (Original post by pineneedles)
    I'm using the idea of kinetic energy being conserved to answer it, but your method seems fine.
    What did you get as your second answer?
    Also, that's a good point...
    [edit
    ] It's a smooth surface, so assuming air resistance is negligible and considering the balls are close together as in the diagram I have, it's not unreasonable to assume velocity on impact is equal to initial velocity.
    I got 1/3 as the final answer. And it said the collision was elastic so that means no energy is lost at all (so no air resistance or anything like that)
 
 
 
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