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Homework Help: Elastic collisions of particles

  1. Sep 18, 2015 #1
    1. The problem statement, all variables and given/known data
    Three particles, A, B, and C, with masses M, 2M, and 3M respectively, lie at rest in that order in a straight line on a smooth horizontal table. The particle A is then projected directly towards B with velocity U.
    Assuming the collisions are perfectly elastic, I need to find the fraction of U that C moves with, immediately after impact.
    2. Relevant equations

    3. The attempt at a solution
    I used the idea that kinetic energy is conserved in a perfectly elastic collision:

    0.5 * M * U^2 = 0.5 * 2M * kU^2
    1 = 2k
    k = 1/2
    Therefore the speed of B after impact is U/2
    0.5 * 2M * (U/2)^2 = 0.5 * 3M *kU^2
    2(U^2/4) = 3kU^2
    U^2/2 = 3kU^2
    1 = 6k
    k = 1/6
    It asks for the answer as a decimal to three significant figures, so I've typed in: 1.67 x 10^-1 but it's not having it. Where have I gone wrong?

    Thanks for your help in advance.
    [edit] I wrote it down wrong.
    Last edited: Sep 18, 2015
  2. jcsd
  3. Sep 18, 2015 #2


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    Total kinetic energy is conserved in an elastic collision. You tried (and got wrong) having the kinetic energy in the first collision completely pass to mass B. With twice the mass and half the speed, what is the kinetic energy?

    In an elastic collision both total momentum and kinetic energy are conserved. You must use both.

    The easy way to do collisions of this sort is in the centre of momentum frame. Before the collision you have two particles coming in with opposite momentums. After they just "reflect" so their momentums just change sign. Then you transform back to the lab frame.
  4. Sep 18, 2015 #3
    Thanks for your reply.

    I'm confused about what you've told me, however. How would I use both the conservation of momentum and kinetic energy to answer this? I can't see how it's necessary. Surely if you know that both momentum or kinetic energy is conserved because it is a series of elastic collisions, you can solve this considering only one of them?
    I tried this again, considering momentum:
    M * U = 2M * kU
    1 = 2k
    We have the second particle B, with half the speed, is that not right?
    2M *U/2 = 3M * kU
    1 = 3k
    k = 1/3
    However I get a different answer for this one. I'm really confused. I think you may have misunderstood the situation I described "Before the collision you have two particles coming in with opposite momentums." I meant all three balls are at rest, then A is projected and hits B, B then goes on to hit C.
  5. Sep 18, 2015 #4


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    You now have the second particle going half the speed of the first one. That means the second particle has the full momentum of the first one before the collision. That means the first one has to have stopped to conserve momentum. But what is the kinetic energy of the second particle? Is it the same as the first one before the collision?

    You must conserve both momentum and energy. You have not done that. The easy way to do that is in the centre of momentum frame.


    But if you don't like the COM frame, fine.

    Before the first collision you have particle mass M moving at U. That means you have a momentum MU, and a kinetic energy 1/2 M U^2. After the collision you will have the particle mass M going at some other speed, call it V1 and particle mass 2M going at some new speed, call it V2. Use positive values to be in the same direction as the original, in case they wind up flipping. You have a total momentum after.

    M V1 + 2 M V2

    And a total kinetic energy after.

    1/2 M V1^2 + 1/2 2 M V2^2

    And the after momentum has to be the same as the before momentum of MU. And the after kinetic energy has to be the same as the before of 1/2 M U^2. That's two equations in two unknowns. Solve for V1 and V2.

    If this is all completely unfamiliar, then you need to go read your notes again. Or your text.

    But it is WAY easier to transform to the COM frame. In that frame, as I said, yada yada, equal but opposite momentums just reflect in an elastic collision.
  6. Sep 18, 2015 #5
    Thanks for your patience, I've never learned about momentum in this much depth before. Some of this is completely new to me. I now understand why I need to combine the two equations in elastic collisions.
    I don't understand the COM you mentioned, that looks too complicated for me? So I'll work with the equations.
    Basically I did this:

    U = V1 + 2V2
    U^2 = 2V2 ^2 + V1 ^2
    V1 ^2 + 4V1V2 + 4V2 ^2 = 2V2 ^2 + V1 ^2
    4V1V2 + 2V2 ^2 = 0
    V2(4V1 + 2V2 ^2) = 0

    Is this alright so far?
  7. Sep 20, 2015 #6
    Please help me, someone.
    This is my latest attempt:

    Where V1, V2, and V3 are the final velocities of particles of masses M, 2M, 3M respectively,

    Momentum is conserved:

    (1) MU = MV1 + 2MV2
    U = V1 + 2V2
    As is KE:

    0.5* MU^2 = (0.5 * MV1^2) + (0.5* 2MV2)
    (2) U^2 = V1^2 + 2V2^2

    Making V1 the subject of equation (1) and subbing it into (2):
    U^2 = (U - 2V2)^2 + 2V2^2
    U^2 = U^2 - 4V2U + 6V2^2
    4V2U - 6V2^2 = 0
    V2(4U - 6V2) = 0

    V2 =/= 0

    4U = 6V2
    V2 = 4/6U
    : The velocity of particle 2 is 2/3 of that of the first before impact, in the same direction.

    Where V4 is the velocity of the particle with mass 2M, after its impact with particle of mass 3M and V3 is still the velocity of particle 3M :
    2M 2/3 U = 2MV4 + 3MV3
    (1) 4/3 U = 2V4 + 3V3


    0.5 * (2/3U)^2 *2M = 0.5 * 2MV4 ^2 + 0.5* 3MV3 ^2
    (2) 8/9 U^2 = 2V4 ^2 + 3V3 ^2

    Rearranging (1) for V4 and subbing into (2) =

    8/9 U^2 = 2(2/3U - 3/2 V3)^2 + 3V3 ^2

    8/9 U^2 = 2(4/9U^2 - 12/5 V3U + 9/4V3^2) + 3V3^2
    8/9 U^2 = 8/9 U^2 - 12/5 V3U + 21/4 V3^2

    12/5 V3 U - 21/4 V3^2
    V3(12/5U - 21/4 V3) = 0
    V3 =/= 0

    12/5U = 21/4 V3
    12/5 * 4/21 U = V3
    48/105 U = V3

    V3 is 0.457 (3 s.f) of that of the velocity of U.

    I know the way I've typed this out is really hard to understand, but if someone could try I'd be super grateful.
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