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Homework Help: Conservation of Momentum Problem

  1. Jan 28, 2012 #1
    1. The problem statement, all variables and given/known data
    A ball of mass [itex]m_1[/itex] travels along the x-axis in the positive direction with an initial speed of [itex]v_0[/itex]. It collides with a ball of mass [itex]m_2[/itex] that is originally at rest. After the collision, the ball of mass [itex]m_1[/itex] has velocity [itex]{v_1}_{x}{\hat{x}}+{v_1}_{y}{\hat{y}}[/itex] and the ball of mass [itex]m_2[/itex] has velocity [itex]{v_2}_{x}{\hat{x}}+{v_2}_{y}{\hat{y}}[/itex]. Consider the following five statements:

    I) [itex]0=m_1{v_1}_x+m_1{v_2}_x[/itex]
    II) [itex]m_1v_0=m_1{v_1}_y+m_2{v_2}_y[/itex]
    IV) [itex]m_1v_0=m_1{v_1}_x+m_1{v_1}_y[/itex]
    V) [itex]m_1v_0=m_1{v_1}_x+m_2{v_2}_x[/itex]

    Of these statements, the system must satisfy
    a) I and II
    b) III and V
    c) II and V
    d) III and IV
    e) I and III

    2. Relevant equations

    3. The attempt at a solution
    I've never learned how to incorporate conservation of momentum into a 3-d plane...
    Last edited: Jan 28, 2012
  2. jcsd
  3. Jan 29, 2012 #2


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    Homework Helper

    This looks like only 2-D to me?
  4. Jan 29, 2012 #3


    Staff: Mentor

    write down the initial total momentum

    then write down the final total momentum

    then group the X components into one eqn and the Y components in another and see whether you can answer the question.

    show your work and we can help
  5. Jan 29, 2012 #4
    Whoops, sorry, I meant 2-d.
    But I think I understand now... You just have to do conservation of momentum for each axis, right? Since the initial momentum is all in the x-direction, the final momentum for the y-components must sum to 0, which is shown in equation III. The initial x-momentum is given by [itex] m_1v_0 [/itex], so the sum of components for the final x-momentum should be exactly that, given by equation V. Final answer (B).
    Last edited: Jan 29, 2012
  6. Jan 29, 2012 #5


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    Homework Helper

    That is correct.
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