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Conservation Of Momentum

  1. May 24, 2008 #1


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    1. The problem statement, all variables and given/known data
    This diagram shows a white snooker ball moving with a velocity of [itex]3.2 \ ms^{-1}[/itex]. The white ball is on a collision course with a stationary blue ball of identical mass of [itex]0.173 \ kg[/itex]. The player is aiming to propel the blue ball into the pocket along the path shown.


    The player succeeds in propelling the blue ball along the path shown. The blue ball has a velocity of [itex]2.77 \ ms^{-1}[/itex] after the collision with the white ball. Calculate the velocity of the white ball after the collision.

    2. Relevant equations
    [itex]P=mv[/itex] (I think)

    3. The attempt at a solution
    As momentum is conserved, I would have momentum before equals momentum after hence...

    [itex](0.173)(3.2) = (0.173)(v) + (0.173)(2.77) \implies v = 0.43 \ ms^{-1}[/itex]

    But this is incorrect. Can someone explain how I would tackle this question and point out my mistake. Thanks in advance. :smile: (The correct answer is [itex]v = 1.6 \ ms^{-1}[/itex].)
    Last edited: May 24, 2008
  2. jcsd
  3. May 24, 2008 #2


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    Homework Helper

    Note that this is a two dimensional problem. In this case you have to break up the momentum of each ball into its x and y components. Apply conservation of momentum in each direction.
  4. May 26, 2008 #3
    This question must be solved in two dimensions. Therefore separate the momentum before and after into x, y components. As a hint since both balls have the same mass, you can cross the mass out of your conservation equations and deal only with velocities. Remember to use the angle given to find the x and y velocity of the blue ball after the collision.
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