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How to calculate momentum at right angle

  1. Jan 5, 2012 #1

    I am trying to figure out how to calculate the momentum of something at a right angle.

    basically a hockey puck is traveling on the x axis and hits a stationary hockey puck. One of the hockey pucks goes up the y axis (puck A) and one hockey puck goes down the y axis (Puck B).

    Am I supposed to use m*v*sin90 ? and m*v*sin-90 ? or do I use cosine? I'm fine calculating with triangles but I don't know what to do when things (forces, momentums) are at right angles
    Last edited: Jan 5, 2012
  2. jcsd
  3. Jan 6, 2012 #2

    Simon Bridge

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    There has to be something else.

    Before the collision you have momentum p in the x dir.
    After collision you have no momentum in the x direction.
    Therefore, momentum was not conserved in the collision.
    Need more information.

    Example - conservation of momentum:
    If the incoming puck was A, and it finishes with momentum p (same as initial) in the +y dir, then puck B would head off with momentum about (1.4)p at 45 degrees... preserving the zero total momentum in the y dir and the p total momentum in the x dir.
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