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Hockey puck momentum/ collision

  1. Nov 30, 2009 #1
    1. The problem statement, all variables and given/known data
    A puck of mass 7 kg moving at 3 m/s strikes an identical puck that is stationery on frictionless ice. After the collision, the first puck leaves with a speed of v1 at an angle of 41.1647 degrees with respect to its original line of motion and the second puck leaves with a speed of v2 at 48.8353 degrees.

    2. Relevant equations
    MaVa + MbVb = MaVa' + MbVb'

    3. The attempt at a solution
    Puck along x-axis:
    MVa = MVa' * Cos 41.1647 + MVb' * Cos -48.8353

    0 = MVa' * Sin 41.1647 + MVb' * Sin -48.8353

    I solved for Va' and got 1.844 (rounded to 3 decimal places), but that answer is wrong. What did I do wrong?
  2. jcsd
  3. Nov 30, 2009 #2
    Figured it out. Calculator error.
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