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Final Velocity (involving momentum)

  1. Nov 25, 2007 #1
    [SOLVED] Final Velocity (involving momentum)

    1. The problem statement, all variables and given/known data

    A 0.329-kg puck, initially at rest on a frictionless horizontal surface, is struck by a 0.222-kg puck that is intitially moving along the x-axis with a velocity of 2.24 m/s. After the collision, the 0.222-kg puck has a speed of 1.24 m/s at an angle of 24 degrees to the positive x-axis. (a) Determine the velocity of the 0.329-kg puck after the collision. (b) Find the fraction of the kinetic energy lost in the collision.

    3. The attempt at a solution

    i tried m1 = 0.329kg ----- m2 = 0.222kg

    m1v1(i) + m2v2(i) = m1v1(f) + m2v2(f)

    v1(f) = [m2v2(i) - m2v2(f) cos theta] / m1
    v1(f) = 0.7471105 m/s (this was wrong)

    I assume I need to find the x and y components and then use vector sumation but I have NO idea how to find the y...
  2. jcsd
  3. Nov 26, 2007 #2


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    Homework Helper

    Let v2(f) makes an angle theta with positivde x-axis. Take the compondents of v1(f) and v2(f) along x and y- axis. Y camponents miust be equal and opposite because v1 is along x-axis. So you can wright two equations. 1) m1v1 + m2*0 = m1v1(f)cos(24) + m2v2(f)cor(theta)
    2) m1v1(f)sin(24) = m2v2(f)sin(theta) . Rewrigth the equation(1) as
    m1v1 -m1vi(f)cos(24) = m2v2(f)cos(theta)....(.3) Square eq. 2 and 3 and add. After solving you will get v2(f)
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