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Total momentum after collision

  1. Dec 6, 2007 #1
    Excuse the Title =X

    1. The problem statement, all variables and given/known data
    A 0.300 kg puck, initially at rest on a horizontal, frictionless surface, is struck by a 0.200 kg puck moving initially along the x axis with a speed of 2.00 m/s. After the collision, the 0.200 kg puck has a speed of 1.00 m/s at an angle of [tex]\Theta[/tex]= 50.0° to the positive x axis.

    http://img265.imageshack.us/img265/5695/811vx2.gif [Broken]

    Determine the speed of the 0.300 kg puck after the collision.

    2. Relevant equations
    m1v1i + m2v2i = (m1 + m2)vf

    3. The attempt at a solution

    total momentum before collision = 0
    total momentum after collision = 0.3 * v1f *sin 50 - 0.2 * v2f *sin φ
    from the law of conservation of momentum ,
    0.3 * v1f * sin 50 - 0.2 * v2f * sin φ = 0
    0.3 * v1f * sin 50 = 0.2 * v2f * sin φ
    where v2f = 1 m / s
    φ = 90 - 50 = 40

    so: (0.3)(v1f)(sin50) - (0.2)(1)(sin40) = 0.5596 m/s ??
     
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Dec 6, 2007 #2

    rl.bhat

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    Homework Helper

    total momentum before collision = 0 This statement is wrong, because 0.2 kg mass is moving with a velocity 2 m/s.
    total momentum after collision = 0.3 * v1f *sin 50 - 0.2 * v2f *sin φ This is only y-components. What about x-components?
     
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