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Conservation of Momentum - Shuffleboard

  1. Jan 24, 2010 #1
    Hi PhysicsForums. :P First time here. I would just like someone to verify whether I did this problem correctly.

    A moving shuffleboard puck has a glancing collision with a stationary puck of the same mass. If friction is negligible, what are the speeds of the pucks after the collision?

    Given variables: v1i = 0.95 m/s, after collision, m1 (puck 1) moves at a 50 degree angle to the horizontal, positive direction in the y (assuming a regular coordinate grid), while m2 (puck 2) moves at a 40 degree angle toe the horizontal, negative direction in the y.


    Attempt:

    Conservation of Momentum (X direction)

    Pi = Pf
    m1v1ix + m2v2ix = m1v1fx + m2v2fx
    v1ix = v2fx-v1fx
    0.95 = v2 (cos 40) - v1 (cos 50)

    Conservation of Momentum (Y direction)

    m1v1iy + m2v2iy = m1v1fy + m2v2fy
    0 = v1fy + v2fy
    -v1 (sin 50) = v2 (sin 40)
    v1 is a vector, so since its moving in a positive direction (according to my grid), it's still positive anyway...
    v2 = v1(sin50)/sin40

    substitute this into the equation I got for C.o.M. in the X

    0.95 m/s = v1 (cos50) + v1(sin50)(cos40)/sin(40)
    0.95 m/s = v1 (cos 50 + sin50(cos 40)/sin(40)

    v1 = 0.95m/s / (cos 50 + sin50cos40/sin40)
    v1 = .6106482 m/s

    substitute this into the equation I had for C.o.M. in the Y

    v2 = .6106482 m/s (sin 50)/sin40
    v2 = .7277422 m/s

    Hopefully I got this right? >_>
     
    Last edited: Jan 24, 2010
  2. jcsd
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