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The conservation of momentum in 2D

  1. Jun 23, 2011 #1
    1. The problem statement, all variables and given/known data
    Two hockey pucks of equal mass undergo a collision on a hockey rink. One puck is initially at rest, while the other is moving with a speed of 5.4m/s. After the collision, the velocities of the pucks make angles of 33 degrees and 46 degrees relative to the original velocity of the moving puck. Determine the speed of each puck after the collision.

    (Assuming the moving puck is object A and the stationary puck is object B)
    Ma = Mb
    Va1 = 5.4m/s
    Vb1 = 0m/s
    Deflection Angle of A = 33
    Deflection Angle of B = 46

    2. Relevant equations
    p(before) = p(after)


    3. The attempt at a solution
    Okay, the first thing I did was broke it into components as it's in two dimensions. For x,

    p(before) = p(after)
    (Ma)(Va1x) + (Mb)(Vb1x) = (Ma)(Va2x) + (Mb)(Vb2x)
    (5.4) + 0 = Va2x + Vb2x
    5.4 = Va2x + Vb2x

    for y,

    p(before) = p(after)
    (Ma)(Va1y) + (Mb)(Vb1y) = (Ma)(Va2y) + (Mb)(Vb2y)
    0 + 0 = Va2y + Vb2y
    Va2y = -Vb2y

    And then I don't know where to go from there. I've thought about trig and using Pythagorean theorem but there's too many variables.
     
  2. jcsd
  3. Jun 23, 2011 #2
    Set up a drawing of the resultant momentum vectors after the collision. Trig and pythagorean theorem might be necessary here but draw a picture first and see what you can deduce from it.
     
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