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Two-d momentum

  1. Oct 14, 2009 #1
    1. The problem statement, all variables and given/known data

    On an air hockey table, two pucks collide with each other. Puck A have .05 kgand inital velocity of 5 m/s in the x direction. It collides with puck B which is initially at rest and has a mass of .1 kg. The collision is not a head on. AFter the collision, puck A flies of the north east direction with velocity of 3 m/s in an angle theta_A above the x axis. And Puck B flies off the south east direction with a velocity of 2.5 m/s in an angle of theta_B below the x axis. Find Theta_ A and theta_B

    2. Relevant equations



    3. The attempt at a solution

    well I have two equations and three unknows I got

    x direction: 0=.05kg*3m/s*sin(theta_a)+.1kg*2.5m/s*sin(theat_B)
    and
    y direction: m1*V=m1*v_1f*cos(theta_A)+m2*v_2f*cos(theta_B)

    please help thanks for your time...
     
  2. jcsd
  3. Oct 14, 2009 #2
    I see only 2 unknowns. Just the 2 directions.
     
  4. Oct 14, 2009 #3
    I think my equation in the y direction is wrong any thoughts?
     
  5. Oct 14, 2009 #4
    You swapped the x and the y directions.

    What you wrote down for the y-direction is valid for the x-direction

    what you wrote for the x-direction is nearly valid for the y-direciton, except
    that you forgot [itex]\theta_A[/itex] is an angle above and [itex]\theta_B[/itex] below the x-axis
     
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