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General Physics (I) Kinetic energy and momentum

  1. Feb 11, 2012 #1
    Betty Bodycheck (mB = 51.1 kg, vB = 22.0 km/h in the positive x-direction) and Sally Slasher (mS = 40.5 kg, vS = 28.0 km/h in the positive y-direction) are both racing to get to a hockey puck. Immediately after the collision, Betty is heading in a direction that is 76.0° counterclockwise from her original direction, and Sally is heading back and to her right in a direction that is 12.0° from the x-axis. What are Betty and Sally's final kinetic energies?

    http://www.webassign.net/bauerphys1/7-p-055.gif

    Kinetic Energy: K=(1/2)mv^2 ........ Momentum: p=mv


    When I attempted this problem I first converted both velocity from km/h to m/s. Next I proceeded to calculate Betty's and Sally's KE: [Betty] (1/2)x51.1x(6.11278)^2= 954 J......[Sally] (1/2)x40.5x(7.7779)^2=1225 J.

    Both answers were wrong, a few step(s) or missing equations are missing I believe but I do not know what. Can anyone please help?

    Thank You.
     
    Last edited: Feb 11, 2012
  2. jcsd
  3. Feb 11, 2012 #2
    The question asks for the KE AFTER the collision. Kinetic energy is not necessarily conserved in a collision.
     
  4. Feb 12, 2012 #3
    I don't quite understand just what you mean. I'm still quite lost on where to start, if you had anymore advice or tips I would greatly appreciate it.
     
  5. Feb 12, 2012 #4
    When objects collide and the collision is not completely elastic, kinetic energy is not conserved. Therefore the initial KE's do not equal the final KE's. However, momentum is conserved. Big hint: Write equations that relate initial momentum in the X and Y directions before and after the collision of Betty and Sally. This will enable you to solve for the velocities which then can be used to determine the kinetic energy after the collision.
     
  6. Feb 12, 2012 #5
    I agree with LawrenceC. This problem is a typical conservation of momentum question. Use conservation of momentum in x directions and y directions to solve.
     
  7. Feb 12, 2012 #6
    Okay, I believe I understand more now. Thanks for the explanation
     
  8. Feb 12, 2012 #7
    No problem.
     
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