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Collision Question

  1. Feb 11, 2007 #1
    1. The problem statement, all variables and given/known data
    A 64.0 g tennis ball with an initial speed of 28.0 m/s hits a wall and rebounds with the same speed. The figure below shows the force of the wall on the ball during the collision. What is the value of Fmax, the maximum value of the contact force during the collision, if the force is applied for ti=17.4 ms?

    [​IMG]


    2. Relevant equations
    mvi = mvf
    Momentum is conserved?


    3. The attempt at a solution


    I am totally clueless as to where to start i dont even know if this really is a momentum question i am only assuming if you guys can just point me in the rite direction. Thanks
     
  2. jcsd
  3. Feb 12, 2007 #2

    cepheid

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    Momentum is *not* conserved. You know this because an external force acts on the ball. Also, you know that the ball starts out going in one direction and then ends up going in entirely the opposite direction. Remember...momentum is a vector. But what do you know? You know that the speed before is the same as the speed after. Therefore you know the change in momentum. Therefore, you know the total impulse that acted on the ball. Can you use this information to answer the question?
     
  4. Feb 12, 2007 #3
    ok thanks for the help, ill work through the problem and post my solution.
     
  5. Feb 12, 2007 #4
    hmm i must be doing something wrong again, this is how i worked it out
    I used impulse = change in momentum so;
    F t = m (V2-V1)
    F = [(0.64kg) (56m/s)]/ (0.0174s)
    F = 2059.77 N

    ==> this is the wrong answer could u plz tell me where im going wrong...thx
     
  6. Feb 12, 2007 #5

    PhanthomJay

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    what you have calculated is the average force acting during that time interval. What's the maximum force (looks like from the graph that there are 3 equal time periods ....a linearly varying force in the first, constant in the second, varying linearly again in the third).
     
  7. Feb 12, 2007 #6
    OK so how do i do that?!
    2059.77 = 2F/3
    F = 3089.66 N
    idk that would be my guess....any suggestions?
     
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