Collision Question

  • Thread starter saralsaigh
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  • #1
saralsaigh
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Homework Statement


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?

http://capa.physics.mcmaster.ca/figures/kn/Graph09/kn-pic0910_new.png [Broken]


Homework Equations


mvi = mvf
Momentum is conserved?


The Attempt at a Solution




I am totally clueless as to where to start i don't 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
 
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Answers and Replies

  • #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?
 
  • #3
saralsaigh
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ok thanks for the help, ill work through the problem and post my solution.
 
  • #4
saralsaigh
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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 please tell me where I am going wrong...thx
 
  • #5
PhanthomJay
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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 please tell me where I am going wrong...thx
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).
 
  • #6
saralsaigh
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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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