Calculating Fmax of Collision Q: Tennis Ball & Wall

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The discussion revolves around calculating the maximum contact force (Fmax) experienced by a tennis ball during its collision with a wall. The ball, weighing 64.0 g and traveling at 28.0 m/s, reverses direction upon impact, indicating that momentum is not conserved due to the external force from the wall. The initial attempt to calculate the force using impulse and change in momentum resulted in an average force of 2059.77 N, but this was identified as incorrect for determining Fmax. Further analysis suggested that the maximum force could be derived from the average force, leading to an estimated Fmax of 3089.66 N. The discussion emphasizes the importance of understanding the difference between average and maximum forces in collision scenarios.
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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


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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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?
 
ok thanks for the help, ill work through the problem and post my solution.
 
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
 
saralsaigh said:
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).
 
OK so how do i do that?!
2059.77 = 2F/3
F = 3089.66 N
idk that would be my guess...any suggestions?
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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