Calculating Fmax of Collision Q: Tennis Ball & Wall

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Homework Help Overview

The problem involves a 64.0 g tennis ball colliding with a wall, rebounding with the same speed, and seeks to determine the maximum contact force (Fmax) during the collision, given a specific time interval for the force application.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the conservation of momentum, noting that it does not apply due to the external force from the wall. There are attempts to relate impulse to change in momentum, with questions about the calculation of average versus maximum force.

Discussion Status

Some participants have provided guidance on understanding the relationship between impulse and momentum, while others are exploring the implications of average force versus maximum force. There is an ongoing exchange of calculations and suggestions for correcting errors in reasoning.

Contextual Notes

Participants are working under the constraints of the problem's parameters, including the mass of the ball, its initial speed, and the time interval for the force application. There is also a reference to a graphical representation of the force during the collision.

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


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?
 

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