Tennis ball hitting a wall and rebounding - contact force

[mathewings]

A 57.0 g tennis ball with an initial speed of 25.2 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=34.8 milliseconds?

I've tried using the formula Force = change in momentum/change in time; after determining change in momentum with m1v1=m2v2.

p1
=m1v1
=(0.057kg)(25.2m/s)
=1.4364 kgm/s

p2
=m2v2
=(0.057kg)(-25.2m/s)
=-1.4364kgm/s

Change in p = -2.8728kgm/s

F = 2.8728kgm/s/0.0348s
= -82.55N

However, this is not the answer. Any help would be appreciated.

 

[PhanthomJay]

A 57.0 g tennis ball with an initial speed of 25.2 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=34.8 milliseconds?

I've tried using the formula Force = change in momentum/change in time; after determining change in momentum with m1v1=m2v2.

p1
=m1v1
=(0.057kg)(25.2m/s)
=1.4364 kgm/s

p2
=m2v2
=(0.057kg)(-25.2m/s)
=-1.4364kgm/s

Change in p = -2.8728kgm/s

F = 2.8728kgm/s/0.0348s
= -82.55N

However, this is not the answer. Any help would be appreciated.

When the ball hits the wall, the tennis ball is compressed something like a spring (it gets squished), then it rebounds back to its circular shape. At the instant of contact, the force is just a hair above 0, then it reaches a maximum at full compression, then reduces to 0 again as the ball leaves the wall. Like a spring, the max force is kx, where x is the maximum displacemnt or compression; the min force is 0 at the uncompressed condition. The force you calculated is the average of these two forces. The max force is twice that.