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.

 

[kyle8921]

Based on your calculations, it seems that you have correctly determined the change in momentum and the time interval of the collision. However, there may be some errors in your unit conversions or in the direction of the force. Remember that the force should be a positive value, as it is the maximum value of the contact force during the collision. Also, make sure that you are using the correct units for mass (kg) and velocity (m/s).

One way to approach this problem is to use the formula F = ma, where F is the force, m is the mass, and a is the acceleration. In this case, the acceleration is the change in velocity divided by the time interval (a = Δv/Δt).

Using the values you have calculated for the change in momentum and time interval, the acceleration can be found as follows:

a = (-2.8728 kgm/s)/(0.0348 s)
= -82.55 m/s^2

Now, using the mass of the tennis ball (0.057 kg), we can calculate the maximum contact force as:

F = (0.057 kg)(-82.55 m/s^2)
= -4.176 N

Since the force should be a positive value, we can take the absolute value to get the maximum contact force:

Fmax = |-4.176 N|
= 4.176 N

Therefore, the maximum contact force during the collision is 4.176 N. It is important to note that this value is an approximation, as it assumes that the force is constant during the entire collision. In reality, the force may vary during the collision.