N (Intro 1)Tennis Ball Rebound: Force on Ball During Collision

[julz3216]

Homework Statement

A 60 tennis ball with an initial speed of 33 hits a wall and rebounds with the same speed. The figure (Intro 1 figure) shows the force of the wall on the ball during the collision. t=6
http://session.masteringphysics.com/problemAsset/1070440/4/09.EX11.jpg

What is the value of , the maximum value of the contact force during the collision?

Homework Equations

p=mv
change in p = F *change in time
F=ma

The Attempt at a Solution

I attempted and got Fmax = 1320

 

[LowlyPion]

Homework Statement

A 60 tennis ball with an initial speed of 33 hits a wall and rebounds with the same speed. The figure (Intro 1 figure) shows the force of the wall on the ball during the collision. t=6
http://session.masteringphysics.com/problemAsset/1070440/4/09.EX11.jpg

What is the value of , the maximum value of the contact force during the collision?

Homework Equations

p=mv
change in p = F *change in time
F=ma

The Attempt at a Solution

I attempted and got Fmax = 1320

Not sure about your units, but I'll go with 60 g for a tennis ball and 33 m/s.

So maybe check your units.

Impulse is Δmv.

But your force varies during the time. So you want to take the integral of the area under the F - t graph they provide. Each square is an Fmax * 1ms. When you figure the area, then you can calculate Fmax by dividing the Impulse you found by the ms factor from the area you found.

 

[julz3216]

When you figure the area, then you can calculate Fmax by dividing the Impulse you found by the ms factor from the area you found.

The graph doesn't provide F values so for area i got ((6+2)*Fmax)/2
For impulse I got .06*33 = 1.98 ...(60 g is right -> .06 kg)

But, I don't know what you mean by dividing the impulse by the ms factor?

 

[LowlyPion]

The graph doesn't provide F values so for area i got ((6+2)*Fmax)/2
For impulse I got .06*33 = 1.98 ...(60 g is right -> .06 kg)

But, I don't know what you mean by dividing the impulse by the ms factor?

The impulse is change in momentum. That would be twice what you show.

Having found the area under the Force function you have that as 4*Fmax in units of N-s right?

So take the Impulse and divide by the time - 4ms to yield the value of your Fmax looks like to me.

Didn't you say F*Δt = I = Δmv ?