# Average force of a tennis ball

1. Sep 8, 2010

### Null_

1. The problem statement, all variables and given/known data

A tennis ball has a mass of 0.057 kg. A professional tennis player hits the ball hard enough to give it a speed of 44 m/s (about 99 miles per hour.) The ball moves toward the left, hits a wall and bounces straight back to the right with almost the same speed (44 m/s). As indicated in the diagram below, high-speed photography shows that the ball is crushed about d = 2.1 cm at the instant when its speed is momentarily zero, before rebounding.

What is the magnitude of the average force exerted by the wall on the ball during contact?

2. Relevant equations

pf=pi+Fnet($$\Delta$$T

3. The attempt at a solution

For this problem, we also had to solve for the average speed from contact to 0, $$\Delta$$T, and mag. of Fgrav.

vavg= 22 m/s in x direction
$$\Delta$$T= 9.5455e-4 sec
mg=.5586 N

I got those right. Now for my attempt at the force.

I know that
pf=pi + Fnet $$\Delta$$T.
and
pavg= m(vavg)
= (.057)(22)
= 1.254

I assumed that Favg=pavg/$$\Delta$$T
and got
Favg= (1.254)/(9.5455e-4)
= 1313.708

That's not right.
Help would be greatly appreciated!

2. Sep 8, 2010

### housemartin

Hello!
I find your solution kinda clumsy ;] force according to Newton's second law is:
F = dp/dt
dp is change in the linear momentum, and dt is time interval in which change in momentum occurred. You assumed that average force is (average momentum)/dt, which is not true. You can find dp directly from data you are given, and to find dt, just use the fact that the ball's center of mass (i guess so ;]) moves 2.1 cm to stop, and 2.1 cm to rebound.

Last edited: Sep 8, 2010