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 52 m/s (about 117 miles per hour.) The ball moves toward the left, hits a wall and bounces straight back to the right with almost the same speed (52 m/s). As indicated in the diagram below, high-speed photography shows that the ball is crushed about d = 2.8 cm at the instant when its speed is momentarily zero, before rebounding.

Question

Making the very rough approximation that the large force that the wall exerts on the ball is approximately constant during contact, determine the approximate magnitude of this force.

What is the average speed of the ball during the period from first contact with the wall to the moment the ball's speed is momentarily zero? 26m/s

How much time elapses between first contact with the wall, and coming to a stop?

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

In contrast, what is the magnitude of the gravitational force of the Earth on the ball? .5586

Answer 1

change in momentum = 2*.057*52 = 5.928 kg m/s

average speed during crush = 52/2 = 26 m/s
so time to stop = .028/26 = .0010769 s
same time to accelerate back so
time force acts = 2*.028/26 = .0021538 seconds

force = rate of change of momentum
= 5.928/.0021538 = 2752 Newtons