Physics

Problem

A ball of mass

m

is dropped vertically from a height

h_1

and rebounds to a height

h_2

. The ball is in contact with the floor for a time interval of

\tau

. a. a Show that the average *net* force on the ball is given by

F = m \frac{\sqrt{2gh_1} + \sqrt{2gh_2}}{\tau}

b. b If

h_1 = 8.0 \text{ m}

h_2 = 6.0 \text{ m}

\tau = 0.125 \text{ s}

and

m = 0.250 \text{ kg}

, calculate the average force exerted by the ball on the floor.

Intuitive explanation

The key to this problem is understanding the collision as an impulse—a large force acting over a very short time. Rather than worrying about the complicated forces during impact, use the change in momentum to connect the velocities just before and after the bounce. The coefficient of restitution (or energy loss) links the heights to these velocities, and the contact time $\tau$ lets you calculate the average force the floor exerts on the ball.

Intuitive explanation

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