Assume you are dropping a ball onto a perfectly hard floor.  It will rebound, but even the "liveliest " ball will not rise back to its starting position.
Why does it bounce back, and why does it loose height on the rebound?

The ball behaves like a spherical spring.  When the ball hits the floor it exerts a force on the floor larger than its weight, and the floor exerts a force on the ball of equal magnitude.  This force and the gravitational force acting on the ball compress the ball.  As long as the compression is small, Hooke's law is satisfied, the force compressing the ball is proportional to the displacement of the ball from its equilibrium shape.  The gravitational potential energy the ball has before it is dropped is converted into kinetic energy while the ball is falling and then into elastic potential energy when the ball is compressed.  But because the material the ball is made of is not perfectly elastic, internal friction converts some of the energy into thermal energy.

The elastic potential energy stored in the ball when it has lost all its kinetic energy is converted back into kinetic and gravitational potential energy.  The thermal energy, however, is not converted back.

The ball on the floor acts like a compressed spring.  It pushes on the floor with a force proportional to its displacement from its equilibrium shape.  The floor pushes back with a force of equal magnitude in the upward direction.  This force is greater in magnitude than the weight of the ball.  The net force is in the upward direction and the ball accelerates upward.  When the ball's shape is the shape it has when it is sitting still on the floor, (just slightly squashed), the net force is zero.  When its shape relaxes further, the net force is in the downward direction.  But it already has velocity in the upward direction, so it keeps on going upward until its speed has decreased to zero.  Because some of its initial gravitational potential energy has been converted into thermal energy it does not regain its initial height.

A particular ball is characterized by its coefficient of restitution, the ratio of its rebound speed to its collision speed just above the surface of the perfectly hard floor.  A perfectly hard floor is a floor that does not move itself.

coefficient of restitution = (outgoing speed)/(incoming speed)

Exact definition: The coefficient of restitution is defined as the ratio of the velocity components along the normal to the plane of contact after and before the collision.

The coefficient of restitution depends on the material the ball is made of and is always smaller than 1.  The coefficient of restitution is a ratio of speeds.  To find the ration of the outgoing to the incoming kinetic energy we use

Kout/Kin = v2out/v2in = (coefficient of restitution)2.

Real surfaces are not perfectly hard.  They distort when hit by the ball.  They store energy themselves, and return some of it to the ball as it rebounds.  "Lively" surfaces, such as a trampoline, store energy very efficiently and return almost all of it to the rebounding object.

#### Problem:

An elastic ball that wastes 30% of the collision energy as heat when it bounces on a hard floor.  To what fraction of its original height will it rebound?   What is its coefficient of restitution?

• Solution:
When the ball reaches its maximum height, all its remaining energy has been converted into gravitational potential energy, U = mgh.  Since the energy has decreases by 30% and m and g have not changed, h must have decreased by 30%.  The ball will rebound to 70% of its original height.
The coefficient of restitution is .

Question:

What happens when a ball with a given coefficient of restitution strikes a smooth, hard surface at an angle from the surface normal an then rebounds?