A ball of mass m is dropped onto a floor from a certain height. The collision is perfectly elastic and the ball rebounds to the same height and again falls. Find the average force exerted by the ball on the floor during a long time interval.

#### Solution

It is given that the mass of the ball is m.

Let the ball be dropped from a height h.

The speed of ball before the collision is v_{1}.

\[\therefore \text{v}_1 = \sqrt{2\text{gh}}\]

The speed of ball after the collision is v_{2}.

\[\text{v}_2 = - \sqrt{2\text{gh}}\]

\[\text{ Rate of change of velocity = acceleration }\]

\[ \Rightarrow a = \frac{2\sqrt{2gh}}{t}\]

\[ \therefore \text{ Force}, F = \frac{m \times 2\sqrt{2gh}}{t} \ldots(1)\]

Using Newton's laws of motion, we can write:

\[v = \sqrt{2gh}, s = h, u = 0\]

\[ \Rightarrow \sqrt{2gh} = gt\]

\[\Rightarrow t = \sqrt{\frac{2h}{g}}\]

\[ \therefore \text{ Total time }= 2\sqrt{\frac{2h}{g}}\]

Substituting this value of time t in equation (1), we get:

F = mg