A tennis ball is dropped on a horizontal smooth surface. It bounces back to its original position after hitting the surface. The force on the ball during the collision is porportional to the length of compression of the ball. Which one of the following sketches describes the variation of its kinetic energy K with time t most appropriately? The figures are only illustrative and not to the scale.

The ball is dropped from rest at say t = 0. It hits the group at  $\mathrm{t}={\mathrm{t}}_1$ when its velocity $\mathrm{\nu }=\mathrm{gt}$ Its kinetic energy in time interval 0 to t₁ is given by

$\mathrm{K}=\frac{1}{2}\mathrm{m}{v}^2=\frac{1}{2}{\mathrm{mg}}^2{\mathrm{t}}^2$

i.e. $\mathrm{K}\propto {\mathrm{t}}^2$ So the slope of the K-t graph increases with time and the graph is not linear as shown in the figure

From $0 \mathrm{to} {\mathrm{t}}_1$ , as the ball is falling down kinetic energy increases.

From ${\mathrm{t}}_1 \mathrm{to} {\mathrm{t}}_2$ , ball comes in contact with ground and kinetic energy is converted into potential energy due to change in its shape.

At $\mathrm{t}={\mathrm{t}}_2$, maximum deformation in shape happens. Kinetic energy becomes momentarily zero and potential energy stored in ball becomes maximum.

From ${\mathrm{t}}_2 \mathrm{to} {\mathrm{t}}_3$ , ball starts recovering its shape and potential energy converts into kinetic energy. As a result, speed of the ball starts increasing.

After $\mathrm{t}={\mathrm{t}}_3$, ball leaves contact from ground and starts moving in upward direction. Due to gravity speed starts decreasing. Since there is no loss in energy the ball reaches the same height from which it was dropped.