Question

A ball is dropped from rest from a height of 12 m. If the ball loses 25% of its K.E. on striking the ground, what is the height to which it bounces? How do you account for the loss in K.E.

Answer 1

Step 1: Given data

The height from which the ball is dropped is $h=12m$

Loss in kinetic energy = $25\%$

Step 2: Calculating kinetic energy

When the ball is at rest, it possesses only potential energy and no kinetic energy. As the ball is dropped, its potential energy is converted to kinetic energy. Therefore, at any point, the loss in potential energy is equal to the gain in kinetic energy. Therefore, the kinetic energy of the ball will be,

$KE=PE=mgh$

where, $m$ is mass, $g$ is the acceleration due to gravity and $h$ is height.

Assume $g=10m/s^2$

Substituting the known values, we get,

$$\begin{gathered} KE=m\times 10m/s^2\times 12m \\ KE=120\times mJ \end{gathered}$$

Since, the ball loses $25\%$ of its kinetic energy on striking the ground, therefore, kinetic energy will be,

$$\begin{gathered} KE=(100-25)\%of120\times mJ \\ KE=75\%of120\times m \\ KE=\frac{75}{100}\times 120\times m \\ KE=90\times mJ \end{gathered}$$

Step 3: Calculating height to which the ball bounces back

When the ball hits the ground, it possesses only kinetic energy and no potential energy. As it bounces back, its kinetic energy starts converting to potential energy. At the highest point, the entire kinetic energy of the ball is converted to potential energy. Hence, the potential energy at the highest point will be equal to the kinetic energy when the ball hits the ground. Therefore,

$$\begin{gathered} 90\times mJ=PE \\ 90\times m=mgh \\ 90\times m=m\times 10\times h \\ h=9metres \end{gathered}$$

Therefore, the ball bounces back to a height of $9metres$.

Step 4: Explaining the loss in kinetic energy

When the ball touches the ground, it produces heat, sound, and vibrations on the floor. The kinetic energy of the ball is converted to different forms of energy and hence there is a loss in kinetic energy by $25\%$.