Question

A ball is dropped from height h on a horizontal floor. If it loses \[60\% \] of its energy on hitting the floor then upto which it will rise after first rebounce is
A. \[\dfrac{h}{5}\]
B. \[\dfrac{{2h}}{5}\]
C. \[\dfrac{{3h}}{5}\]
D. \[\dfrac{{4h}}{5}\]

Answer 1

Hint: The above problem is resolved by using the concepts and application of the conservation of energy. The energy is conserved when the ball is dropped from a specific height, such that we need to find the rebounding height. The height upto which the ball will bounce back will depend on the impact caused by the ball on the floor.

Complete step by step answer:
As the energy loss is, 60%. Then apply the conservation of energy as,
\[mgh = mg{h_1} + \left( {\dfrac{{60}}{{100}}} \right)mgh\]
Here, m is the mass of the ball, h is the height from where the ball is dropped, \[{h_1}\] is the required height and g is the gravitational acceleration.
On further solving the above equation, we get,
\[\begin{array}{l}
mgh = mg{h_1} + \left( {\dfrac{{60}}{{100}}} \right)mgh\\
h = {h_1} + 0.6h\\
{h_1} = h\left( {1 - 0.6} \right)\\
{h_1} = 0.4h
\end{array}\]
On simplifying the above relation as,
\[\begin{array}{l}
{h_1} = 0.4h\\
{h_1} = \dfrac{2}{5}h
\end{array}\]
Therefore, the ball will rebound upto the height of 2/5 times the given height.

Note: To resolve the given problem, one must try to understand the concept and applications of the conservation of the energy. While we are talking about the height, then the energy associated with it is known as the potential energy. Moreover, the potential energy depends on the height, along with the effect of gravity.