Question

A ball is dropped from a height of 10m. If the energy of the ball reduces by 40% after striking the ground, how much high can the ball bounce back?
A. 6 m
B. 4 m
C. 3 m
D. None of the above

Answer 1

Hint: Since the ball dropped from the rest, its initial kinetic energy is zero. Therefore, the total initial energy of the ball is the potential energy. Calculate the lost in the potential energy after striking the ground and then determine the energy remaining in the ball after it strikes the ground. This energy can be used by the ball to bounce back.

Formula used:
Potential energy, \[U = mgh\]
where, m is the mass, g is the acceleration and h is the height.

Complete step by step answer:
We have given that the initial height of the ball is \[{h_1} = 10\,{\text{m}}\]. Since it is dropped from this height, its initial kinetic energy is zero. Therefore, the total initial energy of the ball is the potential energy.
Let us calculate the potential energy of the ball at height \[{h_1}\] as follows,
\[{U_1} = mg{h_1}\]
Here, \[m\] is the mass of the ball and g is the acceleration due to gravity.
Substituting \[g = 10\,{\text{m/}}{{\text{s}}^2}\] and \[{h_1} = 10\,{\text{m}}\] in the above equation, we get,
\[{U_1} = m\left( {10} \right)\left( {10} \right)\]
\[ \Rightarrow {U_1} = m \times 100\]
Now, we have given that on striking the ground, the ball loses 40% of its energy. Therefore, the final potential energy possessed by the ball after striking the ground is,
\[{U_2} = {U_1} - 40\% {U_1}\]
\[ \Rightarrow {U_2} = \left( {m \times 100} \right) - \dfrac{{40}}{{100}}\left( {m \times 100} \right)\]
\[ \Rightarrow {U_2} = \left( {m \times 100} \right) - \left( {m \times 40} \right)\]
\[ \Rightarrow {U_2} = m \times 60\]
Using this amount of energy the ball will bounce back off the ground. Let us calculate the height attained by the ball after striking the ground as follows,
\[{U_2} = m \times 60 = mg{h_2}\]
\[ \Rightarrow 60 = g{h_2}\]
Here, \[{h_2}\] is the height attained by the ball after striking the ground.
Substituting \[g = 10\,{\text{m/}}{{\text{s}}^2}\] in the above equation, we get,
\[60 = \left( {10} \right){h_2}\]
\[ \therefore {h_2} = 6\,{\text{m}}\]

So, the correct answer is option A.
Note:The ball will continue to bounce back until its total energy becomes zero. The reduction in the total energy of the ball after striking the ground can be used to convert it into sound energy and thermal energy to raise the temperature of the ball. We can also calculate the velocity of the ball when it bounces back using the law of conservation of energy.