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# A rubber ball drops from a height $'h'$ . After rebounding twice from the ground, it rises to $\dfrac{h}{2}$ . The coefficient of restitution is:(A) $\dfrac{1}{2}$(B) ${\left( {\dfrac{1}{2}} \right)^{\dfrac{1}{2}}}$(C) ${\left( {\dfrac{1}{2}} \right)^{\dfrac{1}{4}}}$(D) ${\left( {\dfrac{1}{2}} \right)^{\dfrac{1}{6}}}$

Last updated date: 24th Jul 2024
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Hint: The coefficient of restitution is the ratio of the final velocity to the initial relative velocity between two objects after they collide. It normally ranges from zero to one where one would be a perfectly elastic collision. When the ball first hits the ground, its potential energy will be converted into kinetic energy.

Complete step by step solution: At height $'h'$ , the ball has potential energy $PE$ which is equal to:
$PE = mgh$
Here, $m$ is the mass of the rubber ball
$g$ is the acceleration due to gravity;
When the ball hits the ground, the potential energy becomes zero and the kinetic energy $KE$ is maximum. The potential energy is converted in the kinetic energy. Therefore, we can have:
$PE = KE$
$\Rightarrow mgh = \dfrac{1}{2}m{v^2}$
Here, $v$ is the velocity of the ball.
$\Rightarrow v = \sqrt {2gh}$
Let $e$ be the coefficient of restitution of the ball when the ball touches the ground and bounce back.
By definition, coefficient of restitution is given as:
$e = \dfrac{{{v_2} - {v_1}}}{{{u_1} - {u_2}}}$
Here ${v_2}$ is the speed of ground which is ${v_2} = 0$
${v_1}$ is the speed at which the ball rises upwards;
${u_1}$ is the speed at which the ball drops downwards, we have ${u_1} = \sqrt {2gh}$
${u_2}$ is the speed of the ground which is zero;
Substituting these values, we get
$e = \dfrac{{0 - {v_1}}}{{v - 0}}$
$\Rightarrow {v_1} = - ev$
The negative sign indicates that the velocity is in the opposite direction.
Now., the ball rises upwards and again hits the group and rises upwards. Let this velocity of rising upwards be ${v_2}'$ . This velocity will be given as
${v_2}' = - e({v_1})$
$\Rightarrow {v_2}' = - e( - ev)$
$\Rightarrow {v_2}' = {e^2}v$
The kinetic energy thus will be $\dfrac{1}{2}m{\left( {{v_2}'} \right)^2}$
At height $\dfrac{h}{2}$ the kinetic energy is zero and the ball only has potential energy of $\dfrac{{mgh}}{2}$
As energy after rebounding twice must be conserved, thus we have
$\dfrac{{mgh}}{2} = \dfrac{1}{2}m{\left( {{v_2}'} \right)^2}$
$\Rightarrow gh = {\left( {{e^2}v} \right)^2}$
Substituting $v = \sqrt {2gh}$ we get
$gh = {\left( {{e^2}\sqrt {2gh} } \right)^2}$
$\Rightarrow gh = {e^4}2gh$
$\Rightarrow {e^4} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{1}{4}}}$
This is the value of the coefficient of restitution.

Therefore, option $3$ is the correct option.

Note: The energy of the ball must be conserved when it is dropped from some height and hits the ground floor. Velocity is a vector quantity and thus its value is taken as per the direction of velocity. Finally, the potential energy of the block after rebounding twice must be equal to the kinetic energy of the ball as it hits the ground for a second time.