Answer
Verified
91.2k+ views
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.
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.
Recently Updated Pages
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main
What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main
A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main
Other Pages
The vapour pressure of pure A is 10 torr and at the class 12 chemistry JEE_Main
Electric field due to uniformly charged sphere class 12 physics JEE_Main
3 mole of gas X and 2 moles of gas Y enters from the class 11 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Derive an expression for maximum speed of a car on class 11 physics JEE_Main
Velocity of car at t 0 is u moves with a constant acceleration class 11 physics JEE_Main