Question

A plastic ball is dropped from a height of 1m and rebounds several times from the floor. If 1.3 sec elapse from the moment it is dropped to the second impact with the floor, what is the coefficient of restitution?
A. 0.85
B. 0.25
C. 0.94
D. 0.65

Answer 1

Hint: In this case, the coefficient of restitution $e=-\dfrac{{{v}_{2}}}{{{v}_{1}}}$. Find the time taken for the first impact of the ball by using $s=ut+\dfrac{1}{2}a{{t}^{2}}$. Then with this find the velocity before the first impact by using v=u+at. Then calculate the time taken for the second impact after the first impact. Then find the velocity of the ball just after the first impact by using $s=ut+\dfrac{1}{2}a{{t}^{2}}$.
Formula used:
$s=ut+\dfrac{1}{2}a{{t}^{2}}$
$v=u+at$
$e=-\dfrac{{{v}_{2}}}{{{v}_{1}}}$
W=mgh
$K=\dfrac{1}{2}m{{v}^{2}}$

Complete answer:
Let us first understand what is meant by coefficient of restitution (e). Coefficient of restitution tells us about the conservation of kinetic energies of the particles that collide. Or it tells us the extent of deformation (or elasticity)caused in the particles when they collide with each other.
The maximum value of e is 1. When e=1, this means that the collision is perfectly elastic and the particles reform into their original shape and size after the collision. Hence, the kinetic energy of this system is conserved.
When a body collides with ground the value of coefficient of restitution is given as $e=-\dfrac{{{v}_{2}}}{{{v}_{1}}}$, where ${{v}_{2}}$ is the velocity of the body after the collision with ground and ${{v}_{1}}$ is its velocity before collision.
Therefore, we have to find the velocities of the ball before and after collision.
Let us first find ${{v}_{1}}$.
Consider the case where the ball is dropped from 1 m height and it hits the ground. Here, ${{v}_{1}}$ is the velocity of the ball just before hitting the ground.
To find ${{v}_{1}}$, let us first find the time that the ball takes to hit the ground. Let that time be ${{t}_{1}}$.
Let us use the kinematic equation $s=ut+\dfrac{1}{2}a{{t}^{2}}$ …. (i)
In this case , s= -1m, u=0, t=${{t}_{1}}$. The ball is under the force of gravity. Therefore, its acceleration a= -g$=9.8m{{s}^{-1}}$.
Substitute the values in equation (i).
$\Rightarrow -1=(0){{t}_{1}}+\dfrac{1}{2}(-9.8)t_{1}^{2}$
$\Rightarrow 2=(9.8)t_{1}^{2}$
$\Rightarrow t_{1}^{2}=\dfrac{2}{9.8}=0.2040$
$\Rightarrow {{t}_{1}}=0.451s$
This means that the ball takes 0.451 seconds to hit the ground for the first time.
Now use the kinematic equation $v=u+at$ ….. (ii).
In this case, v = ${{v}_{1}}$, u=0, a= -g$=9.8m{{s}^{-1}}$ and t = ${{t}_{1}}=0.451s$.
Substitute the values in (ii).
$\Rightarrow {{v}_{1}}=0+(-9.8)(0.451)=-4.42m{{s}^{-1}}$.
Let us now find the value of ${{v}_{2}}$
For this, let us calculate the time taken for the second impact of the ball after the first impact. Let that time be ${{t}_{2}}$
Therefore, the time taken from the moment the ball is dropped to the second impact wih the floor is ${{t}_{1}}+{{t}_{2}}$.
And it is given that ${{t}_{1}}+{{t}_{2}}=1.3s$.
$\Rightarrow {{t}_{2}}=1.3-{{t}_{1}}=1.3-0.451=0.849s$.
This means that after the first impact, the ball takes a time of 0.849 seconds to hit the ground for the second time.
Let us now use the kinematic equation $s=ut+\dfrac{1}{2}a{{t}^{2}}$ …. (iii).
In this case, s=0, u=${{v}_{2}}$, a$=9.8m{{s}^{-1}}$ and t =${{t}_{2}}=0.849s$.
Substitute the values ion equation (iii).
 $\Rightarrow 0={{v}_{2}}(0.849)+\dfrac{1}{2}(-9.8){{(0.849)}^{2}}$
$\Rightarrow {{v}_{2}}=\dfrac{1}{2}(9.8)(0.849)=4.16m{{s}^{-1}}$
Therefore, the coefficient of restitution is $e=-\dfrac{{{v}_{2}}}{{{v}_{1}}}=-\dfrac{4.16}{-4.42}=0.94$

Hence, the correct option is C.

Note:
We can also find the magnitude of ${{v}_{1}}$ by using a work-energy theorem that says the work done is equal to the change in the kinetic energy of the body.
Here, the work done on the ball is by the gravitational force. Work done by gravitational force is given as W=mgh, where m is the mass of the ball and h is the height from where it is dropped.
This means that W=m(9.8)(1)=9.8g ….. (1).
The change in kinetic energy of the ball is $\Delta K=\dfrac{1}{2}mv_{1}^{2}-0=\dfrac{1}{2}mv_{1}^{2}$.
And $W=\Delta K$.
$\Rightarrow 9.8m=\dfrac{1}{2}mv_{1}^{2}$
$\Rightarrow v_{1}^{2}=19.6$
$\Rightarrow {{v}_{1}}=19.6=4.42m{{s}^{-1}}$.