Question

A tennis ball bounces down a flight of stairs, striking each step in turn and rebounding to half of the height of the step. The coefficient of restitution is
$A)\text{ }\dfrac{1}{2}$
$B)\text{ }\dfrac{1}{\sqrt{2}}$
$C)\text{ }{{\left( \dfrac{1}{\sqrt{2}} \right)}^{\dfrac{1}{2}}}$
$D)\text{ }{{\left( \dfrac{1}{\sqrt{2}} \right)}^{\dfrac{1}{4}}}$

Answer 1

The coefficient of restitution in a collision is the ratio of the speed of an object just after the collision to the speed of the object just before the collision. We can get the speeds of the ball before and after the bounce on the step from the information of the height to which it rises by using the direct formula for the vertical height attained by an object in terms of its initial velocity.

Formula used:
$e=\dfrac{v}{u}$
$v=\sqrt{2gh}$

Complete step by step answer:
We will use the formula for the vertical height attained by an object with an initial speed against the force of gravity before it starts to fall down.
The height $h$ attained by an object with an initial speed $v$ against the force of gravity is given by
$v=\sqrt{2gh}$ --(1)
Where $g$ is the acceleration due to gravity.
As mechanical energy is conserved, the same speed $v$ will be attained by the object just before hitting the ground when dropped from a height $h$ above the ground without any initial velocity.
The coefficient of restitution $e$ during a collision is the ratio of the speed $v$ of the body just after the collision to the speed $u$ of the body just before the collision.
$e=\dfrac{v}{u}$ --(2)
Now, let us analyze the question.
Let us consider the situation where the tennis ball is falling from a height ${{h}_{i}}$ on to a step.
Let the speed of the ball just before hitting the step be $u$.
Let the required coefficient of restitution be $e$.
Just after bouncing from the step, let the speed of the tennis ball be $v$ and let it attain a height ${{h}_{f}}$.
Therefore, using (1), we get
$u=\sqrt{2g{{h}_{i}}}$ --(3)
Also, using (1), we get
$v=\sqrt{2g{{h}_{f}}}$ --(4)
Using (2), we get
$e=\dfrac{v}{u}$ --(5)
Putting (3) and (4) in (5), we get
$e=\dfrac{\sqrt{2g{{h}_{f}}}}{\sqrt{2g{{h}_{i}}}}=\dfrac{\sqrt{{{h}_{f}}}}{\sqrt{{{h}_{i}}}}=\sqrt{\dfrac{{{h}_{f}}}{{{h}_{i}}}}$ --(6)
Now, according to the question, the ball rises to half of its original height after bouncing off of each step.
$\therefore {{h}_{f}}=\dfrac{1}{2}{{h}_{i}}$ --(7)
Putting (7) in (6), we get
$e=\sqrt{\dfrac{\dfrac{1}{2}{{h}_{i}}}{{{h}_{i}}}}=\sqrt{\dfrac{\dfrac{1}{2}}{1}}=\sqrt{\dfrac{1}{2}}=\dfrac{1}{\sqrt{2}}$
Hence, we have got the required coefficient of restitution as $\dfrac{1}{\sqrt{2}}$.
Therefore the correct option is $B)\text{ }\dfrac{1}{\sqrt{2}}$.

Note:
It is not necessary to memorize the formula for the speed attained by a body just before reaching the ground when dropped from a height. It can easily be derived by equating the decrease in the gravitational potential energy written in terms of the height to the increase in kinetic energy written in terms of the speed. The same reason is also valid for the height attained by a body when given a vertical speed. Here, the decrease in kinetic energy is equated to the total increase in gravitational potential energy.