Question

A steel ball falls from a height h on a floor for which the coefficient of restitution is e. The height attained by the ball after two rebounds is
A) eh
B) ${e^2}h$
C) ${e^3}h$
D) ${e^4}h$

Answer 1

Hint:In order to find out the second rebound height we need to find first rebound velocities in both the rebounds. The height will be taken as h in calculating first rebound velocity. But in the second rebound the maximum height attained by the ball in the first rebound will become height for the second rebound velocity, and hence we will find the height in the second rebound.

Complete step by step answer:
Step 1:
We are given:
A steel ball falls from a height h on a floor for which the coefficient of restitution is e.
The coefficient of restitution denoted by e is the ratio of the final to initial relative velocity between two objects after they collide. It normally ranges from 0 to 1 where 1 would be a perfectly elastic collision.
Here it is not specified whether the collision is perfectly elastic or inelastic.
Step 2:
After the first collision:
The velocity of the striking steel ball will be$\sqrt {2gh} $ where g is the gravity and h is the height from which a steel ball falls.
Then velocity rebounded is equal to the product of the coefficient of restitution with the velocity of strike i.e. $e \times \sqrt {2gh} $. Let this velocity rebound be denoted as V.
The maximum height attained by a steel ball after the strike will be$\dfrac{{{{\left( {Rebound{\text{ }}velocity} \right)}^2}}}{{2g}}$ …… (1)
This implies, height attained=$\dfrac{{{V^2}}}{{2g}}$
Substituting the value of rebound velocity we get, $\dfrac{{{e^2} \times 2gh}}{{2g}}$, cancelling out the 2g
Then we get rebound velocity after the first collision equal to${e^2}h$……. (2)
Step 3:
The steel ball falls two times and we are asked to find the maximum height attained by the ball after the second rebound.
Again using the same formula and procedure stated above in step 2
But this time the height will be equal to${e^2}h$
Now, the velocity of hitting the floor second time will be$\sqrt {2g\left( {{e^2}h} \right)} $
Again using same statement of velocity rebound for second term will be, $e \times \sqrt {2g\left( {{e^2}h} \right)} $
Taking out square term outside we get second rebound velocity equal to ${e^2}\sqrt {2gh} $
Let the second rebound velocity be denoted as ${V_1}$
Using equation (1) for second rebound to get the maximum height attained by it, =$\dfrac{{{V_1}^2}}{{2g}}$
Substituting the value of ${V_1}$ we get, height attained=$\dfrac{{{e^4} \times 2gh}}{{2g}}$, again cancelling out common term 2g.
We get height attained by the steel ball after the second rebound is${e^4}h$.

Hence, option D is the correct answer.

Note:We have found out the two rebounds of the ball. When it is not specified whether the collision of the body is elastic or inelastic then we can tell n numbers of height attained by the ball after n number of collisions. The height attained by the ball after n number of strikes is ${e^n}h$ .