Question

A plastic ball falls from a height of 4.9 metre and rebounds several times from the floor. What is the coefficient of restitution during the impact with the floor if 1.3 seconds pass from the first impact to the second one?
A) $0.9$
B) $0.1$
C) $0.7$
D) $0.8$

Answer 1

Hint: The coefficient of restitution is a concept which is ratio of velocity. Hence, in order to find the coefficient of restitution, we have to express the motion in terms of the velocity. Since, the given quantity is time taken for the second collision, it is important to express the velocity in terms of time to derive an expression to calculate the coefficient of restitution.

Complete step by step answer:
When collision occurs between two objects, there is a transfer of momentum from one object to another. If 100% of the velocity of one object gets transferred to the other object without any loss, it is called perfectly elastic collision.
This criterion is calculated by the coefficient of restitution or $e$.
The coefficient of restitution is the ratio of the final velocity after collision to the initial velocity before collision. If the value of $e = 1$, it is a perfectly elastic collision.
Consider a plastic ball falling from a height of $h = 4.9m$. The velocity acquired by the ball during the motion:
$v = \sqrt {2gh} $
where g = acceleration due to gravity, $9.8m{s^{ - 2}}$
When the ball hits the ground and bounces off with a velocity ${v_1}$ , it is equal to the coefficient of restitution times the initial velocity.
$\Rightarrow e = \dfrac{{{v_1}}}{v}$
$ \Rightarrow {v_1} = ev$
$ \Rightarrow {v_1} = e\sqrt {2gh} $
The time taken during the fall is given by –
$\Rightarrow t = \dfrac{v}{g}$
$ \Rightarrow t = \dfrac{{\sqrt {2gh} }}{g}$
$ \Rightarrow t = \sqrt {\dfrac{{2h}}{g}} $
Similarly, the time taken for the bounce after the first collision is –
$\Rightarrow {t_1} = \dfrac{{{v_1}}}{g}$
$ \Rightarrow {t_1} = \dfrac{{ev}}{g}$
$ \Rightarrow {t_1} = \dfrac{{e\sqrt {2gh} }}{g}$
$ \Rightarrow {t_1} = e\sqrt {\dfrac{{2h}}{g}} $
In the question, it is given that the total time taken for the initial fall, the bounce after the first collision and then fall towards the second collision, is equal to 1.3 seconds. Therefore,
$\Rightarrow 1.3 = t + 2{t_1}$
Substituting the values of time,
$\Rightarrow 1.3 = \sqrt {\dfrac{{2h}}{g}} + 2e\sqrt {\dfrac{{2h}}{g}} $
$ \Rightarrow 1.3 = \left( {1 + 2e} \right)\sqrt {\dfrac{{2h}}{g}} $
Substituting the values of h and g, we have –
$\Rightarrow 1.3 = \left( {1 + 2e} \right)\sqrt {\dfrac{{2 \times 4.9}}{{9.8}}} $
$ \Rightarrow 1.3 = \left( {1 + 2e} \right)\sqrt {\dfrac{2}{2}} $
$ \Rightarrow 1.3 = \left( {1 + 2e} \right)$
$ \Rightarrow 2e = 0.3$
$ \Rightarrow e = 0.15 \approx 0.1$
The coefficient of restitution, $e = 0.1$.

Hence, the correct option is Option B.

Note: However, it is impossible in nature for a perfectly elastic collision to occur. There is always reduction in the final velocity after imparting the velocity from one body to another. For example, when a body of a finite velocity hits the stationary ground, it does not bounce back to original height since it is not perfectly elastic collision. There is reduction in the velocity of the body when it arises.