Question

A ball falls vertically onto a floor with momentum p and then bounces repeatedly. If the coefficient of restitution is e, then the total momentum imparted by the ball to the floor is –
$A.\,p(1+e)$
$B.\,\dfrac{p}{1-e}$
$C.\,p\left( \dfrac{1-e}{1+e} \right)$
$D.\,p\left( \dfrac{1+e}{1-e} \right)$

Answer 1

Firstly, we will define the terms, such as the coefficient of the reinstitution and the momentum. Then, step by step, we will compute the initial and the final momentum of the ball. We will then substitute these values to find the total momentum. Finally, we will obtain a series of total momentum.

Formula used:
$e=\dfrac{{{v}_{s}}}{{{v}_{a}}}$
$p=mv$

Complete answer:
The definition of the terms is given as follows:

The coefficient of the reinstitution is the ratio of the velocity of separation by the velocity of the approach.

The mathematical representation of the same is given as follows.

$e=\dfrac{{{v}_{s}}}{{{v}_{a}}}$

Where e is the coefficient of the reinstitution, ${{v}_{s}}$is the velocity of separation and ${{v}_{a}}$is the velocity of approach.

The momentum is the product of the mass and velocity of an object.

The mathematical representation of the same is given as follows.

$p=mv$

Where p is the momentum, m is the mass of the object and v is the velocity of the object.

As the ball keeps on bouncing repeatedly, so, we have the expressions for the total momentum imparted by the ball to the floor per every bounce is given as follows.

For the first time,

The initial momentum just before the ball strikes the floor is, $-p$.
The final momentum after the bounce is, $pe$.
The change in momentum is, $pe-(-p)=p(1+e)$

For the second time,

The initial momentum just before the ball strikes the floor is, $-pe$.
The final momentum after the bounce is, $p{{e}^{2}}$.
The change in momentum is, $p{{e}^{2}}-(-pe)=pe(1+e)$

For the third time,

The initial momentum just before the ball strikes the floor is, $-p{{e}^{2}}$.
The final momentum after the bounce is, $p{{e}^{3}}$.
The change in momentum is, $p{{e}^{3}}-(-p{{e}^{2}})=p{{e}^{2}}(1+e)$

……

This series continues till the ball stops. Therefore, the total momentum is given as follows.

& \Delta p=p(1+e)+pe(1+e)+pe{}^{2}(1+e)+.... \\\ & \Delta p=p(1+e)(1+e+{{e}^{2}}+.....) \\\ & \Rightarrow \Delta p=p\left( \dfrac{1+e}{1-e} \right) \\\ \end{aligned}$$ $$\therefore $$ The total momentum imparted by the ball to the floor is $$\Delta p=p\left( \dfrac{1+e}{1-e} \right)$$ **As, the total momentum imparted by the ball to the floor is $$p\left( \dfrac{1+e}{1-e} \right)$$ , thus, the option (D) is correct.** **Note:** The relation between the momentum and the coefficient of the reinstitution should be known. The method to solve the series should be known, as, in this case, we obtained the series of the total momentum and then we have resolved to the final answer.