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 on the floor till the ball comes to rest is :

• A. p(1 + e)
• B. $\frac{p}{1-e}$
• C. $p(1+\frac{1}{e})$
• D. $p\left(\frac{1+e}{1-e}\right)$

Answer 1

The correct option is D $p\left(\frac{1+e}{1-e}\right)$

The initial momentum of the ball is $p$

The momenta of the ball after each successive collision be $p_1,$ $p_2,$ $p_3,$ and so on till it comes to rest

For 1st collision :

$\frac{p_1-0}{-p-0}=-e$ $⟹p_1=ep$

Thus net momentum imparted to the floor by 1st collision, $P_1^{\prime }=p_1-(-p)$

$\therefore$ $P_1^{\prime }=ep+p=p(1+e)$ .............(1)

For 2nd collision :

$\frac{p_2-0}{-p_1-0}=-e$ $⟹p_2=e{p}_1=e^{2}p$

Thus net momentum imparted to the floor by 2nd collision, $P_2^{\prime }=p_2-(-p_1)$

$\therefore$ $P_2^{\prime }=e^{2}p+ep=pe(1+e)$ .............(2)

Similarly momentum imparted to the floor by each other successive collisions are -

$p{e}^2(1+e),$ $p{e}^3(1+e),$ and so on

Total momentum imparted $P=p(1+e)+pe(1+e)+p{e}^2(1+e)+p{e}^3(1+e)+.........\infty$ term

$P=p(1+e)[1+e+e^2+e^3+............\infty ]$

$P=p(1+e)\times \frac{1}{1-e}$

$⟹P=\frac{p(1+e)}{(1-e)}$