Question

A ball impinges directly on a similar ball at rest. The first ball is brought to rest by the impact. If half of the kinetic energy is lost by impact, the value of coefficient of restitution is

• A. $\frac{1}{2\sqrt{2}}$
• B. $\frac{1}{\sqrt{3}}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

Answer: (C)

$\frac{1}{\sqrt{2}}$

Let $u_1$ and $v_1$ be the initial and final velocities of ball $1$ and $u_2$ and $v_2$ be the similar quantities for ball $2$. Here,m $u_2=0$ and $v_1=0$.
$\therefore$ initial $KE,K_i=\frac{1}{2}m{u}_1^2+\frac{1}{2}m{u}_2^2=\frac{1}{2}m{u}_1^2$
and final $KE,K_f=\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2=\frac{1}{2}m{v}_2^2$
Loss of $KE,△K_i-K_f=\frac{1}{2}m{u}_1^2-\frac{1}{2}m{v}_2^2$
According to question,
$\frac{1}{2}\left(\frac{1}{2}m{v}_1^2\right)=\frac{1}{2}m{u}_1^2-\frac{1}{2}m{v}_2^2$
($\because$ half of its $KE$ is lost by impact)
or $u_1^2=2v_2^2$ or $v_2=\frac{u_1}{\sqrt{2}}$
$\therefore$ Coefficient of restitution,
$e=\left|\frac{v_2-v_1}{u_1-u_2}\right|=\frac{v_2}{u_1}=\frac{1}{\sqrt{2}}$.