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)

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