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 u₁ and v₁ be the initial and final velocities of ball 1 and u₂ and v₂ be the similar quantities for ball 2. Here, u₂ = 0 and v₁ = 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, $\Delta K=K_i-K_f=\frac{1}{2}m{u}_1^2-\frac{1}{2}m{v}_2^2$
According to question,
$\frac{1}{2}\left(\frac{1}{2}m{u}_1^2\right)=\frac{1}{2}m{u}_1^2-\frac{1}{2}m{v}_2^2$
($\because$ 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,
$e=\left|\frac{v_2-v_1}{u_1-u_2}\right|=\frac{v_2}{u_1}=\frac{1}{\sqrt{2}}$