Question

A ball of mass $m$ moving at a speed $v$ makes a head on collision with an identical ball at rest. The kinetic energy of balls after the collision is ${\frac{3}{4}}^{\text{th}}$ of the original. If $e$ is the coefficient of restitution. Then:

• A. After collision, ball which was initially at rest starts moving with velocity $\frac{1+\frac{1}{\sqrt{2}}}{2}v$
• B. After collision, ball which was initially moving, has velocity $\frac{1-\frac{1}{\sqrt{2}}}{2}v$
• C. $e=\frac{1}{\sqrt{2}}$
• D. $e=\frac{1}{2}$

Answer 1

Answer: (A), (B), (C)

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

Given that a ball of mass $m$ moving at a speed $v$ makes a head on collision with an identical ball at rest.
For $e=\frac{\text{velocity of seperation}}{\text{velocity of approach}}$
$e=\frac{v_2-v_1}{v-0}$
$v_2-v_1=ev....\left(1\right)$
now from conservation of linear momentum,
$mv=m{v}_1+m{v}_2$
$v=v_1+v_2.....\left(2\right)$
From $\left(1\right)\left(2\right),$ we get
$v_1=\left(\frac{1-e}{2}\right)v$ and $v_2=\left(\frac{1+e}{2}\right)v$

Given that, $\frac{3}{4}K{E}_i=K{E}_f$
$\Rightarrow \frac{3}{4}\left(\frac{1}{2}m{v}^2\right)=\frac{1}{2}m{\left\{\left(\frac{1-e}{2}\right)v\right\}}^2+\frac{1}{2}m{\left\{\left(\frac{1+e}{2}\right)v\right\}}^2$
$\Rightarrow \frac{3}{4}=\frac{(1-e{)}^2}{4}+\frac{(1+e{)}^2}{4}$
$\Rightarrow 3=2+2e^2$
$\Rightarrow 1=2e^2$
$\therefore e=\frac{1}{\sqrt{2}}$
and $v_1=\left(\frac{1-\frac{1}{\sqrt{2}}}{2}\right)v$