A particle loses $25\%$ of its kinetic energy during head on collision with another identical particle initially at rest. The coefficient of restitution will be

The correct option is C $\frac{1}{\sqrt{2}}$

Applying momentum conservation before and after collision,
$p_{initial}=p_{final}$
$mu=m(v_1+v_2)u=v_1+v_2...\left(i\right)$

We know,
$e=\frac{\text{(velocity of separation)}}{\text{(velocity of approach)}}=\frac{(v_2-v_1)}{(u-0)}\Rightarrow e\times u=v_2-v_1....\left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$

$v_1=\frac{u(1-e)}{2}\text{and}{v}_2=\frac{(1+e)u}{2}$

Given:
${K.E}_{final}=\frac{3}{4}{K.E}_{initial}$
As $\frac{1}{4}$th of initial $K.E$ is lost during collision

$\Rightarrow \frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2=\frac{3}{4}\times \left(\frac{1}{2}m{u}^2\right)$
On substituting the value of $v_1$ and $v_2$, we get
${\left(\frac{1-e}{2}\right)}^2+{\left(\frac{1+e}{2}\right)}^2=\frac{3}{4}(1-e{)}^2+(1+e{)}^2=3\Rightarrow 2+2e^2=3\Rightarrow e=\pm \frac{1}{\sqrt{2}}$
As $e$ always takes $+ve$ value,
$\therefore e=\frac{1}{\sqrt{2}}$