A rod of mass M and length L is lying on a horizontal frictionless surface. A particle of mass m travelling…

From momentum conservation, $P_{\mathrm{i}}^0=P_{\mathrm{f}}$
$mu=Mv \dots \left(\mathrm{i}\right)$
From angular momentum conservation about $O$,
$mu·\frac{L}{2}=\frac{M{L}^2}{12}\omega$
$\Rightarrow \omega =\frac{6mu}{ML} \dots \left(\mathrm{ii}\right)$
From $e=\frac{R.V·S}{R.V·A}$
$1=\frac{V+\frac{\omega L}{2}}{u}$

$V+\frac{\omega L}{2}=u$
$V+\frac{3mu}{M}=u$
$\frac{mu}{M}+\frac{3mu}{M}=u$
$\frac{4mu}{M}=u$
$\frac{m}{M}=\frac{1}{4}$
$x=4$