A rod of mass M and length L is lying on a horizontal frictionless...

From momentum conservation, ${P}_{i}^{0} = {P}_{f}$
$mu = Mv \ldots \left(i\right)$
From angular momentum conservation about $O$,
$mu \cdot \dfrac{L}{2} = \dfrac{M{L}^{2}}{12}\omega$
$⇒ \omega = \dfrac{6mu}{ML} \ldots ( ii )$
From $e = \dfrac{R . V \cdot S}{R . V \cdot A}$
$1 = \dfrac{V + \dfrac{\omegaL}{2}}{u}$

$V + \dfrac{\omegaL}{2} = u$
$V + \dfrac{3mu}{M} = u$
$\dfrac{mu}{M} + \dfrac{3mu}{M} = u$
$\dfrac{4mu}{M} = u$
$\dfrac{m}{M} = \dfrac{1}{4}$
$x = 4$