An alpha- particle of mass m suffers 1- dimensional elastic...

From momentum conservation,

$mv + 0 = - m{v}_{1} + M{v}_{2} . . . ( 1 )$

Using kinetic energy conservation,

$\dfrac{1}{2}m{v}^{2} = \dfrac{1}{2}m{v}_{1}^{2} + \dfrac{1}{2}M{v}_{2}^{2} ⇒ \dfrac{1}{2}m{v}^{2} = 0 . 36 \times \dfrac{1}{2}m{v}^{2} + \dfrac{1}{2}M{v}_{2}^{2} ⇒ 0 . 64m{v}^{2} = M{v}_{2}^{2} ⇒ {v}_{2} = 0 . 8v\sqrt{\dfrac{m}{M}}$

As it retain $36$,$⇒ 0 . 36 \times \left(\dfrac{1}{2}m{v}^{2}\right) = \dfrac{1}{2}m{v}_{1}^{2} ⇒ 0 . 36{v}^{2} = {v}_{1}^{2}$ Hence, ${v}_{1} = 0.6v$
From $( 1 )$, $mv = - m\left(0 . 6v\right) + M\left(0 . 8v\sqrt{\dfrac{m}{M}}\right)$

$⇒ 1 = - 0 . 6 + 0 . 8\sqrt{\dfrac{M}{m}}$

On solving, we get $M = 4m$