An alpha- particle of mass m suffers 1 - dimensional elastic collision with a nucleus at rest of unknown…

From momentum conservation,

$mv+0=-m{v}_1+M{v}_2 ...\left(1\right)$ 

Using kinetic energy conservation,

$$\begin{gathered} \frac{1}{2}m{v}^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}M{v}_2^2 \\ \Rightarrow \frac{1}{2}m{v}^2=0.36\times \frac{1}{2}m{v}^2+\frac{1}{2}M{v}_2^2 \\ \Rightarrow 0.64m{v}^2=M{v}_2^2 \\ \Rightarrow v_2=0.8v\sqrt{\frac{m}{M}} \end{gathered}$$ 

As it retain $36\%$,$\Rightarrow 0.36\times \left(\frac{1}{2}m{v}^2\right)=\frac{1}{2}m{v}_1^2 \Rightarrow 0.36v^2=v_1^2$ Hence, $v_1=0.6v$
From $\left(1\right)$, $mv=-m\left(0.6v\right)+M\left(0.8v\sqrt{\frac{m}{M}}\right)$

$\Rightarrow 1=-0.6+0.8\sqrt{\frac{M}{m}}$

On solving, we get $M=4m$