Explanation:
Let \(M\) be the mass of the nucleus.
Applying conservation of linear momentum,
\(mv = m{v_1} + M{v_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\)
Also, \(\frac{1}{2}mv_1^2 = \frac{{36}}{{100}}\frac{1}{2}m{v^2} \Rightarrow {v_1} = \frac{3}{5}v\,\,\,(2)\)
Applying conservation of kinetic energy,
\(\frac{1}{2}m{{\rm{v}}^2} = \frac{1}{2}m{\rm{v}}_1^2 + \frac{1}{2}M{\rm{v}}_2^2\)
\( \Rightarrow \frac{1}{2}M{\rm{v}}_2^2 = \frac{{64}}{{100}}\frac{1}{2}m{{\rm{v}}^2}\)
\( \Rightarrow {v_2} = \frac{8}{{10}}v\sqrt {\frac{m}{M}} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} (3)\)
Substituting (2) and (3) in eqn. (1)
\(mv = - \left( {\frac{3}{5}v} \right)m + M\left( {\frac{8}{{10}}v\sqrt {\frac{m}{M}} } \right)\)
\( \Rightarrow \frac{8}{5}mv = \frac{8}{{10}}v\sqrt {mM} \)
\( \Rightarrow M = 4m\)