Ans.

Mass of the neutron is \( m \), and the mass of the nucleus is \( M \).

Let us consider \( u_1 \) be the velocity of the neutron, and the velocity of the nucleus before collision is zero, i.e., \( u_2=0 \)

Let , \( v_1 \) and \( v_2 \) be the velocities of the neutron and nucleus respectively after collision.

Since the collision is elastic head-on collision, so according to the conservation of linear momentum

\( mu_1+M\cdot{0}=mv_1+Mv_2 \)

or, \( mu_1=mv_1+Mv_2\tag{1} \)

If \( e \) be the coefficient of restitution,

\( \displaystyle{e=\frac{v_2-v_1}{u_1}}\tag{2} \)

Now the reduction of kinetic energy of the neutron is

Kinetic energy before collision ( \( T_1 \)) – Kinetic energy after collision ( \( T_2 \))

=\( \frac{1}{2}mu_1^2-\left(\frac{1}{2}mv_1^2+\frac{1}{2}Mv_2^2\right) \)

=\( \frac{1}{2}\left[mu_1^2-\left(mv_1^2+Mv_2^2\right)\right] \)

=\( \frac{1}{2(m+M)}\left[(m+M)mu_1^2-(m+M)\left(mv_1^2+Mv_2^2\right)\right] \)

=\( \frac{1}{2(m+M)}\left[(m^2u_1^2+mMu_1^2)-\left((m^2v_1^2+M^2v_2^2)+mM(v_1^2+v_2^2)\right) \right] \)

=\( \frac{1}{2(m+M)}\left[(m^2u_1^2+mMu_1^2)-\left({(mv_1+Mv_2)}^2-2mMv_1v_2+mM(v_1^2+v_2^2)\right)\right] \)

= \( \frac{1}{2(m+M)}\left[(m^2u_1^2+mMu_1^2)-\left({(mv_1+Mv_2)}^2+mM{(v_2-v_1)}^2\right) \right] \)

= \( \frac{1}{2(m+M)}\left[(m^2u_1^2+mMu_1^2)-m^2u_1^2-mM{(v_2-v_1)}^2 \right] \)
[ using equation (1) ]

= \( \frac{1}{2(m+M)}\left[mMu_1^2-mM{(v_2-v_1)}^2 \right] \)

= \( \frac{1}{2(m+M)}\left[mMu_1^2-mMe^2u_1^2\right] \)
[ using equation (2) ]

= \( \frac{1}{2}\frac{mM}{m+M}(1-e^2)u_1^2 \)

therefore the fraction of the kinetic energy reduced is

\( \displaystyle{\frac{T_1-T_2}{T_1}} \)

= \( \displaystyle{\frac{\frac{1}{2}\frac{mM}{m+M}(1-e^2)u_1^2}{\frac{1}{2}mu_1^2}} \)

= \( \displaystyle{\frac{M}{m+M}(1-e^2)} \)