A neutron travelling with a velocity V and kinetic energy E collides elastically head on with the nucleus of an atom of mass numbers A at rest. The fraction of total energy retained by the neutron is

Let mass of neutron = m
Then mass of nucleus = mA 
In head on elastic collision, velocity of neutron (after collision)
${\mathrm{\vartheta }}_1=\left(\frac{\mathrm{m}-\mathrm{mA}}{\mathrm{m}+\mathrm{mA}}\right)\mathrm{\vartheta }=\left(\frac{1-\mathrm{A}}{1+\mathrm{A}}\right)\mathrm{\vartheta }$
Total energy retained by the neutron
$\begin{array}{l}=\frac{\frac{1}{2}{\mathrm{mv}}_1^2}{\frac{1}{2}{\mathrm{m\vartheta }}^2}={\left(\frac{1-\mathrm{A}}{1+\mathrm{A}}\right)}^2\\ ={\left(\frac{\mathrm{A}-1}{\mathrm{A}+1}\right)}^2\end{array}$