Q. A neutron travelling with velocity $u$ and kinetic energy $K$ collides head on elastically with the nucleus of an atom of mass number $A$ at rest. The fraction of its kinetic energy retained by the neutron even after the collision is
A) $\left( \frac{1 - A}{A + 1} \right)^{2}$
B) $\left( \frac{A + 1}{A - 1} \right)^{2}$
C) $\left( \frac{A - 1}{A} \right)^{2}$
D) $\left( \frac{A + 1}{A} \right)^{2}$
Explanation
(a) Here $m_{1} = u,m_{2} = Au,u_{1} = u$ and $u_{2} = 0$
$\therefore\ \ v_{1} = \frac{\left( m_{1} - m_{2} \right)u_{1}}{(m_{1} + m_{2})} + \frac{2m_{2}u_{2}}{(m_{1} + m_{2})} = \left( \frac{1 - A}{1 + A} \right)u$
$\Rightarrow \ \ \ \frac{v_{1}}{u} = \left( \frac{1 - A}{1 + A} \right)$
$\therefore\ \ \frac{K_{final}}{K_{initial}} = \left( \frac{v_{1}}{u} \right)^{2} = \left( \frac{1 - A}{1 + A} \right)^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.