Q. An $\alpha$-particle of mass $m$suffers one dimensional elastic collision with a nucleus of unknown mass. After the collision the $\alpha$-particle is scattered directly backward losing $75\%$ of its kinetic energy .then the mass of the nucleus is
A) m
B) 2 m
C) 3 m
D) $\frac{3}{2}m$
Explanation
(c) $\frac{1}{2}m_{1}u_{1}^{2} - \frac{1}{2}m_{1}v_{1}^{2} = \frac{75}{100} \times \frac{1}{2}m_{1}u_{1}^{2}$
Or $u_{1}^{2} - v_{1}^{2} = \frac{3}{4}u_{1}^{2}$
$or\ \ \ \ \ \ \ \ {\ \ \ \ \ v}_{1} = \frac{1}{2}u_{1}$ .............(i)
Now $v_{1} = \frac{\left( m_{2} - m_{1} \right)u_{1}}{(m_{1} + m_{2})}\ldots.(ii)$
Thus, $\frac{1}{2}u_{1} = \frac{\left( m_{2} - m_{1} \right)u_{1}}{(m_{1} + m_{2})}$
$or\ \ \ \ \ \ \ \ \ \ \ \ \ m_{2} = 3m_{1} = 3m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.