Q. A body of mass $m$ moving with velocity $v$ collides head on another body of mass 2$m$ which is initially at rest. The ratio of KE of colliding body before and after collision body before and after collision will be
A) 1:1
B) 2:1
C) 4:1
D) 9:1
Explanation
(d) KE of colliding body before collision=$\frac{1}{2}mv^{2}$
After collision its velocity becomes
V’=$\left( \frac{m_{1} - m_{2}}{m_{1} + m_{2}} \right)v = \frac{m}{3m}v = \frac{v}{3}$
KE after collision=$\frac{1}{2}mv^{2} = \frac{1}{2}m\left( \frac{v}{3} \right)^{2}$
$= \frac{1}{2}\frac{mv^{2}}{9}$
Ratio of kinetic energy=$\frac{KE_{before}}{KE_{after}}$
$= \frac{\frac{1}{2}mv^{2}}{\frac{1}{2}\frac{mv^{2}}{9}} = \frac{9}{1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.