Q. A body of mass $m_{1}$collides elastically with another body of mass $m_{2}$at rest .If the velocity of $m_{1}$ after collision becomes 2/3 times its initial velocity, the ratio of their masses, is
A) 1:5
B) 5:1
C) 5:2
D) 2:5
Explanation
(b) In elastic collision
$v_{1} = \left( \frac{m_{1} - m_{2}}{m_{1} + m_{2}}\ \right)u_{1} + \left( \frac{2m_{2}}{m_{1} + m_{2}} \right)u_{2}$
If the second ball is at rest ,ie $u_{2} = 0,$then
$v_{1} = \left( \frac{m_{1} - m_{2}}{m_{1} + m_{2}} \right)u_{1}\ $
$\frac{2}{3}u_{1} = \left( \frac{m_{1} - m_{2}}{m_{1} + m_{2}} \right)u_{1}$ $\left\lbrack \because v_{1} = \frac{2}{3}u_{1} \right\rbrack$
Or $2m_{1} + 2m_{2} = 3m_{1} - 3m_{2}$
Or $m_{1} = 5m_{2}$
Or $\ \frac{m_{1}}{\ m_{2}} = \frac{5}{1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.