Question

A body of mass $m_{1}$ collides head on elastically with a stationary body of mass $m _{2} .$ If velocities of $m_{1}$ before and after the collision are $v$ and $-v / 3$ respectively then the value of $m _{1} / m _{2}$ is-

• A. 1
• B. 2
• C. 0.5
• D. 4

Answer 1

Answer: (C)

From conservation of momentum
$P_{i}=P_{f} $
$m_{1} v=-m_{1} \frac{v}{3}+m_{2} v_{2} $
$\frac{4}{3} m_{1} v=m_{2} v_{2} $
$\frac{m_{1}}{m_{2}}=\frac{3 v_{2}}{4 v} \ldots$(1)
Form energy conservation
$\frac{1}{2} m_{1} v^{2}=\frac{1}{2} m_{1} \frac{v^{2}}{9}+\frac{1}{2} m_{2} v_{2}^{2}$
$\frac{1}{2} m_{1} v^{2}-\frac{1}{18} m_{1} v^{2}=\frac{1}{2} m_{2} v_{2}^{2}$
$\frac{9 m_{1} v^{2}-m_{1} v^{2}}{18}=\frac{1}{2} m_{2} v_{2}^{2}$
$\frac{48 m_{1} v^{2}}{18}=\frac{1}{2} m_{2} v_{2}^{2}$
$\frac{8}{9} m_{1} v^{2}=m_{2} v_{2}^{2}$
$\frac{8}{9} \frac{m_{1}}{m_{2}}=\left(\frac{v_{2}}{v}\right)^{2}$
$\frac{v_{2}}{v}=\sqrt{\frac{8 m_{1}}{9 m_{2}}}$
Now in equation (1)
$\frac{m_{1}}{m_{2}}=\frac{3}{4} \times \sqrt{\frac{8 m_{1}}{9 m_{2}}}$
$\left(\frac{m_{1}}{m_{2}}\right)^{2}=\frac{9}{16} \times \frac{8}{9}\left(\frac{m_{1}}{m_{2}}\right)$
$\frac{m_{1}}{m_{2}}=\frac{1}{2}=0.5$