Question

A moving body with a mass m₁ strikes a stationary body of mass m₂. The masses $m_{1}$ and $m_{2}$ should be in the ratio $\frac{m_{1}}{m_{2}}$so as to decrease the velocity of the first body 1.5 times assuming a perfectly elastic impact. Then the ratio $\frac{m_{1}}{m_{2}}$ is

• A. 1/ 25
• B. 1/5
• C. 5
• D. 25

Answer 1

Answer: (C)

5

Answer 2

$v_{1} = \left( \frac{m_{1} - m_{2}}{m_{1} + m_{2}} \right)u_{1} + \frac{2m_{2}u_{2}}{m_{1} + m_{2}} = \left( \frac{m_{1} - m_{2}}{m_{1} + m_{2}} \right)u_{1}$

[As u₂ = 0 and $\left( v_{1} = \frac{u_{1}}{1.5} \right)$ given]

⇒ $\frac{u_{1}}{1.5} = \left( \frac{m_{1} - m_{2}}{m_{1} + m_{2}} \right)u_{1}$ ⇒ $m_{1} + m_{2} = 1.5(m_{1} - m_{2})$

⇒ $\frac{m_{1}}{m_{2}} = 5$.