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Conservation of Momentum

A mass of m...

Question

A mass of $m_{1}$ is a moving with velocity $v$ . It collides head on elastically with a stationary ball of mass $m_{2}$ . The velocity of ball becomes $\frac{v}{3}$ after collision,then the value of the ratio $\frac{m_{2}}{m_{1}}$ is:

0.5

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2

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Solution

The correct option is A $0.5$
Given : $u_{1} = v$ $u_{2} = 0$ $v_{1} = \frac{v}{3}$ $e = 1$

Using $v_{1} = \frac{(m_{1} - e m_{2}) u_{1} + (1 + e) m_{2} u_{2}}{m_{1} + m_{2}}$

$\frac{v}{3} = \frac{(m_{1} - 1 \times m_{2}) v + (1 + 1) m_{2} (0)}{m_{1} + m_{2}}$

$\frac{1}{3} = \frac{(m_{1} - m_{2})}{m_{1} + m_{2}}$
OR $m_{1} + m_{2} = 3 m_{1} - 3 m_{2}$
OR $2 m_{1} = 4 m_{2}$ $⟹ \frac{m_{2}}{m_{1}} = \frac{1}{2}$

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