Q9 / 105 · NTA Abhyas (JEE Main) 2020 · Physics · Center of Mass Momentum and Collision

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

A
$1$
Correct answer
B
$2$
C
$3$
D
$4$

Solution

3 m_{1} v = m_{1} v + m_{2} v

3 m_{1} v - m_{1} v = m_{2} v

2 m_{1} v = m_{2} v

∴ \frac{m_{2}}{m_{1}} = 2

Hence, the correct answer is option: 'B'.

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