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Particle $m_{1}$ experiences a perfectly elastic collision with a stationary particle $m_{2}$ . After the collision (head-on) the particles fly apart in opposite directions with equal speeds, then

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$\frac{m_{1}}{m_{2}} = 1$

$\frac{m_{1}}{m_{2}} = 3$

$\frac{m_{1}}{m_{2}} = 2$

$\frac{m_{1}}{m_{2}} = \frac{1}{3}$

answer is B.

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Detailed Solution

from conservation of momentum

$m_{1} u_{1} = m_{2} u_{2} = m_{1} v_{1} |v_{1}| = |v_{2}| = v$

$m_{1} u_{1} = v (m_{2} - m_{1})$

$e = 1 = \frac{v_{2} - v_{1}}{u_{1} - u_{2}}$

$\vec{v_{2}} - \vec{v_{1}} = \vec{u_{1}}$

$v_{2} + v_{1} = u_{1}$

$2 v = u_{1}$

$m_{1} 2 v = v (m_{2} - m_{1})$

$2 m_{1} = m_{2} - m_{1}$

$3 m_{1} = m_{2}$

$\frac{m_{1}}{m_{2}} = \frac{1}{3}$

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