A mass $m_{1}$ with velocity $v_{1}$ impacts with a mass $m_{2}$ at rest. After the impact, the mass $m_{1}$ comes to rest. Then the coefficient of restitution e should be

e = \frac{m_{1}}{m_{1} + m_{2}}

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e = \frac{m_{2}}{m_{1} + m_{2}}

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

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

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Solution

The correct option is C $e = \frac{m_{1}}{m_{2}}$
$Given: v_{1} = 0 and u_{2} = 0$

$(v_{2} - v_{1}) = e (u_{1} - u_{2})$

$v_{2} = e \times u_{1} . . . . . . (1)$

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

$∴ m_{1} u_{1} = m_{2} v_{2}$

$∴ \frac{m_{1}}{m_{2}} = \frac{v_{2}}{u_{1}} = e$

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