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Conservation of Momentum

A stationary ...

Question

A stationary object explodes into masses $m_{1}$ and $m_{2}$ . They move in opposite directions with velocities $v_{1}$ and $v_{2}$ . The ratio of kinetic energy $E_{1}$ to kinetic energy $E_{2}$ is

A

\frac{m_{2}}{m_{1}}

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B

\frac{m_{1}}{m_{2}}

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C

\frac{2 m_{2}}{m_{1}}

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D

\frac{2 m_{1}}{m_{2}}

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Solution

The correct option is A $\frac{m_{2}}{m_{1}}$
As stationary object explodes into masses $m_{1}$ and $m_{2}$ and move in opposite direction.
According to conservation of momentum
$m_{1} v_{1} = - m_{2} v_{2} \Rightarrow \frac{v_{1}}{v_{2}} = \frac{{- m}_{2}}{m_{1}}$
Now, ratio of KE
$\frac{{K E}_{1}}{{K E}_{2}} = \frac{1 / 2 m_{1} v_{1}^{2}}{1 / 2 m_{2} v_{2}^{2}} = \frac{m_{1}}{m_{2}} {[\frac{m_{2}}{m_{1}}]}^{2} = \frac{m_{2}}{m_{1}}$

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