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Kinetic Energy

A bomb at res...

Question

A bomb at rest explodes into two parts of masses $m_{1}$ and $m_{2}$ If the momentums the two parts be $p_{1}$ and $p_{2}$ then their kinetic energies will be in the ratio of

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

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

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\frac{p_{1}}{p_{2}}

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\frac{p_{2}}{p_{1}}

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Solution

The correct option is C $\frac{m_{2}}{m_{1}}$
By conservation of momentum

$P_{1} + P_{2} = 0$

$⟹ m_{1} v_{1} + m_{2} v_{2} = 0$

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

Ratio of $K . E = \frac{\frac{1}{2} m_{1} v_{1}^{2}}{\frac{1}{2} m_{2} v_{2}^{2}}$

$= \frac{m_{1}}{m_{2}} \frac{v_{1}^{2}}{(\frac{m_{1}}{m_{2}} v_{1})^{2}} = \frac{m_{2}}{m_{1}}$

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