Two bodies A and B having masses m and 4m respectively are moving with equal linear momentum. The ratio of kinetic energies between A and B is _________.

1 : 4

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4 : 1

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1 : 1

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1 : 2

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Solution

The correct option is C 1 : 4
Let $M_{1}$ and $M_{2}$ be momentum of the two bodies moving with velocities $v_{1}$ and $v_{2}$ resp.
Now $M =$ $m \times v$
For A: $M_{1}$ = $m_{1} \times v_{1}$ = $m \times v_{1}$
For B: $M_{2}$ = $m_{2} \times v_{2}$ = $4 m \times v_{2}$

But $M_{1}$ = $M_{2}$
$∴$ $\frac{v_{1}}{v_{2}}$ = $\frac{1}{4}$
$K E_{1}$ = $\frac{1}{2}$ $\times m_{1} \times v_{1}^{2}$
= $\frac{1}{2}$ $\times m_{1} \times v_{1} \times v_{1}$

= $\frac{1}{2}$ $\times M_{1} \times v_{1}$

$K E_{2}$ = $\frac{1}{2}$ $\times m_{2} \times v_{2}^{2}$
= $\frac{1}{2}$ $\times m_{2} \times v_{2} \times v_{1}$

= $\frac{1}{2}$ $\times M_{2} \times v_{2}$

Hence, $\frac{K E_{1}}{K E_{2}}$ = $\frac{v_{1}}{v_{2}}$ = $\frac{1}{4}$

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