Two bodies of masses $3 k g$ and $1 k g$ have equal momentum. Ratio of their kinetic energy respectively is:

1 : 1

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1 : 3

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3 : 1

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2 : 3

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Solution

The correct option is B $1 : 3$
Given ,
$m_{1} = 3 k g$
$m_{2} = 1 k g$
Let velocity be $v_{1} a n d v_{2}$ respectively.
Given, momentum is the same for two bodies.
$\Rightarrow m_{1} \times v_{1} = m_{2} \times v_{2}$
$\Rightarrow \frac{v_{1}}{v_{2}} = \frac{1}{3}$
Now ratio of kinetic energy is :
$\Rightarrow \frac{\frac{1}{2} m_{1} v_{1}^{2}}{\frac{1}{2} m_{2} v_{2}^{2}} = \frac{3}{9} = \frac{1}{3}$

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