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A light body A and a heavy body B have equal linear momentum. Then the KE of the body A is

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zero

greater than that of B

equal to that of B

smaller than that of B

answer is B.

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Detailed Solution

$m_{1} v_{1} and m_{2} v_{2} and \frac{K_{1}}{K_{2}} = \frac{m_{1} {v_{1}}^{2}}{m_{2} {v_{2}}^{2}} = \frac{m_{2}}{m_{1}}$

As $m_{1} < m_{2} K_{2} < K_{1}$

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