If the momentum of the body reduces n times its kinetic energy reduces by

1/n

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1/n²

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n²

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Solution

The correct option is B
1/n²

Momentum of the body = P = mv

Kinetic energy (KE) = $\frac{1}{2}$ m $v^{2}$

KE = $\frac{P^{2}}{2 m}$

If momentum reduces by $\frac{1}{n}$

KE = $\frac{P^{2}}{2 m n^{2}}$

So, Kinetic energy of the body reduces by $\frac{1}{n^{2}}$ times

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