If K.E. body is increased by $100 %$ . Then $%$ change in 'P'.

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Solution

Linear momentum $P = \sqrt{2 m K}$
Final kinetic energy $K_{f} = 2 K$
So, final momentum $P_{f} = \sqrt{2 m K_{f}} = \sqrt{4 m K}$
Percentage change in momentum $\frac{P_{f} - P}{P} \times 100 = \frac{\sqrt{4 m K} - \sqrt{2 m K}}{\sqrt{2 m K}} \times 100$
So, $\frac{P_{f} - P}{P} \times 100 = \frac{\sqrt{4} - \sqrt{2}}{\sqrt{2}} \times 100 = 41.4 %$

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