When the momentum of a body increases by 100%, its KE increases by

400%

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100%

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300%

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none

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Solution

The correct option is A
400%

$\frac{d p}{p} = 100 % B y W o r k - K i n e t i c E n e r g y T h e o r e m, K = \frac{p^{2}}{2 m} d k = \frac{2 p}{2 m} d p \frac{d k}{k} = \frac{2 p}{2 m} \times \frac{2 m}{p^{2}} d p \frac{d k}{k} = 4 \frac{d p}{p} = 4 \times 100 % = 400 %$

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