If the momentum of a body is increased by 100 %, then the percentage increase in the kinetic energy is

150 %

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200 %

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225 %

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300 %

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Solution

The correct option is D
300 %

$E = \frac{P^{2}}{2 m} \Rightarrow \frac{E_{2}}{E_{1}} = {(\frac{P_{2}}{P_{1}})}^{2} = {(\frac{2 P}{P})}^{2} = 4$

$E_{2} = 4 E_{1} = E_{1} + 3 E_{1} = E_{1} + 300 % o f E_{1}$

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Linear momentum of a body is increased by $40 %$ . Then percentage increase in its kinetic energy is:

125

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