The momentum of a body is increased by 50%. The percentage increase in the kinetic energy of the body is ________ %. [2023]

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Solution

Q.
The momentum of a body is increased by 50%. The percentage increase in the kinetic energy of the body is ________ %. [2023]

Ans.
(125)

$Kinetic energy of body = \frac{p^{2}}{2 m}$

$Initial kinetic energy = \frac{p_{i}^{2}}{2 m}$

$Final kinetic energy = \frac{p_{f}^{2}}{2 m} = \frac{{(1.5 p_{i})}^{2}}{2 m} = \frac{2.25 p_{i}^{2}}{2 m}$

$% increase in KE = \frac{2.25 \frac{p_{i}^{2}}{2 m} - \frac{p_{i}^{2}}{2 m}}{\frac{p_{i}^{2}}{2 m}} \times 100 = 125 %$

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