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If kinetic energy of a moving particle increases by 2%, find the percentage change in its linear momentum

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Detailed Solution

$k = \frac{1}{2} {mv}^{2} = \frac{p^{2}}{2 m} \Rightarrow \frac{∆ k}{k} = 2 (\frac{∆ p}{p})$

$∴ \frac{∆ k}{k} x 100 % = 2 x (\frac{∆ p}{p}) x 100 %$

$\Rightarrow (\frac{∆ p}{p}) x 100 % = \frac{1}{2} x 2 % = 1 %$

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