If kinetic energy of a moving body becomes four times its initial kinetic energy, then percentage ch

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5.Work, Energy, Power and Collision

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If the kinetic energy of a moving body becomes four times its initial kinetic energy, then the percentage change in its momentum will be $...\%$

A

$100$

B

$300$

C

$400$

D

$200$

(JEE MAIN-2021)

Solution

${K}_{2}=4 {K}_{1}$

$\frac{1}{2} m v_{2}^{2}=4 \frac{1}{2} m v_{1}^{2}$

$v_{2}=2 v_{1}$

$P=m v$

$P_{2}=m v_{2}=2 m v_{1}$

$P_{1}=m v_{1}$

$\% \,\text { change }=\frac{\Delta P}{P_{1}} \times 100=\frac{2 m v_{1}-m v_{1}}{m v_{1}} \times 100=100\, \%$

Standard 11

Physics

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