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Work Energy Power

JEE Main level, single correct, medium

100 of 183

Question 100 of 183

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

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Answer: (A) $100%$

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

\dfrac{1}{2}m{v}_{2}^{2}=4\times \dfrac{1}{2}m{v}_{1}^{2}

{v}_{2}=2{v}_{1}

P=mv

{P}_{2}=m{v}_{2}=2m{v}_{1}

{P}_{1}=m{v}_{1}

%

change

=\dfrac{\Delta P}{{P}_{1}}\times 100=\dfrac{2m{v}_{1}-m{v}_{1}}{m{v}_{1}}\times 100=100%

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JEE Main 2025 Apr 2 Shift 1 question paper

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