When kinetic energy of a body becomes 36 times of its original value, percentage increase in momentu

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5.Work, Energy, Power and Collision

hard

When kinetic energy of a body becomes $36$ times of its original value, the percentage increase in the momentum of the body will be :

A

$500 \%$

B

$600 \%$

C

$6 \%$

D

$60 \%$

(JEE MAIN-2024)

Solution

$\text { Kinetic energy }(\mathrm{K})=\frac{\mathrm{P}^2}{2 \mathrm{~m}}$

$\Rightarrow \mathrm{P}=\sqrt{2 \mathrm{mK}}$

$\text { If } \mathrm{K}_{\mathrm{f}}=36 \mathrm{~K}_{\mathrm{i}}$

$\text { So, } \mathrm{P}_{\mathrm{f}}=6 \mathrm{P}_{\mathrm{i}}$

$\% \text { increase in momentum }=\frac{P_f-P_i}{P_i} \times 100 \%$

$=\frac{6 \mathrm{P}_{\mathrm{i}}-\mathrm{P}_{\mathrm{i}}}{\mathrm{P}_{\mathrm{i}}} \times 100 \%$

$=500 \%$

Standard 11

Physics

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