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JEE Mains · Physics · STD 11 - Work,energy,power and collision
  • JEE Mains 2024
  • Hard
  • MCQ

When kinetic energy of a body becomes \(36\) times of its original value, the percentage increase in the momentum of the body will be _______.

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Verified Solution

Answer & Solution

Correct Answer

(A) \(500 \%\)

Step-by-step Solution

Detailed explanation

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

\(\Rightarrow \mathrm{P}=\sqrt{2 \mathrm{mK}}\)

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

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

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

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

\(=500 \%\)