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JEE Mains · Physics · STD 11 - Work,energy,power and collision
  • JEE Mains 2023
  • Medium
  • Numerical

The momentum of a body is increased by \(50 \%\). The percentage increase in the kinetic energy of the body is \(...........\,\%\)

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Answer & Solution

Correct Answer

(B) \(125\)

Step-by-step Solution

Detailed explanation

Kinetic energy of body \(=\frac{p^2}{2 m}\)

Initial kinetic energy \(=\frac{p_1^2}{2 m }\)

Final kinetic energy \(=\frac{p_f^2}{2 m }=\frac{\left(1.5 p _{ i }\right)^2}{2 m }\)

\(=\frac{2.25 p _{ i }^2}{2 m }\)

\(\%\) increase in \(KE =\frac{2.25 \frac{ p _{ i }^2}{2 m }-\frac{ p _{ i }^2}{2 m }}{\frac{ p _{ i }^2}{2 m }} \times 100=125 \%\)

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