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 \(...........\,\%\)
Pick an option to check your answer.
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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