JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2021
- Medium
- Numerical
If the kinetic energy of a moving body becomes four times its initial kinetic energy, then the percentage change in its momentum will be \(...\%\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(100\)
Step-by-step Solution
Detailed explanation
\({K}_{2}=4 {K}_{1}\)
\(\frac{1}{2} m v_{2}^{2}=4 \frac{1}{2} m v_{1}^{2}\)
\(v_{2}=2 v_{1}\)
\(P=m v\)
\(P_{2}=m v_{2}=2 m v_{1}\)
\(P_{1}=m v_{1}\)
\(\% \,\text { change }=\frac{\Delta P}{P_{1}} \times 100=\frac{2 m v_{1}-m v_{1}}{m v_{1}} \times 100=100\, \%\)
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