JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
If the kinetic energy of a moving body becomes four times its initial kinetic energy, then the percentage change in its momentum will be \(...\%\)
- A \(100\)
- B \(300\)
- C \(400\)
- D \(200\)
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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