Question:

The momentum of a body is increased by 25%. The kinetic energy is increased by about

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Since \(KE \propto p^2\), if \(p\) increases by \(x%\), then \(KE\) increases by \((1+x/100)^2 - 1\) times 100%.
Updated On: Apr 8, 2026
  • 25%
  • 5%
  • 56%
  • 38%
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The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
Kinetic energy \(KE = \dfrac{p^2}{2m}\), so \(KE \propto p^2\).
Step 2: Detailed Explanation:
New momentum \(p' = 1.25p\)
New \(KE' = \dfrac{(1.25p)^2}{2m} = 1.5625 \times \dfrac{p^2}{2m} = 1.5625 \times KE\)
Percentage increase \(= (1.5625 - 1) \times 100 = 56.25 \approx 56%\)
Step 3: Final Answer:
Kinetic energy increases by approximately 56%.
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