If the momentum of a body is increased by 50%, what will be the percentage increase in its kinetic energy? MCQ with Answer and Explanation

If the momentum of a body is increased by 50%, what will be the percentage increase in its kinetic energy?

A. 50%

B. 125%

C. 225%

D. 100%

Answer: Option B

Solution (By JKExamLibrary)

Kinetic energy (K) and momentum (p) are related by K = p^2 / 2m. If momentum increases by 50%, the new momentum p' = p + 0.5p = 1.5p. The new kinetic energy K' = (1.5p)^2 / 2m = 2.25 (p^2 / 2m) = 2.25 K. This means K has increased by 1.25 K, which is a 125% increase. Understanding quadratic relationships is crucial for such problems.

This question belongs to: Science Physics

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Practice More Physics Questions

Question #1 Report Error

Wire stretched to double length (volume constant). New resistance:

A. R

B. 2R

C. 4R

D. R/2

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Correct Answer: Option C

Explanation:
Volume constant: A' = A/2 when L' = 2L. R' = ρ(2L)/(A/2) = 4ρL/A = 4R. Memory tip: 'Stretch wire: R ∝ L² when volume fixed'. Resistance proportionality frequently tested in competitive exams.

This question belongs to: Science Physics

Question #2 Report Error

A body of mass 2 kg has momentum 10 kg m/s. Its kinetic energy is

A. 20 J

B. 25 J

C. 5 J

D. 50 J

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Correct Answer: Option B

Explanation:
Kinetic energy K = p²/(2m) = (10)²/(2×2) = 100/4 = 25 J. Also, velocity v = p/m = 5 m/s, K = ½×2×5² = 25 J. Formula p²/(2m) is useful for quick computation. Always ensure units: kg, m/s, J.

This question belongs to: Science Physics