Home/
Competitive Exams/
JEE Main & Advanced/
Physics/
Laws of Motion and Friction/
Question 30284
JEE Physics · Hard

Laws of Motion and Friction: Advanced — Momentum Conservation MCQ

Solve this quality-checked JEE multiple-choice question, then review the correct answer and explanation.

Timer-based practiceDetailed answer reviewMobile-friendly flow
Advanced — Momentum ConservationHardQuestion 30284

Question

A bomb (10 kg, at rest) explodes into two pieces: 4 kg moving at 20 m/s. The speed of the 6 kg fragment is:
  1. A
    13.3 m/s
    Correct
  2. B
    30 m/s
  3. C
    20 m/s
  4. D
    8 m/s

Correct answer

13.3 m/s

Explanation

Momentum conservation: 0 = 4×20 + 6×v₂. v₂ = −80/6 = −13.3 m/s. The 6 kg piece moves at 13.3 m/s opposite to the 4 kg piece.

Review Laws of Motion and Friction notesStart timed chapter practice