Newton's Third Law
A man of mass 60 kg is standing on a boat of mass 140 kg in still water. If the man walks from one end of the boat to the other with a velocity of 2 m/s relative to the boat, what is the velocity of the boat relative to water?
Select the correct option:
Solution
0.6 m/s opposite to man's motion
- Principle: In the absence of external horizontal forces (still water, no friction), the center of mass remains stationary, and total momentum is conserved.
- Conservation of Momentum: mmanvman+mboatvboat=0.
- Relative Velocity: Velocity of man w.r.t water vman=vman/boat+vboat.
- Equation:
- Let man's velocity w.r.t boat be +2 m/s.
- 60(2+vboat)+140vboat=0.
- 120+60vboat+140vboat=0.
- 200vboat=−120.
- Result: vboat=−0.6 m/s. The negative sign indicates movement opposite to the man.
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About This Question
- Subject
- physics
- Chapter
- laws of motion
- Topic
- newton's third law
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
0.6 m/s opposite to man's motion
- Principle: In the absence of external horizontal forces (still water, no friction), the center of mass remains stationary, and total momentum is conserved.
- Conservation of Momentum: mmanvman+mboatvboat=0.
- Relative Velocity: Velocity of man w.r.t water vman=vman/boat+vboat.
- Equation:
- Let man's velocity w.r.t boat be +2 m/s.
- 60(2+vboat)+140vboat=0.
- 120+60vboat+140vboat=0.
- 200vboat=−120.
- Result: vboat=−0.6 m/s. The negative sign indicates movement opposite to the man.
This hard difficulty physics question is from the chapter laws of motion, covering the topic of newton's third law. It appeared in the 2025 exam.
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