Newton's Third Law

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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?

RankGuru · Accepted Answer

1. **Principle**: In the absence of external horizontal forces (still water, no friction), the center of mass remains stationary, and total momentum is conserved.
2. **Conservation of Momentum**: $m_{man} \vec{v}_{man} + m_{boat} \vec{v}_{boat} = 0$.
3. **Relative Velocity**: Velocity of man w.r.t water $\vec{v}_{man} = \vec{v}_{man/boat} + \vec{v}_{boat}$.
4. **Equation**:
   - Let man's velocity w.r.t boat be $+2$ m/s.
   - $60(2 + v_{boat}) + 140 v_{boat} = 0$.
   - $120 + 60 v_{boat} + 140 v_{boat} = 0$.
   - $200 v_{boat} = -120$.
5. **Result**: $v_{boat} = -0.6$ m/s. The negative sign indicates movement opposite to the man.

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Newton's Third Law

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