Two people, one of mass 78 kg and the other of mass 59 kg, sit in a rowboat of mass 88 kg. With the boat initially at rest, the two people, who have been sitting at opposite ends of the boat 2.9 m apart from each other, now exchange seats.
How far will the boat move?

The boat moves 0.244 m towards the end where the 59 kg person was at the start of the calculations.

Explanation:

The boat only moves because the centre of mass changes a bit if the two people on opposite ends of the boat exchange seats.

The boat moves a distance of the change in centre of mass

Noting that the weight of the boat acts at the centre of the boat at 1.45m from both ends.

For convention, we call the original position of the 59 kg person as x=0

This means,

59 kg person is at x = 0 m

88 kg of the boat acts at x = 1.45 m from the end of the 59 kg person.

78 kg person is at x = 2.90 m

Centre of mass = X = (Σ mᵢxᵢ)/(Σ mᵢ)

For the initial setup,

X = [(59×0) + (88×1.45) + (78×2.90)]/(59+88+78)

X = (353.8/225)

X = 1.572 m

(Don't forget that this is 1.572 m from the end we designated x=0 m)

When the people exchange positions,

59 kg person is now at the other end of the boat with x = 2.90 m

88 kg of the boat still acts at the centre of the boat at x = 1.45 m

And 78 kg person is now at the end of the boat with x = 0 m

Centre of mass = X = (Σ mᵢxᵢ)/(Σ mᵢ)

X = [(59×2.90) + (88×1.45) + (78×0)]/(59+88+78)

X = (298.7/225)

X = 1.328 m

(This is 1.328 m from the end we designated x=0 m from the start)

So, the centre of mass moves a distance of (1.572 - 1.328) towards the end of the boat we designated x=0 m from the start.

Hence, the boat moves 0.244 m towards the end where the 59 kg person was at the start of the calculations.

Hope this Helps!!!

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