Two persons 60 kg and 80 kg are sitting at the ends of a boat of mass 60 kg and length 4 m. The boat is stationary. If the men now exchange their positions, then:

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The boat moves by 0.4 m

The centre of mass of the two men shifts by $\frac{4}{7} m$

The centre of mass of the two men shifts by 2 m

The boat moves by 0.6 m

answer is A.

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Detailed Solution

Let $m_{1}, m_{2}$ are masses of two persons. m₃ is the mass of the boat.
$m_{1} {\vec{s}}_{1} + m_{2} {\vec{s}}_{2} + m_{3} {\vec{s}}_{3} = 0$
${\vec{s}}_{3}$ is the displacement of boat
Where ${\vec{s}}_{1} & {\vec{s}}_{2}$ are displacements of persons
Give that
$\begin{array}{l} {\vec{s}}_{13} = 4 \hat{i} = {\vec{s}}_{1} - {\vec{s}}_{3} \\ {\vec{s}}_{23} = - 4 \hat{i} = {\vec{s}}_{2} - {\vec{s}}_{3} \end{array}$
From the above equations we get
$\begin{array}{l} 60 (4 \hat{i} + {\vec{s}}_{3}) + 80 ({\vec{s}}_{3} - 4 \hat{i}) + 60 {\vec{s}}_{3} = 0 \\ 200 {\vec{s}}_{3} = 80 \hat{i} \\ {\vec{s}}_{3} = 0 .4 \hat{i} \end{array}$