A boy of 40 kg is standing at a distance of 20 m from the shore on a stationary boat of mass 120 kg. He slowly moves in the direction of the shore on the boat by 8 m relative to the boat. Then select the wrong statement.

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The distance of the boy from the shore is now 14 m

The boat moves back, without thus disturbing the position of the centre of mass of boy+boat.

The centre of mass of the boat remains at same position but the boy has moved forward

The centre of mass of the boat and the boy remain at the same place.

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{120 \times (d + 20) + 40 \times 20}{160} = r_{CM} \\ \frac{120 \times (d + 8 + x) + 40 \times x}{160} = r_{C . M}^{'} \end{array}$
As no external force is acting $r_{C . M .} = r^{'} C . M$
$\begin{array}{l} ∴ 120 \times d + 120 \times 20 + 40 \times 20 = 120 \times d + 120 \times 8 + 120 \times x + 40 x \\ \Rightarrow x = 14 m \end{array}$
The boat has moved back to keep the C.M. at the same place.