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A flat boat of mass 100 kg and length 6m is at rest on the water in a tank. Two persons A and B of masses 70 kg and 50 kg are standing at the extremities of a boat. A is closer to the shore than B. They move towards each other and meet at the centre of the boat. When they have moved to the centre, the centre of mass of the boat will be (Neglect the friction between the boat and water)

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closer to the shore by 3/11 m

farther from the shore by 3/11 m

at the same distance from the shore

perform to and fro motion about the shore

answer is A.

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Detailed Solution

Centre of mass of the system will not change.

Let boat moved by x m away from the shore.

So, taking A as reference point,

$\frac{70 \times 0 + 50 \times 6 + 100 \times 3}{70 + 50 + 100} = \frac{(70 + 50 + 100) (x + 3)}{70 + 50 + 100}$

$\Rightarrow x = - \frac{3}{11}$

Hence, boat moves towards the shore by $\frac{3}{11} m .$

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