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NEET PHYSICSMedium

Two persons of mass 55 kg55 \text{ kg} and 65 kg65 \text{ kg} respectively, are at the opposite ends of a boat. The length of the boat is 3.0 m3.0 \text{ m} and weighs 100 kg100 \text{ kg}. The 55 kg55 \text{ kg} man walks up to the 65 kg65 \text{ kg} man and sits with him. If the boat is in still water the centre of mass of the system shifts by:

A

3.0 m3.0 \text{ m}

B

2.3 m2.3 \text{ m}

C

zero

D

0.75 m0.75 \text{ m}

Step-by-Step Solution

Let the system be the combination of the two men and the boat. Since the boat is in still water, there is no net external force acting on the system in the horizontal direction. When the man walks inside the boat, the forces involved (friction between feet and boat) are internal forces of the system. Internal forces cannot change the state of motion of the centre of mass. Since the system was initially at rest, the centre of mass will remain stationary. Therefore, the shift in the centre of mass of the system is zero.

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