A body is standing at the centre of a boat which is free to move on water. If the masses of the body and the boat are $m_{1}$ and $m_{2}$ , respectively, the boy moves a distance of 1 m forward, then the movement of the boat is :

\frac{m_{1}}{m_{1} + m_{2}}

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\frac{m_{2}}{m_{1} + m_{2}}

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\frac{m_{1}}{m_{2}}

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\frac{m_{2}}{m_{1}}

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Solution

The correct option is B $\frac{m_{1}}{m_{1} + m_{2}}$
The centre of mass of the boat lies at its geometric centre i.e. point O. Also, initially the body is standing at the center of the boat. Thus COM of the body+boat system is at point O.
Now the body moves $1$ m forward. As no external force is acting on the system, thus the COM of the system must not move. To do so, the centre of boat moves backwards by an amount $x$ .
Using $x_{c m} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}$
Or $0 = \frac{m_{1} (1 - x) + m_{2} x}{m_{1} + m_{2}}$
Or $m_{1} (1 - x) + m_{2} x = 0$
$⟹$ $x = \frac{m_{1}}{m_{1} + m_{2}}$

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