Question

Man $A$ and $B$ come to the middle of a boat which is at rest. Then displacement of the boat (neglecting any kind of friction) will be

• A. $15\text{cm}$
• B. $20\text{cm}$
• C. $13\text{cm}$
• D. $10\text{cm}$

Answer 1

The correct option is C $13\text{cm}$

For (man + boat) as a system, there are no external forces acting along horizontal direction.
Hence, position of centre of mass will not change along $x-$direction.
$\Rightarrow \Delta X_{\text{com}}=0...\left(i\right)$
Since, $X_{\text{com}}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}$
Hence, position of centre of mass of the system from the edge of the boat where man $A$ is initially.
$X_{\text{com}}=\frac{50\times 0+40\times 2+60\times 4}{50+40+60}$
$X_{\text{com}}=2.13\text{m}$

Let the boat shift to the right by distance $d$ when the two men come to the middle of the boat.


Now, centre of mass of system will lie at the centre of boat as two men reach the mid point of the boat.
Also, it will be at a distance of $2.13\text{m}$ from the earlier origin taken. (As all three masses are concentrated at mid point now)
$\Rightarrow (d+2)=X_{\text{com}}$
$\Rightarrow d+2=2.13$, from figure
$\therefore d=0.13\text{m}=13\text{cm}$
Boat will shift by $13\text{cm}$ to the right.