Two men $(A$ & $B)$ of mass $50\text{kg}$ and $100\text{kg}$ are standing at the two opposite ends of a boat of mass $200\text{kg}$ and length $L=20\text{m}$. $A$ travels a distance $5\text{m}$ right relative to the boat towards the centre and $B$ moves a distance $15\text{m}$ left relative to the boat and meets $A$. Find the distance travelled by the boat in the water when they meet.

The correct option is B $\frac{25}{7}\text{m}$



Assumed that the distance travelled by the boat is $x$ towards left w.r.t the ground.

Distance travelled by $A$ w.r.t ground $x_1=(5-x)\text{m}$
Distance travelled by $B$ w.r.t. ground $x_2=-(15+x)\text{m}.$
Distance travelled by the boat w.r.t. ground $x_3$ = $-x\text{m}$.

Here, no external force is acting on the system. Hence $x_{com}$ will not change its position i.e $x_{com}=0$

$x_{com}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}$
Here, $m_1$ = mass of man $A$
$m_2$ = mass of man $B$
$m_3$= mass of boat
Then,
$x_{com}=\frac{50\times (5-x)+100\times (-15-x)+200\times (-x)}{50+100+200}=0$
$250-50x-1500-100x-200x=0$
$350x=-1250$
$x=\frac{-25}{7}\text{m}$
Negative sign indicates the boat will move towards right.