Physics - Center of Mass Question with Solution | TestHub

Since there are no external horizontal forces acting on the man-boat system, the center of mass of the system remains stationary.

Let $x_b$ be the displacement of the boat. The displacement of the man relative to the ground is $L-x_b$ (if he walks forward, boat moves backward, so his displacement relative to ground is $L$ relative to boat, minus boat's displacement). Change in position of COM: $\Delta x_{CM}=\frac{m\Delta x_{man}+M\Delta x_{boat}}{m+M}=0$.

So, $m(L-x_b)+M(-x_b)=0$. (Taking displacement of man relative to boat as $L$ and boat displacement as $-x_b$ if man walks from left to right).$mL-m{x}_b-M{x}_b=0\Rightarrow mL=(M+m)x_b\Rightarrow x_b=\frac{mL}{M+m}$.

However, the question asks for the displacement of the boat relative to the ground. If the man walks from the front to the back (or vice-versa), the boat moves in the opposite direction. If we assume the man walks in the positive direction ($+L$ relative to the boat), the boat moves in the negative direction, so its displacement is $-\frac{mL}{M+m}$.