Centre of Mass: Application Problems (2)

Question

A man of mass $m = 60\text{ kg}$ stands on the left end of a boat of mass $M = 120\text{ kg}$ and length $L = 6\text{ m}$ floating on still water. He walks to the right end of the boat. By how much does the boat shift relative to the water? Assume no friction between boat and water.

Solution — Step by Step

Why This Works

Internal forces (the man pushing against the boat with his feet) cannot move the centre of mass. The boat moves backward, the man moves forward, and the system’s CM stays put. This is the same principle behind rockets, recoil, and walking on a frictionless floor (you can’t).

The fraction $m/(m+M)$ tells us how much of the relative motion shows up as boat displacement. Heavier boat → boat barely moves and man moves nearly the full $L$.

Alternative Method

Use the explicit CM formula. Place the origin at the original man-end. Initially, $x_{\text{cm}} = (m \cdot 0 + M \cdot L/2)/(m+M)$. After walking, the boat’s centre is now at $L/2 - d$ and the man is at $L - d$:

$x_{\text{cm}}' = \frac{m(L-d) + M(L/2 - d)}{m + M}$

Set $x_{\text{cm}} = x_{\text{cm}}'$ and solve — same answer, more steps.

Common Mistake

Using $d = ML/(m+M)$ instead of $d = mL/(m+M)$. Quick check: in the limit $M \gg m$, the boat barely moves, so $d$ should go to zero — that requires $m$ in the numerator.