Question
A man of mass stands at one end of a boat of length on still water. He walks to the other end. By how much does the boat shift?
Solution — Step by Step
The water is frictionless (or frictionless ice — same idea). No horizontal external force acts on the man-boat system, so the centre of mass cannot move horizontally.
Whatever the man moves to the right, the boat moves to the left, in just the right ratio.
Let the boat shift by (to the left). The man moves relative to the boat, so his ground displacement is (to the right).
Setting the COM displacement to zero:
The boat shifts by .
Why This Works
Conservation of COM position is one of the highest-yield shortcuts in mechanics. Whenever the system has no external horizontal force (boats on water, astronauts in space, two-block Atwood with horizontal symmetry), we can solve in one equation.
For our numbers: . Done in one line.
Alternative Method — Track Each Object Separately
We could write Newton’s laws, find the friction between the man and boat (via the man’s acceleration), then integrate. That gives the same answer but takes ~5 minutes vs 10 seconds.
For JEE-style timed papers, the COM shortcut is non-negotiable.
60-second rule: For man-on-boat problems, write instantly. The fraction is “weight of mover / total weight” times the relative displacement.
Common Mistake
Students often write (the full length) for the boat shift — confusing relative motion with ground motion. The man walks relative to the boat, not relative to the ground.
Another trap: ignoring the boat’s mass and assuming the boat shifts by exactly the same amount the man walks. That would only happen for a massless boat — which is rarely the case.
JEE Main 2024 used a near-identical problem with two children walking toward each other on a boat. Same template, just two movers. COM analysis gives the answer in three lines. This is a recurring favourite — appeared in JEE Main 2018, 2019, 2022, 2024.