Linear momentum conserved on a boat? A man of mass $m$ is standing at one end of a stationary, floating barge of mass $3m$. He then walks to the other end of the barge, a distance of $L$ meters. Ignore any frictional effects between the barge and the water.
If the man walks with a velocity $v$, what is the average velocity of the barge?
I tried solving this problem keeping the center of mass in the same place. According to that, the barge moved $L/4$ meters to the left in the same time that the man crossed the barge. This would indicate that the average velocity of the barge is $-v/4$.
But when I calculate the average velocity using conservation of linear momentum (setting $0 = mv+3$ (velocity of the barge), I get average velocity of the barge is $-v/3$.
Why are these two answers different?
 A: From what I can tell, the important distinction to make here is whether the man is moving with a speed v relative to the barge or relative to the surface of the water.
If he is moving with a speed v relative to the barge, and we are assuming there is no friction between the barge and the water, then the barge must be moving along at an average speed of -v underneath him. I assume that this is not the case.
Alternatively, if he moves with speed v relative to the surface of the water: the speed of the barge would be $-\frac v3 $.
I'm not totally sure how you solved the problem the first way, the center of mass changes as the man walks across the barge.
A: I think that there is a mistake in your second strategy: if you want to use the conservation of momentum as seen by a reference system solidal to the platform, you have to consider that the velocity of the man is not $v$, but $v + V$, where $V$ is the velocity of the boat. If you do so, you obtain the same result as the first strategy, i.e. $V=-v/4$.
