Perhaps it is to do with moments? As he's on one end it tips the boat slightly. The boat is 2m long, but when tipped the horizontal distance is changed. Therefore moving to the other side will have the same angle so he only moves that distance.

http://www.sparknotes.com/physics/linearmomentum/conservationofmomentum/problems.html

>>131834
I don't think you are supposed to take that into consideration.
>>131839
This will help. Thanks.

To be clear, we're assuming frictionless water etc etc.

As the man walks left, he exerts force on the boat to the right (his feet have to push on the boat to propel himself leftward).

Intuitively, consider a simpler case where he weighs the same as the boat: 25kg. Then if he walks 1m to the left, the boat will be pushed 1m to the right. If he weights 50kg, then now he weighs twice as much as the boat, and so the boat should move twice as far as he does.

And in the actual problem, he weights 75kg or 3 times what the boat weighs. This means that however far he moves, the boat moves 3 times as far.

So if he moves 1m to the left, then the boat would need to move 3m to the right. But obviously in the middle of this, he's going to run out of boat to walk on. The total distance he can move is limited - the distance he moves on the top of the boat plus the amount the boat moves to the right on the water cannot be more than 2m, or he'll walk off the edge.

So call the distance he walks x. Then the amount the boat moves is 3x. And x is limited such that 2m = x + 3x. Therefore, x = 0.5m.

The actual physics term involved here if you wanted to work this out more explicitly is "work." The work done moving the person to the left must equal the work done moving the boat to the right. Work is force times distance, and the force is proportional to the mass, which is how you get the 3x ratio of boat movement to person movement to begin with.

>>131847
That's a good explanation unfortunately it still doesn't feel very intuitive but thank you for the help. Much appreciated.