physics
posted by sara on .
A man with a mass of 50 kg stands up in a 66kg canoe of length 4.0 m floating on water. He walks from a point 0.75 m from the back of the canoe to a point 0.75 m from the front of the canoe. Assume negligible friction between the canoe and the water. How far does the canoe move? (Assume the canoe has a uniform density such that its center of mass location is at the center of the canoe.)

Use the fact that the center mass of the man and canoe (together) does not move, as seen from shore. This is a result of total momentum conservation.
Initially, the CM is 1.25*50/116 = 0.514 m back of the center of the canoe. After moving, it is 0.514 m from forward of center.
For the CM to stay in the same place (as seen from shore), the canoe must move 2 x 0.514 = 1.028 m backwards. 
thanks