Posted by **sara** on Wednesday, February 15, 2012 at 4:30pm.

A man with a mass of 50 kg stands up in a 66-kg 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.)

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**drwls**, Wednesday, February 15, 2012 at 4:53pm
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.

- physics -
**sara**, Wednesday, February 15, 2012 at 5:36pm
thanks

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