A man with a mass of 52 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.)
CM does not move.
52*(4-1.5)+66*X=0
solve for X
this didn't work
To find out how far the canoe moves when the man walks, we need to consider the conservation of momentum.
The initial momentum of the system (man and canoe) is zero because they are at rest. When the man starts walking, he exerts a force on the canoe in the backward direction. According to Newton's third law of motion, the canoe exerts an equal but opposite force on the man. This force causes the man to move forward, and the recoil force causes the canoe to move backward.
Knowing the initial momentum is zero, we can equate the final momentum of the system to zero as well. The total momentum of the system is given by the sum of the momentum of the man and the momentum of the canoe.
Let's denote the displacement of the man as x and the displacement of the canoe as y. According to the law of conservation of momentum:
(mass of the man * velocity of the man) + (mass of the canoe * velocity of the canoe) = 0
Since the mass of the canoe is given as 66 kg and the mass of the man is given as 52 kg, and the final velocity of both the man and the canoe is zero, we can write:
(52 kg * v) + (66 kg * (-v)) = 0
Simplifying the equation:
52v - 66v = 0
-14v = 0
v = 0
This implies that the final velocity of both the man and the canoe is zero, indicating no net momentum after the man walks.
Therefore, the displacement of the canoe, y, is zero. So the canoe does not move.