Question

A 40-kg canoe carrying two people is at rest on the surface of a lake. One of the people is a 50-kg person who is sitting 1.5 m to the left of the center of mass (CM) of the canoe. The other person has 90 kg and is sitting 0.5 m to the right of the CM of the canoe.

At one point in time, the two people rearrange themselves so that the 90-kg now sits 1.0 m to the left of the CM of the canoe, and the 50-kg person is sitting 0.75 m to the right of the CM of the canoe.

By how much, and in what direction, did the canoe move as a result of this rearrangement? Ignore any friction between the canoe and the water.

Answer 1

1.67 m towards right

Step-by-step explanation:

Let the position of the center of mass of the canoe be at the origin

`m_(c)` = mass of canoe = 40 kg

`x_(c)` = position of center of mass of canoe = 0 m

`m_(l)` = mass of person on left = 50 kg

`x_(l)` = position of center of mass of person on left = - 1.5 m

`m_(r)` = mass of person on right = 90 kg

`x_(r)` = position of center of mass of person on right = 0.5 m

Position of center of mass of the system is given as

`x_(cm) = (m_(c) x_(c) + m_(l) x_(l) + m_(r) x_(r))/(m_(c) + m_(l) + m_(r)) \\x_(cm) = ((40) (0) + (50) (- 1.5) + (90) (0.5))/(40 + 50 + 90) \\x_(cm) = - 0.167 m \\`

After the two people rearrange their positions, we have

`m_(c)` = mass of canoe = 40 kg

`x'_(c)` = position of center of mass of canoe

`m_(l)` = mass of person on left = 50 kg

`x'_(l)` = position of center of mass of person on left = - 1 m

`m_(r)` = mass of person on right = 90 kg

`x'_(r)` = position of center of mass of person on right = 0.75 m

Position of center of mass of the system remains at the same location and is given as

`x_(cm) = (m_(c) x'_(c) + m_(l) x'_(l) + m_(r) x'_(r))/(m_(c) + m_(l) + m_(r)) \\- 0.167 = ((40) x'_(c) + (50) (- 1.5) + (90) (0.5))/(40 + 50 + 90) \\x'_(c) = 1.502 m`

Distance traveled by the canoe is given as

`d = x'_(c) - x_(c) = 1.502 - (- 0.167)\\d = 1.67 m`

direction of movement : towards right.