### Motion with no external force

When no net external force acts on a system, the center-of-mass is stationary.

One or more parts of the system can move, but other parts must move in response to keep the center-of-mass in the same place.

#### Example

A 90 kg man sits at one end of a 2.4 meter canoe that has a mass of 30 kg. If the man moves to the opposite end of the canoe, how far does the canoe move?

1. 0.6 m
2. 0.8 m
3. 1.8 m
4. 7.2 m

There are several ways to attack this problem. Perhaps the simplest is to say that for each distance d the man moves, the canoe must move 3d in the opposite direction for the center-of-mass of the system to remain in the same place. Therefore, for each distance d the man travels relative to a fixed reference point, he moves 4d relative to the canoe. The 2.4 meter distance is relative to the canoe, so

4d = 2.4 m and d = 0.6 m relative to a fixed reference point.

The canoe moves 3d, 1.8 m, in the opposite direction relative to the reference point.

A formal solution using the center-of-mass equation looks like this. Taking the origin to be where the man is initially, and using subscripts c for canoe and m for the man, we get:
Xcom =
 xcmc + xmmm mc + mm

With xc = 1.2 and xm = 0, this gives:
Xcom =
 1.2 * 30 + 0 * 90 30 + 90
=
 1.2 * 30 120
= 0.3 m

If the man moves a distance dm to the right his position is now xmf = dm. The canoe moves a distance dc to the left so its new position is xcf = 1.2-dc.

The center-of-mass equation is now:
Xcom =
 xcf mc + xmf mm mc + mm
0.3 m =
 (1.2-dc) mc + dm mm mc + mm

Here you can use the relationship dc + dm = 2.4 m to reduce the equation to a single unknown and solve for dm.

Plugging in dc = 2.4 - dm gives:

dm = 0.6 m, which agrees with what we got above.

Conservation of momentum is a third approach. There is zero momentum before and after the man moves, so there must be zero net momentum while he moves. The momentum he has in one direction must be cancelled out by the canoe's momentum in the other direction.