Motion of the Center of Mass showed that with no outside force, a system's center of mass coasts at constant velocity. Center of Mass by Integration located the center of mass of anything. Now the general rule: what external forces do to it.
Why it works
For each particle $i$, $m_i\vec{a}_i = \vec{F}_i^{\text{ext}} + \sum_j \vec{F}_{ij}$, where $\vec{F}_{ij}$ is the force particle $j$ exerts on particle $i$. Add over all particles. The left side is $\sum m_i\vec{a}_i = M\vec{a}_{\text{cm}}$, since $M\vec{r}_{\text{cm}} = \sum m_i\vec{r}_i$ differentiated twice. On the right, every internal force $\vec{F}_{ij}$ appears with its third-law partner $\vec{F}_{ji} = -\vec{F}_{ij}$, so they cancel in pairs. Only the external forces survive.
Key idea
$$\sum \vec{F}_{\text{ext}} = M\vec{a}_{\text{cm}}.$$ The center of mass moves like a single particle of mass $M$ acted on by the external forces alone, however the parts push, pull, spin or break apart.
Worked example: An exploding shell
A shell fired at $20\ \text{m/s}$ and $45^\circ$ over level ground breaks into two equal halves at the top of its path. One half stops dead and falls straight down. Where does the other half land? (Ignore the air.)
- The center of mass keeps following the original parabola: it lands at $R = v_0^2\sin 90^\circ/g = 400/9.8 = 40.8\ \text{m}$.
- The first half falls from the top, landing at $R/2 = 20.4\ \text{m}$. Both halves start the fall with zero vertical velocity, so they land together.
- $\tfrac{1}{2}(20.4) + \tfrac{1}{2}x_2 = 40.8$, so $x_2 = 61.2\ \text{m}$.
Answer: $61.2\ \text{m}$ from the launch point.
Worked example: Walking on a boat
A $60\ \text{kg}$ person walks $4\ \text{m}$ from one end of a $140\ \text{kg}$ boat to the other. How far does the boat move? (Ignore water drag.)
- No external horizontal force, and everything starts at rest, so the center of mass stays put.
- If the boat moves back $d$, the person moves forward $4 - d$ relative to the water: $60(4 - d) = 140d$.
- $d = 240/200 = 1.2\ \text{m}$, opposite to her walk.
Answer: $1.2\ \text{m}$ backward.
Common mistake
Including forces between the parts of the system when finding $\vec{a}_{\text{cm}}$. The spring between two linked carts, the tension in a rope joining them, and the push of a person on a boat all cancel in pairs. Check by drawing the boundary of the system: only forces from something outside it go in $\sum\vec{F}_{\text{ext}}$. The internal forces still matter for how each part moves, just not for the center of mass.