The center of mass of a system is its mass-weighted average position. It turns out to move in the simplest way possible, and that gives a second view of momentum conservation.
Why it works
Take two objects. Their center of mass is at $x_{\text{cm}} = \dfrac{m_1x_1 + m_2x_2}{m_1 + m_2}$. In a short time each object moves $v\Delta t$, so the center of mass moves with velocity $$v_{\text{cm}} = \frac{m_1v_1 + m_2v_2}{m_1 + m_2} = \frac{p_{\text{tot}}}{M}.$$ If the total momentum is constant, so is $v_{\text{cm}}$. More generally, $\Sigma F_{\text{ext}} = M a_{\text{cm}}$: only outside forces can change how the center of mass moves.
Key idea
The center of mass moves as if all the mass were there and every outside force acted on it. With no net outside force, it moves in a straight line at constant speed (or stays at rest), no matter how the parts collide, push apart or explode.
Worked examples
Worked example: Velocity of the center of mass
A $3.0\ \text{kg}$ glider moves right at $1.2\ \text{m/s}$ and a $1.0\ \text{kg}$ glider moves left at $0.80\ \text{m/s}$. Find $v_{\text{cm}}$.
- $$p_{\text{tot}} = (3.0)(1.2) + (1.0)(-0.80) = 2.8\ \text{kg·m/s}.$$
- $v_{\text{cm}} = \dfrac{2.8}{4.0} = 0.70\ \text{m/s}$ to the right, before and after they collide.
Answer: $0.70\ \text{m/s}$ right
Worked example: Walking on a raft
A $60\ \text{kg}$ person walks $4.0\ \text{m}$ along a $140\ \text{kg}$ raft floating at rest. How far does the raft move?
- Person plus raft start at rest with no outside horizontal force (ignore water drag), so the center of mass stays put: $60\,\Delta x_p + 140\,\Delta x_r = 0$.
- She moves $4.0\ \text{m}$ relative to the raft: $\Delta x_p = \Delta x_r + 4.0$.
- $60(\Delta x_r + 4.0) + 140\,\Delta x_r = 0$, so $\Delta x_r = -\dfrac{240}{200} = -1.2\ \text{m}$: backward $1.2\ \text{m}$.
Answer: $1.2\ \text{m}$ backward
Worked example: A shell that bursts
A firework shell would land $100\ \text{m}$ from its launch point. At the top it bursts into two equal pieces; one stops and falls straight down. Where does the other land?
- The burst is internal, so the center of mass still lands at $100\ \text{m}$.
- The top is halfway, so the first piece lands at $50\ \text{m}$. Both pieces start falling from the same height with no vertical speed, so they land together.
- $100 = \dfrac{50 + x_2}{2}$, so $x_2 = 150\ \text{m}$.
Answer: $150\ \text{m}$ from the launch point
Common mistake
Thinking an explosion changes the path of the center of mass. The explosion's forces are internal pairs, so they cancel in $\Sigma F_{\text{ext}}$. Only gravity (an outside force) acts before and after, so the center of mass keeps its parabola. Check: the mass-weighted average of the pieces' landing points (if they land at the same time) is where the unexploded shell would have landed.