For a system of particles with masses $m_i$ at positions $\mathbf r_i$, the center of mass is

$$\boxed{\mathbf R=\frac{1}{M}\sum_i m_i\mathbf r_i},\qquad M=\sum_i m_i.$$

It is the mass-weighted average position of the system.

Differentiating gives the center-of-mass velocity

$$\mathbf V_{\rm CM}=\frac{d\mathbf R}{dt} =\frac{1}{M}\sum_i m_i\mathbf v_i.$$

Since the total momentum is

$$\mathbf P=\sum_i m_i\mathbf v_i,$$

we obtain

$$\boxed{\mathbf P=M\mathbf V_{\rm CM}}.$$

Thus the total momentum determines the translational motion of the system as a whole.

Motion under external forces

Differentiating again and applying Newton's second law to each particle gives

$$M\mathbf A_{\rm CM}=\sum_i\mathbf F_i.$$

Internal forces cancel in pairs when they satisfy Newton's third law, leaving

$$\boxed{M\mathbf A_{\rm CM}=\mathbf F_{\rm ext}}.$$

The center of mass therefore moves as if the system's total mass were concentrated there and acted on by the net external force.

Internal forces can produce complicated relative motion without changing the center-of-mass motion. If the net external force is zero,

$$\mathbf A_{\rm CM}=\mathbf0,$$

so the center of mass moves with constant velocity even while the parts of the system collide, separate, rotate, or rearrange.

Example

Two masses lie on the $x$-axis: $m_1=1,\mathrm{kg}$ at $x_1=0$ and $m_2=3,\mathrm{kg}$ at $x_2=4,\mathrm m$. Their center of mass is

$$X=\frac{(1)(0)+(3)(4)}{1+3}=3,\mathrm m.$$

The center lies closer to the heavier mass because the average is weighted by mass.

If their velocities are $v_1=5,\mathrm{m/s}$ and $v_2=-1,\mathrm{m/s}$, then

$$V_{\rm CM} =\frac{(1)(5)+(3)(-1)}{4} =0.50,\mathrm{m/s}.$$

Equivalently, their total momentum is

$$P=(1)(5)+(3)(-1)=2,\mathrm{kg,m/s}=MV_{\rm CM}.$$

Center-of-mass motion separates the translation of an entire system from the internal motion of its parts. This idea is fundamental in collisions, rigid-body mechanics, and two-body dynamics.