An object of an ordinary size which we call a “macroscopic” system contains a huge number of atoms or molecules. It is out of the question to attempt to use the laws we have discussed for a single particle to describe separately each particle in such a system. There are nevertheless some relatively simple aspects of the behavior of a macroscopic system that (as we will show) follow the basic laws for a particle. We will be able to show that in any multiparticle system each of the mechanical quantities relevant to the state of the system consists of two parts:
 One part in which the system is treated as though it were a single particle (with the total mass of the system) located at a special point called the center of mass. This part is often called the CM motion.
 Another part was describing the internal motion of the system, as seen by an observer located at (and moving with) the center of mass.
We consider a system composed of N point particles, each labeled by a value of the index I which runs from 1 to N. Each particle has its mass m_{i }and (at a particular time) is located at its particular place r_{i} .
The center of mass (CM) of the system is defined by the following position vector
Center of mass 
\(r_{CM}=\frac{1}{M}\sum_{i}m_{i}x_{i}\)

Where M is the total mass of all the particles.
This vector locates a point in space — which may or may not be the position of any of the particles. It is the massweighted average position of the particles, being nearer to the more massive particles.
As a simple example, consider a system of only two particles, of masses m and 2m, separated by a distance. Choose the coordinate system so that the less massive particle is at the origin and the other is at x = as shown in the drawing. Then we have m1 = m, m2 = 2m, x1 = 0 , x2 = .
We find from the definition \(x_{cm}=\frac{1}{3m}\left ( m.0+2m \right )= \frac{2}{3}l\)
The location is indicated on the drawing above.
As time goes on the position vectors of the particles r_{i} generally change with time, so in general the CM moves. The velocity of its motion is the time derivative of its position:
\(v_{CM}=\frac{1}{M}\sum_{i=1}^{N}m_{1}v_{1}\)
But the sum on the right side is just the total (linear) momentum of all the particles. Solving for this, we find an important result:
Total momentum of a system 
\(P_{tot}=Mv_{CM}\)

This is just like the formula for the momentum of a single particle, so we see that: The total momentum of a system is the same as if all the particles were located together at the CM and moving with its velocity.