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 behaviour 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 multi-particle 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 centre 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 centre of mass.

**Center Of Mass**

We consider a system composed of N point particles, each labelled 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 centre 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 mass-weighted 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.