Conservation of Momentum

We start by defining and explaining the concept of the center of mass for the simplest possible system of particles, one containing only two particles. From our work in this section we will generalize for systems containing many particles.

Before quantifying our idea of a center of mass, we must explain it conceptually. The concept of the center of mass allows us to describe the movement of a system of particles by the movement of a single point. We will use the center of mass to calculate the kinematics and dynamics of the system as a whole, regardless of the motion of the individual particles.

If a particle with mass m₁ has a position of x₁ and a particle with mass m₂ has a position of x₂, then the position of the center of mass of the two particles is given by:

Thus the position of the center of mass is a point in space that is not necessarily part of either particle. This phenomenon makes intuitive sense: connect the two objects with a light but rigid pole. If you hold the pole at the position of the center of mass of the objects, they will balance. That balancing point will often not exist within either object.

Now that we have the position, we extend the concept of the center of mass to velocity and acceleration, and thus give ourselves the tools to describe the motion of a system of particles. Taking a simple time derivative of our expression for x_cm we see that:

Thus we have a very similar expression for the velocity of the center of mass. Differentiating again, we can generate an expression for acceleration:

With this set of three equations we have generated the necessary elements of the kinematics of a system of particles.

From our last equation, however, we can also extend to the dynamics of the center of mass. Consider two mutually interacting particles in a system with no external forces. Let the force exerted on m₂ by m₁ be F₂₁, and the force exerted on m₁ by m₂ by F₁₂. By applying Newton's Second Law we can state that F₁₂ = m₁a₁ and F₂₁ = m₂a₂. We can now substitute this into our expression for the acceleration of the center of mass:

However, by Newton's Third Law F₁₂ and F₂₁ are reactive forces, and F₁₂ = - F₂₁. Thus a_cm = 0. Thus, if a system of particles experiences no net external force, the center of mass of the system will move at a constant velocity.

But what if there is a net force? Can we predict how the system will move? Consider again our example of a two body system, with m₁ experiencing an external force of F₁ and m₂ experiencing a force of F₂. We also must continue to take into account the forces between the two particles, F₂₁ and F₁₂. By Newton's Second Law:

Substituting this expression into our center of mass acceleration equation we get:

Again, however, F₁₂ = - F₂₁, and we can sum the external forces, producing:

Let M be the total mass of the system. Thus M = m₁ + m₂ and:

This equation bears a striking resemblance to Newton's Second Law. In this case, however, we are not speaking of the acceleration of individual particles, but that of the entire system. The overall acceleration of a system of particles, no matter how the individual particles move, can be calculated by this equation. Consider now a single particle of mass M placed at the center of mass of the system. Exposed to the same forces, the single particle will accelerate in the same way as the system would. This leads us to an important statement:

We have derived a method of making mechanical calculations for a system of particles. For simplicity's sake, however, we only derived this for a two- particle system. A derivation for an n particle system would be quite complex. A simple extension of our two particle equations to an n particle system will suffice.

Previously, M was defined as M = m₁ + m₂. However, to continue the study of center of mass we must make this definition more general. If there are n particles in a system, M = m₁ + m₂ + m₃ + ^... + m_n. In other words, M gives the total mass of the system. Equipped with this definition, we can simply state the equations for the position, velocity, and acceleration of the center of mass of a many particle system, similar to the two-particle case. Thus for a system of n particles:

These equations require little explanation, as they are identical in form to our two particle equations. All these equations for center of mass dynamics may seem confusing, however, so we will discuss a short example to clarify.

Consider a missile composed of four parts, traveling in a parabolic path through the air. At a certain point, an explosive mechanism on the missile breaks it into its four parts, all of which shoot off in various directions, as shown below. What can be said about the motion of the system of the four parts? We know that all forces applied to the missile parts upon the explosion were internal forces, and were thus cancelled out by some other reactive force: Newton's Third Law. The only external force that acts upon the system is gravity, and it acts in the same way it did before the explosion. Thus, though the missile pieces fly off in unpredictable directions, we can confidently predict that the center of mass of the four pieces will continue in the same parabolic path it had traveled in before the collision.

Such an example displays the power of the notion of a center of mass. With this concept we can predict emergent behavior of a set of particles traveling in unpredictable ways.

We have now shown a way to calculate the motion of the system of particles as a whole. But to truly explain the motion we must generate a law for how each of the individual particles react. We do so by introducing the concept of linear momentum in the next section.