# The Physics Philes, lesson 32: Save the Momenta!

*In which laws of motion are revisited, momentum is conserved, and forces are classified.*

Last week was a big week for me on my journey to understanding physics. I started trigonometry, my first for-real math class in 10 years. *Ten years*. So I was a little busy doing further review and readings for class. Nevertheless, I did manage to wedge in a little physics into my study time. This week I present to you a quick post on conservation of momentum.

We know from the last couple of weeks what momentum is (at least, I hope we do). But what happens if we have two interacting bodies? Let’s enter our mind labs for a quick thought experiment. Let’s imagine that we have two particles floating out in space and touch each other. According to Newton’s third law of motion, the forces the particles exert on each other are equal in magnitude and opposite in direction. That means that the impulses and changes in momentum are also equal and opposite.

That’s pretty easy to understand. Let’s say that our two hypothetical particles is a system. The forces the particles exert on each other are called internal forces. Forces exerted on the particles by some object outside the system are called external forces. When there are no external forces, we say the system is an isolated system. In our hypothetical two-particle system, there are no external forces, so it’s isolated.

Newton’s second law of motion motion in terms of momentum says that the net force acting on a particle equals the time rate of change of momentum of the particle. In math it looks like this:

In our hypothetical system, the net force exerted on particle A by particle B and the net force exerted on B by A, the rates of change of the momenta are

The momentum of each particle can change, but the changes will be related to the each other by Newton’s third law of motion. Since the forces will always be equal in magnitude and opposite in direction, the sum of those forces will be zero. To wit:

The rate of change of the momenta is equal and opposite, so the rate of change of the vector sum the momenta of the two particles is zero. The total momentum of the system is the vector sum of the momenta of the individual particles, indicated by a capital P. The time rate of change of the total momentum P is zero, so the total momentum of the system is constant, even if the individual momenta of the particles that make up the system changes.

If there are external forces involved, they must be added along with the internal forces, If this happens, in general, the total momentum will not be constant. However, if the vector sum of the external forces is zero, then the forces don’t contribute to the sum and we’re back to zero again. More officially, if the vector sum of the external forces on a system is zero, the total momentum of the system is constant.

This, boys and girls, is the simplest form of the principle of conservation of momentum. And you know what’s pretty sweet about it? We don’t really have to know much about the internal forces that act in a system. We can still apply this principle. Kinda cool, right?

That is basically all I know about conservation of momentum. Next week we’ll do a sample problem or two so we can see how it applies. Until then, if you spot a mistake or something that needs clarification please leave a comment.

*Featured image credit: mikemol*

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It’s also worth mentioning that in non-conservative systems, momentum could be conserved in one direction and not in another. Even if the net external forces on a system is nonzero, if it’s zero in some direction, then momentum is conserved in that direction.

Think, for example, about a particle with mass m which moves in a 2D plane, with a constant external force (say, gravity). Let’s call the position of the particle, which is a vector, q:

q = ix + jy

Here, i and j are the usual 2D basis unit vectors.

Let’s say the external force is:

F = -mgj

Here, g is the strength of the gravitational pull, m is the mass of the particle, and j is the unit vector.

Clearly, momentum is not conserved. If the particle starts at rest, then it will accellerate down due to gravity.

However, there is no force acting in the i direction, only the j direction. So momentum

isconserved in that direction; m dx/dt is constant, but m dy/dt is not.Thanks! That’s a nuance I was not aware of. Makes perfect sense, though