A thought on definition of momentum

Question

Well, this is a simple, basic and I think even silly doubt. The first time I saw the definition of momentum as $p = mv$ I started to think why this is a good definition. So I've read the beginning of Newton's Principia where he said that momentum is a measure of quantity of motion.

Well, this started to make sense: if there's more mass, there's more matter and so there's more movement going on. If also there's more velocity, the movement is greater. So it makes real sense that quantity of motion should be proportional both to mass and velocity.

The only one thing I've failed to grasp is: why the proportionality constant should be $1$? What's the reasoning behind setting $p = mv$ instead of $p = kmv$ for some constant $k$?

Thanks in advance. And really sorry if this doubt is to silly and basic to be posted here.

Answer 1

It's a good question! Physics is all about linking your intuition to science, so it's good that you're thinking about this. The statement that momentum should be proportional to mass and velocity is intuition. Like elfmotat says, you can choose your constant as long as it's consistent with units, I guess.

If you want another reason, consider the time-derivative of the equation $p=mv$. It's Newton's Second Law! $F=ma$. In other words, $F=\frac{dp}{dt}$, which is a great reason for k to be 1.

Answer 2

If you're using consistent units, for example kilograms for mass, meters/second for velocity, and kilogram*meters/second for momentum, then for simplicity it is natural to choose $k=1$ as our definition of momentum. You can choose other constants, but it makes calculations unnecessarily sloppy. Any nonzero choice of constant $k$ is valid because it preserves important properties like momentum conservation.

If, on the other hand, we're working with inconsistent units then a non-unity proportionality constant is required! For example, if we decided to measure momentum in units of kilogram*meters/second, mass in milligrams, and velocity in meters/second, then there's a required factor of $k=10^{-6}$.