# The Physics Philes, lesson 33: I’ve Got the Momentum!

*In which a gun recoils, a crappy diagram is drawn, and kinetic energy is differentiated.*

Last week I explained the basics of conservation of momentum. Today we’ll do a problem to demonstrate this basic principle. I know you’re excited, so let’s get to it.

We know that if a system has no external forces, momentum in the system is conserved (OK, it’s more nuanced than that, but for our purposes let’s say it’s so.) For our problems today we’re going to assume that there are no external forces.

First, let’s look at the recoil of a rifle. Let’s say that a rifle has a mass of 3.00 kg fires a bullet with a mass of 5.00 g horizontally with the ground at 300 m/s. What is the recoil velocity of the rifle? What is the final momentum of the rifle and the bullet?

For this problem we’re going to assume that any horizontal force exerted by the person shooting the gun is negligible. The only forces we have to worry about are the forces exerted by the bullet and the rifle, thus the total horizontal momentum of the system (i.e. the bullet and rifle) will be the same before and after the shot is fired.

What do we know? We know that the mass of the rifle is 3.00 kg, the mass of the bullet is 0.00500 kg (or 5.00 g), and the final velocity of the bullet is 300 m/s.

We’re looking for the recoil velocity of rifle and the final momentum of both the bullet and the rifle:

Here is are before an after sketches of our problem:

Remember that, since we’re only looking at forces acting horizontally, we only need to use the x-component equations. First, let’s find the recoil velocity of the rifle.

Conservation of the x-component of total momentum gives:

All we have to do is plug in the numbers and solve for the rifle’s velocity:

See that negative sign? That means that the recoil is in the opposite direction of the bullet. That makes sense. Who ever heard of a gun following a bullet out of the shooter’s hand? Now, what about the final momentum of the bullet and the rifle? To find the momentum of an object, we use the following equation:

(But we’ll use the x-component version, remember.) Luckily, we have those values for both the bullet and the rifle so, again, all we have to do is plug in the values.

See that? The bullet and the rifle have equal and opposite momenta! But check out what happens when we find out which has the biggest kinetic energy:

Hmm…interesting. Just because the rifle and the bullet had equal and opposite momenta does not mean they have equal and opposite kinetic energy. Far from it, actually. The ratio of the kinetic energies is 600:1, which is the inverse of the ratio of the masses. This is, I’m told, always the case in recoil situations. In any case, it’s important to not conflate momentum and kinetic energy. This problem shows that they are not interchangeable.

Well, boys and girls, that’s all I have time for today. Stay tuned next week for more momentum in the context of collisions.

*Featured image credit: State Library of South Australia*

## 1 Comment

I’m surprised that you didn’t just copy the first bad drawing of the gun instead of making two bad images of guns. I want to believe that you thought you would draw it better the second time, failed and decided… whatever! 🙂

On a more serious note, thanks for the post. It’s a good refresher and it also brings back good memories of learning this in high school and messing up the momentum question in the final exam. 🙁