Guys, in the last couple of videos, we saw how to use conservation of momentum to solve problems where objects are interacting. Interactions between objects usually fall into two broad problem types. We've already seen the first one, which is called a collision. In the time you have a collision, the basic setup of these problems is you have objects that are moving towards each other and they're going to collide. We saw a couple of different outcomes of this, and we'll talk about that in a later video.

What I want to talk about in this video is the second type of problem, which I like to call a push away problem. We're going to see a couple of examples of this. Some really common examples include objects that are thrown or like the recoil of a gun, which is actually the one we're going to solve below. We'll also see things that are pushing off of each other like ice skaters on a frozen lake, or astronauts in space, or something like that. You might even see some problems involving fireworks explosions. But the general setup, the reason I like to call these push away problems, is because objects are initially together. And then what happens is they're going to push away from each other, and then afterwards they're going to be moving in opposite directions. And really, this is just because of action-reaction. But what's fortunate for us is that regardless of the problem type, whether it's collisions or push away problems, we're going to solve all problems by using conservation of momentum. Any problems involving interacting objects can be solved by using conservation of momentum because momentum always has to be conserved. That's a sacred law of the universe. Whether it's collisions or pushaway, momentum has to stay the same from initial to final.

Alright. So, we have a 4-kilogram sniper rifle that's shooting a 5-gram bullet. Basically, what's happening here is I've got this little gun like this. So that's my little gun, and I've got the bullet inside. This is before, but then afterwards, what happens is that the sniper rifle actually fires the bullet. I've got my gun like this and then I've got the bullet going away like this. I've got the mass of the gun which is 4 kilograms and I've got the mass of the bullet which is 0.005 kilograms. Now, what happens is, we know that the initial velocity of the "before" case, before the bullet is actually fired out of the barrel, is actually going to be 0. The initial speed for both of these objects, the bullet sitting inside the gun and the gun not moving anywhere. So both of these objects have our initial speed of 0. So, we have

V v2 final = 600 meters per secondI want to figure out what is *v1 final*. What is the recoil speed of the gun? So that's the first step, drawing the before and after. Now we're at my enter my conservation of momentum:

Let's look at each of our terms. We know the masses of all the objects. Now I just need to know the speeds. Well, like we just said, the initial speeds for both of these objects, the bullet and the gun, are just equal to 0. What does that mean? It just means that for both of the terms on the left side, they're actually both going to equal 0. So I end up with 0 momentum on the left side. In fact, this is going to happen in a lot of your push away problems. In most of your push away problems, objects are going to be initially at rest, which just means that the total momentum of the system initial is equal to 0. So that actually means that the conservation of momentum equation is going to simplify for us. Let's check out how. So we have:

0 = m1 v1 final + m2 v2 finalIf you start out with 0 momentum, you have to end with 0 momentum. So how do I solve for this *v1 final*? Basically, I can just move this to the other side like this. And what you'll end up with is:

If you have zero momentum initial and final, and you have these things that are moving in opposite directions, it means that the magnitude of those real momentums have to be the same. They're just in opposite directions. That's what this negative sign means. So this is going to be the general setup. Negative m1 v1 final is negative m2 v2 final here. So we can solve for this *v1 final*. and we can say that *v1 final* is equal to negative. Now we're just going to plug in the numbers.

If you go ahead and work this out, you're going to get -0.75 meters per second. So as expected, what happens is that the gun has actually been recoiled to the left. The bullet was traveling to the right at 600 meters per second, and so it has some momentum in this way. So the gun has to recoil backward and its speed is 0.75 meters per second. Basically, what happens is if the gun or if the bullet has gained 10 momentum to the right, the gun has to recoil and gain 10 momentum to the left. I'm just using 10 as an example there. Right? So that's the speed. It's -0.75.

Alright. So that's it for the problem. I have one last sort of conceptual point to make, which is that momentum is conserved like we had in this example here only if your system is isolated. So we've seen that word before. Isolated just means that all the forces that are acting in your system or in your problem are internal. So remember, we defined this system as the gun plus the bullet. And if you look at the forces, what's happening is that the gun exerts a force on the bullet to the right. That's what shoots it out of the barrel. But because of action-reaction, the bullet has to exert a force backward on the gun. Now even though there are two forces here if you've defined your bubble to be the gun and the bullet, then these forces are internal, which means that momentum is going to be conserved. If I had an external force, if I actually put my hand against the gun with the bullet inside and pushed it, that's an external force that's coming from outside of the system. And therefore, momentum is not going to be conserved.

In your problems where you define your system as both of these interacting objects, these forces are usually going to be internal, which means your momentum is going to be conserved. That's what we can use our conservation of momentum equation.

Alright. So that's it for this one. Let me know if you guys have any questions.