11. Momentum & Impulse

Push-Away Problems

# "Push-Away"Problems

Patrick Ford

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Hey guys. So in the last couple videos we sought to use conservation momentum solve problems where objects are interacting well, interactions between objects usually fall into two broad problem types. We've already seen the first one, which is called a collision anytime you have a collision. The basic setup of these problems, you have objects that are moving towards each other and they're going to collide. So 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. So we're gonna see a couple of examples of this. Some really common examples include objects are thrown or like the recoil of a gun, which is actually the one we're going to solve down 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 what happens is they're going to push away from each other and then afterwards are 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 gonna solve all problems by using conservation of momentum. Any problems involving interacting objects can be solved by using conservation momentum because momentum always has to be conserved. That's a sacred law of the universe. Whether it's collisions or push away, momentum has to stay the same from initial to final. So, let's take a look at our problem here. I'm gonna go ahead and skip this for just a second. We're gonna come back to it in just a second. All right, so, we have a four kg sniper rifle that shooting a five g bullets. So, we're gonna go ahead and draw our before and after. Basically, what's happening here is I've got this little gun like this, that's my little gun, and I've got the bullet that's inside. Now, this is before. But then afterwards, what happens is that the sniper rifle actually fires the bullets? Right, So, I've got my gun like this, and then I've got the bullet that's going away like this. So, I've got the mass of the gun which is four kg, and I've got the mass of the bullets that's inside. This is equal to 0.005. Now, what happens is we know that the initial velocity of the before a case before the bullet is actually fired out of the barrel is actually gonna be zero, right? The initial speed for both of these objects, the bullets sitting inside the gun and the gun is not moving anywhere. So both of these objects have an initial speed of zero afterwards. What though is I have the two finals equals to 600 m per second like this. What I want to do is I want to figure out what is V one final. What is the recoil speed of the gun? So that's the first step drawing the before and after. Now. We were at my my conservation of momentum. M1 V1 initial plus M two V two initial equals M one V one final plus M two V two final. All right. So, we're looking for here is we're looking for how do we solve for this V one final right here. So let's take a look at each of our terms, you 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 is just equal to zero. What does that mean? It just means that for both of the terms on the left side they're actually both going to equal zero. So I end up with zero momentum on the left side. In fact, this is going to happen in a lot of your push away problems and most of your push away problems objects are gonna be initially at rest, which just means that the total momentum of the of the system initial Is equal to zero. So that actually means that the conservation of momentum equation is going to simplify for us. Let's check out how, so we have zero equals M one V one final plus M two V two final. If you start out with zero momentum you have to end with zero momentum. So how do I solve for this? V one final? Basically I can just move this to the other side like this and what you'll end up with is negative M one V one final equals M two V two final. If you have zero momentum, initial and final and if you have these things that are moving in opposite directions, it means that the magnitude the momentum's have to be the same. They're just in opposite directions. That's what this negative sign means. So this is gonna be the general setup negative M1 view and final is negative em to the two final here. So we can solve for this V one final and we can say that V one final is equal to negative. Now we're just gonna plug in the numbers M two is equal to 0.5 Your V two final is 600 you're gonna divide the mass, which is four. If you go and work this out, you're gonna get negative 0.75 m per second. So as expected, what happens is that the gun has actually been recalled to the left. The bullet was traveling to the right at 600 m per second and so it has some momentum in this way. So the gun has to recall backwards and its speed is 0.75 m per second. Basically what happens is if the gun or if the bullet has gained 10 momentum to the rights, the gun has to recoil and gained 10 momentum to the left. I'm just using 10 as an example there. Right? So that's the speed. It's negative 0.75. All right. 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 problems are internal. So remember we define the system as the gun plus the bullets. 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 backwards on the gun. Now, even though there's two forces here, if you define your bubble to be the gun and the bullets, 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 a 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. All right. So in your problems where you define your system as both of these interacting objects, these forces are usually gonna be internal, which means your momentum is going to be conserved. That's why we can use our conservation momentum equation. All right. So that's it for this one. Let me know if you guys have any questions.

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