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Anderson Video - Momentum- Collisions in 2D

Professor Anderson
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>> We should all do like bug eyes or something on the count of three. Everybody just goes what? [ Laughter ] All right. Are we still rolling? >> Okay. You guys good? Everybody feeling all right? Hello class, Professor Anderson here. Let's talk a little bit more about conservation of momentum. And specifically, let's talk about these collisions in two dimensions. So we of course live in a three-dimensional world. But a lot of the time you can sort of simplify the problem to two dimensions. So when you're driving around in your car on the surface of the Earth, it's really like you're driving in two dimensions, okay? Because you're sort of constrained to move on the surface of the Earth. You typically don't launch your car up off the ground, and you don't go down underneath the road. You're on the surface of the road. It really is like two dimensions, okay? Billiards is another example of two-dimensional collisions, right? They are constrained to move around the surface of the table. They don't -- they typically don't jump up off the table, unless you're doing some trick shots. Okay? So they stay in these two dimensions. So two dimensions really governs a lot of what we're worried about with collisions. And let's take a look at the car collision example. And let's set it up in the following. Let's say that we have a car here that has mass two. It's moving in this direction at speed V2. And then we have another car here that is mass M1. It's moving at speed V1. And these two things are going to come together and collide right there. Okay? And when they come together and collide, let's say they stick together. Okay? This is a pretty good approximation to what happens in a real car collision, okay? When cars collide, they tend to sort of stick together because the bumpers get deformed. Things bend. There's a whole bunch of energy lost in that collision. And so it is truly an inelastic collision. So after they collide, they're going to head off in some new direction and they are going to be stuck together. There's car 1. There's car 2. They are stuck together and they're moving at some final speed, V sub F. So this is what it looks like in general. And whenever you draw something in general like this, you should always identify a coordinate system. So it makes sense to use this as our X and use this as our Y. And if that's the case, then we need to identify some angles. Let's call this angle Phi. And just for kicks, let's call this angle theta. Usually, we write theta relative to the X axis. But just as an example, let's use that as our theta. Okay? So this is the before picture. This is the after picture. And now we can write down conservation of momentum. It tells us that p initial = p final. And those are vectors. And each one has two elements in it. Okay? And we have an m1 and an M2 on the left side. We have an m1 and an M2 on the right side. We have X components. We have Y components. We have a lot in there. So let's expand this a little bit. Let's do the X components first. P1 inintial in the X + p2 initial in the X = P1 final in the X + p2 final in the X. And let's do this part explicitly. So M1V1 is the momentum of the initial guy. But we want the X component of it. Right? It's coming down at this funky angle. It's partially in the X; it's partially in the Y. So we have to worry about what is in the X direction. So when I think about this velocity vector right there, V1, what can I say? Well, V1 is kind of coming down like that. And that's going to have some component in the Y and some component in the X. And so this side of the triangle is the X. This side of the triangle is the Y. And if that is angle V and that's our right angle, then I know that this one is going to be cosine. So right here I need a cosine of Phi. That's this first term, P1 initial in the X. What about P2 initial in the X? What is the momentum of this guy in the X direction? Chris, what do you think? >> It's going to be MV sine theta. >> Okay, we're looking at this one right here. >> Oh, sorry. The X = 0. >> X = 0. Perfect. Very nice. Okay? Now let's worry about the right side of this equation. Well, I can still treat these as independent particles. And so I can write this as M1 times V final. And now I have to figure out what the X component of this thing is. Hm. Let's take a look at this V final. V final is heading up in that direction. That is composed of two things. Vertical and horizontal. This is my right angle. This is the angle that we defined as theta. So this is going to be V final times what? It's going to be sine of theta. And this sine over here is V1 cosine -- sorry, V final, cosine theta. So on this right here, I want V final sine theta. And then we have to add P2, but that's just M2V final sine theta, and so on. This right-side we can in fact combine the two. Which makes sense, they're stuck together. Okay? So let's rewrite this equation. M1V1 cosine p = m1+m2 v final sine theta. Everybody okay with that part so far? We're just looking at the X component now. Okay, let's deal with the Y component. The Y component, it's going to look very similar to this. We're just going to change the X's to Y's. P1 initial Y + P2 initial Y = P1 final Y + P2 final Y. Okay, it's really the same equation, we just changed all the Xs to Ys. And now we need to think about P1 initial and the Y. All right. M1V1, we know that's got to be in there somewhere. We used cosine before, so we probably want to use sine -- that makes sense, that's this side of the triangle, right? And sine is -- let's see, sine of Phi right? That was our angle. Okay? That looks pretty good. Is there anything else I want to add to this? Anything else I should put in here? Grab the mic. Anything else I want to put into this first statement right here? >> I actually don't know [laughs]. >> Okay. Well, it looks pretty good, right? We used a cosine before, so we're going to use a sine. But is this car, M1, is this heading in the positive X direction? Or the negative X direction? >> It's the negative direction, so we should add a negative sine. >> It's heading in the positive X direction, which we're already took into account, yes. And so now we're looking at the Ys. So is it moving in the positive Y or the negative Y? >> Negative Y. >> Yeah. Absolutely. So really, we need to put a minus sign right there. And that just means that this arrow is pointing down. Good. Okay, P2 is M2V2 and that car is moving directly in the Y direction, so we don't have to worry about any angles, and it's positive number right there. All of that = M1 + M2 and we're going to multiply that by VF. And we go back to this triangle right here, the sine was the horizontal in that case, the way we defined our theta. And so the vertical is going to be cosine theta. Okay. And now we have the problem set up and we can box it. This is our X components. This is our Y components. And now you can put those together and you can solve for whatever you want. Okay? Questions about that one? Everybody okay with that? Everybody good? Okay.