Physics: Collisions in Two Dimensions

This lesson has not been reviewed.

Please purchase the lesson to review.

We've been talking about collisions in one dimension. If you use conservation of momentum, it connects what's happening in the beginning to what's happening in the end. If the collision is elastic, then you have a second equation, conservation of kinetic energy, and that really allows you to completely predict what happens in the end, given what was going on at the beginning. So that's a very nice situation. The real world is not one-dimensional. All collisions aren't just trains running back and forth on straight tracks.
What happens when we move into two dimensions or three dimensions? First of all, two dimensions are enough. Once we've talked about two dimensions, going to three dimensions doesn't really add anything new at all. Two dimensions barely add anything new, but there are a few little details that are worth looking at. So here's the general two-dimensional problem: supposing that you have a ball. You might imagine that you're looking down at a billiard table. That's a nice image of a two-dimensional collision, although it's somewhat restricted, because billiard balls all have the same mass. So here I have m[1] and m[2] and the initial ball is coming in with v[1] initial and we're in two dimensions now. You don't have to strike head-on. What's going to happen? It would be nice to be able to predict, using our basic conservation principles, what the final situation is going to be like.
Now, let me draw you a picture of the final situation after this collision. You look at this and you say, "Oh, no, it's a disaster!" I've got m[1] and m[2] and m[1] is flying off in this direction, m[2] is flying off in this direction. When you look at this picture, you'll say, "It's a hopeless story!" Don't panic, look at the picture and think - I've drawn an awful lot of circles and arrows here. Think about what exactly we would like to know. I would like to know v[1f], this arrow's length and this arrow's angle, . So that's two things. I would also like to know v[2]'s velocity vector length and its angle, . So there are really four things I'd like to know, two lengths and two angles. Everything else that I've drawn on the picture was to guide the eye, because I tend to work in components when I'm working problems in 2-D. So, for instance, here's the y component of v[1f] and, if you look at the picture, this is just the magnitude of v[1f] times the sin of . So that's the formula from a little geometry. Here's the y component and the x component. So if I know those two things, that's equivalent to knowing everything there is to know about v[1f]. So there's two quantities here and two quantities here, v[2] x and v[2] y. So, once again, whichever way I like to think about it, four quantities in the final state that I would like to be able to deduce.
So what have I got to work with? It's a collision problem, Newton's Second Law is surely hopeless. I don't really know about billiard ball forces. It's going to be conservation laws. As long as there's no net external force, then I know that total momentum will be conserved. Let me write down the equation of conservation of total momentum. You've seen this equation many times now, ball one has momentum m[1]v[1i] coming in, and this is ball two's initial momentum. Here's ball one's outgoing momentum and here's ball number two's outgoing momentum. It doesn't have to be a ball, we're just talking about flat, two-dimensional things. I should mention that, in a real billiard game, they're not flat, they can roll. And that's a slightly extra story we'll come to later in the course. But fundamentally, this equation is certainly absolutely correct, no matter what's going on, even in a realistic billiards game, if there's no net external force.
So what is this equation telling me? Initial momentum equals final momentum. It's a vector equation, so I think of components. I can just put a little sub x everywhere. That's one equation that tells me about how the x components are related. Then I could look at the y components, put a y subscript everywhere. So I've got really two equations. This momentum equation is really two hidden. So I've got two equations. And remember I had four unknowns. So even if this isn't billiards, even if there's some loss of energy, things are getting deformed, there's noise and crash sounds, I can still write down two equations, which at least relate the quantities in the end. In other words, if I don't look at this one at all, but merely measure x and y components of this object, I can use the momentum equations to completely predict the x and y components of the other object. So momentum conservation alone is very powerful and you've basically always got that, as long as you don't have to worry about external forces or, to be more specific, if they're small enough that force times t is small during the time of the collision. So we can't quite solve for everything, given the initial conditions, but if you measure one of the outgoing particles, conservation of momentum tells you about the other, no matter what.
If the collision is elastic, if kinetic energy is conserved, that's good news. It's another equation. What's the next equation? Conservation of energy. It's simple to write down; m[1]v[1i]^2, that's the initial kinetic energy of ball one and ball two, here's the final energy of ball one. Notice what does this mean, v[1f]? That's this vector squared. It's the length of the final velocity of object number one squared. And finally, here is kinetic energy of ball number two.
By the way, I should have mentioned this, you might be thinking this is not the most general problem in the world, because this ball was at rest. And I just assumed that for simplicity. Remember you can always do this problem in the reference frame, in which this ball is at rest and this ball is moving in the x direction. So I really am, in a realistic sense, solving the most general problem.
Well, I'm not quite solving it. I've got three equations, the two vector equations, the two components from momentum conservation, and conservation of energy. Three equations, but remember there were four unknowns. So too bad, that's correct. I cannot completely predict, given only v[1i] and v[2i] ;that information alone is not enough to completely predict what's going to happen. It's almost enough. Three equations and four unknowns means I only need to make one observation in the final situation. For instance, maybe I only look at the angle . That's all I need to do, and then I can predict everything else. These two components, I can figure out , I can figure out the length of this, I can figure out all of that stuff, but I need one more piece of information.
Sometimes, people will talk about the extra piece of information as residing in the initial conditions. Knowing the initial velocity of the two objects, that's what I've been talking about, if you look at the picture, there is another piece of information that I might be given. It's given the technical name the impact parameter. It's how far off to the side this collision occurred. And that extra piece of information is also enough, but that's an even more complicated algebra story.
So, look, these equations are, in principle, solvable and, in practice, they're a bit messy, especially since this one's quadratic. So you could do it, especially if the conditions are chosen to make your life as simple as possible. At least you know that you can relate a lot of the final quantities.
The situation of a totally inelastic collision is always our favorite. Totally inelastic means the objects stick together. It's very different from what we were just looking at. In that case, this final picture is totally wrong. If the two objects stick together, I don't have two objects flying apart, I've got a big blob going off in some direction. There's only the x component and the y component of the final situation to worry about, only two unknowns. And we have our two momentum equations no matter what, the x components and the y components of conservation of momentum. So we really can solve in the totally inelastic case for everything that we would want to know. Let me just do an example.
Here's a car crash. Car number one is moving directly to the right, car number two is moving up the page in the y direction. We're given v[1] and v[2], and then smash! They crash together and they stick. It's a totally inelastic collision. The moment after the collision, I would like to predict what are the direction and the magnitude of velocity of the big mess. I should point out in this picture you see them, they've scraped across the highway and ended up somewhere else. Remember conservation of momentum is only true if there's no net external force. Of course, after the crash, there's going to be friction of this think scraping across the road. That's an external force and, yes, they'll come to a halt. Momentum won't be conserved anymore. But just before and just after the collision, it's a very brief period of time, the external forces will be negligible, in general, and we can use conservation of momentum. So let's just write down the equations and see what they tell us.
Remember car number one was moving to the right, so our x component equation says m[1]v[1x], which is the whole thing, plus m[2]v[2x]. This is not moving in the x direction, so that's zero. And that's equal to - remember they stick together, so the mess has m[1] plus m[2], and then we've got v[f] in the x direction. That's one component of conservation of momentum. And, in fact, already if I told you v[1x], I could solve for v[fx]. So I already know what the x component of the final velocity is. Here is the mess cruising along with v[f], and I can now deduce what this piece is.
Let's look in the y direction. Particle number one has no y momentum, and so I get zero for that term. Particle number two has m[2]v[2] in the y direction, and that's going to equal, by conservation of momentum, to the final total momentum in the y direction, m[1] plus m[2]v[fy]. So this equation alone solves for v[fy]. I'm done. That's usually the case. It was the case in one dimension, it's the case in two dimensions, and it's the case in three dimensions. Totally inelastic collision, you can completely predict the final situation, given the initial. Of course, you can also work it backwards, given the final v[fx] and v[fy], you can work out what's v[1x] and v[2y]. The police are often interested in working that way. They look at the final conditions and figure out who was speeding.
So totally inelastic collisions are the nicest ones. It's funny, they seem like the messiest. They're the ones where everything is all smashed up and there are all sorts of complicated forces, and yet, they're the ones that you can really solve for everything. Total elastic collisions, those are almost as nice. Unfortunately, you've got three equations and four unknowns, that's in two dimensions, so you can't quite solve for everything, but, given one extra piece of information you can solve for all the rest. And the general case that's neither elastic nor inelastic, you've just them momentum equations. So you can certainly learn something, you can relate some of the final variables, but you have to measure a little bit more, in order to make quantitative predictions in that sort of general worst case scenario.
And in three dimensions, really it's all the same story. You get one more equation from the z component of momentum and everything else that we've said just follows through exactly the same.
Momentum
Elestic and Inelastic Collisions
Collisions in Two Dimensions Page [3 of 3]