Physics: Elastic Collisions in One Dimension

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Particles come together, they react in some way, and then, in the end, the two of them are doing something. This is a collision. A collision is just a generic word that refers to a situation of particles coming together, and then later on you ask, given what happened in the beginning, what happens in the end. It's a central question in lots of branches of physics and engineering. If you want to know the answer, you might just think it's Newton's Second Law. If you know about the forces, you can work it out, but forces can be complicated. They can depend on time. You might not know exactly what they are and, if the collision is violent or brief, which is kind of the way you think of the word collision, it might be awfully difficult in any practical sense to use Newton's Second Law.
What you use if there's no net external force, of course, is conservation of momentum. Conservation of momentum tells you, given the initial conditions, something about the final conditions. Unfortunately, it doesn't tell you the whole story. Imagine two train cars coming towards one another and boom, there's a collision and they stick together, dead halt. Here's another situation: two train cards coming together, but they've got rubber bumpers, bounce. They go flying apart. Very different situations, and yet the initial conditions were identical. So, apparently, conservation of momentum alone doesn't predict all the details of the final answer. What did conservation of momentum tell us? Initially, they were coming together, equal and opposite momenta. Afterwards, I guarantee you if there were no important external forces, the total momentum of the system will still be zero. They could both be sitting still, that's zero total momentum, or they could be heading out back to back. That's still zero total momentum. So momentum conservation does give us some information. It relates the final motion of the two, but it doesn't tell the whole story.
Conservation of energy, together with conservation of momentum, at least in one dimension in this train collision story, will tell us what's going on. If you have conservation of energy, then we're going to see you can work out what are the final conditions.
There's a word for a collision where energy is conserved, it's called elastic. Elastic, you, I don't know, think of elastic bands, but it just means energy is conserved. And since we're talking one dimension, there are no potential energies to worry about. I'm talking what's happening just before a collision and what's happening just after a collision. So elastic collisions really means conservation of kinetic energy.
Here's a little super ball. I bounce it off this little table. There's a collision in this story; it's the collision with the table. And at the moment just before it's heading down, at the moment just after the collision is heading up, you can tell that it's nearly elastic, because it comes back up to almost the same height. There's an external force, gravity, in this story, but right here around the collision, it's just above and then hitting the table, and then just above. Gravity is irrelevant during the collision. This is very close to an elastic collision.
There are other kinds of collisions in the world, like when the two trains came together and stuck. That's called an inelastic collision. Here's another identical-looking super ball, but it's very inelastic. In fact, this is a totally inelastic collision. When you fall and you stick - falling isn't part of the story. In a collision when the two objects stick, that defines a totally inelastic collision. And you can have any degree of inelasticity, from totally elastic to totally inelastic.
When the situation is partly inelastic, what's going on? We seem to be losing some kinetic energy. In fact, you're losing a lot, you're losing as much as possible when it's totally inelastic. Where is that energy going? Well, mostly likely it's going into heat. There's lots of internal friction, this thing is getting deformed, and molecules are breaking bonds. Somehow or another, we're taking the kinetic energy and getting rid of it, at least as far as we're concerned in this macroscopic world of ours. Microscopically, energy is conserved, it's just going into heat energy or chemical energy of bonds or something.
If you're given that a collision is elastic, typically that means good solid rubber balls or good springs that conserve kinetic energy, then we can solve in one dimension for the final conditions. How do we do that? What equations do we have? Conservation of momentum. Imagine, in a collision, you've got mass m[1] and a mass m[2]. We'll do the memo with these carts on the air track, m[1] and m[2]. And the initial momentum coming in, that's the left-hand side of the equation, is m[1]v[1i], that's the momentum of particle number one, plus the momentum of particle number 2 in the beginning, that's the total initial momentum, and that should equal to the total final momentum, that's conservation of momentum. This is the momentum of particle number one afterwards and the momentum of particle number 2 afterwards. It's a vector equation. We're working in one dimension right now, so I wouldn't need to put all these arrows on the top, but I would have to bear in mind that v's can be positive or negative. So you really need to know exactly who's coming which way at the beginning, and then looking at the signs in the end will tell you who's going which way in the end.
I think of this as one equation and two unknowns, v[1f] and v[2f]. I suppose you could be given the final information and asked for the initial, but I just think of this as not enough information yet. Conservation of momentum alone doesn't tell you completely, microscopically what v[1] is doing and what v[2] is doing. It only tells you about this combination. Combination of energy, kinetic energy, that's what you need to solve this story. If the collision is elastic, that means that the initial kinetic energy is going to equal to the final. Initial kinetic energy is m[1]v[1i] squared plus m[2]v[2i]squared. This is the kinetic energy of particle one and of particle two in the beginning, this is particle one's kinetic energy at the end and this particle two's kinetic energy at the end. It looks like a mess. It is a mess. It's two equations in the two unknowns v[1f] and v[2f]. In principle, it's solvable. It's practice, when an equation is quadratic and the other is linear, it's a little bit tricky algebra. It's not that bad, but I'm not going to go through it in front of you. If you like this kind of stuff, try it yourself. It's sort of a fun exercise to see if you can solve for v[1f] and v[2f] in terms of v[1i] and v[2i].
I'm just going to tell you the answer and even the answer is messy and the equations are not so attractive-looking. And the generic general answer has got lots of terms. Let me make one little simplification. Supposing that v[2i] is equal to zero and this one is at rest, then the equations will at least look pretty reasonable in the end. In any physical situation, you can pick your coordinate system. So you can always choose the coordinate system, where particle number two is at rest. So, in a certain sense, I am really telling you the general answer to this problem. You might think of this as sort of one-dimensional billiards. Pool is a game where it's very nearly elastic collisions, kinetic energy is conserved to a very good approximation, although billiards is two-dimensional and I'm working in 1-D only. And here's the answer: v[1f], expressed in terms of v[1i]. Remember I'm assuming v[2i] is zero. m[1] minus m[2] divided by m[1] plus m[2]. The next equation, v[2f], is 2m[1] over m[1] plus m[2]v[1i].
Can we learn anything from these equations? It's nice to know that you can solve them. It's nice to know that they're there if you need them. And when you stare at messy equations like this, especially after you've gone through some hard algebra, it's worthwhile just looking at them and thinking, "Can I make some sense out of this?" For instance, suppose m[1] equals m[2], equal masses. What does this formula tell me? It says v[1f] is going to be zero. So that means equal masses, this one comes in, it's going to go boom, come to a halt. What's going to happen to v[2]? If the two masses are equal, 2m over 2m, v[2f] equals v[1i]. So one's coming it, it should stop, and the other one should just continue on its way with the initial one's same old velocity. Let's just try it.
This one's at rest, coming in, this one stops and that one heads off with the original speed. So the equation works, that's nice to see it happen in reality.
How about it in the limit where m[1] is significantly larger than m[2]? So a mass, let me replace it with a smaller mass. Now, this isn't that much smaller, so I'm going to look at the equations and pretend it's much smaller. m[1] is the little one now. So this says if m[1] is little and m[2] is big, this one goes negative. It's going to be negative some number times v[1i], so I'm claiming that when the little one hits the big one, this one's at rest, this one's going to bounce backwards. And what's going to happen to the other one? Well, this is positive. m[1] is smaller than m[2], so it's a little bit hard to tell without plugging-in numbers, but I certainly know that it's going to go forward, so let's just take a look at that. Is this one really going to bounce? It's heading backwards, not very fast, but it's heading backwards, just like the formula says.
Let's do one other case, where we make m[2] be the small one. So m[2] is small, so this gives me m[1] minus a small over m[1] plus a small, v[1f] should be fairly close to v[1i], but a little bit less. And that makes some sense to me; when a big object comes piling into a little object and the collision is elastic, I expect the initial one to continue on forward, just slow down a little bit. And v[2f], well, if m[1] is rather - sorry, m[1] is the big one. If m[1] is big, I expect v[2f] to be somewhat bigger than v[1i], so this one's going to go springing off to the right and this one should continue on with a slightly reduced speed. Now, let's try it. Sure enough.
Equations really work. They have to, it's just conservation of momentum and conservation of energy in one dimension. As long as the net external force is zero, that's the critical requirement to use conservation of momentum. As long as energy is conserved, here I have nice elastic springs, so that I'm not losing very much energy to internal friction or heat, then you could use conservation of energy and momentum. In one dimension, you can really solve for v[1f] and v[2f]. You really know the whole story.
When we get to two dimensions, we'll see that it's a little bit more complicated, but not conceptually. Conceptually, I can add all sorts of complications to the story. If it's an elastic collision, conservation of momentum and energy together is going to teach me an awful lot. I don't have to worry about the details of the internal forces, I can just write down the equations.
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Momentum
Elastic and Inelastic Collisions
Elastic Collisions in One Dimension Page [1 of 3]