Physics: Inelastic Collisions in One Dimension

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In any collision of any kind, if the net external force is equal to zero, momentum is conserved. It's important to realize. That means it can be a really complicated collision. There can be loud noises and flying glass and internal friction like crazy. That would mean that energy is not conserved, but momentum will be conserved, so you really can figure out something about any kind of collision using conservation of momentum.
If energy is also conserved, that's like two rubber balls bouncing or springs bouncing, that's called an elastic collision and you can use conservation of energy together with conservation of momentum to give you very strong predictive power. But what about in the real world, when there's almost inevitably some kind of lost heat? In the real world, not very many collisions are exactly elastic. It's an idealization, like friction-less surfaces or perfect springs. This is a pretty elastic collision between the Ping-Pong ball and the solid block, but if I drop the Ping-Pong on the table, it kind of peters out pretty quickly. This is clearly not elastic, you can tell. Energy is being lost in the collision. And here is my sad ball, it's essentially perfectly inelastic. So there are different degrees of how inelastic a collision is. This is a little inelastic and this is totally inelastic.
You might think that the more inelastic the collision is, the more complicated the story is, there's more internal forces and more bending and noise, and that's correct. A totally inelastic collision, like when two trains just come together and smash and make this big old blob, it's the most complicated situation, in terms of all the internal stuff going on. But from the outside world, it's a great simple situation. A totally inelastic collision of two particles, we can solve exactly. We don't need to worry about any of the details. We can figure out what the final condition is going to be, given the initial. It's just conservation of momentum.
Imagine that object number one with mass m[1] has velocity v[1]. And imagine that it's running in one dimension and colliding with m[2] with velocity v[2]. Of course, if there's going to be a collision in this picture, v[1] had better be bigger than v[2] so it catches up.
This is the before picture. And what's the after picture? If it's totally inelastic, that means they stick. So afterwards, we've got a blob and it's got total mass, m[1] plus m[2]. It's the total mass and there will be some final velocity, v[f]. Totally inelastic collision, where they stick, means that there's only one unknown at the end, because they stuck together so there's really only one final velocity, not two. That's the key to solving this problem. Just write down conservation of momentum. Initial momentum equals final momentum and the result is m[1]v[1] plus m[2]v[2] is equal to m[1] plus m[2] times v[f]. Now, v[2] could be negative. They could have been heading towards one another, in which case I would have to be sure to plug-in a negative number here in this equation. And this is simple, just solve for v[f]. We've got the answer. v[f] is m[1]v[1] plus m[2]v[2] divided by m[1] plus m[2]. So there's our equation for the final velocity in a totally inelastic collision.
Let's just work a quick example. So there's a train, and here's the train cruising along. It's got a velocity of a train and a mass of a train, and it's on the railroad tracks and we've got trouble coming up. So the tracks are busted and Superman's there to save the day. Superman's smart, so he knows about conservation of momentum. In the old days, way back in the 1930's, Superman was not magic. Superman mostly obeyed the laws of physics. I don't think that's true anymore. If you watch the movies, he's flying, it's magic, there's no physics explanation. But in the old days Superman didn't fly, he jumped. He was just really strong, that's all. So Superman jumps, that means he's in free fall. He's following a parabolic path. I think what he's going to want to do is make the top of his parabolic path just be right at the spot where he reaches the train, so I guess he's got to start down below. And what does he want to do? Superman wants to come in, here's m[S] and v[S], and Superman wants to work it out so that he's going to whack into the train and stop it. So I didn't draw the picture very well, because I'm guessing that the mass of the train is much bigger than the mass of Superman. I'm not totally sure how much Superman weighs, but he can walk around on floors without breaking them, so I'm figuring he doesn't weigh as much as a train. We've got to figure out v[S], that's v[Superman] in the x direction, and you can see from the arrow that it's going to have to be in an opposite direction. Let's just plug into this equation. I want m[T], v[T] and, in this case, v[S] in the x direction is negative, minus m[S], v[S] should equal to zero. I want the total system to end up at rest and Superman's going to stick. If Superman bounces, it just makes it worse for the train, because they were coming this way and now they're going that way, even more momentum transfer to the poor passengers.
So, simple equation. The minus sign is because of the direction of arrows and I'm writing down components here, and you can solve for v[S]. It's just m[T] over m[S] times v[T]. So you work that out. When you're doing parabolic motion, your horizontal component of velocity never changes. So Superman jumps up in the air and he's got exactly the correct final sideways v[S] and that will always be preserved. His vertical motion will be affected by gravity.
So what's going to happen here? Conservation of momentum, you'd think Superman saved the day. But, in fact, what's the difference between this collision and a big brick wall here? The train piles into the brick wall and comes to a dead halt. There is no difference. I think the people on this train are going to be pretty unhappy that Superman tried to save them in this way.
The physics of what I just said is change in momentum is force times t. And the amount of time during this collision is short. Wham! They just come together, so t is small. p is p[f] of the train minus p[i] of the train. Now we're focusing on one little piece of the story, not the whole system, and so momentum is not conserved for the train alone. The train feels forces that are going to be big forces. If Superman were smart, he would stand on the tracks and use friction. That would be an external force. And then momentum would not be conserved and you would have p is force of friction times t and you could scrape for a much longer time. You could always use momentum principles to save the day, but you've got to think fast.
Let me work one more example for you of conservation of momentum. Supposing that we have a totally inelastic collision, m[1] traveling with v[1]. It's an icy day and here's a car at a stop sign. And right after the stop sign, there's a hill going up. m[2], this car is a rest, this car is traveling forwards and it's totally icy, so there's no friction on the surface here, and these cars are going to smash together. They're going to stick, it's totally inelastic. What's going to happen?
Now, for the brief moment of the collision, nothing much happens. It's a very short amount of time. They barely start moving forward, so I can use conservation of momentum from just before the collision till just after. I cannot use conservation of momentum for the whole story, because, once the wreck starts sliding up the hill, there's an important external force, gravity, which will slow everybody down and change the momentum of the system.
This is one of those physics problems where, yes, you can use conservation of momentum and you just have to be careful and use it just for the collision itself. During the brief moment of the collision, m[1]v[1] plus zero, that's the initial momentum, is going to equal m[1] plus m[2] times v[f]. And so you can solve for v[f]. And now, how high up the hill does it go? Well, from now on, you've just got this mass of car and it's sliding without friction up a hill. You don't use conservation of momentum now, because there's this external force. What do you use? Conservation of energy. We have a certain initial kinetic energy at the moment right after the crash. m[1] plus m[2]v^2, where v is v[f]. The word `f' here is now a funny one. You can choose anything to be final. First, it was final right after the collision, and now we have a new real final, which is up the hill. So m[1] plus m[2]v[f]^2, that's the initial total energy, kinetic plus potential. There is no potential right now, because we're at ground level. And that's going to equal, when you reach the top of the hill, when you come to a halt, to zero kinetic energy plus total mass, m[1] plus m[2]gh and you could solve for the height.
Sometimes in physics problems you just use conservation of momentum, like in the Superman story. Sometimes you can combine conservation of momentum for one part of the story and conservation of energy for another part of the story and you've got to think, "When can I apply conservation of momentum? When there's no net external force. When can I apply conservation of energy? When there's no internal friction or loss of energy, like during this horrible car crash." Conservation of momentum alone is all you need when you have totally inelastic collisions. It's kind of surprising. You would think with all that noise from a totally inelastic collision, all that mess, that the physics would be too complicated for us to solve. And, in fact, it's a really very easy case. So it's nice to see problems when objects stick together.
Momentum
Elastic and Inelastic Collisions
Inelastic Collisions in One Dimension Page [2 of 2]