 All right, we've been looking at collisions. We've found a couple things. One is that if we look at the collisions between any two things, whether they're moving towards each other or not, doesn't matter. We came to understand that if we look at each one of them individually, and we're talking here about the type of collision where they hit and for some reason they stick together and go with just a single speed, that's the only type of collision we're talking about now. We'll expand on that a little bit. We came to understand that if we look at the individual object, whatever it is, then we have to take into account that individually, each of them experiences an external force, external to themselves. Of course, they've hit something and that something's gonna exert a force on them. But we didn't know anything about that force. We also saw, if you remember, we need to know how long that force was acting. Remember, there was a DT in there? What was that? There's two over there. Was that, it was over here? My theater. Was that it? It was a terrible noise. I'm gonna go back and watch the tip. I'm gonna find out what that was. I don't know who did it. We couldn't know what the force was that was in there. We couldn't know how the time for how long it acted was in there. But then if we put the two together as a system, by, well, we just put an imaginary boundary around them and said that's our system, then that made any of the forces of collision internal forces. It was an action-reaction pair, the one on the other anyway. So those were equal in office. They canceled internally. And we didn't need to worry about that. And we came up with this business that momentum would then be conserved. By that, I mean the momentum of the system. Before collision, there's a certain momentum characteristic to the motion of the objects in the system. After collision, there's also a momentum characteristic to the two objects in the system. And that momentum between the two is exactly the same. And so that allowed us to calculate that final velocity v prime without knowing anything about the forces that caused the changes in speed there. We were able to skip all that stuff. However, that doesn't mean there aren't collisions. It's useful to pay attention to what that force is. And so we're gonna do that now. Anytime we have an external force acting on something and that force is unbalanced, which means this sum doesn't go to zero, then we expect there to be some acceleration. So hopefully that's not new, because that should look pretty friendly and familiar to you. Let me back up a little bit and just put in the definition for ourselves of velocity. I mean to have acceleration. And now I'm just gonna do a little simple algebra. I'm going to multiply through by that dt. Simple differential algebra, I guess you'd call it. And I don't know that. That's kind of an incomplete form there because we don't really deal in lengths of time that are dt. Remember, those are time periods that are infinitesimally small. But let's go ahead and integrate over that time. We'll just go from zero to some time t and we'll integrate over, well this has got to integrate over v, so it'll be v1 to v2. Let's make the other simple step that's very useful to us. We'll assume that the forces are constant and that the mass is constant so we can go ahead and finish these integrals. So forces are constant, they come out. That leaves me with the integral of dt, which is, but I'll go ahead and just leave it as delta t just in case, in fact, I guess that could be a little more complete. We could say t1 to t2 just to give us the option that t1 might not be zero. It doesn't necessarily need to be. And then this integrates to m delta v. That m delta v, we're familiar with that because that's what we've been working with for two or three days. That's the change in momentum. So no great shakes there really, except we just took something we had anyway and went a step farther with it. We did have this f delta t before. Remember, that was when we first started with the collisions, but we had to come up with a way to ignore that because we didn't know what the forces were and we didn't know what the delta t's were. So we kinda said, let's see if we can get around that somehow and we got around it by putting the two objects together as a system. Now I'm coming back to it, because what kind of physicists would we be if we ignored forces when they're there? All right, so let's take a look again in a very personal way at the two sides of this equation. One side there and one side there. Look at those in a very personal way and see if that doesn't help us understand what's going on with these forces and this time here. So I'm gonna start with this first, m delta v or if we were a little more complete m times v two minus v one. So let's make this real personal. Let's say that m is you. Some of you make a pretty good mass. Some of you are a little more globular than others. Especially in your approach to life. Some of you are not so globular. No, what are you? You're a blob? Do you strive to be one? All right, so let's say this is what's an appropriate mass of a human being in kilograms, give or take. 90, pretty globular that is. That's a little big, we'll say seven. So we'll have m equals 70 kilograms. Oh 90, you were thinking of me. Yeah, sure, there's all that steel. So we'll say that then let's pick some speeds that people typically go. Let's say this is you and you're in your car. How fast are you going in your car? Well, most of you go like, let's see, on the streets of Queensbury in the 60s, and on the freeway in the 90s, thank you for confessing. So we'll say some reasonable speed, like I think the freeway speed out here is 65. So we'll say that 65 miles per hour. Real quick, make that into meters per second, please. Unit conversion cell phone app. They have those, they must have them. Some student made those. Lens check, no, lens on this girlfriend. He said, oh, I'll act like I'm using my cell phone app, but I'm really just calling my babe to see how she's doing. How you doing babe? That's me, your blogular, Bill, just called you a blog. So that's my girlfriend? No, I'm saying I'm being a mom. Showing me not, well, you have to show me not a blog. Say, you're not gonna take this. You just got a tap out shirt on your knucklehead. You don't say that to people with tap out shirts. You're already being so messy. Yeah, God, watch it, Bill. You don't even have to wonder if it was Jack. Yeah, Bill. I might as well have a sign on his forehead and say, go ahead and take me on. Who's got a V2 for me? Anybody? I mean, V1, V1, that's the speed we're starting with. What? There is a unit conversion app. What is it, is it free? Maybe like $29 for unit conversion app. What do apps cost, anyway? There's a couple, 99 cents or something at the app store. That should be free. Joe, I'm sorry, were you trying to talk to me? Not something? What was it? I guess, why not a unit per second? We're okay, let's make it 30. Allen, 29 was close enough. We'll just call it 30, because you're not paying attention to your speed anyway. That's why you're usually going too fast and much more attention on text messaging. All right, so there we go. There's you in your car, presumably, going highway speeds. So you have some momentum, just mass times velocity. We know what the momentum was. Sooner or later, you need to come to a stop. Who doesn't? You gotta go get some coffee or charge your cell phones and get back out on the highway and do some more texting. So sooner or later, you gotta come to a stop. So that's the whole site one I called there. Figure out, we'll quit for me, what the change in momentum is then. To bring you from highway speeds to a stop, that's a change in momentum of what? Just calculate it real quick. You could just about do it in your head, I bet. Now it did in head, it acted like it did in head. Then he's gonna, then he checked it anyway and then Tyler said 70 times 30 in your head. All right, so, no, sorry, you're wrong. No, sorry, you're wrong. What's the answer? What is the change in momentum? I guess, well, what's the change in momentum? You don't even need your calculator for this. God, you wasn't yelling at me. That's what usually happened. What'd you say? I think, yeah, negative 2100, Newton seconds. Kilograms times meter per second, is that a Newton second? Well, yeah, it is because you know that's gonna be a Newton second. They're equal, so that's gotta be a Newton second and you can check it and it is. So if you're ever traveling at highway speed and you ever come to a stop, it's going to take that, it's gonna be a change in momentum of that amount. Just take that away and move it down here so I can do the other side now. Bless you. We just did side one. No matter what, you're going highway speeds, some time you gotta come to a stop. That's the change in momentum, no matter what. That's side one, all done. Even though it's the right hand side there, I'll call that side one. Any questions about that side? Couldn't really be whole lots more straightforward than that. That's a different side. That's why side two. But there's a whole lot of different physics going on there even though they're equal. This side is already set. We've discussed it and we've put it to bed. You were going highway speeds. You came to a stop. Your change in momentum's minus 2,100 Newton seconds. No discussion, non-negotiable has got to be the same because they're equal. So this side has got to be, oops, forgot the delta D. This side has got to be minus 1,100 Newton seconds. Nothing more to discuss about it. Why negative? Yeah, because if you're moving one way, some force has gotta work against you to bring you to a stop. That's the only way it's gonna work. You need to decelerate, so you need a force to do that. And that's gonna be a negative force if your velocity is possible. The force has gotta be in the opposite direction back. Also non-negotiable, so that's the negative. So no big deal there. We were talking about that kind of stuff weeks and weeks ago. So this side's set, we understand the negative. Of course we understand the units. So let's look at these two things. There's two major ways this could happen. Let's say, we'll call them A and B. A, you come to a nice leisurely stop. You see your sign for your exit. You come on the exit ramp, you roll a little bit. You see the light up there's red, so you don't even have to put on the brakes because the exit ramp's kind of going up the hill. Maybe that all takes you, let's say 20 seconds. That's fair, drive along the highway. You know it's time to get off now. You're not going 65 anymore. It's time to prepare to get off and all of that takes about 20 seconds. Give or take a little bit. How much force required to bring you to a stop, to change your momentum by this set amount, how much force is required to change your momentum in about 20 seconds? Negative 105 newtons. Negative, again, meaning it's opposing your motion. So, it takes about 105 newtons to bring you to a stop. That's not a lot of force, really, that. Remember, a newtons, give or take about an apple, that's about the force required to lift 100 apples. And you can bring yourself to a stop in the space of about 20. Now, if we threw the car in there too, we're gonna need more force, of course, because we have more mass, we have more momentum that needs changing. But I'm just worried about you. That way we can talk about whether you're in a big, gigantic tanker truck or on a motorcycle or running along the highway at that speed. Here's option A. I'm sorry, option B. Option A, take a while to do it. Slow down nicely, get all the chance to get a few more text messages in there. Let's see, let's come to a stop a little more abruptly. Let's say you are texting, not paying a whole lot of attention. You hit the bridge above it. Same change in momentum, isn't it? But now it happens in, let's say, 0.02 seconds. That's about how long it's gonna take you to come to a stop if you hit the bridge. Now what's the force? Still negative, yeah, we understand what that is. 105,000, yeah, the time period went down by 1,000, factor of 1,000. The force is gonna go up by a factor of 1,000. You're not gonna survive that. That's why crashing kills you because it happens so quickly that the force required to do the change in momentum has gotta go way up to compensate for the fact that time went way down. You come to a stop very, very quickly as in a crash, you're gonna get killed. Except you come to a crash, you hit the bridge in a relatively modern car, a car 15 years old or younger. What's gonna happen? Airbag's gonna go off. Why does that save your life? All the airbag does, the only point of the airbag is to make Delta T greater. Buy enough that the force goes down by enough and you can survive. You might be damaged, but you might be alive, which is a lot better than dead for most people. That's the only point of the airbag is to increase Delta T to decrease F. That's the only point of the seat belts that we've had for 40 or 50 years. I remember when I was a kid, cars didn't have seat belts. My mom would tell us that she felt if she got in the head on crash, she'd put out her arm and save our lives. And we were standing there on the vinyl bench seat of our Bel Air station wagon because it's fun to stand there while you're driving along. I mean, if she even hit the brakes, we'd have been through the windshield, but she was gonna save our lives and put her arm out. I don't think so. If she was gonna go like this, that's what I think, but we never got to that. But the seat belt, all they point was to extend this. Now the airbag extends this even more, especially the early seat belts. They were rigid. Now the ones nowadays, you know, you can move forward a little bit in them. They have a little bit of time to take up the slack. Even have now built-in structure in the car where they're attached, where it'll kind of deform for a little bit before it brings you to stop. All of that, just to increase delta T. Motorcycle helmet, that's the only point to it. That's why there's foam inside of it to extend delta T when you splat your skull on the pavement as all motorcycle riders end up doing sooner or later. It's only to increase the delta T to bring down F to some survivable limit. This side's all set by, no matter what, this is the only side that you can do anything about. So you put on your seat belt and you buy a car with airbags. And if the airbags ever pop, you go get them replaced. They don't just stuff them back in there and duct tape it over. Joey. Because you might need that airbag and you're not gonna know when. Just to increase delta T. What? I didn't know if they'd pop. Because the sunglasses you were wearing are now on the other side of your head and there's the sunglass shaped hole right between you. So if the airbag is released, you can't like the fact. No, I think most cars, you can't even start. I believe you can't drive them until you go get it replaced. But if you go to some cheap auto bodies place, they will sometimes try to sneak a used one in. And I don't think you wanna do, that's not a good place to get corners. But it's pretty easy to tell if they pop. Because there they are. You'll be saying that wasn't there a second ago. You'll be saying that wasn't there a second ago. Don't know much about the cars, huh? I do, I've never done the cars. I don't know what an airbag does. Oh man, you just said that? You don't ever say I've never been in the next. Well, I mean I was last week on Wednesday. Oh, but other than just last week. Oh, and then there was the week and two months ago. Oh, and four times last year. No, I just mean the time you had me up on the floor and then, like, we were on. What was the delta T? Survived it? The delta T was big enough. And you're here to Pester Phil. Okay, I have another question. When the airbag comes out, does it let it blow up? Or does it go down? No, well, I just saw a slow motion video you can go on YouTube and just put an airbag and it'll, one of the choices will be a very slow motion video and they'll show it. And I think the time even runs with it. But it comes out and it's kind of a loose looking pillowy type thing. I guess that's because it's shaking around and it's happening very quickly and it stays inflated for a little bit but then it deflates all within a second or two. If you see somebody in a crash by the time you cross the street and get to them, that airbag will be down. They'll be going, thank God I had an airbag because look and they'll pull out a sheet of calculations and you can check them. And you can say, where's your negative sign? It's a professional, a professional, a survivor, a general. Yeah, off-dand I don't know what it was. It's something like a couple thousand newtons. But it also does have, you can stand very high pressure or forces. I mean, it's a very high force but it's for such a small time which is, of course, being a different product here, gotta be some other different problem. You can do that. But I don't know off-hand what survival is. Am I even gonna be in the textbook on that? Maybe I should read that. Well, you don't have to set it with yourself, though. Well, that was for the climbers. Yeah, that's all. That one is, that's it. Well, I don't know, it might be different if they're taking a rope wrapped around your waist as opposed to a blunt force to the head. I don't know, some. Well, you asked what was the limit? And you can give us a little report next week. Just see if we can get to Google it and see if you find something. And if you were to do this more realistically, we'd be integrating each separate force. Yeah, by going through this integration, what we've said is what's the average force? In actuality, a collision force, might look something like this, where maybe this is the .02 seconds. In actuality, the collision force might look something like that. It would come up peak and then drop off and all of this very quickly. By us looking at the average force, we're saying, well, let's essentially just have a value that gives us the same area under there. Because remember, this came from the integral of FDT, which is the area under a force time graph. So by saying we're using the average area, we're saying, I'm sorry, an average force, we're saying, oh, let's just have something with the same area as the two. Because that's what that side number two is. It's actually the area under a force time graph. And in fact, if you remember, when we first started talking about kinetics, didn't I say there would be three methods to solve kinetics problems? Did I tell you that? I think I did. What was the first one we discussed? Remember, the simplest of all of them, it's F equals MA. If we know the acceleration, we can find out what forces will give us that. If we know the forces, we can know what the acceleration will be. So it was really good. This was really good for general type kinetics problems or very good for constant force problems. Brainly well, for that, we didn't loss in lots of those problems. What was the second method we used for solving kinetics problems? It actually came from F equals MA. So they're not different. It was just a different way to look at the same thing that made it more usable for certain types of problems. The work energy equation came right from F equals MA, so they're not exclusive of each other and just allowed us to look at problems a little bit differently and in fact became much more useful for certain types of problems where the force might not be constant. What type of problems? Position dependent problems. Like springs, where you are attached to a spring is very much a position type problem and if any force there is that changes with position, it springs, but they were pretty easy to handle. All we had to do was figure out the potential energy for and after, where the spring was, all done with it. So good for position dependent problems. In fact, remember the work? Remember how work was defined? Remember, that was some time ago. How was it? In fact, its full definition was the sum of the forces dotted with DS which means it was very useful if the forces change with position. When they change with position in a very predictable way, like springs do, we didn't have to deal with this integral very much. But this integral is the area under the force position curve. So if we had some force position curve and we had the force varying, however it does, whatever the problem could possibly be, look at the area under that and we'd know what the work was. You wouldn't have to actually do the integral if you could just look at the graph and figure out the area. So we could very easily handle position dependent problems even if the force was position dependent. This is our third method right here. This side, well we know what that side is. That's the change in momentum. This side here, a force applied for a certain amount of time. Here we did force applied for a certain amount of distance. Now we're doing force applied for a certain amount of time. It would make sense that that's an important thing to look at, a useful thing to look at sometimes. We just saw it here. Could mean you live or die. But how long a force is applied, forever how long it is applied, that kind of makes sense that that works. You could have sat at home and even come up with that and just thought about it a little bit. This is called the impulse. The amount of impulse given to something is the area under the force time curve. And that's our third method. The impulse momentum equation. There it is right there for constant forces, constant mass or problems that are time dependent. Especially if the force is time dependent. And there's our third and last method for solving our kinetics problem. Doesn't mean that the other methods wouldn't work at all. For example, if we had some problem where I told you a force was applied for a certain amount of time, what's the change in velocity? You could go back here and say, oh, it sounds like a constant acceleration problem. I'll solve it this way and you can do just fine. Or you can solve it this way. It's kind of already broken out for you as all the differences. But if we did have a problem where the forces were time dependent, all we need to do is take the area under the graph and we'd know what the impulse was. Once you know what the impulse is, you know what the change in momentum is. So always cool to come to class and leave with something that might save your life. Any other classes that happened today? They took from you, didn't they? They didn't give, they took. I give, huh? You've been asked for in the chemistry class. Did they let you take it, though? Did they let you actually have it? No. No, see? No, it was supposed to be more pure than anything you could get in the store to. It's a hard thing. Your's back, fire. Yeah, there's a little. It gave you a head edge. That's the shape of it. You made it. Is that why you wouldn't take it? In fact, right? What was it? Black. Turn purple on the test, I was supposed to. So there goes your career with bear or et cetera and bummer. OK. All right, let's continue on then. So we've got a whole new method, the one. It's not that the others don't save your life. We could have said with the first one, we need to keep the acceleration low enough so the forces are low enough. And we would have come to the very same conclusion. But when you look at the impulse momentum method, I think it just becomes so much more obvious why seat belts and airbags and the like are so very useful. All right. So let's do a little bit more, a little bit more than with collisions because we can either do them as systems where we don't have to worry about the forces. Or if we need to, we can look at them individually because you colliding with something that doesn't move, that has no changing momentum, is going to give us a very useful problem anyway. So far, we've only looked at one-dimensional collision. And so far, we've actually only looked at one kind of one-dimensional collision. It's called a perfectly inelastic collision that we looked at. I didn't give you that name, but there it is. It's the type of collision where the two objects stick together and their subsequent velocity is the same because they're now really, in essence, a single object. They've gone from being two independent objects to a single object to the collision. The object will have some velocity v prime. And in fact, that might be our unknown in the problem. We have the two objects moving towards each other in some way. They collide. They stick together perfectly. And we want to figure out what the velocity is afterwards. Or, as might happen in the crash, we know one of the velocities before. Maybe that's a car that had one of those black boxes in it. The other one didn't. But we know what the after-collision velocity was. So the unknown becomes the second car's velocity before collision. We want to know if he gets the ticket. Either way, these problems are such that there's a single unknown. Or we need how many equations to solve that problem? We need one equation to solve a one unknown problem. What was that one equation to solve these problems? What was that one equation to solve these problems? Momentum, conservation of momentum. The momentum was conserved because as they collided and we look at them as system, the forces are internal. We didn't have to worry about them. That's exactly what we did. One unknown, one equation required, no sweat. We've got it. There it is right there. Remember Monday, I think it was, we also did the kinetic energy calculation with some of those problems. We'd already solved the problem. And then I said after we'd solved it by that, I mean we found the velocity we didn't know. I said calculate the change in kinetic energy. What was the change in kinetic energy? And I hope you do too, but what was it? If I asked you what was the change in momentum, you'd say, oh, yeah, it's zero. Now I ask you what was the change in kinetic energy? We did that calculation, we did it twice I think. Two different calculations we did it. What was the change in kinetic energy? Was it the same for the two different collisions we did? We did one where the rearward object hit the one in front and they were both going the same direction anyway, like a rear end on the highway. We also redid it with the very same speeds only they were going towards each other. Did we get the same change in kinetic energy? For those two, no, we didn't. In fact it was quite a bit different. Did we, what could we say about the change in kinetic energy though? Can we give some general statement to this, something easy I can write down here about change in kinetic energy of perfectly elastic collisions? No, no, just tell me. Look at the two delta K's we calculated. Both of them were positive, negative, zero, what? Delta K, what do I write next? Delta K, was it negative? That's what I'm asking, delta K. Everyone here? Delta K was less than zero both times. We lost kinetic energy both times. Here, we're gonna do one more. One more of those calculations real quick. No calculators on this. Mass M, moving with velocity V. Mass M, same mass, moving with velocity V. They hit, what's the final velocity of that two? Huh? They'll need units when it's zero. Zero for feet per second, or furlongs. What was it, furlongs per fortnights? Our favorite. Subsequent velocity is zero. Any disagreement? Any dispute? Any wondering how Alan came up with it that quick? Was he just guessing and hoping it was right? Because he's kinda sneaky. Maybe he was doing that. Everybody came up with zero as well? 10th of a decimal place? What was the change in kinetic energy? Was the change in kinetic energy zero, less than zero. In fact, an absolute maximum less than zero. We lost all the kinetic energy in that collision. Right? Actually, two times one half MV squared ended up with zero. We lost the maximum possible amount we could have lost. However you want to say it. We could say I guess, I don't know what, delta K2 equals zero is a possibility. I don't know, I don't want to put that there because it's not always true. But it certainly is true that we could lose all the kinetic energy. So I don't know how to write that down there. We just have to keep that in mind as a possibility. All right, so that's the one collision we've looked at in some detail there. So let's look at another one. Here's some mass. Oh look, Samantha got a new car after she crashed her dog one last week. Yeah, so there's some mass. We'll just make up some numbers here. It's a very lightweight car, it's only three kilograms. It's moving with some velocity, let's say four meters per second. Here's some other mass. We'll say four kilograms. Some velocity V2, we'll say two meters per second. They don't stick like they used to. They bounce off each other this time, like billiard balls do. Or like your fist in that guy's nose got in front of you, Cumberland's the other day. Remember him? Yeah, remember that guy? Yeah, so they hit, they bounce off. This is M1 now going some velocity V1 prime, this one bounces off, going with some velocity V2 prime. In fact, one of them could even keep going in the same direction, doesn't lose all of its forward speed, just loses some of it, we don't know. But let's say that's what we observe, how many unknowns. We don't know either one of those. Before we calculate the final velocity, we only had one unknown because they stuck together conveniently. Now we have two unknowns. We need two equations. Did anybody suggest one equation? Nope, you can do even better than that. And maybe this will help you. We looked at an equation, we did that, we could solve the problem then when we did this. I just did it again. What's one of the equations we could use? Now you can do better than that. Because he already said that. So you can certainly do better than copying something that's already been disallowed. Conservation, it's a collision within a system. No external forces would momentum be conserved. You bet. That's not two equations, that's one equation. In two dimensions, it'll be two equations, but in one dimension, it's only one equation. So we're halfway there. Second equation, go to eBay, go to Equations, go to Extra Equations, make a bid quick. Get us an Extra Equation. Well, it's certainly not an inelastic equation. And you called it a what? This is an elastic collision. When they bounce off, this is an elastic collision. You might even think that if we have perfectly inelastic collisions, we might have perfectly elastic collisions. Perfectly inelastic collisions. Nobody got that reference, that literary reference there? What was it? Practically perfect in every way. You have cheese going in. What, cheese? Oh, you don't get that one either. Cultural vacuum on students nowadays. But if I was talking about some z-diggy-diggy low wrapper guy at your door, yeah, I know about him. I didn't like him on Facebook. But Mary Poppins, chow over in your head in a flying umbrella, we should have just Google, ready to go here so we can Google something. And then you go, oh, now I get it. It's right there, it's on Google. The only repository for any cultural information I've got anymore. We do have such a thing as perfectly elastic collisions. When we do, and there are imperfectly elastic collisions. In fact, almost every collision in real life is imperfectly elastic. Most things do hit and bounce off each other. Even cars, rarely do cars hit and stick together like that. They usually hit and kind of skid off in their own directions with a bunch of flying metal and glass and stuff. Those are imperfectly elastic collisions. But there are perfectly elastic collisions and billiard balls are very, very close to that. Atomic collisions are very, very close to perfectly elastic. And when we have a perfectly elastic collision, we get our second equation. And it's as simple to write as this one is. Anybody want to take a guess what it is? It does. It is, conservation of kinetic energy. So we could do a collision like this, measure the velocities after the collision, check how much kinetic energy there was before and after, we'd know how perfect or imperfect that collision was. We don't have imperfectly elastic, imperfectly inelastic, but we do have imperfectly elastic, perfectly elastic collisions. They also have two unknowns. Actually, there's three, yeah, there's three unknowns actually in a imperfectly elastic collision. Three unknowns. Two of the unknowns are the subsequent velocities, but momentum is still conserved because there's no external forces. When we take the two as a system, no matter what kind of collision it is. The third unknown is what's delta K? All we know is we lose kinetic energy in an imperfectly elastic collision. So we need a third equation. An imperfectly elastic collision is called the coefficient of restitution. And I'm not gonna define it here because we're not gonna use it. It has to do with how well certain things bounce off each other. You know if you take a tennis ball and drop it, it only comes up to a certain height. That's because of its coefficient of restitution. Yeah, that's a little scripty. If you take dynamics next year, which I teach in the spring, we'll actually come up with that and do some problem building. It's experimentally determined just like coefficient of friction was. The only way to find it out is to do a test. You take some things and you bounce them off each other and see how far they rebound. Then you can know the coefficient of restitution. It's not quite, but it's kind of like how much kinetic energy do we lose? Because if you've got a tennis ball and you drop it and it only returns to 80% of its height, that kind of makes sense there and that all goes together. If we take a tennis ball and don't drop it on this floor, but we drop it on a carpet, it bounces entirely different. So it has to do just like friction did with the two things interacting. Coefficient restitution also has to do with whatever two things it is that are hitting each other. Is it a two ivory pool balls hitting or is one, your arm fall off? Is one a billiard ball and one a patella or something? That's a different collision. So we have those three possible collisions that we look at in our class. Let's add to it a little bit. Kind of a reverse collision, but it's still characteristics of the very same collision we've been looking at. So let's see what you can make of this. Imagine a ballistic rocket launch. You know what that means? When I say ballistic, no, never really not. What's it mean when you go ballistic? What's it, sorry, Phil? What's it mean when Samantha goes ballistic? Nothing, it just means the sun's up, doesn't it? I got your name right at least. What's ballistic mean in terms of projectile motion? Yes, if a projectile is a ballistic projectile, it's gonna have a very much different trajectory than does a non-ballistic one. But what do you see in the object that tells you whether it's ballistic or not? A ballistic trajectory is one that is not powered. Most rockets are powered through a great portion, if not all of their flight, because their engines are burning. But if you take something like a bullet where there's a huge explosion in the rifle barrel to shoot it out, and then after that it's not powered anymore and it just coasts and it's actually a freefall problem. So let's take this to be a ballistic rocket launch. So there's a huge explosion to get it going and then it's going for what? Very same type of thing fireworks are. They're ballistic, big explosion in the mortar thing and then after that they're unpowered. There's a lot of pretty sparks falling off and stuff, but that's just so you're impressed. All right, so here's a trajectory of some kind. Let's see, I think I have some details we can put to it. Fast, we're going here. Wow, where'd that problem go? I guess I don't have that problem. Must have been my favorite physics class I gave that to. Oh, well, we'll make up some stuff. Nope, there it is. I'm under Fat Brother and all, there it is. All right, so we have a nice trajectory launch here. So here's some of the details. Turns out at the very peak of the lot of this ballistic trajectory, right when it has a velocity up there. Oh, we don't have that, but we do have the height at a height of 2.5 times 10 to the fourth meter. Right at the very peak there, maybe it is a simple firework or maybe it was something tragic that happened, but there was an explosion on board that blew it into two perfect halves conveniently. That's, you know, all physics problems are nice and neat, perfect like this. Blew it into two perfect halves. Right here when it had that velocity, right, no vertical velocity at that point. Right there, it blew into two perfect pieces. One half the mass in each piece happened in such a way that one piece fell straight down below the explosion where the peak was. The other piece with Fat Brother-in-law on board landed where? If you're right, you may go. That's what I want you to figure out. Let me make sure you don't need something else if I have it here. Yeah, we're, yep. The launch velocity was such that the launch velocity was five times 10 to the third. Wait, that's, though. Nope. We're not doing this. Huh? Yeah, thanks. Don't start calculating because you got nothing to calculate. You can do a little bit with trajectory, but then what are you gonna do about the explosion part? So let's think about that a little bit. We've got this thing that explodes. It causes, well it clearly causes an acceleration on the two halves because here the halves were together but still whatever half was that went this way had a very clear, sharp change in velocity. It was going horizontal and then suddenly it's going straight down. There had to be a huge force on that. A significant, well that was the explosion. That's what shot it straight down from the going sideways. How big was the force on that piece that did that? In fact, since we're talking about collisions, we don't want just the force, we also want however long that force acted. So how big was the force and how long did it act during the explosion to make the first half do that? You don't have to answer because you already know where the second part is. It's in the ocean. I didn't ask about the second half, I asked about the first half so far. That just look at this one half. Like it would help we'll draw the rocket as two halves. There's a pink half and a blue half, a boy half and a girl half. And it went up to here exploded and the pink half went down there, went straight down. What's the force on the pink half that made it do that? We know what its velocity is there. Then suddenly it's got this velocity down there, it lands there, the force. So force can be V. Times the change in the velocity by the change. Now I asked what's the force and what's the time. I didn't ask what the mass and the velocity are. I asked you what the force and the time are. Remember for the collision, our collision equation really is the impulse momentum equation. The impulse side has force and time on it. That's our collision equation. What is the force and the time on this? It's the mass time. Okay, that's this side. Right, that's gotta be equal. Well, we can do a little bit more with that. What is it that we did in collisions to solve collisions before we had this full impulse momentum equation? What did we do with the collision that made the whole thing such that we didn't even care what the force and the time were? Make it all one system. Well, that's easy here. It starts out as one system. Our magic system boundary there. After collision, and it breaks into two halves, can we still treat it as one system? Yes. Sure. Why not? When we first did collisions, we had them apart to treat it as a collision, I mean as a system, and then they stuck together. This is just the reverse. They're already stuck together, then they come apart. But we treated them as a system before when they were apart. Let's treat them as a system now when they're apart. So I don't know where this other piece goes. I don't know where the blue piece is. We'll just draw it somewhere for the sake of argument. But it's still a system. So if we look at this system, whether they're actually touching or not, we look at the system of these two halves, now what's the force and the time during which that force acts? We're talking about here's our system. Our system undergoes an explosion and our system now is there. It's bigger, but it's still just those two pieces. Nothing went in or out of that system boundary. The force of the explosion is the department of any direction. Yeah, something on board blew up. No one, my bad brother-in-law, he probably burped and that was sufficient. Just figure out why I don't get Christmas cards anymore from them. I got it. He's watching the videos. Yes he is. I had to stop all the mother-in-law jokes because I knew she was gonna watch them. So what was the force and the amount of time during which it acts? Or this. This is really, this is an essence. The first collision we looked at run backwards, isn't it? We had two things apart. They hit and stuck. Here's two things stuck. They come apart. So what's the force and the delta T that force acts? We don't need to know. We didn't need to know first. Why would we need to know now? Is the force that shoots that thing down internal to the system? Because if it is, what's the impulse part? Zero. What external force would? Well, there's the gravity that pulls it down, but during the explosion, remember when we're talking about these collisions, we're talking about the instant before and the instant after. During the instant of the explosion, gravity doesn't have any time to act. It's so quick. So we're not gonna worry about that. Maybe it drops a little tiny bit. We're not gonna worry about it. That explosion is an internal force. Didn't come from outside, came from inside, whatever it was. So the force of the explosion, we can call it a, yeah we'll do it as a vector, but the force of the explosion on the system, well, was internal. We only do external forces. Want me to calculate those things, don't we? Conserve. Last thing to take for the weekend, we'll pick it up back on Monday. It's not gonna be easy to calculate this. If this, the explosion was such that it lost all its velocity and just dropped from there. Let's say it didn't have any vertical velocity either after the explosion. It just came to a stop and then fell. Maybe we could figure out, but here's the bigger thing we can use and it's gonna be very useful in two-dimensional equations collisions. Momentum is conserved. We don't even, necessarily, have to calculate the momentum of the individual pieces. Remember, the way we did it before, we did the momentum of the system before equals, oh I know what, I used the little tick mark, equals the momentum of the system after. Another way to write this would be the momentum of the center of mass. Remember we were looking at that? Was that Monday or was that a week ago? We were looking at center of mass. How to find out the center of mass of a system? The momentum of the center of mass does not change. Does the mass of the system change? If the mass of the system doesn't change, then the velocity of the center of mass doesn't change either. So if we can figure out where the center of mass is and what it's doing, we can find where the other piece goes and it turns out it's not in the ocean. Darn good guess though. That's where I would've looked first. So that's what we'll do on Monday. We'll look at the center of mass and figure out what happens with this problem. Boy, if that's not a incentive enough to come back, what is? In fact, you might not even want to leave. That's why I guess you do. Looks like you do. Fine, I didn't leave.