 All right, if I remember, we had a problem on last Wednesday that we just had gotten partway through. And that was this idea of a ballistic rocket of some kind was shot such that right at the peak, it underwent an explosion and broke into two halves. The explosion and the halves of no orientation and everything was such that one piece immediately came to a stop. So it was kind of at the back end of whatever this rocket was. So it went and exploded. Instead of going forward anymore, it immediately came to a stop and then, of course, just dropped right from there. The other piece, and the two pieces were equal in mass, the other piece continued on such that it went somewhere else and we're not sure where. Unfortunately, this is the kind of thing they did a couple of years ago with the Columbia explosion. Remember that? The one the space shuttle was coming back and started to break up over California, I think, and then really broke up over Texas. And so they had to look at the pattern of where things were scattered to try to figure out where pieces were going to be. And they were looking over a two and a three state area for pieces. And by pieces, they really meant pieces, as gruesome as some of that was. And I understand they had a lot of astronauts who were doing the looking themselves, a lot of astronauts around the ground, because they understood that some of what could be found could be very emotional. They're also worried about people gathering stuff up for souvenirs and stuff. People do that. So they had a lot of astronauts out there trying to help and that was very, very emotionally trying, as you can imagine, doing this forensic work. But this is the kind of thing they looked at. They had to look at it. But based on the size and the pattern they knew where to look for a better chance of finding stuff and where to not bother looking, because it just wasn't a good chance of finding stuff there. So we're keeping it simple. Two pieces after explosion, I gave you, I think I only gave you two things. One is at explosion, the altitude was 10, 2.5 times 10 to the minus four meters. And I think I gave you the horizontal component of velocity only, is that right? That was the only two things I gave you. And that was five times 10 to the third, 5,000 meters per second. My question to you then was, remember this was a ballistic projectile. We talked about that, right? Remember what that means? When I say it was a ballistic projectile? Yeah, there was no thrust or anything to make it fly the way it was. It was just a simple, just like the projectile motion you'd see when you threw a ball. I don't know how many you could throw 5,000 meters per second to an altitude of 25,000 meters itself, but I could. All right, so the question was, what half of it lands here? Where's the other half land? That was my question. This is like the collisions we were looking at only in reverse. Instead of two things coming together to become one and from that we figured out what was going on after the collision, this is one thing then becomes two. And we want to figure out what happens after a collision. So you probably sell a weekend thinking about it to the exclusion of all else. And how'd you get? How far did you get on it? Pretty far. Pretty far. You're ready to go look. What's the deal? What governs what we're doing here? Usually when we say what governs, we mean what physics, principle, and or equation. The user equations come from our principles. What we've been looking at for the last couple of days. Let's start there. You gotta figure it's gonna apply since I gave it to you during one of those days. We're looking at conservation of momentum. Where did that come from? And does it apply here? The conservation of momentum come from. We actually got to a two ways. I remember the first thing, we were just laying out collisions and the like and what happens during them. But then I took it a step farther on Wednesday and that also came right to conservation of momentum. Put you right down on Wednesday. We had conservation of energy. Couldn't expect it to be conserved here as I'll talk about in a second. Sometimes conservation of energy is very useful to us. Remember though it wasn't in those type of things where two things collide and stick. At least kinetic energy was not conserved, wasn't. And that was really all the only type of energy we had. We just had two things moving. We didn't look at anything in a gravitational field until now. So energy was not conserved earlier. This is kind of the same thing running backwards. It's one thing that now becomes two. There were, we got to this from the fact that there were no outside forces. But then I formalized that on Wednesday. I actually gave you an equation that links the outside forces if there are any with the momentum. And since we sometimes don't have any outside forces as we did in our collisions, then the momentum was conserved and it all came out of that very same equation that I gave you on Wednesday. Well? Post momentum. Huh? Post momentum. Do we do that on Wednesday? Yeah. Did I just do it for when all by itself? We had the impulse momentum equation. We actually got to part of it with our first work on collisions and then got to the rest of it on Wednesday. What was the impulse momentum equation? What's the impulse? The sum of the forces times the amount of time they work. However long those forces are being applied is the impulse. Well, that's just the area under a force time graph if we happen to have it. If we have a force that changes with time, that integral is simply the area under it. For most of our things though, the forces are constant then we just finish that integral by pulling the forces out. That's the left-hand side. What's the right-hand side? Well, more completely, the change in momentum. And our first collisions were such that there were no outside forces. That's what this is here. Any outside forces. There were no outside forces. So we had right from the start with our collisions that momentum was conserved. Now we can do a little bit more completely if we need to. If there are outside forces and we have some idea how long they apply then we can figure out how much momentum is going to change. Or if we can observe how much momentum changes we can figure out how much impulse there was that brought that about. And then we talked about individual car crashes in that light. And so hopefully you all went home with your seatbelts on. And your airbags intact. Joey, did you? I don't believe you. All right, so we have this ballistic projectile that then explodes. No reason this shield shouldn't govern it. This actually governs anything we've been talking about because remember that came directly from F equals MA. Anyway, it's not like this works and that doesn't. It's just F equals MA in a slightly different form that's a little more useful to us. It allows us to see why very small times and crashes lead to very big forces that kill people sometimes. So we have this projectile that goes up here and explodes. What is the impulse side? Who said that? Johnny, you said that? You want to vent to it now? Taking it back. Taking it back. It's got weight to it. That's an outside force. But remember what we're talking about in these problems here is the instant of collision. When we were talking about the two cars hitting, it was the instant before collision to the instant after collision. Very, very little time went by, if any at all. So in that time, gravity's not gonna act in any significant way. So let's just ignore it. Because remember, we're only talking about the instant of these collisions. So are there any outside forces? What about this gigantic explosion? Something on board blew up, an oxygen tank was leaking or that astronaut brother in law belched or something would pretty easily blow our ship apart. So we have this thing go along and there's some internal explosion on board but does that count as an external force? And we need to take into account then we can figure out what's going on for the rest of this. That internal explosion is really fundamentally no different than our car collision run backwards because whatever explosion there is, it's equally felt by the two halves. It's internal to the ship and since it's equal on either side it's gonna be equal in office and canceled. So that's as good an example of an internal set of forces as we're gonna see. So we can take then for an internal, something catastrophic happened on board. We can take that the internal forces are zero. Now that's of course not the case if it had been shot by something, some kind of ground air missile. But that's not what we're talking about here. Momentum must be conserved then for this problem. Momentum of what must be conserved? Well you said the system, you said center mass. Which one's right? Oh you switched now? You buy that? Or is that like him trying to jump on board your wonderful answer to try to make his answer all shiny? Is it the center of mass of the system? Let's put it to the rest of everybody. He said momentum of the system was conserved. He said momentum of the center of mass was conserved but then tried to hop on board over here because he thought maybe that sounded better. Then now he's back pedaling and scrambling and saying but they're the same thing. Well, system momentum is conserved, center of mass momentum is conserved. If you don't know which one's right who do you like better, John better or Alan better? Do the aplosometer, you don't want to do that because you don't know if you might not be next, huh? I can taste a green on. Which one of them's right? Momentum, the system momentum is conserved, the center of mass momentum is conserved. Both, if you like John better but you're afraid Alan can beat you up. That's okay, worse things than that make people vote for president. A lot worse than that. Phil, what do you think? And the system is conserved or momentum of the center of mass is conserved? They're both right, there's no difference because we get to all the same thing anyway. In fact, if you remember when I first showed you how to calculate the location of the center of mass what did I do with that equation? You remember, took the time derivative of it and you got the velocity of the center of mass equation which was just conservation of momentum. So whatever momentum the system has going into this is the same momentum it has coming out of that. We look at it the way Alan put it, the momentum of the center of mass. What does the center of mass do in this whole thing? Let's say, remember this is all ballistic so after the explosion there's no applied power in any way for these things. Let's say that this piece has had enough time to drop down to there. Where's the other piece? So let's see, maybe if we, let's draw a big picture of the two instantly after explosion. So here's their projectile, their path, the trajectory takes them up to there. Let's see, the pink ones, the one that drops straight down. The blue one is the one, what does it do? Immediately after explosion. This one is brought completely to a stop and then starts to drop from there. So this one is brought completely to a stop, what happens to the upper one? Maybe it would help the pictures work more like this so that when the explosion goes off it makes it look like the pink one comes to a stop. Blue one, what's it do? It continues on in one way. They start with, we'll call it 2M so each one of them has 1M after explosion. They had how much momentum just before explosion? They had zero momentum or 2MV, whatever, you said VX even? Just before explosion, right here, explosion at an ambient, what was their speed? This remember is the launch speed, I gave you the X component, the launch speed. What's their X velocity at this moment? Zero? At this instant, they still have velocity of VX. Remember that's what we learned about projectile motion. The X velocity does not change. There's nothing to change it. So right before collision, I mean explosion, the system momentum was just what you said, 2MVX. Just what John said, then there's an explosion, what's the momentum after explosion? What's the momentum after explosion? 2MVX. There was nothing that would change it. Remember this is the system momentum. System momentum is the same afterward because there was no external force to change that momentum. So afterwards, and we've used the designation prime, signify after the event, it's also going to be 2MVX. Comfortable with that? What would change it? You have to, well the governing equation, put it back up there. If this is zero, this must be zero. If that's zero, those two must be the same. No matter what your mind tells you, your mind hasn't taken physics. It's just now taking physics. So this piece, this pink piece is brought to a stop. Remember I said the explosion was such that it was instantly brought to the stop. It's kind of the back half of the craft. So when it blows up, the explosion brings it to a stop. Bless you. What then's the velocity of the other piece that keeps going? Well do it, give it to us in terms of V. Give it to us in terms of V's, A's. Oh, VX? Nope. What's V to V? In the X direction? Now that piece starts to drop. This piece will go where? What will its velocity be? The blue piece, the pink piece has a velocity right after explosion of zero. What's the explosion of the blue piece right after explosion? It's got to be two VX. If the system momentum is two MVX, one of the M's has zero velocity. The other must have the two V velocity to give them the same momentum before and after. So this piece, and in what direction? It's got to be in the same direction. Remember this is a vector quantity. So we have to have not only the same direction on the momentum, or magnitude on the momentum, we've got to have the same direction too. So the blue piece will now have two VX to it. So that yellow was just before explosion. The pink and the blue is just after explosion. So let's let a little bit of time go by like we were doing here. The pink one's dropped down to here. The blue one is going to start to drop, isn't it? Where will it be in relation to this one? This one started its fall with zero velocity. At the same instant this one had horizontal velocity of two V. Where will the pink one, the blue one be in relation to this one? Is there anything we could tell? Same height from the ground, why is that? They have the same vertical acceleration. So they're always going to be at the same height. When the explosion went off there at the same height, they're going to always stay at the same height. So the blue one will be somewhere on that same level. Sometime later the pink one will have fallen to here. The blue one will be somewhere on that same level because they have the same vertical acceleration. And they start with the same vertical velocity. They've got to always be at the same vertical height. It couldn't be any other way. Can we figure out where? Well, let's go back to what Alan said, where he said the center of mass, the momentum of the center of mass would be conserved, which is the same thing as saying the momentum of the system is conserved. Because there are no external forces, the momentum is conserved, whether you look at it as the system or the center of mass. Well, where does the center of mass go then? When they're together, each of them and the center of mass are all doing the same thing. Then they reach up here an explosion such that this momentum of the center of mass does not change. So what does the center of mass do after the explosion? It takes the very same path it would have taken as if there'd never been an explosion. Because there was nothing done to alter the path of the center of mass, no external force. So the center of mass does exactly what it was gonna do before. So if at this height, there's the pink piece, same mass as the blue piece, where's the blue piece? Twice as far away from the center of mass path, such that the center of mass does just what it was gonna do before. Because I said center of mass. It's not funny. Did you write down center of mass? Yeah. You're good, because that's what I said. You'd better write down what I say. Equidistant on either side of the center of mass are the two pieces. Because they have the same mass. If they didn't have the same mass, the center of mass wouldn't be right in between. But it is. When the pink piece reaches here, where's the blue piece at the same instant? Now that wasn't funny. He's not even laughing. He'll laugh at anything. What was he listening? That's why he's not laughing, he wasn't listening. When the pink piece is now here, where's the blue piece at the same instant? Twice as far away. There's the scientific way to do it. It's going to follow a path, something like that. So if this is L, from the launch to where we find the first piece, where's the second piece going to be? A total of three L away. Because we know that the center of mass would have done right in between, would have made a nice symmetric launch, would have been two L in base. And so half that extra will locate the blue piece and land with the blue part. That could be useful because it also works the other way around if we have a collision. There's nothing saying we couldn't shoot these two pieces and they join here. That's what they do to fall in a collision or in an explosion, which is a backwards collision. The path of the center of mass in the absence of any external forces like our collisions are, the path of the center of mass will remain unchanged. Our system changes because it was just this, whatever the object was, and then now the system's a lot bigger and the system's a lot bigger. But the path of the center of mass does not change. There wasn't anything to change it. When there are external forces, not under the chapter heading of collisions, no. However, if you have a collision between two objects, just look at one of them. And then you'll be able to see what you can do. Exactly, we can do that. Look at, well, I could let you. I could let you figure out exactly where that is because you know enough about projectile motion to figure out what L is from what you were given here. I hope we did projectile motion months ago now, not even weeks. So we'll do a collision problem now and then I'll show you how we can do the no external, the forces aren't, there are external forces. All right, so let's look at a collision as could typically happen on a city street where the streets are at 90 degrees to each other. So here's one car, 1,200 kilograms. Lides with another car that was going 90 degrees to it, 3,000 kilograms. So it must be a Cadillac or something, a Hummer. What else is that big? 50 kilometers per hour. How to know that, you know some of these cars have black boxes on them now so we can get the speed and impact. Skid marks go off at about 59 degrees. What was the speed of this car? You know it already? No, do we need a distance on the skid mark? What the wheelbase of the car is? Eight miles. Well, not horizontal, we're looking from above. This is, these are city streets here. But 59 degrees to the original direction of this car where the skid marks. Whether that's north, south, or east, west or something doesn't matter. Just as long as you know that's 59 degrees. So what was the speed of that car? We call that one, and that two. Blinded and stuck together for the most part, which they might not actually do, but they could certainly hit kind of be side by side each other and going the same direction now. Spoken to anybody since you got yourself home? Be careful about texting or something coming into work because sometimes you go right across in front of where a professor's sitting watching it and then drive into the parking lot and the professor could just pull up to you and say, hey, saw your text in the back there, your fat head, risked my life. No, who's gonna be watching? Not that you guys are fat heads. You know what to do? I tell you, no, okay. So you kind of, that's like moving a checker's piece but keeping your hand on it. You know, I'm in with the bag up here. Phil, Mike, you guys agree? Phil, how you doing? I used to the spring hat, yeah. You're in the spring bonnet, Easter bonnet. But what did you say to that? It's over there, I'll give you some. I'm gonna give you some, you don't need. You're in here, so I'm gonna give you some. I'm in here, so I'm gonna give you some. I'm in here, so I'm gonna give you some. I'm in here, so I'm gonna give you some. You know the direction here, but not the magnitude. However big that is determines that 59 degrees. If that guy's moving faster than a certain speed, it'll be less than 59 degrees. If he's moving less than a certain speed, it'll be greater than 59. He's moving such a speed that, what is that 59 degrees? Did you use it? Len, did you use it? No. No? This is gonna be very interesting. Did you use it? No? Did you use it? No, I didn't. Did you? Phil, did you use it? Did you use the 59 degrees? What is the 59 degrees? Phil, what's it tell you now in this problem? Sure, it's the skid marks, but what's it tell you in this problem? No, in terms of the collision, remember the collision is just two things. If you don't wanna draw the cars, you draw those little masses, what's the 59 degrees in terms of the problem that we need to solve now? What's the equation we have that works here? What's the equation? Of course, you can set up a right triangle. Hell, anytime you have 90 degrees, you can set up the right triangle. Momentum. What about momentum? Momentum is not an equation. Well, darn it, that's an equation. I don't know, you've gotta write it down. That's a definition. Whoa, three passenger units. No, no, no. I don't wanna start putting numbers in until I have some place to put numbers. I can't just do mass times velocity, I won't have anything. I only have mass and velocity for one thing. What goes that? Don't calculate something just because you can. Calculate something because you need to, it takes you somewhere, Phil. I don't want numbers because I don't even know where to put numbers. There's nowhere here to put any numbers. Nobody's given me anything. It's a collision. Therefore, what? Momentum is conserved. That's something I can write down and do something with. Momentum of what? The system. The system. Now it wasn't hard, was it? Still don't have any place to put the 3,000, whatever's. Momentum of the system, what's conserved? So let's figure out the momentum of the system before the collision. What is it? And these are vectors because V1 and V2 are in completely different directions. Plus M2 V2. Now we have all of that one. We don't have the magnitude here. We do know the direction, we just don't know the magnitude. Well, that's what we're supposed to find. What else? Equals. Good, guess as any. I'm one. Now they stick together. They're one object. The mass of that one object is M1, M2 times... Times what? V prime. V prime of velocity after collision. And remember, this is a full vector equation. We don't know the magnitude of V1. We're supposed to find that. Is there anything else we don't know? The direction or the magnitude or both? What do you mean? What's the direction? You told me you didn't use that. I thought you did. I said you didn't use it. You didn't use it, sorry Mike. You didn't use it, a couple others said so too. You said you didn't use it. You were mad at me for something. You used it. We don't know the magnitude of V prime. We weren't asked to find it. So we may not need to find it, but we don't have it anyway. But we do have the direction. And yes, you do need that because it's a full vector equation and you're gonna need that now. So now what? Now there's a place to put some numbers before you guys, 3,000, put it somewhere. Quit. What do I write in for V1? That's not enough. No, vector. How do I do that? It's a vector. Don't we kind of have a nice orthogonal coordinate system being laid out here itself? Let's use X and Y direction or I, J vectors. Unit vectors. So V1, I. May sound trivial. That's gonna help in a second. What's V2? Sorry, M2, V2. One of 50 kilometers per hour. If we do this in 50 kilometers per hour, then that'll be in 50 kilometers per hour, which will instantly know which one's speed without having to make any conversions. 50 kilometers per hour. What? J. J. That's the momentum before collision. The momentum of the system afterwards. What's that look like? 1 plus M2 is what, 4,200. What's the velocity afterwards? We don't know the magnitude, but we know the direction. X component, some Y component that we can put in here. V prime, we don't know what it is. Times cosine 59 for the X direction, 59 I, plus some of the system, and they're equal. Why are they equal? Of course, there's big internal forces on each one, but no external forces. Now how do we solve it? Now what do we do? It's on this side, and better match the I components on that side, or they're not equal. Does that make sense? So, in the I direction. 1200 kilograms V1. So we know it's going to be kilogram kilometers per hour. Must equal the I components on the other side. 4,200, cosine 59, true or false? Mike, you okay with that, or? He does not. Oh, I just did, but that's the way I asked. I just asked true or false. Notice two unknowns though, so you need another equation. So we looked at the J direction. 3,000 times 50 equals 42 V prime sine 59. V prime sine 59. How many unknowns? One unknown. Oh, 4 and 4,200. So you can get V prime from there. We didn't need it, but it was something we have to come up with because we're going to need it there to find V1. So what's the speed after collision? 41.7, little under 42. Now you can use that in the first equation to find V1. Remember, all vector equations are really two equations so we can solve them for two unknowns. So V1 is coincidence, that happens to be the car you were driving. 75 kilometers per hour on city streets. What is? John was correct, he's an adult. So for like one hit a ramp and did like a digs a hazard. Yeah, you just do the very same thing only in three direction. These become three dimensional vectors and you do I, J, and K directions. Now maybe you looked at it some other way than that but hopefully that's what you came up with either way. Is that what you got, Mike? If I walk back there, I'd see that. Optional units, isn't that nice of you? All right. Okay, but then any questions with this before I do something else with it? Now what we could do too is if we didn't know either one of these speeds we could have measured the length of the skid marks if we'd known the coefficient of friction could have backed that up to get the velocity immediately after collision and then use that to get the two velocities. Now we could see if either one of them was locked clearly me and vehicle two was doing a nice, safe job. Got it. Go on out there, Alan. Party, you're gonna run like, huh? You're gonna run like, I don't think so. Stop sign, stop sign, you didn't see it but there was, that's why you ran it. You didn't see it. Okay, John though, you asked a different question a little bit ago and then I said what we could do after we did this one. What'd you ask, remember? Yeah, let's factor in some external questions. Okay, do the same problem. Just look at one of the vehicles now. Vehicle one. We now know its speed to be what, 75. Round it off a little bit, make it easy. Gets in a collision, goes off at 59 degrees and going this speed as well, right after collision. Same collision, we're just looking at one car instead of both, spending about its speed after the collision. Yeah, it was this, that's what that speed was, that was the two of them together now. So it's going, we'll call that 42. Sure. This car could actually be one of them. Oh no, no, I'll put three dimensions in it. Well, this car doesn't necessarily go in that direction, does the center mass of the system it? No, no, no, well, yeah, except that's what I said. I said they stuck together. So if they stuck together, then wherever they go, that's where the center mass goes because they're one of the same. The question then is, what was the impulse? Is it zero? Is it zero? Because I said that in collisions there are no external forces. If there are no external forces, there's no impulse. Is that true here? If so, we're all done. Then you vote for that, and let's go. Is the impulse zero here? And if not, why not? This car most certainly had an external force on it. It came from car two, which we've pulled out of this problem though it's not letting anything else change. So how can we find out what the impulse was on two? I'm sorry, on one, change in time. What's the force? Yeah, that's not the impulse side. That's the momentum side. But aren't they equal? Yeah, so you don't know. So the impulse, we don't know either part of, but we do know it's got to be the same as the change in velocity because the same momentum, because that's what it causes. Notice impulse itself is a vector. So it's gonna have direction and magnitude. Now that we know, hang on, we gotta fix things here because now that we know that car was speeding, we know what color it was, too, which is red. What? Does that mean you have to do a separate impulse, right? It doesn't matter, it's a vector. No, this is a vector, whether it's two or even three dimensions, this holds. You know this side in two dimensions because you know its velocity coming in and you know its velocity going out. So that must equal the impulse. So to find this side, you don't do anything over here. You do this side, because they're equal. What was the mass? 1200 kilograms, and I'd much rather be in that class. Listen, you know anybody in that class, you know? Something, is it chemistry? Geology? It could be geology. I've taken geology, it's not that fun. What's the change in velocity? It's not just the magnitudes, these are vectors. So it's V2 minus V1, remember V after minus V before. V after is 42 kilometers per hour times 49 I, sorry, 59, well, 59 J. That is all V2, sorry, V prime for car one, V1 prime. Full thing, V1 prime after collision. We subtract from it before collision, 75 I. Then we add to that I component. We'll bring it up here now and we'll just take the impulse. I'm car one, who's got this part? 1200, 42, cosine 59 minus 1275 I. The whole thing. I want it here, and then another piece of it down there. They both get the 1200 from this, that, and then multiply by 1200 once each one. I need a couple of people to do this because I don't have a number written down doing this in the response to John's question. Thank you. I'll write it down, but I want to see if somebody can confirm that 20,000 is close enough? Yeah. Okay, 20,000, and that's kilogram, kilometer, hour. I want to change that to Newton's, just because that is a more normal force. Speed is fine kilometer per hour, but that's pretty goofy for Newton's, I mean Newton's seconds. Anybody else confirm that? The J part, which is 1242, sine 59. Even the I's and the J's added together? No. Oh, yeah. It's not negative. I bet it's negative. Knowing what I'm looking for, I bet it's negative. You got what? Negative 64,000. 42 dynacost by 59, minus 75. Gotta be minus. That's negative. Then times 1200. $4,000 question. That's what just happened to you. That's the I part. That's the I part. That's the I part. That's the I part. That's the I part. And it's negative. What's the J part? Which has got to be positive. There's no negatives there. $43. Somebody? Joey, you got it? Yeah. Okay. Let's see what this means. At least according to the equation, that's the impulse. So right at the collision, right there, right at the collision, R1 saw a force or an impulse that was minus 64,000 i plus 4,000 j. So let's see. Minus 64,000 those units. Plus 43,000. So it's going to be about that size. Give or take a little bit. Would a force like that, or an impulse like that, would an impulse like that, which is the same direction as the force because the time is going to change the direction? Would an impulse like that cause an object to go like that? Yeah. Wouldn't you think so? If we just hit it from the side, it might do that too. But the fact that it was essentially a dead object in the x direction that it hit also gave us enough for this whole thing then to be about 59 degrees that we're doing. That's all you need to do when we have external forces. If we know the change in momentum, we can find out what the impulse was. If we knew maybe from, you know maybe at that intersection there was a video camera running, we can get an estimate of how long they were actually in the collision, and in this split second, we could then estimate what the total force was. Total impact, remember, we're only looking at the collision itself. As people, we consider the skid out of it as part of the collision, but as physicists looking at a collision, we're only looking at that instant collision, which is a very, very short time, very, very high force system because the impulse is going to be pretty big. You would have no more impression of something. You really don't need to. Well, no, it sounds to me like you're dying for a problem. No, no, it's too late. We can't get out of it now. We've got five minutes here. It's so small. We have time for another question. No. No more questions. Here we go. Two cars. That one's going that fast. We're switched, so just to make sure they don't get worse than anywhere else. So this is V2. V1 is like that. V1, 900 kilograms. Sandy cars before, 1200 kilograms. Just got just driving home from the auto body shop after the first collision. What a loser. V1 is 14 meters per second. They hit, skid off in that direction. You can even hear them. Oh, they skid it. Yeah, they skid it. No, no, they skid it before the collision. That's what I thought I just heard. That means these velocities are different. You have to change the whole problem. They skid, hit and skid. Skid marks are 17.4 meters long and the road and weather conditions were such that the coefficient of friction that be kinetic or static? Kinetic. Kinetic, because the tires are moving in relation to the road surface. 0.85. Find V2. Here we don't have that information, so there you go. When you get that, you can leave. Either way, whatever these are, the velocity of the center of mass wouldn't change. The velocity of the center of mass of the system wouldn't change. So if we have to know what V1 was, what the two masses are, we know where the center of mass is, possibly figure out that, we'll see how you do it. Million dollars to whoever comes with the answer on Wednesday. And that's a million dollars for every one of you who get it. So if somebody gets it, sell it to the other people for 500,000 we all make out. Unless you're wrong.