 We thought for our kinetics problems, those problems where we're no longer looking just at the kinematics, we're looking at how we affect those kinematics. How do we get the acceleration we want of those type of questions? There were three solution methods that we've come across now, not in any particular order. Well, I guess the first one makes sense we start with that one because the other two follow right from that one. The first one was just simply f equals ma. That's the one you would find in the first two days of chapter three that are called on the schedule just kinetics. I think there's linear kinetics and then curvilinear kinetics. So that essentially just means one d and two d kinetics. Straight forward problems like given acceleration, find out what the forces must be. Given certain forces, find out what the acceleration will be. This works best for constant force, constant acceleration problems, which we did a lot in physics one. However, in this class we have come to see some other problems to do besides constant acceleration problems. And if this acceleration is not constant, then the force isn't constant either. So we did go over a couple of the different possibilities for variable acceleration. Acceleration is the function of time, or of velocity, or of position. I spent a good part of that. They vary early in the term doing that if I remember second or third day even. But so it works for general problems where it fits into one of those. Clear categories without, I guess, falling into one of the others. Then we have the work energy equation, which works very good for position dependent problems. The work is dependent upon the distance traveled, but certainly position flavor to it. And the gravitational potential energy and the spring potential energy both have very much to do with the position of the object within the problem. Then we have the impulse momentum method. Remember where that was best used? Time dependent problems. Because the impulse is the integral of the forces with respect to time. So it's great for forces that change with time. The momentum part of it that usually doesn't have a time component to itself. So it could mean that the forces change with time. It could mean too that there's just a time component in the problem. And it's very direct to solve it that way. Could be solved that way, because there's a time component in there as well. This is dv dt. And that fact is how we got down to this. We just multiply it through by the dt and then integrate it. So they're not mutually exclusive, but I'm not trying to trick you and make one of them kind of obscure. So it'll probably be something like three problems on the test in that order. The easy way to write it, easy way for you to do it. I'm not trying to get you tripped up, but I do wanna see if you can do the three different. Understand that they really got what you can see is that the three different parts. Fair enough? Then we'll leave off angular momentum and then what we're gonna talk about today, which is impact. Okay, off we go. Impact, what we call in physics one, collisions is a collision between two objects that generally takes place in very short time periods. Because of that, it's most easily looked at as an impulse problem. Remember our impulse equation, both momentum equation looks something like that. So it's quite easily looked at in this way to see the kind of things that are happening. Because the impact time is so very small usually, well let's see how that's going to affect things. We'll just take a very simple case, an object coming in with some velocity, impacts a wall, and then of course rebound off of that with some other velocity. This object obviously has some mass to it, some velocity, I don't need to call it one, we don't have a two, and this will be prime after collision. Clearly the change in momentum is at least in a quantitative sense in terms of how we're calculating this equation, the change in momentum has nothing to do with the time. There's some mass and there's some velocity coming in and some velocity coming out. That's what it is, whatever it is, at least in terms of us analyzing what's going on here. However, if this, for these very same changes in velocity, if delta t and the dt, the time period is very, very small, then the forces involved must be very, very big to get the same change in velocity we might find. So just as in terms of us analyzing it, we observe some mass coming in at a certain speed, going out at a certain speed, then those things are set. So we come to look at this side of it and see, well, if it's in very, very small times, then the forces gotta be very, very big because otherwise we wouldn't get the momentum we happen to be observing. So what we're gonna look at is this, is mostly this size, this size right now to see if we can figure out what's going on with that bit. A little bit of stuff we need to define first just so we can understand some things as we analyze what's going on here. We'll look at basically two types of collisions, what we call a direct central impact. This central, let me look at that word first. Because once we have it, we're very easily done with it. Central essentially means that the objects, the center of mass and the two objects are in line. By in line that's as defined by the surfaces in contact during impact. For example, if we have two circular objects collide, their centers of mass are perfectly lined up during the collision. If we have a circular object that collides with the wall, kind of like I had there, the center of mass of the object is obvious where that is, but the center of mass of the wall could be anywhere who knows where that is. And once they collide, and we have this geometry set up by the surfaces as they collide, the center of mass of the wall might have nothing whatsoever to do with that. So we're not gonna necessarily look at those type of collisions other than nice simple ones where we're looking at a circular object colliding with the wall. So for our purposes, this essentially means two circular objects hitting like two pool balls, two hockey pucks, those type of problems that are fairly commonly looked at. This term direct is our first type of those collisions. And this is where the velocities are directly in line with the centers of mass as well themselves. So the center of masses travel exactly along the same line as each other, and they collide and rebound along that very same line. That's a direct central impact. We'll also look and we'll spend more time with just because they're a little more involved, an oblique central impact. Central, you already know what that means. It means that essentially for our purposes two round objects colliding. Oblique means that they are coming in and going out at angles that are somewhat off of whatever the line is that's joining their centers of mass, the instant of collision. So we have that type of thing where then they collide and bounce off of each other. And our problem's going to be to determine what those rebound velocities are, or if those are known to determine what the approach velocities are. For either one of these collisions, the instant of collision, and it doesn't matter whether it's a direct or an oblique central impact, that this will be true, because once they collide at that instant, either one of these impacts look the same. At the instant of impact, that will define our coordinate system based upon those surfaces in contact. The, as drawn here, the vertical coordinate system, that's tangential to the surfaces in contact. So we'll call that the tangential direction. The horizontal component, as drawn here, we'll call the normal because it's normal to these surfaces in contact. So all of our velocities and components thereof will be defined in that coordinate system. It's hopefully obvious that when these things collide, they might not necessarily collide always nicely lined up like that. They might collide like this. And then at that instant impact, we can have this orthogonal coordinate system, but it may just not be as nicely horizontal and vertical as that first one was drawn. Doesn't matter. That's the nature of tangential and normal components as we saw them when they were first introduced. And we take them as they come. All right, simple straightforward definitions. The main one you need to pull this is our coordinate system there. So I'll start the analysis with the direct central impact. I won't redo what we're doing here for the oblique central impact because it's all just the same. It's sort of like our usual steps from one D to two D. Do a lot of the stuff, setting it up at one D and then the extension of two D is almost trivial and just means we have a little bit more to calculate, but not necessarily anything hard. So here's our situation. We have two objects, M1 and M2, traveling relatively speaking towards each other. They might be traveling absolutely towards each other where one's traveling in the opposite direction of the other, or it could just be that the behind one is going fast enough to overtake the front one and there will be a collision, an impact of some kind. And since it's a direct impact, they'll always be traveling along that one line and will not leave from it. So that's our situation to set up the problem. We have this faster object overtaking this slower object and there's gonna be a collision. Some little bit of time later, they actually do make contact and there's even a little bit of deformation as this happens. In fact, it's the deformation and then the recovery of the object from that deformation that gives the rebound. So that's why we're gonna have to look at it. So for some very short period of time, they come into contact. During that time, they're traveling in some common velocity. So that B sub C, we don't know what it is. It's hard to observe what it is. It's pretty easy to observe the incoming velocities. It's not very easy to observe that common velocity. Remember, these contact periods are very, very short. But if we look at each individual piece, so here's one moving at some velocity, Vc, this common velocity, it feels an impact force from the other object that I'll call F sub D because this is a deforming the force. This is still the first half of the impact, as you will. The pieces have come into contact. They're still moving towards each other somewhat. They're deforming each other. And so piece one feels a force from piece two, and likewise, piece two feels an equal and opposite force exerted on it by piece one. That's what's gonna cause the changes of speeds, or that's part of what's gonna cause the changes of speeds. So the D there stands for deformation. A little bit later, they're still together, but now they're starting to rebound off each other. In fact, this is some very, very short period of time, delta T, during which there's contact. Outside of that time, before that and after that, they're no longer in contact with each other. So this is the entire collision time here that we're talking about. But they have moved a little bit down, because of this common velocity of the thing that they have. And so if we look at that, we have the same type of thing. Now, one is trying to recover its shape. Because of that, it will be pushing on piece D, and D pushes back on it with something we'll call FR for the force during the restoration phase. And again, these are equal in opposite. Remember, this all happens very quickly, very short. And then of course, they've now completely rebounded off of each other, and are moving with whatever velocity they now have. So we'll put the after collision velocities with a little prime on it. Of course, now V1 must be less than V2. V1 prime must be less than V2 prime, otherwise they wouldn't move apart. Phase we call the morning after. And this is when you'll see each of them maybe having a little cigarette, having a little smoke here, kind of relaxing in the after glow. Do be dealing with marriage, so she's the only one who gets that. What? I was just wondering about the restoration part. What is that? What is that? It's a velocity right there. It's still Vc. Vc, it's different. Well, it may change a little bit. Remember, this happens so quickly that it's very difficult for us to see what's happening at these speeds. It may or may not be constant. I didn't see that. We don't even really care. So for the most part, they're still moving at the same velocity, but we don't care what that is. It's gonna be problematic to find it anyway. You need very high speed photography to do much with it. So a couple of things are true during this brief contact period. During that very, very brief Delta T. Couple things are important to us to remember. The velocities, if I had that, I mean the V1, V2, V1 prime, V2 prime, they change greatly. A big change in momentum. You can most easily see that in the example where they're approaching each other, they collide and they rebound off each other. Not only could the magnitudes change, but the directions change, which is even a greater change in velocity than just having the magnitudes change, just then directions flip as well. That's a huge change. Position changes very little. Mostly as a consequence of that Delta T being so small that happens so quickly they just don't have a lot of time to go anywhere. We're also going to neglect any friction. That could come in two forms. During the collision, these contact surfaces could actually slide over to each other. Generally the nature of perfectly circular objects colliding, that's very close to truly negligible. But we're also, oh, and there's also friction between the molecules themselves during this actual change in shape. We'll also neglect other non-impulsive forces. Glasses on, I'm so glad to see us stop squinting at me. I still can't read what that says. During the... Is it my handwriting or your glass? I thought I'm saying it's handwriting. Maybe you have them on upside down. Oh, did you take any borrow yours? Just for a minute. Is the handwriting that bad or? It's a fuel contact. Fuel contact. I don't know if it's like that. It's a fuel. It's like it goes up fuel. Oh man, you can even see it says brief on the screen. Other non-impulsive forces. That means those forces that we wouldn't put over on the impulse side. Things like generally gravity doesn't go in over there. Well that's the big one, I guess. We typically don't. I guess that goes kind of with this position doesn't change very much. If the position doesn't change, height can't change very much. Gravity wouldn't be a part of it. These are the only things we're worried about. These impact forces between the two objects. Nothing else is gonna be a great concern for us. A big consequence comes because of that. If there's no impulsive forces, then what? What's the impulse momentum equation then become? If there are no impulsive forces in the problem. If there are no impulsive forces, this is zero, this is zero. Momentum for the system is conserved. So we're gonna need that as we move to solve these problems. It's not true individually because this is a very large impulsive force in terms of what number one experience is. But in terms of the two together, those two become internal equal and opposite forces and they just cancel. So in terms of the system, those forces do not add any impulse and therefore the momentum is unchanged. So what we need to look at next is then what these objects are undergoing individually and they're a bit different. We'll take a look at one. It's the same business for two all on our purposes. So here's the object one during the deformation phase. Give or take the first half of the collision itself. So the impulse it feels, and this will be from time zero to some intermediate time I guess during which there's deformation. After that time, then there's restoration and we'll do that part in a second. So during this time of deformation, there's a negative impulse applied to object one. Negative remember because it's moving, the deformation force is to the left. A change in momentum in object one, which will be easier to write out at least m one times vc minus v one. Remember v one was at some approach velocity. It's now undergoing a deforming force from contact with the other object, reducing its speed to some intermediate value of vc. Some little bit of time later, they start to rebound from each other. It's actually a physical molecular restoring of the object to original shape. We'll say it's still moving the same speed vc or at least average speed during collision. And so now there's an impulse from tc to whatever the final time of collision is. The total time from zero to t is the delta t we were talking about, the collision time itself, which is very short remember. And this looks very much the same. It's the restoration force that's causing, remember this is the force it's feeling from the other object as the other object restores its shape. Well, and force it feels because it's exerting a force on that other object as well. This will cause a change in momentum. I'll put a prime on it because this is after the collision or at least in the second half of it if you will. And that's going to be m one times v one prime, the velocity it has as it comes out of the collision minus the vc it went into the restoration phase with. Okay, so far, a lot of stuff. We don't know, we don't know what these forces are. They're not easy to measure. Happens very quickly. We don't necessarily know what this common velocity is. It's not, hopefully it's not too obvious yet why we even want to know it. In fact, we're going to be done with it very shortly here. But we still have a little bit of ground work to lay. So we're going to define here the coefficient of restitution, efficient of restitution tells us something about how two objects collide with each other. Given the little symbol E or the symbol little E and it's defined as the restoration impulse, what this is right here. This left hand impulse term on our impulse momentum equation divided by the deformation impulse. The same thing only in the first equation. This thing here is, those are both impulse terms. And so that's the restoration impulse over the deformation impulse. So between the same time one that's shown over here that we don't want to deal with those. It's a little easier deal at the other sides. So we've got delta G, that's the, that's what the restoration impulse is equal to. That's over delta G from that's a little bit easier than having the integrals in there with forces we don't know. Those are harder to measure. We did have to make our VC had a high speed camera. The mid-busters, they can do it with their high speed cameras and the black and white stripes they always use on everything to very accurately gauge the speed of something. All right, but we have even another term for those two things. And obviously the masses will cancel and so we get V one prime minus VC over VC minus V one, the original velocity. So it's pretty clean. If only we knew what that VC was, it's pretty clean. There's not a whole lot to it. So let's see if we can do something about that VC. So we have here coefficient of restitution. It's now down to V one prime. I'm just rewriting that over here. VC over VC minus V one. Make sure you get your prime, right? Make sure I get the prime, right? Remember that. Anything prime is after collision. So the speed of one after collision with two. Speed of one before the collision with two. Okay, so far. Yeah, even with or without glasses, it's okay. All right, we can do the very same thing for object two. And since they're the same objects and the same restorations and the same times are going on, they're going to be equal, which will give us then V two prime minus VC over VC minus V two. If we just did the exact same thing over again, I don't think there's any point to actually do that because we'll get down to just do exactly right there. However, what it means is that we can take this right here and eliminate VC. An exercise in algebra, of course. But again, it's a bit VC, which we don't know, which we could measure, but it would be a pretty crappy measurement. And we don't even really care about it, especially since when we do so, when we eliminate VC, put everything back together, we get a coefficient of restitution that's very nice and neat, has only to do with the approach velocities and the departure velocities, which are very easy to measure. And it comes out to be when you do all the algebra you're welcome to if you want. You're lucky for punishment. V two prime minus V one prime, V two minus V one. Pretty simple. Very easy to measure, coefficient of restitution. This minus sign is just an outcome of the algebra. It allows us to keep these indices in the same order, which is just a little simpler. Plus we can even reduce this to a slightly different form, one we've used before. We've had this type of thing when we talked about relative velocities, velocity of one object relative to another. We have exactly this form in it. So if we use that same format, we can make it at least easier, more condensed symbolically, where if you remember, this subscript meant the velocity of two with respect to one, and it was found exactly this way. This is exactly how we defined it. So the coefficient of restitution is the ratio of the departure velocities over the approach velocities. The minus also, of course, allows us to keep this as a positive number. So a couple things about that coefficient of restitution. It's greater than or equal to one, sorry, less than or equal to one, greater than or equal to zero. We can't have a collision where we come out to be negative. We can't have a collision, at least not any kind of real collision of just simple things colliding each other, where this is greater than one. That means they'd have to be picking up relative velocity from somewhere. If E equals one, it's what you've heard of before as a perfectly elastic collision, which also means there's no loss of energy during the collision. If you remember from the physics one lab where we ran carts along the air track, there was always a loss of energy in those things. It can be zero, which is a perfectly plastic collision. What would that mean? The only way for this to be zero is if the numerator, the two velocities are equal. If the two velocities are equal, that means that these things stuck together when they hit. If these were two blobs of clay and they hit, stuck together, we have a coefficient of efficient restitution with zero. And in this case, the loss in energy is a maximum. Everything else, especially in real life, is somewhere in between the two. If you'd ever like to read a fascinating paper, remember the, I'm sure you've played solitaire on Microsoft Windows, but before Windows 7, remember XP and before, what happened when you played solitaire and you won? The cards would then start peeling off and they bounce. And you can take that bounce and calculate the coefficient of restitution of a playing card, which is a fascinating paper written by a very famous engineer who has trouble taking anything seriously. So he wrote that paper and then found out in Windows 7, the cards don't do that. You know what they do now? They explode, they hit the thing and they explode. They've changed them from a nice bouncy material into a brittle material and they fracture. It's really a pain. So that's a darn good paper. Maybe I'll post that paper on Angel for you. That's a darn good paper. All right. Oh, one other thing that's very important. Actually, two other things very important about this coefficient of restitution. Absolutely vital. It depends on the two materials of the two objects involved. Two glass spheres bounce off of each other very differently than two steel spheres would. Even if the masses and the incoming velocities were the same, the outgoing velocities would be very different. And it's only determined through experiment. You cannot do calculations and figure out what the coefficient of restitution is going to be for particular materials. So it actually doesn't make sense to talk about the coefficient of restitution of playing cards because it also depends upon what the surface they were bouncing on is, but Microsoft wouldn't tell him what that was. All right. Good enough. So we can take this a little bit farther. Again, this is so much fun. Got rid of that VC. That's good. We got rid of the forces that were involved. That's good. We got rid of Delta T. All those things that are hard to determine, we got rid of. So let's take our analysis a little bit of a step farther and look at oblique central impact. Now, during the collision, nothing changes in terms of what we just did. No matter what they're doing, before and after, during the collision, things are exactly the same. That's kind of bad, man. We need, here's the easy way to do it. Draw your two circles, then draw in and see. How's that, Jake? Yeah, these circles are hard to do when you've got to do them together. All right, central. So we don't need to look at the collision itself. You can't tell from this picture whether this is a direct central impact or oblique central impact. What does change, of course, is whatever's going on beforehand as some object of mass one approaches at some velocity, v one, some mass, some second mass here approaches at whatever its velocity is, each with some approach velocity, which means it's a vector, so we need magnitude and direction on these things. So notice these angles aren't really known until the moment of collision because in the moment of collision, our coordinate system is defined. Once we have the coordinate system, then we measure the angles. So, well, you know how all the problems go, is how to do these things. And then there's some rebound of these things that we're then gonna try to determine. So that's v one prime is rebounded some angle. Theta one prime, remember prime is everything after collision. And the masses may or may not be equal. Depends on how bad a mode I am when I write the problem. That's v two prime at some angle theta two prime. So that's our collision setup. And we don't need to do that restoration impulse, deformation impulse, all that stuff. We don't need to do all that again because the moment of collision is exactly the same for the direct impact that we just did as it is here for this impact of this oblique impact as we now look at it. So for the problems we're gonna do here with these oblique central impact type problems, they're always gonna be set up such that you know enough coming in that there are gonna be four unknowns to solve. For example, if we know the incoming velocity, actually what we need to know is the incoming momentums. We need to know the mass and the velocity vectors. If we know those coming in, then there'll be four unknowns out of the problem that we need to find. The magnitude and the angle of each of the two velocities, there's always gonna be four unknowns for our problem. For the problems we do here. In real life, there may be more, but that's real life, we're not there. I have a problem, for example, there could be a problem where you come upon some car crash scene. You know what the masses of the vehicles were, you know what the velocities after the crash were. From that, you'd try to back out the velocities. Before the crash, either way, they're still, for our purposes, four unknowns, which means, of course, we need four equations. Not in any particular order. Here are the four equations. We've already gone over them, but we're gonna have to pull them together because they went by in such a way that you didn't really see them properly. All right, number one. For the system, there are no impulsive forces. There are no impulsive forces in the n direction. Actually, there aren't any in the t direction either, but we need to separate that out here because we'll use the vector, no impulsive forces in the t direction. We'll use it in a more useful way in a second, in the next two. So no impulsive forces in the n direction for the system, which means the change in momentum of the system in the n direction is zero. Another way to write that would be that M1, V1n, which is the incoming component on the first object in the n direction. So that's this little piece here. This is V1n, but for the system. So we have to add on to it M2, V2n. So over here, V2n, we're gonna equal M1, V1n, plus M2, is that right? Did I get all my subscripts right? All the ones are together, the n's, okay. So there's our first equation. And you can imagine in problems where these two masses are the same, it gets even simpler because the mass divides out. But there's equation, our first equation. No, this is the, if you rearrange this, so we have the momentum of one after minus the momentum of one before, plus momentum of two after, minus momentum, that would equal zero. These are the same, just algebraically different. This is delta G cis before, this is not delta, sorry. That's G cis before, this is G cis, right? And they're equal, so the change is zero. All right, that's for the system. They collide, individually there's forces in the n direction, but on the system, there are no forces in the n direction. Notice since we say no impulsive forces, that means this could be happening in space with gravity acting, but it's not gonna change anything. It'll change things if we let them travel very far, but in the very short periods we're looking at here, they don't travel very hard. All right, so here's the second equation. Individually, and we did the system, now we're doing an individual, there's no impulsive forces, the t direction. There most certainly is in the n direction individually, but in the t direction there's no impulsive forces. When these two collide, they're only going to feel that restoration and deformation force, and that force is only in the n direction. Remember the n direction is defined by the geometry of the collision itself. There are no forces on either one of them in the t direction. So we can say delta G1, t equals zero. Doing the same kind of thing here where we expanded that out. Or V1, t equals V1, t prime. Because when you put the momentum before, it equals the momentum after, the mass is canceled. We know then that whatever component of velocity either of them has in the tangential direction will not change. So V1, t going in is going to equal V1, t coming out. Because there's nothing to change it. There's no force in that direction. The force is the only thing that's going to change the momentum, and momentum and velocity are hand in hand here. The third equation, well, the very same thing for piece two. Oh, that's pretty easy for those two things. It may get much easier with that. We already know tangential velocity is. So you know these kind of problems where we're giving the incoming velocity, we know the magnitude and the angle. Well, it's trivial then to use the trigonometry. Find that, you're halfway done. You already know two of the four unknowns. Just write out the box there with those things. Hardly anything to do. Need four equations. The fourth equation is we'll either need to find or we'll know ahead of time the coefficient restitution which involves these very same unknowns. V2 prime minus V1 prime over V2 minus V1. So in these problems we're going to need to know the coefficient of restitution. Now, be careful, it could be stated in the problems such as in a perfectly elastic collision such and such happen. Or in a perfectly plastic collision such and such happen. In those words, you've just been told the coefficient of restitution. So you have to be careful sometimes. Other problems, we're going to have to, we're going to expect it to be given. Four examples, we'll do a problem. If you notice, by the way, these two automatically take care of the delta G for the system in the tangential direction. It's also zero. That's why we expressly laid it out in the in direction so that we can get a useful equation here. If we did delta G of this whole system, these would all be vectors and wouldn't be any more useful than this. We do also, though, have to realize that because this is true, two and three are true, then we need this to be done in the normal direction only. In the tangential direction, it just won't meaning it. In the tangential direction, E is identically one anyway. So it's not that it isn't true in the T direction. It's just that in the T direction, because of these two, the coefficient of restitution to automatically, identically equal to one. All right, so we'll do out a problem here. So here's an object, 30 feet per second, 30 degrees from the horizontal, and we'll set up the collision so that that's exactly how they collide and define our coordinate system. Convenient how things in life always line up for us like that, just as we need. Here, much deeper angle. So that's at 60 degrees, 40 feet per second at 60 degrees. We can even similar on this for this first step. M1 equals M2, and coefficient of restitution for these two materials, known ahead of time to be 0.9. And we wanna find then the rebound velocity of each. We wanna find the velocity V1 prime and the velocity V2 prime, okay? Unknown's four equations, but we have the four equations. So we're just gonna work through and make sure we got the right numbers. All right, well, my old equations are gone now. But notice in the equations we need the four components. We might as well figure those out ahead of time and then we've got them right there. So we need V1n, V2n, V1t, and V2. All right, fifth trivial, so I'm not gonna make you do that. 25 feet per second for that one, 15 for this, 20 and 34, six. Trigonometry, you're not sure where those come from, God help you in your second year in engineering. I can't help you. Notice how badly deformed they are after the collision. They're just so lumpy, hurt. All right, so let's try the equations. My old doleman order, what was the first one? Delta G for the system. Ah, I bite, check your notes. What'd you write down? You didn't write down Delta G for the system, you wrote Delta G in the normal direction for the system is zero. Bit easier here because the masses are the same. So that means for us V1n plus V2n equals V1n prime plus V2n prime. Because the masses divide out. Don't forget the masses would normally be in there. But in this case, M1 equals M2, so they divide out. We've got those, oh, hang on, I did, I made a goof here that we're gonna need to be careful with. V2n is not 20, minus 20. Because it's in the minus n direction. So that would have been disastrous. So we get, let's see, V1n. So we get 25 minus 20 feet per second. I need to smooth your notes, not a big one. This is not 25, it's 26. Sorry about that. So I'm gonna make a mistake at the board. All right, so there's equation one. Equation two, what was that one? Initial direction is zero. Well that's the one, and especially with the masses being equal, V1t equals V1t prime. Remember, if the masses aren't equal, they wouldn't divide out. When the masses are equal, that's just what happens. In fact, maybe to emphasize that point, I'll write it down with the masses in there and then we'll cancel. So V1t equals M1V1 after, in the tangential direction. Well we've already got those. That's what, 15. There we go. One of the velocities after collision, one of the four we need is already done. We know that V1 in the tangential direction coming out is 15 meters per second as it was going in. And same thing for number two, because of the masses. Well in this equation, I guess as a masses couldn't help but cancel. It's the first equation that they might not cancel. 34.6. And the last equation, and I'll leave you with the two hours of algebra it would take to solve them. Remember, this is in the normal direction. We know it's 0.9, experimentally determined, given that as part of the problem. So it equals 0.9. Has our last two unknowns in it. We just found two of them in the first part. Let's see, V2n is minus 20. V1n is 26, it was 0.9 in it. Equation one here has two unknowns in it. Equation four here has the same two unknowns in it. It's a matter of just solving it then. I'll give you these. Because we're right at the end, and it's just algebra that's left now. V1n prime equals minus 17.7 when we solve this set of equations. The minus, yeah, makes sense from the picture. We'd expect one to be going in the minus n direction. And V2n prime comes out to be 23.7. And when you put those together, that gives you rebound velocities of 23.2, 40.3 degrees, I believe, 41.9 at 55. The minus signs have got to mean what they mean. One quick question, get out of class question, except that I don't know the answer, because I'm only asking it right now. What was the change in energy? Just see if you all get the same thing. It's a very quick calculation for you. We're gonna see if you all get the same. What energies, what energy are we talking about? Sorry, kinetic, because remember, we're going just during the moment, instant before and to the instant after collision. So there's no gravitational potential energy chain, no elastic median here. So all we're saying is that K1 plus K2 equals K1 prime plus K2 prime. So there's a little bit of a trick still left in here. So I want to make sure if I get the right. You know the answer already? Nobody's doing it. No, Frank's calculating. Frank's the only one running the calculator. How come nobody else is? Are you done? Do a beat? Not playing along, you're mad at me. Sure you do. Do this. Let's see. One half M1 V1 squared plus one half M2 V2 squared. That's K1 plus K2 before collision. Equals one half M1 V1 prime squared. You know square to prime as a double prime. That would make sense. All the one halves cancel. Remember the masses were equal in this particular problem? They cancel, but what velocity do you square? Yeah, this is V1, this is V1 prime. There's V1, there's V1 prime. Are those the numbers that go in here? Kinetic energy, remember, has no directional component to it. This is not a vector. Kinetic energy is not a vector. It only matters the speed and the mass. So you, we know that, oh, sorry. These aren't gonna be equal. There's gonna be some loss here plus delta E loss. Because they're not gonna be equal of that. We're so used to them being equal. So then you should find whatever delta E loss is. I don't have it here. So I'll trust you to do those numbers, can't I? If it loses 206 joules, we need a unit fix because we divide it through by M. So this is delta E loss per kilograin. Because we divide it through by M so we have to do that on all terms, right? Anybody here take thermo? Had thermo yet, Frank? Nobody else did last spring. When we divide by M, we typically go to a lower case because that then means energy per kilogram. You'll do that in thermo. Okay.