 We did a little bit on angular momentum on Monday. We're going to step back to just the business we needed to do for linear momentum. That's all that's going to apply here. We're going to look at impact and collisions. Make it the right color chalk. We're going to look at impact and collisions. We did a little bit of that in physics one as most of this stuff has been. But we're going to of course look at it in greater depth. For the simplest kind of collisions we can look at, we might look at a solid, immovable, incredibly massive barrier into which we'll run one of these objects, these type of things that we throw around. In particular, moving at some velocity and it has some mass that's going to hit that barrier and then rebound with some other velocity. Sort of the tradition in these impact type things is we tend to put a little prime mark on those things that happen after the impact, after the collision. I'm not real fond of it. If you're not careful of that it starts to look like a superscript or an exponent. So be careful with it. There are places where we actually have to square those. So you have the prime squared. If you don't write it carefully, as with a lot of the other stuff we write, if you don't write it carefully it's not going to show what you needed to show. So to figure out something about the change in velocity for that object because it's not in any way necessarily true that V equals V prime and magnitude. In fact, in real collisions, and you know this from any time you've done this very type of collision of dropping a tennis ball or a golf ball on the floor, that never comes back up to the same height, which you would have to do if it had the same velocity going in, the same speed going in as the speed had coming out. So we can figure out something about the change in momentum. We might want to make this a vector equation where in this case we could get by with just pluses and minuses there, but we have that on the right-hand side, the momentum side of the impulse momentum equation. This is the momentum of whatever the object is itself. Of course, the momentum of the wall is essentially zero because it doesn't have any change in velocity, but this momentum is caused by the impulse that the ball receives due to some force exerted on it by the wall itself. And this is over whatever period of time is that part of the collision. And these are typically, as you can imagine, very, very short collision times, often then very, very high forces applied. And in fact, there's a lot of study that goes on looking at these, and they're typically kind of big things where they occur in tenths and not even hundreds of a second. What we'll usually do is not look at the peak force or the entire integral, but we'll use an average force to represent that. However, it's tough to know what those forces are. There's not really any way you can calculate. We can calculate what the average force is because we can figure out something about the change in velocity if we have some idea of what this time period is, but that's very quick, too. And so if we really want to know some of the details here, just what the impulse was, the area under the curve to get the average, we're going to need some kind of experimental data of some kind to be able to finish that. However, very shortly, we're going to see how we'll get rid of that entirely. Since these forces and these times are difficult to know, well, John could do them with some of this gizmo, and if we could do it if we drop your laptop, we could do a little test that way. Sure. Bounce of dew for the door. Well, great. Maybe we'll just launch it in and do that as a little lab experiment. All right, but look, we're going to be done with those forces shortly anyway, just because they're hard to determine. You need some very good experimental equipment to determine it. So we're going to have basically two types of collisions. A direct central impact. That's where we have two objects, both circular, originally heading towards each other. Now, those are relative velocities. It could be that one is still and the other is moving with a grade and a philosophy. All we care about is they have some relative motion towards each other. There's an impact and then there's relative motion after collision between the two. The instant they collide, our coordinate system is defined at those at that moment. Once they collide, we've got the original path direction and perpendicular to that. And that's true with the geometry and for our definition. At the interface of contact, we have our other coordinate direction defined at that time. So we'll call this the normal direction and that the tangential direction tangent to the two objects themselves, the surfaces of the two objects. And orthogonally define instantaneously upon impact between the two. So we'll look at those in some detail. We'll also use a subscript notation to delineate the two velocities between each other, the two objects. And then after collision and rebound, whatever they might do, then we'll use also this prime notation. So they'll come in at each other with v1 and v2, collide, rebound with v1 prime and v2 prime. And remember, these may be relative velocities. All we're concerned with is they're moving towards each other for an initial collision. The only other collision we'll look at is that of an oblique central impact. That meant simply that the centers of mass of the two objects are falling along the same path and that path is also defined at the moment of collision by that normal direction. Oblique central impacts are those where the objects are coming in at some angle with some speed such that those are not co-linear. So oblique refers to the nature of approach of the two. However, just as before, as these come in at some instant in time, they'll be in contact during the impact and at that moment then is to find our coordinate system and all angles are in reference to that coordinate system. So we'll have approach velocities with magnitude and direction. We'll have initial approach angles and we'll use our subscript to designate one of the two of them and then there'll be some velocity after collision that we designate with a prime and we do the same thing with the angles. So this would be theta one prime that's the angle of the first object after collision and we have theta two prime there. I will look at both of those in some detail. The deal with these being central collisions is by definition taken care of by the fact that these are circular objects that are hidden so that the instant they're in contact, their centers of mass are both right on the normal axis. So that's the nature of the central impact. It's mostly the direct and the oblique that we need to separate what we've got here. And then there are other impacts that can be looked at that are not central impacts where the objects are of different sizes and shapes but we're not going to go into the detail needed when others were just going to spend one day on these. All right, so we'll step through a direct central impact in some detail and use that to sort of set things up for us as we go through these. So we have two objects moving with independent velocities and could be in any direction but it's necessary of course that v1 be greater than v2 otherwise they'll never impact. So we don't want to study non-impact impacts. They're kind of boring. So we have v1 with a great enough velocity such that it overtakes v2 and this is just the setup. This is the before collision period and we only need to be a split second before that to be able to set up the early part of the problem. Sometime later v1 is caught up to v2 and they're now in some kind of contact with each other where there's enough of a force between the two that there's actually a deformation phase where the objects themselves are actually deformed. At this instant they're moving with some common velocity because they're in contact, they're essentially one object. Some little bit of time later, usually very small, they're then in still in contact but now they're trying to restore their deformed shape. We're using essentially elastic materials, at least somewhat elastic and pool balls and automobiles all have some nature of this where they're still moving with some common velocity but now they're trying to restore their shapes and it's actually this restoration that pushes them apart as they try to reform to their original shapes. This takes place in some very, very short period of time after which they're now restored to their original shape. They each have some subsequent velocity and I'll draw them like that where typically v2 then now is departing from v1 with its greater velocity. So we have those four phases before collision, deformation, restoration, this is known as the morning after. Joey, sometimes followed by an embarrassment, maybe, maybe, maybe then smoke a cigarette. All right, what we're interested in then of course is this collision period there because that's what sets everything up. We assume that the velocities are constant so we don't care anything about this. What we need to look at is the impact itself. During this very, very short collision period the velocities might change significantly. Could certainly be that one is coming in and moving to the right after the collision is moving to the left of vice versa so that can be a significant change in the velocity. We're going to assume that during this time the position change is very little. It's such a short period of time that even at this common velocity which usually is not, might not be all that great. If we have two pull walls coming in towards each other the common velocity could actually be zero and then the position change could be almost nothing. And then we'll also neglect friction and that means any slipping of the surfaces of the two objects over each other as it might happen in either imperfect direct collisions or in any other oblique collection, collisions anyway, we'll also overlook other non-impulsive forces. Those forces that don't contribute to the impulse side of the impulse momentum equation and if you remember those were external type forces. What this means for us then is the momentum of the system is conferred and served. If there are no impulse forces there's no impulse to the side of that side of the equation and thus the impulse, the momentum of the system is conserved. We can use our prime notation on that. Alright so that will help us solve some of it won't help us solve all of what's going on so we need to bring in some other parts here especially as we start to look at the oblique collisions. They're a little bit more involved because more is going on we have more unknowns we're going to need more equations and nothing else. So we're going to focus now on this deformation and restoration phase during deformation. So that's where everything's going on that's where we're going to need to focus to figure out what the deal is that's going on. So let's look at one of the objects here comes in it's colliding now with that other one because of some very large force deforming force due to its collision with the other piece the other object in the problem. So this will be M1 it was moving with some velocity V1 but at this instant is moving with the common velocity of the two as a single unit a single system. Now the impulse this one receives of course will be that deformation force for however much time it acts which is we'll call it on the order of delta t over 2 half of the phase is deformation the other half is the restoration. The only other thing we need to tidy this up is that the force is opposing this common velocity so there'll be a minus sign in there. And that will tell us then what the change in momentum of this first object would be a little simpler it's M1 times V1 prime minus V1 the original V1 oh sorry not V1 prime let's see Vc that's it's it comes in with velocity V1 it acquires velocity Vc during the collision so that's the before and after speed. All right so that's our that's our setup of the deformation. The problem is we don't know much about this over here we don't know what there is to integrate we don't know what the average velocity is we don't necessarily know what that delta t is. That's part of the problem there so we're going to have to address that. So after that it goes into the reformation phase where it reforms to its original I think I called that restoration on Reformation. Restoration. Restoration. I like that word restoration. All right the object is trying to go from its deformed state back to its original state in rebound so I look something like this where it's still moving with the common velocity of the two I do that the wrong side that's object two this is object one we're looking at still moving with some common velocity and still receiving some kind of force it's trying to restore its original shape it's pushing off on the second object so it's the second object pushing back on it so it still looks like that in the same direction and the integral of impulse momentum is going to still look very much the same only we have this restoration force rather than a deformation force and there's nothing to say that those should be equal and so this will be the change in momentum of object one after the or in the second half of the collision where it now acquires its new rebound speed after having gone into the restoration phase with this combined velocity that it started with what we're going to do is compare those two now by defining the coefficient of restitution coefficient of restitution we'll define it as well we'll give it the little symbol little e and define it as the restoration impulse divided by the deformation impulse and compare the two in a lot of the things we looked at in physics one those two would have been equal therefore the momentum was perfectly conserved in a perfectly elastic collisions that's just for the case we'd have but in more realistic collisions we need to look at other parts of them so we have this restoration impulse divided by this deformation impulse both of those integrated over the approximately the same time period but it's easier to work with the momentums so we have the change in momentum after impact divided by the change in momentum before impact I guess since we're only dealing with one here we want one dose as well but we've got more detail on those two since we have these velocities so this would be v1 prime minus vc over vc minus v1 where the m's cancel the masses cancel because it's the same massive of the same object shouldn't be too big a stretch to see that if we did the same analysis for the other object we'd get exactly the same kind of thing this force is coming from the second object and so if we look at the second object we're going to see this very same force same magnitude just directed in the other way and so we would also then go through the second object and get v2 prime minus vc over vc minus v2 since the forces are the same since they're equal in opposite pairs the time they're in contact of course is the same because one is in contact with two at the same amount of time these two impulses are the same these two are definitely then equal it's not uh not necessarily even an approximation we're making anybody see a obvious trouble with this though the trouble is what normal is vc v1 and v2 and v1 prime and v2 prime are very easy to get that we could have done in fact we did if we did the linear air track and physics one which I think I did with some of you maybe I didn't do that with anybody in this class we'll be using the linear air track it's very easy to get those the velocities of the individual objects before and after collision but it's very difficult to know what the velocity is during collision so we need to eliminate that somehow which is fine because we don't really care about it so with a little bit of algebra we should be able to come up with this we can combine these so that the coefficient of restitution is defined as the opposite of v2 prime minus v1 prime over v2 minus v1 or if you remember we had a notation for relative velocities we can write v2 slash v1 prime over v2 slash v1 and those can be quite easily measured um and are not a great experimental difficulty so this coefficient of restitution depends on the two materials no great surprise how a tennis ball two tennis balls would bounce off each other would be very different than how a tennis ball and a golf ball would bounce off each other experimentally determined only you cannot give me two different materials two different balls and I can't come up with what the coefficient of restitution would be I have to run a test like just like friction was if you remember for a perfectly elastic impact and it's also true that energy is conserved in those type of collisions both objects have an initial kinetic energy both have a final kinetic energy you know perfectly elastic collision the there's no loss in energy between the two perfectly inelastic collision this is the one like what happened if one of the objects was a lump of clay and they stuck together then became one single object so I don't know if we should call that perfectly inelastic or imperfectly inelastic but in that case we have a relative velocity between the two of zero in the numerator and then of course he is zero itself no relative velocity between the two after collision because they stick together in a perfectly inelastic collision and in this case the change in kinetic energy is a maximum it's not all lost because they're still quite possible that they're moving after the collision and there would still be some residual kinetic energy but that's the maximum amount that's lost during collision all right so let's do a quick sample problem just to warm up with it if you're today using that so imagine an object there two pounds dropped from a height of six feet onto a 10 pound plate that's sitting immovably on the floor to find out what the coefficient of restitution would be for this ball on that plate whatever those materials are remember it depends upon the two materials we do this test and measure the rebound height of the ball so it drops from six but then rebounds and we'll say it rebounds to five feet so drop it from six feet measure out far it rebounds and from that we can get the coefficient of restitution for these two objects remember what we're concerned with is the velocities just before impact and just after impact those we can get assuming uh freefall neglecting air resistance we can find those two velocities if this is object one and this is object two v one we can get from simple freefall analysis you can use the work energy equation you can just use f equals m a constant called acceleration and you you'd be able to come up with that just comes from one of the constant acceleration equations that v one is dependent upon the height from which it was dropped so that's this part here v two the speed of the plate before collision is zero it's not moving it's just sitting there on the floor v one prime then we get from the rebound height two gh prime and v two prime is still zero and so we can get the coefficient of restitution from the two of those so i want you to go through that just because there's a tricky little bit in and if you're not careful v two prime minus v one prime the minus sign allows us to keep those subscripts in the same order and for us in this problem v two and v two prime are zero simple calculation just to make sure you get something right on it there's one tricky little part in it just want to make sure you get what you got chris you got it david just just depends now you get down to a point where you want to see right here for students in the week tell them doesn't we're getting we're getting some disagreement here all right let's see we're not agreeing what do you get for v one give 19.6 and that's to be for a second i assume and for v one prime not the same sign on those remember this all came from the change in momentum which means we have to take into account direction so we have to know that those two are in opposite directions so we get minus v one prime but there's another minus sign there units are the same so they'll just cancel then we have a minus here and another minus there so that's a minus a minus 19.6 that's four total minuses so it's a positive number something like uh 0.91 or 92 something like that but a positive number coefficient of restitution is a positive number most uh coefficients like these in science are meant to be positive sometimes they artificially add a negative sign to produce that but in this case when you solve for the coefficient of restitution eliminating the constant velocity it comes out that way anyway and allows you to keep these in the same order two minus one two minus one which is just a lot less chance of a mistake same problem with a slight change to it same problem drop it from the same height onto the same plate so two pound ball drop from six feet onto a ten pound plate however the ten pound plate is on uh shocks so now we've already tested the plate we know what the coefficient of restitution is between the ball and the plate just because there's springs on here doesn't mean that changes it's dependent upon the two materials but not dependent upon what the materials are doing themselves so the coefficient of restitution remains the same still dropping it from six feet with this information we should be able to estimate the rebound height of the ball how many unknowns now because we're going to need that many equations how many unknowns we know the coefficient of restitution because we've run that test that involves the velocities to approach velocity to two before collision velocities two after collision velocities the before collision velocity of the ball is not going to change it's still the six foot drop with that ball so that's known what about before collision velocity of the plate it's also known it's just sitting there nothing's happening to it after collision we don't know the rebound velocity there's no reason it should be the same as this in fact it couldn't be because some of the momentum of the ball has been transferred to that of the plate so there's going to be a after collision velocity of the plate so we have two unknowns but right now we only have that one equation the the plate is two so before collision the plate is not moving so coefficient of restitution is known v1 v1 is known because that's no different we don't know either of those two velocities so we need another equation go to equation.com get out your credit card that'd be awesome download immediate download if you're on the internet god that was great it should be easy to be an undergraduate student that was the case what's our other equation what do you think it might but notice i haven't even given you the spring constant now sometimes i forget to give you stuff until later in the problem but not this time it has to do with the collision itself and it's in fact an equation i've already given you yeah the momentum is conserved there are no outside forces there are no outside non-impulsive forces yet gravity is an outside force but it's not an it's an impulsive force not a non-impulsive force our conservative non conservative if you wish so the momentum of the system will be conserved so we can use that as our next part of the equation so let's see a massive one v1 there's the momentum of the ball just before collision with the momentum of the plate just before collision but of course the plate's not moving it has no momentum and they're both going to have some velocity after collision and thus some momentum after collision we don't know either of those so we have two equations and two unknowns now it's just algebra left i don't expect you to go through it so v1 prime turns out to be well it was 17.9 is it going to be greater than less than or equal to that this time with this plate on the springers it's going to be less it's it's no different than if you uh drop this onto it you know if you did this test onto a board on the floor and then put the board on your bed and dropped it it's just not going to rebound this time you know that so it's rebound velocity is minus 6.3 feet per second if if down is positive up as negative i guess we did the opposite that over there so we're sticking with that i'll put take that minus sign off and after velocity or after collision the plate is actually moving down with uh again i got those reversed the ball is 11.7 the plate is the minus 6.3 my minus sign was okay it's the plate's moving down now that we have this 11.7 we can then figure out the rebound height comes out to be something like uh uh something just under two feet so it had to rebound into five now it rebounded to something under two notice though doesn't matter what the spring constants are which is interesting i don't think that's obvious from the start but it's immaterial it would certainly affect how far that plate rebounded but its initial velocity the instant after collision is independent of the springs and spring constants themselves all right any questions before we move on to the oblique central impacts all right things are a little bit different we can use a lot of what we just came up with in the oblique central impacts we just can't use at all and we'll see why in a second so we have two objects moving obliquely at each other with some initial velocities such that they'll collide and do so and the instant of contact might look something like that which defines our cornered directions from which we measure the angles so we have v1 coming in at theta one v2 coming in at theta two they collide and they rebound in some direction prime to prime at some angle theta one prime and theta two prime so let's assume this much assume that the masses are known in fact are we even going to take them to be equal not necessarily no but they're known now of course once we establish everything we can we can do other things later and the problems where maybe those aren't known but all the other stuff is the before collision velocities are also known and since these depend upon the directions they have to be known as full vectors and we'll assume that coefficient of restitution is also known that whatever these objects are in collision we've gone through the test where we let them go through a direct central collision and from that we calculate the coefficient of restitution once we've got that for those two materials we don't need to do it again how many unknowns now i heard two i heard three do i hear three and a half i hear four i hear three or four six why not there's four we don't know the magnitude or direction of either the velocities after collision so we don't know the magnitude or the angles or we don't know the two components in that coordinate system either way we don't know enough but we in fact we don't know anything about the after collision vectors four unknowns v1 prime as a vector v2 prime as a vector that's four unknowns whether you do it magnitude and direction or whether you do it as n component and t component either way we don't know two of those so we need four equations one of the equations is already known it's the coefficient of restitution equation however there's a fairly obvious concern with that i hope in that this is not a vector equation so these are not the vectors it's not even the magnitude what we have to do is realize that this only applies for the velocity components in the normal direction only use the normal components in those in those calculations why is that daily they don't collide in against each other in the tangential direction they only collide on each other in the normal direction so we only have that normal component and we apply it to it so if you if it's a helpful make sure that the ends might even be on the individual velocity components parts themselves what's a another equation we could use David I'm thinking the tangential component of v1 and d2 well there's even a more obvious one than that so let's do that one first and we'll dig a little deeper the momentum of the system is conserved that's not going to help us in the tangential direction so we need to apply it in the normal direction to get any useful information out of it in the tangential direction there are no forces in the tangential direction so there's not going to be any changing momentum in the normal direction tangential direction um tangential momentum is identically zero not not not the tangent the change in tangential momentum is identically zero the momentum's aren't zero but the change in tangential momentum is identically zero there are no forces no no there are no forces in the tangential direction so there's no change in momentum in the tangential direction the tangential impulse is zero so the tangential change in momentum is zero so that's no help to us we have to look at it in the normal direction anyway so we have to set this up so it'd be m1 v1 in the normal direction plus m2 v2 in the normal direction equals m1 v1 prime normal and m2 v2 prime normal yeah normal direction only because the tangential direction is not going to give us any information you calculate the change in tangential momentum you're going to get let's see uh delta g no let's see uh g tangential before is identically equal to d tangential after so you're just not going to get it and they're both both uh i don't know how to put it uh change in is identically zero i don't know they they couldn't be anything else but equal so there's just nothing we can get out of that so that's why we have to do these in the normal direction only because there's normal forces only so there's uh the two magnitudes in the normal direction with the two magnitudes in the normal direction we don't have anything about the tangential direction though we need that to get the full vector flavor of those two after collision vectors we can get the normal components of those two vectors with these we can't get the tangential components with those two so what are we going to do what what uh what could possibly work well this one's not quite as obvious sorry yeah the tangential momentum for each individually is conserved this is for the system in the normal direction for the individual pieces in the tangential direction momentum is also conserved in fact uh when you do that the masses will cancel and that shows exactly what christius said that v one in the tangential direction is equal to v one prime in the tangential direction so those are the easiest ones and it also applies for the second project as well there are no tangential forces on the balls individually as well as collectively which we just used and so there's our four equations and four unknowns if we if we took a look at just the uh see these two here these also mean that e in the tangential direction is identically zero which is why we have to use the normal normal components over here so uh there's no sense looking at the coefficient of restitution in the tangential direction it's identically zero all right i think we need a problem to take us into the weekend all right so here's what's known for this problem and then you have to figure out what's unknown two objects is the same mass just as you would do if you're playing pool v one is 30 feet per second at 30 degrees that's theta one so i guess that's the thing v two is 40 feet per second at 60 degrees nice round numbers like always appears in nature never have any decimal values in real life and coefficient restitution point nine no we've got four equations four unknowns some of them are very easy to use and you can finish in just seconds others are going to take a little bit of time to go through particularly those two tangential equations figure out the tangential velocity components and you know those don't change so somebody should volunteer to do those take the rest of the day off and that will take us into the weekend to get out of class early question which is having pizza on saturday afternoon pleasure to make in question two masses are equal that helps things a bit makes it a little easier if you want to do the two easy ones first just to get warmed up i don't blame you you'll stay here all the way through on the surface just come up with the four unknown component directions no surprise that'll completely define the after collision velocity vectors i'll put up the easy ones because i get to go first individual tangential momentums is conserved the tangential components of both are conserved and doing the trick to figure out what those are 34 6 for the second that's just the tangential components those are the same say what so you can set up now two equations for the two last unknowns the normal components system momentum is conserved in the normal direction it's conserved to in the tangential direction that's not useful to us in finding an out and then use the coefficient of restitution in the normal direction because that's the only direction in which we have any collision it's the normal components that make them collide the tangential the bonus don't make them collide that whole thing tangential components looking for the normal components actually i have the first more interesting Travis now that box you want to be slow so you got three minutes to finish up and get out of class early Chris you going early should be able to tell don't forget your minus sign if you want to pick that as the positive direction then there are minus signs on some of these and that's as important as the values themselves and then you can also tell from your answer if you've got the same the appropriate velocities if that's our positive direction then we should expect a negative normal component on one kind of positive normal component on two don't forget those minus signs as important as the magnitude themselves Phil you got it let me come over this initial velocity is positive that component this initial velocity component there is negative if if you use the same coordinate direction I've got and why not yeah by erasing them oh I did yeah sorry Travis do as I do as I say not as I do okay you can go over one minute extra weekend don't waste it one more minute down at Rowdy's Roadhouse and Tavern Dive Phil yeah no even more significant figures with him you're close is that a component direction don't forget the component direction is all just out of the matter though right as long as you get your minus sign right all right I'm in a good mood I'll give them to you v2 well then don't look v2 normal prime v1 let's see v1 normal prime that's negative minus 17 second if you didn't want to hear that answer don't look if you don't want to hear this answer don't look either 237 but I won't tell you the units that way you won't know what to write down all right as long as you set the equations up it's just algebra that's left you know it's not the point of our class here we gotta get the figures right not necessarily the the algebra just sometimes takes a little time okay