So, let's return just quickly to a few issues from the last example problem we did,

just to highlight where we're going. I'm not going to resolve the problem by any means, but I do want to look at the collision, right? We had a bullet with some velocity v naught that we wanted to find. There was the block of wood with velocity zero. This is our initial, and then at the end, they're combined.

This is our final, and it has some net velocity v final. When I did the problem, I apologize. I did not actually draw this picture, and I had a few questions afterwards, so I wanted to make sure we got the picture drawn.

Because it's a collision, we conserved momentum. So, when we looked at the initial momentum, we had the mass of the bullet times v naught.

That's the momentum of the bullet, plus zero. That was the momentum of the block of the wood because it was not moving. And our final momentum, it was a single object, so it had a combined mass of the two times velocity final.

So, we went from two objects to one object. So, if we have a collision where things stick, that's what we're going to have. We're going to have two objects going to one object, and you want to keep track of that. If we have an explosion or launch in a rocket, we'll do a couple of those today,

then we go the other way. We start with one object, and we end up with two. The other type of collision is where two things hit, and they bounce off each other. And so, they don't stick, so we start with two, and we end up with two. So, our three basic collisions

or explosions are two objects going to one object. When that happens, you will have an inelastic collision.

We will lose kinetic energy, and we will talk about that. We will have one object going to two objects. This is what we refer to as an explosion. And the final one is our two objects going to two objects,

and this will either be elastic or inelastic. And though we defined elastic collision as conserving mechanical energy, almost all the time,

the elastic versus inelastic is a question of whether or not the change in kinetic energy is zero or whether it's not zero. Now, there are a couple of ways, things you want to pay attention to.

So, if two objects stick, then it's always inelastic. And what you would typically do is you use momentum conservation to find your velocities, and then

if you wanted to compute the energy lost, and again keep in mind that this is a strange phrase, which is why I'm putting it in quotes, right? The energy lost is not really lost. It goes into other types of energy. It's almost always kinetic energy lost, and I don't know why we don't just say that,

but I'm using the common phrase that everybody else uses. Notice it's also energy lost, so it's not change in energy. Okay? If your kinetic energy decreases, that will be a positive energy lost. So, it's kind of a weird thing, okay? So, it's often the initial minus the final

because it's how much you lost, not what the change was. But you would take these velocities and just plug into whatever you had initially and subtract whatever you have at the end, and often at the end it might be some object that's combined.

This is just an example. It's not a formula. You might have an initial single object, so this is one object moving, sticking, and both together. You might have had two objects at the beginning. Then you would have 1 half mv, m1vi squared plus 1 half m2v2i squared,

and you would subtract from it whatever you have at the end. Any questions on that? Now, if you have, say, a more common one, two objects, mass A,

mass B, they collide, maybe they bounce off each other, so this is my before, and this is my after,

now there's kind of two possibilities. The problem may tell you it is elastic or inelastic, and then you know kinetic energy is conserved or it's not conserved. That's the easy one. It might not say anything, but it might ask you, is it elastic?

How much energy is lost?

A danger there, too often students just see two objects bouncing off, assume it's elastic, solve the problem using that, and since you've assumed it, you force it to be elastic, you get an energy loss of zero when, in fact, it wasn't elastic and energy was lost,

and you'll get the problem wrong. So if they're asking you, is it elastic or how much energy is lost, you cannot assume it's elastic. That means there must be enough information to only use that momentum's conserved, and then you find these and do what we did before

and then compute your change in kinetic energy, and using that you would then give your answer. Either it's zero, in which case it's elastic, or it's not zero, in which case energy is lost. So those are the scenarios in collisions, and we're going to be going through some of those today.

Questions on that? Yes? Say again? So this type of question might be on Friday's exam. It's in the homework. Everything I do today might be on Friday's exam

because this also is included in the homework that's due Wednesday. Everything on Wednesday is kind of review. So pieces of that, anything that reviews Chapter 7, 8, and 9 might be on Friday. Everything else is for the final. This leads us to an easy warm-up clicker question.

So this is referring to the block problem, okay? So this is the bullet hitting the block, the two of them sticking together, and then moving.

The question is, how would you find the kinetic energy lost in this collision? There is no energy lost because it's always conserved. Subtract the final initial kinetic energy, find the force from the impulse, and compute the work. The problem cannot be done. Excellent. Seventy percent of you got it right. This is exactly what you do. When you have a collision, particularly where they stick

and you know it's an elastic, finding the kinetic energy lost is a matter of you've solved for both velocities, which you can do using momentum conservation. You then plug it in for kinetic energy before and kinetic energy after, and look at how much is lost. So it's fairly straightforward. So now let's take a little segue.

We did a bunch of collisions. The other thing that can happen, right, is I have an explosion. And a standard sort of explosion is a rocket ship launching, and then you launch the rocket ship and it goes up, and the fuel goes down.

And the before situation is you have the mass of the rocket plus the mass of the fuel. And typically, you have a velocity equals zero. It's just sitting there. So your initial momentum is zero. After, you have the mass of the rocket times its velocity,

and the mass of the fuel times its velocity because they're separate. And so we can say p initial equals zero equals p final, which is the mass of the rocket times the velocity rocket

plus mass of the fuel times velocity of the fuel. So we get the fact that they must move in opposite directions and depending on the mass of the fuel relative to the rocket and the speed of the fuel, that gives me the speed of the rocket.

Pretty straightforward. Now, notice this is a simplified version because in a typical launch of a rocket, if I'm going vertical, what is the one other thing I do have to worry about? What else is around? Gravity. And typically, is that sort

of launch instantaneous or does it take time? It takes time. So there's a lot of fuel coming out over a finite amount of time. So depending on where you want to know the velocity, if you want to know it, say, here's our rocket again

on the ground and then here's our rocket right up in the air and all the fuel is gone and you know how fast it's going, right? In the absence of gravity, you could figure out what the speed of the rocket is and then you can ask, okay,

I know mg of the rocket and it was going up in the air for some distance delta x and you could figure out the impact of gravity on it and recalculate the new speed. If you want to find it at some time in between, this is real, you really have to worry about things

like the impulse being an integrated force dt and work out the details. So or there's other ways you can do it that get even more complicated using fun differential equations. But if you're basically looking at two key instance or if you're launching something horizontally

where gravity doesn't matter, it works. So we can demonstrate this by using our cool rocket which is going to have a lot of air under high pressure. The air will shoot out the bottom, that's our fuel and our rocket will launch. We all ready?

Assuming I don't get too tired. Trust me, it really does launch. I've been assured of this. Now I'm starting to get nervous, particularly

since the bike pump is getting very hot. Whee! Oh, there it goes, whew, okay. That was definitely the hardest demo we've done all year.

Okay, now what's nice about that is you, whoa, well there comes all the dust, you really get to see that the launching of the rocket is just conservation of momentum and one of the key things is, right,

in this case though that rocket has a bunch of air in it, we'll do it in terms of magnitude now, right, because we know they went in opposite directions. The speed of the rocket is really given by that.

Notice this is often much greater than the mass of the fuel, right. I mean the fuel is just air and I mean air is pretty light in the rocket, I mean this isn't the heaviest of rockets but there's enough plastic here, it's heavier. And so what it really comes down to is getting the velocity of the fuel to be quite large

and that's why we really pumped it up to a high pressure. Now if you switched, and how many of you have ever done this with water, made a water bottle rocket? Most people have if they've been in like, you know, Girl Scouts, Boy Scouts or various other things. Yeah, it usually happens there, okay, if you did Boy Scouts

and you get into a water bottle rocket, go back and complain. With water, you really increase the mass, right, of your fuel a fair amount and then you can get some really high launches, it goes quite well. We can also do this, same basic idea, horizontally with a cannon, okay.

So of course we have to ask a clicker question about our cannon just to make sure we're all on the same page. So right here, we have our cannon ball and here we have our cannon. We'll load our cannon up and then we'll fire it

and the question is when you fire it, does the cannon stay in place? Does the cannon itself moves backwards? Does the cannon itself moves forward or will it depend on how fast this object is launched, whether it moves forward, backward or stays in place? So go ahead and vote. Excellent. So the cannon's got to move backwards.

This is our basic recoil problem. So just like with any good, you know, Civil War era cannon, we load it with liquid nitrogen. Having primed the cannon, we then have

to get the cannon ball in.

That was fun, okay, heavier cannon ball

and maybe we use a little more liquid nitrogen.

We need a quick volunteer to hold the back end of my cannon here. Come on down, because I was starting to freeze my hand off.

Liquid nitrogen, but there you go. So the cannon clearly moves backwards at least. I will say the best time I ever did this and I haven't been able to repeat it, is I got enough liquid nitrogen, got that far enough in, they don't like me to do this, that the top froze on the bottle.

So when it launched, the water sprayed everywhere. That was kind of cool. And it gave it a little extra kick as the water came flying out the back. But, man, I am out of shape. Those are tiring demos. So there you go. That's the rocket. But it's important to look at this in a little bit of detail.

So if you see this on a problem, you don't make some of the common errors. So what I want you to do, here's the cannon, right, and here's the cannonball. Let's suppose this whole thing is up 30 degrees from the horizontal.

I want you to tell me, okay, figure out quickly on a piece of paper how you might figure out what the speed of the cannonball is if you know how fast the cannon is going backwards. So see how you would figure out how fast the cannonball is going if you could figure

out how fast the cannon is moving backwards and whether or not there's any problem with that calculation. So try that for a few minutes here. So another way to think of it, you're given the mass of the cannonball, the mass of the cannon, and you're given the velocity of the cannon

after launch backwards, okay? Question?

And explosions, we'll mention that. Notice, consider the time of the explosion to be basically zero. Let me ask you this, has anyone come across a problem with this problem?

You should have a problem with this problem. No, no, you just needed the velocity, turns out, but there's something else that's going on here. What direction is the cannonball traveling when the collision is over? Up at what?

30 degrees. When the collision's over, what direction is the cannon traveling? Straight back in this fairly realistic problem. And here's where we run into trouble, okay? Certainly delta P, which equals J, which is F average delta T, will be zero if delta T is small enough.

The problem is, when I'm sitting on a rigid surface here with my normal force, unless I tell you the actual recoil velocity of the cannon at 30 degrees, you're not going to be able to solve the problem.

Because what's happened is by telling you the cannon's going horizontally, I haven't really made delta T zero because this normal force kept it from going into the floor. So this is one of those tricky cases where you have to pay attention to the information that you're given,

because it is sometimes done this way in problems, where this is really sitting on the floor, so even though delta T is very, very small, it's never exactly zero, and the normal force can act very quickly, so you never see the velocity down. You only saw the horizontal recoil velocity. So from a recoil problem,

you could only get the horizontal momentum. So let's just do that real quickly, because what happens is I need to worry about X and Y, right? And in my initial state, V equals zero, so P total is zero,

okay, in my final state, I have M of the cannon times the velocity of the cannon, and I have the cannonball times the velocity of the cannonball, but now in my XY coordinate system,

the velocity of the cannonball is that way, and if I haven't given you the true cannon velocity, if I just give you that, right, when I go to write down my X and my Y equation in the X direction, I'm okay.

I have zero equals mass of the cannon times the X component of the velocity of the cannon plus mass of the cannonball times its X component, but in the Y direction, I have zero equals some unknown number plus mass of the cannonball times velocity of the cannonball in the Y, and it's this that makes conservation

of momentum not useful for this problem, even though it looks exactly like one. So either I would have to tell you the cannon recoils down at 30 degrees with some velocity, then you would know that and you'd be okay, right, or I tell you something

about F and delta T to get the impulse, and you actually focus just on the cannonball and use F on the cannonball delta T equals delta P of the cannonball.

So that's where you have to make decisions. Yes? No, because you really, this problem and the information I gave you gives you no information about the vertical recoil of the cannon.

You truly have three unknown things. You have the cannon's velocity in the Y direction and then the cannonball's velocity in the Y in X direction. So there's really nothing that allows you to actually get any further on this problem. If I'd given you some information, say, about friction or about the normal force and the time

of the collision, then you could do stuff. Then you would have things you could work with. But with just this information, you're stuck. And that was just kind of a warning problem. Now, very commonly, though, you might have a slightly different problem.

You might have been told where the cannonball lands. Now, in this case, and you might be asked for the X recoil of the cannon.

Now, notice you have enough information to find the cannonball velocity because you can do the projectile motion problem. And when I have the cannonball velocity, since I have both X and Y by definition, I can do that problem.

I now have enough information to go in that direction. So there I would use conservation of momentum because I am allowed to conserve momentum in just one direction. Even though it wasn't conserved in the Y direction because of the normal force, or I don't know enough about it, that's okay. I can still find the X by using conservation of momentum.

I know these get a little tricky, and that's why I wanted to kind of point this out. Hopefully it will clarify as you do the problems not confused. Questions? Yes. Right. So here what you did in this problem,

if I told you a cannonball was launched from a cannon, went 200 meters, landed at the same height as the cannon, it was launched at 30 degrees, what was the recoil velocity of the cannon in the X direction? Then you would use projectile motion to solve for that, and then you would use conservation of momentum in the X direction to solve for that.

So there, notice the difference. Here, these were both unknown because I gave you no other information, okay, and this was unknown. Because you really, the zero is not the actual velocity

of the cannon in the Y direction after the launch. It is actually moving down, it just gets stopped real quickly by the ground. So by only measuring its X component, you're missing a piece of information, its Y component. If what you want to do is solve for the full velocity of the cannonball. Notice, if I give you this, I can find that.

I just have no idea what that is. This is the one I can't find, okay? Yes. Right. If you know the X component

and you know the angle it's supposed to recoil at, why is that, why is using trig not enough to find this? Because you don't know the total. You only have one side of the triangle and the angle. You need two sides of the triangle to use trig. Yes, you do. Sine is opposite over hypotenuse, cosine,

I mean sine is adjacent over, wait, no. Sine is opposite over hypotenuse, cosine is adjacent over hypotenuse. Pythagorean theorem has three sides squared. All of my formulas with my triangle involve three numbers.

Oh, shoot, I gave you the angle, didn't I? Oh, I am being so silly. Does that really work? Yeah, I guess that really works. Dang. Hell, I still have my A, but yes, that really works.

Okay. I thought I cleverly came up with an example that would confuse you and now I have thoroughly confused you. A good way to end the quarter. Thanks, everyone's saying, ah, well you got three days to recover from my confusion. But, however, I would like to say from momentum conservation my point still holds.

So if I hadn't given you the angle, you would still be in trouble. And that's, if you notice, yes, I had given you the angle, which was bad. But in this conversation, I totally ignored the fact that I'd given you the angle and that's kind of what I meant. I apologize, that was really bad of me. So, everything I said up to now was true

if you didn't know the angle. When you watch the video, it'll all make sense because you can play it backwards from here. Any other questions that don't show I'm wrong? I'm impressed I made it this far. Any other questions?

I may have to just go back and edit that part of the lecture so I look more intelligent. Excellent. Now, what we want to do next, having done this, is what I consider algebraically the hardest part of the course

and really I don't do a lot with it on the exams. But it's important for you to see it and that is what do you really end up doing with elastic collisions. And here's the challenge why it's algebraically hard. In an elastic collision, now, notice you have to be careful,

listen very carefully, in an elastic collision as opposed to an inelastic collision, yeah, it's almost worse than molar and molal, molarity and molality which I can't either say. I will have an object A and it'll have some velocity A,

VA initially, I will have an object B and it will have some velocity VB initially, they will collide, all sorts of things can happen. I'm going to draw one possible case, velocity A,

object A will have some final velocity and velocity object B will have some final velocity. So this is before and this is after. Notice I drew them both going to the right afterwards, that can happen, they can both collide and be going opposite directions after,

they can both be collide and be going to the left after. However, this is one case in which we are dealing with reality, what will not happen after the collision is that object A will be to the right of object B and going faster. Okay, objects in the real world do not go through each other unless they shatter it.

So if you have a collision where object A goes through object B, a bullet through a card, a bullet through an apple, a bullet through a block of wood, that will never be an elastic collision, that will be an inelastic collision because you've destroyed stuff, you've lost kinetic energy

in the destroying of stuff. So when I'm talking about elastic collisions, I can never have that on the other side of that. Notice, any questions on that? Okay, notice I have four unknowns.

Now, I have basically two equations, it's an elastic collision, so the change in kinetic energy is 0 and notice how many variables I have. Lots of room here to make a typo

after the whole angle debacle, hopefully we won't. Now, the textbook uses primes for final velocity. I prefer to actually write the I and the F because I don't know,

I get lost with the primes, that's just me. So, that's the kinetic energy formula, it's conserved, potential energy is conserved, I mean potential. Momentum is conserved because it's a collision, keep in mind these are vectors,

you can see how boring this is already as you fall asleep while I write. There, so I have four unknowns, two equations. So in this collision, if I give you two of the unknowns which is usually what happens, you can solve for the other two.

Now, if I'm doing this in two dimensions, so if I have a collision where say I'm shooting pool and the pool ball comes in, hits another pool ball and they go off that way and that way, then I need X and Y

and for momentum I need to use components. So I do gain another equation but I'm dealing with X and Y components as unknowns as well. So it's fundamentally the same issue. It's roughly two unknowns, two equations with four potential unknowns.

So I have to give you enough information and what makes it worse is the algebra is very nasty because these are quadratic. So they tend to be long, they tend to be ones you have to be really, really careful with your algebra. There are some homework ones where I have you practice these full nasty ones.

What I want you though to focus on for the exam is really the special case of 1D because if I'm moving only in one dimension, all I have to worry about is say that X component.

And now life is just a little bit easier because if I'm clever, first of all, it always helps to quickly multiply this equation by two because everything has a one-half, then you don't have to deal with your one-halves.

So if I write down my two equations and I'm doing it in 1D, what I really have is I can rearrange things. So if I want all the mass of A on one side, the momentum equation looks like this and notice the difference.

What's the difference between the left-hand side and the right-hand side? Anyone, it's actually important. On the left, which is first?

Initial, and on the right, which is first? Final. That's just because I chose to put the As on the left and the Bs on the right. Nothing special, but that's important. The minus signs are different. Same thing with my kinetic energy.

Now, can anyone tell me why that might be useful, what I just did? There's two things you should notice about that. By writing the equations that way, I've gained two things. What have I gained?

We did this way back with some of our kinematics problems for solving them, and I purposely wrote them right under each other. What can I now do with those two equations and get rid of? I can divide them, and what goes away? The masses, right?

So if I divide the bottom by the top, the left side, I can cancel my MAs, and on the right side, I can cancel my MBs. Now, how many of you remember your cool factoring formulas from Algebra 1? How many remember what A squared minus B squared is?

Yeah, you can admit it in public. It's cool. Being a math geek is now cool. It was just announced, sorry. What is A squared minus B squared? A lot of whispering. A plus B times A minus B, okay?

And the same thing over here. So notice, when I divide, I get to cancel that, and oops, I don't know why I wrote a B there. See, I told you there would be a typo, and I get to cancel that. So I actually get a new equation,

which if I reorganize as before, so this is one of the few equations that's not on my formula sheet, but it is useful to know. It's not on the formula sheet because it's an incredibly specific situation.

If I have an elastic collision in one direction, one dimension, I get to use this. This is the equivalent of conserving my kinetic energy, and then I also get to use my momentum equation that I had before.

Keep in mind that everything in here has an actual direction now, right? So it might be plus or minus, okay? But now I have two equations and two unknowns that are much easier equations to use because they're not quadratic. So these are the ones you would actually want to use

if it was elastic and one dimension. So it's a rare moment where there is an extra equation that's worth knowing. Any questions on that? Yeah. So all I did here was I took my kinetic energy

and my momentum equations, right? And since I had a difference of two squares, I factored it, and then I took this equation and divided by that equation, so the minuses and the masses canceled out, and all I was left with were the pluses, and that was my error there. That's a plus two, okay?

It's a pretty easy equation to use. I'm not really going to do any examples with it. This is a very small part of it, but you do need to know it. Now, before you pack up, because we still have five minutes, as a preview to our review of the class on Wednesday, as a preview to studying for the final, it is very important

as you go back over everything to keep in mind, I think, the following questions. Now that you've seen every part of the course, you want to be able to think in terms of motions.

Somebody name for me some motions we've used in the course. Well, kinematics is a physics. We've used it most often with what motion? Projectile motion. What's another motion that we've had? Circular motion.

What's another motion that we've had? Now, some of these I haven't used the name before, but I'm just, collision and explosions, and there's one more I never gave a name to, so I'm just going to tell you. It's all our motion along a surface.

Right? Down a ramp, up a ramp, along a flat horizontal plane. All of our motions on surfaces, which either had friction or were frictionless. If you can identify the motion, that's a big step towards knowing what you need

to do to solve the problem. On the other side, some physics we used was constant acceleration versus centripetal acceleration. Right? Two different types of accelerations, which is basically kinematics. Kinematics was describing motion.

The other thing we did, and keep in mind, this had to do with situations that were instantaneous. Instantane, yeah, we'll just stop there, was we used Newton's law, F equals MA, because it just had

to do with the instant. The other physics we used a lot was when we were worried about a before and an after. We had some change in the state of the system, okay? We took some version of work, change in energy, or momentum impulse, and these became our conservation laws

when appropriate. And really, on this one page, if you add in all the definitions that we needed to make this work, and if you think about it, right here we basically had three definitions, position,

velocity, acceleration. Right here, we had, I don't know, I'm trying to think how many forces we had. You know, we had four or five definitions for all the different forces. Energy, we had four energies, and momentum we have one definition, what momentum is. That's kind of those two sheets reorganized

in another useful way. And as you review problems, as you study, keep in mind the challenge for the final is to be able to put two or three of these together in one problem, and that's why recognizing what you did at each individual problem, what were the key words? When I said a lecture, we know it's a this type of problem because of this, that's going to really help you a lot.

So Wednesday, a great review lecture with a lot of cool demos, Friday your test, have a good week.