 I'm here at the Space Science Center in Houston for some literal rocket science. Now, we're not going to get into all the details of rocket science in this course, but I will say this. Rocket science is fundamentally based on the idea that to go someplace, you have to leave something behind as the line goes from the movie Interstellar. And in fact, what they're talking about is momentum. And momentum is going to be the subject of what we talk about next. Particularly, it's changes in momentum and generating those changes that can get you from point A to point B, even if you have nothing to push against. So let's do some rocket science here at the Space Science Center in Houston. We're going to learn about the last major significant fundamental concept that we're going to encounter in this course. And that is the concept of momentum. The key ideas that we will explore in this section of the class are as follows. First, we will define what is the momentum of an object. We will compare and contrast that with the energy concept that we have just concluded learning about. And we're going to use momentum to revisit Newton's second law of motion to understand the implications from the laws of motion for this quantity of a physical body momentum. We will learn about the relationship between momentum and force. And we will learn about momentum as a conserved quantity of nature. And finally, we will look at applications of momentum when velocity is changing and when mass is changing. And this will be a new angle on looking at motion in the context of this class. So we've been exploring energy and we've been exploring motion and we've seen how these concepts are intertwined. So what could possibly be missing? Well, before we look at the history of the idea of momentum let's kind of pause and use this moment to reflect on what we have learned about motion so far in this course. First, motion can be described using the framework of space and time. A state of motion is described by the velocity of an object. So if you want to talk about the state of motion of an object you should be talking about its velocity. Velocity is the time rate of change of position, the change in position with respect to the change in time. And here I write the instantaneous quantity of velocity to remind you of the role that calculus has played in helping us to understand at instance of time what the motion is. Now changes in that state of motion, changes in velocity are given by accelerations. And we write that as a vector A which is the time rate of change of velocity and using calculus one can rewrite this as the second derivative of position. That is the time rate of change of the time rate of change of position. Now it is force that causes acceleration. We explored this in the context of Newton's laws and encountered the first laws of nature that you typically see in a class like introductory physics. And then we began to explore what is traded from one body to another or lost from one thing and given up someplace else as states of motion begin to change. We considered gravity and springs and friction and drag and what is and isn't associated as quantities of energy with each of those things. We've certainly seen that motion has an associated energy and that we call kinetic energy. And it's given by this formula K equals one half times the mass times the speed squared. And we've contrasted energy concepts with force concepts. Energy is a scalar quantity and when only conservative forces are in play in a situation mechanical energy, the sum of kinetic and potential energy is conserved. In general all energy is conserved in a closed and isolated system. Even when non-conservative forces are present. And we have seen how mechanical energy can be traded into thermal energy or internal energy in a system that isn't given back. We can determine all information about the degree of motion of an object by applying conservation of energy. And in addition we saw that in the situation where one has conservative forces, it is possible to relate force and changes in potential energy by looking at how the potential energy of a system changes with spatial position. And this is represented by the vector f and its relationship to the negative of the so-called gradient of potential energy u. But changes in the degree of motion. What we're talking about there is changes in magnitude, length, sort of size of motion. But we've lost directional information. If I talk about kinetic energy, there is no direction that the kinetic energy points. If a car is moving east at 55 miles an hour and that same car turns around and goes west at 55 miles an hour, it has identical kinetic energy in both cases. And yet something has changed. It's direction and therefore its velocity have changed. Is there any quantity, maybe even one that could be conserved like energy, but that somehow provides us with this crucial missing directional information when internal quantities like energy change. The answer will be that momentum is the thing that we seek. Now momentum is in many ways a far older idea than energy, at least energy as we think of it now in a modern context. It's impossible to do a fair treatment of the philosophy and mathematics and eventually physics that leads us to a current modern understanding of momentum, but I think it's worth giving at least a couple of highlights here to show you the diversity of thought that has been brought to bear on this subject. While it is certainly true that ancient philosophers including some from Greece are recorded as having thought about the relationship between motion and mass and the sort of degree of force that this can then impose on something by colliding with it, in many ways it's the individual depicted here on the left, Abnesina, also known in more Latin or Greek founded cultures as Avacena who lived from 980 to 1037 and was many things. He was a polymath for sure, that meant he did many things and he did them very well. He was a physician, he was a philosopher, he was a mathematician, he was an astronomer and this would have been in now what we call Uzbekistan where he was born or Iran where he died. It was he that hit on the right idea although like many people prior to the renaissance that occurred in Europe much later than this it didn't necessarily involve doing experimentation to verify thought. Something could be logically or philosophically true without necessarily being physically true. He hit on the idea that a body whose state of motion is changed must obtain or perhaps in slowing down give up some persistent quantity of motion. And what distinguished him from the earlier Greek thinkers on all of this was that the earlier Greek thinkers had the misconception that the state of motion preferred by objects, material bodies in the universe was one of rest. But of course we know from having looked more carefully at nature and then visited that observational set in the context of Newton's laws that in fact the natural state of motion in the universe is one of constant velocity changed only by the action of an external force. And it was in fact Abnusina who hit on this key idea although it's not clear that he did experiments to actually check this. He hit on the idea that there is a persistent quantity of motion and it can be removed only by resistance to that motion in external force. He wrote these ideas down. These ideas were certainly widely read outside of the culture in which he lived. Others certainly would have read this idea and certainly would have had their thinking influenced by such ideas later. He was renowned as a remarkable physician and philosopher and would have influenced absolutely the people that came after them that would have had access for instance to his published works. So skipping far forward in history in that intermediate time between Abnusina's work and much later say in the context of Galileo or Newton's lives many would argue about this quantity of motion. Philosophers would call it things like impetus and they would correctly surmise although they would do so without experimentation that it was related to the product of weight and speed. René Descartes would argue that it was independent of direction depending only on speed. Now we understand that what probably he was really more talking about although again it's not clear that anyone was making a distinction between these things at the time was what we now call energy, kinetic energy. I mean kinetic energy certainly is independent of direction depending only on speed and mass. Galileo would pick up on this idea in his own work calling it impetus. Gaffrey Leibniz whom we've mentioned before in the context of the discovery of calculus would argue that velocity and not speed was at the heart of this quantity. So Leibniz was one who in that same kind of turbulent era of thinking as science was developing into a forefront way of knowing about the universe. Leibniz would understand or at least argue that it was direction and magnitude that played a role in this internal quantity of motion. But it was finally Newton who ultimately included all of this in a really complete mathematical treatment. So most famously of course the Principia and found its change to be related to force. And this is the key thing we will visit in this lecture because whatever this thing is, this thing we call momentum, its change has a relationship to force. And it would ultimately come to be understood as the product of quantitative material, mass, inertia for instance, with velocity, directional information about the change of space with respect to time. So let's define momentum. Momentum is a relatively friendly and approachable thing. Although it can get quite complicated as we begin to think about how changing momentum can actually work in real physical situations. Momentum is defined as follows. Building on these development of ideas, culminating in the mathematical ideas of Newton and those that would follow, we define momentum with the symbol p, a lower case letter p, for instance when we're talking about individual particles, it's a vector quantity, so it has a vector hat over it. And it's equal to inertial mass m times velocity v, including directional information. So momentum p is a vector, v is a vector. Both of these sides of these equations make sense. And of course we can relate v to changes in position with respect to time by putting in the first derivative of a position vector r that starts at the origin and points to where the object is currently, the change in that vector with respect to time. We can see that this has both magnitude and direction. Its magnitude is the product of mass and speed. Its direction is given by the direction of the velocity vector. It is a vector quantity. So just as an exercise, let's do a little comparison for the same physical object in the same state of motion. Let's compare its kinetic energy and its momentum. So I like to think about an electric car. I own one, so I'm a little biased in that regard. So let's say we have an electric car with a mass of 1,700 kilograms and it's traveling eastward at 55 miles an hour, kind of that little example I gave you before, but let's put a mass into this now. Okay, well we can convert 55 miles per hour to meters per second. It's about 24.6 meters per second or thereabouts. So great, we've got everything in MKS units and we can now go ahead and do things like calculate kinetic energy in MKS units. So the kinetic energy of the car, k subscript car, is one-half times its mass times its speed squared. So one-half times 1,700 kilograms times the quantity of 24.6 meters per second all squared. And you should find that this is equal to 510,000 joules, about half a million joules. That's a lot of kinetic energy. We've explored relating energy or in quantities of energy to something a bit more physical, something we can relate to like heating up water if you've ever made coffee or tea, ever made soup, anything that involves having to bring water to a boil or near the boiling point. You know it takes time, it takes a lot of energy and it takes time to get water to march up to its boiling point. And that's because water has a very high specific heat capacity. It's about four joules required to raise one gram of water, one degree Celsius. So this is 510,000 joules if we had some process for turning the kinetic energy of a car into pure heat energy that we could put into water, this would really heat up water very quickly. Now what's the momentum of the car? So let's assume, let's define, that eastward is along the x direction in the positive x direction. So then we can write down the momentum of the car. The momentum of the car is the product of the mass and the speed and the direction in which the car is moving, which is the positive i hat direction. So I need only put in m and v and I'll get a number out of this. So this will help us to understand the units of momentum, which I've always found a little bit awkward, but it will also give us a sense of comparison between these things. So in terms of momentum, the car has a numerical quantity of momentum of 42,000. Now what are the units on that? Well, they're kilograms times meters divided by second squared, mass times speed, kilograms meters per second. All right, so we see that these are both big numbers. Well, great. What does that mean? It's moving in the positive i hat direction, 42,000 kilograms meters per second. How can we humanize this? We'll come back to that question in a little bit, especially when we start thinking about changes in momentum, changes in the state of either velocity or the quantity of mass, and what that can do to a physical system, perhaps a car carrying a person, for instance. Before we do that, though, let's expand our thinking to more than just one particle-like object. You know, a moment ago I was essentially engaging in that old exercise from the first part of this course where I take a car and I think about all its mass being compressed at a point. And we now know from our previous look at the center of mass that that's okay as long as I compress all the mass to the center of mass. I can think about the center of mass of all the Avogadro's number of worth of atoms in the electric car all moving together at once. So on the heels of our look at how to describe collections of particles or extended material bodies, we can then also determine the total momentum of a large collection of particles. It would just be obtained by taking one particle, getting its momentum, taking another particle and getting its momentum and adding those up, taking a third particle and doing the same thing, getting the sum and doing this for all n particles in the system. Now in a car, n is about Avogadro's number or so, but maybe it's only three or four particles that we're having to worry about here. Maybe it's a bunch of pool balls on a pool table, a game of billiards, and you just have to track the motions of the balls, treating each ball like an individual particle. Fine. You sum them up and in mathematical summation notation we have the sum from particle label i equals one to n of the vector pi for each particle. Now just to sneak ahead, if it were true that momentum can be a conserved quantity, well this would be quite useful because whatever momentum a system began with, we could isolate it and close it off and then know that once we've achieved that the momentum inside the system does not change, then we would know exactly the momentum the system would end with. Okay, that's a potentially useful thing, but we're not quite there yet. So before we get into that, let's consider what it means for momentum to change. So let's go back again to the comfortable example of a single particle with some mass m. What does it mean if it's momentum changes? So let's consider a small change in momentum, which I'll denote dp, meaning tiny change in momentum, in some equally small amount of time, dt, meaning a tiny change in time. So let's fix the mass of the particle m so that it can never change. This particle is indivisible. It can be sped up, it can be slowed down, but you can't break its mass up into any other pieces. Fine. So it's an indivisible particle. m is a fixed, it's a constant. It doesn't change in time. Okay, well, in that case, I can start thinking about time rate of change of momentum. To do that, I would write down the first derivative with respect to time of the momentum vector. Well, plugging in the definition of momentum for a single particle, this is the time rate of change of the product of the mass and the velocity. But the mass is a constant. It doesn't change in time. And so using the rules of calculus, we can take that constant, pull it out of the action of the derivative, move it comfortably off to the left-hand side, and deal with what's going on in the velocity vector. So let's do that. I've moved mass off to the left side, I still have the first derivative with respect to time, and now I've made another substitution. I've put in the fact that velocity is itself the first derivative with respect to time of position. Well, the derivative with respect to time of the derivative with respect to time of position is this very definition of the second derivative of the position with respect to time. The derivative of the derivative is the second derivative. So I can write this more compactly as m times the second derivative of r with respect to time. Well, this is a wonderful tangle I've created for myself. What does it mean to have the second derivative of coordinate location with respect to time? Why? That's the very definition of acceleration. What I've written here is that the time rate of change of momentum is ultimately equal to mass times acceleration. The second derivative of position with respect to time is acceleration. So what we have learned is that this quantity here is actually equal to mass times acceleration, which from Newton's second law of motion must be equal to the sum of the forces acting on this particle. So we have learned that changes in momentum with respect to time are related to net forces acting on this particle. Newton's second law of motion has just been totally recast in terms of the momentum concept, and in fact this was something that Newton himself had done. This is in fact profound. Momentum should only ever change under the influence of a net external force. That, this equation tells us, is what makes momentum change. This internal quantity that has both direction and magnitude carried by a body is only altered by forces. So let's reconsider a system of particles. Given this recasting of Newton's second law, we can relate the changes in momentum, the center of mass, and external forces on a system of particles. It must be true that if I consider changes in the momentum of an entire system of particles, the sum of all the individual momentum with respect to time, that that must be equal in the end to the total mass of the system times the second derivative with respect to time of the position of the center of mass of the system. This is just an extension of the center of mass concepts from the previous material in this course. So for a complicated system of particles, all of which may be moving around, colliding with each other, we need only determine the acceleration of the center of mass of the system, whose total mass is m, which is the sum of all the masses of the things that make up the system. Okay, well, this must then need to be equal to the sum of all the external forces on the system. After all, f equals ma tells us that acceleration, even of a system of particles, would be caused by a whole bunch of external forces adding up in some way to act on the system. We'll come back to this in a bit because this bears on the conservation of momentum, but before we do that, I want to humanize the concept of momentum a little bit more. And I want to talk about changes in momentum and another concept, it's only a small extension of what we've been doing already called impulse. Now, we are pretty familiar with the world around us and thus the everyday idea that when two objects collide with each other, changes in the state of motion are possible. Imagine bouncing a basketball on a basketball court. You throw the ball at the floor, that's what you're doing when you're dribbling. The ball hits the floor and it rebounds back up toward your hand. You catch it with your hand again and you push it back down to the floor. You are changing the momentum of that ball constantly. Your hand changes the momentum vector to point down. The floor changes the momentum vector when it collides with the ball to point up and you repeat and that's dribbling. A softball bat makes contact with a softball and the direction and the magnitude of the ball's velocity is changed. Also, if you've ever been a batter yourself, either in cricket or baseball, any sport where you have to strike something with a stick, field hockey is another good example. You know that striking that object also affects the motion of the bat or the stick. So you also have a sense that it's not just you putting energy into changing the state of motion of the ball, but the action of the ball on the stick also affects the state of motion of the stick. We know, therefore, because direction and magnitude of velocity can both be changing in these situations, there must be some acceleration present and that implies a force. So let's see if we can understand the force associated with these kinds of changes in momentum. So let's define this thing called impulse. Impulse is simply a statement of the change in momentum between two times. So again, think about striking a softball with a softball bat. The ball is heading toward the bat. The two meet. The ball then rebounds off the bat even as the bat continues to follow through on the swing. A good batter knows to follow through on the swing and not stop when they hit the ball. Give the momentum as much of a chance to work its change on the ball as possible. The ball is incoming toward the bat before the strike and after the collision it's outgoing away from the bat. There's a change in momentum and it happens very quickly. It's within a second of time that that change occurs. So let's define a quantity that's associated with that change. So if I take the momentum at a later time and I subtract off the momentum at an earlier time, I can write this as delta p. This is the same delta. We've been writing this whole time, delta x, delta t, delta e, all of this stuff. Let's define that as this quantity j, a vector, which is defined as impulse. Now, it would be true that the larger the change in momentum, the larger the impulse. That's because of the definition here. So let's consider the consequences that such changes, especially large ones, could have in light of the connection between momentum changes, dp dt, and force f. So let's begin by considering small changes in momentum over very short periods of time. We can relate impulse to the time-dependent force that could be causing accelerations, and we can do all of this using Newton's second law, recast using momentum. So let me start by writing down Newton's second law in terms of force and momentum. dp dt equals f. Now, f is a force. It could be a sum of many forces, and I'm allowing it to have a time dependence. Maybe the force doubles with each instant of time that we consider in this problem. Now, that's a time-dependent force. It's not constant, and one would have to take that into consideration. I can recast this equation by moving the chunk of time, dt, to the other side, in preparation for doing an integral. I'm going to add up all the little changes in momentum, dp. I'm going to add up all the little products of the force and the changes in time, and I'm going to see what I get. So let's do that. I then construct this integral. I'm going to sum up all the little dps. I'm going to sum up all the little products of f and dt. Now, I have to have limits for these integrals. I'm going to have to sum from a minimum to a maximum value. So I'm going to have a minimum momentum. Maybe that's the momentum I start with at time 1, p1. And I'm going to have a maximum momentum. Well, that's the momentum I end with at the time t2. Let's call that p2. So I have two times, t1 and t2, and two corresponding momenta, p1 and p2. So let's go ahead and consider these definite momenta and times as limits on the integral. So we're going to go from an indefinite integral, a sum without limits, to a definite integral, a sum with limits, and we arrive at a formal definition of impulse. The integral from the minimum momentum, p1, to the later momentum, p2 of dp, well, that's just going to give us p, the momentum evaluated at the endpoints, p2 and p1. And so we wind up with p2 vector minus p1 vector. Well, by definition, that's the delta of p. That's the difference, the change in momentum from time t1 to t2. It's the later momentum minus the initial momentum. And that is, by definition, the impulse. So finally, we arrive at a formal definition of impulse in terms of force and time. That is the integral from the earlier time t1 to the later time t2 of this product of the force, which could be a function of time and dt. Now, of course, I have to know the functional form to really do this integral formally. That would be specific to a problem. Maybe the force decreases quadratically as a function of time. I'll have some t squared or t to the minus 2 dependence in this. But what we learn from this, no matter what, the bigger the change in momentum over short times, the bigger the force that we're talking about, that can be very bad. If we, for instance, knew the average force and the average time over which it acts, we could say that the impulse on average is just the average force times delta t, the time difference between the two. So without knowing the exact function that describes force as a function of time, let's take a look at an example, a car accident. This is a very human example, something that can happen involving a change in momentum. And as a result, it's very interesting to look at the risks to a human being under the conditions of such changes in momentum. So let's say that our electric car from earlier moving at 55 miles an hour with a mass of 1700 kilograms accidentally crashes into a solid wall. The wall or the barrier that it strikes cannot move. It's firmly rooted to Earth and it can't move at all. Well, in those cases, and we can see this by looking at videos of crash tests where you put human analogs, so-called crash test dummies inside of a vehicle, either seat belted in or not seat belted in, a car with airbags or without airbags, and you ram the vehicle with its analogs for human passengers into barricades at various angles, head-on, glancing blow, side impact. And you can see that this causes tremendous damage to a vehicle moving at even modest highway speeds. Even at 25 to 30 miles an hour, you can do tremendous damage to a car at 55 miles an hour as we'll see, things can get lethal. So let's assume that the car strikes the barrier and it will compress, it will crumple, and it will come to a complete stop. It will lose all of its speed. Well, from an energy perspective, we have a sense of what's going on. The car has kinetic energy, but the kinetic energy is given up to deforming the car during the collision. It goes into altering the chemical bonds internally in the vehicle, but this changes the internal energy of the car. It goes into heating, metal, the barrier, and so forth from friction. That kinetic energy can be dissipated in many ways, but the bottom line is that it takes you from a state of motion, non-zero motion, to a state of zero motion. So we'll take the mass of the car again to be 1,700 kilograms, and the initial speed of the car to be 55 miles an hour, 24.6 meters per second. The change in momentum is easily computed. The impulse, which is the change in momentum, is the later momentum minus the earlier momentum, so we're considering the later time to be just after the crash when the car has stopped and the earlier to be the speed of the momentum of the car before it strikes the barrier, just before it strikes the barrier. So we have a situation where we have no momentum at the end and momentum mv at the beginning, and so we wind up with a negative change in momentum, a decline in the momentum of the vehicle. This makes sense. So all the momentum we had earlier, 42,000 kilograms per second, is now depleted from the car during this change. It comes to rest after the accident. So let's see if we can turn this into some other information that will give us a sense of, say, what could happen to the vehicle or what could happen to the passenger inside the vehicle. Now, one thing we need to know is how long does an accident like this take? Well, you can look at data for measurements of speed and time in crash tests, and you'll find out that this compression, this deceleration to zero speed, takes about 0.05 seconds. This is incredibly fast. This is why in an accident, people often have a very hard time remembering what happened, it happens fast, they may be unaware of the conditions under which the accident occurs, and by the time all the damage is done, almost no time has passed. Now, using this information that an average time over which this accident will occur is about 0.05 seconds from the moment just before striking the barrier to coming to a complete stop or a nearly complete stop after compressing the vehicle, we can solve for the average force. So the impulse will be related to the average force times the time over which the average force is acting. We can rearrange and solve for the average force, and we find out that the average force that's causing this change has a magnitude of 840,000 Newtons. Ouch, is about the nicest thing I could say. 840,000 Newtons acting on a vehicle is a lot of force. Let's take a look at what this would do to a passenger. Again, we're trying to come up with a human-scale idea of what changes in momentum over short periods of time mean. Now, a person inside the car might have a mass of 75 kilograms. The car is 1,700 kilograms. A person is far smaller than that, maybe 75 kilograms or whatever abouts. Now, let's assume that they're seat-belted into the car. Now, that's important because when you're seat-belted into the car, you are part of the car. And so if the car slows, you slow. And if the car stops, you stop. And if the action of the seat-belt is fast enough to hold you in place, your motion will really be defined by what the car's motion is doing. You are part of the car. You are a collection of atoms that is part of the larger collection of atoms of the vehicle itself. So, as a human, because you're bound to the vehicle, you will experience the same time duration required to stop. But because your mass is different, you will experience a different change in momentum. So, a human being's mass is far less than that of the car's and thus there's a different force acting on you. So, let's repeat this exercise. If you plug in the numbers, mass is 75 kilograms, average time to stop is about 0.05 seconds. This assumes no airbag. And I'll talk about the airbag in a moment. But this assumes no airbag that basically you are stopping at exactly the same rate as the vehicle. And so, the force on you is 37,900 Newtons or thereabouts. So, what does that mean? Well, let's use Newton's second law to figure out what that means. If you do F equals MA and you figure out what the acceleration of the driver experiences in, it has a magnitude of 486 meters per second squared. Compare that to the acceleration due to gravity, 9.81 meters per second squared. This is a multiple of 50 times the acceleration due to gravity or a g-force of 50 g's. That is a lethal acceleration. If you take a look at a chart of acceleration versus human health outcome, 50 g's is capable of killing you. So, you can see why a car accident into a fixed barricade at 55 miles an hour is an extremely dangerous situation. Car accidents at highway speeds are bad and at 70 miles an hour the problem is even worse. Run this calculation for highway speeds that are more typical in Texas and many parts of the United States now. 70 miles an hour, take a car down to 0 miles an hour in about 0.05 seconds, take a human down to 0 miles an hour in about 0.05 seconds and see what kind of g-force, what multiple of g we're talking about. You'll find it's well in excess of 50. It's well beyond what's required but this is also why airbags exist and why they're so useful. What does an airbag do? An airbag is an explosive device that sends a pillowy cushion out at you filled with air. It shoots it out at you from the steering column. So, as you are being decelerated toward the front of the car as the car compresses, the airbag's job is to expand outward faster than 0.05 seconds to get out to a distance and catch you and then gently decompress, slowing the time it takes for you to come to 0 miles per hour. So, if you watch a test involving a crash test dummy and an airbag, watch how long it takes the car to come to essentially 0 speed versus how much longer in comparison it takes the crash test dummy to come to essentially 0 speed. It can extend 0.05 seconds up as far as 1 second. So, instead of the time the car is being stopped, the human is stopped over a much longer time. So, the delta T goes from 0.05 seconds to 1 second. This reduces the average force on a human from 37,900 newtons or so down to only 1,900 newtons, which still sounds bad, using Newton's second law, you'll find out that that's a deceleration of about 25 meters per second squared and as multiples of g goes, that's only about 3g. That's extremely survivable. It won't feel good, but it shouldn't be too dangerous either. So, the data backs up that airbags play a valuable role in reducing harm to the passenger. So, maybe this will give you a more human level of sense of scale of what momentum and changes in momentum mean on the person's sizes of things. Now, let's go back to something I put on pause in order to more humanize momentum and go back to Newton's second law and think about changes in momentum with respect to time and force and what it means if we have a closed and isolated system. So, if we observe no net external force that acts on a system, that is, if it's closed and isolated from such forces, then it must be true that f is zero and if f is zero, dp dt for any system of particles is also zero. So, we've learned something quite profound if we can isolate and close off a system so that we have no unaccounted for external forces that can act on a system that it must be true that even as time changes the total momentum of the system will not. The momentum would be conserved and we can write this as the total momentum of a system of particles where the system is closed and isolated is a constant which implies that the momentum at any initial time, Ti, is the same at any final time, Tf. Pi equals Pf that's not a magnitude of momentum that also includes the direction of the net momentum of the system. This is insanely useful for understanding not only motion, but degree of motion in a system and this is how we're able to make predictions about not only the degree of motion of a system but its directionality as well. This is the law of conservation of linear momentum and when conditions are right for applying this you will absolutely find this an invaluable tool in addition to the conservation of energy for understanding physical situations. This is a concept that is absolutely as useful and on equal footing with the conservation of energy that the net forces on a system sum up to zero along in fact any coordinate direction if it's not true that there are zero net forces in Y and Z but there are zero net forces in X then you have some profound statement about X the momentum along the X direction will be conserved along that direction. It gives you a handle for solving difficult problems when something is conserved. I talked about any other the mathematical physicist in the lecture on the conservation of energy we can understand the conservation of momentum from other theorems as well energy was conserved according to another's theorems because the laws of nature they have a continuous symmetry under time linear momentum is conserved in nature because the laws of nature also have a continuous symmetry in space small changes in the equations of motion or the laws of nature in space do not affect the outcome of the system so if I shifted the whole universe continuously by teeny tiny amounts of space the universe according to another's theorems wouldn't notice it and as a result of that momentum is conserved this is an incredible statement so there's a deep connection between mathematical symmetries in laws of nature and conservation laws so let's look at kinetic energy and momentum. Kinetic energy is energy of motion. Momentum is a quantity of motion of magnitude and direction so we have two concepts related to moving bodies about things that are inherent to those moving material bodies we have kinetic energy we now also have linear momentum p vector equals m times v vector so let's consider how these two interplay with one another in solving problems and understanding natural phenomena because we can use both of them to tackle difficult problems in all cases I want you to consider for now an ideal closed and isolated system so if a system of particles that is material objects move and interact in such a way that the total kinetic energy remains unchanged that is energy doesn't change form from kinetic to anything else in any significant way or at all then this is known as a system that is undergoing elastic collisions so I can show you an example of this for instance I can take a frictionless air track this is like an air hockey table but in one dimension you can place an object a cart that has a little v shape that fits over the track it rides on a cushion of air and if the track is well balanced it won't accelerate if you leave it very still on the track if you impart momentum to the cart and then stop acting on it it's got little bumpers on each end and it will bounce off the end of the track and then change its direction of motion changing its momentum and then it will bounce off the other end of the track and change its direction of momentum but notice that the magnitude of momentum doesn't appear to be changing the velocity of this cart whose mass isn't changing the velocity may be changing direction but the speed is not changing so it looks like kinetic energy is conserved and it also looks like the magnitude of momentum is conserved even if during these collisions with the walls the momentum is changed but nonetheless the total momentum is not changing so this is a system that's an example of one that's undergoing elastic collisions now if instead such a system changes its total kinetic energy over time for instance energy has lost a potential energy in some way or goes into other forms like thermal energy then the system is said to be undergoing inelastic collisions so the car accident I showed you before is a good example of an inelastic collision kinetic energy is clearly transformed during the collision to the compression of the car the kinetic energy is absolutely lost in this collision another good example is two bodies that are able to move but they stick together during the collision process and move off as one body after that this is an extreme that is known as a totally inelastic collision so if you start with two things and they collide and stick together and now you have one big thing at the end that is a totally inelastic collision in these cases kinetic energy is absolutely lost from the system and these objects are now locked together but what's important to recognize is that while total kinetic energy may change during inelastic collisions total momentum does not total momentum in a closed and isolated system never changes even if kinetic energy changes using these ideas together you can imagine how useful it will be for setting up and solving problems so let's take a look at collisions so I've talked about elastic collisions one where the kinetic energy doesn't change in the system over time and inelastic collisions one where the kinetic energy can change in the system over time now for all such collisions in a closed isolated system it will be true that the initial momentum total is equal to the final momentum total closed and isolated system so for elastic collisions it will also be true that the final kinetic energy of the whole system is equal to the initial kinetic energy of the whole system that is the kinetic energy total of the system will never change the sum of one half mv squared for all particles at any time will be the same even if their v's are changing over time now inelastic collisions we have a different case in this case we have kinetic energy as a total can change so for instance it could decline over time as a result of collisions or maybe it could be increased over time as a result of explosions within the system that break objects apart in this case final and initial kinetic energies will not be the same and we will have to rely on momentum to help us deal with the problem but one thing that's very helpful here will be that the velocity of the center of massive such systems will not be affected by collisions and that will give us a constant phenomenon to consider even so that's something we'll try to exercise but something to keep in mind so let's take a look at an example of an elastic collision specifically imagine having two masses that are equal so I could arrange this for instance with two carts on this frictionless air cushion track they're both of mass m the same mass at least if the manufacturer has done a good job and I can construct a situation where the initial velocity of the first mass is non-zero and it's moving in the positive x direction so we'll denote this as v with a subscript one comma i for the initial velocity of the first mass but I have a second mass also m that's initially at rest the first mass collides elastically with the second what I mean by that is that momentum is conserved but also kinetic energy is conserved by this collision so what would I predict will be the final velocities v1 final and v2 final along the x direction well we know a couple of things for instance we know that the kinetic energy shouldn't change with time whatever the total kinetic energy was at the beginning that will be the total kinetic energy at the end well the kinetic energy at the beginning is just the kinetic energy of the first mass the second mass has none it's not moving the total kinetic energy at the end well we don't know what the velocities are going to be so it's going to be one half mv1 final squared plus one half mv2 final squared it's going to be some sum like this well one half and m appear on all terms so it cancels from both sides of the equation and I'm left with just a relationship between the sums of the squares of velocities on one side and the square of a velocity on the other so the square of the initial velocity of mass one is equal to the sum of the squares of the final velocities of one and two whatever they are we're going to try to see if we can make a prediction about what those velocities are going to be that's what we get from the conservation of kinetic energy in this elastic collision situation from the conservation of momentum we get a slightly different equation we have that the velocity of the first object is non-zero but the second object is zero at the beginning of the problem and then after the collision we don't know what the velocities are going to be they could be non-zero so we have that mv1 initial is equal to mv1 final plus mv2 final and again the m is common in all three terms and so it cancels from both sides and we get a different statement about the velocities not only is the sum of the squares of the final velocities equal to the square of the initial velocity of the first cart or mass the sum of the velocities of the final masses must be equal to the velocity of the first mass or cart so great if I define what the initial velocity is if I make it a definite number like 10 centimeters per second or something like that and I want to predict what the final velocities are I have two unknowns v1f and v2f and I have two equations I can solve this this is a solvable problem I can make definite predictions given an initial velocity for the cart where the second one is initially at rest I can make definite predictions about the velocities of the two carts after the collision so let me remind you of the velocity relationships that we've gotten from conservation and from momentum conservation we can do a substitution where we take the fact that v1 final is equal to v1 initial minus v2 final I get that from the conservation of momentum equation I can then plug that substitution into the kinetic energy based equation and I will find the following that v1 initial squared is equal to this substitution squared plus v2 final squared and if I expand out that polynomial of v1 minus v2 minus v2 final all squared I get this beast here I invite you to do some algebra gymnastics group terms together, cancel up terms that have opposite signs between them but otherwise are the same and at the end of this rearrange and solve for v2 final and if you do that you'll find that v2 final is equal to v1 initial so what this tells us is that the final velocity of the second cart or mass is equal to the initial velocity of the first cart or mass and if we plug that result into the momentum equation we find out that that means that after the collision the first mass has no velocity whatsoever, it's not moving is this what really happens is it true that if I take a cart of mass m and I slam it into another cart of equal mass that's initially at rest that the second cart gets all the momentum let's take a look and see what happens in fact it is in fact we see that the first cart transfers all of its momentum to the second one which takes off rebounds off the end of the frictionless track, smacks into the first one kicks it off again and then comes back now the track is not perfectly balanced so gravity is accelerating these carts a little bit it's slightly inclined and of course this isn't a perfectly isolated and closed system, there's turbulent flow from the air that's flowing up through the holes in the track and so forth this is not a perfect ideal closed system but it's fairly close to one and it's really remarkable to see this experiment happen in reality let's take a look at one more example before we close out this lecture and that's the example of rocketry rocketry is a slightly different case than the one we've been considering already in the case of the colliding carts the mass of cart one and the mass of cart two were the same the whole time they don't shed mass but what's a rocket? a rocket is a large vessel containing fuel the fuel mass and the rocket mass together form the mass of the whole vehicle you ignite the fuel chemically for instance you do some kind of chemical reaction and this increases pressure in the vessel which spits mass out the back end of the rocket you're taking the original mass of the rocket and you're declining it essentially by throwing mass out the back end so the secret to rocketry that is launching an object from the earth say and putting it up into space it's not the force of the exhaust against air or the ground I mean after all rockets continue to accelerate even long after their exhaust plume isn't touching the ground anymore and it's not because that exhaust plume is pushing on the air rockets work just as well in space rockets are used to reposition things in space all the time there's no air so it's not pushing against anything so how do rockets work? well rockets work by taking advantage of the fact that ejecting mass out the back of a vessel is also a change in momentum remember momentum is mv and I originally considered the case where m was constant and v could change but m could change and v could also change and you could have two changes you could have an m change and a v change and those would also be changes in momentum that can cause an impulse and that results in a force changes in momentum with respect to time exert forces so in one dimension if we think about a rocket as moving along only one dimension we're going to have a situation where the force the force that's experienced by the vessel as it ejects mass out the back will be related to dp dt which is the time derivative of the product in m and v but now we have a situation where m and v can change and so we have a chain rule situation m is a function of time, v is a function of time so I have to do the chain rule on this product and I wind up with v dm dt plus m dv dt this is the result of doing the faithful execution of the chain rule assuming that m is a function of time and v is a function of time so if speed is changing, momentum is changing we've seen that already that this term on the right is not a surprise but if mass is changing with time then momentum is also changing and force can result from either kind of momentum change rockets take advantage of this more to the point a rocket fuel system is a closed and isolated system especially in space so it's not so much that there are external forces on the system that cause changes in momentum it's that changes in mass cause changes in the speed of the rocket so throwing mass out the back causes the speed of the vessel to change in response rockets take advantage of this they consume fuel at some rate and that rate can be given as the negative of the change of the mass of the rocket with respect to time after all when you kick mass out the back you still have the same total mass the exhaust plume and the rocket all together have the same mass but the vessel itself has less mass than it did before because you kicked exhaust out the back so you're expelling mass out the back that's DMDT this in response in a closed system will cause a DVDT for the remaining object so let's apply this let's use conservation of momentum so we have a closed and isolated system it's fuel starts burning the fuel and ejecting mass out the back let's see what happens conservation of momentum will apply for a closed and isolated system so we know that the sum of external forces will be 0 and whatever the total momentum at the beginning before it ejects a mass out the back is that will be equal to the total momentum at the end after the mass has been ejected so we have a system where we have a rocket of initial mass big M moving out of velocity initially relative to us we're viewing this whole process from the rest frame outside the rocket it then expels some mass so it's losing mass DM is a negative quantity and that exhaust that ejected mass leaves at a velocity of u relative to us so from the perspective of the vessel the fuel is moving away from the rocket from our perspective the fuel might still be moving in the direction of the rocket so we can see that it's operated from it as time goes on so this is a kind of relative frame problem as well and I'll come back to that in a moment so ejecting this matter out the back will cause some change in the velocity of the rocket so it starts at v but then after ejecting mass DM it will be at v plus dv and again that's relative to us the outside observer so we can write this conservation equation as the original momentum the mass of the original rocket times its original speed v must be equal to the negative of DM remember DM is a negative decline so this is actually a positive momentum here times the velocity of the exhaust products and then the mass of the rocket is now changed it's m plus DM remember DM is a negative change so this actually is a decline in the total mass of the rocket and v plus dv now all of these speeds have been measured relative to a stationary observer but the rocket has a speed relative to its own exhaust products and that's given by the relative velocity of the rocket to its exhaust products is v plus dv minus u so using this information in the above equation we can then substitute we can substitute in rearrange and we get if we move terms around we can get mv equals mv plus dv plus DM plus dv minus u and v plus dv minus u is exactly the velocity of the rocket relative to its exhaust products and so if we rearrange this one more time we will find that 0 is equal to mv v plus DM times v rel so that is that the change in velocity of the rocket times the original mass of the rocket plus the loss of mass from the rocket due to fuel times the relative velocity of the rocket and the fuel are related to each other and we can in fact solve for the change in velocity in terms of these other things now if we divide both sides by a small change in time if we think about changes in velocity with respect to time then we have dv dt on the left side that's acceleration and we get an equation that relates the acceleration of the rocket the original mass of the rocket including its original fuel complement is how fast it can eject fuel out the back and the velocity of the rocket relative to its own fuel so that would be for instance the speed with which the exhaust shoots out the back of the rocket that would be measured for instance with respect to the rocket so somebody standing on the rocket can kind of look and see oh all the exhaust is shooting out the back is going out at 100 meters per second that's a v rel this little exercise plugging in for the fuel burn rate r gets us to one of what's known as the rocket equations the acceleration of the rocket times its original mass including its fuel is equal to the rate it burns fuel times the speed of the fuel once it's ejected relative to the rocket that's a lot to process let's actually do an example and see what we can learn from this can a fire extinguisher be used to turn me into a rocket so using the above information how would I do this well I want to make it I can't go into space I wouldn't be able to breathe and it's expensive to get there anyway so instead I'm going to try to reduce friction so I'm going to reduce the fact that the ground wants to resist my motion by sitting on a cart with wheels and I'm going to hold the fire extinguisher and I'm going to try to keep my my center of gravity as low as possible so that the fire extinguisher doesn't tilt me over which would be funny for you but painful for me now the empty mass of a fire extinguisher is about 18 kilograms or so but a fully loaded fire extinguisher has about 10 kilograms of compressed carbon dioxide in it so we have about a 28 kilogram fire extinguisher if you add the mass of the cart and my mass and the mass of the fully loaded fire extinguisher together that's the mass of our rocket that's big M and that comes out to be about 120 kilograms or there about so that's our rocket mass so there we go there's our big M fire extinguisher including its fuel and me and the cart 120 kilograms now fire extinguishers can discharge their contents in about 4 or so seconds so the rate of discharge is 10 kilograms of CO2 ejected in 4 seconds or 10 kilograms over 4 seconds which is 2.5 kilograms per second when you fire a fire extinguisher you don't have a lot of time to use it your aim must be true and you're not going to get a lot of time out of it put fires out quickly so that's okay so the so-called burn rate of fuel is 2.5 kilograms per second from a typical fire extinguisher that you might find in say fondant science building now the contents are expelled at speeds that are about 20 meters per second so if I lock the fire extinguisher in place and I measure the speed with which the exhaust products shoot out the back that comes out to be something like 20 meters per second that is the velocity of the exhaust relative to the rocket because the fire extinguisher will be fixed to the cart and I will be fixed to all of that I'll be holding all of this down and so the speed of the exhaust relative to me is going to be 20 meters per second so we can make a prediction what's the acceleration of me and the fire extinguisher in the cart as I blow CO2 out the back well plug in the numbers the acceleration will be the burn rate of CO2 over the original mass of the rocket times the relative speed of the exhaust to the rocket and you'll find out that this is maybe an underwhelming 0.4 meters per second squared that doesn't sound very exciting 0.4 meters per second squared but let's assume that that's a constant acceleration and I can get that over the time required for the exhaust products to completely be exhausted from the fire extinguisher what does this mean it means I could travel roughly the front of the room which is a distance of 6 meters in Fondren Science 123 our typical lecture room here in the physics department I could expect to using the equations of motion that we learned earlier in the class at this constant acceleration rate of 0.4 meters per second squared I could expect to go 6 meters in about 5 seconds well that doesn't sound so bad so if I sit on a cart with a fully loaded fire extinguisher and I brace myself around the fire extinguisher and aim it backward and pull the trigger on the fire extinguisher and I can get that thing to go for 4 or 5 seconds of exhaust I should be able to push myself 6 meters almost the full distance across the front of the lecture room I say let's try it and see what happens fire extinguisher on a cheap cart with friction in the wheels and a person like me can do fun things you can see how rockets work so fantastically well it took a long time to perfect the art of rocketry and to this day it's still something where especially if you're going to put a human payload inside a rocket you have to be extremely careful but nonetheless we as a species have become quite effective at going somewhere by leaving something behind and ultimately that is what rockets do they get you some place by forcing you to leave something behind by changing the mass of the object you can change its speed and even in a closed and isolated system in outer space if you can find a way to do this if you can find a way to alter mass you can effect speed via the conservation of momentum it is a remarkable fact of nature and it works incredibly well let's review the key ideas that we have explored in this section of the course we have defined the momentum of an object and we have used that quantity to revisit Newton's second law of motion in doing so that's the relationship between momentum and force specifically forces cause changes in momentum with respect to time and if there are no changes in momentum with respect to time it means there are no net external forces on the system and as a result it's possible for momentum to be a conserved quantity now that's of course in a closed and isolated system but those are typically the systems we'll try to construct or consider when we're thinking about solving problems we've taken a look at a couple of applications of momentum and specifically momentum conservation in such systems we looked at the elastic collision of two equal masses and saw the remarkable fact that momentum is transferred from one colliding object entirely into the next and then back again in a sort of lovely little dance we've also looked at the case where mass is changing and seen that in a closed and relatively isolated system when mass is changing it's possible to alter the velocity of the remainder of the system in such a way to develop the entire field of engineering known as rocketry all of this depends on the momentum concept, changes in momentum and an understanding that momentum can be conserved under certain conditions