 Hi, I'm Zor. Welcome to Unisor Education. I would like to continue the topic of momentum and impulse. The previous lecture was about momentum of motion, and this will be about the impulse of the force and how they are related to each other. Now, this lecture is presented as part of the course called Physics 14 on Unisor.com. I suggested to use the website for watching this lecture and any other lecture of the course for many different reasons. One reason is every lecture has very detailed notes. Another reason is that the website contains a prerequisite course, which is mass for teens, and especially the vectors and calculus must be really perfect if you would like to master the Physics 14. And also it has exams, the website, and it's also free and no advertisement. Alright, so back to the impulse of the force. Now, let's consider a very simple situation when you have a uniform motion of the object of mass A under certain constant force F, which basically gives it an acceleration A. Now, from the second Newton's law, we know this. I'm using vectors, but whenever it's uniform motion, it doesn't really matter, because it's only one dimensional anyway. Okay, anyway, we do have this. Now, what does it actually mean in terms of the speed? Well, speed is increasing. Since we have a constant acceleration, speed is increasing. And let's say at moment T1, speed was V1, and at moment T2, speed was V2. Now, the increase in speed, which is V2 minus V1, is acceleration times the time difference, right? That's the basically definition of constant acceleration. It's the division of the difference in speed by difference in time. So that's why we have this. Now, obviously, if we multiply by M, both sides, we can combine these into force. So it would be force times T2 minus T1, and this would be MV2 minus MV1. Now, this, as we know, is the momentum of motion. It's increment of the momentum of motion during period of time from T1 to T2 under the constant force F. Now, obviously, this can be written in slightly more, well, scientific or mathematical form, which will basically allow us to drop the requirement of the constant force. Because if I will make this interval of time infinitesimal, I will have this, and I will have this. Now, obviously, now force can be a function of T, and the speed can be the function of T of the time. And they do not necessarily have to be like a uniform force can be changing, but during the infinitesimal moment incremented time, we will have an infinitesimal increment of the moment of motion. And this would be true even for variable force. Okay, so now this is increment of momentum of the motion. Now, this is an impulse which acted during this time because there is this particular force. So this is basically a definition of what impulse is. So impulse is the product of the force times the time interval, and that gives you the increment of the momentum caused by this force acting during this time. Okay, so that's basically all about definitions which we can express it in this form or we can express it in a slightly different form, well, actually in more than one slightly different form. Now, the same thing can be done like this, and this is also function of T. Now, this is basically a derivative, right? So you can write it this way. That's just a different notation of a derivative, this versus this. But we also can do from here an integral formulation of this equality between increment of the momentum and the impulse of the force. Because if I will integrate this from, let's say, moment T1 to T2, what happens? Well, this is the full differential. So after the increment, after integration, it will be mv of T2 minus mv of T1. That's on the left side because it's full differential. On the right side, I will leave whatever it is. So this is an integral formulation of this equality. And again, this is an increment in momentum but not in an infinitesimal period of time T. But during a substantial from T1 to T2, any substantial interval, as long as we know how the function behaved, I mean the force function behaved, we can integrate it by time and that would give me the increment in momentum of motion. In a particular case of function f of T, let's say this function was constant in one period and then constant in another period. That's kind of easier to integrate because it's really a sum of two. So if my function f of T is equal to, let's say, f1 on the physical constant, on the period let's say from 0 to T1 and it's equal to f2 on the time period from T1 to T2, then this integral for an entire period from 0 to T2 is basically a sum of two. So it's integral from 0 to T1 and now the function is a constant. Plus integral from T1 to T2 of another constant, which is f1 times 0 minus T, sorry, T1 minus 0, which is T1 plus f2 T2 minus T1. So this is the impulse of the function exhorted from T1 to T2 and this is the impulse exhorted by the force from the period from 0 to T1. So in any case that and that would be in this case from 0. So in this case, all I'm saying is that the increment of the entire moment of motion in this particular case can be represented as sum of impulse on one and impulse on another period of time. So basically that's it about definition of what impulse of force is and equivalence of the increment of the momentum with impulse exhorted by the force during the time we are measuring. And again, this impulse is additive. It's basically like if you are adding one teaspoon of sugar into your tea and then another, the result will be exactly the same as if you will add two together at the same time. So this editiveness is very, very important because each teaspoon of sugar contributes certain sweetness and two teaspoons will sequentially or consecutively applied to tea will give exactly the same sweetness as if you just put them together and do it in one time. Same thing with this, your sweetness of momentum of motion will increase first by one impulse and then another or we have, if we can, we can combine them together and that basically gives you exactly the same thing. So if I apply only this one, f1 t1 to my motion, I will have increment from t1 minus m v of 0, right? This is what this guy does. Then, having already my momentum reached this point, I can apply this thing and I will get m v of t2 minus m v of t1. And if I will add them together, this will cancel. And again, I will have an entire increment of my momentum or sweetness of the tea. So that's the editiveness of impulse applied to increment of the momentum. Now, I have a couple of examples where I can actually demonstrate it. One example is, well, basically exactly what I just did right now with the function which represents the force actually is a constant on one period and then constant on another period. Same thing, basically. I'll just use different notation. So let's say you have one force which acts during the time t1. So that's basically t1 from 0 to t1, if you wish. But anyway, this is the number of seconds this particular function acts upon certain object. And then during another number of seconds, the t2 number of seconds, this function which is equal to constant f2 acts on the same object. So let's just analyze the first increment of momentum and then the second. So the first increment of momentum would be m times v of 0 to m v of 0 plus f times f1 times t1, which is equal to m v of t1. So we have this is the old momentum at point 0. This is additional impulse which we give to this which is increment and that would be equal to my momentum at time t1. And then during the time t2 after that, I will have from m v of t1 to m v of t1 plus f2 t2. And that's my m v of t2. Again, we have this particular impulse applied to already achieved before momentum of motion and this is my result. So I can do it in two steps. Now at the same time, I can do it in one step basically combining f1 t1 plus f2 t2 and that would be my combined motion which is m v of t2 already minus m v of 0, obviously. So this is again just an illustration in a different kind of way of editiveness. And then I would like to solve using this to solve one particular problem which kind of has a pretence to be from the real life, but it's not really. Anyway, I would like to measure what kind of resistance the air presents to the flying airplane. Now, how can I do it? Let's assume you have a plane. It has engine. Now, what engine of the plane does? There are certain input coming into the combustion chamber coming whatever the fuel is, some kind of liquid. Then it burns here and the gases are pushed back. That's basically how jets are flying, right? So some liquid is coming into this combustion chamber and the gases are flying back. And that's what pushes the plane forward. Now, why is it pushing the planes forward? Well, there is a third Newton's law, right? We are pushing gases this way and the third law gives us the reaction force which is equal. And so the whole plane is moving forward just because of this reaction force. Okay, now let's assume that my pumps which are pumping liquid, whatever the liquid fuel is, are working at capacity mu kilogram per second. Well, that's basically the same kind of amount of mass which is going out every second, right? Whatever we're pumping in burns and as gases is pushing back with a relatively high speed, right? So, during certain time t, second, we are pumping mu times t mass of the fuel, right? Now, let's imagine that these gases are coming out with certain speed v out. Then this is the momentum of the gases which are coming back, right? Now, this momentum should be equal to the force which we are applying to the gases or gases applying to us times the same t. So this is the equivalent of the momentum and the impulse. So that's how we get the force f. And by the way, the t now is canceling so basically f is equal to mu times v out. So this is amount of t which we are pumping in a unit of time and this is the speed the gases are actually going out. So this is therefore the force which goes this way. Now, the plane goes with a constant speed. Now, where is this particular force then? What is it wasted for? Well, it's basically wasted for resistance of the air. So there is the same by magnitude force of resistance of the air and that's how we are traveling with a constant speed. When the airplane is already in route to something. I mean we are not talking about initial going out from the ground. I mean there are many different forces which are acting in those times but when it's already on the course for a long time, it goes with a uniform speed and basically the fact that the speed is constant, there is no acceleration means that this force and this force are equal in magnitude and opposite in direction. So this is just one of the example of how you can calculate something like air resistance knowing basically your engine parameters of the plane. Okay, that's it for today. This is a relatively short lecture because the concept is relatively simple but there are problems and there are exams and everything is in unizord.com and I do suggest you to, well you can read first the notes for this lecture obviously and then if there are any problems presented, I will explain certain problems in the lectures but I always ask you to try to solve these problems yourself first and then listen to whatever my solutions are. Alright, that's it. Thank you very much and good luck.