 Hi, I'm Zor. Welcome to Unisor Education. We will switch to the next topic on my list of momentum and impulses, which is called conservation of momentum. Now, the conservation of momentum obviously is the consequence of the definition of what is momentum and what is impulse, etc. And I'm going to demonstrate it right now. Now, this lecture is part of the Physics 14 course presented on Unisor.com. I suggest you to go to the website Unisor.com to watch this lecture because the lecture has very detailed notes, exams. It has certain educational functionality for people who would like to study under supervision or self-study. And it's free and there are no advertisements. So, conservation of momentum. Well, let me remind you that momentum is defined in a differential form as the increment of momentum is equal to the force applied to my object of mass m and speed v during the time dt. Infinite decimal interval of time. Or in the form of derivatives it would be derivative of m v of t by dt is equal to f. Alright, now, what does it mean? It means that if there are no forces at all, my momentum is constant. That's very important. The momentum is constant if there are no forces applied to this particular object. Well, obviously because it's mass constant, the speed is also constant since there is no force, there is no acceleration. So, the speed is constant. So, momentum is constant. Okay, this is kind of a simple thing. What's a little bit more important, interesting, but maybe complicated is, let's just switch from one object to a system of objects. So, let's consider you have, well, two objects or three or n, doesn't really matter. Now, what happens if there are no external forces acting on these objects? Well, then the only action which can really change something is their action against each other. For instance, they collide or something like this happens. Now, what's the result of it? Well, whenever, let's just consider collision as the most kind of simple interaction between two objects. Well, it produces certain force, right? When they collide, they act on each other with certain force during certain period of time. So, there is a force which this object is acting upon this object, let's call it f, during some time delta t. This is the interval of collision. But at the same time, this object acts on this one with exactly the same force by magnitude but directed in opposite direction. So, there is a minus f during the same, basically, contact, the same collision time. Which means that their impulses are added together, are equal to zero. So, the total impulse of the system is not really changing. Which means that the total momentum, sum of this momentum plus this momentum also should not change. So, we do have certain change in how these two objects after collision behave. I mean, they might change the speed, change the erection, whatever happens with their momentum. Well, yes, these changes do happen, but the sum of these momentum as vector, vector sum, obviously, should not change at all. Because the total of impulses is exactly zero. Now, force is obviously a vector. So, that's why we have change of momentum of this cause this. Change of momentum of this is caused by this. And if we will add them together, these two will nullify each other. Which means my momentum of the first one plus momentum of the second one as function of t, obviously, is constant. So, that's the most important part of this. And that is the conservation of momentum. So, if there is a system of objects, and that's really important for the system of objects, now we are kind of complicated the picture. There is a system of objects, and there are no external forces. Let's say you are on a spaceship, which is moving somewhere in open space, far from any planet, any gravitation field, etc. It just moves on inertia by itself. So, no external forces are applied to this. Now, inside the spaceship there are maybe people or objects moving left and right, etc. But the total momentum of the whole thing should not change. Which means no matter how they move inside, the spaceship will continue moving in exactly the same direction. Because every action which is happening inside that ship has a reaction opposite in direction and equal in magnitude. And that's why the total impulse is always zero. So, that's a very important law. It's called the law of conservation of momentum. It follows basically from the definition of the momentum. We have fine momentum to be equal to impulse, and the impulses are nullifying each other. And that's why we have this conservation of momentum. Or if you wish conservation of impulses, it's basically the same thing in this case. And that's basically it as far as the theory is concerned. Now, let's talk about practice. So, I would like to solve a couple of problems using this particular law of conservation of the momentum. Here is my first. Let's consider you are playing in a billiard. This is your ball. And this is another ball. So, you hit this one. Now, this ball hits this one. And this ball goes this way. So, it's not exactly a head-on collision. It's under some kind of an angle. So, they are actually going into their own directions. Now, let's consider this is alpha. This is a positive angle alpha. And this is, let's say, it's minus beta. Beta is a positive angle, but this is minus because we are going clockwise. Counterclockwise is a positive direction and clockwise is negative. Now, what do we know? Well, we know this speed. Speed we have hit this particular ball. So, now we have two balls. Let's put this is black and this is white. So, this is initial position in black and this is final position. So, knowing this particular speed and knowing these two angles, I would like to find out magnitudes of these two balls after the collision. So, before the collision, I have this is V and this is zero. After the collision, whatever was zero becomes V1, whatever was V becomes V2. So, I would like to determine what are the speeds of these two balls after the collision. This is actually a very typical problem on conservation of momentum because the conservation of momentum is all you need to solve this particular problem. Now, how can I express my conservation of momentum? Well, I will take the momentum of these two before collision and after the collision, I will equate them to each other and see what happens, right? That's basically the logic of this. So, now what was my momentum before? Well, let's consider the mass of this is m. Now, I didn't say it actually, but I presume that the masses of both balls are the same. That's very important obviously, right? So, this is also m. Now, m is not actually given, but it will basically cancel out at the end. So, it doesn't depend. As long as they're equal to each other, the speeds will depend only on the initial speed of this ball, all right? All right. So, now my initial momentum is P, let's call it zero. It's equal to mv plus m0, so it's mv. Which means my final momentum should also be equal to the same thing, right? Okay. So, if this ball has the speed v1 and angle alpha, now obviously my momentum are vectors. Now, this is a horizontal component and I will put the little letter x. Now, the vertical component is equal to zero, right? This is an x, this is a y. So, since my initial momentum is along the x-axis, now this initial momentum has only the x-coordinate and y-coordinate is zero. Well, let's now consider what will be my final. Now, P1x. Now, what is P1x? So, P1 is a vector of momentum of the system after collision. Well, now what is the x-component of this guy? Well, the x-component would be v1 times cosine of alpha. So, it's m times cosine of alpha. That's the piece of this guy. Now, this also guy contributes certain impulse in the x-direction. So, which one? m times cosine of minus beta, right? Well, we know that cosine is an even function, so plus or minus doesn't really matter. So, let's just put beta. Now, what's my final y-component? Okay, now this is the y-axis. Now, from this v1, I will get this component is v1 times sine of alpha. So, it's m sine of alpha. Now, this is opposite direction, right? So, that's m sine minus beta. And obviously, sine is an odd function, so instead of minus here, I will have minus here. Well, that's okay. And what do I know? Well, I know that this is equal to mv and this is equal to zero, right? Because if two vectors are equal to each other, their x-component must be the same and their y-component must be the same. So, this vector has coordinates m cosine m sine of alpha. This vector has components m cosine beta and minus m sine beta. So, if I'm saying that the sum of these two vectors is equal to this vector, it means that my x-coordinate and my y-coordinate separately should be equal. So, basically I have a system of two equations which allow me to find basically my v1 and v2. Okay, except I made a little mistake here. I forgot to put v1 here and v2 here. These are unknown, obviously. So, this is my system of two equations with two unknown. v1 and v2 are unknown. I didn't see them before, that's why I kind of was surprised myself. Okay, now how can we solve this system? Well, come on, that's just a plain system of two equations, linear equations with two unknown. And first of all, I obviously cancel the m, m, m, m, m and zero also, right? So, my new system of two equations with two unknowns will look like v1 cosine alpha plus v2 cosine beta equals v. Now, v1 sine alpha minus v2, yeah, v2 sine beta is equal to zero. Okay, so as you see, it doesn't depend on the masses. That's first, I mean, I promised you that and that's obvious. So, from here, well, I think the simplest thing is let's just multiply this thing by what? Let's say multiply this by sine beta and this one by cosine beta and add them together. What will happen then? This multiplied by sine will be v2 cosine beta and sine beta and this will be minus sine beta and cosine beta. So, if I will add them together, v2 will cancel completely. v1 would be cosine sine beta cosine alpha plus cosine beta sine alpha equals 2 v times sine beta and this is zero, right? So, I multiply left and right by this multiplier and add together and that's what I've got. Now, this is, by the way, sine of alpha plus beta, if you remember trigonometry, right? So, from here, we get v1 is equal to v sine beta divided by sine alpha plus beta. That's the answer. So, that's the component which went that way. This is the magnitude of this particular ball. Now, what's the magnitude of another one? Well, we can do a very similar trick. We will multiply this by sine beta and cosine beta. It will be cosine alpha sine, no, I mean alpha. So, this one will be and then subtract. So, if I multiply this and this by this and then subtract, you will have only this piece, right? Or, alternatively, I can use this and have v2 is equal to v1 times sine alpha divided by sine beta, right? From this equation. So, v2 sine beta should be on the right and then I divide by sine beta. So, whatever happens, it will lead to the same result, obviously. And from this, this is actually easier. So, it's v1 which is this one. I divide by sine beta which can cancel this but multiply by sine alpha. So, it will be this. So, that's the solution. So, using the momentum and the conservation of the momentum, I had the momentum before collision and momentum after the collision. Both are vectors. They have x-coordinate and y-coordinate. And that's how I have arrived to two equations with two unknown variables. And solution is obviously trivial. Okay. I have another problem. Okay, another problem is also two-dimensional thing. I have two objects, object A and object B. Object A goes this way, object B goes this way. So, they are going on the perpendicular trajectories. Now, at moment, now here they have met at zero, at origin of co-ordinates, okay? They met and they stick to each other. Now, as a result, my combined object, which I call AB, would go somewhere here. This would be my AB. Let's consider that they are made of Play-Doh. So, they are moving to each other, stick to each other and continue, obviously, to some direction. Now, what I know is, I know that this object is more massive. So, MB is equal to MA by how many times? Four times. I know that. But this object is faster. So, VB is equal to my, okay, it's three times. So, VA is faster than VB by three times. But the mass of VA is four times less than massive B. So, I know just this. What else I know? I know the magnitude of this speed of the object after the collision. Now, collision is non-elastic, which means they are not bouncing of each other. They are speaking to each other. It's non-elastic collision. So, I know the VB. It's a magnitude. I don't know the angle or anything like that. I just know the magnitude. But I have to know speeds of these guys. VA and VB are unknown. So, I'm going to do exactly the same thing as before. Namely, let's assume that I will consider this object first and I will check its impulse. So, its impulse is PA and it has two coordinates, right? Coordinate X and coordinate Y. Now, coordinate X is MA times VA because it's moving this direction. Now, coordinate Y is equal to zero, right? There are no movement of this object in the Y direction, only in the X direction. How about this object? Well, this object has zero X component. It moves only along the Y axis. And along the Y axis, it has MB times VB. Now, VA and VB are unknown. We don't know them. Masses, we don't know them. But what I can say right now, now this is the vector and this is a vector. I know that the sum of these two vectors has a magnitude given to us. Now, what is the sum of these vectors? Well, obviously, it has X component and Y component. So, the PAB X component is equal to sum of X components, which is MAVA and PAB Y component is sum of Y components, zero plus MBVB. So, this is my sum of these two momentums. And it's supposed to be equal to momentum of this, right? Because they are together right now. So, they have the same speed and the mass is a combined mass. So, let me just make a little easier here. Instead of MB, I can put 4MA, right? And instead of, let's say VA, I will put 3VB. This reduces basically the number of variables that I have, right? So, what is the mass of combined object? The mass of combined object is the sum of these masses. And in terms of the MA, this is MA, this is 4MA. So, when we combine together, the mass is equal to 5MA, right? So, I know basically in terms of MA, I know the mass of this, mass of this which is 4 times greater and mass of this which is 5 times greater because it's a combination of MA and 4MA. So, this is known thing. Okay, that's good. And the speed, we also have absolute value. That's given, that's the only thing which is given. Which means I can combine the knowledge which I have about components of sum of these. And since I know that the sum of these are supposed to be equal to this, I will make an equation. I don't need this anymore. My magnitude of this particular vector is equal to square root of x-component square plus y-component square, right? That's the magnitude of the vector. That's Pythagorean theorem. I know x-component, I know y-component, so this is diagonal. Obviously, it's Pythagorean theorem. So, it's 9MA squared VB squared plus 16MA squared VB squared root. And that's the magnitude of this vector, right? And it's supposed to be equal to magnitude of this vector of the momentum of the combined object Mb, which is its mass, which is 5MA, and speed. And that's the only thing which is given. What do we have from here? Well, obviously, 9 plus 16 is 25, so it's 5MAVB, right? So, it's 5MAVB from which we cancel this and we have the answer VB is equal to VAB. Well, maybe a little unusual that the magnitude of this is exactly the same as magnitude of that. But, well, direction is different. That's number one. Number two, I obviously chose certain numbers to have an easier calculations with square roots, etc., right? So, this is the answer. And since I know the VB, I know that the A is 3 times faster, which is 3VAB. And that's my answer. So, knowing the magnitude of this speed and knowing the proportion between the masses and the speeds, absolute speeds of these two objects, I can actually find their corresponding velocities. Okay, again, what's here? What to take from this particular problem? All you have to do is carefully calculate the momentum of the system, combined all the momentum of each particular object together before whatever happens. In this case, it's non-elastic collision and after. So, before we had two objects in this case, but after we had one object. So, we calculated basically the sum of these two momentum, this one and this one, and equated it to the momentum of the system after the collision. And that's the result of it. So, that's the only thing which you have to really understand how to use the law of conservation of momentum. Before and after whatever happens, you calculate separately, you equate to each other, and from here you basically derive whatever is necessary to derive. All right, that's it. Thank you very much. I do suggest you to go to theunisor.com and read whatever the notes for this lecture is, just to repeat yourself. And if you want, you can probably try to just forget whatever I just said, solve these problems yourself and see if you will get the same results. Also a very helpful exercise. All right, that's it. Thanks very much and good luck.