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### Course: Mechanics (Essentials) - Class 11th>Unit 8

Lesson 3: Can you push a car from inside?

# Momentum conservation derivation

Let's derive the conservation of momentum mathematically (Mathematical Proof) from Newton's third law. Created by Mahesh Shenoy.

## Want to join the conversation?

• if two ball are moving in same direction we take both velocities positive
but if two balls are moving towards each other then we take one of the velocities in negative sign??
(0 votes)
• Yes.
As a vector quantity, velocity is sensitive to direction. Just as you said if both balls are moving in the same direction, their direction is positive. However, if one ball is moving in the opposite direction, its sign becomes the opposite of the other ball (-) in this case.
(2 votes)

## Video transcript

- [Instructor] When things collide with each other, like say when this blue coin hits this white coin, we see that their speeds change after collision. If you look at it one more time, notice before the collision, the blue coin was faster than that white coin. Here I've shown the picture of this before collision. But notice what happens after collision. The white coin becomes faster than the blue coin. Again, here is the picture of this after collision. But what's interesting is that even though their speeds have changed after collision, their momentum stays conserved, what we mean by that is, if you calculate its total initial momentum of these two coins before the collision, that will equal it's total momentum after the collision. And the goal of this video is to try and prove that. We'll mathematically prove this statement. And before we do that, if you're not familiar with this concept, then we'll talked a lot about this, with few examples in a previous video called Conservation of Momentum. So it will be a great idea to go back and watch that and then come back over here. Anyways, if you are ready, then let's go ahead and prove this. Whenever we want to prove something, the first obvious question we might have is, where do we start? We can start by trying to write this mathematically. What do we mean by total initial momentum? Well remember, momentum is just mass into velocity. So total initial momentum would be, mass into velocity of this coin before the collision plus mass into velocity of this coin. And total final momentum would be, again, the same thing, after collision. So since there will be masses and velocities, let's first go head and give names to them. So, let's say, this blue coin, let me call that as coin number one. And let's call its mass as m one. Similarly, let's say the white coin is coin number two and let's call its mass as m two. And because their velocities are changing, let's name that as well. Let's say that the initial velocity of that first coin is u one before the collision. And after the collision it becomes v one. And similarly for the white coin, before the collision, let's say it's initial velocity is u two. And after the collision, let's say it becomes v two. Okay, so the first thing, we'll have to do now is write this equation, what we need to prove, mathematically in terms of m one, u one, and everything. And you know what will make this derivation great? If we can do it together. So in between, you may have to pause and try some steps yourself. So it'll be nice if you could have a pen or a pencil and a paper with you. All right, so our first step is to write this statement mathematically. And so, can you try that yourself first? See if you can write this statement using these terms mathematically first. Go ahead, pause the video and give this a try. All right, so total initial momentum would be, the mass into initial velocity of the first coin. That's going to be m one u one, plus mass into initial velocity of the second coin. That's going to be m two u two total initial. And that will be equal to mass into final velocity of the first coin, that is m one v one, plus mass into final velocity of v two. So this what we need to prove. Awesome. Next question we might have is, how to prove that? Where do we start? Well in any derivation, to figure out how to start, we need to know what concept needs to be applies over there. So, what concepts need to be applied for momentum conservation? Turns out, that momentum conservation actually comes from Newton's third law. And so for me, this is the most important step in the derivation. If we remember that Newton's third law is responsible for momentum conservation, then we can start and we can derive everything. So let's see how. What does Newton's third law say? It says that, "For every action "there is equal and opposite reaction." And these actions and reactions are forces, right? So where does force come into the picture over here? Ah, during collision. When the blue coin goes and hits this white coin, it pushes that, it puts a force on that. So, the secret to deriving this is to look at the collision. All right, so, let's look at the collision instance. So let's say the blue coin hits the white coin this way and puts a force on it. And we'll give a name to that as well. Let's call that as f two. And at the same time, two will push back on one. And so they'll be a force on our first coin as well. And let's call that force f one. Now what's Newton's third law telling us? It's saying that f one must be equal and opposite to f two. And so that's where we can start. So again, great idea to pause the video and see if you can write this statement that f one is equal and opposite to f two mathematically. Because that's the starting point. Again, give this a shot. All right, so f one is equal and opposite to f two. So we can write this as f one is equal and opposite, how do we write opposite? Well, mathematically, we usually write opposite as negative. So it'll be negative f two. Okay, what next? Well, we have an equation with force, but I want to bring mass and velocity into the picture. So, can we think of any connection between forces and masses and velocities? I'm pretty sure you know this one. Newton's second law, one of the most famous equations in physics. F equals MA. So we can write Newton's second law for both these forces. We can substitute that. So if we do that, f one, we can write that equals m one times a one. That's for the first coin. And similarly, for f two we can write it is equal to m two times a two. So we have brought in the mass into the picture. We are slowly coming there, that's nice. But what we don't want is acceleration. I don't want A and I want velocities. So the next thing we can think is, is there a connection between acceleration and velocity? And I think you know that. Acceleration is rate of change of velocity, isn't it? So again, great idea to pause the video and see if you can substitute for a one and a two in terms of velocities. Again, give it a try, pause the video and give this a try. Okay, hopefully you've tried. So we'll get m one times a one is the rate of change of velocity of the first coin. So, to concentrate better, let's dim the white coin. So we will get rate of change of velocity of the first coin which is final velocity minus the initial velocity. That's the change. And rate of change means you have divided by time. And so, let's say that time is T. That T is the time it took for this blue coin to go from initial to final. And if you're wondering why I'm writing T and not t one, I'll come back to that a little bit later. Anyways, that will be equal to negative m two times now to calculate a two, let's dim the blue ones, the blue coin. Okay, so that will be equal to m two times a two, which is final minus initial. That's v two minus u two divided by time T. So, why am I writing time T both, why not t one and t two? Well, the reason is, these two times are exactly the same. I mean, again, if we bring everything back, this time is the time it takes for them to change their velocity, right? And when are they changing their velocities? It's during the collision, right? It's during the collision, forces act on them and they change their velocity. So this time is actually the collision time. But collision time, it must be the same for both of them, right? I mean, if one is in contact with two for let's say, five milliseconds, then two will also be in contact with one for exactly that same time, five milliseconds. And so that's why the times must be the same. And so, that means we can cancel them. And now, look, we are brought in everything that we wanted into the picture. We have masses and we have initial and the final velocities, which means the physics is done. Now all we have to do is algebra and we'll get this. And again, good time to pause the video and see if you can do the rest of the derivation and get what we want. Go head, pause the video and try. All right, so, final steps, all we have to do is just simplify that little bit of algebra. So if we open the brackets, we'll get m one v one minus m one u one equals, if you open the brackets here, now be a little bit careful with the signs, we'll get negative m two v two, negative m two v two. And we'll negative m two into negative u two. That becomes plus, right? Minus times minus is plus. So that's get plus m two u two. And all you have to do now is rearrange this equation, such that all the U terms are on one side and all the V terms are on the other side. And I'm pretty sure you can do that. And if you do that, we get what we want. And there we have it. So we have derived what we wanted. We found that the total final momentum should equal total initial momentum. And so just to quickly summarize, how did we do this derivation? Well, we first wrote what we wanted in a mathematical form, because we are deriving mathematically. Then we asked ourselves, what concept do we have to apply? Which is the most important step in our derivation and that is the Newton's third law. Once we understood that, we started with Newton's third law in the mathematical form and we just kept substituting and then we did some algebra and we derived this.