Linear momentum and impulse

0:00- [Instructor] Let's talk about momentum and impulse. So we can start directly with Newton's second law, which says that the net force acting on object equals mass of the object times the acceleration of that object. But we know acceleration is rate of change of velocity, so I can write this as m times dv over dt.

0:20But if mass is a constant, I can put it inside the derivative, right? So I can now write the net force equals d/dt of mv. This is interesting because now we can rethink of Newton's second law as the net force acting on object equals the rate of change of this quantity.

0:44Let's call this quantity p. Then we can say that net force acting on object equals rate of change of P, and this is what we call momentum. So momentum P is just the product of mass times velocity of an object. Notice that it's a vector quantity and it has the same direction as that of velocity.

1:09And what's the unit of momentum? Well, it's just the unit of mass times unit of velocity, so it's kilogram meters per second. So this equation says that the net force acting on an object equals the rate at which its momentum is changing. Notice that this is the same thing as f equals ma, so this is literally Newton's second law.

1:30It's written slightly differently. But let's get some intuition for this. This is saying that if I were to put a small force on an object, like for example if I were to push a baseball with a little bit of force, then its momentum changes very slowly. This arrow marker represents its momentum, okay?

1:50On the other hand, if I were to whack that baseball with a bat, putting a much bigger force on it, now its momentum changes very quickly, right? That makes intuitive sense, right? How quickly the momentum of an object changes purely depends on net force acting on it.

2:08But here's where things get really interesting. We can now say that if an object's momentum is changing, there must be a net force acting on it, right? That's another way to think about it. All right, so net force equals rate of change of momentum. We have written the Newton's second law in differential form.

2:24Is there a way we can write the same equation in an integral form? Yeah, let's try and do that. How would we go about doing that? Well, first of all, we can multiply both sides by dt. So I would get net force times dt equals dp, right? Now if I want to write this in integral form, I just have to integrate on both sides.

2:44But what would be the bounds of the integral? Well, on the right hand side, I'm integrating with respect to momentum. So the bounds would be some initial momentum to some final momentum. Let's call it as P1 to P2. On the left hand side, I'm integrating with respect to time, so it'll be from some T1 to T2.

3:01Okay, the left hand integral depends upon the force function, but what about the right side integral? Here, right side, I'm just integrating one with respect to P. What's that? Well, that's just P and with the bounds P1 to P2, so it's gonna be P2 minus P1. In other words, the right hand side is the change in momentum.

3:23And so now I can write that this integral equals change in momentum. So we give this integral a name, we call it the impulse and the letter, the symbol that we use is J, capital J. Notice that capital J is also a vector quantity, it has the same direction as the net force, and what we find is that the impulse equals the change in momentum.

3:53This equation is called the impulse momentum theorem. Now we will dig deeper into what this equation is telling us, but before we do that, let's do a quick unit check. What do you think is the unit of impulse? Well, the unit of impulse is going to be the unit of force that is Newtons times seconds.

4:10But because of the impulse momentum theorem, I can also say the unit of impulse has to be the unit of momentum, which is kilogram meters per second, which means these two units must be the same. It will be a great idea to pause the video and just confirm that.

4:24All right, remember that Newtons is basically kilogram meters per second square. So the second cancels and you end up with kilogram meters per second. But onto the main question, what exactly does this equation tell us? Well, let's take a concrete example. Suppose you're catching a fast moving baseball, how do you catch it?

4:44Well, you naturally retract your hand like this, right? But why? You probably know because it hurts less. I mean, if you were to catch a ball this way without retracting your hand much, it would hurt a lot. But why? Why does retracting your hand reduce the pain?

5:01The think in terms of impulse and momentum. I want you to pause the video and compare the impulse delivered in the first case versus the second case, in which case more impulse is delivered. Pause and think about this. All right, since I know that impulse equals change in momentum, I just have to figure out in which case the momentum change is bigger.

5:19Well, in both cases, the initial momentum is the same because we're talking about the same ball, the same scenario, right? And what's the final momentum in both cases? It's again zero because in both cases they have come to a rest, which means the change in momentum in both cases are the same.

5:34So in both cases, the impulse is exactly the same. Interesting. But now think about this: over here because you retracted your hand, the time over which that impulse was delivered, this time, that was much higher compared to over here. So that time is much higher over here compared to over here.

5:56So what about the net force that's delivered? Here's where we can do a simplification. Let's assume that the force, the net force acting over here is a constant. It's not really, but let's assume that. If the net force is a constant, then look at what happens to this integral.

6:14We can pull out the net force outside the interval and then you'll just get integral of dt, which is just gonna be delta T, T2 minus T1. In other words, if a force is a, the net force is a constant, impulse is just the product of net force and the time interval over which that net force is acting, right?

6:32So now, that product must be exactly the same because we know the impulse is the same, but over here, delta T is much higher than over here. Therefore, in order for the product to be the same, the net force over here must be smaller compared to over here. That's the only way the product can stay exactly the same.

6:53And as a result, you can see the net force delivered over here is smaller compared to over here. So we can now get the full picture. In both cases, the change in momentum is exactly the same. So the impulse delivered is exactly the same. But over here, when you retract your hand, you are increasing the time over which that impulse is delivered.

7:15Because you increase the time, the net force acting reduces. That's why your hand hurts less here compared to over here. Does that make sense? But of course, we made a big assumption just to simplify the math over here. We made the assumption that the net force is a constant, which is not really true.

7:39I mean think about it: before the ball comes in contact with your glove, the net force is zero, at least in the horizontal, I'm only considering horizontal forces, ignoring vertical forces, okay? Then the moment it comes in contact, the net force, there is some net force, so the net force is actually increasing.

7:55And then eventually the net force again goes down to zero once the ball has come to rest. So the net force is not really a constant, it's changing with time. So now, how do you think about this? How do we visualize it? How do we get an intuition behind, the difference between the changing net force over here versus over here?

8:17Well, for that, let's resort to graphs. What if we were to plot a graph of the net force versus the time? Let's see what this and what information the graph can give us. Again, let's first start with a constant net force just to get an intuition and then we'll talk about these two cases, okay?

8:37So if the net force is a constant, some constant value and I wanted to calculate the impulse or the change in momentum from some T1 to T2, how would I do that? Well, we saw that if the net force is a constant over that time interval, then the impulse would just be the product of net force and the time interval.

8:58But what does that graphically represent? What is net force times the time interval in the graph geometrically? That's just the area, right? So that means geometrically, impulse represents the area under the force time or the net force time graph. And by the way, over here, we only consider the magnitude of the net force, don't worry too much about the direction over here.

9:23So that area gives us the impulse. But at this point, I have a confusion, haven't we learned somewhere before that the area under the force graph is work done? Well actually, no. I mean, if we recall what work done is that is, or at least the net work done on object, that is the line integral of the net force, .dr.

9:53You're integrating over here the net force with respect to position. So work is the area under the force position graph. Impulse is the area under the force time graph. We don't really have to memorize any of this, you just look at the equations and it tells us that.

10:14And while we are at it, remember the work energy theorem? It says that the net work done on an object equals the change in kinetic energy. Well now, we have the full picture. What we notice is that if you integrate the force with respect to position, that gives you change in kinetic energy.

10:31That's work energy theorem. But if you integrate the net force with respect to time, that gives you the change in momentum, that's impulse momentum theorem. Isn't that amazing? We have the full picture in front of us. But of course, let's come back to our example.

10:51When you're catching a ball, the net force is not a constant, but we don't have to worry about it anymore because we have found graphically how to calculate the impulse. We just have to calculate the area under the graph. This will work even if the net force is not a constant.

11:08All right, so now let's consider the case where you're catching the ball without retracting your hand. Where it hurts a lot, okay? So what would the graph of the force versus net force versus time look like over here? Well, let's see. We know that before contact, the force is pretty much zero, there is no net force.

11:25And then we also know that once the ball has come to a rest, again the net force goes down to zero. So it should rise and then come back down. And so something like this, the graph would look like. And the impulse delivered or the change in momentum, that would be the area under this graph, right?

11:46Okay, now let's think about this particular case. What would the graph for this one look like? Well, since you're retracting your hand, you are increasing the time over which that force is acting. This graph would be much wider, but, and here's the important part, the area under this graph should be the same as this one because the impulse is the same.

12:10So if the graph goes wider, its peak must be much shorter or much lower, otherwise, you wouldn't have the same area. So notice the difference between the force function between these two. Here, the peak force is much higher, substantially higher than the peak force over here.

12:34That's the reason it hurts more. I mean let's appreciate that we are actually dealing with very complicated real world scenarios where forces are changing with time and yet we can now get some intuition behind what's going on, that is pretty awesome, that's why I absolutely love the impulse momentum theorem.

12:55So the whole idea is that when you're delivering impulse, if you extend the time over which that impulse is delivered or extend the time over which the momentum is changing, the graph widens and as a result, the peak force drops. People do the same thing in boxing.

13:12When your opponent is punching you, there's something called riding the punch. You move your face along, in the direction of the punch so that you ride it, you take that punch over a long time interval so that the peak force delivered by that punch would be much smaller than if you didn't do that.

13:30This is the reason why cars and vehicles have crumple zones. So if at all there was an unfortunate event of a crash, the momentum would change more slowly. So the peak force reduces. Airbags do the same thing. I mean you can see this impulse momentum theorem being applied in action almost everywhere, in almost all the collision cases that you would find.

13:53Okay, now we'll do one last graph. What if we were to plot a graph of momentum versus time of an object? And let's just take an example, just a random example. Let's say the graph look like this. The final question for us is, what does the slope of this graph represent?

14:12Why don't you pause the video and think about it? All right, we know that the net force acting on object equals rate of change of momentum. And rate of change is basically slope of the tangent at a particular point. In other words, the slope of a tangent at any point over here represents the net force at that particular point.

14:32So let's analyze this particular graph and see what's happening. So initially, the graph is pretty much flat, therefore, the slope is pretty much zero, so net force over here is zero. Then look, the graph starts sloping downwards. So you start having a negative slope.

14:48Let's not worry about the signs, but what's important is it becomes steeper, so the net force is increasing. The net force is increasing, increasing, oh my god, this is the maximum steepness you get over here, you have a maximum net force, magnitude is very high and then the net force reduces, reduces, goes back to zero.

15:05Hey, can you see, we are actually talking about a situation very similar to this, right? So this situation is kind of sort of similar to what we saw, either this one or this one, okay? Now, here's the final question. What if this particular graph represented this situation?

15:26When you are catching a ball, retracting your hand, okay? The momentum is changing over a long time. What would the graph look like for this situation? The momentum time graph look like for this situation? Final, question, final time to pause the video. Okay, so the initial momentum would be the same.

15:44The final momentum would also be the same as over here, zero. So the graph would go from here to here, pretty much the same kind of graph. It's just that it'll all happen much quicker, so the graph would look like this. And so if you now over here look at the slopes, oh, the slopes are pretty high, it's pretty steep over here.

16:03So you can see this slope over here, which is the net force at this point is much higher than the slope over here, net force over here. So this pretty much represents a case like this. So slope of the momentum time graph gives you the net force and we can see that from this equation.

16:22And the area under the force or the net force time graph gives us the impulse or it gives us the change in momentum. Again, we don't have to memorize any of this. Integrals always gives us areas and derivatives always gives us slopes.