Guys, in this video, I'm going to introduce you to another physical quantity that's related to momentum that you're absolutely going to need to know, and it's called impulse. So let's take a look here. Basically, when an object experiences an impulse, which we'll use the letter J for, it experiences a change in momentum. So the way we rewrite this is J is equal to Δp. I'm going to come back to this in just a second here. What I want to do is actually want to sort of rearrange this equation to be a little bit more useful. So remember that p is just m*v and Δ just means final minus initial. So really Δp is really just equal to m*vfinal minus m*vinitial here. Now, how do we actually get from J to ΔP? This actually is going to be done in your textbooks. I'm going to show you this really quickly. It really just comes from Newton's second law, which remembers f = m*a, but we're going to rewrite this in terms of momentum. I'm going to show you real quickly. So we're going to use the definition for a, which remember the acceleration is just Δv over Δt. But now what I can do is in these problems, the mass is always going to remain constant. So I can actually bring it inside of the Δ sign and that's perfectly fine. So I have Δ(m*v) divided by Δt. Now you should know by now that actually this m*v is really just your momentum. So we can rewrite this and we can say that f is equal to Δp over Δt. So there are 2 ways we can actually now, sort of express or write Newton's second law. We can say f = m*a, but the way that Newton originally wrote it was that f is actually equal to Δp over Δt. So it's change in momentum over change in time. So what I can do here is I can actually rewrite this new expression now, and I can get back to impulse. So I have f is not equal to m*a. It's just equal to Δp over Δt, and now I'm just going to move the change in time over to the other side. And what I come up with is I come up with f*Δt is equal to Δp. This thing here on the left side is actually what is the definition of an impulse. So J is really just f*Δt. The way I like to think about this is if you look up the definition for impulse, an impulse is like a very sudden thing that happens over a very short amount of time. That's exactly what's going to happen in your problems. You're going to have, in physics, forces that act over some change in time and that's where an impulse is. So really what happens is we're just going to write this equation, J = f*Δt, and then you could write either one of these sort of, but I like to write like this, using basically these three terms, J, f, Δt, and the change in the momentum. So the units that we're going to use are either going to be Newton-seconds and it really just becomes, that comes from force times time, or if we're talking about change in momentum, then the units are just going to be the same as momentum, kilogram meters per second. Let's go ahead and work out a problem here. So here we have a crate that is initially at rest. So the initial velocity is 0, and we're going to push it now. We're going to push it with this 100 Newton force, that's an applied force, for a certain amount of time. It's going to be 8 seconds here. This is going to be a Δt. And in part A, I want to calculate the impulse that I'm delivering to the crates. So what happens here is in part A, I want to calculate the impulse which is J. And remember, I'm always going to write it as f*Δt, and this is also related to m*vfinal minus m*vinitial. J is really just f*Δt, but it's also the change in the momentum. We'll talk about that in just a second here. So if you look through my variables, what I have is I have f and Δt. So I can actually just use this first part here, this first equals sign to calculate the impulse. So your J is just equal to the force which is 100 times the time, which is 8, and this is equal to 800 Newton-seconds. And that's the answer. That's the impulse, 800. So let's take a look at part B now. In part B, we want to calculate the crate's speed after 8 seconds, and we want to use impulse to do this. So basically, what happens is that once you've you exerted, you know, this force over Δt of 8, the box is going to be over here and because you've pushed it over some time, it's going to have some final velocity here, vfinal. How do we figure that out? Well, we're going to write our impulse equation, J = f*Δt and this equals m*vfinal minus m*vinitial. We want to find this vfinal here, so we just have to figure out everything else. Well, remember that the box initially starts with an initial_velocityvi of 0. So actually, there is no initial momentum, so m*vfinal - m*vinitial = 0. Now remember that we just calculated what f*Δt is. This is just 800. So what we can actually say is that 800, which is the impulse, is equal to m*vfinal. So now I'm just going to move the m over to the other side, and I have that vfinal is equal to 800 divided by 50. And if you work this out, you're going to get 16 meters per second. Alright? So that's how you use both sides of the impulse equation to figure this out. In all these problems, you're going to be given 3 or 4 out of or or 3 out of these 4 variables, f, Δt, m, and some combination of the velocities. And as long as you have 3 out of those 4, you can always figure out the other one by using this equation here. Alright? So I have one last point to make here, which is that we're actually going to see a lot of similarities between impulse and momentum and work and kinetic energy, which we've seen before. So impulse kind of relates to momentum the same way that work relates to kinetic energy. Remember that J is equal to it's defined as the force times Δt. In the same way that the work is defined as f times Δx. Now we actually used f, you know, d cosine theta, but the sort of simplified version is if you have force and displacement in the same directions f*Δx. So both of these things sort of defined as force times a change in either displacement or time, and what they cause is the is f*Δt causes a change in your momentum. Right? So we saw in this problem here that this impulse causes a change in your momentum. In the same way that the force actually caused a change in the kinetic energy. So there are lots of similarities there, some pretty interesting things. So that's it for this one, guys. Let me know if you have any questions.
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11. Momentum & Impulse
Intro to Impulse
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