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*v_{final} minus m*v_{initial} 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*v_{final} minus m*v_{initial}. 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, v_{final}. How do we figure that out? Well, we're going to write our impulse equation, J = f*Δt and this equals m*v_{final} minus m*v_{initial}. We want to find this v_{final} here, so we just have to figure out everything else. Well, remember that the box initially starts with an initial_velocity_{vi} of 0. So actually, there is no initial momentum, so m*v_{final} - m*v_{initial} = 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*v_{final}. So now I'm just going to move the m over to the other side, and I have that v_{final} 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.

# Intro to Impulse - Online Tutor, Practice Problems & Exam Prep

### Impulse & Impulse-Momentum Theorem

#### Video transcript

You throw a 100-g ball with 30m/s. If the ball is in your hand for 0.2s during the throw, a) calculate the impulse

you deliver to it. b) Calculate the average Force that you exert on the ball.

$J=3,000\operatorname{kg}$•m/s, $F_{avg}=15,000N$

$J=3\operatorname{kg}$•m/s, $F_{avg}=15N$

$J=0.2\operatorname{kg}$•m/s, $F_{avg}=1N$

$J=2\operatorname{kg}$•m/s, $F_{avg}=10N$

### Impulse between Bouncy Ball & Wall

#### Video transcript

Hey, guys. Let's take a look at this problem here. So I have a bouncy ball or a rubber ball that's going to strike off of a wall. Let's go ahead and draw that out real quick. So I have this bouncy ball like this, the m is equal to 0.15, and initially, it's going with some speed, which is 40 meters per second to the right. Then it strikes the wall like this, and then afterward, it's going to rebound and bounce backward. So then after the collision or after the bounce off of the wall, it's going to be going down to the left. And because it's going to the left, when I write the final velocity here, it's not going to be 45 meters per second. It's going to be negative 45 meters per second. If you have opposite directions and velocities, you're going to have to pick a direction of positive and stay consistent with that throughout the problem. So let's take a look at part a. In part a, we want to calculate the impulse, which is j, delivered to the ball from the collision with the wall. Alright? So that's j equals, and we have an equation for that, f times delta t. But this is also equal to the change in momentum, which is m(v_{final} - v_{initial}). Alright. So we just basically look at all the variables, which ones we have and which ones we don't. So we have the force. We're told here in the second part, the average force is 410 newtons during the bounce. That force is also directed to the left like this. So this is our force here. So we actually don't have what the change in time is. In fact, that's actually what we're going to calculate in part b. So we actually can't use f times delta t, but that's okay because we have mass, we have the final velocity, and we have the initial velocity, So we can calculate the impulse by using that side of the equation. Alright. So let's get to it. So j is going to equal. We have the mass, which is 0.15. Now I have to be careful here, so I'm going to write a little bracket. My final velocity is negative 45 meters per second. So this is negative 45 - the initial velocity, which is just 40. So keep track of your minus signs. What you actually end up getting here is negative 12.75 kilogram meters per second. That's the answer to part a. That's the impulse. Notice how it's negative, and that makes sense because the impulse is actually going to point to the left, because it's basically going to cause a change in momentum to the left like this. So your impulse should be negative. Let's take a look at part b. In part b, now we want to calculate how long or the amount of time that the ball is in contact with the wall. So basically, from here to here, the initial to final, the force is acting over a very small delta t in order to create an impulse that acts to the left, and that's what we want to figure out here. So we know that the force that the wall exerts on the wall during the bounce is going to be 410, but in you know, to stay consistent with our directions, because it points to the left, I'm actually going to write a negative sign here. It's going to be negative 410 newtons. So we actually have what our impulse is. We just calculated that in the last part. We have this. And now we basically want to use the other side of the impulse equation which is f times delta t. So f times delta t here, we want to figure out what's this delta t. We actually have what the force is, so we can go ahead and do that. So we know this is negative 12.75 equals negative 410 times delta t, and all you have to do is just do a division. So basically, our delta t is going to be Well your negatives are going to cancel out because there are 2 of them. This is basically going to be 12.75 divided by 410, and you should get a delta t of 0.031 seconds. Alright. So it's about 31 milliseconds for the bounce. That's really it. So let me know if you guys have any questions.

You catch a 0.6 kg ball initially moving with 10 m/s. Calculate the impulse delivered to the ball during the catch.

$16N\cdot s$

$6N\cdot s$

$0.6N\cdot s$

$60N\cdot s$

### Car Colliding with Wall

#### Video transcript

Guys. Let's take a look at this problem here. So we have a 1200 kilogram car that's unfortunately going to collide with a wall. This is our 1200 kilogram car like this. It has an initial speed of 20. Unfortunately, it's going to hit a wall and over the course of the collision, it actually crumples through a distance, a displacement. So basically, the point of impact like this, it's going to crumble over distance delta x of 1 meter. What I want to do in part a is I want to figure out how long the collision lasts. So that's a time. So how long the collision lasts is actually going to be a delta t. So I know the distance that it deforms or crumples over, but I want to figure out what is the delta t here. So let's take a look. Which equation are we going to use? Well, so far, we actually only need one equation that deals with delta t, and that's the impulse equation. So let's go ahead and write this out. We have J=f⋅Δt=Δp=mV_{final}−V_{initial}. So let's take a look at all of our variables. What happens is we don't know the force. We don't know the time. That's actually what we're looking for here. And we do have the velocity, the mass. We do have the final velocity and initial velocity. So what happens is even if you could figure out the impulse by using this m times V initial minus V final, V final minus V initial to figure out the impulse, you'd still be stuck because you actually still have 2 unknowns in this problem. You have f and delta t; they are both unknown. So we wouldn't be able to solve for delta t using the impulse equation. So what else can we use? Well, fortunately, what the problem tells us is that we can assume the acceleration during the collision is constant. What that means is you can actually go back and use our old motion or kinematics equations to solve for delta t. Right? So that's really all we're going to do. We're just going to go ahead and solve for delta T by using another set of equations that we've seen before. So how do we figure out this delta T here? Well, we have to figure out 3 out of 5 variables. So I have my delta x, V initial, V final, a, and t. So this is what I'm looking for here. It's the whole reason I came over here to try to find delta t. So let's write out all I need is 3 out of 5. I have my delta x is 1, I have my initial speed is 20, final speed is 0. The a actually I don't know, so I'm just going to use that as my ignored variable, and that's perfectly fine because I have 3 out of 5, and I can write an equation that deals with or that solves for delta t. Now what happens is remember, when you have the acceleration as your ignored variable, the equation that you're going to use is actually equation number 4. Now remember, some professors don't necessarily like you using this equation, so I'm going to put an asterisk here. Just make sure your professor does allow you to use this equation. If they don't, you still can solve this. Basically, you just have to solve using equation number 2, what the acceleration is. Then you can plug it into either one of equation number 1 or 3 to figure out the time. We're just going to skip that and we're going to go ahead and use the equation number 4 here. Alright. So let's do that. So equation number 4 says, this is your delta x equals this is the initial velocity plus final velocity divided by 2 times the time. Just going to go ahead and plug this in my numbers. This is 1 equals 20 + 0 divided by 2. So that's that. And then we have times delta t. So So basically, I have 10 = 10t. And so therefore, your delta t is equal to 0.1 seconds. Alright. So that's actually how you figure out delta t here. Not by using impulse, but actually by going back and using kinematics to solve these kinds of problems. Sometimes you'll have to do that. So now that we figured out delta t is equal to 0.1 seconds, now let's take a look at the second part of our problem. In the second part, we want to figure out the magnitude of the average force that the wall exerts on the car during the collision. So basically now, if we go back to our impulse equation, we actually know what this delta t is, And the only variable that we're looking for now is f. And because we only have one variable, now we can go ahead and solve for this. Right? So we only have one unknown variable. So j for part b is going to be f times delta t. And so this is going to be, let's see, delta p equals mV_{final}minusV_{initial}. Just going to write that out again. Now we're looking for the average force, and we know what delta t is. We also know what m, V final, and V initial are. So basically, I'm just going to move everything over to the other side, or this delta t over to the other side, And my f is just going to equal Let's see. I've got the mass which is 1200 times the final velocity which is 0. The initial velocity was 20, and then we're going to divide that by the delta t, which is 0.1 seconds. If you go ahead and work this out, what you're going to get is 240,000 newtons. Because we were only looking for the magnitude of the force, technically, actually, it was supposed to be negative because of this negative sign right here. We could just leave it as positive because all we're looking for is the magnitude. So this is really our answer, 240,000 newtons. It's a massive force because it acts over a very, very short period of time to stop a heavy car that's moving at 20 meters per second. So you need a massive amount of force to do that. Alright, guys. So that's it for this one. Let me know if you have any questions.

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