Intro to Impulse - Video Tutorials & Practice Problems

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1

concept

Impulse & Impulse-Momentum Theorem

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6m

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Hey guys, so in this video, I'm going to introduce you to another physical quantity that's related to momentum that you're absolutely gonna need to know. And it's called impulse. So let's take a look here basically when an object experiences an impulse, which will use the letter J. For it experiences a change in momentum. So what we rewrite this is jay is equal to delta P. I'm gonna I'm gonna 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 delta just means final minus initial. So really delta P is really just equal to M. V. Final minus M. V. Initial here. Now, how do we actually get from J? Two delta P. This actually is gonna be done in your textbooks. I'm going to show you this really quickly. It really actually just comes from Newton's second law, which remembers F equals M. A. But we're going to rewrite this in terms of momentum. I'm gonna show you real quickly. So we're gonna use the definition for a which remember the acceleration is just delta V over delta T. But now what I can do is these problems, the mass is always going to remain constant. So I can actually bring it inside of the delta side and that's perfectly fine. So I have delta Mv divided by delta T. 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 delta P over delta T. So there's two ways we can actually now sort of express or write Newton's second law, we can say F equals M. A. But the way that Newton originally wrote it was that F is actually equal to delta P over delta T. So it's changing momentum over changing time. So what I can do here say, I actually I can actually rewrite this new expression now and I can get back to impulse. So I have F. Is not equal to Emma is just equal to delta P over delta T. And now I'm just going to move the change in time over to the other side. And when I come up with is I come up with F times delta T. Is equal to delta P. This thing here on the left side is actually what is the definition of an impulse? So J. Is really just F times delta T. The way I like to think about this is if you look at the definition for impulse, an impulse is like a very sudden thing that happens to a very short amount of time. and that's exactly what's going to happen in your problems you're gonna have in physics forces that act over some change in time and that's what an impulse is. So really what happens is we're just going to write this equation, jake was F delta T. 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 delta T. And the change in the momentum. So the units that we're gonna use are either gonna be newton seconds. And it really just becomes uh that comes from force times time. Or if we're talking about changing momentum, then the units are just gonna 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 zero and we're gonna push it now, we're gonna push it with this 100 m 100 newton force. That's applied force for a certain amount of time. It's gonna be eight seconds here. This is gonna be a delta T. And in part, I want to calculate the impulse that I'm delivering to the crates. So what happens here is in part, I want to calculate the impulse which is J. Remember I'm always going to write it as F delta T. And this is also related to M. V. Final minus M. V. Initial jay is really just F delta T. 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 or to have is I have F and delta T. So I can actually just use this first part here. This first equals sign. To calculate the impulse. So you're J is just equal to the force, which is 100 times the time which is eight. 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 creates speed after eight seconds and we want to use impulse to do this. So basically what happens is that once you've exerted this force over delta T. Of eight, the box is gonna be over here. And because you've pushed it over some time, it's gonna have some final velocity here. V final. How do we figure that out? Well, we're gonna write our impulse equation, jay equals F delta T. And the Sequels mv 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 of zero. So actually there is no initial momentum. So in the final and initial zero. Now remember that we just calculated with delta T. Is this is just 800. So we can actually say, is that 800? Which is the impulse is equal to M Times the final. So now I'm just gonna move the em over to the other side and I have that the final is equal to 800 divided by 50. And if you work this out, you're gonna get 16 m/s. All right, so that's what you sort of both sides of the impulse equation to figure this out. And all these problems, you're gonna be given three or four out of three out of these four variables F delta T. M. And some some combination of the velocities. And as long as you have three out of those four, you can always figure out the other one by using this equation here. All right, 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 jay is equal to it's defined as the force times delta T. In the same way that the work is defined as F times delta X. And we actually used F. D. Co signed data. But the sort of simplified version is that if you have forced displacement in the same direction, F times delta X. So both of these things sort of sort of defined as forced time to change in either displacement or time. And what they cause is the F. Times delta T. 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's a lot of similarities there. Um some pretty interesting things. So that's it for this one. Guys, let me know if you have any questions.

2

Problem

Problem

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.

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 the 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 m/s to the right, then it strikes the wall like this and then afterwards it's going to rebound and bounce backwards. So then afterward 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 gonna be 45 m/s, it's going to be negative 45 m/s. If you have opposite directions and velocities, you're gonna have to pick a direction of positive and stay consistent with that throughout the problem. So, let's take a look at part a and part. We want to calculate the impulse which is J. Delivered to the ball from the collision with the wall. All right, 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 minus V. Initial. All right. So we just basically take a look at all the variables which ones we have, 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 forces also directed to the 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 a that's actually 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 final velocity and we have initial velocity. So we can calculate the impulse by using that side of the equation. All right, so let's get to it. So J is gonna equal we have the mass which is 0.15. Now, I have to be careful here, so I'm gonna write a little bracket. My final velocity is negative 45 m per second. It is negative 45 minus the initial velocity, which is just 40. So keep track of your minus signs when you actually end up getting here is negative 12.75 kilogram meters per second. That's the answer to part. 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 and part B. Now we want to calculate how long the amount of time that the ball is in contact with the wall. So basically from here to here, the initial to final, the forces acting over a very small adult A. 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 the that the Wall exerts on the ball during the balance is going to be 410 But continue to stay system with are consistent with our directions because it points to the left, I'm actually going to write a negative sign here. It's gonna be negative 410 Nunes. 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 have 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 you 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, your negatives are going to cancel out together as two of them. This is basically going to be 12.75 divided by 410. And you should get a delta T. Of 0.31 seconds. All right, so it's about 31 milliseconds for the bounce. That's really it. So let me know if you guys have any questions

4

Problem

Problem

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

A

$16N\cdot s$

B

$6N\cdot s$

C

$0.6N\cdot s$

D

$60N\cdot s$

5

example

Car Colliding with Wall

Video duration:

5m

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Hey guys, let's take a look at this problem here. So we have a 1200 kg car that's unfortunately gonna collide with a wall. So this is our 1200 kg 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 displacement. So basically the point of impact like this, it's gonna crumple over a distance delta X. Of one m. 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 that sort of deformed or crumples over, but I want to figure out what is the delta T here. So let's take a look which equation we're going to use. Well so far, we actually only 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 equals F, times delta T, which equals the change in momentum, Mv final minus 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 of the mass, We do have the final velocity initial velocity. So what happens is even if you could figure out the impulse by using this, M times V initial minus the final of the final days of the initial to figure out the impulse, you would still be stuck because you actually still have two unknowns in this problem. You have F and delta T. They're both unknown. So you 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 we can actually go back and use our old motion or kidney. Matics equations to solve for delta T. Right? So that's really all we're gonna do. We're just gonna go ahead and solve for delta T. By using another set of equations we've seen before. So how do we figure out this delta T here? What we have to write three of us? We have to figure out three out of five variables. So I have my delta X, the initial, the final A and T. So this is what I'm looking for here. It's the whole reason that came over here is try to find delta T. So let's write out all I need is three out of five. I have my delta X. Is one. I have my initial speed is 20. Final speed zero. The A. Actually, I don't know. So I'm just going to use that as my ignored variable and that's perfectly fine. I have 3 to 5 and I can write an equation that deals with that souls 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 four. Remember, some professors don't necessarily like you using this equation. 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 two what the what the acceleration is. Then you can plug it into either one of equation number one or three to figure out the time. We're just gonna skip that and we're gonna go ahead and use the equation number four here. Sorry, let's do that. So equation number four says this is your delta X equals this is the initial velocity plus final velocity divided by two times the time. Just go ahead and plug plug in my numbers. This is one equals 20 plus two, divided by two. I'm sorry, 20 plus zero divided by two. Since that's and then we have times delta T. So basically I have 11 equals T. And so therefore your delta T is equal to 0.11 2nd. Alright, so that's actually how you figure out delta T. Here. Not by using impulse but actually by going back and using cinematics to solve these kinds of problems. Sometimes you'll have to do that. So now that we figure out delta Z is equal to 0.1 seconds. Now, let's take a look at the second part of our problem. 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 the only equation. The early variable 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 gonna be, let's see delta P equals M. V final minus the initial. Just gonna write that out again. Now we're looking for the average force and we know what delta T. Is. We also know an M. V. Final and V. Initial are basically, I'm just gonna move everything over to the other side or this delta psi over to the other side and my f. Is just an equal. Let's see, I've got the mass, which is 1200 times the final velocity, which is zero. The initial velocity was 20. And then we're gonna divide that by the Delta T, which is 0.1 seconds. If you go ahead and work this out, what you're gonna get is 240,000 newtons. Because we were only looking for the magnitude of the forest, technically actually was supposed to be negative because of this negative sign. Right here, we can 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 actually a very, very short period of time to stop a heavy car. It's moving at 20 m per second. You need a massive amount of force to do that. Alright guys, so let's look at this one. Let me know if you have any questions.

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