11. Momentum & Impulse

Intro to Momentum

1

concept

## Intro to Momentum

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Hey guys, so in this video I'm gonna introduce you to a new physical quantity called momentum. We're gonna be talking a lot about momentum in the next couple of videos. So it's a good idea to get a good conceptual understanding of what it is, how we calculate it, and also how we solve problems with it. Let's check this out, momentum is really just a physical quantity that objects have. It's kind of like energy and it's really just related to an objects mass and it's velocity. So objects that have mass that are moving with some velocity have momentum. The equation for the letter that we use is lower case P. Don't ask me why and it's really just M. Times V. P equals Mv. Very straightforward. So let's talk about the units. The units really just come from these two letters M and V. Right? The units for mass or kilograms and units for velocity are meters per second. So momentum actually doesn't have a fancy unit, like a jewel or watts or newton or anything like that. It's just kilogram meters per second. So if you ever forget, you can always just get back to it by using Mv. So like I said before, we're gonna be talking a lot about momentum's and an objects momentum, how it gets changed or transferred. And so it's a good idea to get a good conceptual understanding of what it is momentum conceptually is really just a measure of how difficult it is to stop moving objects. So let me back up for a second cause we've talked about a similar idea, which is inertia. Remember that inertia was how hard it is to change an object speed. And if you remember inertia was really just a was really just your mass. Mass is a measure of inertia. If you have lots of mass or if you have like 100 kg then you have to push really hard in order to change your speed. Versus if you had like 10 kg or something like that. So momentum is kind of related to inertia except that only just applies to moving objects. So really if you look at the equation for momentum there's actually two ways you can have lots of momentum. So you can have lots of mass, right? Lots of inertia. Therefore it's harder to stop you. But you could also have lots of speed and it's also harder to stop you. Right So let me go ahead and actually work out this problem here. Um Just so I can show you what I mean by this. So here we have a truck and a race car that are moving. So I'm gonna draw this little flat surface like this. So I've got a 4000 kg truck that's the mass. And it's moving to the right with some speed. So this is V. T. I also have a 800 kg race car that's moving to the left with some V. R. What I wanna do is I want to calculate the momentum. Right? This is just the plural word for momentum of both vehicles here. So I'm gonna calculate P. T. And then P. R. Just by using the equation here. So let's talk about the truck first. Well, if you think about this, the truck right really has some momentum because it has mass and it's moving to the rights. Now, what I want to point out here is that momentum is actually a vector. We can see from the equation here that if you have P. M. And V, what happens is that your velocity is an arrow, right? That's a vector. And if your momentum depends on velocity and momentum is also a vector and it just points in the same direction as your velocity. So momentum is always gonna point in the same direction as an object velocity. So if you have knowledge of moving to the right with some velocity, then its momentum is also going to be to the right like this. So when we calculate the momentum for the truck, we're really just gonna use the mass of the truck, times the velocity of the truck, which we know is equal to 10 m per second. So what I'm gonna do is I'm gonna choose the right direction to be positive and therefore my velocity is gonna be positive. So I'm just gonna calculate this is gonna be 4000 times the velocity of 10 And I have the momentum is really just equal to 40,000 kilogram meters per second. So let's talk about the race car. Now, now, the race car is moving to the left, which means that it's actually gonna have a negative velocity. We're told that the race car is 50 m per second. So VR is actually gonna be negative 50 m per second. Signs are gonna be very important in these kinds of problems. So once you pick a direction of positive, you're gonna stick to it. So what happens is if you have an object that's moving in the direction of negative, then the velocity and the momentum are both going to be negative here. So V. R. Is negative 50. And when we calculate the momentum of the race car, we're just gonna use the mass of the race car times the velocity of the race car. So this is going to be 800 times negative 50. So now, when you work this out, the momentum of your race car is gonna be negative 40,000 kg meters per second. So that's how we got the same number. And we also got a negative sign here. So, if you take a look here, we actually have the same number that we calculated for the momentum of the truck versus the momentum of the race car. And really, this just points out, I'm just trying to point out here that it's actually just as difficult to stop both of these moving objects here because they have the same magnitude momentum. So, for the truck, the truck has a high mass, right? It's M. Is very high, but its speed is only 10 m per second. It's not moving that fast, so the race car is actually much lighter. It's only 800 kg. The M is isn't as high, but the speed is actually much greater than the trucks. So, depending, so basically between the two, it's actually just as difficult to stop the race car moving as it is to stop the truck. Alright, so, hopefully, that makes sense. Guys, let me know if you have any questions.

2

example

## Momentum in 2D

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Hey guys, let's take a look at this problem here. So I have a two kg object like this and I'm given what its velocity is, Its velocity is sure I'm told is 10 m per second and it's some angle above the horizontal, 37 degrees. So let's take a look at the problem here. And the first part just calculate the objects momentum. Remember that's just P. And we have this equation for P. P uses equal to Mv notice that we don't have to do any vector decomposition, there's no sign and co signs just yet. All we have to do is multiply the mass times the velocity and the momentum will point in the same direction. So our p vector is just gonna be the massive two times the velocity which we have and that's just 10. So momentum vector is just 20 kg meters per second. And that's the answer. So one we can kind of understand this, is that the velocity vector is 10 this way. And basically, because the mask is to the momentum vector is going to be exactly double in the same exact direction here. So this pea is just gonna be 20. If the mass were three, the momentum would be three times as big and so on and so forth. So we have this momentum vector, I'm just gonna call this is 20 for now. That's the answer. So let's take a look at part B. Now, part B. Now we actually want to figure out the components, the horizontal components and vertical components of the velocity and the momentum. So basically that's just V. X. And V. Y. And then we want to figure out PX and py so let's go ahead and do that. Well DX and Dy are very straightforward, right? We have the magnitude the 10 m per second and the angle. So we've done this a million times before. The extra six is gonna be 10 times the Kassian of hoops, 10 times the coastline of 37. And you've got eight, it's eight m per second. So this is V. X. Here, the Sequels Eights and V. Y. Is just gonna be 10 times the sine of 37 you get six. So that's the component there. So you basically just have a um V. Y equals six. So this is basically a 68, 10 triangle. Right? What about PX and py? Well there's actually two ways you can kind of understand and make sense where PX and Py are. Well, one thing we can do is we can say, well if P in two dimensions equals M times V in two dimensions. And if we want P. X we're just gonna multiply mass times V. X. P Y. Is just gonna be M times V. Y. Right? So if P equals M V P X equals M. V. X and so on and so forth. Another way you can kind of think about this is that p is just equal to if this is a vector then the X component is gonna be the magnitude times the cosine of the angle. That's 37 degrees same thing's gonna happen with the y axis. So p. Y. Is just gonna be p times the sine of 37. We can use the same angle here because again these these vectors are gonna point exactly along the same direction. So that means that they're thetas are going to be the same. There's two different ways you can think about and you're gonna get the right answer for both of them. Let's check this out. So we know the mass is just too and the V. X. That we just calculated over here is just eight. So if you work this out, what you're gonna get is 16, you're gonna get 16 kg meters per second. So basically what happens is that the X component of this momentum here, this P. X. That we just calculated is 16. Notice how it's just twice whatever this was the same way that this was twice whatever the velocity was. So another way you can think about this is we could just use P. X equals 20 times the cosine of 37. And you're gonna also get 16 here, let me go ahead and just scoot this down a little bit. So notice how you just get the same answer for both calculations here, you'll get 16 either way. So that's what PXE PXE is you do the same exact thing for py so this is gonna be two times the six and you'll get the 12, that's kilogram meters per second. Or you can just do um You could just do P. Y. is equal to 20 times the sine of 37 and you'll get 12 as well either way you get the right answer. Alright so those are the components eight and six and then we have PX and PY. So basically just finish off this little triangle here, we've got the Y. Component of my momentum and it should be twice what the velocity is, what P. X. Is 12. So notice how this is 10, this is 20 this is eight and this is 16 and then this is six and 12. Basically everything has scaled twice in each direction. That's about this one. Guys let me know if you have any questions.

3

Problem

ProblemHow fast would you have to throw a 150-g rock for it to have the same momentum as a 10-g sniper rifle bullet traveling at 900 m/s?

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Additional resources for Intro to Momentum

PRACTICE PROBLEMS AND ACTIVITIES (8)

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- CALC A particle of mass m is at rest at t=0 . Its momentum for t>0 is given by px = 6t² kg m/s ,where t is ...
- CALC A 500 g particle has velocity vx = −5.0 m/s at t=−2 s .Force Fx= (4−t²) N , where t is in s, is exerted o...