Intro to Momentum - Online Tutor, Practice Problems & Exam Prep

Intro to Momentum

Guys, in this video, I'm going to introduce you to a new physical quantity called momentum. We're going to 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 object's mass and its velocity. So objects that have mass, that are moving with some velocity, have momentum. The equation for it or the letter that we use is lowercase 'p'. Don't ask me why. And it's really just \( m \times v \), \( p = m \times v \). 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 are kilograms, and the units for velocity are meters per second. So momentum actually doesn't have a fancy unit like a joule or a watt or a 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 \( m \times v \).

Like I said before, we're going to be talking a lot about momentums and an object's 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 because we've talked about a similar idea which is inertia. Remember that inertia was how hard it is to change an object's speed. And if you remember, inertia was really just your mass. Mass is a measure of inertia. If you have lots of mass, or if you have like 100 kilograms, then you have to push really hard in order to change your speed versus if you had like 10 kilograms or something like that. So momentum is kind of related to inertia, except it only just applies to moving objects.

So really, if you look at the equation for momentum, there are actually two ways you can have lots of momentum. So you could 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, 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 going to draw this little flat surface like this. So I've got a 4000 kilogram truck, that's the mass, and it's moving to the right with some speed. So this is \( v_t \). I also have an 800 kilogram race car that's moving to the left with some \( v_r \).

What I want to do is I want to calculate the momenta. Right? This is just the plural word for momentum of both vehicles here. So I'm going to 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 some mass and it's moving to the right. 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, then momentum is also a vector and it just points in the same direction as your velocity.

So momentum is always going to point in the same direction as an object's velocity. So if you have an object 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 going to use the mass of the truck times the velocity of the truck, which we know is equal to 10 meters per second. So what I'm going to do is I'm going to choose the right direction to be positive and, therefore, my velocity is going to be positive. So I'm just going to calculate. This is going to be 4000 times the velocity of 10, and I have the momentum is really just equal to 40000 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 going to have a negative velocity. We're told that the race car is 50 meters per second, so \( v_r \) is actually going to be negative 50 meters per second. Signs are going to be very important in these kinds of problems. So once you pick a direction of positive, you're going to 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 going to 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 going to be negative 40000 kilogram meters per second.

So, notice 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? Its \( m \) is very high, but its speed is only 10 meters per second. It's not moving that fast. So the race car is actually much lighter. It's only 800 kilograms. The \( m \) isn't as high, but the speed is actually much greater than the truck's. So depending on, you know, 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.

Momentum in 2D

Guys, let's take a look at this problem here. So, I have a 2 kilogram object like this, and I'm given what its velocity is. Its velocity, as I'm told, is 10 meters per second, and it's at some angle above the horizontal, 37 degrees. So let's take a look at the problem here. In the first part, all I want to do is just calculate the object's momentum. Remember, that's just \( p \), and we have this equation for \( p \). \( p \) is just equal to \( m \times v \). Notice how we don't have to do any vector decomposition. There's no sine and cosines 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 going to be the mass of 2 times the velocity, which we have, and that's just 10. So, the momentum vector is just 20 kilogram meters per second, and that's the answer. So one way you can kind of understand this is that the velocity vector is 10 this way, and basically because the mass is 2, the momentum vector is going to be exactly double in the same exact direction here. So this \( p \) is just going to be 20. If the mass were 3, the momentum would be 3 times as big, and so on and so forth. So we have this momentum vector. I'm just going to call this as 20 for now. That's the answer.

So let's take a look at part b now. In 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 \( p_x \) and \( p_y \). So let's go ahead and do that. Well, \( v_x \) and \( v_y \) are very straightforward. Right? We have the magnitude, the 10 meters per second, and the angle. We've done this a million times before. \( v_x \) is just going to be 10 times the cosine of 37, and you've got 8. That's 8 meters per second. So this is \( v_x \) here, this equals 8, and \( v_y \) is just going to be 10 times the sine of 37, and you get 6. So that's the component there. So you basically just have \( v_y \) equals 6. So this is basically a 6-8-10 triangle.

What about \( p_x \) and \( p_y \)? Well, there's actually 2 ways you can kind of understand or make sense of where \( p_x \) and \( p_y \) are. Well, one thing we will do is we can say, well, if \( p \) in 2 dimensions equals \( m \times v \) in 2 dimensions, then if we want \( p_x \), we're just going to multiply mass times \( v_x \). \( p_y \) is just going to be \( m \times v_y \). Right? So if \( p \) equals \( m \times v \), \( p_x \) equals \( m \times v_x \), 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 going to be the magnitude times the cosine of the angle. That's 37 degrees. The same thing is going to happen with the y axis. So \( p_y \) is just going to be \( p \) times the sine of 37. We can use the same angle here because, again, these vectors are going to point exactly along the same direction, so that means that their thetas are going to be the same. So there are 2 different ways you can think about it, and you're going to get the right answer for both of them. Let's check this out. So we know the mass is just 2, and the \( v_x \) we just calculated over here is just 8. So if you work this out, what you're going to get is 16. So you're going to get 16 kilogram 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 could think about this is we could just use \( p_x \) equals 20 times the cosine of 37, and you're going to also get 16 here. So we go ahead and just scoot this down a little bit. So notice how we just get the same answer for both calculations here. You'll get 16 either way. So that's what \( p_x \) is. We can do the same exact thing for \( p_y \). So this is going to be 2 times the 6, and you'll get the 12. That's kilogram meters per second, or you can just do \( p_y \) equals 20 times the sine of 37, and you'll get 12 as well. Either way, you'll get the right answer.

Alright. So those are the components at 8 and 6, and then we have \( p_x \) and \( p_y \). 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 when \( p_x \) is 12. So notice how this is 10, this is 20, this is 8, and this is 16, and then this is 6, and 12. Basically, everything has scaled twice in each direction. That's it for this one, guys. Let me know if you have any questions.

How 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?

• A. 60 m/s
• B. 400 m/s
• C. 40 m/s
• D. 600 m/s