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.