Hey, guys. So now that I know the basics of completely inelastic collisions, I want to show you another type of problem where you have a mass that's added onto a system that's already moving. So let's go ahead and check this out. The problem we're going to work out down here, we have a sled that's already moving at some speed, and a box gets dropped onto it, so you're adding mass to a system that's already moving. Basically, the idea is that they're very similar to completely inelastic collisions. Whenever this happens, both objects in the system, right, whatever you had before like the sled plus the new mass like the box, have to be moving with the same final velocity. So let's go ahead and take a look at our problem, and we'll come back to this in just a second here. So we have this 70 kilogram sled and a 30 kilogram box. The sled's already moving to the right with 10 meters per second, and then what happens is the box gets dropped onto it. In part a, we want to figure out the final speed of the system. So let's draw our diagrams for before and after. This is the before. What does the after look like? Well, basically now, this sled is moving to the right, but the box is on top of it. So the box is on top of it like this, and because they're on top of it and they basically become one system or one object, they're both moving with the same v final. And that's what we want to figure out here. So let's go ahead and take a look at our energy or, sorry, our momentum conservation equation. So we're going to use \( m_1 v_{1 \text{ initial}} + m_2 v_{2 \text{ initial}} \), and we can use our shortcut for completely inelastic collisions. We know that we can group together the masses because they're both going to have the same \( v_{\text{final}} \). That's what we want to figure out here. So I'm going to just call this, let's see. I'm going to call the sled 1 and the box 2. Alright? So let's take a look. So we're going to have our sled, that's 70. That's \( 70 \times 10 + 30 \times \) some speed here equals \( 70 + 30 \times v_{\text{final}} \). So what goes inside of this parenthesis right here? What's the initial speed of the block? Well, what happens here is that this block gets dropped vertically onto the sled, which means that it has some initial y velocity that I actually don't know. But it turns out it doesn't matter that I don't know it because the box is only dropping vertically, which means it doesn't contribute any x momentum to the system. Basically, if all the velocity is just vertical, what we can say is that \( v_{2x} \) is just equal to 0. It doesn't contribute any horizontal momentum to the system. So what we can do is in our momentum conservation equation, we can just cancel this out and there's actually 0 initial speed. So this actually makes our equation even simpler because now we can do is figure out \( v_{\text{final}} \). So now what we can do is this is 700, and when we divide the 100 from the other side, this becomes our \( v_{\text{final}} \), and this is going to be 7 meters per second. So that's our answer. So it turns out what happens is that you have a sled that's initially moving at 10, and then you're adding mass to the system, and then the final velocity is going to be less. It's going to be 7 meters per second. This should make some sense to you because momentum conservation, \( p = mv \), says that in order for momentum to be conserved, if your mass is increasing, then in order for you to have the same \( p \) as a system, your \( v \) has to go down. If your mass increases, your speed has to decrease proportionally so that you keep the same momentum. Alright? So that's what's happening here. Alright. So now let's go ahead and take a look at parts b and c. So in part b, what we want to do is calculate the change in the momentum of the box. So that's going to be \( \Delta p_2 \). So what's \( \Delta p \)? Remember, it's just going to be \( m_2 (v_{\text{final}} - v_{\text{initial}}) \). It's \( v_{\text{final}} - v_{\text{initial}} \) here. So what we can do is you can group this together, \( m_2 (v_{\text{final}} - v_{\text{initial}}) \) here. Okay. So we have the 30 kilogram box, and then what's the \( v_{\text{final}} \)? The \( v_{\text{final}} \) is just the 7 meters per second that we just calculated here. So 7 minus what's the initial speed of the box? Well, remember, it's just 0. So what ends up happening is you got a change of momentum of 210. Let's do the same exact thing now for part c, except now we want to calculate the change in momentum of the sled. So this is going to be the same exact equation. I'm just going to skip ahead here. It's going to be \( m_1 (v_{\text{final}} - v_{\text{initial}}) \). So \( m_1 \) here is going to be 70, not the 30, but now we're going to use the final velocity, which is again 7. What's the initial velocity of the sled? Remember the sled was actually moving already at 10 meters per second, that's the initial. So it turns out what happens is you're going to get negative 210 kilogram meters per second. So if you take a look at these two numbers here, they're the same but they're opposites and these two things are related. Basically, what we've had here is a situation where we've had momentum conservation, but one object has gained momentum, and that means that the other one has to lose the same amount. For momentum to be conserved, if one object gains momentum, the other one has to lose the same amount, and the opposite's true. If one object loses momentum, the other one has to gain the same amount. There's always basically a transfer or an exchange of momentum. The sled has to do some work in accelerating the box to the final speed of 7 meters per second, and so the sled loses some speed, the box gains that speed. Alright? So that's it for this one, guys. Let me know if you have any questions.

11. Momentum & Impulse

Adding Mass to a Moving System

11. Momentum & Impulse

# Adding Mass to a Moving System - Online Tutor, Practice Problems & Exam Prep

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### Adding Mass to a Moving System

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Problem

ProblemA 40-kg skater runs parallel to a 3-kg skateboard. Both are moving to the right at 10m/s. The skater jumps on the board, and they move continue moving right (i.e. no change in direction). Calculate the final speed of the system.

A

0.7 m/s

B

3 m/s

C

6 m/s

D

10 m/s

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