Adding Mass to a Moving System - Video Tutorials & Practice Problems

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Adding Mass to a Moving System

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Hey guys, so now they know the basics of completely inelastic collisions. I want to show you another type of problem where you have massive attitude onto a system that's already moving. So let's go and check this out and the problem, we're gonna work out down here. We have a sled that's already moving 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 kg sled and 30 kg box. The sleds already moving to the right with 10 m per second. And then what happens is the box gets dropped onto it. In part, 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 is the after look like? Well, basically, now the sled is moving to the right, but the box is on top of it, so the boxes on top of it like this, and because there it's on top of it, 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, take a look at our energy or sorry, momentum conservation equation. So we're gonna use M one V one initial plus M two V two 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 final and that's what we want to figure out here. So, I'm gonna call this um let's see. I'm gonna call the uh I'm gonna call the sled one in the box to All right, so let's take a look. We're gonna have our sled. That's 70 at 70 times 10 plus 30 times some speed here equals 70 plus times V final. So, what goes inside this parentheses right here? What's the initial speed of the bloc? Well, what happens here is that this block gets dropped vertically onto the sled, which means that it has some initial. Why 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 just vertical. Well, we can say is that V two X is just equal to zero. It doesn't contribute any horizontal momentum to the system. So we can do is in our in our moments and conservation equation. We can just cancel this out and there's actually zero initial speed. So this actually makes our equation even simpler because now we can do is figure out the final. So now we can do is this is 700. And when we divide the 100 from the other side this becomes RV final and this is going to be 7m/s. So that's our answer. So it turns out what happens is that you have a slip 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 seven m per second. This should make some sense to you because momentum conservation P equals M V. Says that in order for a momentum to be conserved. If your mass is increasing, then in order for you have the same P as a system, your V has to go down. If you're mass increases your speed has to decrease proportionally so that you keep the same momentum. All right, So that's what's happening here. All right. So now let's go ahead and take a look at parts B and C. So in part people want to do is calculate the change in the momentum of the box. So that's going to be delta P two. So what's delta P. Remember? That's just gonna be V M two V final minus M two. V initial its voice final minus initial here. So we can do is you can group this together. M two V final minus V initial here. Okay, so we have the 30 kg box and then what's the v final? The v final is just the seven m per second that we just calculated here, so seven minus what's the initial speed of the box? Remember? It's just zero. So what ends up happening is you gotta change the momentum of 210. Let's do the same exact thing now for part C. Except now you want to calculate the change of momentum of the sled. So this is gonna be the same exact equation. I'm just gonna skip ahead here. It's gonna be M one V final minus V initial. So M one here is going to be 70 not the 30 but now we're going to use the final velocity which is again seven. What's the initial velocity of the sled? Remember? The sled was actually moving already at 10 m per cent and that's the initial. So it turns out what happens is you're gonna get negative to 10 kg meters per second. So if you take a look at these two numbers here, they are 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 amounts for momentum will be conserved If one object gained momentum, the other one has to lose the same amount and the opposite is 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 and accelerating the box to the final speed of seven m per second. And so the sled loses some speed. The box gains that speed. All right, So that's it for this one. Guys let me know if you have any questions

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Problem

Problem

A 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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