Guys, in this video, we're going to take a look at a specific type of collision that you're going to see probably pretty often in your homework and tests called a completely inelastic collision. Let's go ahead and take a look, and we'll do a quick example. The whole idea is that in some problems, you're going to have two objects that collide and stick to each other. Some keywords that you might see are lodged, embedded, or sticks to. For example, in our example down here, we have a 1 kilogram block that's going to collide with and stick to a 9 kilogram block. This is going to be a completely inelastic collision. The main idea here is that, after the collision, if both objects are stuck together, they're going to move together with the same final velocity. That's what sort of defines these types of problems, that afterwards, the velocity of both objects is going to be the same.

Let's take a look at our example here. We have a 1 kilogram block that collides with a 9 kilogram block that's initially at rest. We have the initial speed, and basically, what we want to do is figure out the final speed of the system here. So, let's go ahead and go through the steps. We have our before and after diagrams. This is going to be before and after. Basically, what happens now is that you're going to have these two blocks, right? These two blocks are actually going to move together sort of as a system. We're just going to imagine that these two objects are stuck with Velcro or something like that, and they're going to move together with some final velocity, and that's what we want to calculate.

Now, what happens is we have this 1 and this 9 kilogram block that effectively just become one big block like this. Let's go ahead and take a look at our momentum conservation. We have mv1initial + mv2initial = mv1final + mv2final. We want to figure out basically what is the final speed of the system. Our mass one is traveling at 20 meters per second, and our 9 kilogram block is initially at rest. That means that the initial speed of this guy is actually equal to 0, so that term just goes away. What about the final velocity? Well, both objects, the 1 and the 9 kilogram block are both going to be moving together with that same speed. Since v1final equals v2final, we're actually going to simplify our conservation of momentum equation.

The general form is going to be mv1vm2vv2. On the right side, we can combine both the masses, and we say both masses m1+m2 are going to be moving at the same final speed like this. So, we simplify and solve for this. Now, when you go ahead and simplify, you'll get 20 over 10 which is vfinal=2 meters per second. What happens is initially, you have this 1 kilogram block moving at 20 meters per second. So, once it hits the 9 kilogram block, and they're both moving, effectively, the mass has increased by a factor of 10. So in order for momentum to remain conserved, if your mass increases by a factor of 10, your speed has to decrease by a factor of 10. Notice how this 10 kilogram block is now moving at 2 meters per second, and momentum is conserved. So that's basically it for this one guys. Let me go ahead and show you a couple more examples.