11. Momentum & Impulse

Completely Inelastic Collisions

# Completely Inelastic Collilsions

Patrick Ford

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Hey guys in this video, we're gonna take a look at a specific type of collision you're going to see probably pretty often in your homework and tests called a completely inelastic collision. Let's go and take a look and we'll do a quick example. So the whole idea is that in some problems you're gonna have to objects that collide and they're gonna stick to each other. So some key words you might see are lodged or embedded or sticks to, for example, in our example, down here, we have a one kg block that's going to collide with and stick to a nine kg block. This is it can hold a completely an elastic collision. And the whole idea here is that after the collision, if both objects are stuck together, they're gonna move together with the same final velocity. So that's sort of sort of defining characteristics of these type of problems. Is it afterwards the velocity of both objects was gonna be the same. Let's take a look at our example here, we have a one kg block that collides with a nine kg block that's initially arrest. We have the initial speed and basically we want to do is we want to figure out the final speed of the system here. So let's go and go through the steps. We have our before and we have we're gonna draw after diagrams, that's gonna be before and after. So basically what happens is that now you're gonna have these two blocks, Right? So these two blocks are actually gonna move together, sort of as a system. So we're just going to sort of, you know, imagine that these two objects are stuck with velcro or something like that. They're gonna move together like this. So they're gonna move together with some final velocity like this, and that's what we want to calculate. So now what happens is we have this one and it's nine kg block. That effectively just become one big block like this. So that's what happens. So let's go ahead and take a look at our momentum conservation. We have M one V one initial, so M one V one initial plus M two V two initial equals M one V one final plus M two V two final. We want to do is we want to figure out basically what is the final speed of the system. So let's go ahead and check this out. What happens is we have the I'm gonna call this M. One and I'm gonna call this one M. Two. So we can go ahead and replace the mass one and two with the numbers. We have one time something. The speed plus nine times something equals one time something plus nine times something. Now we just figure out what goes inside the parentheses here. So our mass one is traveling at 20 m per second, it's 20 And there are nine kg block is actually initially arrest. So that means that the initial speed of this guy is actually equal to zero. So that means that this term just goes away. What about the final velocity? Well, basically what happens is that that's what we're trying to figure out. But both objects to one and the nine kg block are both going to be moving together with that same speed. So this is the final like this, in fact, that's gonna be the defining characteristic. That's what's common about all these types of problems. So because the V one final equals V two final, what we're gonna do is we're actually to simplify our conservation of momentum equation. What ends up happening is the general form is going to be M one V one, M T V two and on the left on the, on the right side, we can actually sort of combine both the masses, We can say both masses M one and M two are gonna be moving at the same final speed like this, so this is exactly what we have right here. So notice how we just have the final and the final here. So this actually can let us simplify our equation because we can say that this is going to be 20 on the left side from here and this once you combine the one and the nine, once you add these masses together, it's as if you have this sort of one block that's at 10 kg, it's traveling at the final. So when you go ahead and you simplify your, you solve for this, you're gonna get 20 or 10 which is the final and that's going to be two m per second. So what's happened here, what happens is we have a final speed of two. So initially what happens is you have this one kg block that's moving at m per second. So what happens is when a one kg block is moving, You have a speed of 20 m/s afterwards, once it's hit the nine kg block and they're both moving. Now, what happens is effectively the mass has sort of increased by a factor of 10, it's gone from one and now the system actually is 10 kg. So in order for momentum to remain conserved, if your mass increases by a factor of 10, your speed has to decrease or shrink by a factor of 10. So notice how this 10 kg block is no movie about two m per second and momentum is going to be conserved. So that's basically it for this one. Guys, let me go ahead and show you a couple more examples

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