Completely Inelastic Collisions

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

Hey guys got a classic example here for you, where a bullet gets fired inside a block of wood and we want to figure out the original speed of the bullets. Now, what happens is the bullet is going to embed itself into the block. So we know this is a completely an elastic collision. Let's go ahead and write out some diagrams. So we've got, let's see um let's say, I've got this speeding bullet like this, that's speeding along this way. I have the mass of this bullet is 0.01. It's going to strike a block of wood, which is five kg. That's my M2 and then eventually it's going to sort of embed itself inside. So this is the before And in the after case, what happens is now we have the block that has the bullet that's stuck inside of it. And these things are both moving together with a final velocity of 0.6. That's what we're told right here, is that the bullet would system immediately is moving at 0.6. All right. So this is basically what the after case looks like we want to do is we want to figure out the initial speed of the bullets. So if this is M. One we really want to do is figure out what is V one initial. So I need to figure out what is this. V one initial right here. So to do that, I'm gonna go ahead and move onto the second step. Gonna write out my momentum conservation before I do that. I just want to point out that this V two initial is also going to be zero because the block of wood was stationary. So let's take a look at part A. So for part A we're gonna write our momentum conservation. M one V one initial plus M two V two initial equals M one V one final plus M two V two final. Now, I'm just gonna write all the masses. So again this is going to be 0.1 plus. This is gonna be five equals 0. plus five. Okay, What we can do here is we can actually use the shortcut because we know that if in a completely an elastic collision, these two velocities are going to be the same. So instead of writing it this way, I can actually say this is going to be 0.01-plus five. Right? Both of these things are going to combine together as a single mass and they're both gonna be moving at some final velocity, which I already know. Uh it's really this this initial velocity here that I'm actually interested in. So what's the initial velocity of the bullets? So what goes in here? What's the initial velocity of the five kg block? What's actually stationary? So, you know, it's going to be zero. So basically that's what we have here. We know the final velocity of the whole system is actually going to be 0.0 point six. So if we go ahead and simplify the left and right sides, this is just going to be 0.01 V1 initial equals 5.01 times 0.6. So if you multiply, you're gonna get 0.01, The initial equals, and then you're gonna get 343.006. So now what happens when you divide, you're gonna get viewing initial is equal to 301 m per second. And that sounds about right. For a speeding bullet, you know, it's about like 700 miles an hour or something like that. So that is the speed of the bullets. So it's traveling at 301 m per second, lodges in the wood and they both moved together at 3010.6. All right, So now what we're gonna do in part B is we want to figure out how much kinetic energy gets dispersed inside of this, an elastic collision. Remember that completely inelastic collisions, actually, all type of an elastic collisions do not conserve kinetic energy. So in order to figure out how much kinetic energy is lost, we want to figure out delta K delta case just K final minus K initial. Right? So it's how much energy you have now versus how much energy you had before. So let's go ahead and calculate those individually. What's the kinetic energy final? We're gonna look at the after case. Right. So, we basically have these two objects that are moving together with a combined speed of 0.6. So you're K final is going to be one half. And what we can do is we can say that this was someone and this is M. Two. But when they combined together their masses, big M. Is actually just going to be M one plus M two, which is just 5.1 That's what we calculated over here. So it's going to be one half of big M times V final squared, which is gonna be one half times 5. Times This is going to be 06 square. Which end up getting for K. Final is you end up getting 0. jewels. So just under one jewel. What about K. Initial? Okay. Initial. We're going to look at the before case. What's the only object that has kinetic energy here? Well, if the block is stationary, has no kinetic energy, that doesn't contribute anything to the system. The only kinetic energy in the before cases, actually the kinetic energy of the speeding bullets. So here I'm going to use one half and I'm going to use M. One the little M. Notice how for the after case I use the big M. And then this one I'm using just little M. One uh times if you want initial, which is the speed we just calculated. So this is gonna be one half. This is going to be 0.1 And the initial speed that we just calculated was You're gonna square. That what you end up getting here is going to be getting about 453 jewels. So these are remember just the kinetic energy, initial and final. Now what we have to do is subtract them. So the delta K, which is really what I'm interested in here, is going to be my final minus initial. So it's 0.902 minus jewels, which end up getting is negative 452. jules. So it loses, basically the system loses almost all of its kinetic energy that basically that energy goes in from the bullet that basically embeds itself into the wood gets dissipated as heat and sound and other things, but he loses a ton of energy during the collision. All right, so listen to this one, let me know if you have any questions.

Hey everybody. So in this problem we have a car and a truck that's colliding now, luckily nobody's hurt, but we do have the cars that stick together and lock together during the collision. So this is a completely an elastic collision. And we're gonna calculate the magnitude and direction of the final velocity after colliding. So let's get into the steps, we're gonna write some, we're gonna draw some diagrams from before and after. So here's what's going on here. So I've got this car like this and this car has a mass of 1000 and it's traveling east at 20 m per second. So in other words, it's a velocity kind of looks like this and I'm gonna call this basically this is car A so it's via is going to equal 20 m per second. Now, what's happening here is that it slams into a truck that's moving north. So what I'm gonna do is I'm gonna draw a little X and y axis like this. Alright, so I've got this velocity here, this truck has a mass of 3000 and it's moving totally north, right? So basically vertically up like that, that's north and that's east. So its velocity vector is going to look like this. I'm going to call this one be And this VB here is equal to 10 m/s. So once they lock together, what's going to happen. Well, if you draw the after diagram, you can kind of sort of intuitively guess what's going to happen afterwards, The two cars, the trucks, trucks, whatever going to stick together, They're going to have a combined mass of 4000, right? 2000 3000. And then they're gonna move off in which direction? Well, you imagine this, right? If you have a car that's moving like this and a truck this movie like this, once they stick together, they don't move totally to the right or perfectly up. They move sort of a combination of the two directions. They sort of go off in the diagonal. So here's what happens here. This truck is is going to go off. These two cars are gonna go off like this and this is my V final. And that's what I want to calculate when you calculate the magnitude of that vector. But I also want to calculate the direction as well. Right? So I want to also want to calculate the angle with respect to the X axis. Okay, that's really what we're trying to calculate here is V final and theta. So how do we do that when we actually just go back to using using, you know, normal vectors. If I want to calculate the magnitude of this vector, then I need its components. Right? I'm gonna need basically the X components and I also need the Y component. So the tricky thing here is that I want to calculate the final but I have to first calculate with Vfx and V F Y r. How do we do that? Let's move on to the second step here, which is going to write momentum conservation equations. So what I've got here is I've got M a v a initial plus, MB vb initial equal. And then remember these two things combined so we can combine their masses and sort of use that shortcut equation uh an A plus and B times the final. Now, are we done here? Well, actually, no, because for all the all the collisions that we've seen so far, we've only had them in one dimension and one access. We've had a car moving to the right and truck moving to the left or whatever to objects that are basically just moving along the line. But in this problem we have a two dimensional collision because one is moving sort of left and right and the other object is moving up and down. And here's the idea for a two dimensional collision. We're still gonna write momentum conservation equations. But now we're just gonna do it for both the X and the y axis. It's just like any other two D problem that we've encountered when we did motion in two D. When we did forces in two D. We basically just wrote a bunch of equations in the X and the y axis, We broke them down and then solve them separately. That's exactly what we're gonna do here with momentum. So what I'm gonna do is I'm going to write equations for the X and Y. Conservation of momentum. So this is going to be a the initial plus and bVB initial equal M A plus MB V final. But the only thing that's different here is that I'm gonna conclude this is the X axis and this is the y axis and all the velocities are sort of implied that you're just gonna look at the X and y axes respectively. Alright, so then how do we do this? Well let's just jump right into the X axis because we're going to go ahead and plug values and solve So anyway, remember that's just the mass of the car. That's going to be the 1000 now, V A X initial. So in other words, the velocity components of this car that lies along the X axis. Well this just lies like this. So it's basically just gonna be 20 like that. Now the second term, this is going to be M V which is 3000 but what about the velocity in the X axis? Remember this truck is still moving at 10 m per second, It's still going but it doesn't have a component in the X axis. Right? That the component of the velocity is only in the y axis here. So basically what happens here is that there is no X components and it's zero. So that just means that we can actually just cancel it out in our equation because it goes away on the right side. We're just gonna combine the masses. Remember this is 4000 V final X. So that's how you calculate the V final, basically what you're gonna get here, is that 20,000 is equal to 4000 the final X. And so your V final X is going to be five m per second. Alright, So now that we have that component here, basically what we can do is we can say this final, this v final is going to be the square root of V final X squared plus V final y squared. So, we just need those two components, like we said. Uh So when we just have this one now, we just need to go ahead and calculate the V final squared And to do that, we're gonna look at the Y-axis, we're gonna do the same music thing, right? So it's gonna be 1000. What about the via initial in the Y axis via remember just points along this, right? It's just 20 m per second to the right. What's the Y component of this velocity? It's zero. It's just kind of like what happened to the VB over here. So this is going to be zero here. So we can cancel this house And this is going to be plus Times 10. This is going to equal 4000 times the final in the Y axis. So what you're gonna get here is you're gonna get 30,000 equals 4000 V final Y. And you get that V final Y is equal to 7500 I'm sorry. 7.5, 7.5 m/s. All right. So now that we've got the two components of here, of these the v final. Now we can just go ahead and put them together and actually calculate the v final here. So this v final, I'm just going to bring this down here. The final is going to be the square roots of the five squared plus 7.5 squared. And if you work this out, what you're gonna get here is you're going to get 9. m/s. All right? So that's the one term. This is going to be 9.5 m/s. Sorry, 9.01. I'm sorry. 9.01. Now, what about the angle? Well, to get the angle we just use basically the inverse tangent, Right? So it's gonna be the tangent inverse Of the Y component over the x component. So in other words, we figured out that this was equal to five, this was equal to 7.5. And so this is going to be seven point 5/5. And when you work this out, what you're gonna say what you're gonna get is 56.3 and that's gonna be degrees 56.3°. And that is your final answer. Right? So these are both your numbers 56.3°. Alright, guys, so that's this one, let me know if you have any questions.

Hey everybody. So let's take a look at our problem here. So, this one's kind of interesting because we have three different objects, a bulletin two blocks and two different collisions that are happening. But basically what we wanna do is we want to calculate the speed of the second block after the bullet embeds itself into the second one. So let's go ahead and stick to the steps. We're gonna draw a diagram from before and after. So this is what's going on in this problem. We have a bullet like this, it's flying and it's gonna hit towards a block, it's gonna head towards a block. So I'm going to call this M1. This has a mass of 0.01 And the block has M2, which I'm gonna call 0.5. But what happens is the book, the bullet actually keeps on going after it's passed through the first block. So it keeps on going through the first block and actually embeds itself in a second block. I'm gonna call the M three And this equals 0.5 kg as well. And basically, once it embeds itself in there, the whole entire system is going to be moving off with some velocity. I'm gonna call this v at this point. All right. And that's really what we want to calculate the tricky thing here is that we have three different objects and two different collisions. So if you're not careful, what happens Is that if you use M1 V1, initial and MTV to initial the final will be the initial for the second collision and that can be really confusing. So, we're gonna do is we're gonna do something that we've used in other sort of chapters like motion and forces, which is we're going to label the different events that are happening. So, here's what I'm gonna do. I'm gonna call this A this one is B and this one is C. So A is actually going to be so A is the point where it's before the first collision. So this A B and C do not refer to different objects that refer to different sort of events or different points in time. I want to be really clear about that. So it's before the first collision. Now, point B is after the first collision, right? It's when the bullet exits the first block and keeps on going towards the second one. So it's after the first, but it's before the second collision. And then finally point sees when the bullet embeds itself into the second block. And so that's after the second collision. So, really what we want to calculate is this final velocity here and I'm gonna call this velocity. See that's really what we're interested in. So, how do we do that? Well, we're gonna we're gonna again stick to the steps, move on to the second thing, we're gonna write the conservation of momentum equation. If I want to calculate the VC, then I'm gonna pick the interval that basically includes, see so in other words, I'm gonna include the, I'm gonna have the interval from B to C. Right, that's the interval after the first collision, but before the second one and then when the bullet finally embeds itself. So let's set up our equation here for B- two c. So what I'm gonna do is again keep just keep in mind that we have three different objects. So really, what's going on? Is that the bullet, which is M one is going to collide with the third object, M three. So here's how our equation is going to get set up, It's going to be M one V one B, right? That's the initial from B to C Plus M three, V 3, B equals. Now, what happens is when the bullet embeds itself, that is a completely an elastic collision. So what happens here is that from B to C is completely in elastic? Whereas the first collision from A to B is just an elastic one, because nothing gets stuck together, right, the bullet exits and both of these things sort of like, you know, they go apart, right? They don't actually stick together. Okay, so this is gonna be again using our shortcut M one plus M three and this is gonna be V C. So this is really my target variable is what is the final velocity of the bullet plus block combo. So that's what I'm looking for here. So if you look through your variables, what you're gonna see is that all these masses are given to us. So can we actually just solve this by using this interval? Well, if you look at this, what happens is we're going to hit 0.01 and then V one B is going to be the velocity of the bullet after it exits the first block. Right? So this is gonna be V one B. Now, what is that velocity? If you look through the problem, you might see that this 430 m per second. But remember that's the initial speed of the bullet. In other words, the V one A is going to be 430, but what happens is after it exits the first block, it's gonna change, right? There's gonna be an exchange of velocity, there's gonna be some exchange momentum and V one B is definitely not going to be 430 m per second, but we actually don't know what that is, so I don't know what this is yet and I'm just going to leave it blank for here. Now, what about M three V three B? Well, basically this is just the initial velocity of the second block before the collision, in that case the third, the second block is stationary. So in other words, that term actually just goes away because that term is zero, So we have zero on the left side and then on the right side we have the two velocity, the two masses combined 0.01 0.5 V C. So there's only one variable that we really need and that's this V one B over here. It's the velocity of the block after it exits the first block. Or sorry, the velocity of the bullet after it exits the first block. Now, how do we go find that? Remember we in this problem, we have two different collisions, two different sort of intervals and so to go and find this V one B. I'm gonna have to go and use the other interval which is gonna be the A to B interval. Alright, so let's set up that equation now, now that equation is gonna look like M one V one A plus M. Two V two A equals M one V one B plus M2 V two B. Remember we can't use the shortcut like we did over here because this collision is not completely an elastic, the two masses will not stick together. Alright, so that's the important part. So really this V one B here is actually gonna be this V one B and that's again why we didn't use initials and finals because it can get really confusing. So let's go ahead and start plugging in the values and solving So this is going to be 0.1. What about this V one A. We actually know what that is. That's just the initial velocity of the bullet, which is the 4 30 m per second. So 4 30 plus M two V two A. So basically that's the mass and velocity of the first block of this one. And remember both blocks are at rest. So what happens here is you're just going to get 0.5 but then this cancels out because this is equal to zero. So basically the whole term goes away, Then we have a 0.01 and then V1 B. That's exactly what we're trying to find here. Plus. And then we've got 0.5. And then what about V to be? It's basically the velocity of the block after the collision. What number is that? Well, if you look through your variables, what's going to happen here? Is that the speed of the first block immediately afterwards is 5.6 m/s. So, that's basically v to be all right, that's what's going to go over here. The 5.6. So now we can do is we can solve this equation for V one B and that's exactly what we're missing in the B two C interval. Okay, So if you plugged all that stuff in, what you're gonna get is 4.3 on the left side, This is going to equal 0.01 V one b plus. And then this works out to 2.8 when you bring this over to the other side, which you're going to get here Is you're gonna get 1.5 is equal to 0.01 v. one b. And then finally v. one b. is equal to 15. I'm sorry, 150 m/s. Alright, so now that we have this 1 50, that's what we stick into this equation over here. So this becomes the 1 50 now we can go ahead and solve for V. C. This is going to be zero point uh where you work this out. This is gonna be 1.5 again equals 0.51 V. One. Sorry, VC. And now if you work this out, what you're gonna get is that VC is equal to 2.94 m per second. Alright, So it's a little bit of you know, sort of upfront work, getting organized, labeling everything. But once you do that, the rest of the math becomes pretty straightforward. All right, so that's your final answer. It's 2.94 m per second. Let me know if you have any questions