9.4 2D Collisions

All right, you know, no, I'm not all right. In our last lesson, we were looking at one-dimensional collisions. These were a little bit easier to solve mathematically because we were only dealing with one direction. We either had an object moving and hitting another object head-on, resulting in two head-on collisions, or we could have one object moving in one direction and hitting another object moving in the opposite direction, like a rear-end collision. We also looked at scenarios where a moving object hits a stationary object, and then the objects either stick together or separate into two parts. We did lots of different examples, but they all used the same basic formula: mass a‚ÇÅ times velocity a‚ÇÅ plus mass b times velocity b equals mass a‚ÇÇ times velocity a‚ÇÇ plus mass b times velocity b. Now, what's changing is that we're still going to be dealing with the same concepts, but instead of just plugging the numbers into the formula and solving it, we have to consider vectors because we're dealing with two dimensions. So, what's going to happen now is that we'll have one object moving at an angle and colliding with another object at an angle, and the results of the collision could either make the objects bounce apart or stick together, among other possibilities. As soon as we add a second dimension, things become more complicated. We have to be careful with our diagrams, drawing the vectors accurately, and doing our calculations correctly. Before we start looking at an example, let's make sure we understand the terminology and how we should draw these diagrams. The first example says that we have two moving pool balls with the same masses that are going to glance off each other. Glancing means it's like a sideswipe or a bouncing collision. It states that they strike each other at an angle of 30 degrees. So, let's say one ball is going down at an angle and the other ball is going upwards or horizontally. The specific direction doesn't really matter, but the key is that there's a 30-degree angle between them. Typically, in this scenario where the masses and speeds are similar, the balls would bounce apart. The top ball would bounce and go at one angle, and the bottom ball would do the same. Let's erase the incorrect diagram and redraw it correctly. Let's suppose the red ball goes down at an angle, and the blue ball goes up. And as they collide, the red ball would go back down and the blue ball would likely bounce back up. However, this doesn't necessarily mean that they would bounce back again at 30 degrees. If one ball is way faster or has more mass, its momentum will dominate the collision, and the resulting angle may not be 30 degrees. We won't know the exact angle until we do the calculations. Now, let's try one more example. It says that we have a ball that is going to strike a stationary ball, and then they both move off with an angle of 60 degrees between them. In this case, let's suppose we have one ball coming in and hitting a stationary ball. After the collision, we would have the red ball moving in one direction and the blue ball moving in a different direction. The angle between them would be 60 degrees after the collision. The idea is that's sort of how we want to draw our diagram, so we don't necessarily have to draw the balls and the collisions and all that, but we really want to be careful with drawing our vectors properly. So let's do an example. The key thing that we all remember is our total momentum is what we're dealing with rule, and we're working with momentum again just like we did with one dimension. But we have to make sure that our x and y components before and after equal, not just necessarily total momentum because of the angles. So let's draw our first one here. It says we got a 1.2 kilograms red ball moving to the right at 17.1. So let's draw that. We got 1.2 kilograms, 17.1 meters per second. And it says it's gonna strike a stationary one, so that one's not moving, okay? And then, so that's our before. Now, our after. It says the red ball is gonna go up at 23 degrees above the horizontal, and 13.5 meters per second. The question is asking what the blue ball is doing. So just using common sense, it's likely going to bounce up, probably our blue, and it's gonna go down somewhere. But we don't know how fast or at what angle. So to solve these, what we got to work with is we have to deal with momentum only. We don't want to be working with speeds and masses and all that. So let's calculate the momentum of the original one. So P equals MV. So we've got 1.2 times 17.1, which gives us an answer of 20.52 kilograms meters per second. We don't need to worry too much about the units because we're dealing with momentum all the way through. So that's our only initial momentum we have, just a horizontal vector of 20.52. Alright, so now the trick is how do we connect that to the momentum afterwards. So what we want to do is we want to now change our... I'm gonna erase these diagrams or some good colors. So this one was at what angle? And that's what the blue one is, down at some sort of angle. And that's it. Now, let's... we know our momentum of the red ball afterwards. That was given to us. It was a speed of 13.5 and it's still 1.2 kilograms. So we can calculate its momentum. So let's go mass times velocity. So 13.5 times the mass of 1.2, and that'll give us an answer of 16.2. Let me erase our speed, actually. What I'll do, I'll redraw it. Okay, so that red ball that was shooting up at an angle, so I'll draw that one. It has a momentum of 16.2, and we know the angle is going to be 23 degrees. Well, now our next step is let's figure out its components. So we can't deal with angles. We want to deal with horizontal and vertical components. So we'll have a horizontal component... I'll draw that in as red, and then we're also going to get a vertical component. I'll draw that one in as green, okay? So we need to calculate those, so just use regular trig. So we'd have sine 23 is our Y direction, or 16.2. So we get an answer of roughly 6.3. And then we want to do the same thing with the x-direction, so we want to go cosine 23 equals x over 16.2. So cosine 23 times 16.2 gives us an answer of 14.9. Okay, so in our diagram, let me just erase so I have a little more room here. So we have 6.3 going up, and then we'd have 14.9 horizontally. Okay, so let's go back to our original setup, and I'm gonna go to a new page here that's blank. So let's go before again, we have 20.52. I'm going to redraw that. So our vector to start was a momentum of 20.52, and then after we had 14.9 to the right and we had 6.3 up. Hey, so that was our final red ball's vectors. And then the only other thing we had left was to calculate the blue ball's vectors. We estimated the blue ball would go downwards at some sort of angle, and it doesn't really matter. You can use some common sense to estimate it, but we don't have to get it exact. The math will sort of take care of itself. So if we go back to this diagram, let me just kind of show you here. So we got a horizontal vector and a vertical vector for both for the one that we're looking for. So if we can calculate those components, we then should be able to figure out what our final answer should be. So off our diagrams, what we need to do then is we need to calculate the vertical and the horizontal components of that last vector, and then we should be good. So using conservation of momentum, let's go horizontal first. We got 20.52 as our initial that has to equal our final, we got 14.9 in the final direction. Plus, we'd have the one that we're looking for. So we just have to basically solve for X. So go 20.52 minus 14.9, and that gives us an answer of 5.62. And do the same thing for our vertical direction. Okay, what was our vertical motion to start with? Zero, there wasn't any. So we have zero to start with, and we have 6.3 up. So that means our Y has to be negative 6.3, otherwise, we wouldn't get it equal to zero. And that's it. So we've got our two components figured out. Let me kind of recatch this again. So that's 6.3. Three going down and we can write it as a negative. You want? Sometimes it's a good idea so you don't forget. And we have fourteen point five point six. Let's keep it around it to one decimal, so 5.6 going to the right. So now we're good to go, we're ready to solve for our final answer. So we need to figure out our hypotenuse and we want that angle. So if we just do the Pythagorean theorem, that'll give us our hypotenuse, which will be 8.4. And let's use tangent to figure out our angle. So our angle would be 6.3 opposite over adjacent. The tangent of 6.3 divided by 5.6 gives us an angle of about 48 degrees. Okay, so you just got to be careful. Now, is that our final answer? If we go back to our question, the original question was to determine the final velocity of the blue ball. So what we just figured out was the momentum. Right, this was our momentum, so that's in kilograms meters per second. So we've got our momentum and our angle. Now the only thing we've got left to do is figure out the velocity. Based on our formula, our velocity would be momentum divided by mass. All we have to do is take that 8.4 divided by the mass of the blue ball, which was 2.31 in our original question. So when we divide that out, 8.6 divided by 2.31, we get a final answer of 3.7. So then our final vector answer should be 3.7 meters per second as our speed, and we're doing it at 48 degrees south of East. And that would be our complete final answer. So you can see we're still doing the same thing, we're still using P equals and B. What's different though now is we're doing just the x-direction components. They have to equal each other, and then just our y-direction components. And usually, as long as we have one triangle, we can add and subtract to figure out whatever the remaining one is. Let's do one more example and then that'll be it. So in this example, we've got a 1.2-kilogram red ball moving downwards. So we've got 1.2 kilograms. It's moving at 10 meters per second. So if we go 1.2 times 10, that will give us the momentum of 12. And then our blue one, it says it strikes the blue ball. That's 2.31, and it's moving at 15. So that gives us a momentum of 34.65. Okay, let's go to the after effect. So after, we have the red ball, so it's still 1.2, but it says it's moving at 13.5 now. So that gives us a momentum of 16.2. The question is, what is the velocity of the blue ball? So we're gonna have to figure out the momentum of the blue one. Once we have the momentum, we should be able to figure out the velocity and the angle, just like we did in the last question. With this question, we're gonna have to do the same thing. We're gonna need to figure out the x-components, the horizontal components of all of the different vectors, and then we're also gonna have to figure out the vertical component. The red one is going down and over. Let's do that in blue. Okay. We got the one going down and to the right. The blue ball went up and to the right, our red ball, the way the diagram is, is up and to the right, and our blue ball is downwards. So what we want to do then is calculate all of our vector components for all three triangles and then add or subtract them to combine them and get the new one. So on this page, I have a new diagram, let's use that. So we've got our original, and the notes here have the momentum of the red ball in the y-direction. To get that, we can just actually calculate it. So if we go sine 30 times 12, that'll give us six, and if we go cosine 30 times 12, that gives us ten point three nine, so let's round it up to ten point four. Now let's do the same thing for the blue ball, before and after. So we'd have cosine 40 times thirty four point seven, that gives us twenty six point six. And let's do the y-direction the same way, so it'll be sine 40 times thirty four point seven, and that gives us twenty two point three. Okay, so that's our initial before vector. We have six down and ten point four to the right. And then for our second ball, the blue ball, it's twenty two point three up and twenty six point six to the right. Why are the blue numbers quite a bit bigger? Well, remember, it was the larger ball and it was going quite a bit faster, so we have more momentum on that blue ball. So it does make sense that our momentum after should be larger as well. Now, for our after vectors, we're gonna do the same thing. We've got twenty-five degrees, so let's do the y-direction first. We'll have sine twenty-five times sixteen point two, so that gives us six point eight. And cosine twenty-five times sixteen point two gives us fourteen point seven. Okay, so we have all of our vectors. In the previous page, let me just go back to the previous page. So we have all our same numbers. We used thirty four point seven for this one instead of thirty four point six five, but it's close enough. We don't need to get too picky with our rounding. So we have everything right. When we look at our diagram, we assume that the blue one would go downwards and to the right, and you can draw it like that, it's totally fine. But what we're gonna find out is that actually it doesn't go in that direction after all once we do our components. So let's do our total before. So our total before, we've got ten point four to the right and twenty six point six to the right. When we add these together, that's gonna give us 37 to the right. And if we add our vertical components, we've got twenty two point three up and six going down. So twenty two point three plus negative six gives us sixteen point three up. So that's our net results for before. That means our after has to be the same, right? We have to get the exact same answer. So once Again, if we kind of think of it that way, we got our 10.4 plus 26.6, that's gonna equal our 14.7 to the right plus our unknown, so let's call it X. So we'll get our 37 minus 14.7, that gives us an answer of 22.3. So we know our horizontal one has to be 22.3 because our 22.3 plus 14.7 is the same thing as 26.6 plus 10.4.Now vertically, we want to do the same thing. So before, we've got -6 plus 22.3, and that's going to equal 6.8. That gives us 16.3 on the left side. Let's subtract the 6.8, and when we solve that, we get a Y of 9.5. So what does that mean? Well, that means if we got a Y of 9.5, that means we're going upwards, not downwards. Okay, so that's one thing you gotta be careful about and you don't really know that until you actually do the math.So let's go back to this diagram again. It was drawn as though that blue one after would move downwards. But if we think about it, the initial blue one has way more momentum. It had a momentum of 35, almost, the red one only was 12. So, and it's going out at steeper angles when they collide because it has way more momentum. It's gonna likely just go straight through, and the red one will be the one that gets bounced off. Okay, so it doesn't make sense that our blue ball one ends up going above the horizontal instead of back down. So I think we're good calculation wise.So now, our final step, just like the previous example, we've got our final vectors, we just need to figure out what is that side, what is that angle. So once again, we can just use trig and the Pythagorean theorem, and we'll find the angle as well.So we get 24.2 at 23 degrees. So just like our last question, remember, if we go back to the original question, it was asking what is the velocity, not momentum. So once again, to get our velocity, we need to divide by the mass. So 24.2 divided by 2.31, we get a final answer of 10.5, approximately. So that means we'll have 10.5 meters per second, and our angle is 23 degrees above. So we'd say 23 degrees north of East, and that will be our final answer.So let's go back and kind of see if that makes sense. So 10.5 going at 23. So if we go back to our original one, the red one was 1.2 kilograms, this one was 2.31, so the blue one was quite a bit larger, and the red one was going down at 10, the blue one was going at 15, so it was quite a bit faster. And then after the collision, the red one was 13.5 and our blue one didn't go downwards, it ended up going upwards, and we calculated 10.5. So, if you think about it, the blue one was going faster. It hits the red one, which shoots the red one a little bit quicker, so it lost some speed. It gave the red one some speed but lost some speed itself, so it drops down to ten point five. So, that seems realistic. And if we look at the angles, right? We figured we calculated that the blue one ended up moving upwards, and when we do the calculation, it's 23 degrees. So, it ends up being almost in line with the red one. This kind of goes back to what we talked about when you have car accidents. If a big, huge semi has a lot of momentum, it's almost going to end up going pretty much straight through, and the red one just kind of gets pushed into the same path. So, it's almost the same thing. The blue one's momentum in the direction kind of dominates, so it's going to plow through, and the red one gets sort of bounced off and in the scene. That's it for this question. There are a few homework questions you can try, and we've got the online assignments as well.