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11. Momentum & Impulse

1

concept

9m

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Hey guys. So in previous videos we've seen how to solve motion problems by using conservation of energy. And in the last few videos we've seen how to solve collision problems by using conservation of momentum, we're gonna combine those two things in this video because some problems are gonna involve a collision and then they'll involve motion with changing speeds or heights after the collision. So what I have, what I mean by that is that in this problem we're gonna work out down here, we have a crate that collides with another crates, so that's a collision in the first part. And then what happens is they're gonna stick together and they're gonna both rise up this inclined plane. So they're gonna change some speeds and heights after the collision. So this is gonna be the motion parts. Now we know to solve these problems individually. But really what happens is that to solve these problems are actually use both conservation equations, You're gonna use conservation of momentum during the collision parts and conservation of energy during the motion parts. These problems are super important and there's a lot of different variations that you're likely to see on a homework or test and they can get very messy if you're not careful. But what I'm gonna do in this video, I'm gonna show you a step by step, process and system so you can avoid any confusion and get the right answer. So let's go ahead and check out our problem here. So we have is what we said a collision and then followed by motion. So let's go ahead and stick to the steps here. What we're gonna do is we're gonna draw diagrams for our problems, but then we're gonna label points of interest. Here's what I mean by this. In the collision part of our problem, we have an initial and a final in the momentum equation, we have M one V one initial, M two, V two final. Things like that. But you also have an initial and final during the motion parts in the energy equation, we have like K initial and you final and things like that. So what happens in these problems is that the final part of the collision part of the problem is actually the same as the initial of the motion part of the problem. So if you're not careful, you're gonna have eyes and s floating around anywhere everywhere and that can get very messy and potentially confusing. So to avoid this problem, I'm gonna take a similar approach to what I did with projectile motion. I'm basically gonna call these things these points of interest Hsbc's and so on. Instead of eyes and f. So what happens is the different points of interest of our problems are right before the collision, it's right before the 20 hits the 30. So, I'm gonna call that point a the next point of interest is what happens after the collision and they stick to each other. So I'm gonna call that point B here. So point B is right after the collision. What's interesting about point B is that's also the same point is when the motion part starts is basically where they are at the bottom of the incline and they're gonna start moving up. So point B. Is after the collision, but it's also where the motion part starts and then eventually they're going to rise up and eventually stop once they hit the top of the incline. So the last point is where the motion part ends and I'm gonna call that point C. So what's our target variable? We're looking for how high the crates travel before they stop. So basically they're gonna travel some vertical distance or some heights. I'm gonna call that why? But that's gonna be why at point C. Here. So this is what we're looking for in our problem, how do we solve for that? We've already drawn the diagrams and label the points of interest, but now we're going to go ahead and start writing some equations. So the second step, you're gonna write out both your momentum and energy conservation equations here. So we're gonna use the conservation of momentum for the collision parts and then we're gonna use conservation of energy for the motion parts. So let's go ahead and do that. So what I'm gonna do here is I'm gonna use M. One V. One except instead of initial and final. I'm gonna use this according to the letters that I've just discussed. And one view and A. M. Two V two A equals M. One V one B plus M. Two V two B. And now for the energy equation I'm going to use K. Initial that's K. B plus you initial plus work done non conservative equals K final plus you final here. So that's both of the equations. Now we just have to figure out which one to start off with. So hopefully you guys realize that if we're trying to figure out why C. Then we're gonna start with the equation that includes point C. Which is our energy conservation equation. So let's go ahead and start expanding out the terms here. Do we have any kinetic energy at B? Remember bees after the collision? So after these two things collide they're both moving up the inclined plane with some speed. So they have kinetic energy. However do they have potential energy? Well, actually, no, because this is the point where they're still at the bottom of the incline. So here what I'm gonna do is I'm gonna call this the ground level Y equals zero. And therefore there's no potential energy here. There's also no work done by non conservative forces because you're not doing anything, there's no friction. What about K. C. K. Final? What happens is when the blocks go up the incline aren't you gonna stop? That's where their maximum height is that's going to be the speed is zero and therefore there's no kinetic energy. So basically all of it gets converted to gravitational potential, which is where are Y. C. Comes from? So let's go ahead and expand out the terms. So for KB we're gonna have one half M V B squared. But hopefully you guys realize that this is actually a completely an elastic collision because one crate sticks to another one and they both move at the same speed. So what happens is this is a completely in elastic collision. So what happens is I'm not gonna use little M I'm actually gonna use Big M. So here, what happens is that big M is equal to basically both the masses are gonna stick together as one. So one big M V B squared is equal to big M G Y C. So that's our target variable and we can actually see that the masses are gonna cancel. So even if you didn't know that they stick together, that would have been okay because they cancel out. So let's go ahead and write an expression for R. Y. C. I'm gonna move the G to the other side. And basically with this one half, Y C just becomes V B squared over two G. So this is our target variable, Y C. And I'm always ready to start plugging everything in. The problem is I don't know what this VB is this velocity which is after the collision. So how do we figure this out? We'll remember point B is the part where r is the point where our motion starts. But it's also the same point is where the collision ends. So if I'm stuck here solving for VB, hopefully you guys realize they can actually solve this by using our conservation of momentum equation because I have VBS on this equation as well. In fact this will very commonly happen in your problems. You're gonna start off by using one of these equations, but then you're gonna have to go to the other one to fully solve the problem. So that's what we're gonna do here. We're just gonna use our conservation of momentum equation now. So what I'm gonna do is I'm gonna call this object one and two and we'll start plugging in all the numbers that we have. So this is gonna be 20 times the initial speed of 40 plus 30 and then times the initial speed of zero. So this via here, this V two A is equal to zero because this block is initially at rest, so there's no momentum here. So then what happens afterwards, we already said this is a completely an elastic collision. So basically what happens is that these masses combined and these two velocities have to be the same. So what I'm gonna do is I'm gonna add the masses 20 plus 30 and then this is just gonna be V B. So this is what I'm looking for here, if I can figure out this VB, I can plug it back into this equation. So to solve for this, you're gonna do 800 divided by the 50 on the other side and you're gonna get VB equals 16 m per second. Alright, so now we're just gonna finish off the calculation here. We're gonna plug this because now we know VB this is 16 squared divided by two times 9.8. And if you go to work this out, you're gonna get 13.1 m. So what happens is these two things stick to each other and they're both gonna rise a distance height of 13.1 m above the incline. Notice how this problem? The angle of the incline never factored into the problem. And that's because in conservation of energy this potential energy doesn't depend on the angle, just depends on the heights. All right, so, I have like that's really what there is uh that's really all there is to the problem. I have a couple more conceptual points to make here. So, some of you may be wondering why I was able to use conservation of energy even though this was a completely an elastic collision. And the idea here is that energy isn't conserved. You're gonna lose some energy during the collision. That's in this interval from A to B. But once they stick together, energy is conserved afterwards. So basically travels one object and energy is gonna be conserved in the interval from B to C. So we're totally okay with using the conservation of energy here. Alright, so only if the work done by non conservative forces is zero if you have no friction or any work done by you or something like that. Alright, so the last thing I want to point out is that we actually did one sort of specific example of this type of problem. We saw a collision where they both travel up an incline plane afterwards. But a lot of these problems are gonna have a couple of different variations or situations that you might see, but you're always gonna use this system whenever you see these kinds of problems, you could have a collision where they go up against a spring like this, you could have a collision in which both objects encounter some friction afterwards or you could have a collision and then both objects basically swing up like this as a pendulum. Notice what's common about these is in all of these situations you have changing speeds or heights after the collision. So we're gonna use all the steps that we just discussed here, You're gonna write out your conservation equations and you're gonna start working down where your target variable is. Alright, so that's it for this one. Guys, let me know if you have any questions

2

Problem

A 300g bullet is fired horizontally into a 10-kg wooden block initially at rest on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.6. The bullet remains embedded in the block, which then slides 35 m along the surface before stopping. What was the initial speed of the bullet?

A

492.5 m/s

B

20.3 m/s

C

677 m//s

D

697 m/s

3

example

9m

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welcome back everybody. So in this problem we have a bullet that is being fired into the bottom of a wooden block, so it's fired upwards, it basically keeps on going after it passes through, so the bullet continues rising, but having transferred some of its momentum to the block, the block now starts rising as well and basically we want to calculate how high does the block reach, so this is gonna be a maximum height. What I'm gonna do here is I'm gonna just call this some y max value, that's really what we're interested in. So in this problem we have a collision that is followed by some motion. So we're gonna go ahead and stick to our steps for solving these kinds of problems, these conservation problems with energy. So we're gonna have our diagrams, we're gonna label our points of interest. So in this problem, what happens here is before the collision, you have the block that's being fired upwards and then it hits the block and then that's when the block starts moving and then finally it goes upwards and it reaches its maximum height. So there's basically three points of interest here, A. B and C. The only thing that's different is that most of our problems up until now have been mostly horizontal, but in this problem we have basically purely vertical motion. That's the only thing that's really different here. So let me go ahead and just draw some stuff out, remember that point A is going to be before the collision point B. Is going to be after the collision. This is also where the motion of the block starts and then finally when it rises up to point C. That's where the block ends, its motion. Alright. Just so it's kind of consistent with the other diagrams are sort of timelines that we've set up. Alright, so let's go ahead and uh take a look. So if we want to calculate the Y max, we're gonna do that by writing out our momentum and energy conservation equations. So remember that from A to B. You have to use conservation of momentum because this is a collision. So we're gonna use conservation of momentum here and then we're gonna use conservation of energy for the B. Two C. Interval. Okay, so let's write this out first. Want to do what I want to do is label out these masses. So the bullets, this is gonna be my M. One which is gonna be six g. 6 g is going to be 0. kg. The mass of the block, I'm gonna call this M two is equal to 1.2. So I've got 1.2 over here. So I'm going to write out my momentum conservation. So I've got M one V one A plus M two V two A equals M one V one B Plus M2 V two be right, that's initial and final the A. And the B. Right so that we can confuse their. Alright, one thing I also did here is I noticed that the blocks actually don't stick together so we can't use the shortcut of sort of combining everything into one object. Remember the bullet after it passes through the center of the block it continues moving upwards. So these things are separate objects that don't basically become one. All right. So um let's look at the B2C interval because now we're going to use conservation of energy. So remember initial to final. So we have to use kinetic initial plus potential initial and the work done by non conservative forces equals kinetic plus potential final. Okay, so what are we looking for? We're looking for the maximum height. So in other words we're looking for a variable that happens at point C. Where the block ends its motion. So naturally the best interval, the best equation to start with is by using the B two C interval. Right? So let's expand out the terms. But first of all, what are we going to consider as our system? Remember, this is kinetic and potential. But first for conservation problems you have to define what your system is. Do we consider the bullet plus the block? Do we just consider the block only? Well, the best thing here is actually just to consider just the block. So I want you to write this out here. So the system is only the block and the reason for that is remember in energy conservation problems you can always pick your system as long as you're consistent with the terms. What happens is after the collision and the bullet basically just becomes separate object. We don't really care about it anymore. We're only really just considering or interested in the maximum height of the block. So we only have to worry about the energy conservation of the block. Alright, so that's really important here. Make sure you keep that in mind. Okay, so what is the um what kind of terms can we, can we cancel out here? Um Well let's see in the initial, is there any kinetic energy? Well, what happens here is at point B. This is after the collision between the bullet and the block. That's when the block starts moving. So there's definitely some kinetic energy here. What about any potential? Remember that spring or gravitational potential? We don't have any springs here but we do have things moving in the vertical axis. Well, this potential energy just depends on where we choose are zero points. And remember with energy problems, you always want to pick the lowest point to be your zero points. So here where the block is on the floor, I'm just gonna consider that my Y equals zero. And so what that means is that there's no potential energy in my calculation there's also no work done by non conservative. There's no friction or anything like that. Right? What about here at point C. Is there any kinetic energy. I remember once the block travels from B to C and stops and ends its motion. That's when it's at the maximum height. So there is no kinetic energy because V. C. Is equal to zero. Once the block reaches its maximum height, there's no velocity. But now there's definitely some gravitational potential because the block has risen some distance and there's this is really what we're trying to find here. This is really the Y maximum, that's really our target variable here. Alright, so then when we expand out our terms, what you're gonna see here is that you're going to have one half of M two V two B squared. Right? What happened? B squared? That's kinetic energy equals gravitational potential energy. Mg. Y. So it's M two G times Y max. So, this is my target variable here, which you'll also notice is that the masses will cancel out because they're on both sides of the equation. So really I'm really close to just being able to plug in everything and solve the problem is that I actually don't have with this V two B is that's the velocity of this block. Right after the collision with the bullets. So, right after the collision, the bullet, the block picks up some of the speed from the bullet and it's going to go upwards like this with V to be, I don't know what that is yet. And so I can't go ahead and finish out this equation, But this is basically just going to be 1/2 V to be One half of something squared divided by 9.8 is going to equal your y maximum. Right? So all I have to do is just figure out what's going to go inside of this equation and to do that when we get stuck in one of our intervals. And these problems were always going to go to the other interval here because remember V2B is also in my momentum conservation. So that's why I have to use both of them. All right, so let's go ahead and plug in some values and then just go ahead and solve. So M1 is gonna be the mass of the bullet. That 0.006 times V one A. That's the initial speed of the bullet. Remember the initial speed of the bullet is going to be 800 m per second. That's what it's fired at. So, it's 800 Now its mass to master is the 1. what's the initial speed of the block? That's V two a. Well, it's actually just nothing because again, at point at point a when the bullet is still being fired upwards before the collision, there is no initial velocity for this block, it's equal to zero. So that term goes away. Alright, so then we have is we have 0.6 and then V one B. That's going to be the speed of the bullet after the collision here. So what happens here is that this bullet exits and it keeps on going upwards. I'm gonna call this V one B and it's emerging moving upwards at 1 50 m per second. So that's V one B over here, it's 1 50. Alright, so that's for the bullets. Actually let me go ahead and write that in blue. So this is going to be V1B Which is 150, So you plugged in and it's gonna be 150 and this is gonna be mass to which is going to be the 1.2 times V to be. Remember I came over here because I needed what this V two B is. If you look to your variables, this is the only one that's missing. So I'm just gonna go ahead and simplify some stuff, This is going to be 4.8. When you work this out, this is gonna be 0.9 plus 1.2. V two B. Alright, so pretty complicated equations. But they work out and they basically simplified to just a couple of numbers. Alright, so then when you work this out and you move the 0.9 over to the other side, what you're gonna get is 3.9, and then when you divide by the 1.2, that's going to be V two B and when you work this out, this is going to be 3.25 m per second. Okay, remember we're not done yet. This is not our final answer because this is just the velocity of the block right after the collision. Now, we just plug it back into This equation over here. So, this is going to be my three points 25. Now, when you work this out, this is gonna be the final answer, because this is your Y max, and this is gonna be a final maximum height of 0. m. Alright, that's pretty reasonable. Right? A bullet was really, really small mass, but it has a lot of momentum because it's being fired very, very quickly at this 800 m per second when it hits something like a block, which is, you know, 20 times heavier. Um It gives a little bit of speed and it rises a little bit of distance, but nothing too crazy. All right, so that makes sense. Um let me know if you have any questions

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