11. Momentum & Impulse

Collisions & Motion (Momentum & Energy)

# Collision Problems with Motion/Energy

Patrick Ford

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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

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