Hey, guys. So for this video, I'm going to show you another type of collision with motion and energy type problem, which is where you have a collision and then some interaction with a spring. So in our example that we're going to work out down here, we have a crate that collides with another crate. That's the collision part. And then afterwards, these crates are going to stick together and they're going to compress up against the spring. So that's the motion or the energy part of the problem. There's really nothing new to this. We're just going to use conservation of momentum for the collision, that's what we always do, and then conservation of energy for the spring parts. Alright? So there's nothing new. We've seen how to do these kinds of problems before. Let's go ahead and take a look. So we have the initial speeds of the crates, and we want to go ahead and draw the diagrams and then label the points of interest. Right? So before the crates actually collide, that's before the collision, that's going to be part a. After the collision, it's going to be part b and that's also right where the motion starts. That's right when the block start compressing up against the spring. And then finally, when they reach that maximum compression distance, that's where the motion ends. That's going to be part c. So what we really want to find here is we want to find the maximum compression distance of that spring. So before the collision, what happens is that the spring is just relaxed like this, and then afterwards, when these two crates combine and they compress against the spring, now the tip of the spring is over here. What we want to find is this distance over here. What is this x? How far did it actually compress? Alright.

So let's go ahead and take a look at the second step. We're going to write our momentum energy conservation equations. So we're going to go ahead and do that over here, and we're going to use conservation of momentum for the collision and then conservation of energy for the motion parts. So I've got m_{1}v_{1a} + m_{2}v_{2a} = m_{1}v_{1b} + m_{2}v_{2b}. And then for the energy part from b to c, I've got Kb+Ub+work non-conservative=Kc+Uc.

So now we have our two conservation equations. I just figured out which one I'm going to start with, and hopefully you guys realize that what I'm trying to actually find here is what's the maximum compression distance, which is basically what is the distance when you reach part c where the motion ends. So if we're trying to find x_{c}, that's going to come from the potential energy term inside of my conservation of energy equation. So that's what I'm going to start with. So let's go ahead and expand out each of the terms, figure out what I can cancel. So do we have any kinеtic enеrgy initial at b? Well, yes. Because these two crates are going to stick together and they're both going to start compressing the springs, they have to have some kinetic energy. Is there any potential energy? Well, remember that this potential energy is gravitational plus elastic potential energy. And right here at point b, that's where the spring is still relaxed. It hasn't started compressing yet. So there's no gravitational potential and there's also no elastic potential energy. Remember your elastic potential energy is 1/2kx^{2}, but it hasn't started compressing yet. There's also no work done by non-conservative forces because you're not doing anything and there's no friction. What about k final? So what happens, I'm sorry, there's no potential еnеrgy here. So what about k_{c}? Well, hopefully you guys realize that the maximum compression distance of the spring is going to happen where the blocks have stopped moving. So when the, the spring is maximally compressed, the velocity here for both the crates is going to be 0, and so therefore, there's no kinetic еnеrgy. And so therefore, all of this sort of potential еnеrgy happens because you have some, you have some compression distance like this. Right? So we're going to go ahead and expand out our terms. So our kinetic energy at b is going to be 1/2, and then we're going to use little m or we're going to use big M. What happens is these two crates are going to combine together. They're going to stick to each other. So this is a completely inelastic collision. And so what happens here for both the crates is that your big M is equal to m_{1}+m_{2}. They stick together. So we're going to use 1/2 big M v_{b}^{2} = 1/2 k x_{c}^{2}. So there's our target variable. That's what we're looking for here. And unlike what happens in other problems, our mass is not going to cancel. We have m on one side and k on the other. We can cancel out the one halves and then we can write an expression for this x_{c} term here. So let's go ahead and do that. So x_{c}, when I basically move the k over and I take the square roots, I'm going to get the square roots of m over k times x, times v_{b}^{2}. So this is going to be the square roots of I've got the big mass which is really just the combination of 10 and 30, so it's going to be 40 here, divided by k. My compression of my spring constant was equal to 500. That's what I was given here. So all I have to do is just figure out what is my velocity. So notice how I don't have the velocity after the collision and where the motion starts. So what happens here is I need to go figure out this v_{b} by looking at my conservation of energy equation because I have v_{b}'s over here. Alright? So once I can figure that out, that's what I'm going to plug into this equation right here. So let's go ahead and do this conservation of momentum. So I'm going to have 10 times 20, if I call this object 1 and object 2. So 10 and 20 plus this 30 kilogram block is actually initially at rest. So v_{2a} is equal to 0 up here. So I have 0 over here, it cancels out, and this is going to equal Well remember, if this is a completely inelastic collision, then both of these velocities here are actually going to be the same. So I can group them together, and I can basically use my completely inelastic collision shortcut? So I'm going to use 10 +30 and they're both going to have the same speed of v_{b}. So that's what I'm looking for here. And so this is going to end up being 200 divided by I'm going to move this 40 to the other side. That's going to equal v_{b} and you're going to get 5 meters per second. So that's the number that we plug into this equation over here. So if you just plug in this square root of 40, you know, times whatever, you're going to get the maximum compression distance of 1.41 meters. And that's the final answer. So, again, you're just using conservation of momentum and energy to solve your problem. That's it for this one, guys. Let's move on.