Hey, guys. Hopefully, you were able to take a look at this one on your own. Maybe try it out. But this is going to be a little bit different than the problems that we've seen before. We've got a 5 gram, 3 microcoulomb charge. It has some initial speed, and it's moving away from some negative charge. And then, basically, we're supposed to figure out how far can this thing travel before it stops. So I want to figure out just sort of like what's going on here. So let's go ahead and draw a quick diagram. Now, I've got this negative 5 microcoulomb charge, and then I've got another 3 coulomb or 3 microcoulomb charge. I know the mass of this charge is 5 grams, and I know the initial speed of this is equal to 20 meters per second. So what ends up happening is that as this positive charge is moving away from this negative charge, the force on it wants to pull it back towards the negative charge because these two things are opposite charges. Opposites attract. Right? Now I know what the initial distance is between these two objects, so that our initial is equal to Let's see what this What is that? Five centimeters? Yeah. It's 5 centimeters. So what ends up happening is that as this thing is flying out with some velocity, the force is eventually going to bring it to a stop at some later distance over here. Now, I know at this point be while it stops, the final velocity is going to be 0. And basically, what I need to figure out is what is the horizontal distance, or just horizontal, I'm guessing. What is this distance here that it can travel before it actually stops at this point? So if I can kinda think about this, this will be the final distance, \( r_f \). So that means that this \( x \) distance, if I can write it as an equation, is basically just going to be, I've got \( x \) is equal to \( r_{\text{final}} - r_{\text{initial}} \). And so this really is my target variable. How far can it travel before stopping? I know what the initial distance is. Now I just need to figure out what the final distance is. So in all these cases, I've got this final distance, initial distance. I've got speeds, masses, and I'm also talking about energy. So what I need to use is I actually need to use energy conservation in order to solve this problem. So what I'm going to do is look at all of my energies in the before and after case. So we know that our energy conservation formula is the initial kinetic energy plus any initial potential energy, which we know is going to be electric potential energy. Now there's no work done by non-conservative forces. We know that already. And then that's going to equal the final kinetic energy plus the final electrical potential energy. Okay. So we've got that the work done by non-conservative forces is 0. The electric force is conservative. And we know that when this thing finally stops, is there anything we can cancel out? Well, let's see. We've got an initial kinetic energy because we know the object is moving and we also have an electrical potential energy because we have 2 charges and they're separated by some distance, so there's always potential energy. Now, what happens is when this thing stops at this moment that's right out here, we know that the kinetic energy is equal to 0, but there still is some final potential energy. Okay? So basically, we can go ahead and expand out all of these terms. I know that this initial kinetic energy is going to be \( \frac{1}{2} m v_0^2 \). Now what happens is I've got to do the, the initial, kinetic or sorry. The initial electrical potential energy, which is \( \frac{k q_1 q_2}{r_{\text{initial}}} \), and then that's going to equal the final electric potential energy, \( \frac{kq_1q_2 }{r_{\text{final}}} \). So in other words, I'm actually looking for this \( r_{\text{final}} \) because then I can take this \( r_{\text{final}} \) and plug it back into this formula and \( x \) distance is that it travels before stopping. So that's basically the game. I'm going to just try to figure out what this \( r_{\text{final}} \) is. But what happens is, I can't go around and like start manipulating this by flipping these fractions or flipping these formulas because this is like, this is an addition, and I've got a formula that's a fraction in here. It's going to get really really ugly. Fortunately, what I can do though, is I know what the mass is, I know what the initial velocity is, I know what all of these constants are and including the distance. So rather than try to algebraically manipulate it, just start plugging in numbers for this stuff. Just reduce it down to like a simple number and then worry about the algebra later. So basically, I'm just going to plug in this really long mess right here. We've got 5 grams, so that's going to be 0.005. Now, we've got the initial velocity is 20, so then we're going to square that. Now, you've got plus, and we've got 8.99 times \( 10^9 \). Now the 2 charges that are involved. We've got 3 microcoulombs, so that's going to be 3 times \( 10^{-6} \). And a, let's see. Negative 5 times \( 10^{-6} \). And now you've got the distance, which is point 0, that's point 5. That's in centimeters. Right? So, that's the point 5 meters, and then we've got these formulas right here. We we we know what these charges are. Okay. So, basically, if you plug all of this stuff in, that should equal what \( \frac{k q_1 q_2}{r_{\text{final}}} \) is. And, basically, if you plug all of this stuff in really into your calculator, you're just going to get one big number, or doesn't have to necessarily be big. But this is actually going to be negative 1.7 joules. And that's going to equal \( \frac{k q_1 q_2}{r_{\text{final}}} \). So now, what we can do is now we can actually manipulate what this \( r_{\text{final}} \) is. And basically, what's just going to happen is this \( r_{\text{final}} \) goes up to the top, this negative 1.7 comes down to the bottom, and then we can just go ahead and plug everything in again. So that's why it's going to be easier to do that than actually work through all the algebra in that big step over there. Okay? So we've got the final distance is going to be \( 8.99 \times 10^9 \). Now I've got the 2 charges, which are, let's see. I've got 3 times \( 10^{-6} \), negative 5 times \( 10^{-6} \). By the way, there are some shortcuts that you could take in manipulating some of this algebra, but, yeah. So we've got this right here, and then we've got this negative 1.7. Okay? And now, if you plug all of this stuff in, you should actually get 0.0793. But remember, this is going to be in meters. So what this actually represents is this actually represents 7.93 centimeters. So this either one of these expressions, if you got these if you got this, would be correct, but basically, just this is just a different way of expressing this. But what we can do is we can stick this number. I'm just going to go ahead and stick with the centimeters because we're dealing with centimeters, and then go back and plug that into this formula over here. So now what my \( x \) distance is, what my real target variable, how far can this thing travel before it stops, is going to be 7.93 centimeters minus the initial distance of 5 centimeters. So that just means that this thing travels 2.93 centimeters before stopping. So this right here is actually our final answer, that 2.93 meters/cm. Okay? Let me know if you guys have any questions. This energy conservation stuff can actually become really useful in solving some problems. Sometimes you'll have to do it. So make sure you'\re comfortable with this. Go ahead and watch the video again if you didn't understand everything. Drop me a link, or comments, or a question, anything like that, and I'll see you guys later.