Hey, everyone. So, hopefully, you got a chance to work this problem out on your own. So you've got these two blocks of equal mass. They're going to undergo a head-on elastic collision. We want to calculate the magnitude and direction of the final velocities after colliding. So let's go ahead and work this out step by step. So the first thing we're going to do is just draw a quick little sketch of what's going on. So we've got these two blocks that are heading toward each other. So I'm going to call this one block A and this one block B. We've got that \( v_{A_{\text{initial}}} \) is equal to 5, and then we've got \( v_{B_{\text{initial}}} \) like this is equal to negative 3. Right? It's to the left, so it picks up a negative sign. And, ultimately, they're going to collide. And then afterward, we want to figure out, basically, where are they going and how fast. So after the collision, block A and B, they're going to have sort of unknown speeds. Right? They could be both going to the right or they could be going off in different directions or to the left. So we can't draw any arrows, but, basically, we want to calculate what are the final velocities of these two blocks. So let's move on to the next step. We're going to write our equations for conservation of momentum and elastic collisions. Remember, ultimately, we want to solve a system of equations. We're going to have to write both of these. So for the conservation of momentum, this is going to look a little bit different than this because instead of \( m_1v_1 \), we're really going to have \( m_A v_A \). Right? That's the object. \( m_A v_{A_{\text{initial}}} + m_B v_{B_{\text{initial}}} = m_A v_{A_{\text{final}}} + m_A m_B v_{B_{\text{final}}} \). Right? And then, over here, what we've got is we've got our elastic collision equation. Remember, this is the only one that we can use. We can only use this equation for elastic collisions, and it's basically that \( v_{A_{\text{initial}}} + v_{B_{\text{initial}}} = v_{A_{\text{final}}} + v_{B_{\text{final}}} \). Right? Remember, they look different. And so on and so forth. Okay. So we're basically just going to go ahead and start solving the system of equations by plugging in some numbers. So what happens is in the first problem here, usually, we would start plugging in the masses of each of these objects, but we actually don't know what they are. But that's fine because the mass of A is actually equal to the mass of B because they're equal mass. Alright. So they're equal mass, so that means that \( m_A \) is equal to \( m_B \). What that really means here is we can actually cancel out the \( m \) term from the entire equation. Remember, if you have the same number that goes through all of your terms, you can just cancel it out completely. So in other words, the mass actually really doesn't matter in this problem or at least in the equation. So, then let's go ahead and start plugging in some initial values. So at \( v_{A_{\text{initial}}} \) is going to be 5. The \( v_{B_{\text{initial}}} \) is going to be negative 3. This equals \( v_{A_{\text{final}}} + v_{B_{\text{final}}} \). And when you simplify this, what you're going to get here is you're going to get 2 equals \( v_{A_{\text{final}}} + v_{B_{\text{final}}} \). Alright? This is the first equation that we're going to need solve our system of equations because, remember, this is the one that has 2 unknowns. So let's look at the other equation, the elastic collision equation. Remember, if we start plugging in our values, \( v_{A_{\text{initial}}} \) is 5, \( v_{A_{\text{final}}} \) is unknown, \( v_{B_{\text{initial}}} \) is negative 3, and then \( v_{B_{\text{final}}} \) is unknown. Okay? So we also have these same 2 unknowns in this problem. And so what we want to do is, again, we want to sort of add something to this first equation so that one of the terms will cancel out, and then, basically, you're left with one unknown. Alright? So we want to sort of stick another equation down here, so that we can cancel out one of the terms. Okay. So what happens is when you bring the negative 3 over to the other side, it becomes a positive, and you end up with 8. And then when you move the \( v_{A_{\text{final}}} \) to the other side, it picks up a negative sign. So in other words, you get 8 equals negative \( v_{A_{\text{final}}} + v_{B_{\text{final}}} \). Alright? Now, again, this is where we would have to either multiply this equation to get one of these terms to cancel out, but we actually don't have to do that here because in this equation, we have a \( v_{A_{\text{final}}} \), and we also have a negative \( v_{A_{\text{final}}} \) in this equation. So we don't actually have to multiply this by anything just to make the numbers line up. So we can actually just go ahead and stick this equation right down in this box. So we're going to add this to, let's see. This is 8 equals negative \( v_{A_{\text{final}}} + v_{B_{\text{final}}} \). Notice how now when you stack these two things on top of each other and then you add them down, the \( v_{A_{\text{final}}} \) will just cancel out. And what you'll end up with here is you'll end up with 10 on one side, and you'll end up with 2 \( v_{B_{\text{final}}} \) on the other. Alright. So now that we sort of eliminated one of the equations, we're just going to go ahead and solve. So this final velocity here, this \( v_{B_{\text{final}}} \) for B is going to be 5. So now let's move on to the last step here, which is this is one of our target variables. This \( v_{B_{\text{final}}} \) here is actually going to be going off to the right like this. So this \( v_{B_{\text{final}}} \) equals 5, and we got a positive number, so it points to the right. The last thing we have to do is we just have to plug the first target variable into any of the other equations to then solve for the other missing variable. So in other words, we can stick this \( v_{B_{\text{final}}} \) into either one of these two equations to solve for the other one. It really just your preference. They're both pretty much the same. I'm just going to go ahead and go with the first one here. So then, basically, if we rewrite equation number 1, what you're going to get is the 2 is equal to, \( v_{A_{\text{final}}} + v_{B_{\text{final}}} \), but we actually know that that's 5 already. So in other words, this is just going to be 5 like this. Okay? So, we just go ahead and solve, and we're just going to get that \( v_{A_{\text{final}}} \) is equal to negative 3 meters per second, and that is your other final answer. So in other words, \( v_{A_{\text{final}}} \) is going to be moving to the left at negative 3. Alright? So now I want you to notice that something interesting happened in this problem. So we actually had these two blocks of equal mass. They collided, and what happens is the \( v_{A_{\text{initial}}} \) for object A was 5, and then the \( v_{B_{\text{initial}}} \) for object B was negative 3. But afterward, their velocities basically switched. So now, \( v_{A_{\text{final}}} \) is negative 3, whereas \( v_{B_{\text{final}}} \) is 5. So in other words, the \( v_A \) for object A became the \( v_B \), the final velocity for object B and vice versa for the for the second block. So in other words, when you have these two objects of equal mass and they undergo elastic collisions, one pro tip that you can use is they actually trade or exchange velocities. So they trade or exchange. Basically, they just swap the initial and final velocity between both objects. Alright, guys. So that's it for this one.