Hey, everybody. So let's take a look at our problem here. This one's kind of interesting because we have three different objects, a bullet and two blocks, and two different collisions that are happening. Basically, what we want to do is we want to calculate the speed of the second block after the bullet embeds itself into the second one. So let's go ahead and stick to the steps. We're going to draw a diagram from before and after. This is what's going on in this problem. We have a bullet like this. It's flying and it's going to hit towards a block. It's going to head towards a block. So I'm going to call this \( m_1 \). This has a mass of 0.01 kg and the block has \( m_2 \), which I'm going to call 0.5 kg. What happens is the bullet actually keeps on going after it's passed through the first block. It keeps on going through the first block and actually embeds itself in a second block. I'm going to call the \( m_3 \), and this equals 0.5 kilograms as well. And, basically, once it embeds itself in there, the whole entire system is going to be moving off with some velocity. I'm going to call this \( v \) at this point. Alright? And that's really what we want to calculate. The tricky thing here is that we have three different objects and two different collisions. So if you're not careful, what happens is that if you use \( m_1v_1{initial} \) and \( m_2v_2{initial} \), the final will be the initial for the second collision, and that can be really confusing. So what we're going to do is we're going to do something that we've used in other, sort of chapters like motion and forces, which is we're going to label the different events that are happening. So here's what I'm going to do. I'm going to call this a, this one is b, and this one is c. So a is actually going to be so a is the point where it's before the first collision. So this a, b, and c do not refer to different objects. They refer to different sort of events or different points in time. I want to be really clear about that. So it's before the first collision. Now point b is after the first collision. Right? It's when the bullet exits the first block and keeps on going towards the second one. So it's after the first, but it's before the second collision. And then finally, point c is when the bullet embeds itself into the second block and so that's after the second collision. So, really, what we want to calculate is this final velocity here, and I'm going to call this velocity c. That's really what we're interested in. So how do we do that? Well, we're going to stick we're going to, again, stick to the steps, move on to the second thing, and we're going to write the conservation of momentum equation. If I want to calculate the \( v_c \), then I'm going to pick the interval that basically includes c. So in other words, I'm going to include the I'm going to have the interval from b to c. Right? That's the interval after the first collision but before the second one, and then when the bullet finally embeds itself. So let's set up our equation here for b to c. So what I'm going to do is, again, keep, just keep in mind that we have three different objects. So, really, what's going on is that the bullet, which is \( m_1 \), is going to collide with the third object, \( m_3 \). So here's how our equation's going to get set up. It's going to be \( m_1 v_1b \). Right? That's the initial from b to c. Plus \( m_3v_{3b} \) equals now what happens is when the bullet embeds itself, that is a completely inelastic collision. So what happens here is that from b to c is completely inelastic. Whereas, the first collision from a to b is just an inelastic one because nothing gets stuck together. Right? The bullet exits and both of these things sort of like you know they go apart. Right? They don't actually stick together. Okay. So this is going to be, again, using our shortcut, \( m_1 + m_3 \), and this is going to be \( v_c \). So this is really my target variable is what is the final velocity of the bullet plus block combo? So that's what I'm looking for here. So if you look through your variables, which what you're going to see is that all these masses are given to us. So can we actually just solve this by using this interval? Well, if you look at this, what happens is we're going to hit 0.01 and then \( v_{1b} \) is going to be the velocity of the bullet after it exits the first block. Right? So this is going to be \( v_{1b} \). Now what is that velocity? If you look through the problem, you might see that this 430 meters per second, but remember, that's the initial speed of the bullet. In other words, the \( v_{1a} \) is going to be 430, but what happens is after it exits the first block, it's going to change. Right? There's going to be an exchange of velocity there. There's going to be some exchange of momentum, and \( v_{1b} \) is definitely not going to be 430 meters per second. But we actually don't know what that is. So I don't know what this is yet, and I'm just going to leave it blank for here. Now, what about \( m_3v_{3b} \)? Well, basically, this is just the initial velocity of the second block before the collision. In that case, the third the second block is stationary. So in other words, that term actually just goes away because that term is 0. So we have 0 on the left side and then on the right side we have two velocities the two masses combined, so 0.01, 0.5, \( v_c \). So there's only one variable that we really need, and that's this \( v_{1b} \) over here. It's the velocity of the block after it exits the first block or, sorry, the velocity of the bullet after it exits the first block. Now how do we go find that? Remember, we in this problem, we have two different collisions, two different sorts of intervals. And so to go and find this \( v_{1b} \), I'm going to have to go and use the other interval, which is going to be the a to b interval. Alright? So let's set up that equation now. Now that equation is going to look like \( m_1 v_{1a} + m_2 v_{2a} = m_1 v_{1b} + m_2v_{2b} \). Now remember, we can't use the shortcut like we did over here because this collision is not completely inelastic. The two masses will not stick together. Alright? So that's the important part. So, really, this \( v_{1b} \) here is actually going to be this \( v_{1b} \), and that's, again, why we didn't use initials and finals because it can get really confusing. So let's go ahead and start plugging in the values and solving. So this is going to be 0.01. What about this \( v_{1a} \)? We actually know what that is. That's just the initial velocity of the bullet, which is 430 meters per second. So 430 plus \( m_2v_{2a} \). So, basically, that's the mass and velocity of the first block of this one. Now, remember, both blocks are at rest. So what happens here is you're just going to get 0.5, but then this cancels out because this is equal to 0. So, basically, the whole term goes away. Then we have a 0.01 and then \( v_{1b} \), that's exactly what we're trying to find here, plus. And then we've got 0.5. And then what about \( v_{2b} \)? It's basically the velocity of the block after the collision. Now what number is that? Well, if you look through your variables, what's gonna happen here is that the speed of the first block immediately afterward is 5.6 meters per second. So that's basically \( v_{2b} \). Alright? That's what's going to go over here, the 5.6. So now what we can do is we can solve this equation for \( v_{1b} \), and that's exactly what we were missing in the b to c interval. Okay? So if you plug all the stuff in, what you're going to get is 4.3 on the left side. This is going to equal 0.01\( v_{1b} \)+, and then this works out to 2.8. So when you bring this over to the other side, what you're going to get here is you're going to get 1.5 is equal to 0.01\( v_{1b} \). And then finally, \( v_{1b} \) is equal to 150 meters per second. Alright? So now that we have this 150, that's what we stick into this equation over here. So this becomes the 150, and now we can go ahead and solve for \( v_c \). This is going to be 0 point, we work this out. This is going to be 1.5 again equals 0.51\( v_1 \) or, sorry, \( v_c \). And now if you work this out, what you're going to get is the \( v_c \) is equal to 2.94 meters per second. Alright? So it's a little bit of, you know, sort of upfront work, getting organized, labeling everything, but once you do that, the rest of the math becomes pretty straightforward. Alright? So that's your final answer. It's 2.94 meters per second. Let me know if you have any questions.