Welcome back, everybody. So in this problem, we have a bullet that is being fired into the bottom of a wooden block. It's fired upwards. It basically keeps on going after it passes through, so the bullet continues rising. But having transferred some of its momentum to the block, the block now starts rising as well. Basically, we want to calculate how high the block reaches. So this is going to be a maximum height. What I'm going to do here is I'm going to just call this some ymax value. That's really what we're interested in. In this problem, we have a collision that is followed by some motion. So we're going to go ahead and stick to our steps for solving these kinds of problems, these conservation problems, with energy. We're going to have our diagrams. We're going to label our points of interest. Before the collision, you have the bullet that's being fired upwards, and then it hits the block, and that's when the block starts moving, and then finally, it goes upwards and reaches its maximum height. There are basically three points of interest here: a, b, and c. The only thing that's different is that most of our problems up until now have been mostly horizontal, but in this problem, we have basically purely vertical motion. That's the only thing that's really different here. Let me go ahead and draw some stuff out. Remember that point a is going to be before the collision. Point b is going to be after the collision. This is also where the motion of the block starts. And then, finally, when it rises up to point c, that's where the block ends its motion. Just so it's kind of consistent with the other diagrams or sort of timelines that we've set up.
If we want to calculate the ymax, we're going to do that by writing out our momentum and energy conservation equations. Remember that from a to b, you have to use conservation of momentum because this is a collision. So we're going to use conservation of momentum here, and then we're going to use conservation of energy for the b to c interval. First, I want to label out these masses. The bullet, this is going to be my m1, which is going to be 6 grams. 6 grams are going to be 0.006 kilograms. The mass of the block, I'm going to call this m2, is equal to 1.2 kg. So I've got 1.2 here. So I'm going to write out my momentum conservation. So I've got m1v1a + m2v2a = m1v1b + m2v2b Right? That's the initial and final, the a and the b. One thing I also did here is I noticed that the blocks don't stick together, so we can't use the shortcut of combining everything into one object. Remember, the bullet, after it passes through the center of the block, it continues moving upwards. So these things are separate objects. They don't become 1.
Let's look at the b to c interval because now we're going to use conservation of energy. Remember, initial to final. So we have to use KEinitial + PEinitial + workdone by nonconservative forces = KEfinal + PEfinal. What are we looking for? We're looking for the maximum height. So in other words, we're looking for a variable that happens at point c where the block ends its motion. Naturally, the best equation to start with is by using the b to c interval. Let's expand out the terms. But first of all, what are we going to consider as our system? Remember, this is kinetic and potential, but first, for conservation problems, you have to define what your system is. Do we consider the bullet plus the block? Do we just consider the block only? Well, the best thing here is actually just to consider just the block. Write this out here. The system is only the block. The reason for that is, remember, in energy conservation problems, you can always pick your system as long as you're consistent with the terms. What happens is after the collision, and the bullet just becomes separate objects, we don't really care about it anymore. We're only really just considering or interested in the maximum height of the block. So we only have to worry about the energy conservation of the block. That's really important here. Make sure you keep that in mind. What kinds of terms can we cancel out here? Let's see. In the initial, is there any kinetic energy? Well, what happens here is at point b, this is after the collision between the bullet and the block. That's when the block starts moving. So there's definitely some kinetic energy here. What about any potential? Remember that spring or gravitational potential? We don't have any springs here, but we do have things moving in the vertical axis. This potential energy just depends on where we choose our 0 points. And remember, with energy problems, you always want to pick the lowest points to be your 0 points. So here, where the block is on the floor, I'm just going to consider that my y = 0. And so what that means is that there's no potential energy in my calculation. There's also no work done by nonconservative forces. There's no friction or anything like that. What about here at point c? Is there any kinetic energy? Well, remember, once the block travels from b to c and stops and ends its motion, that's when it's at the maximum height. So there is no kinetic energy because vc = 0. Once the block reaches its maximum height, there's no velocity. But now there's definitely some gravitational potential because the block has risen some distance, and this is really what we're trying to find here. This is really the ymax. That's really our target variable here. Then when we expand out our terms, what you're going to see here is you're going to have 12m2v2bsquared, right, 12bsquared, that's kinetic energy, equals gravitational potential energy, mgymax. So this is my target variable here. What you'll also notice is that the masses will cancel out because they're on both sides of the equation. So, really, I'm really close to just being able to plug in everything and solve. The problem is that I actually don't have what this v2b is. That's the velocity of this block right after the collision with the bullet. So right after the collision, the bullet, the block picks up some of the speed from the bullet, and it's going to go upwards like this with v2b. I don't know what that is yet, and so I can't go ahead and finish out this equation. But this is basically just going to be 12v2b, 12somethingsquared divided by 9.8 is going to equal your ymax. All I have to do is figure out what's going to go inside this equation. And to do that, when we get stuck in one of our intervals in these problems, we're always going to go to the other interval here, because remember, v2b is also in my momentum conservation. That's why I have to use both of them.
Let's go ahead and plug in some values and then just solve. So m1 is going to be the mass of the bullet, that's 0.006 times v1a. That's the initial speed of the bullet. Remember, the initial speed of the bullet is going to be 800 meters per second. That's what it's fired at, so it's 800. Now what's m2? Mass 2 is 1.2. What's the initial speed of the block? That's v2a. Well, it's actually just nothing because, again, at point a, when the bullet is still being fired upwards, it's before the collision, there is no initial velocity for this block. It's equal to 0, so that term goes away. So then what we have is we have 0.006 and then v1b. That's going to be the speed of the bullet after the collision here. So what happens here is that this bullet exits and it keeps on going upwards. I'm going to call this v1b, and it's emerging moving upwards at 150 meters per second. So that's v1b over here. It's 150. So you plug that in, it's going to be 150, and this is going to be a mass 2, which is going to be 1.2 times v2b. So remember, I came over here because I needed what this v2b is. If you look through your variables, this is the only one that's missing. So I'm just going to go ahead and simplify some stuff. This is going to be 4.8 when you work this out. This is going to be 0.9 + 1.2v2b. So pretty complicated equations, but they work out and they basically simplify to just a couple of numbers. So then when you work this out and you move the 0.9 over to the other side, what you're going to get is 3.9. And then when you divide by the 1.2, that's going to be v2b. And when you work this out, this is going to be 3.25 meters per second. Remember, we're not done yet. This is not our final answer because this is just the velocity of the block right after the collision. Now we just plug it back into this equation over here. So this is going to be my 3.25. Now when you work this out, this is going to be the final answer because this is your ymax, and this is going to be a final maximum height of 0.54 meters. That's pretty reasonable. Right? A bullet was really, really small mass, but it has a lot of momentum because it's being fired very, very quickly at this 800 meters per second. When it hits something like a block, which is 20 times heavier, it gives a little bit of speed and it rises a little bit of distance, but nothing too crazy. That makes sense. Let me know if you have any questions.