11. Momentum & Impulse

Collisions & Motion (Momentum & Energy)

# Bullet-Block with Height

Patrick Ford

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welcome back everybody. So in this problem we have a bullet that is being fired into the bottom of a wooden block, so 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 and basically we want to calculate how high does the block reach, so this is gonna be a maximum height. What I'm gonna do here is I'm gonna just call this some y max value, that's really what we're interested in. So in this problem we have a collision that is followed by some motion. So we're gonna go ahead and stick to our steps for solving these kinds of problems, these conservation problems with energy. So we're gonna have our diagrams, we're gonna label our points of interest. So in this problem, what happens here is before the collision, you have the block that's being fired upwards and then it hits the block and then that's when the block starts moving and then finally it goes upwards and it reaches its maximum height. So there's 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. So let me go ahead and just 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. Alright. Just so it's kind of consistent with the other diagrams are sort of timelines that we've set up. Alright, so let's go ahead and uh take a look. So if we want to calculate the Y max, we're gonna do that by writing out our momentum and energy conservation equations. So remember that from A to B. You have to use conservation of momentum because this is a collision. So we're gonna use conservation of momentum here and then we're gonna use conservation of energy for the B. Two C. Interval. Okay, so let's write this out first. Want to do what I want to do is label out these masses. So the bullets, this is gonna be my M. One which is gonna be six g. 6 g is going to be 0. kg. The mass of the block, I'm gonna call this M two is equal to 1.2. So I've got 1.2 over here. So I'm going to write out my momentum conservation. So I've got M one V one A plus M two V two A equals M one V one B Plus M2 V two be right, that's initial and final the A. And the B. Right so that we can confuse their. Alright, one thing I also did here is I noticed that the blocks actually don't stick together so we can't use the shortcut of sort 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 that don't basically become one. All right. So um let's look at the B2C interval because now we're going to use conservation of energy. So remember initial to final. So we have to use kinetic initial plus potential initial and the work done by non conservative forces equals kinetic plus potential final. Okay, so 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. So naturally the best interval, the best equation to start with is by using the B two C interval. Right? So 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. So I want you to write this out here. So the system is only the block and 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 basically just becomes separate object. 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. Alright, so that's really important here. Make sure you keep that in mind. Okay, so what is the um what kind of terms can we, can we cancel out here? Um Well 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. Well, this potential energy just depends on where we choose are zero points. And remember with energy problems, you always want to pick the lowest point to be your zero points. So here where the block is on the floor, I'm just gonna consider that my Y equals zero. And so what that means is that there's no potential energy in my calculation there's also no work done by non conservative. There's no friction or anything like that. Right? What about here at point C. Is there any kinetic energy. I 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 V. C. Is equal to zero. 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 there's this is really what we're trying to find here. This is really the Y maximum, that's really our target variable here. Alright, so then when we expand out our terms, what you're gonna see here is that you're going to have one half of M two V two B squared. Right? What happened? B squared? That's kinetic energy equals gravitational potential energy. Mg. Y. So it's M two G times Y max. So, this is my target variable here, which 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 with this V two B is that's the velocity of this block. Right after the collision with the bullets. 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 V to be, 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 1/2 V to be One half of something squared divided by 9.8 is going to equal your y maximum. Right? So all I have to do is just figure out what's going to go inside of this equation and to do that when we get stuck in one of our intervals. And these problems were always going to go to the other interval here because remember V2B is also in my momentum conservation. So that's why I have to use both of them. All right, so let's go ahead and plug in some values and then just go ahead and solve. So M1 is gonna be the mass of the bullet. That 0.006 times V one A. That's the initial speed of the bullet. Remember the initial speed of the bullet is going to be 800 m per second. That's what it's fired at. So, it's 800 Now its mass to master is the 1. what's the initial speed of the block? That's V two a. Well, it's actually just nothing because again, at point at point a when the bullet is still being fired upwards before the collision, there is no initial velocity for this block, it's equal to zero. So that term goes away. Alright, so then we have is we have 0.6 and then V one B. 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 gonna call this V one B and it's emerging moving upwards at 1 50 m per second. So that's V one B over here, it's 1 50. Alright, so that's for the bullets. Actually let me go ahead and write that in blue. So this is going to be V1B Which is 150, So you plugged in and it's gonna be 150 and this is gonna be mass to which is going to be the 1.2 times V to be. Remember I came over here because I needed what this V two B is. If you look to your variables, this is the only one that's missing. So I'm just gonna go ahead and simplify some stuff, This is going to be 4.8. When you work this out, this is gonna be 0.9 plus 1.2. V two B. Alright, so pretty complicated equations. But they work out and they basically simplified to just a couple of numbers. Alright, so then when you work this out and you move the 0.9 over to the other side, what you're gonna get is 3.9, and then when you divide by the 1.2, that's going to be V two B and when you work this out, this is going to be 3.25 m per second. Okay, 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 three points 25. Now, when you work this out, this is gonna be the final answer, because this is your Y max, and this is gonna be a final maximum height of 0. m. Alright, 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 m per second when it hits something like a block, which is, you know, 20 times heavier. Um It gives a little bit of speed and it rises a little bit of distance, but nothing too crazy. All right, so that makes sense. Um let me know if you have any questions

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