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California State Polytechnic University, Pomona



Problem 104 Easy Difficulty

A 12.0 kg shell is launched at an angle of $55.0^{\circ}$ above the horizontal with an initial speed of 150 $\mathrm{m} / \mathrm{s}$ . When it is at its highest point, the shell exploded into two fragments, one three times heavier than the other. The two fragments reach the ground at the same time. Assume that air resistance can be ignored. If the heavier fragment lands back at the same point from which the shell was launched, where will the lighter fragment land how much energy was released in the explosion?


$5.33 \times 10^{5} \mathrm{J}$


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Video Transcript

{'transcript': "so once again, we have a new problem. This time we have a shell, uh, that's launched on a project toe. So this is a projectile. And and And the shell is launched at an angle of 55 degrees. So this angle right here, uh, data happens to be, uh, 55 degrees. That's what happens. And And we're also told that the mask of the shell is given us 12 kg and also the initial velocity. This initial velocity of the shell, uh, is given us 150 meters per second. So that's the information we're given. And so, you know, once the explosion happens, it reaches the highest point. So the fragment, uh, the shell Sorry, which is the highest point. And then something happens. It explodes into two fragments. Um, one of them. Okay, so you have a light of fragment and behavior fragments. Uh, they actually do reach the ground at the same time. We don't have air resistance or no air resistance. Okay, so in this problem, we don't have any resistance. The heavier fragment kind of goes back. I mean, you know, that's unusual, but it's what it is. So the heavier fragment lands back at the same point where the shell was launched. So you can imagine an explosion happening. And then I think it's kind of like some type of reverse momentum. Uh, where, uh, it just goes back. It goes back and lands back right here. So, you know, we have two fragments. The behavior one which we're going to call A. This is a So, you know, the mass of A is three times the mass of B. You know, that's that's what we're given in this problem. They're telling us that the mass of a is three times the mass of D. Uh, and the the initial the initial months of the, uh the initial mass of the, uh, shell. We could just call it m. You know, we're gonna call it M and M happens to be 12 kg, as you can see. So this one is going to come back and land right there and and the second one continues all the way up until the end like that. And this is B and B happens to have a mass, uh, M b. Now, if the total mass is Emma plus MB, then it means that the mass of hair, which is three times the mass of B plus the mass of B has to be the total mass, which is 12 kg. And and so we can compute the mass of B, which happens to be, uh, you know, this is 12 kg equals to four MB, and then we divide both sides by four. And so the mass of B happens to be to read 0.0 kg. And we can always tell what the mass of A is because it's three times the mass of be. So it's 9 kg. So those are pieces of information critical pieces of information that were given in the problem, Um, our goal or the intention. So this is the information that's given and we want to find want to find two things in this problem. Just two things. The first thing we want to find out is that since we know the heavier fragment is gonna land back right there, uh, we want to find out you know where Well, well, the light up fragment mhm land. And then the second thing we're looking at is how much energy will be released in the explosion. Okay, How much energy? Mhm is released mhm in the explosion. So that's, uh, that's something we're looking at. So, you know, it's, uh yeah, It's, uh, you know, at some point, we're going to use the law of conservation of momentum, and also, we have to use the equations of projectile motion. Okay, so on the second page on the second page, we know that this is the initial velocity of the problem. And you can always decompose this one into the x components of the I, uh, cosign theta and also decomposed that into the white components of the Y. It goes to, uh, the I, uh, Cynthia to that's the initial velocity. And then there is motion going on. So, you know, this thing is gonna go up until the maximum height. So we want to know what happens at the maximum height v X, The X is the initial time schools and data, which is 100 and 50 m per second, um, 150 m per second times sine of 55. So we get, you know, round about 100 and 22 point uh 87 meters per second, and then at the maximum height, we can use the equations of motion. Remember, V y equals two. Uh huh. So this is V y initial, and then this is V Y final. Okay, that's V y final. I have to write it. Well, so this is V Y final. So v y final equals to, uh, of the y initial. Okay. And then remember these acceleration due to gravity. So, um, minus minus, Uh, GT okay. Minus duty. Because, you know, gravity is pulling stuff downwards. Uh, the at the maximum height at the maximum height. Mm hmm. What? The final velocity at the maximum height is going to be 0 m per second because it stops doesn't continue going up. And so this is zero equals to a V Y initial minus gt saw V y initial. Um, obviously has to be, um Well, the vertical component becomes zero at the top, so V y initial has to be equal to G time. Steve, we want to solve for the time taken to reach the maximum height, and we're gonna call that, uh, t marks becomes equal to the y initial of G. Question is, why do we need the t marks. We need the t marks because it's going to help us determine this horizontal distance from this point from this point right here, up until that point. So that distance, which is kind of like half the range Um, let's call it X max. No. Yeah, X marks. I'm gonna call this x marks, So X marks equals two. Um, the time, the time or T marks times the V X So T marks is V y initial of g times V X, which is V I cosign data. You know, that's what we have in this problem. Uh, the distance, the X distance for that. And so two compute this distance from right here up until right here. That's part of the problem. As you can recall, we wanted to find where it's gonna land and also the energy expanded. So that's, uh, v y initial happens to be, uh, v I mhm Signed data times V. I cosign data all over G. So V, I, um uh of initial is 150 m per second. So this is 150 m per second. We have two of them. So is this one and that one right here, Uh, so it's going to be squared, and then sign of data. Data is 55 degrees, Uh, time score sign of data, which is also 55 degrees. And then we divide that by G, which is 9.8 meters per second. 9.8 m per second. I'll rather meters per second squared. Okay, meters per second squared. Um, we plug in the numbers. We have to be very careful. This is 1 50 squared times sign of, um, 55 times course sign off 25 and then divided by 9.8. Um, so we get a value for X marks. X marks. Is this distance right here? That's part of the problem we're looking for. So X marks becomes equivalent to 1000 and 78 0.73 m. That's roughly the distance for X marks. And then the other part of the problem is we want to get the remaining distance. Okay, We want to get the remaining distance. We're going to do that on the next page. So, you know, just like a quick sketch of the, uh uh, projectile Mhm. So you know this. It goes like that. There's an explosion that happens. Part of the fragment goes back, and then the second part of the fragment continues. So we found out, um, we found out what this distance is. We're calling it X marks. And now we want to find out the remaining distance. Um, which we're gonna call X range because it happens at the maximum range using the law of conservation, of momentum, using the law of conservation, uh, of momentum. We get to see that the initial momentum of the system which is the mosque of the shell times the X velocity of the shell it calls to the final. So m a V A in the, um Via in the X direction plus MB, we be in the X direction. Our goal is to get the value for VB because that's what this fragment is. We have two fragments. A n B. This is B. That's our target. Uh, we're told that eight tons back and goes back, so it means that it's going to have an opposite velocity. So this is mass of a times negative of V X. Like that uh, plus months of B v bx. We're simplifying the problem. So we get to see that the mass of the shell velocity of the shell plus massive a velocity of X. This one moves to this side the right, the left side and becomes a plus. As you can see, uh, equals two months of b of a velocity of being the X. So the velocity of being the X happens to be massive. The shell, its philosophy is the momentum of the shell, and then the momentum of a also divided by the mass of B. Um, uh, this becomes 12 kg, mhm times the velocity of, um, the velocity of XVI ex. Uh, we had to get that Vieques. Let's see where the X is right here. Well, this is 1 50. So or that was V. Y. That wasn't the X. This is V. Y. Got to check that. Check that out. I just want to make sure this is V Y. Okay. This one right here is V y or wait. The V X is 1 50 m per second times co sign of 50 or rather 55. And that gives us, um, that gives us 86 meters per second. You know, round about 86 m per second. That's that's the information that we get from right there. Uh, so that's the number we're gonna plug in in this part. 86 we're looking for 86. So this is 86 made us for a second and then plus M e, there's 9 kg. If you go back, you see em. A right here happens to be 9 kg. I'm gonna plug it in times the velocity of the times the velocity of X again. So it seems that we're repeating that we could have simplified that problem anyways, but that's fine. And then we want to divide that by massive B, which is just 3 kg. So this is gonna help us get the velocity of B. Um, So in the next page, the velocity of B is in the X direction happens to be a 602 m per second. We use the law of conservation of momentum. Uh, once it explodes, so it comes their explodes. One fragment goes back, the other fragment goes up until the end. So we want to find this distance X range and remember distance from a high school algebra is the space Samos Speed times time. Okay, speed. Uh, times time. We did get what the time was from before, so the time it takes to the time it takes this is 602 things. Wolf. Mm. 12.5. That speech. Yes. So the time it takes to get this fragment from right here up until the center is the same time it takes to get right there. We did get the first time. Which was this one right here. Uh, this was our initial time. Where is the initial time? This one right here. T marks. So T max is V y initial of a G. So we want to compute that. So t is V y initial of G v y. Initial, as you can see from right here is 1 22.87 are meters second. Divide that by 9.8 m per second squared, and we get the number for that as 1 to 2.87 Divided by 9.8. That's about 12.54 seconds. Since the distance is speed, times, time, X range, which is a distance, it's going to be equivalent to the speed that we got for B, which is zero. So x range mhm equals two vb x Times T, which is six or 2 m per second times 12.54 seconds. So we get approximately 7547 That's the distance. Seven 547.7375 47 0.73 So we'll call the X range is 75 47 point 73 m. So now we want to get the total x total. Remember, X total equals two x marks. X marks is this distance right here plus X range, uh, X range And I'm using range because it's the it gets to the to the maximum. I'm using x X marks because this is the marks height, and this is the range. I mean, you could choose to use anything you want, but, uh, you know, this is just the way I'm doing it. So 1078.73 1078 and 73 plus 75 47.73 Both of these are meters and so we're able to get a distance. We'll do that on the on the other page where it's but there's most peace. Um, yeah, right, Yeah, yeah. Mhm. Yeah. So you know, when we add these to the x marks and the X range, the X total becomes equivalent to so, plus 10, 78.73 So the total becomes 8626.45 So 8000. There's a lot of rounding going on. So sometimes you might not get the exact answer 0.46 m. That's the first part of the problem where they were asking, you know how where will the light of fragment land? So we've dealt with that. And then now we want to find the energy released. So the energy released. Okay, the energy released would be pretty much the change in kinetic energy of the system. So the final kinetic energy minus the initial Connecticut, the final kinetic can adjust the two fragments. Um, so you know, one half m a V A. Um, that's that's the final. The A, um X Remember of the ex? Uh, was right here. V X is right there. So we're using that and then plus one half M b v bx minus the initial is when the shell was intact together of the X one half the X squared. All these are squared because it's a kinetic energy. So, you know, we could simplify it. Let's pull out the one half from all over and then we have, uh we have m a and M B, which is, um, Emma is the big one from 9 kg. Okay, Yes. 9 kg times V a x go back. V x was given us negative v x so v a x, his negative VX. And, uh, you know, that happened to be what? The X That's the velocity in the X direction. Mhm. Mm. That becomes nine times 86 and then three times 602. Yeah. Um, so this has got to be 12. So 9 kg Times of the day X is right here. 86. Yes. So, times 86 um, meet his second squad like that, plus MBS 3 kg times. Um v bx, if you can recall, you know, this is this is minus, but, you know, the 86 is a minus, but when you squared, it's always gonna be a plus. So, you know, we don't have a problem with that. So this is 3 kg times VB X, which happens to be 602 m per second. You just second, uh, you know, that's the initial part. This one, too has to be squared, and then you subtract Oh, you know what? We don't have to get that bracket yet. And then you subtract because this one half was taken out. The total mass of the shell is 12 kg of 12.0 kg times V x V X is the 86 meters per second. And then we have to square that That's how we do the problem. Just plugging in the numbers. Um, and so our final final answer for that, the energy expended Let's do nine times 86 squared, Yeah. Plus, um, three times 602 squared gives us a number minus 12 times 86 squared. Yeah, and then divide by two. Um, this gives us a final answer, as so we do it in the next page. The energy released becomes 5332 512 jewels. Or you could say 532.512 killer jewels. That's the energy released in the process. So once again, you know you have a projector. It goes up until the maximum high it splits into two fragments. One is a the other one is B. So we want to find the distance that be travels fast parties. We have to get X marks the distance from this position up until the middle. We use the equations of motion times, times velocity. But we have to get the time using another equation of motion. Uh, this one right here. Uh, so we saw for time we plug it into x marks and get a value for x marks, which is this distance from here to there. And then in the next part, we have to get the X range, which is the remaining portion of the distance we still use the same time. But then now we have to use get the velocity of bvb ex uh, from the previous page, you could see that to get vb X, we had to use, uh, the law of conservation of momentum. The initially equals to the final. Uh, we saw for VB X. Plug it into the X Range and and get the, uh, the value for the X rays. So we have those two values. We add them up, helps us get the total distance traveled by Fragment B. And then after that, we want to get the energy released. That's just the change in kinetic energy. The initial of final minus the initial and the initial is when the shell was still intact. The final is when you have those two fragments, we simplify that problem and get the final energy released. Hope you enjoyed the problem. Feel free to send any questions or comments or, uh, ideas. And, uh, we'll be glad to respond and have a wonderful day. Okay, thanks. Bye."}