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A 20.0 -kg projectile is fired at an angle of $60.0^{\circ}$ above the horizontal with a speed of 80.0 $\mathrm{m} / \mathrm{s}$ . At the highest point of its trajectory, the projectile explodes into two fragments with equal mass, one of which falls vertically with zero initial speed. You can ignore air resistance. (a) How far from the point of firing does the other fragment strike if the terrain is level? (b) How much energy is released during the explosion?

(a) $849 \mathrm{m}$ from the firing point.(b) $1.60 \times 10^{4} \mathrm{J}$

Physics 101 Mechanics

Chapter 8

Momentum, Impulse, and Collisions

Moment, Impulse, and Collisions

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so once again, welcome to new problem. This time we have a projector. So there's a surface right here. And given that there exists our projectile, uh, that's going to be fired a certain distance and ah, you know, it ends up going that way, and at the maximum height, for some reason, the projectile explodes and part of it falls down on DH. Then the remainder goes all the way up until the end. So, you know, this is this is the scenario we have for this projector. The initial mass off the projectile before it explodes. Um happens to be twenty kilograms. Remember? They're two fragments, the fast one who were going to call this one. And then the second one was going to call it B. And so are the obviously, you know, when it goes on and explodes, you're gonna have half off half off and half off B because they split. So this is the information that we've given another piece of information that's critical in this problem is that the angle off launch off the projector happens to be sixty degrees aunt. Also, we have on initial velocity off off the velocity of the projectile at the beginning happens to be, um, eighty million meters per second. So that's the velocity. This's the information will given They're two parts to The problem put is requesting that we, um the way. Find out this distance from the origin up until where the projectile stops, the second piece stops or peace. Be, uh, assuming this is the marks in on point right here. So half off this will be are over too. And then the's also of path taken by the scent of muss. And so, you know, it's almost like this projectile is being launched again. So this distance right here will be another are over too. And then this one also will be another are over too. So there's three of them. The last of the initial velocity off the projectile that drops zero meters per second. Oh, so so imagine. You know you have an initial launch of eighty minutes per second. It gets the maximum height Ah, in the middle off the projector and explodes into two fragments. So from meant, a drops with initial velocity, zero meters per second and then from mint be continues all the way up until the end. So you concede that from the beginning Up until the end, If you sum up these threes or are over two plus are over two plus are over too. You get three are over to remember our is thie distance taken from the initial position up until where the center ofthe moss off the projectile ends. So if you split that into the middle, get r ver to an Arab too, and then the other projector pushes all the way. So part, eh? Okay, put a We want to find the distance. Three are three over to R. That's what we want to find on then in. But we want to find the energy released during the process. So this is kind of like saying, you know, like, what's the change in Connecticut energy of this process? And this is usually the final kinetic minus the initial kinetic energy one day when the fragments explode. So, you know, this is this is what we have. Uh, this is the information that we've given, and so are ver. Tu is the projectile is the distance. They have half the distance, the range on DH. Then we're keeping the centre ofthe Mass. You know, from this point, up until ah, the end point, The scent of mosses. Your are okay. And then we split that into and to get the distance traveled by the second projectile, which is be who you want to do three over to our has to equal free over too times the formula for the range, which is the square. Sign off to data, um, over the acceleration due to gravity. So this one becomes three of the squared. Sign off to beta over to G with plug in the numbers we get, we can pull up three over two. That's fine. And then the V squared his eighty meters per second squared that then the sign off too. Times sixty, which is the initial angle. Sixty degrees. And then we divide that by two our times nine point eight a media spurssecond square which is acceleration due to gravity. So there the distance we get through your three over to our becomes equivalent to round about eight hundred eighty four meters. That's but ill the problem which is finding this distance from the initial position up until the position where the second fragment falls. Um, this second part of the problem were required to get the change in kinetic energy. Or rather, what's the energy expended in the process? Remember, changing kinetic energy becomes the initial or rather, the final Connecticut and you minus the initial kinetic energy, the initial clerical. It is a little bit straightforward because it's just one half the mass times the velocity squared. But then when it comes to the ah, final kinetic energy just to go back a little bit, remember, if you want to get the the the velocity, you're thinking about the horizontal velocity. Okay, so you know, this is this is the projectile has two components. This velocity is V and then you have an angle data. So you almost thinking about VX? The ex is the call Sign off data. So that's what you're looking for in terms of kinetic energy. And and so this initial Connecticut and just simply becomes one half the mass before it splits. Times ve co signed data and we square that. And so this is one half of him v squared co sign squared data. That's the initial kinetic energy. The final kinetic energy. The final kinetic energy happens once the frogman split. So they're two parts. They're two parts to this whole problem. There's one that goes down. You can see the initial velocity zero meters per second. So that means the kinetic energy for this one, it's going to be zero. But then the other one proceeds, and we want to find out what the kinetic energy of that is. So the second fragment, the fast fragment has a muscled him over to the second one also has a massive movement, too. So this zero plus one half the mass of the second one times the horizontal velocity off the second one, which is which is gonna be, um, which is going to be, you know, we have to We have to be very careful with the algebra one doing those problems. So the, um, the projector, you know, house too has to go up until up until the end, so it's going to be twice the previous velocity. So this one is becomes twice times V of X, look twice times V of X. So this one remember V of X is the horizontal velocity. So this is twice times of the co sign off data. That's what it is. And then we still have to square that, um the simplification of this problem, remember, Two times two becomes one fourth, and then we have m and we few square that you get for v squared. Co sign squared off data thes two. We're gonna cancel out. So we have a clean value for the what? The the final final Lost E for this problem. Final of no final kinetic energy, if you call it So m b squared sine squared data. You know, that's that's what we have in this case on DH. Then we're going to go to the next page where we have to get the change in Connecticut Energy, which is the final kinetic energy minus the initial kinetic energy. The final we found out was M V squared cosign scored data and B squared co science. What they learned then the initial had one half on it right here. So one half and the squared and this word co sign squared off data. You can see this part that part exactly the same. You can either one there if you want to do that. And so that one minus half is just one half. So the velocity looking for his M v squared co sign squared data like that. And so if you plug in the numbers, you get one half the mast. Initial muss, if you can recall, was twenty kilograms. So you come back and plug that in twenty point zero kilograms and then the velocity was eighty meters of seconds per seconds are not for second squared per second squared that then you have co sign squared off the either, which is sixty degrees. You plug in those numbers the kinetic energy expended or the change of kinetic energy becomes one point six zero times ten to the fourth. Jules, hope you enjoyed the problem. Feel free to send any questions or comments and have a wonderful day. Okay, thanks. Bye.

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