🎉 Announcing Numerade's $26M Series A, led by IDG Capital!Read how Numerade will revolutionize STEM Learning Numerade Educator ### Problem 64 Easy Difficulty # In a volcanic eruption, a 2400 -kg boulder is thrown vertically upward into the air. At its highest point, it suddenly explodes (due to trapped gases) into two fragments, one being three times the mass of the other. The lighter fragment starts out with only horizontal velocity and lands 274$\mathrm{m}$directly north of the point of the explosion. Where will the other fragment land? Neglect any air resistance. ### Answer ##$-91.3 \mathrm{m}\$

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Moment, Impulse, and Collisions

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{'transcript': "Okay, So here we are, given a question about a volcanic eruption in which a 24 100 kilogram boulder is sent vertically upward and then explodes into two fragments. So in this question, the boulder it is 2400 kilograms, and then at its highest point, it explodes into two fragments. Now the question gives us that one fragment is three times the mass of the other fragments. So using some pretty simple laugh here, we can conclude that we have an 1800 kilogram fragment and a 600 kilogram Friday. Now, the next important piece of information that we're given is that the lighter fragments starts out with Onley, horizontal velocity and lands 274 meters directly north of the point of explosion. So but we can conclude from that is that the 1800 kilogram fragment is gonna land directly south from point of explosion because duty, Newton's third law, every action and every force has an equal and opposite force. So that means that since the force exerted on the 600 kilogram fragment is forcing it to the north than the 1800 kilogram fragment is going to be forced to the south and that sensitive. It was a horizontal for some 600 Newton fragment that it is also 600 kilogram fragment that is also a Onley horizontal force exerted on the 1800 kilogram fragment. Now from there, we want to see not just the direction that it goes in. But actually how far from the point of explosion does this 1800 kilogram fragment land so do to the conservation of momentum? We know that the mo mentum before the explosion should be equal to the momentum after the explosion. Since they're actually no outside forces acting on the boulder at the specific instant in terms of horizontal because gravity is working on it vertically, there is no horizontal external forces on the boulder. So what that means is that the momentum before the explosion horizontally is gonna be equal to the momentum after the explosion horizontally. No, it says in the question that before the explosion it was the boulder was only moving vertically, meaning that the horizontal velocity would have been zero. So since Mo mentum is equal, the mass times velocity. This means that there was actually zero horizontal momentum before the explosion and from there we can conclude that there was zero horizontal momentum after the explosion between the two fragments combined. So what that means is that the momentum of the 600 Newton fragment is equal in offices to the mo mentum of the 1800 kilogram fragment. So we know that the bolder the the fragments started out with only horizontal velocity after the explosion. So because of that, we can actually conclude that they took the same amount of time to fall to the ground because everything accelerates due to gravity at the same rate of 9.81 meters per second squared. So if we're neglecting air resistance than actually, both objects will take the exact same amount of time to fall to the ground. No, if they take the same amount of time to fall to the ground. And we know that the lighter fragment travel 274 meters, then from there due to the conservation of momentum since the 600 kilogram fragment is 1/3 the mass of the 1800 kilogram fragment, then we know due to the conservation of momentum, that is gonna have three times the velocity of the 1800 kilogram fragment. So now we can say that the 1800 kilogram fragment had 1/3 the velocity of the 600 kilogram fragment and fell for the exact same amount of time. Now distance is equal two velocity times times. So, since the velocity is 1/3 of that of the lighter fragment and the time is the same in the distance is gonna be 1/3 of that of the lighter fragment, which therefore would make it 1/3 of 274 which would make it 91.3 meters south of the point of explosion just"}

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Moment, Impulse, and Collisions

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