INT A 1500 kg weather rocket accelerates upward at 10 m/s² . It explodes 2.0 s after liftoff and breaks into two fragments, one twice as massive as the other. Photos reveal that the lighter fragment traveled straight up and reached a maximum height of 530 m. What were the speed and direction of the heavier fragment just after the explosion?

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Hey, everyone. So this problem is working with conservation of momentum. Let's see what it's asking us during a military test. A 1200 kg missile is launched with an acceleration of m per second squared in a vertically positive direction. However, the test fails and the missile blasts three seconds after it's launched into two pieces before reaching its final destination. One of the broken pieces was thrice as large as the other or three times as large as the other. The heavier piece moved vertically up to a height of m determine the velocity of the lighter piece. Our multiple choice answers here are a negative 164.2 m per second. B 144.0 m per second. C negative 348.7 m per second or D 269.9 m per second. OK. So while this is a conservation of momentum problem, we're also going to need to recall some of our basic kinematics equations as well. There are a few pieces to this. So, so it's a little bit tricky but stick together, stick with me, we will get through it and figure out how to find this velocity of the lighter piece. So the first part of the scenario is the initial conditions at the blast. So we have this rocket that at first is kind of going up in the air and then it breaks apart right into multiple pieces. So in the scenario, all right, during the condition of the blast itself, we have momentum conserved. But during that three seconds from when we blast off to when we break up, that is going to be modeled using our kinematics equations. So we can recall that V F is equal to V I plus A T and D or distance is equal to V I plus V F divided by two multiplied by T. So our initial velocity is zero m per second. Our acceleration was given to us and the problem as 12 m per second and we know the time from blast to breaking apart is three seconds. So we can plug that in to find our final velocity or the velocity when the, the final velocity for this part of the problem is actually going to be the initial velocity when the rocket breaks apart. OK. So initial velocity zero m per second plus our acceleration, 12 m per second squared multiplied by our time three seconds. And that is 36 m per second. And then we also want to find the height that the rocket rises before it breaks off of the ground. And so that's what we're going to find with D. So D equals again, initial velocity, zero m per second, final velocity we just saw for 36 m per second, all divided by two and then multiplied by time of three seconds. And that is 54 m. So that tells us at the time a blast when the rocket breaks apart our initial conditions, we have a speed of the rocket of 36 m per second at a height of 54 m above the ground. Now, we're going to move into the second part of the problem which is where our conservation of momentum will come in. We can recall that the conservation of momentum says that the initial momentum is sequel to the final final p final momentum where our momentum is given as mass multiplied by velocity. So our initial condition is right at the time of the blast and our final condition for this kind of part two of the overarching um arc that this broken apart piece that these broken apart pieces take. This final momentum is right at the right after the blast. So before we start experiencing the external forces of of gravity, um as those pieces fall to the ground. So we are looking at this middle part just around the blast as as its own separate system here. So initially, we have the whole missile rocket, whatever we wanna call it here. I guess I actually called it a rocket here, but the problem calls it a missile. So we will call it a missile. So we have our initial momentum for the missile is going to be the mass of the missile times, the initial velocity of the missile is equal to the final, which is again, those two parts right after they break apart. So that'll be the mass of the heavy peace times the velocity of the heavy piece kind of after the blast plus the mass, the light piece times the velocity of light piece after the blast. And so if you'll recall what we're looking for in the problem, the the final answer is this here, the velocity of the light piece after the blast. So we go through our variables here, the mass of the missile was given to us and the problem is 1200 kg. We know the initial velocity of the missile we just solved for 36 m per second. Now, as far as the mass of the heavy and the light pieces, we know that together they equal 1200 kg. And we also know that the heavy piece weighs three times as much as the light piece. So we can write that as three times the light piece or multiplied by the light piece, sorry, three multiply by the light piece plus the mass of the light piece is equal to 1200 kg. So the mass of the light piece is 300 kilograms, which means the mass of the heavy piece is 900 kg. OK. So what we don't have? So we're solving for the velocity of the light piece, but we don't have the velocity of the heavy piece. We do know that that heavy piece reached a height of 1430 m. So we can go back to our kinematics equations to solve for this velocity of the heavy piece. And then we will be able to solve for the velocity of the light piece, which is what we're ultimately after. OK. So this is kind of scenario part three of the scenario where we're looking at just the heavy piece after it has split away from the, the rocket and is raising rising to that final height of 14 1430 m. So back to the kinematics equations, we'll use V F squared equals V I squared plus two ad and so our final velocity is going to be zero m per second. That's at that peak. When it finally reaches that meters, it will have a velocity of zero m per second before it starts um falling back to the ground. Our initial velocity is that philosophy we're solving for the H B our acceleration. We're just in a brief um free rise. I guess there's no external forces working on it. It's gravity constant negative 9.8 m per second squared. And our distance, it would be easy to think that our distance is exactly what was given to us in the problem for 1430 m. But we are actually looking for the distance after the blast. So it's important to subtract the distance at which the blast occurred, which we calculated to be 54 m. So the distance in this problem is 1376 m. OK? So we rearrange to solve for V I. So we'll have V I squared equals negative. So V F squared is zero. So we'll subtract the two multiply by acceleration and distance to the other side. So V I squared equals ne negative two multiplied by negative 9.8, sorry 9.8 m per second squared multiplied by 1376 m. We plug that in and take the square root and we get 164. m per second. OK. Now, we actually have everything we need to go back to this equation here and solve for our velocity of the light piece after the blast. So we'll plug in our known values mass of the missile 1200 kg multiplied by that velocity of 36 m per second is equal to the mass of the heavy piece. 900 kg multiplied by the velocity of the heavy piece which we just saw for 164. m per second plus mask, the light piece 300 kg multiplied by the velocity of the light piece at the blast. When we plug all of this into our calculator, we get velocity of the light piece after the blast of 348 sorry of negative 348.7 m per second. So that is the answer to this problem when we look at our multiple choice answers that aligns with answer choice C. And so that negative sign is telling you that after the missile breaks apart, that heavy piece is going vertically in a vertically positive direction or going upwards. And then the lighter piece is because it's negative, it's moving in a vertically negative or downward direction. So that's all we have for this one. We'll see you in the next video.

Verified Solution

Key Concepts

Newton's Second Law of Motion

Conservation of Momentum

Kinematics of Projectile Motion