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A fireworks rocket is fired vertically upward. At its maximum height of 80.0 $\mathrm{m}$ , it cxplodes and breaks into two picces, one with mass 1.40 $\mathrm{kg}$ and the other with mass 0.28 $\mathrm{kg}$ . In the explosion, 860 $\mathrm{J}$ of chemical energy is converted to kinetic energy of the two fragments. (a) What is the speed of each fragment just after the explosion? $(b)$ It is observed that the two fragments hit the ground at the same time. What is the distance between the points on the ground where they land? Assume that the ground is level and air resistance can be ignored.

(a) $14.3 \mathrm{m} / \mathrm{s}$(b) $=4.04 \mathrm{s}$

Physics 101 Mechanics

Chapter 8

Momentum, Impulse, and Collisions

Moment, Impulse, and Collisions

University of Michigan - Ann Arbor

Simon Fraser University

Hope College

University of Winnipeg

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problem. 8.97. We have a fireworks rocket that it's launched vertically. At its maximum height of 80 meters, it explodes into two pieces with the mass is shown and the the chemical energy that's released too break it apart has a total of 860 Jules, that's all converted in the kinetic energy of the two pieces. So what we want to do is find the speeds at which they're moving. And we also are told they're observed to land the same time, which means that they'd shoot off horizontally and we want to know how far apart they are when they land. So we want to apply the conservation of energy and the conservation of momentum. So you have. We're gonna call the heavier piece they and the lighter beast. So this is what we're We're told here that the total kinetic energy of the two pieces is our 860 Jules bee and Vee vee, we're going to call the velocities right after the speeds right after the explosion, not 80 sixed rules 860. So we know that also m a b a equal negative m b. He'd be so we can use this two. Eliminate the A in this equation, say, by solving it for B e a. Then we have an expression on the B B and what that looks like when we're do some simplifying and so forth. 1/2 and b b b squared times one plus the ratio of the two masses. Mike, this is equal to 860 Jules, and it was okay because this comes from putting expression here for via into this equation. So the B 71.6 meters per second Then you just sort of used this knowing Bebe and I find that B A is 14.3 meters per second for part B. Each of the pieces has its horizontal speeds that we just found and the vertical speeds initially are zero. And so they're going to accelerate down because of gravity. We'll have the positive white direction pointing down this time. So the initial part initial white component of the velocity is zero. We know that initial fight is 80. Well, now the issue height of zero, but, uh, a change in the high, his 80 meters regardless and so we want to find the time at which this happens, and then we can put that take that time and put it, multiply it by the two speeds to get the distances they travel, and then they just add those together. And that will tell us how far apart they are. So felt. Why peoples? We would have the initial speed here, that zero and so that it's just a 1/2 80 squared thing. You know, A is equal to G and then solving this, we find the tea. When the things hit the ground, it's 4.4 seconds. So X and X b are just their respective ex speeds. Times this 4.4 seconds and so ex move or a rather was 57 point eight meters be moves 200 89 meters. And so then the distance between them is just the sum of their distances. They travelled because they travelled in off visit directions. That's 300 47 meters. So we see that, uh, I saw that aim being more massive, moved more slowly and so didn't move this far of a distance, and similarly with a the opposite is true. It's lighter, and so it goes faster and for

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