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Professional ApplicationThe Moon's craters are remnants of meteorite collisions. Suppose a fairly large asteroid that has a mass of $5.00 \times 10^{12} \mathrm{kg}$ (about a kilometer across) strikes the Moon at a speed of 15.0 $\mathrm{km} / \mathrm{s}$ . (a) At what speed does the Moon recoil after the perfectly inelastic collision (the mass of the Moon is $7.36 \times 10^{22} \mathrm{kg}$ )? (b) How much kinetic energy is lost in the collision? Such an event may have been observed by medieval English monks who reported observing a red glow and subsequent haze about the Moon. (c) In October $2009,$ NASA crashed a rocket into the Moon, and analyzed the plume produced by the impact. (Significant amounts of water were detected.) Answer part (a) and (b) for this real-life experiment. The mass of the rocket was 2000 $\mathrm{kg}$ and its speed upon impact was 9000 $\mathrm{km} / \mathrm{h}$ . How does the plume produced alter these results?

(a) $v ^ { \prime } = 1.02 \times 10 ^ { - 6 } \mathrm { ms } ^ { - 1 }$(b) $K E _ { \text {Inas } } = - 5.63 \times 10 ^ { 20 } \mathrm { J }$(c) $v ^ { \prime } = 6.79 \times 10 ^ { - 17 } \mathrm { ms } ^ { - 1 }$ and $K E _ { \text {low } } = - 6.25 \times 10 ^ { 9 } \mathrm { J }$

Physics 101 Mechanics

Chapter 8

Linear Momentum and Collisions

Moment, Impulse, and Collisions

University of Michigan - Ann Arbor

University of Washington

University of Winnipeg

Lectures

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So for this question, we have an asteroid colliding into the moon in a perfectly an elastic collision. And a perfectly an elastic collision is one where the two objects will stick together at the end. So let's make a diagram of our asteroid moon system before and after the collision. So before we have an asteroid with some mass m sub for massive asteroid, which is given to us as five times 10 to 12 kg, and it's moving with some initial velocity and call that V sub A, which is given to us, is 15 kilometers per second and it's colliding into the moon with some call that EMC them given to us as 7.36 times 10 to 22nd kilograms. And since the moon is stationary, we know that the velocity of the moon initially call that piece of them must equal serial meters per second and then after the collision, the two objects are going to stick together and move as one mass. So we'll have the asteroid in the moon stuck together and they'll move together with some velocity. Call that v prime. So the mass of the two objects stuck together call it and total will be equal to the mass of the asteroid plus the mass aluminum. So we're trying to find the speed at which the moon recoils after the collision. So we need to make use of the conservation of momentum principle, which basically says that the initial momentum of our system must equal the final momentum of our system, where P momentum is equal to the mass times velocity. So the initial momentum of our system is equal to the momentum of the asteroid, plus the momentum of the moon. And that's equal to the momentum of the moon and the asteroid step together. So we can say that the momentum of the asteroid is equal to the massive asteroid times The velocity of the asteroid plus the mass of the moon and the initial velocity of the moon is equal to the mass of the asteroid plus the mass of the moon times v prime. And since we know that the moon isn't moving and has a velocity of zero initially, this whole term will actually go to zero. So then we only have the mass of the asteroid times The philosophy of the asteroid initially making up our initial momentum of the system. And then if we divide that by the mass of the asteroid plus the masts moon will get RV prime, which is the speed at which the moon will recoil. So plugging in numbers will have five times 10 to the 12th and our velocity of asteroids given in kilometers 15 kilometers per second. But we actually need to get this an S I units meters per second. So recall that one kilometer is equal to 1000 m, so therefore 15 kilometers per second, equal to 15,000 m per second. And that's the value we want to use for our velocity. It will divide that by the mass of the asteroid. Plus the mass of the and that will get us are the prime which is equal to 1.2 times 10 to the negative 6 m per second. So that's part a for part B. We're looking for the amount of kinetic energy lost. So how much kinetic energy as well? So what we need to do is find a change in kinetic energy. Delta K A is equal to the final kinetic energy, minus the initial kinetic energy where the formula for kinetic energy is equal to one half mast, pencil velocity squared. So for the final, kinetic energy will have one half and recall that our asteroid and moon are moving together as one object. So we'll need to include the mass of both the asteroids and the moon times the primary found squared minus one half mass of the asteroid times the initial velocity of the so Plugging this into our calculator, we get that the change in kinetic energy is negative 5.63 times 10 to 20 jewels, and this negative sign indicates that energy has been lost. So for part to see, we have a rocket colliding into the moon so we can draw a before and after picture of our collision. So before we have a rocket with the given mass of 2000 kg and it's moving at some initial velocity called the advice of our 900 kilometers per hour, and it's climbing into the moon with a given mass of 7.36 times 10 to 27 kg and again the moon stationary. So we know that the initial velocity of the moon must be zero and this is also a perfectly inelastic collision. So the rocket and Moon will stick together at the end and move together with some final velocity v prime. We can say that the total mass is equal to the mass of the rocket, plus the mass of the moon. So again we need to use the conservation of momentum. And since the moon isn't moving initially, we know that the initial momentum is just made of the momentum from the rocket and that is equal to the momentum of the rocket and we once they've stuck together. So we have the mass of the rocket times. The initial velocity of the rocket is equal to the mass of the rocket, plus the mass of the moon times the prime and then to solve for B prime. We just need to divide over the mass of the rocket plus the mass of the moon. And now we can start plugging in numbers. So now we have 2000 kg, is the mass of the rocket, and the philosophy of the rocket is given as 9000 kilometers per hour. But we need to convert this into meters per second In order to do that, we need to multiply by 1000 because there are 1000 m in one kilometer and then divide by 3600 because there are 3600 seconds in an hour and that gets us to a philosophy of 2500 m per second, which is the velocity we need to use. And then we needed some of the masses. So 2000 plus 7.36 times 10 to the 22nd we find the dispute in which the moon recoils is 6.79 times 10 to the negative 17 m per second. We also need to find how much kinetic energy is lost in this collision. So we need to find the change in kinetic energy. Delta K E is equal to the final kinetic energy of a system minus the initial kinetic energy where again kinetic energy is equal to one half mass times velocity squared. So we'll have for the final kinetic energy one half mass of the rocket, plus massive balloon times V prime squared minus one half mass of the rocket times the initial velocity of the rocket square. So that's one half 2007.36 times 10 to the 22nd and the V primary files of 6.79 times 10 to the negative. 17 squared minus one half, 2000 times our initial velocity in meters per second. So 2500 square. And we get that. The change in kinetic energy is negative. 6.25 times 10 to the ninth jewels where again the negative indicates that energy and then lost, and that's our final answer.

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