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A $\mathrm{C} 6-5$ model rocket engine has an impolse of 10.0 $\mathrm{N}$ . for 1.70 $\mathrm{s}$ , while buming 0.0125 $\mathrm{kg}$ of propellant. It has a maximum thrust of 13.3 $\mathrm{N}$ . The initial mass of the engine plus propellant is 0.0258 $\mathrm{kg}$ , (a) What fraction of the maximum thrust is the average thrust? (b) Calculate the relative speed of the exhaust gases, assuming it is constant. (c) Assuming that the relative speed of the exhaust gases is constant, find the final speed of the engine if it was attached to a very light frame and fired from rest in gravity-free outer space.

$$\begin{array}{l}{\text { (a) } f=44.2 \%} \\ {\text { (b) } v_{e x}=800 \mathrm{m} / \mathrm{s}} \\ {\text { (c) } v=530 \mathrm{m} / \mathrm{s}}\end{array}$$

Physics 101 Mechanics

Chapter 8

Momentum, Impulse, and Collisions

Moment, Impulse, and Collisions

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In this question, a rocket engine has an impulse off. 10 Newtown Stein, second for 1.7 seconds while burning a mass off 0.0 125 kg off proper land. The maximum trust achieve it is 13.3 Newton's and the total mass that if the mass off the engine plus the mass off propellant before burning anything is 0.0 258 kg. By knowing that we have to answer free items, the first one asks about what fraction off the maximum trust is the average trust. To answer that, we begin by calculating what is the average trust, and we can do that by using this relation. This relation tells us that the impulse is equal to the average force applied in this case, the average trust. Multiply it by the time during which that force is applied in our case, 1.7 seconds. Then, using that equation, we can say the following 10, which is the impulse is equals to the average force times the time, which is 1.7 seconds. Then the average force is given by 10, divided by 1.7 which is approximately five 0.88. The reform, The average trust is off 5.88 Newtons. Now. The fraction is given by the average trust divided by the maximum trust, which is 13.3 Newton's. By solving this calculation, we get a fraction off approximately zero point 44 to which is equivalent to 44.2% in case you are interested. But this is sufficient as an answer for the first item in the second item, we have to calculate what is the relative speed off the exhausted gas is with respect to the rocket. For that, we have to use this equation which relates the acceleration off the rocket or, in this case, the engine off that rocket with the velocity off the expelled gas, the mass off the rocket, the variation in the mass off the rocket on the time interval during which that variation of court notice that I'm talking about the mass off the rocket. But keep in mind that I'm actually talking about the mass off that engine which were treating in this question. Okay, knowing that we have, the acceleration is equal to minus that velocity divided by the mass off the rocket times the variation in the mass of the rocket divided by some interval of time. We begin by sending the mass to the other side, multiplying then the mass times. The acceleration is equal to minus the velocity off the gas is times the variation in the mass of the rocket, divided by some interval of time. Notice that the mass times the acceleration is actually the average force that is acting on the rocket. So we have that the average force is equal to minus the velocity off the gas times the variation in the mass divided by the time now sending delta T to the other side. We have the average force multiplied by some interval of time being equals two minus the velocity of the gas times the variation in the mass of the rocket Notice that this is just the impulse. Just take a look at this equation. Then we know that the impulse is equal to minus the velocity off the gas times the variation in the mass off the rocket. Then we can finally calculate what is that velocity? So the velocity off the expelled gas is equal to the impulse, which is 10 mutants second, divided by the variation in the mass off the Rocket Delta M, which is given by the mass off repellent 0.0, 125. Notice that we also have a minus sign here, but the variation in the mass off the rocket is also negative because it's losing mass. The reform. We have a minus sign and another minus sign inside the variation in the mass. Then at the end we have no minus sign. By serving this calculation, we gotta velocity off 800 m per second. And this is the answer to the second item off this question. The last item we are asked to assume that the rocket or the engine was at rest, and then it started burning field after burning that fuel. What will be the velocity off that rocket? For that we can use this equation, which tells us that the final velocity is nothing but the initial velocity, plus the velocity off the gas is that are being expelled times the logarithms off the initial mass off the system, divided by the current mass off the system, Then using this equation you get the following The final velocity is equals to the initial velocity, but the initial velocity zero Because we are assuming that the rocket is starting addressed, then the final velocity is just the velocity of the Gaza's times. The luxury item off the initial mass divided by the final mass plugging in the values that we have, we got the following The velocity off the gas is is 800 m per second and then you take the luxury item off the initial mass which was 0.0, 258 divided by the final mass. The final mass is the initial mass 0.0, 258 minus the mask off propellants that was burned which is 0.0, 125. Then we get the final velocity is 800 times the longer it um off 0.0, 258 divided by 0.0, 133. And solving this calculation results in approximately 530 m per second. And this is the answer to the final item off this question

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