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Suppose the first stage of a two stage rocket has total mass $12,000 \mathrm{kg}$ , of which 9000 $\mathrm{kg}$ is fuel. The total mass of the second stage is $1000 \mathrm{kg},$ of which 700 $\mathrm{kg}$ is fuel. Assume that the relative speed $\boldsymbol{U}_{\mathbf{e x}}$ of ejected material is constant, and ignore any effect of gravity. (The effect of gravity is small during the firing period if the rate of fuel consumption is large) (a) Suppose the entire fuel supply carried by the two-stage rocket is utilized in a single-stage rocket with the same total mass of $13,000 \mathrm{kg}$ . In terms of $v_{e x}$ what is the speed of the rocket, starting from rest, when its fuel is exhausted? (b) For the two-stage rocket, what is the speed when the fuel of the first stage is exhausted if the first stage carries the second stage with it to this point? This speed then becomes the initial speed of the second stage. At this point, the second stage separates from the first stage.(c) What is the final speed of the second stage? (d) What value of $v_{\mathrm{ex}}$ is required to give the second stage of the rocket a speed of 7.00 $\mathrm{km} / \mathrm{s}$ ?

(a) $v=1.37 v_{e x}$

(b) $v_{1}=1.18 v_{\mathrm{ex}}$

(c) $v_{2}=2.38 \mathrm{vex}$

(d) $v_{e x}=2.94 \mathrm{km} / \mathrm{s}$

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{'transcript': "We have a test rocket that weighs 22,540 kg. It has a velocity that's given by this expression here, when it starts, um at t equals zero. When the engines fire, we have an acceleration of 1.5 m per second squared. And after one second we have a velocity of two m/s. Now, um these two conditions here allow us to find these coefficients A and B. So velocity at one second tells us that A plus B has to be um uh, two. I should probably not put units on here because we're um Yeah, no, that's all right. So let's see here we have then that equals and then acceleration. We could take a time drivers over that and get this. So, the acceleration at zero is just A. And that is um That is a Yeah. Um 0.5 m per second squared. And then that tells us that B or 1.5 and then that tells us that he has to be 0.5. All right. So let's see here. Um Yes. Now we have yeah. And this should be a second square. That's what I was. All right. So now they tell us that at four seconds. Um Well, they want us, they want what is uh let's see what So we found those coefficients at four seconds after ignition. What is the acceleration of the rocket? Well, we can just now that we know A and B. We can take the time to reboot of that. Here's the acceleration here. Plug in A and B. And plug in teeth was four and we get 5.5 m per second squared. And now we need to figure out what thrust does the burning feelings on on it, assuming no air resistance. Um, Yeah, so the thrust is the is gonna doing a free by diagram of here and again. I should put acceleration in here. So we have some acceleration there. So we have doing a little after a little louder probably get that the force equals mass times acceleration of the rocket plus the gravitational acceleration. Yeah. Now I am obviously we could just plug this into here. Um I assume that they meant that the force at four seconds and so that tells us that we have 3.89 times 10 to the 4th Newton's and then they asked us to to give it as a function of or in terms of the weight of the rocket, so we can just divide by the way to the rocket here. So the Um that is 1.56 times the weight of the rocket. So obviously because it's greater than the weight of the rocket, it's going to be accelerating upwards and that's, in fact what, you know, it is doing because we have a positive acceleration here. Did they ask us for the acceleration? Um, the force at, at ignition. So we know that that is a acceleration is just coefficient capital A. Here, which is 1.5 m per second square. So we can plug in capital A. Here at G two. It multiplied by the mass. And we get that, that is 2.87 times 10 to the fourth Newton's."}

University of Michigan - Ann Arbor