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A rocket is fired in deep space, where gravity is…

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Problem 54 Hard Difficulty

A rocket is fired in deep space, where gravity is negligible. If the rocket has an initial mass of 6000 $\mathrm{kg}$ and ejects gas at a relative velocity of magnitude 2000 $\mathrm{m} / \mathrm{s}$ , how much gas must it eject in the first second to have an initial acceleration of 25.0 $\mathrm{m} / \mathrm{s}^{2}$ .

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Video Transcript

So we have a rocket which has a mass of 6,000 kilograms and shoot shoots his jet fuel. It has the speed of 2,000 meter per second. We won. We wanted to have an acceleration off 2.5 25 meters per second squared. We want to know how much mass it ejects per second to achieve this acceleration. Now we know that the rocket fuel is accelerated downwards because of the compassion on dive generates a reactionary force on the rocket upwards on these two forces will be equal on this force generates the moment, the acceleration and we know that force is mass times acceleration and for this case it will be 6,000 kilograms times the acceleration, which is 25. I know that as the rocket consumes fuel, it pushes it out and it's mass decreases. Which means this 6,000 is only It's not a good approximation if we consider the rocket after it has been running for some time, but only if you consider it in the 1st 1 second, we can neglect the amount off mass that has already been pushed out, and we can say that the rocket still has almost 6,000 kilograms mass. Now we also know that force is there is a raid of change of momentum Forces Delta P by the ducky on we can instead your speed the propulsion scenario to find a copy. Now, this jet fuel most advil us the velocity off 2,000 meters per second, which is the velocity. And let's say the rocket pushes out mass m then the changing moment and would be the final moment of minus the initial Wyndham off the gas off the jet fuel. The final moment is simply masters velocity and the initial moment of zero. Because before Sorry, this is a minus. Before the propulsion, the fuel was simply sitting inside the rocket. So really, due to the rocket, the velocity initially was zero and so happy. I'm sorry. Uh, yes, Elope is M. Which means Delta P by Delta T is M V by Delta T on M beidle Daddy, it's simply the mass released, but a second hence, this is tell the m vital daddy the amount ofthe mass released per second times. The velocity off the propulsion off the jet fuel coming down pens. We know that Emma gives you 25 times 6,000. This is equal. Do we know? We want to find the tilde and the amount of mass ejector in the 1st 1 second. So that's still I am divided by the time period is one second times velocity. Being already long is 2,000 meters per second. Now I simply rearranging the towns we see that delta am is 6,000 divided by 2,025 6 1,000 divided by two thousand three on three times stratified 75 kilograms. Hence the rocket spent 75 kilograms of fuel in the 1st 1 second. So what? You 25 meter per second? Know that this is an approximation? Because as soon as the rocket So uh which has some off its fuel out its mass falls from 6,000 on becomes something smaller. But this is an approximation. What problem

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