## $$75.0 \mathrm{kg}$$

#### Topics

Moment, Impulse, and Collisions

### Discussion

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GK

Gwapong K.

November 28, 2020

##### Christina K.

Rutgers, The State University of New Jersey

LB
##### Farnaz M.

Simon Fraser University

##### Aspen F.

University of Sheffield

Lectures

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### Video Transcript

in this question, a 6000 kg rocket is traveling in. The space for gravity is negligible. We don't know what is the velocity of that rocket. All we know is that at some point it started toe object feel with a velocity off 2000 m per second. And by doing that for one second, that rocket managed to achieve an acceleration off 25 m per second squared. Then, knowing that what was the mass off feel that was ejected during that Second? To serve this question, we have to use the principal off momentum conservation. That principle tells us that the net momentum before dejection is equals to the net momentum. After the ejection before the injection on Lee, the rocket was moving and it was moving with a velocity V. So the momentum was the mass off the rocket times its velocity V after the injection, the velocity of the rocket is the prime. So the momentum off the rocket is the man of the rocket times v prime. But we now have the mask off field MGI that is moving in the other direction with velocity V g. This is why we have a minus sign here because the gas moves in an opposite direction with respect the rocket. Now what we do with this equation is the following. We can solve this equation for Delta v the change in the velocity of the rocket. And we do that as follows. We begin by sending this term to the left hand side and this term to the right hand side. By doing that, we get M g V G is equals to m R V prime minus M r times V. Then on the right hand side, we can factor m r to get the following M r times v prime minus v Notice that the Prime Minister is just the change in the velocity of the rocket Delta v The reform we get the following daughter V is equals to m g times v. G divided by M r, which is the mass off the rocket. Okay, now that we know this equation, we can plug in this Delta V into this equation for the acceleration. So the acceleration is given by Delta V divided by Delta t. We know that the acceleration is 25 m per second squared. We know that Delta V is equals to m g times v g divided by M r. And we also know that Delta T is equals to one second. So we have one here, and Delta V, divided by one, is just delta V. So we have this equation now plugging in the values that we know we got the following the mass off gas is unknown. The velocity off the gas is 2000 m per second on. Then we divided by the mass off the rocket, which is 6000 kg. Now we cut the zeros and we can also simplify dividing by two both here and here. Then what you get is the following the mass off gas that was ejected is equals to 25 times three. All I had done was sending these three, which was dividing the mass of gas on the other side. So now it's multiplying. And then we got the final answer, which is that the mass of gas that was ejected is 75 kg on. This is the answer to this question.

Brazilian Center for Research in Physics

#### Topics

Moment, Impulse, and Collisions

##### Christina K.

Rutgers, The State University of New Jersey

LB
##### Farnaz M.

Simon Fraser University

##### Aspen F.

University of Sheffield

Lectures

Join Bootcamp