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1. Jul 13, 2017

1. The problem statement, all variables and given/known data
A rocket starts from rest in deep space. The rocket structure and systems weigh a total of 500kg and it has 500kg of fuel initially. When the engines are fired, fuel is ejected from the end of the rocket at a speed of 100m/s relative to the rocket.

How fast will the rocket be going when all the fuel is used up?

2. Relevant equations

F=DP/D
P1=P2 (Intial momentum equals final in absence of any external forces)

3. The attempt at a solution
The solution of this problem shows a mass being added up into the rocket and the mass being ejected. Then later on, the solution takes the mass at end m+dm and applies integral. How come be a such a large change modelled by dm or dv, i.e a small change? Also, how is the mass being added upto the rocket? Please answer these two questions descriptively. The picture is attached below

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Last edited by a moderator: Jul 13, 2017
2. Jul 13, 2017

### haruspex

The large change can be analysed as the sum of a large number of small changes. That is the whole basis of integral calculus.

3. Jul 13, 2017

Fine. And why is there an increase in mass of rocket?

Last edited by a moderator: Jul 13, 2017
4. Jul 13, 2017

### scottdave

Actually it is not an increase. They show m + dm, but dm can be a negative number as well. It turns out that the limits of the integral start at 1000 and goes down to 500, which tells us that m does decrease (each dm is decreasing the mass). Note that it shows a -dm for the exhaust (but exhaust mass is increasing).

5. Jul 13, 2017

### MartinCarr

I think the mass should be the other way around, but them I may have missed something.. The question is what the rocket velocity will be after 500Kg of fuel has been used,not what the exhaust gas will be (which I would assume zero).

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• ###### Screenshot 2017-07-13 19.52.18.png
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6. Jul 13, 2017

### scottdave

7. Jul 13, 2017

### haruspex

I do not understand your objection to the original solution. It appears to be the same as yours (and gets the same answer).
The last bit of fuel will have a speed 100-69.3=30.7m/s in the rest frame.

8. Jul 13, 2017

### scottdave

I thought this was resolved. Just so it's clear on the mass, the 500 kg rocket plus 500 kg fuel = 1000 kg starting mass. The ending mass (after 500 kg fuel is used) is 500 kg. That is why the integral goes from 1000 kg to 500 kg.
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