How Fast Will the Rocket Travel After Using All Its Fuel?

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Asad Raza
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Homework Statement


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?

Homework Equations



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

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 modeled 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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haruspex said:
The large change can be analysed as the sum of a large number of small changes. That is the whole basis of integral calculus.
Fine. And why is there an increase in mass of rocket?
 
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Asad Raza said:
Fine. And why is there an increase in mass of rocket?
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).
 
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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MartinCarr said:
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).
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.
 
MartinCarr said:
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).
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.