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

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The discussion centers on calculating the final velocity of a rocket after it has consumed all its fuel. The rocket starts with a total mass of 1000 kg, consisting of 500 kg of structure and 500 kg of fuel, with fuel ejected at a speed of 100 m/s. Integral calculus is used to analyze the changes in mass and velocity, where the mass decreases as fuel is consumed. Clarifications are made regarding the notation of mass changes, emphasizing that the mass of the rocket decreases while the exhaust mass increases. Ultimately, the focus remains on determining the rocket's velocity after all fuel is expended.
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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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Asad Raza said:
How come be a such a large change modeled by dm or dv, i.e a small change?
The large change can be analysed as the sum of a large number of small changes. That is the whole basis of integral calculus.
 
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.
 
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