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

[Asad Raza]

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

 

[haruspex]

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.

 

[Asad Raza]

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?

 

[scottdave]

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).

 

[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).

 

[scottdave]

It looks like you did it correctly. Take a look at this, as well. https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation

 

[haruspex]

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.

 

[scottdave]

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.