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Please help in this momentum problem

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  1. Jul 13, 2017 #1
    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. jcsd
  3. Jul 13, 2017 #2

    haruspex

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    The large change can be analysed as the sum of a large number of small changes. That is the whole basis of integral calculus.
     
  4. Jul 13, 2017 #3
    Fine. And why is there an increase in mass of rocket?
     
    Last edited by a moderator: Jul 13, 2017
  5. Jul 13, 2017 #4

    scottdave

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    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).
     
  6. Jul 13, 2017 #5
    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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  7. Jul 13, 2017 #6

    scottdave

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  8. Jul 13, 2017 #7

    haruspex

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    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.
     
  9. Jul 13, 2017 #8

    scottdave

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    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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