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Velocity of a single-stage rocket

  1. Sep 14, 2014 #1

    B3NR4Y

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    1. The problem statement, all variables and given/known data
    A single-stage rocket is in deep space coasting at vrocket,i = 2.4 × 103 m/s. It fires its engine, which has an exhaust speed of vexhaust = 1.3 × 3 m/s.
    What is the rocket's speed vrocket,fafter it has lost one-third of its inertia by exhausting burned fuel? Assume that the fuel is ejected all at once.

    2. Relevant equations

    [itex] p = mv [/itex]

    3. The attempt at a solution

    I honestly am not sure where to begin. I know that [itex] m_{f} = \frac{2}{3} m_{i} [/itex], but what do I do with vexhaust? The chapter the problem is from is on collisions and such, but I don't really see a collision here. This reminds me of a problem we had once where a person lost in space throws a wrench some "v" to change their course, but I can't remember how the problem was set up :S help, please
     
    Last edited: Sep 14, 2014
  2. jcsd
  3. Sep 14, 2014 #2

    rcgldr

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    Instead of a collision, by assuming the fuel is ejected all at once, you have something similar to an instant explosion or an instant pulse. Using the rockets initial velocity as a frame of rereference, the fuel is sent away in one direction, and the rocket is sent in the other direction. You're given the mass and the velocity of the fuel relative to the rocket's intial velocity.
     
  4. Sep 14, 2014 #3

    B3NR4Y

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    Okay, so in terms of a conservation of momentum:

    [itex] m_{i} v = \frac{1}{3} m_{i} v_{exhaust} + \frac{2}{2} m_{i} v_{rocket, f} [/itex]

    and I use this to solve for vrocket, f, right?
     
  5. Sep 14, 2014 #4

    gneill

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    Very often these problems are easier to solve if you first translate to the center of [STRIKE]mass[/STRIKE] momentum frame, then change back to the "lab frame" at the end. The nice thing about the center of [STRIKE]mass[/STRIKE] momentum frame is that the total momentum is zero.

    In the COM frame the rocket is at rest before it ejects its lump of exhaust. The exhaust is kicked one way at its given speed while the remainder of the rocket is kicked in the opposite direction. So it's a classic conservation of momentum setup with a known velocity for one of the masses. Easy!

    To convert back to the "lab frame" just add back your initial velocity to each component's COM frame velocities.

    [Edit: Note, I edited my brain fart: I meant center of momentum frame when I said mass frame]
     
    Last edited: Sep 14, 2014
  6. Sep 14, 2014 #5

    B3NR4Y

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    Attempting the problem now that I am home, My answer was 2950 m/s for the final velocity of the rocket, which off the top of my head makes sense because the velocity increased (as the act of firing the rocket is intended to do). I started off by assuming I was correct above and [itex] m_{i} v = \frac{1}{3} m_{i} v_{e} + \frac{2}{3} m_{i} v{R,f} [/itex] and "plugging in all the values I know, I got [itex] m_{i} 2.4x10^{3} \frac{m}{s} = \frac{1}{3} 1.3x10^{3} \frac{m}{s} + \frac{2}{3} m_{i} v_{f,R} [/itex]. I noticed that the mi terms could be factored out of the right side and cancelled, and I multiplied out by 3 to get rid of the fractions. I then subtracted velocityexhaust from both sides and divided by 2 to achieve my answer.
     
  7. Sep 14, 2014 #6

    B3NR4Y

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    Oh, that does sound a lot easier. We learned about systems and choosing them last chapter, the system I "chose" included the rocket and space around it, but I suppose a system of only the rocket would make the problem easier, there is a problem similar to this one in the practice workbook, I will look at it and see if I can do it

    Thanks.
     
  8. Sep 14, 2014 #7

    gneill

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    Yeah, note I edited my message. I meant "center of momentum frame", not "center of mass frame". Although they are the one and the same if there's only one object :smile:
     
  9. Sep 14, 2014 #8

    B3NR4Y

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    Hate to bother you but my answer of 2950 m/s was wrong and the reason it gave was "Not quite. Check through your calculations; you may have made a rounding error or used the wrong number of significant figures." This doesn't make sense to me because I only used the numbers they gave me, and never rounded anything.
     
  10. Sep 14, 2014 #9

    gneill

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    I suspect the problem lies with having to incorporate the initial velocity with the velocity of the exhaust when you work in the initial frame of reference; the exhaust speed is relative to the rocket, not the "lab frame". So your momentum equation was not correct for that frame.

    Give it a try in the COM frame. The rocket is initially at rest and then 1/3 of it is kicked one way and 2/3 the other. The speed of the exhaust is then relative to that frame, not the rocket. What's the new speed of the rocket in that frame?
     
  11. Sep 14, 2014 #10

    B3NR4Y

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    I got the final velocity of the rocket (in the COM Frame) to be 650 m/s, and to convert back to the "lab frame" I just add the rocket's cruising velocity to that to achieve an answer of 3050 m/s. Is that correct?
     
  12. Sep 14, 2014 #11

    gneill

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    Looks good to me!
     
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