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The expelling of fuel from a rocket (Momentum)

  1. Oct 20, 2008 #1
    1. The problem statement, all variables and given/known data
    A 4.00kg model rocket is launched, expelling 63.0g of burned fuel from its exhaust at a speed of 565m/s. What is the velocity of the rocket after the fuel has burned. Hint: Ignore the external forces of gravity and air resistance.

    2. Relevant equations
    Conservation of Momentum?

    3. The attempt at a solution
    I tried using the equation for the conservation of momentum by doing M1Vi=M-0.063Vf... but that did not work. I am stuck.
  2. jcsd
  3. Oct 20, 2008 #2


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    Hi tyro008! :smile:

    I think you've assumed that 565 m/s is the fuel's actual speed … but the question probably means the speed relative to the rocket.
  4. Oct 20, 2008 #3
    Do you mean that this is the rocket's speed?
  5. Oct 20, 2008 #4


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    No … the rocket starts at speed 0, and so the fuel is at speed 565.

    But later the rocket is at speed v, say, and the fuel is at speed (565 - v). :smile:
  6. Oct 20, 2008 #5
    so i did
    (rocketmass X 0m/s) + (fuelmass X 565 m/s) = (rocketmass X v) + (fuelmass X (565 - v)
    would this be right??
  7. Oct 20, 2008 #6
    (this would be to find v)
  8. Oct 20, 2008 #7
    answer - 8.89m/s...Is this right?
  9. Oct 20, 2008 #8
    hey yeah it is!! thanks so much to all you guys for helping me, i REALLY appreciate it!! :]
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