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Homework Help: Rocket Problem-Need Help

  1. Nov 2, 2009 #1
    1. The problem statement, all variables and given/known data

    A small rocket of mass 1ton is moving horizontally with a velocity of 600 m/s in the x direction.The total force on the rocket is zero.10 kg of jet gas is ejected at a velocity (in the x direction) of -6000 m/s with respect to the ground.The ejection takes place over a time interval of 1 second.

    a) What is the velocity of the rocket after the jet gas is ejected?
    b) What is the force (vector) that the jet exerts on the rocket during the ejection?

    2. Relevant equations

    P=m.v and the conservation of momentum

    3. The attempt at a solution

    i dont know what to do because of the given statement "The ejection takes place over a time interval of 1 second". So i would be glad if you can clarify this mathematically and help me to add it into the equation about the conservation of momentum.(think its something to do with derivative)?
  2. jcsd
  3. Nov 2, 2009 #2


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    Conservation of momentum applies always, and in this case you only need cosider the initial & final states

    To get the time invloved, look at impulse & how its related to momoentum.

    Force is related to the derivtive of velocity through Newtons 2nd, though i don't think its required here
  4. Nov 2, 2009 #3
    thanks for the info but i still have questions about the problem;

    Impulse means the change of momentum in a given time which is basically expressed in the form I=m.ΔV or Force times t.So if i try to find the initial momentum of the rocket it is gonna be: 600.1000=(Momentum for the final situation) so i know the initial momentum but i dont know how to add the impulse to the equation to calculate the final situation that i wrote in paranthesis.
  5. Nov 2, 2009 #4


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    no worries,
    if you know the intial momentum of the whole system, by conservation of momentum, the final will be equal. You can work out the momentum of the gas, which should lead you to the final momentum of the rocket

    then relate the momentum change to the force in b) through the impulse realtion

    note that impulse is defined as F.t and is equal to the change in momentum (or the integral of F.dt if you're using calculus)
  6. Nov 2, 2009 #5
    So for the answer of a)

    i calculated the initial value of momentum which is 600.1000 and that equals to the final value which is 990.V-6000.10.So therefore 600.1000=990.v-6000.10


    which is the new speed of my rocket =) am i correct?
  7. Nov 3, 2009 #6


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    close, if momentum was a scalar qunatity, but momentum has a direction attached, so you must honour the negative in the expelled gas

    think of it this way, based on your numbers, the rockets cruising along (+ve x dir'n) shoots out heaps of gas behind (-ve x dir'n) and slows down (in the -+e x dir'n)?

    you'd probably expect it would speed up hey?
  8. Nov 3, 2009 #7
    lol yea i should have noticed that :/, it should speed up the rocket.... i forgot to put a minus there;

    so with the same calculations but with a minus infront of the momentum of the jet gas my new velocity for the rocket is 666.66 m/s.

    and for the answer of b) i used the equation of:

    F=m.a (m=the mass of the gas)

    so since the ejection takes place over a time interval of 1 second;

    acceleration for the gas= -6000m/s-0 /1s so acceleration is equal to -6000 m/s2

    therefore the force vector on the gas is 60.000 N

    Correct me if im wrong :) thanks for the help so far...
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