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Rocket momentum problem

  1. Sep 18, 2013 #1
    Before I ask the question, just want to say hi to everyone and introduce myself. Name's Derrick and I'm a sophomore engineering physics major at WPI. I'm currently in an Intermediate Mechanics course. Onto the problem:

    1. The problem statement, all variables and given/known data
    A rocket in free space that starts at rest with total mass M ejects exhaust gas at a given speed u. What is the mass of the rocket(including unused fuel) when its momentum is maximum? What is the mass when its energy is maximum? [use E=.5mv^2 for energy]


    2. Relevant equations
    Mv=P
    E=.5mv2
    v=v0+uln(m0/m) (maybe?)


    3. The attempt at a solution
    Mv=0
    -mexhaustu+(M-mexhaust)v=0
    (M-mexhaust)v=mexhaustu
    v=(mexhaust/(M-mexhaust))u
    I'm stuck here. Do I take the derivative of v?
     
  2. jcsd
  3. Sep 18, 2013 #2

    gneill

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    Staff: Mentor

    Hi Derrick, Welcome to Physics Forums.

    Why not look up the Tsiolkovsky rocket equation and start there? Use the formulas for momentum and kinetic energy and a bit of calculus to maximize things as required.
     
  4. Sep 18, 2013 #3

    rude man

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    I would start with the relationship F = dp/dt where F is the force applied to the rocket and p its momentum.

    You know dp is the same (except for sign) of the rocket and the expelled mass. The change in momentum of the expelled gas is easy to express, therefore so is F.

    Then F = (d/dt)[(M + m)v] and you can get a diff. eq. giving v(m). M is the mass of the rocket alone and m is the on-board mass of the expelling matter. Remember, dm/dt ≠ 0.

    Then momentum of the rocket is (M + m)v(m) and you can maximize this w/r/t m.

    Same idea for maximizing 1/2 (M+m)v2.

    Good problem! Toughies over there at Worcester Poly, huh? :smile: Seems more a junior-level course than a sophomore. Or I'm over-complicating the solution.
     
  5. Sep 19, 2013 #4
    Thanks for the help guys. This course is a little harder than I would have expected it to be, but it is a sophomore course. There is a continuation of this course, but I don't think I'm taking it...yet
     
  6. Sep 19, 2013 #5

    arildno

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    Dearly Missed

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