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Rocket momentum problem

  • Thread starter derrickb
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  • #1
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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:

Homework Statement


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]


Homework Equations


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


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?
 

Answers and Replies

  • #2
gneill
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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.
 
  • #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.
 
  • #4
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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.
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
 
  • #5
arildno
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