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Rocket motion - conservation of momentum

  1. Jan 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the speed v of the rocket when the mass of the rocket = m. The rocket starts from rest at with mass M. Fuel is ejected at speed u relative to the rocket.

    2. Relevant equations

    m1v1 = m2v2

    3. The attempt at a solution
    In the textbook, it starts off with a moving rocket with mass m and speed v. The fuel is given the mass (-dm) which is positive. So after a short time dt, the mass of the rocket changes to m+dm and speed v+dv. The mass of the ejected fuel is (-dm) and since it ejects with speed u relative to the rocket travelling at speed v, the ejected fuel travels at speed v-u, which can be positive or negative depending on which of v or u is larger.

    Writing down the equation of conservation of momentum:

    mv = (m+dm)(v+dv) + (-dm)(v-u)

    which then leads to

    m dv = -u dm

    After a few steps, we get:

    v2-v1 = u ln(m1/m2)


    That all looks fine and understandable. However, it mentions in the book that I am free to define dm to be positive, and then subtract it from the rocket's mass, and have dm get shot out the back. So I have decided to try it.

    Writing down the equation of conservation of momentum:

    mv = (m-dm)(v+dv) + dm(v-u)

    which then leads to

    m dv = u dm (note that at this point, the equation is already different from before)

    Moving the variables around and integrating v from v1 to v2, m from m1 to m2 as before, I obtained:

    v2-v1 = u ln(m2/m1)

    which is clearly wrong because m2<m1, so

    ln(m2/m1) < 0

    but

    v2 > v1


    So what went wrong there?
     
  2. jcsd
  3. Jan 9, 2012 #2

    Delphi51

    User Avatar
    Homework Helper

    Well, thanks for an evening's entertainment! I could not wrap my mind around it. Clearly dm has to be negative in order for the value of m to diminish. But it is natural to think of dm as positive when we write (m-dm). Conflict! I found clarification here
    http://ed-thelen.org/rocket-eq.html
    Some explanations use one m for the rocket and another m for the propellant.
    Anyway, it seems one must use different letters for the m in dm and the M for mass of the rocket. Then dM = - dm and it all works out.

    How clever of the author of your book to make us really think!
     
    Last edited: Jan 9, 2012
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