Rocket motion - conservation of momentum

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SUMMARY

The discussion focuses on the conservation of momentum in rocket motion, specifically analyzing the relationship between the rocket's mass and speed as fuel is ejected. The key equation derived is v2 - v1 = u ln(m1/m2), which highlights the importance of correctly defining mass changes during fuel ejection. The confusion arises from the treatment of the differential mass dm, which must be negative to reflect the diminishing mass of the rocket. Clarification is provided through the use of distinct variables for the rocket's mass and the ejected fuel mass.

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Homework Statement


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.

Homework Equations



m1v1 = m2v2

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 traveling 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?
 
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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:

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