Rocket motion - conservation of momentum

AI Thread Summary
The discussion centers on the conservation of momentum in rocket motion, specifically how to calculate the rocket's speed as fuel is ejected. The initial approach uses the equation mv = (m+dm)(v+dv) + (-dm)(v-u) to derive the relationship between the rocket's speed and mass. A misunderstanding arises when redefining dm as positive, leading to an incorrect conclusion about the rocket's speed increase. The resolution involves recognizing that dm must be negative to reflect the diminishing mass of the rocket. Ultimately, clarity is achieved by distinguishing between different mass variables in the equations.
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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!
 
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