Gang:
I know I'm going to get killed for this answer, but here
goes. I think I know what the writer of the question was
thinking about. If you have a rocket with a mass, M, and a
batch of gas, m, which you are about to throw out the back
of the rocket at a time when the rocket is moving with a
velocity V along some reference axis, the conservation of
momentum for the throwing-out of the gas can be written as:
(M + m)V = MU + mu.
where U is the final velocity of the rocket and u is the
final velocity of the gas. The algebra here is simple and
the solution for the final velocity of the rocket is:
U = V - (m/M)(V - u).
It is clearly true that the final velocity of the rocket is
the same as the initial velocity of the rocket if V = u.
This is the answer to the above question, but it is not
really very useful. What is usually fixed with respect to
a rocket engine is the velocity of the exhaust gas with
respect to the rocket. What I am saying here is that as
long as the rocket can maintain a relative velocity of the
exhaust with respect to the rocket, (u - V), the velocity
of the rocket will continue to increase as mass is thrown
out into the exhaust. WBN
Barlow Newbolt
Department of Physics and Engineering
Washington and Lee University
Lexington, VA 24450
Young man if I could remember the names of all of
these particles I would have become a botanist
Enrico Fermi
Telephone and Phone Mail: 540-463-8881
Fax: 540-463-8884
e-mail: NewboltW@madison.acad.wlu.edu