I apologize to the list for prolonging this thread and I beg its
indulgence for one last post (from me anyway.) Out of deference to Brad
(and, I'm sure, many others) I will refer to projectiles. I'd also be
grateful if anyone else might be interested in actually performing the
exceedingly simple calculations that are necessary to put this baby to
bed.
Modified problem statement: 10 kg of a 50 kg object initially at rest on a
frictionless surface is to be ejected "to the rear" at a relative speed of
5 m/s. This can be accomplished via 1 ejection of a 10 kg projectile, 2
ejections of 5 kg projectiles, 5 ejections of 2 kg projectiles, and so on.
There are (apparently) two possible interpretations of the projectile
velocity specification. The one I like is that the projectile's speed
relative to the *remaining* mass is 5 m/s ("preference 1"). Frankly, I
thought everybody agreed on this. But I think I understand A. R. Marlow
to prefer it to mean that the projectile speed relative to the
*pre-ejection* frame of the mass should be 5 m/s ("preference 2"). It is,
in any case, trivial to accommodate either preference. For instance, if a
single 10 kg projectile is used, then under "preference 1" the
projectile's velocity is 4.00 m/s and the remaining 40 kg moves off at
1.00 m/s in the opposite direction. Under "preference 2," the projectile's
velocity is 5.00 m/s (by definition) and the remaining 40 kg moves off at
1.25 m/s in the opposite direction (giving a relative velocity of 6.25 m/s
between the two pieces.)
Here's what I get for different ejection procedures:
Ejection procedure Final velocity of 40 kg remainder
# and unit mass preference 1 preference 2
1 x 10 kg 1.000 m/s 1.250 m/s
2 x 5 kg 1.056 m/s 1.181 m/s
5 x 2 kg 1.091 m/s 1.141 m/s
10 x 1 kg 1.103 m/s 1.128 m/s
... ... ...
10000 x 0.001 kg 1.116 m/s 1.116 m/s
Notice that, as the mass of the projectiles gets smaller, *both* cases
converge nicely to the answer provided by the standard rocket problem (as
derived in many texts and with which I and my results have always fully
agreed), that is, (5 m/s)*ln(50/40) = 1.116 m/s. This certainly should
not be surprising: As the mass of the projectiles gets smaller, there is
an increasingly negligible difference between the pre- and post-ejection
velocities of the remaining mass and, therefore, between the definitions
employed under the two preferences.
If, instead, you'd like to know how many .001 kg projectiles must be
ejected to reach 1.000 m/s (the original modified question), I don't think
it is much of a leap from here to see that it is going to be about 9000
under either preference. Again, the exact answer--either way--is 9064.
As Van would say (where is Van these days?),
Peace.
John
----------------------------------------------------------------
A. John Mallinckrodt email: mallinckrodt@csupomona.edu
Professor of Physics voice: 909-869-4054
Cal Poly Pomona fax: 909-869-5090
Pomona, CA 91768 office: Building 8, Room 223
web: http://www.sci.csupomona.edu/~mallinckrodt/