Re: momentum before force

["JACK L. URETSKY (C) 1996; HEP DIV., ARGONNE NATIONAL LAB, ARGONNE, IL 60439" <JLU@hep.anl.gov>]

Hi Gene-
	You write:
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I did the calculation you suggest using the equation Fext = m dv/dt + v
dm/dt. The resulting expression in both frames is dv/dt = (Fext - v
dm/dt)/m.  However, in the frame of the road the initial value of v is
greater than zero while in the frame moving with a constant velocity equal
to the initial velocity of the wagon the initial value of v is zero. 
Thus, the presumption that Fext = m dv/dt + v dm/dt leads to a
contradiction.  (Acceleration is a Galilean invariant so the initial
acceleration should be the same in both frames.)

I also did the calculation using the rocket equation Fext - u dm/dt = m
dv/dt, where u is the velocity of the sand relative to the wagon.  If we
let the direction of motion of the wagon be the direction of increasing x
then the x component of u is zero (assuming the hole is through the bottom
of the wagon and not through either the back or front).  Solving for the
acceleration using this formula gives dv/dt = Fext/m, a result which is
the same in both reference frames. 
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	But does your result make sense?  Let's modify the problem
a bit.  We'll turn of the engine and let the little red wagon be
coasting along a level sidewalk with constant velocity V.  Then
at time t=0 the famous massless plug falls out of the bottom of
the wagon and the sand begins to drain out as before.  Does the
wagon accelerate?  If so, how much?  Do you get the same
value in both reference frames?
	How does the result change if V = 0 at t=0?
				Regards,
					Jack