In an ideal perfectly rigid rocket or cannon projectile structure in
vacuum, that should be so. In the real world where the structures will
be to some degree elastic, removal of the accelerating force will cause
the structure to revert to its unstressed shape, which, depending on
design details, could then impart a force to your ball-bearing or
mercury droplet.
There's also the elasticity of the ball-bearing (or even, given
surface-tension, of the mercury droplet) which could also cause a
rebound once the overall acceleration ends.
As for the specific case of a projectile within a cannon barrel, that
depends on whether it is being accelerated for its entire time within
the barrel. This depends on the quantity and burn-rate of the
propellant charge. Typically the answer is yes, accelerated the whole
time, because calculating the propellant charge to still be producing
considerable positive pressure on the base of the projectile as it
reaches the muzzle tends to make best use of a given length of barrel.
It is however possible to use a small fast-burning charge such that
pressure has dropped off enough that the projectile is decelerating by
the time it reaches the muzzle.
On 12/1/2015 8:35 AM, John Dom wrote:
Just checking assumptions, please advise.
Am I right assuming a mercury droplet or loose ball-bearing will not
move forward inside a rocket in vacuum the instant the motor stops? It
then only might start floating.
Nor will such droplet move forward inside a cannon projectile after it
leaves the muzzle since on that instant push force has stopped. As to
during the short residence time inside barrel, it might fly forward I
reckon.
jd