Consider a cylinder filled partially with a liquid (e.g. water). The cylinder is sealed, and is at held at room temperature (e.g 298K). At equilibrium (or when no external disturbance is imparted to the system), the liquid in the cylinder exists in equilibrium with its vapor at the vapor pressure of the liquid, applicable at room temperature.
Suppose that the bottom of the cylinder is fitted with a piston. Note that the system is still sealed. Now imagine that the piston moves upwards. The liquid column moves upwards, the volume the vapor occupies decreases, but the vapor still exists at the vapor pressure of the liquid (assuming that thermodynamic equilibrium can be achieved fast enough).
What's interesting is when the piston move downwards. If the piston moves at a slow speed, the liquid column should stay "on top of" the piston (without any "gap" between the column and the piston). Now, if the piston moves downwards fast enough (or when the downward acceleration is high enough), the liquid column shall "leave" the piston, and a "packet" of vapor shall be formed between the piston and the liquid column. (Please refute the former claim if you think it's wrong.) Why? if the "friction" between the cylinder walls and the liquid column is negligible, and the liquid column is already in a state of free-falling, there is no mechanism to "pull" it downwards further anymore.
Now, imagine that the piston moves upwards again. Then the liquid column shall "collide" with the piston. Will it "stick" to the piston, as in an inelastic collision, or recoil, as in an elastic collision?
P.S. After thinking for a while, I think it is not at all an easy question to answer. Now whether the liquid will "recoil" (not "slosh", which implies that the liquid changes shape) depends on how momentum is transferred from the piston to the liquid column. Please refer to the youtube video for the "beer bottle trick". If the cavity formed at the bottom is a "vacuum", which is probably the case, then when the liquid "smash back", the atmospheric pressure shall press on the liquid column, and most probably it will not "recoil". On the other hand, if the cavity formed is gas-filled, then the kinetic energy from the liquid (which is trying to "smash back" on the glass bottle) may be dissipated to the gas (in the original gas-filled cavity), through the formation of many tiny cavitations (claim: the latter claim is not sound; I am just guessing). For as long as the impact is "dampened", a recoil will not happen.
Indeed there are many ways in which a liquid may dissipate energy, because it's formless/shapeless, and there's viscosity in the picture. The more "flexible" it is to energy dissipation, the less likely it is to recoil.