Kinetic energy of a variable mass particle If a particle's mass is a continuous differentiable function of time, $m(t)$,
and its position is also a continuous differentiable function of time, $x(t)$, what is the expression of its kinetic energy? Does $\frac{1}{2}mv^2$ still hold?
 A: Yes; if you integrate the correct form of Newton's 2nd law (see for example http://en.wikipedia.org/wiki/Variable-mass_system) you find the final kinetic energy is $\frac{1}{2}mv^{2}$.
Just for fun I worked out what happens if you naively use the normal form of the 2nd law:
$$\vec{F}_{net\,ext}= \frac{d\vec{p}}{dt}= \frac{d}{dt}[m(t)\vec{p}(t)]=m(t) \frac{d\vec{v}(t)}{dt}+ \vec{v}(t)\frac{dm(t)}{dt}$$
A particle cannot undergo internal displacement and hence cannot have an associated potential energy.  Thus the work done on the particle by external forces goes into kinetic energy: $$\Delta K = m(t)\int\frac{d\vec{v}}{dt}\cdot d\vec{s}+ \vec{v}(t)\cdot \int\frac{dm}{dt} d\vec{s}$$
By changing the variable of integration the first integral becomes $$m(t)\int\vec{v}\cdot d\vec{v}  = \frac{1}{2}[m(t_{f})v^{2}(t_{f})-m(t_{i})v^{2}(t_{i})]$$ Similarly the second integral becomes $$\vec{v}(t)\cdot \int\vec{v}(t)\,dm=m(t_{f})v^{2}(t_{f})-m(t_{i})v^{2}(t_{i})$$
Thus the kinetic energy of the particle should be $\frac{3}{2}m\,v^{2}$.  However the second integral doesn't contribute to the kinetic energy of the particle, but rather to the extra mass the particle lost or gained during the time interval, which is exactly what the added term in Wikipedia's definition accounts for.
A: The equation for kinetic energy still holds even if both mass and velocity are functions of time. Same for something like momentum. Energy and momentum at a specific time don't care about a particle's history. Think about if you were told an object was 8kg and falling at 10m/s. This is an instantaneous snapshot of the object, and you should have no trouble telling me how much kinetic energy it has. The object's history doesn't enter into the calculations.
Mass as a function of time only gets difficult if you're calculating something like force, which is $F= \frac{d}{dt}(mv)$.
