Can you cover a non-singular algebraic variety over an algebraically closed field by pairwise disjoint singular subvarieties?

Oops after a little thought $\mathbb{A}^2$ can be covered by cuspidal cubics. There is at most two values $t$ such that a point $(x, y)$ lies on $(y-t)^2=x^3$ so by axiom of choice we can make this into a partition.

However we can ask for a map from a non-singular variety to a possibly singular variety such that all fibers are singular.