= Solution
Add a uniform <electric field> satisfying $\mathbf E=-\mathbf v_0\times\mathbf B$. Then the <Lorentz force> $q(\mathbf E+\mathbf v_0\times\mathbf B)$ vanishes for the desired <velocity>. With the axes used above, $\mathbf E=E\widehat{\mathbf y}$, perpendicular to both $\mathbf v_0$ and $\mathbf B$, with
$$
\boxed{E=Bv_0=B\sqrt{2U(q/m)_0},\qquad (q/m)_0=\frac{E^2}{2UB^2}.}
$$
The crossed fields form a <velocity selector>. All particles must first have traversed the same <electric potential difference> from rest: that is what converts selection of <speed> into selection of <charge-to-mass ratio>. Equal <charge-to-mass ratios> yield equal <speeds> even if the separate masses and charges differ. For other ratios, the initial transverse <force> is $q(E-vB)$ and is nonzero.
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