If A moves down the incline by 1 metre, B shall move up by $\frac{1}{2}$ metre. If the speed of B is v then the speed of A is 2v.

From conservation of energy:

Gain in K.E. = loss in P.E.

$\frac{1}{2}{\mathrm{m}}_{\mathrm{A}}{\left(2\mathrm{v}\right)}^2+\frac{1}{2}{\mathrm{m}}_{\mathrm{B}}{\mathrm{v}}^2$

$={\mathrm{m}}_{\mathrm{A}}\mathrm{g}\times \frac{3}{5}-{\mathrm{m}}_{\mathrm{B}}\mathrm{g}\times \frac{1}{2}$

Solving we get

$\mathrm{v}=\frac{1}{2}\sqrt{\frac{\mathrm{g}}{3}}$