The blocks A and B shown in figure have masses M_A = 5 kg and M_B = 4 kg. The system is released from rest. The speed of B after A has travelled a distance 1 m along the incline is
 

If A moves down the incline by 1 m, B shall move up by $\frac{1}{2}\mathrm{m}$

If the speed of B is v, then the speed of A will be 2v

From conservation of energy, 

Gain in KE = loss in PE
       $\frac{1}{2}{\mathrm{m}}_{\mathrm{A}}(2\mathrm{v}{)}^2+\frac{1}{2}{\mathrm{m}}_{\mathrm{B}}{\mathrm{v}}^2={\mathrm{m}}_{\mathrm{A}}\mathrm{g}\times \frac{3}{5}-{\mathrm{m}}_{\mathrm{B}}\mathrm{g}\times \frac{1}{2}$

Solving, we get 

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