The blocks A and B shown in the 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 g = 10 m s - 2

Step-by-step solution

If A moves down the incline by 1   m , B shall move up by 1 2   m . If the speed of B is v then the speed of A will be 2 v . From the conservation of energy: Gain in K.E. = loss in P.E. 1 2 M A 2 v 2 + 1 2 M B v 2 = M A g × 3 5 - M B g × 1 2 Solving we get, v = 1 2 g 3 = 5 6