1. The problem statement, all variables and given/known data:There is a wedge of mass kept on smooth surface. M and block of mass m is sliding in the depression of the wedge.And there is wall in left of the wedge.At which instant velocity of the wedge will be maximum?(surface of contact between wedge and block is smooth) 2. Relevant equations But the horizontal component of normal force is balanced by the wall reaction and vertical force is balanced by normal reaction on wedge .So no motion of wedge.So conservation of energy as no dissipative force is there. blocks potential energy is converted into it's kinetic energy completely. 3. The attempt at a solution: But when it reaches at this point After this point the When the block reaches at height h and the block height is achieved at this point,Vertical force is still balanced but horizontal force is not balanced by any force as wall is not on right hand side. So block accelerate.Now the block kinetic energy at mid point lets say 1/2 mv^2 of the depression is not completely converted to it's potential energy.As some goes in kinetic energy of the wedge. My teacher says velocity of the wedge will be maximum at the below instant I am not getting why?As can be seen from the image no net force is acting on thee wedge at this instant.Because the only force acting is in vertical direction which is balanced by normal force.I think velocity of the wedge should be maximum at As in this the only force acting is in horizontal so maximum force ,maximum acceleration hence maximum velocity.