In the above image let us assume that block slides along the plane of the wedge and the wedge moves towards left (on a smooth surface).
I know that the velocity of the block can be found by simply using conservation of energy and conservation of momentum (in the horizontal direction)
My question is as follows:
When the block slides to the bottom most point it's velocity in the horizontal direction is the horizontal component of it's velocity minus the velocity of the wedge. This results in a decrease of Kinectic Energy (With respect to the case of a fixed wedge). I know that this energy went to the kinetic energy of the wedge but I am unable to understand how. I know that while conserving mechanical energy of closed systems of two or more blocks there is usually an action-reaction pair of forces which decreases one objects energy while increasing the others. Which forces are acting in the horizontal direction in this case?
(Edit: New Question) When writing the equations for conservation of energy at the bottom most point (in order to find the velocity of the wedge) why do we write the horizontal component of the velocity of the small block as v cos(a)-V ? v:velocity of block with respect to wedge a:Angle of inclination of the wedge V:velocity of the wedge.
I know that the block is on the wedge but can’t understand what is it about the interaction between the wedge and the block which requires us to account for the velocity of the incline while calculating the horizontal velocity of the block. An analogy of a person sitting in a car can be used to better understand my question. A person sitting in the car at any point has the velocity as well as the acceleration of the car due to the normal reaction between the person and the car. What is the ‘Normal Reaction’ in this case. (Note:The wedge is frictionless)