Relative motion of wedge and block Through my physics tutor I came to know that suppose a block is sliding on a wedge with velocity $v$ with respect the wedge and now suppose the wedge starts to move with velocity of any amount then still with respect to wedge the velocity of block will be $v$ only?
How is this possible? I mean, will the velocity of wedge not have any effect on the velocity of block?
Please explain.
 A: 
how is this possible, I mean will the velocity of wedge not have any
  effect on the velocity of block?

The movement of the wedge will have an impact on the velocity of the box with respect to the wedge, whatever the velocity of the box is, when the wedge starts to move (accelerates). But it will not have an impact when it is already moving at any constant velocity (with respect, say, to the ground). To help understand this, consider the following more familiar example of relative motion.
Suppose you were on a train. You see a ball rolling on the floor of the train (maybe some kid started it rolling). The velocity of the ball as you observe it (which is with respect to you and the train) will be the same whether the train is sitting at the station, or moving at any constant velocity with respect to the tracks. 
However, suppose the ball is rolling along the floor while the train is stopped at the station. Then the train suddenly starts moving (accelerates) with respect to the tracks. Now you will see the ball start to slow down because of its inertia. It will no longer be moving with the same velocity with respect to you or the train. Or if the train is initially moving at constant velocity and suddenly brakes (decelerates). Now the ball will speed up with respect to the train, again because of its inertia.
It's the same with the block and wedge. The wedge can be moving at any constant velocity with respect to the ground and the velocity of the box with respect to the wedge, whatever it is, will be the same. However, when the wedge starts to move (accelerates) the velocity of the box with respect to the wedge will not be the same.

thanks a lot. your explanation really cleared all the doubts except
  one that if the block/box is moving with velocity 10m/s wrt wedge
  which is at rest . now the wedge starts to move with velocity 8m/s
  then still velocity of block will be 10m/s or due to the instantaneous
  acceleration it will change?

"starts to move with a velocity of 8 m/s" doesn't tell me what the acceleration is.  If you say the wedge is initially moving at a constant horizontal velocity, or is at rest,  and then experiences a constant acceleration such that it attains a new constant horizontal velocity of 8 m/s after 1 second, then I know the acceleration is 8 $\frac{m}{s^2}$. 
Now, assuming the wedge is accelerating in the same horizontal direction as the horizontal component of the velocity of the box, then during the period of acceleration of the wedge the box will have a horizontal component of deceleration of 8 $\frac{m}{s^2}$ with respect to the wedge. In other words, the velocity of the box with respect to the wedge will be changing.
Complicating the whole scenario is the fact that in order for the box to initially have a constant velocity down the wedge of 10 m/s the downward force on the box parallel to the surface of the wedge must be equal to the upward kinetic friction force that the wedge applies to the box. When the wedge accelerates the friction force may go below the maximum static friction force and the box may then stop.
Hope this helps.
