there is some confusion to me in the case of "motion of block on a frictionless wedge"
Below is a simple diagram!
Let us consider a situation as above in which there is a block of mass $m$ moving with velocity $v$ in positive $x$ direction on a frictionless wedge as given above.
Now since block doesn't have any velocity in $y$ direction initially so it moves horizontally as long as it is on table But when at the instant it encounters the wedge it it gains some some velocity in upward direction and starts to move upward (i.e in positive $y$ direction) with velocity decreasing with time. it approaches some height and after that comes back down the wedge.
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What gives the block that upward velocity (i.e which force(s) changes the velocity in $y$ direction.) because of which it goes to some height and after that comes back?
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What is the velocity of Block (net velocity which is along wedge). At the Instant it just starts to slide on the wedge?
Best Answer
When the block collides with the wedge, there will be a normal impulsive force from the wedge, in a direction perpendicular to the surface of the wedge.
If you notice, this normal impulsive force has a component in the vertical direction. This is what provides the block with the vertical velocity.
As for your second question, the magnitude of the velocity will not change. This is because work has not been done by any of the forces(Normal force and gravity) during the time the block starts sliding on the wedge.
The normal force doesn't do work because its direction is perpendicular to the displacement.
Gravity won't do work because there is no sufficient displacement in the vertical direction, in the small time it takes to just start sliding on the wedge.
So the speed of the block at the base of the wedge will be the same as it was on the horizontal plane. Of course, it will decrease as the block goes up on the wedge.
Note: this is assuming the block is particle like. If it has finite dimensions, there will be rotational motion when it reaches the wedge.