What Happens First ?

Taking the two blocks and the rope as our system. Now for this system taking the forces in the horizontal direction ( or say horizontal component ) , there is the only horizontal component of force (if possible) of the reaction force of pulley on the rope in the right direction.

Case 1 : right block first hits the table if this is the case then at the time of hitting the centre of mass of our so called system has moved leftwards relative to its initial position which is impossible.

so the left block must first leave the table so that centre of mass is moving rightwards only.

Simple solution. Let the right block come down by a certain height, and let it make an angle alpha with horizontal. For the heavier block its a non uniform circular motion. $Tcos(\alpha ) =mg$ also $Tsin(\alpha)= horizontal component of acceleration$

Solving acceleration of right hand block is $mgtan(\alpha)$

For left block acceleration is T itself which is $mg/cos(\alpha)$

Compare both you will find the acceleration is greater for all left block always. Basically u can cut both the cos quantity as it will be perfectly valid for all angles<=90, and at 90 its a collision time, so this approximation can safely be done.

Other ideas include energy considerations and string constraints.

Let us consider a particular time =t;
Since no information is given about friction ; we assume friction to be zero.
Here let angle made by the rope of the falling block with the vertical be $\theta$
{ Since I am Lacking the tools , I am unable to make the diagram }
let tension in string be T

[Block 1] -----------> T

T cos $\theta$ <----- [Block 2]

As we can see both the blocks has to move equal distance relatively.
and also initial horizontal velocity is zero;

Hence the block with more acceleration will take less time

Block 1 is acted upon by more tension hence more acceleration so block 1 will take less time.

Since the force which is applied by pully to the system of two blocks has a component towards right hence center of mass will shift towards right. So before right block hit the table left block reach to end .

the right block will start moving downwards while the left one will start moving rightwards. now we can see that the same force i.e. M*g acts on both the blocks. the left block has to travel a dist of L to reach the pulley but to reach the table the right one has to travel a dist =( (L^2 + H^2) )^(1/2) .... where H can be considered the height of the initially suspended block from the point where it is supposed to strike the table! hence the left block has to travel less distance .

There's no force acting on right block other than gravity. It will go straight down until left block reaches pulley. Problem would be interesting if friction was involved.