Chain Please Don't Fall!

Let Left part of chain as system at time t=t. mass of left part is :

$m\quad =\quad (\cfrac { L }{ 2 } +\cfrac { y }{ 2 } )\lambda \quad \quad \quad \\ \\ \cfrac { dm }{ dt } \quad =\quad \cfrac { \lambda }{ 2 } \cfrac { dy }{ dt } \quad \\ \quad \quad \quad \quad \quad \\ \quad \quad \quad \quad \quad =\quad \cfrac { \lambda }{ 2 } v\quad \\ \quad \quad \quad \quad \quad =\quad \cfrac { \lambda }{ 2 } \sqrt { 2gy }$ .

By writing dynamics equation of left chain system :

${ F }_{ ext }\quad \quad -\quad { F }_{ thrust }\quad =\quad ma\quad \quad (\quad \uparrow \quad as\quad positive\quad )\\ \\ \because \quad a\quad =\quad 0\quad \\ \\ \because \quad { F }_{ thrust }\quad =\quad { U }_{ relative }\left| \cfrac { dm }{ dt } \right| \\ \quad \quad \quad \quad \quad \quad \quad =\quad \lambda gy\quad \quad \quad \quad .\quad .\quad .\quad .\quad (1)\\ \\ \quad { F }_{ ext }\quad =\quad T\quad -\quad mg\\ \quad \quad \quad \quad \quad \\ \quad \quad \quad \quad =\quad T\quad -\quad (\cfrac { L }{ 2 } +\cfrac { y }{ 2 } )\lambda g\quad .\quad .\quad .\quad .\quad (2)\\ \\ \boxed { T\quad =\quad \cfrac { \lambda g }{ 2 } (L\quad +\quad 3y\quad ) }$ .

Let V be the velocity of the whole chain center of mass and v - the point B velocity.

Then the whole chain impulse $MV$ equals the impulse of the moving part of the chain :

$M/L (L/2 - y/2)v i.e., MV=M/L(L/2 - y/2)v$ .

Differentiating both parts of this equation one gets (with the use of Newton's second law)

$Mg - T = Mg/2 - Mgy/2L - Mv^2/2L. As v^2=2gy, T = Mg/2 + 3Mg/2L. Hence a + b = 5$ .