Use sum of product of vectors T and V

$Let~ the~ distance~ of~ m ~~from~ O~ be ~~ S_m~ and ~from ~the~ pulley~ be~S_M.\\By ~ Pythagoras~ S_m^2+d^2 = S_M^2.~Differentiating~S_m*v = S_M*V,\\where~~ v ~is~ the~ velocity~of~~m~and~V~the~rate ~ of~shortening~ of ~ S_M.\\ But~this~V~is~also~the~velocity~of~~M.\\ \therefore at~final~position~v^f=V^f*\sqrt{ \dfrac{S_m^f}{S_M^f} }=V^f*\sqrt{\dfrac{d/2}{(d/2)^2+d^2}}.\\ \therefore v^f =\sqrt5 *V^f........(1)$
$Only~M~losses~PE~falling~through~H~while~KE~gain~by~both~M~and~m.\\H=shortening~of~ S_M=\sqrt{d^2 + d^2} - \sqrt{(d/2)^2 + d^2} .\\\therefore H=d*(\sqrt2 -\dfrac{\sqrt5}{2}).. ...............(2)\\KE~gain=PE~loss~\implies~\dfrac{1}{2}MV^2 + \dfrac{1}{2}mv^2 =M*g*H\\From~(1)~and~(2),\\\dfrac{1}{2}M(V^f)^2+\dfrac{1}{2}m*(V^f*\sqrt5)^2=M*g*d*(\sqrt2 -\dfrac{\sqrt5}{2})\\ \implies~(V^f)^2=\dfrac{M*g*d(2\sqrt2 -\sqrt5)}{M+5m}..\therefore~a+b=2+5=\boxed{ \Large 7 }$

Let the distance from m to point O be x, and the displacement of M from its original height be y.

By energy conservation, $Mgy = \frac{1}{2}m(x')^2+\frac{1}{2}Mv^2$ . By Pythagoras, distance between m and pulley is $\sqrt{x^2+d^2}$ , so $y = d\sqrt{2} - \sqrt{x^2+d^2}$ .

Also, $v = y' = \frac{d}{dt}( d\sqrt{2} - \sqrt{x^2+d^2}) = (x^2+d^2)^{-1/2} x x'$ , so $x' = \frac{v\sqrt{x^2+d^2}}{x}$ .

Substituting $x'$ and $y$ into the energy equation, we obtain $v^2 = \frac{2Mg(d\sqrt{2} - \sqrt{x^2+d^2})}{M+(1+(\frac{d}{x})^2)m}$ .

Since $x = \frac{d}{2}$ , we have $v^2 = \frac{Mgd(2\sqrt{2}-\sqrt{5})}{M+5m}$ and the answer is $\boxed{7}$ .

I used the concept that work done by tension forces in a two connected body motion is zero overall. Also using constraint relation that ,velocity of two particles in the direction of string are equal at all instants. Using these two equation i could solve for velocity of block without using calculus.