Physics momentum

Using conservation of linear momentum, let v' be the velocity of the centre of mass of the system (bodies of A and B).
mv = (m+m)v'
i.e. v' = v/2

Using conservation of angular momentum, let w' be the velocity of the two masses about the centre og mass.
(mL^2)w = (mL^2).w' + (mL^2).w'
where 2L is the length of the string
w is the initial angular velocity of mass A
hence, w = 2w'
i.e. w' = w/2 = (v/L)/2 = v/2L
but w' = v"/L
where v" is the rotational speed of each mass about the centre of mass
thus, v"/L = v/2L
i.e. v'' = v/2

When the string is turned through a right angle, the velocity of the centre of mass and the rotational speed of each mass is perpendicular to each other.
Hence, the resultant speed = square-root[v'^2 + v"^2]
= square-root[(v/2)^2 + (v/2)^2] = v/[square-root(2)]

Therefore, each mass has a speed of v/[square-root(2)]