A block A of mass m is in equilibrium after being suspended from...

Let the velocity of the combined system just after the collision is& $v '$, then using conservation of linear momentum we get

$mv = 2mv′$,

$⇒ v′ = \dfrac{v}{2}$

Initial elongation of the spring is

$x = \dfrac{mg}{k}$

Now, using the conservation of mechanical energy we get

$\dfrac{1}{2}2m\left(v '\right)2 + \dfrac{1}{2}kx^{2} = 2mgx$

$m\left(\dfrac{v^{2}}{4}\right) + \dfrac{m^{2}g^{2}}{2k} = \dfrac{2m^{2}g^{2}}{k}$

$v = \sqrt{\dfrac{6mg^{2}}{k}}$