The upper ball (A) maintains contact with the vertical wall as long as the normal reaction force $N_A > 0$. The moment the ball leaves the wall, the normal reaction vanishes: $$ N_A = 0 $$

Consider the horizontal forces on the upper ball. Since the ball is constrained to the vertical axis (until it leaves), its horizontal displacement is zero ($x=0$), and thus its horizontal acceleration is zero *just* before leaving.

Let $C$ be the compressive force in the rod (pushing the balls apart). The horizontal component of this force pushes the upper ball against the wall. $$ \Sigma F_x = N_A – C \sin \theta = 0 $$ $$ N_A = C \sin \theta $$ For $N_A$ to vanish (become zero) while $\theta \neq 0$, the force in the rod $C$ must vanish. $$ C = 0 $$ Therefore, the Tensile/Compressive force in the rod vanishes. This matches option (a).

Now consider the horizontal forces on the lower ball (B). The floor is frictionless. The only horizontal force acting on ball B is the horizontal component of the rod’s force $C$. $$ F_{net, x} = C \sin \theta = m a_B $$ Since we established that $C = 0$ at the moment of separation: $$ m a_B = 0 \implies a_B = 0 $$ Therefore, the Acceleration of the lower ball vanishes. This matches option (b).