Two blocks of mass 3 kg and 6 kg respectively are placed on a smooth horizontal
surface. They are connected by a light spring. Initially the spring is unstretched
and the velocity of 2 m/s is imparted to 3 kg block as shown. Find the maximum
velocity of 6 kg block during subsequent motion.

Initially the 3 kg mass is given a KE = 1/2mV² = (0.5)(3)(2)² = 6 J
Eventually, when this system reaches "Dynamic Equilibrium" this energy is distributed between the KE of two masses which must both travel at the same speed (V) because they are connected. So at Equilibrium:
6 = KE of 3 kg mass + KE of 6 kg mass
6 = 1/2(3)V² + 1/2(6)V² = 1.5V² + 3V² = 4.5V²
V² = 6/4.5
V ≈ 1.15 m/s ANS  by KUSH PRANAV (FIITJEE)

its wrong cause the initial KE is distributed not only as the KE of the the bodies but also as PE of spring extension...