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

KUSH PRANAV
21 Points
7 years ago
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)
ABC
11 Points
4 years ago
its wrong cause the initial KE is distributed not only as the KE of the the bodies but also as PE of spring extension...