NTA Abhyas JEE Main2020PhysicsCenter of Mass, Momentum and CollisionPractice
Two blocks A and B of masses 3 k g and 6 k g are connected by a massless spring of force constant 1800 N m - 1 and then they are placed on a smooth horizontal surface. The blocks are pulled apart to stretch the spring by 5 cm and then released. What is the the relative velocity (in m s - 1 ) of the blocks when the spring comes to its natural length?
Correct answer
1.5
Step-by-step solution
Let μ is reduced mass of the two blocks, then μ = m 1 m 2 m 1 + m 2 = 3 × 6 3 + 6 = 18 9 = 2 kg Let v r be the relative velocity of the two blocks, then from conservation of mechanical energy we get 1 2 μ v r 2 = 1 2 k x 2 v r = x k μ = 5 × 10 - 2 1800 2 ⇒ v r = 1 . 5 m s - 1