NTA Abhyas JEE Main2020PhysicsCenter of Mass, Momentum and CollisionPractice
Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m s - 1 to the heavier block in the direction of the lighter block. The velocity of the centre of mass in m s - 1 is 2 n , then n =
Correct answer
5
Step-by-step solution
v CM = m 1 v 1 + m 2 v 2 m 1 + m 2 10 × 14 + 4 × 0 10 + 4 = 10 × 14 14 = 10 m s - 1 = 2 n ⇒ n = 5