NTA Abhyas JEE Main2020PhysicsRotational MotionPractice
Two uniform discs A and B of equal radii but having different masses 1   kg and 2   kg respectively, are connected (at centres) by a massless spring of spring constant k = 1000   N   m - 1 and then placed on a sufficiently rough horizontal surface. If initially, the system is released from rest with the spring being compressed by 10   cm , then the speed of A when the spring comes back to its
Options
- Av A = 40 9   m   s - 1
- Bv A = 20 9   m   s - 1
- Cv A = 20 3   m   s - 1
- Dv A = 40 3   m   s - 1
Correct answer
A. v A = 40 9   m   s - 1
Step-by-step solution
k x - f = M a f R = 1 2 M R 2 a R a = 2 k x 3 M f = k x 3 So the friction forces acting on A and B are always equal in magnitude and opposite in direction. Hence, the net force acting on the system is always zero. By energy and momentum conservation. (When spring in its natural length) m A v A = m B v B v B = v A 2 3 4 m A v A 2 + 3 4 m B v 2 B = 1 2 k x 2 3 4 m A v A 2 + 3 4 m B v A 2 4 = 1 2 k x 2 9 v A 2 8 = 1 2 × 1000 × 0 . 1 2 v A = 40 9   m   s - 1