NTA Abhyas JEE Main2020PhysicsWork, Power and EnergyPractice
A block of mass M = 2 k g with a semicircular track of radius R = 1.1 m rests on a horizontal frictionless surface. A uniform cylinder of radius r = 10 c m and mass m = 1.0 k g is released from rest from the top point A . The cylinder slips on the semicircular frictionless track. The speed of the block when the cylinder reaches the bottom of the track at B is g = 10 m s - 2
Options
- A10 3 m s - 1
- B4 3    m   s - 1
- C5 2    m   s - 1
- D10   m   s - 1
Correct answer
A. 10 3 m s - 1
Step-by-step solution
Let the speed of the block be v . Then from conservation of momentum, the velocity of the cylinder will be 2 v in the opposite direction m = M 2 Now from conservation of energy m g h = 1 2 M v 2 + 1 2 m 2 v 2 , h = R - r = 1 m so v = 10 3 m s - 1