NTA Abhyas JEE Main2020PhysicsCenter of Mass, Momentum and CollisionPractice
A small smooth disc of mass m and radius r moving with an initial velocity v along the positive x -axis collided with a big disc of mass 2 m and radius 2 r which was initially at rest with its centre at origin as shown in the figure. If the coefficient of restitution is 0 , then the velocity of the larger disc after the collision is
Options
- A8 27 v  i ^ - 2 2 27 v  j ^
- B8 27 v  i ^ - 2 27 v  j ^
- Cv 3 i ^
- D2 2 27 v  i ^ - 8 27 v  j ^
Correct answer
A. 8 27 v  i ^ - 2 2 27 v  j ^
Step-by-step solution
The larger disc will move along line of impact As e = 0 , so velocity of larger disc v ′ = mv cos  θ m + 2 m = v cos  θ 3 Velocity of larger disc = v ' cos  θ   i ^ - v ' sin  θ   j ^ = v 3 cos 2 θ   i ^ - v 3 sin θ   cos  θ   j ^ ( from the figure we can calculate cos θ = 2 2 3 , sin θ = 1 3 ) = 8 27 v  i ^ - 2 2 27 v  j ^