NTA Abhyas JEE Main2020PhysicsRotational MotionPractice
A uniform rod of mass m and length L is hinged at one end and free to rotate in the horizontal plane. All the surface are smooth. A particle of same mass m collides with the rod perpendicular to the length of rod with a speed V 0 . The coefficient of restitution for the collision is e = 1 2 . If hinge reaction during the collision is zero then the value of x is:
Options
- ANo such value of x is possible
- Bx = L 2
- Cx = 2 L 3
- Dx = L
Correct answer
C. x = 2 L 3
Step-by-step solution
Angular momentum conservation about the hinge m v 0 x = m v 1 x + m L 2 3 ω v 0 x = v 1 x + ω L 2 3 ...(i) e = ω x - v 1 v 0 = 1 2 ...(ii) Impulse of hinge J = m ω L 2 + m v 1 - m v 0   J = m ω L 2 - m ω L 2 3 x   J = 0 x = 2 L 3 It is independent of 'e'