A bar of mass M = 1 . 00 kg and length L = 0 . 20 m is lying on a horizontal frictionless surface. One end of the bar is pivoted at a point about which it is free to rotate. A small mass m = 0 . 10 kg is moving on the same horizontal surface with 5 . 00 m s – 1 speed on a path perpendicular to the bar. It hits the bar at a distance L 2 from the pivoted end and returns back on the same path with speed v . After

Step-by-step solution

Before: After: Applying conservation of angular momentum about point O , we get m v 0 L 2 = M L 2 3 ω - m v L 2               . . . i As the collision is elastic, e = 1 ⇒ velocity of separation(after collision) velocity of approach(before collision) = 1 ⇒ L ω 2 - - v v 0 = 1 ⇒ v = v 0 - L ω 2             . . . ii Using equation i and ii , we can write m v 0 L 2 = M L 2 3 ω - m v 0 - L