A uniform rod of length L rests on a smooth horizontal table. A particle moving on the table perpendicularly strikes the rod at one end and stops. Determine the distance travelled by the centre of the rod by the time it turns through a right angle. (Note that if the mass of the rod is four times that of the particle, the collision is elastic.)

Step-by-step solution

Let the mass of the particle be m and its initial velocity be v . Let the mass of the rod be M . Since the particle stops after the collision, by conservation of linear momentum, the velocity of the centre of mass of the rod v_c is given by: mv = Mv_c v_c = mv M By conservation of angular momentum about the centre of the rod: mv L 2 = I Substituting the moment of inertia of the rod I = ML^2 12 : mv L 2 = ML^2 12 = 6mv ML The time t taken by the rod to turn through a right angle ( = 2 ) is: t = = /2 6mv ML = ML 12mv