## 1998, 8 Marks

A wedge of mass m and triangular cross-section ( AB = BC = CA =  2R )is moving with a constant velocity (-v$$\vec{i}$$) towards a sphere of radius R fixed on a smooth horizontal table as shown in the figure.
The wedge makes an elastic collision with the fixed sphere and returns along the same path without any rotation. Neglect all friction and suppose that the wedge remains in contact with the sphere for a very short time ∆t during which the sphere exerts a constant force F on the wedge.
(a) Find the force F and also the normal force N exerted by the table on the wedge during the time ∆t.
(b) Let h denote the perpendicular distance between the centre of mass of the wedge and the line of action of F. Find the magnitude of the torque due to the normal force N about the centre of the wedge during the interval ∆t.
Solution : This is simply not justified to tell that since the collision between the fixed sphere and the wedge is elastic so the wedge will return with the same speed in opposite direction. The laws of physics suggest that the velocity of its CM is 1200 degree backwards with initial velocity, so it will have a tendency to rotate after collision. It will not rotate while returning is only possible when it makes a perfectly inelastic collision with the surface beneath it and hence loose some velocity i.e. the kinetic energy. Though its collision with the fixed sphere is elastic and it retains kinetic energy but to avoid a possible rotation it loses some of its kinetic energy to the surface beneath it with which its collision is perfectly inelastic. All the solutions every where is wrong. I discussed this with Dr H C Verma and he agrees.