Two smooth spheres A and B, of equal radius but masses m and M, are free to move on a horizontal table. A is projected with speed u towards B which is at rest. On impact, the line joining their centres is inclined at an angle theta to the velocity of A before impact. If e is the coefficient of restitution between the spheres, find the speed with which B begins to move. If A.s path after impact is perpendicular to its path before impact, show that tan ^(2) theta = (eM-m)/(M +m).

Two smooth spheres A and B, of equal radius but masses m and M, are fr

1.	Two smooth spheres A and B, of equal radius but masses m and M, are free to move on a horizontal table. A is projected with speed u towards B which is at rest. On impact, the line joining their centres is inclined at an angle theta to the velocity of A before impact. If e is the coefficient of restitution between the spheres, find the speed with which B begins to move. If A.s path after impact is perpendicular to its path before impact, show that tan ^(2) theta = (eM-m)/(M +m).
Answer» SOLUTION :When B is struck by the impulse J, it BEGINS to MOVE in the direction of J as shown in the . Along the line of centres, we apply (a) Conservation of linerar momentum , i.e., `m u cos theta MV - mv …(1)` (b) Law of resistution, i.e., `eu cos theta = V +v ...(2)` Solving EQS. (1) and (2) we get `v = ((eM - m) u cos theta )/( (M +m))and V = ((1 + e ) m u cos theta)/( (M + m ))` Hence, `tan phi= ( u sin theta)/( v ) = ((M + m ) tan theta)/( (e M - n ))` Butthe paths of A before and after IMPACT are at rigth angles, therefore,` cot phi = tan theta.` Hence, `((eM - m ))/( (M + m ) tan theta ) = tan theta` `therefore tan ^(2) theta = (eM - m )/( (M + m ))`