Two particles each of mass m are connected by a light inextensible string and a particle of mass M is attached to the midpoint of the string. The system is at rest on a smooth horizontal table with string just taut and in a straight line. The particle M is given a velocity v along the table perpendicular to the string. Prove that when the two particles are about to collide :
(i) The velocity of M is `(Mv)/((M + 2m))`
(ii) The speed of each of the other particles is `((sqrt2M (M + m))/((M + 2m)))v`.

Explanation

(i) `COLM rArr (m + m) xx 0 + Mv = (M + 2m)v_(1) v_(1) = ((M)/(M + 2m))v`
(ii) `COME rArr (1)/(2) MV^(2) = [(1)/(2) m(v_(1)^(2) + v_(2)^(2))] xx 2 + (1)/(2) mv_(1)^(2)`
Net velocity `v_(0) = sqrt(v_(1)^(0) + v_(2)^(2)) = v (sqrt(2M(M + m))/((M + 2m)))`