A smooth ball A of mass m is attached to one end of a light inextensible string, and is suspended form fixed point O. Another identical ball B, is dropped from a heigh h, so that the string just touches the surface of the sphere.

`{:(,"Column I",,"Column I",),((A),"If collision between balls is completely elastic then speed of ball A just after collision is",(P),(3m)/(5) sqrt(2gh),),((B),"If collision between balls is completely elastic then impulsive tension provided by string is",(Q),(sqrt6gh)/(5),),((C ),"If collision between balls is completely inelastic then speed of ball A just after collision is",(R ),(6m)/(5) sqrt(2gh),),((D),"If collision between balls is completely inelastic then impulsive tension provided by string is",(S),(2sqrt(6gh))/(5),),(,,(T),"None of these",):}`

Answer

(A) `rarr` (S), (B) `rarr` (R ), (C ) `rarr` (Q), (D) `rarr` (P)

Explanation

For(A) `v_(0) = sqrt(2gh), sin theta = (R )/(2R) = (1)/(2)`. By definition of `e, e = 1 = (v_(1) sin theta + v_(2))/(v_(0) cos theta)`

Let impulse given by ball B be N. then by impulse momentum theorem
`N = m(v_(2) + v_(0)costheta)` & `Nsin theta = mv_(1)`
`rArr v_(1) = (2v_(0) sin theta cos theta)/(1 + sin^(2) theta) = ((2sqrt2gh) ((1)/(2)) ((sqrt3)/(2)))/(1 + ((1)/(2))^(2)) = (2sqrt(6gh))/(5)`
For (B) Impulsive tension `= N cos theta = ((mv_(1))/(sin theta)) cos theta = mv_(1) cottheta = (6m)/(5) sqrt(2gh)`
For(C ) For completely inelastic collision `e = 0`, so `v_(1) sin theta + v_(2) = 0 rArr v_(1) = (v_(0) sin theta cos theta)/(1 + sin^(2) theta) = (sqrt6gh)/(5)`
For (D) Impulsive tension `= Ncostheta = ((mv_(1))/(sin theta)) costheta = mv_(1) cottheta = (3m)/(5)sqrt(2gh)`