Two identical spheres P and I lie on a smooth horizontal circular groove at opposite ends of diameter. Pis projected along the groove and at the end of time t, impinges on Q. If e is coefficient of restitution, the second impact will occur after a shortest time of

Two identical spheres P and I lie on a smooth horizontal circular groo

1.	Two identical spheres P and I lie on a smooth horizontal circular groove at opposite ends of diameter. Pis projected along the groove and at the end of time t, impinges on Q. If e is coefficient of restitution, the second impact will occur after a shortest time of
Answer» `(2 pi t)/(e )` `(pi t)/(e)` `(t)/(e)` `(2t)/(e)` SOLUTION :LET r be the radius of the above . As P and Q lie at opposite ends of diameter, therefore, distance covered = `1/2 (2 pi r) =pi r` `:.` Speed of P before impact , `u = (pir)/t` Speed of Q before impact = 0 Let v and v. be velocities of P and Q respectively after impact. `:. v. - v = e(u = 0) = eu` Thus velocity of Q relative to P after frist impact = eu Let the next impact occur after time t. s. This would happen when Q completes one round more than P in time t.. Distance travelled in one round = `2 pi r` `:. t. = ("Relative distance")/("Relative velocity") = (2 pi r)/(v. - v) = (2 pi r)/(v. - v) = (2 pi r)/(eu) = (2(UT))/(eu) = (2t)/e` .