Two smooth spheres A and B are at opposite sides of a diameter of a circular smooth tube of radius R. The tube is placed on horizontal floor. At t=0 sphere A moves with constant angular velocity w and B with 2w in the clockwise direction.During collission the velocities are exchanged. a) Find the time taken for first second and third collission. b) Find the distance travelled by the sphere A before second collission

Dear ajit ,

you can apply the concept of relative velocity in case of angular motion also. so here the angular velocityof b w.rt a is w in clockwise direction.

therefore

a) for first collision - time taken is the time when the body b moves a full circle w.r.t to a if we are applying relative concepts ....... so  w_relative   = v_relative/R ; w_relative   =  R/Rt₁ ;    t₁  =1 / w_relative

therefore  t₁ =  1/w

for third collision : consider the scenario when the first hit takes , the velocities are exchanged i.e the angular velocities are also exchanged . so now A has 2w and B has w as angular velocity. so now in the same time A will hit B from behind. So , taking symmetry third collision shud take 3t₁   as time. therefore  t₃ = 3t_1.  i.e  3/w

b)before first collision the angular velocity of A was w . so distance travelled  =  velocity * t₁

   =   wR * 1/w =    R

  between first and second collision -- distance travelled  =  2WR*1/w =  2R

 so total distance travelled  =  3R

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