if a mass of ball is m/2 and that of a rod is m the length of the rod is 20 cm the ball is set on the same level as the lower end of the rod and then released. If pulleys and strings are light, friction negligible how soon will the ball be opposite top the upper end of the rod?

Explanation

Ball:

`2T-(mg)/(2)=(m)/(2)a` …..(i)
Rod:

`mg-T=mxx2a` …..(ii)
`(i)+2(ii)implies(3mg)/(2)=(9ma)/(2)`
`impliesa=(g)/(3)`
Acceleration of ball to reach the upper end of the rod
`L=(1)/(2)a't^(2)=(1)/(2)g t^(2)`
`t=sqrt((2L)/(g))=sqrt((2xx0.2)/(10))=0.2s`