Two men of masses M and `M+m` start simultaneously from the ground and climb with uniform accelerations up from the free ends of a massless inextensible rope whick passes over a smooth pulley at a height h from the ground.
(a) Which man reaches the pulley first?
(b) If the mam who reaches first takes time t to reach the pulley, then find the distance of the second man from the pulley at this instant.

Answer

ligher man; `(m)/((M+m))[(g t^(2))/(2)+h]`

Explanation

Let `a_(1) and a_(2)` be the acceleration of the two men in upward direction and T the tension in the rope. Then,
`T-Mg=Ma_(1)implies a_(1)=(T)/(M)-g` ..(i)
and `T-(M+m)g=(M+m)a_(2)implies a_(2)=(T)/(M+m)-g` ….(ii)
`:. a_(2) lt a_(1)`
Hence, the lighter man will reach the pulley first.
(b) We can find `a_(2)=(ma_(1)-mg)/(M+m)`
The lighter man ascends a distance h in time t with acceleration `a_(1)`.
Hence, `h=(1)/(2) a_(1)t^(2)` ...(iii)
Let s be the distance travelled by the heavier man in this time t, then
`s=1/2 a_(2)t^(2)=(t^2)/(2)[(M)/(M+m)a_(1)-(mg)/(M+m)]`
`=(T^2)/(2(M+m))[M((2h)/(t^(2)))-mg]`
`=(1)/(2(M+m))[2Mh-mg t^(2)]`
the distace of he second man form the pulley is
`h-s = (1)/((M+m))[Mh+mh-Mh+(m g t^2)/(2)]`
`=(m)/((M+m))[(g t^(2))/(2)+h]`