After falling from rest through a height `h`, a body of mass `m` begins to raise a body of mass `M (M gt m)` connected to it through a pulley. Determnethe time it will take for the body of mass `M` to return to its original position

Options

A. `(2m)/(M_m)sqrt((2h)/g)`
B. `(2m)/(M-m)sqrt((2h)/g)`
C. `(2m)/(M-m)sqrt(h/g)`
D. `m/(M-m)sqrt((2h)/g)`

Answer

B

Explanation

`u=sqrt(2gh)` is the velocity of `m` just before collision, and `v` is the velocilty of system `M+m` just after collision, then `(M+m)v="mu"`
`implies v=(msqrt(2gh))/(M+m)`
`a=((M-m)/(M+m))g`
`t=(2v)/a=((2m)/(M-m))sqrt((2h)/g)`