A sphere of mass `m` , moving with velocity `V`, enters a hanging bag of sand and stop. If the mass of the bag is `M` and it is raised by height `h`, then the velocity of the sphere will be

Options

A. `(M + m)/(m) sqrt(2 gh)`
B. `(M)/(m) sqrt(2 gh)`
C. `(m)/(M +m) sqrt(2 gh)`
D. `(m)/(M) sqrt(2 gh)`

Answer

A

Explanation

By the conservation of linear momentum
Initial momentum of sphere = Final momentum system
`m V = (m + M) v_(sys)`
If the system rises up to height `h` then by conservation of energy
`(1)/(2) (m + M) v_(sys)^(2) = (m + M) gh`
` implies v_(sys) = sqrt(2 gh)`
Subtituting this value in equition (i)
`V = ((m + M)/(m)) sqrt(2 gh)`