A gun of mass `M`. fires a shell of mass `m` horizontally and the energy of explosion is such as would be sufficient to project the shell vertically to a height `'h'` . The recoil velocity of the gun is

Options

A. `((2m^(2)gh)/(M(m+M)))^(1/2)`
B. `((2m^(2)gh)/(M(m-M)))^(1/2)`
C. `((2m^(2)gh)/(2M(m-M)))^(1/2)`
D. `((2m^(2)gh)/(2M(m+M)))^(1/2)`

Answer

A

Explanation

If `E` is the energy of explosion, then since it is just sufficient to carry the bullet ot a height `h` so,

We have `E=mgh`. Now if `v` an `V` are the respective vellcities of the bullet and the gun, respectively then from the law of conservation of linear momentum
`mv+M(-V)=0impliesv=(MV)/m`............i
Also, `E=1/2mv^(2)+1/2MV^(2)`
`implies mgh=1/2m((MV)/m)^(2)+1/2MV^(2)` [from eqn i]
`implies 2mgh=(M^(2)V^(2))/m+MV^(2)`
`:. V=[(2m^(2)gh)/(M(M+m))]^(1//2)`