A loaded spring gun of mass M fires a bullet of mass m with a velocity v at an angle of elevation `theta`. The gun is initially at rest on a horizontal smooth surface. After firing, the centre of mass of the gun and bullet system

Options

A. `(mv_(0))/(M)`
B. `(mv_(0)cos theta)/(M)`
C. `(mv_(0)cos theta)/(M+m)`
D. `(mv_(0)cos 2 theta)/(M+m)`

Answer

C

Explanation

Muzzle velocity `=v_(m//g)=v_(0)`
Along `x-` direction `,`
`v_(m(x))=v_(g(x))=v_(0)cos theta`
By momentum conservation`:(M+m)(0)=m (v_(0)cos theta-v)-Mv`
`rArr v=(mv_(0)cos theta)/(M+m)`