A gun is mounted on a stationary rail road car. The mass of the car, the gun, the shells and the operator is 50 m, where m is the mass of one shell. Two shells are fired one after the other along same horizontal line in same direction. If the muzzle velocity of the shells is 200 m/s, then find the speed of the car after second shot.

Explanation

Let `vec(u)` be the muzzle velocity of the shell and `vec(v)` be the velocity of the car after first shot.
`p_(i)=0 " "` (1) [initial momentum of the system]
`p_(f)=49 m vec(v)+m(vec(u)+vec(v)) " "` (2) [final momentum of the system]
`because p_(i)=p_(f)`
`rArr 50 m vec(v)+m vec(u)=0`
`rArr vec(v)=-(vec(u))/(50) " "` (3)
Negative sign shows that `vec(u)` and `vec(v)` are oppositely directed.
For the second shot, before second shot momentum of the system is
`p_(1).=49 m vec(v) " "` (4)
If `vec(v).` be the velocity of the car after second shot then
`p_(f). = 49 m vec(v)+m vec(u) " "` (5)
From COM we get
`49 vec(v).+vec(u)=49vec(v)`
`rArr 49 vec(v).=49 vec(v)-vec(u)=-(49)/(50)vec(u)-vec(u)` as `vec(v)=-(vec(u))/(50)`
`rArr vec(v).=-vec(u)((1)/(50)+(1)/(49))`