Saved Bookmarks
| 1. |
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. |
|
Answer» Solution :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))` |
|