Question

I understand how to solve the problem, I am confused on one aspect. Why isn't the final momentum expressed as pf=-Mv_g+mv_s? Why isn't the final momentum of the gun negative since the gun and the shell are going in opposite directions?

Consider a gun of mass M (when unloaded) that fires a shell of mass m with muzzle speed v. (That is. the shell's speed relative to the gun is v.) Assuming that the gun is completely free to recoil (no external forces on gun or shell), use conservation of momentum to show that the shell's speed relative to the ground is v/(1 + m/M).

Answer 1

(a) To apply conservation of momentum lets choose a inertial reference frame through out

the process: before and after trigger. The frame on the ground would do it.

(b) Before the trigger the total momentum P~ = M ~vg + m~vs = 0 since ~vg = ~vs = 0 from the

ground.

(c) After trigger, the shell and the gun moves with ~vs and ~vg, respectively in the ground

R.F. But the conservation of momentum forces

P~ = 0 = M ~vg + m~vs, therefore, M ~vg = −m~vs (Eq.(1)).

(d) Therefore, the velocity of the shell from the gun is ~v = ~vs − ~vg (Eq.(2)). Solving Eq.(1)

and (2) will give you

vs = Mv/(M + m)

= v/(1 + m/M)