Why do we subtract the velocity of the gun barrel to find the velocity of bullet in classic gun barrel problems? The very subtle conceptual problem I'm having is that assuming the gun barrel is at rest...and then it fires the bullet by means of an mini explosion of gunpowder let's say then it is delivering the force of the explosion to both the gun barrel and bullet simultaneously. Thus the velocity obtained by the bullet should be completely given by the explosion of gunpowder. Then why do we subtract the velocity vector of barrel in order to find bullets velocity w.r.t ground?
 A: The clearest way to tackle this question is to remember that the momentum is always zero.
If the gun is floating in mid-air, then the total momentum of "gun plus bullet" is zero before the gun is fired and zero after the gun is fired.
So if you know the forward velocity of the bullet, you can multiply by the mass of the bullet to get the forward momentum of the bullet. This is the same as the backward momentum of the gun. Then you divide by the mass of the gun to get the backward velocity of the gun.
You do not subtract the backward velocity of the gun from the forward velocity of the bullet. 

So what happens if the gun is heavier, or tied down firmly?
If you work out the velocities for gun and bullet, for different masses of gun - and if the gun is tied to a rock, the "mass" of the gun is the weight of the gun plus the mass of the rock - you find that the total kinetic energy after the explosion is $$\frac12(1+\frac m M)mv^2$$ where $m$ is the mass of the bullet, $M$ is the mass of the gun, and $v$ is the velocity of the bullet. (The velocity of the gun is deduced from these three numbers: it is not a separate variable).
The heavier the gun is, or the more solidly the gun is tied to a large heavy object (a rock, or even the whole Earth), the smaller $(1+\frac m M)$ will be, so that, for a fixed energy of explosion, $v$ will be larger.
The higher bullet velocity from a gun that is firmly fixed does not come from an extra "impulse" from the thing the gun is fixed to. It comes from a higher effective "mass" of the gun-plus-fixture.
A: Because in accordance with Newton's 3rd Law of motion, the barrel recoils in the opposite direction to the shell or bullet. The faster the barrel recoils, the slower the bullet which comes out of the muzzle. The speed of the recoiling barrel varies tremendously, but it will be much slower than the bullet because it has much greater mass. There are various forms of recoil absorber which slow down the barrel's movement even further. I was once in charge of a 90mm anti-tank gun which was so firmly dug in that the barrel was unable to recoil at all, in which case it would have been unnecessary to subtract the velocity of the barrel from he velocity of the shell. High velocity is particularly desirable in the case of an anti-tank gun. And then there are recoilless rifles, which as the name suggests, don't recoil at all, in which case I suppose you'd have to subtract the velocity of the backblast gases which are ejected at very high speed in the opposite direction to the shell. Newton's 3rd Law has to be obeyed.
