Why momentum is conserved when a gun fires? (conceptual question)

[revix]

I understand that conservation of motion comes from the action and reaction pairs of newton's third law. When it is triggered, two forces appear that cancel when analyzed as a system. My question is how is it that momentum is conserved if before the shot there was no force in the system and afterwards there was. How can I analyze it conceptually through equations?

 

[phinds]

Actually, it's a bit more complicated than two forces, for example, if you shoot a handgun. There is the force pair of the explosion sending the bullet out of the gun pushing the gun backwards, BUT ... that recoil creates a force pair between the hand holding the gun and the handle.

 

[phinds]

Well, you have the right idea, but as I said above, it's a bit more complicated than that because the bullet/gun are NOT a closed system. The gun, whatEVER kind it is, is being held by something, a hand, a tank turret, a shoulder, SOMETHING, and that something has to be taken into consideration to get a closed system.

 

[PeroK]

I understand that conservation of motion comes from the action and reaction pairs of newton's third law. When it is triggered, two forces appear that cancel when analyzed as a system. My question is how is it that momentum is conserved if before the shot there was no force in the system and afterwards there was. How can I analyze it conceptually through equations?

The key concept is momentum, which is the product of an object's mass and velocity. Note that velocity and momentum are vector quantities. I.e. they have a magnitude and a direction. Momentum is often denoted by the letter $p$:$$\vec p = m \vec v$$Where the arrow denotes a vector quantity.

For a closed system, momentum is conserved. Which means it doesn't change over time. In your example, you have an initial state, where we can take the gun and bullet to be at rest. In this case the momentum of the system is zero. You also have a final state where the bullet is moving in one direction and the gun is recoiling in the opposite direction. By conservation of momentum we know that the total momentum of the system remains zero. Hence the momentum of the gun is equal and opposite of the momentum of the bullet. We can write this as:$$m_b\vec v_b + m_g \vec v_g = 0$$or, equivalently$$m_b\vec v_b = - m_g \vec v_g$$Note that also immediately after the gun is fired, both the gun and bullet will interact with the environment. Especially the gun is likely to impact something to slow or stop its motion. At this point we no longer have a closed system of just the two objects. The bullet also will be slowed by the air, affected by gravity and possibly impact an external object. And the momentum has spread about the environment in ways that make it hard to track.

 

[Ibix]

My question is how is it that momentum is conserved if before the shot there was no force in the system and afterwards there was.

What do you mean by "afterwards there was force in the system"? And are you considering the realistic case of a gun that is mounted or held, or the more abstract case of a gun floating free in space?

Let's consider a gun floating in space that fires itself - perhaps a radio controlled trigger. There are only forces on the system for the tiny fraction of a second while the bullet accelerates down the barrel. Before that there are no forces, and after that there are no forces. At all times the gun and the bullet have equal and opposite momenta. Initially they are zero; they grow at equal rates due to the Third Law; when the bullet leaves the barrel they have constant equal and opposite momenta. The total momentum is therefore zero at all times.

Ok so far?

As phinds points out, most guns are either held or mounted, and the analysis depends a bit on how that is treated. A typical way to do it is to treat the Earth, and anything fixed to it, as an immovable object, so its momentum is always zero. This actually leads to a violation of momentum conservation because you are neglecting the momentum of the gun/Earth combination, but as long as you know it's an approximation and the failure of momentum conservation is due to that approximation, that's fine. A more complete way to do it is to consider the Earth as an extension of the gun, and have the Earth recoil slightly when the gun fires. So the gun is just floating in space because that's what the Earth is doing (you are welcome to work out the recoil speed of the Earth, $M_E=6\times 1^{24}\mathrm{kg}$, when a 1g bullet is fired at $300\mathrm{ms^{-1}}$). Of course, when the bullet lands it will give its momentum back to the Earth, so there's no final momentum change in this case.

Hope that makes sense.