Momentum Conservation

There is a compressed spring between them, but a rope is holding them together.  Since the carts are in equilibrium, the total force on each of them is 1($).  if the rope is cut, however, the compressed spring exerts an 2($) force on each of the objects.

Newton's third law (the action-reaction law) tells us about the forces on the carts that they are equal and 3($)

• Total momentum of the system before the rope is cut = 1($)
• Total momentum of the system while the carts are moving = 2($)

The total 1($) in any isolated system before any 2($) is equal to the total momentum after the 3($)

The system must be isolated from other forces

When a gun fires a bullet, the gun and the bullet can be considered an isolated system, at least until the gun recoils back against the shooter's shoulder and the shoulder exerts an outside force on the gun.  Before the "event" - the firing of the bullet - the gun and the bullet are at rest so their total momentum is zero after the bullet is fired, it has momentum.  (Although its mass is fairly small, its speed is great, so its momentum is considerable.)  if momentum is to be conserved, something must gain momentum in the direction opposite that of the bullet.  The gun does just that.

1. How does the mass of the gun compare to that of the bullet? 1($)
2. After the firing, how does the momentum of the gun compare to that of the bullet? 2($)
3. Therefore, how does the velocity of the gun compare to that of the bullet after firing? 3($)