Q.A gun of mass $4\ \text{kg}$, initially at rest, fires a bullet of mass $0.04\ \text{kg}$ with a muzzle velocity of $300\ \text{m/s}$. Find the recoil velocity of the gun.

Before firing, both the gun (mass $M = 4\ \text{kg}$) and the bullet (mass $m = 0.04\ \text{kg}$) are at rest, so the total momentum of the (isolated) gun-bullet system is zero. By the law of conservation of linear momentum, the total momentum must still be zero immediately after firing: $MV + mv = 0$ where $v = 300\ \text{m/s}$ is the bullet's (forward) muzzle velocity and $V$ is the gun's recoil velocity. Solving for $V$: $V = -\frac{mv}{M} = -\frac{(0.04\ \text{kg})(300\ \text{m/s})}{4\ \text{kg}} = -\frac{12}{4} = -3\ \text{m/s}$ The negativ …

Set the total initial momentum of the at-rest gun-bullet system to zero, and require it stay zero afte …

• Forgetting to include the gun's own (much larger) mass, and instead just quoting the bullet's velocity as the recoil velocity. …