# A hunting rifle fires a bullet of mass 0.450619 kg with a velocity of 849 m/s to the right. The rifle has a mass of 4.37 kg. What is the recoil speed of the rifle as the bullet leaves the rifle?

This question deals with Newton's third law of motion and the conservation of momentum. Newton's third law states that when a force is applied to one object, an equal force is applied to a second object in the opposite direction. People tend to summarize this law by saying that "every action has an equal and opposite reaction." The law also deals with momentum and the idea that the momentum within a closed system is conserved. If the gun and bullet were not moving at the start, then the total momentum was zero. If the bullet goes forward, in order for the total momentum to remain a value of zero, then the gun must move backward (negative velocity) to conserve the original momentum. The formula for momentum is mass times velocity.

P = mv

P = .450619 x 849

P = 382.57553 kg-m/s

Now that you've found the momentum of the bullet, rewrite the original formula so that you are now solving for the velocity of the weapon.

V = P/m

V = 382.57553/4.37

V = 87.545887 m/s to the left