Gun owners are often interested in recoil velocity, but they aren't the only ones. There are many other situations in which it's a useful quantity to know. For example, a basketball player taking a jump shot may want to know his or her backwards velocity after releasing the ball to avoid crashing into another player, and the captain of a frigate may want to know the effect the release of a lifeboat has on the ship's forward motion. In space, where frictional forces are absent, recoil velocity is a critical quantity. You apply the law of conservation of momentum to find recoil velocity. This law is derived from Newton's Laws of Motion.

#### TL;DR (Too Long; Didn't Read)

The law of conservation of momentum, derived from Newton's Laws of Motion, provides a simple equation for calculating recoil speed. It's based on the mass and speed of the ejected body and the mass of the recoiling body.

## Law of Conservation of Momentum

Newton's Third Law states that every applied force has an equal and opposite reaction. An example commonly cited when explaining this law is that of a speeding car hitting a brick wall. The car exerts a force on the wall, and the wall exerts a reciprocal force on the car that crushes it. Mathematically, the incident force (F_{I}) equals the reciprocal force (F_{R}) and acts in the opposite direction: F_{I} = - F_{R}.

Newton's Second Law defines force as mass time acceleration. Acceleration is change in velocity (∆v ÷ ∆t), so force can be expressed F = m (∆v ÷ ∆t). This allows the Third Law to be rewritten as m_{I}(∆v_{I} ÷ ∆t_{I}) = -m_{R}(∆v_{R} ÷ ∆t_{R}). In any interaction, the time during which the incident force is applied is equal to the time during which the reciprocal force is applied, so t_{I} = t_{R} and the time can be factored out of the equation. This leaves:

## Sciencing Video Vault

**m _{I}∆v_{I }= -m_{R}∆v_{R}**

This is known as the law of conservation of momentum.

## Calculating Recoil Velocity

In a typical recoil situation, the release of a body of smaller mass (body 1) has an impact on a larger body (body 2). If both bodies start from rest, the law of conservation of momentum states that m_{1}v_{1} = -m_{2}v_{2}. The recoil velocity is typically the velocity of body 2 after the release of body 1. This velocity is

**v _{2} = -(m_{1} ÷ m_{2}) v_{1}.**

## Example

**What is the recoil velocity of an 8-pound Winchester rifle after firing a 150-grain bullet with a speed of 2,820 feet/second?**

Before solving this problem, it's necessary to express all quantities in consistent units. One grain is equal to 64.8 mg, so the bullet has a mass (m_{B}) of 9,720 mg, or 9.72 grams. The rifle, on the other hand, has a mass (m_{R}) of 3,632 grams, since there are 454 grams in a pound. It's now easy to calculate the recoil speed of the rifle (v_{R})in feet/second:

**v _{R} = -(m_{B} ÷ m_{R}) v_{B} = -(9.72 g ÷ 3,632g) • 2,820 ft/s = -7.55 ft/s.**

The minus sign denotes the fact that recoil speed is in the opposite direction to the speed of the bullet.

**A 2,000-ton frigate releases a 2-ton lifeboat at a speed of 15 miles per hour. Assuming negligible friction, what is the recoil speed of the frigate?**

Weights are expressed in the same units, so there is no need for conversion. You can simply write the speed of the frigate as v_{F} = (2 ÷ 2000) • 15 mph = 0.015 mph. This speed is small, but it isn't negligible. It's over 1 foot per minute, which is significant if the frigate is near a dock.