MechSimulator

Gun Recoil Simulator

Newton’s third law in a gunshot — momentum, recoil velocity & energy, felt recoil, muzzle rise and recoil systems

Mode
Units
Display Controls
Show
2 t = 0

Σ Live equations — with the numbers of the shot on screen
User Guide — Gun Recoil Simulator
1 Overview

The Gun Recoil Simulator shows Newton’s third law at work in a gunshot. The powder gas pushes the bullet forward and the gun backward; the simulator integrates that shot and then what the gun meets afterwards. It has eight guns (9 mm pistol, .22 rifle, 5.56 mm rifle, .308 rifle, 12-gauge shotgun, .50 BMG rifle, 12-pounder field cannon and 105 mm howitzer), three ways of holding them, and four modes: Simulate, Explore, Practice and Quiz.

2 Getting Started

Follow the numbers: 1 in Set up the shot, pick a gun from the dropdown (the .308 hunting rifle is the default), choose what holds it and set the bullet, speed, powder and gun mass; 2 press ➤ Fire (or Space on the canvas). The shot plays in slow motion and the canvas says by how much. When it finishes, drag the time slider to go back to any instant. 3 Work through the Try this challenges, which tick themselves.

3 Simulate Mode — what holds the gun
  • Free-recoil sled — the gun is clamped to a carriage on low-friction rails, so nothing resists it. It shows conservation of momentum on its own: the gun leaves the sled’s light gate at V = (m v + β C v)/M, the free recoil velocity. An arrester at the end of the rail catches it afterwards.
  • Mechanical shoulder (small arms) — the butt (or, for the pistol, the grip) bears on a padded load cell backed by a spring-mounted body mass, the way recoil test rigs model a shooter. The force graph shows the push on the shoulder; change the pad and its peak changes while its area (the impulse) stays the same. The support hand is a rotational spring, so the gun also rotates: muzzle rise. Rounds 1–5 fire a string at the set interval.
  • Recoil system (cannons) — the 12-pounder rolls back on its wheels against friction from the ground you choose; the 105 mm howitzer’s barrel slides back in its cradle against a constant-force hydraulic buffer set by the recoil length, then the recuperator runs it out again.

Detail A (top left, shown while the gas is acting) is a magnified section of the chamber and bore: the gas colour shows the pressure, and the arrows are the third-law pair — the gas pushing the bullet forward and the breech backward. The momentum bars (top right) show bullet, gas, gun and what holds it; they always add to zero. Underneath, the left graph is the shot itself (momentum against time, in milliseconds) and the right graph is the recoil (velocity, shoulder force or recoil distance).

4 Explore Mode

Twenty-four concept cards in six groups: Third law & momentum, Free recoil, Inside the barrel, Felt recoil & muzzle rise, Recoil systems, and Units & standards. Most carry a worked example with the simulator’s own numbers.

5 Practice & Quiz

Practice draws from fourteen kinds of problem: momentum, recoil velocity with and without the gas, recoil energy, energy sharing, average force from impulse, the steady force of rapid fire, cannon roll-back, buffer force, bullet speed from a recoil sled, muzzle rise, the Lagrange pressure ratio, the gun mass for a recoil limit and the muzzle brake factor. Answers within 2 % are marked correct; Show Solution gives the working. Practice problems are set in SI units. Quiz asks six multiple-choice questions drawn from a pool of twenty-one, with the options shuffled, and scores you with up to three stars.

6 Key Concepts & the model

Momentum: m v + (gas) = M V. For the .308 rifle (9.7 g at 860 m/s, 2.85 g of powder, 4.1 kg): V = (8.342 + 1.75 × 0.00285 × 860) / 4.1 = 3.08 m/s and E = ½ M V² = 19.5 J (14.4 ft·lbf). In the bore the gas moves with the bullet (Lagrange), carrying C v/2; the breech pressure is Pb(1 + C/2m). The pressure curve is a model shape that rises to the cartridge’s rated maximum pressure and falls like an adiabatic expansion, scaled so the bullet leaves the barrel at exactly the published muzzle velocity. After the bullet leaves, the remaining gas momentum C v (β − ½) is delivered as a short exponential pulse. The mechanical-shoulder and body constants are illustrative, so use the force figures to compare, not as a measurement of a real shooter.

7 SI vs Imperial

The Units switch shows bullets and powder in grams or grains (1 gr = 64.79891 mg), cannon shot in kg or lb, speeds in m/s or ft/s, momentum in N·s or lbf·s, energy in J or ft·lbf, force in N or lbf and pressure in MPa or psi. Your choice is remembered across the site. Example in Imperial: the .308 fires a 150 gr bullet at 2 822 ft/s from a 9.0 lb rifle; free recoil 10.1 ft/s and 14.4 ft·lbf.

8 Tips & Other Features

Display Controls (top left of the canvas) switch the force arrows, Detail A, momentum bars, labels, smoke and the graphs; turn the graphs off and the scene fills the whole canvas. Your choices are remembered. Right-click (or long-press) the canvas to save an image, copy the readings or fire. PNG also saves the view. The Live equations panel works the numbers of the shot on screen, and Show Calculations (beside Fire) opens the full step-by-step working for the gun and set-up you have chosen: the momentum balance, the free recoil velocity and energy, the force of the gas on the breech, and then how far the gun goes back and the force on whatever stops it (shoulder, wheels on the ground or recoil buffer), each closed-form estimate set beside the simulated value. Add ?gun=howitzer (or pistol, rimfire, carbine, bolt, shotgun, heavy, cannon) to a link to open a particular gun.

Gun Recoil and Newton’s Third Law

Every physics course uses the recoiling gun to explain Newton’s third law, and this gun recoil simulator lets you watch it happen instead of only reading about it. When the powder burns, the gas pushes on everything around it. It pushes the bullet forward and the breech of the gun backward. The bullet and the gun get equal and opposite momenta, so the total stays zero, exactly as it was before the shot. The gun moves slowly because it is heavy; the bullet moves fast because it is light.

The Recoil Velocity Formula

Momentum conservation gives the textbook result V = m v / M: a 4.1 kg rifle firing a 9.7 g bullet at 860 m/s recoils at 2.03 m/s. Real guns kick harder than that, because the propellant gas is thrown out of the muzzle too. The industry method (SAAMI) adds the gas momentum as the charge mass times an effective gas velocity of βv, with β = 1.75 for rifles and 1.5 for handguns and shotguns:

For the .308 rifle the gas raises the momentum from 8.34 to 12.63 N·s and the recoil energy from 8.5 J to 19.5 J. The table shows the free recoil of every gun in the simulator, computed by its own engine with a bare muzzle.

GunProjectileMuzzle velocityRecoil momentumRecoil velocityRecoil energy
9 mm pistol (0.95 kg)8.0 g360 m/s3.07 N·s3.24 m/s5.0 J
.22 LR rifle (2.3 kg)2.6 g370 m/s1.04 N·s0.45 m/s0.2 J
5.56 mm rifle (3.4 kg)4.0 g940 m/s6.42 N·s1.89 m/s6.1 J
.308 rifle (4.1 kg)9.7 g860 m/s12.63 N·s3.08 m/s19.5 J
12-gauge shotgun (3.4 kg)32 g370 m/s12.84 N·s3.78 m/s24.2 J
.50 BMG rifle (13.6 kg)42 g890 m/s61.68 N·s4.54 m/s139.9 J
12-pounder cannon (1070 kg)5.6 kg439 m/s3203 N·s2.99 m/s4.79 kJ
105 mm howitzer (500 kg recoiling)14.97 kg472 m/s7951 N·s15.90 m/s63.2 kJ

Impulse, Felt Recoil and Muzzle Rise

The gas gives the gun its momentum in about a millisecond, so that force is huge. The shoulder then takes the same momentum away over tens of milliseconds, so its force is much smaller: Favg = Δp / Δt. A soft recoil pad lengthens the push and lowers the peak, but the area under the force–time curve, the impulse, cannot change. Because the bore sits above the shoulder, the recoil force and the shoulder’s reaction form a couple that lifts the muzzle; its angular impulse is p·h, which is why straight-line stocks rise less.

Inside the Barrel

The simulator follows the shot inside the bore with the Lagrange approximation used in interior ballistics: the gas speed grows linearly from the breech to the bullet, so the gas carries momentum C v / 2 and the breech pressure is Pb(1 + C / 2m), higher than at the bullet base. The pressure peaks at the cartridge’s rated maximum (427 MPa for .308 Winchester) and the bullet leaves in about a millisecond. When it does, the remaining gas rushes out of the muzzle and gives the gun its last push, the gas aftereffect. A muzzle brake turns part of that gas sideways and back and so reduces the recoil.

Cannon Recoil and Recoil Systems

A 19th-century field gun had no recoil system: the whole gun rolled back, d = V² / 2μg, and the crew pushed it back after every shot. A modern howitzer lets only its barrel recoil, against a hydraulic buffer whose nearly constant force R = M V² / 2L stops it in the recoil length, and a recuperator runs it out again. In the simulator the 105 mm howitzer’s buffer takes about 56 kN over 1.07 m, against a gas force on the breech of about 2 MN.

Who Uses This Simulator?

High school students meeting Newton’s third law, momentum and impulse (GCSE, IGCSE, AP Physics 1, IB, CBSE); undergraduates in dynamics and mechanics of machines working recoil, energy and buffer problems; and teachers who want the classic “recoil of a gun” example to move on the board.

Frequently Asked Questions

How does Newton’s third law explain gun recoil?

When the powder burns, the gas pushes the bullet forward and pushes the gun backward with the same pressure. The forces come in equal and opposite pairs, so the total momentum stays zero: the forward momentum of the bullet and gas equals the backward momentum of the gun. The gun moves much more slowly because it is much heavier.

What is the formula for recoil velocity?

Ignoring the gas, V = m v / M, where m and v are the bullet mass and speed and M is the gun mass. Including the propellant gas (SAAMI method), V = (m v + beta C v) / M, where C is the powder charge and beta is about 1.75 for rifles and 1.5 for handguns and shotguns. The free recoil energy is E = M V squared / 2.

Why does a heavier gun kick less?

For the same cartridge the recoil momentum is the same, so a heavier gun recoils more slowly: V = p / M. The recoil energy is E = p squared / 2M, so doubling the gun's mass halves the energy the shooter has to absorb.

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