Misplaced Pages

#122877

95-434: Any launching system (weapon or not) generates recoil. However recoil only constitutes a problem in the field of artillery and firearms due to the magnitude of the forces at play. Gun chamber pressures and projectile acceleration forces are tremendous, on the order of tens to hundreds mega pascal and tens of thousands of times the acceleration of gravity ( g's ), both necessary to launch the projectile at useful velocity during

190-402: A n t . {\displaystyle m_{A}v_{A}+m_{B}v_{B}+m_{C}v_{C}+...=constant.} This conservation law applies to all interactions, including collisions (both elastic and inelastic ) and separations caused by explosive forces. It can also be generalized to situations where Newton's laws do not hold, for example in the theory of relativity and in electrodynamics . Momentum

285-418: A Galilean transformation . If a particle is moving at speed ⁠ d x / d t ⁠ = v in the first frame of reference, in the second, it is moving at speed v ′ = d x ′ d t = v − u . {\displaystyle v'={\frac {{\text{d}}x'}{{\text{d}}t}}=v-u\,.} Since u does not change,

380-512: A momentum density can be defined as momentum per volume (a volume-specific quantity ). A continuum version of the conservation of momentum leads to equations such as the Navier–Stokes equations for fluids or the Cauchy momentum equation for deformable solids or fluids. Momentum is a vector quantity : it has both magnitude and direction. Since momentum has a direction, it can be used to predict

475-445: A shock absorber . Energy in firing a firearm comes in many forms (thermal, pressure) but for understanding recoil what matters is kinetic energy , which is half mass multiplied by squared speed. For the recoiling gun, this means that for a given rearward momentum, doubling the mass halves the speed and also halves the kinetic energy of the gun, making it easier to dissipate. If all the masses and velocities involved are accounted for,

570-752: A 1 kg model airplane, traveling due north at 1 m/s in straight and level flight, has a momentum of 1 kg⋅m/s due north measured with reference to the ground. The momentum of a system of particles is the vector sum of their momenta. If two particles have respective masses m 1 and m 2 , and velocities v 1 and v 2 , the total momentum is p = p 1 + p 2 = m 1 v 1 + m 2 v 2 . {\displaystyle {\begin{aligned}p&=p_{1}+p_{2}\\&=m_{1}v_{1}+m_{2}v_{2}\,.\end{aligned}}} The momenta of more than two particles can be added more generally with

665-404: A 8 g (124 gr) bullet of 9×19mm Parabellum flying forward at 350 m/s muzzle speed generates a momentum to push a 0.8 kg pistol firing it at 3.5 m/s rearward, if unopposed by the shooter. In order to bring the rearward moving gun to a halt, the momentum acquired by the gun is dissipated by a forward-acting counter-recoil force applied to the gun over a period of time during and after

760-402: A collision. For example, suppose there are two bodies of equal mass m , one stationary and one approaching the other at a speed v (as in the figure). The center of mass is moving at speed ⁠ v / 2 ⁠ and both bodies are moving towards it at speed ⁠ v / 2 ⁠ . Because of the symmetry, after the collision both must be moving away from the center of mass at

855-432: A gas-operated gun, the bolt is accelerated rearwards by propellant gases during firing, which results in a forward force on the body of the gun. This is countered by a rearward force as the bolt reaches the limit of travel and moves forwards, resulting in a zero sum, but to the shooter, the recoil has been spread out over a longer period of time, resulting in the "softer" feel. A recoil system absorbs recoil energy, reducing

950-435: A longer or shorter distance to bring the car to a stop. However, for the human body to mechanically adjust recoil time, and hence length, to lessen felt recoil force is perhaps an impossible task. Other than employing less safe and less accurate practices, such as shooting from the hip, shoulder padding is a safe and effective mechanism that allows sharp recoiling to be lengthened into soft recoiling, as lower decelerating force

1045-465: Is a vector quantity, possessing a magnitude and a direction. If m is an object's mass and v is its velocity (also a vector quantity), then the object's momentum p (from Latin pellere "push, drive") is: p = m v . {\displaystyle \mathbf {p} =m\mathbf {v} .} In the International System of Units (SI), the unit of measurement of momentum

SECTION 10

#1732859579123

1140-413: Is a good example of an almost totally elastic collision, due to their high rigidity , but when bodies come in contact there is always some dissipation . A head-on elastic collision between two bodies can be represented by velocities in one dimension, along a line passing through the bodies. If the velocities are v A1 and v B1 before the collision and v A2 and v B2 after,

1235-599: Is a measurable quantity, and the measurement depends on the frame of reference . For example: if an aircraft of mass 1000 kg is flying through the air at a speed of 50 m/s its momentum can be calculated to be 50,000 kg.m/s. If the aircraft is flying into a headwind of 5 m/s its speed relative to the surface of the Earth is only 45 m/s and its momentum can be calculated to be 45,000 kg.m/s. Both calculations are equally correct. In both frames of reference, any change in momentum will be found to be consistent with

1330-479: Is also conserved in special relativity (with a modified formula) and, in a modified form, in electrodynamics , quantum mechanics , quantum field theory , and general relativity . It is an expression of one of the fundamental symmetries of space and time: translational symmetry . Advanced formulations of classical mechanics, Lagrangian and Hamiltonian mechanics , allow one to choose coordinate systems that incorporate symmetries and constraints. In these systems

1425-466: Is an inelastic collision . An elastic collision is one in which no kinetic energy is transformed into heat or some other form of energy. Perfectly elastic collisions can occur when the objects do not touch each other, as for example in atomic or nuclear scattering where electric repulsion keeps the objects apart. A slingshot maneuver of a satellite around a planet can also be viewed as a perfectly elastic collision. A collision between two pool balls

1520-552: Is dissipating the kinetic energy of the recoiling gun mass. A heavier gun, that is a gun with more mass, will manifest lower recoil kinetic energy, and, generally, result in a lessened perception of recoil. Therefore, although determining the recoiling energy that must be dissipated through a counter-recoiling force is arrived at by conservation of momentum, kinetic energy of recoil is what is actually being restrained and dissipated. The ballistics analyst discovers this recoil kinetic energy through analysis of projectile momentum. One of

1615-605: Is equal to the instantaneous force F acting on it, F = d p d t . {\displaystyle F={\frac {{\text{d}}p}{{\text{d}}t}}.} If the net force experienced by a particle changes as a function of time, F ( t ) , the change in momentum (or impulse J ) between times t 1 and t 2 is Δ p = J = ∫ t 1 t 2 F ( t ) d t . {\displaystyle \Delta p=J=\int _{t_{1}}^{t_{2}}F(t)\,{\text{d}}t\,.} Impulse

1710-495: Is explained by the law of conservation of momentum, and so it is easier to discuss it separately from energy . Momentum is simply mass multiplied by velocity. Velocity is speed in a particular direction (not just speed). In a very technical sense, speed is a scalar (mathematics) : a magnitude; while velocity is a vector (physics) : magnitude and direction. Momentum is conservative: any change in momentum of an object requires an equal and opposite change of some other objects. Hence

1805-407: Is known as Euler's first law . If the net force F applied to a particle is constant, and is applied for a time interval Δ t , the momentum of the particle changes by an amount Δ p = F Δ t . {\displaystyle \Delta p=F\Delta t\,.} In differential form, this is Newton's second law ; the rate of change of the momentum of a particle

1900-468: Is measured in the derived units of the newton second (1 N⋅s = 1 kg⋅m/s) or dyne second (1 dyne⋅s = 1 g⋅cm/s) Under the assumption of constant mass m , it is equivalent to write F = d ( m v ) d t = m d v d t = m a , {\displaystyle F={\frac {{\text{d}}(mv)}{{\text{d}}t}}=m{\frac {{\text{d}}v}{{\text{d}}t}}=ma,} hence

1995-466: Is numerically equivalent to 3 newtons. In a closed system (one that does not exchange any matter with its surroundings and is not acted on by external forces) the total momentum remains constant. This fact, known as the law of conservation of momentum , is implied by Newton's laws of motion . Suppose, for example, that two particles interact. As explained by the third law, the forces between them are equal in magnitude but opposite in direction. If

SECTION 20

#1732859579123

2090-422: Is the center of mass frame – one that is moving with the center of mass. In this frame, the total momentum is zero. If two particles, each of known momentum, collide and coalesce, the law of conservation of momentum can be used to determine the momentum of the coalesced body. If the outcome of the collision is that the two particles separate, the law is not sufficient to determine the momentum of each particle. If

2185-449: Is the kilogram metre per second (kg⋅m/s), which is dimensionally equivalent to the newton-second . Newton's second law of motion states that the rate of change of a body's momentum is equal to the net force acting on it. Momentum depends on the frame of reference , but in any inertial frame it is a conserved quantity, meaning that if a closed system is not affected by external forces, its total momentum does not change. Momentum

2280-556: Is the angle above the aim angle at which the bullet leaves the barrel, t f {\displaystyle t_{f}} is the time of travel of the bullet in the barrel (because of the acceleration a = 2 x / t 2 {\displaystyle a=2x/t^{2}} the time is longer than L / V b {\displaystyle L/V_{\text{b}}}  : t f = 2 L / V b {\displaystyle t_{f}=2L/V_{\text{b}}} ) and L

2375-574: Is the angle of rotation of the barrel axis "up" from its orientation at ignition (aim angle). The angular momentum of the gun is found by integrating this equation to obtain: I d θ d t = h ∫ 0 t F ( t ) d t = h m g V g ( t ) = h m b V b ( t ) {\displaystyle I{\frac {d\theta }{dt}}=h\int _{0}^{t}F(t)\,dt=hm_{\text{g}}V_{\text{g}}(t)=hm_{\text{b}}V_{\text{b}}(t)} where

2470-417: Is the distance the bullet travels from its rest position to the tip of the barrel. The angle at which the bullet leaves the barrel above the aim angle is then given by: θ f = 2 h m b L I {\displaystyle \theta _{f}={\frac {2hm_{\text{b}}L}{I}}} Before the projectile leaves the gun barrel , it obturates the bore and "plugs up"

2565-438: Is the mass of the propellant charge, equal to the mass of the ejected gas. This expression should be substituted into the expression for projectile momentum in order to obtain a more accurate description of the recoil process. The effective velocity may be used in the energy equation as well, but since the value of α used is generally specified for the momentum equation, the energy values obtained may be less accurate. The value of

2660-592: Is the momentum of the firearm and p p {\displaystyle p_{\text{p}}} is the momentum of the projectile. In other words, immediately after firing, the momentum of the firearm is equal and opposite to the momentum of the projectile. Since momentum of a body is defined as its mass multiplied by its velocity, we can rewrite the above equation as: m f v f + m p v p = 0 {\displaystyle m_{\text{f}}v_{\text{f}}+m_{\text{p}}v_{\text{p}}=0} where: A force integrated over

2755-529: Is the muzzle velocity of the projectile and α {\displaystyle \alpha } is approximately constant. The total momentum p e {\displaystyle p_{e}} of the propellant and projectile will then be: p e = m p V 0 + m g α V 0 {\displaystyle p_{e}=m_{p}V_{0}+m_{\text{g}}\alpha V_{0}\,} where m g {\displaystyle m_{\text{g}}\,}

2850-428: Is the perpendicular distance of the center of mass of the gun below the barrel axis, F ( t ) {\textstyle F(t)} is the force on the gun due to the expanding gases, equal and opposite to the force on the bullet, I {\textstyle I} is the moment of inertia of the gun about its center of mass, or its pivot point, and θ {\displaystyle \theta }

2945-402: Is the product of the mass and the acceleration of the projectile and propellant gasses combined, reversed: the projectile moves forward, the recoil is rearward. The heavier and the faster the projectile, the more recoil will be generated. The gun acquires a rearward velocity that is ratio of this momentum by the mass of the gun: the heavier the gun, the slower the rearward velocity. As an example,

Recoil - Misplaced Pages Continue

3040-409: Is the product of the units of mass and velocity. In SI units , if the mass is in kilograms and the velocity is in meters per second then the momentum is in kilogram meters per second (kg⋅m/s). In cgs units , if the mass is in grams and the velocity in centimeters per second, then the momentum is in gram centimeters per second (g⋅cm/s). Being a vector, momentum has magnitude and direction. For example,

3135-429: Is transmitted into the body over a slightly greater distance and time, and spread out over a slightly larger surface. Keeping the above in mind, you can generally base the relative recoil of firearms by factoring in a small number of parameters: bullet momentum (weight times velocity), (note that momentum and impulse are interchangeable terms), and the weight of the firearm. Lowering momentum lowers recoil, all else being

3230-455: Is unchanged. Forces such as Newtonian gravity, which depend only on the scalar distance between objects, satisfy this criterion. This independence of reference frame is called Newtonian relativity or Galilean invariance . A change of reference frame, can, often, simplify calculations of motion. For example, in a collision of two particles, a reference frame can be chosen, where, one particle begins at rest. Another, commonly used reference frame,

3325-459: The Franck–Hertz experiment ); and particle accelerators in which the kinetic energy is converted into mass in the form of new particles. In a perfectly inelastic collision (such as a bug hitting a windshield), both bodies have the same motion afterwards. A head-on inelastic collision between two bodies can be represented by velocities in one dimension, along a line passing through the bodies. If

3420-420: The barrel after projectile exit is vented rearward though a nozzle at the back of the chamber, creating a large counter-recoiling force sufficient to eliminate the need for heavy recoil mitigating buffers on the mount (although at the cost of a reduced muzzle velocity of the projectile). The same physics principles affecting recoil in mounted guns also applies to hand-held guns. However, the shooter's body assumes

3515-400: The barrel recoils backward, then is dissipated via hydraulic damping as the barrel is returned forward to the firing position under the pressure of the compressed air. The recoil impulse is thus spread out over the time in which the barrel is compressing the air, rather than over the much narrower interval of time when the projectile is being fired. This greatly reduces the peak force conveyed to

3610-413: The barrel upon the gun (recoil force), which is equal and opposite to the force upon the ejecta. It is also determined by the counter-recoil force applied to the gun (e.g. an operator's hand or shoulder, or a mount). The recoil force only acts during the time that the ejecta are still in the barrel of the gun. The counter-recoil force is generally applied over a longer time period and adds forward momentum to

3705-462: The barrel, in order to minimize any rotational effects. If there is an angle for the recoil parts to rotate about, the torque ( τ {\displaystyle \tau } ) on the gun is given by: τ = I d 2 θ d t 2 = h F ( t ) {\displaystyle \tau =I{\frac {d^{2}\theta }{dt^{2}}}=hF(t)} where h {\textstyle h}

3800-419: The barrel. And then to properly design recoil buffering systems to safely dissipate that momentum and energy. To confirm analytical calculations and estimates, once a prototype gun is manufactured, the projectile and gun recoil energy and momentum can be directly measured using a ballistic pendulum and ballistic chronograph . The nature of the recoil process is determined by the force of the expanding gases in

3895-404: The case of zero-recoil, the counter-recoil force is smaller than the recoil force but lasts for a longer time. Since the recoil force and the counter-recoil force are not matched, the gun will move rearward, slowing down until it comes to rest. In the zero-recoil case, the two forces are matched and the gun will not move when fired. In most cases, a gun is very close to a free-recoil condition, since

Recoil - Misplaced Pages Continue

3990-411: The charge is ignited, about half of the recoil impulse is applied to stopping the forward motion of the barrel, while the other half is, as in the usual system, taken up in recompressing the spring. A latch then catches the barrel and holds it in the starting position. This roughly halves the energy that the spring needs to absorb, and also roughly halves the peak force conveyed to the mount, as compared to

4085-432: The common ways of describing the felt recoil of a particular gun-cartridge combination is as "soft" or "sharp" recoiling; soft recoil is recoil spread over a longer period of time, that is at a lower deceleration, and sharp recoil is spread over a shorter period of time, that is with a higher deceleration. Like pushing softer or harder on the brakes of a car, the driver feels less or more deceleration force being applied, over

4180-422: The condition for free-recoil is t r ≪ t cr {\displaystyle t_{\text{r}}\ll t_{\text{cr}}} , while for zero-recoil, F r ( t ) + F cr ( t ) = 0 {\displaystyle F_{\text{r}}(t)+F_{\text{cr}}(t)=0} . For a gun firing under free-recoil conditions, the force on the gun may not only force

4275-496: The conserved quantity is generalized momentum , and in general this is different from the kinetic momentum defined above. The concept of generalized momentum is carried over into quantum mechanics, where it becomes an operator on a wave function . The momentum and position operators are related by the Heisenberg uncertainty principle . In continuous systems such as electromagnetic fields , fluid dynamics and deformable bodies ,

4370-401: The constant α is generally taken to lie between 1.25 and 1.75. It is mostly dependent upon the type of propellant used, but may depend slightly on other things such as the ratio of the length of the barrel to its radius. Muzzle devices can reduce the recoil impulse by altering the pattern of gas expansion. For instance, muzzle brakes primarily works by diverting some of the gas ejecta towards

4465-427: The details below. Request from 172.68.168.226 via cp1108 cp1108, Varnish XID 760503886 Upstream caches: cp1108 int Error: 429, Too Many Requests at Fri, 29 Nov 2024 05:52:59 GMT Conservation of momentum In Newtonian mechanics , momentum ( pl. : momenta or momentums ; more specifically linear momentum or translational momentum ) is the product of the mass and velocity of an object. It

4560-418: The ejecta, and do not alter the overall momentum of the system, they do involve moving masses during the operation of firing. For example, gas-operated shotguns are widely held to have a "softer" recoil than fixed breech or recoil-operated guns. (Although many semi-automatic recoil and gas-operated guns incorporate recoil buffer systems into the stock that effectively spread out peak felt recoil forces.) In

4655-530: The equality of the momenta of the gun and bullet have been used. The angular rotation of the gun as the bullet exits the barrel is then found by integrating again: I θ f = h ∫ 0 t f m b V b d t = 2 h m b L {\displaystyle I\theta _{f}=h\int _{0}^{t_{f}}m_{\text{b}}V_{\text{b}}\,dt=2hm_{\text{b}}L} where θ f {\displaystyle \theta _{f}}

4750-914: The equations expressing conservation of momentum and kinetic energy are: m A v A 1 + m B v B 1 = m A v A 2 + m B v B 2 1 2 m A v A 1 2 + 1 2 m B v B 1 2 = 1 2 m A v A 2 2 + 1 2 m B v B 2 2 . {\displaystyle {\begin{aligned}m_{A}v_{A1}+m_{B}v_{B1}&=m_{A}v_{A2}+m_{B}v_{B2}\\{\tfrac {1}{2}}m_{A}v_{A1}^{2}+{\tfrac {1}{2}}m_{B}v_{B1}^{2}&={\tfrac {1}{2}}m_{A}v_{A2}^{2}+{\tfrac {1}{2}}m_{B}v_{B2}^{2}\,.\end{aligned}}} A change of reference frame can simplify analysis of

4845-479: The expanding gas generated by the propellant combustion behind it. This means the gas is essentially contained within a closed system and acts as a neutral element in the overall momentum of the system's physics. However, when the projectile exits the barrel, this functional seal is removed and the highly energetic bore gas is suddenly free to exit the muzzle and expand in the form of a supersonic shockwave (which can be often fast enough to momentarily overtake

SECTION 50

#1732859579123

4940-583: The firearm and projectile are both at rest before firing, then their total momentum is zero. Assuming a near free-recoil condition, and neglecting the gases ejected from the barrel, (an acceptable first estimate), then immediately after firing, conservation of momentum requires that the total momentum of the firearm and projectile is the same as before, namely zero. Stating this mathematically: p f + p p = 0 {\displaystyle p_{\text{f}}+p_{\text{p}}=0} where p f {\displaystyle p_{\text{f}}}

5035-518: The firing rate. The modern quick-firing guns was made possible by the invention of a much more efficient device: the hydro-pneumatic recoil system. First developed by Wladimir Baranovsky in 1872–5 and adopted by the Russian army, then later in France, in the 75mm field gun of 1897 , it is still the main device used by big guns nowadays. In this system, the barrel is mounted on rails on which it can recoil to

5130-765: The following: p = ∑ i m i v i . {\displaystyle p=\sum _{i}m_{i}v_{i}.} A system of particles has a center of mass , a point determined by the weighted sum of their positions: r cm = m 1 r 1 + m 2 r 2 + ⋯ m 1 + m 2 + ⋯ = ∑ i m i r i ∑ i m i . {\displaystyle r_{\text{cm}}={\frac {m_{1}r_{1}+m_{2}r_{2}+\cdots }{m_{1}+m_{2}+\cdots }}={\frac {\sum _{i}m_{i}r_{i}}{\sum _{i}m_{i}}}.} If one or more of

5225-456: The force is between particles. Similarly, if there are several particles, the momentum exchanged between each pair of particles adds to zero, so the total change in momentum is zero. The conservation of the total momentum of a number of interacting particles can be expressed as m A v A + m B v B + m C v C + . . . = c o n s t

5320-420: The gas expansion. By using internal baffles , the gas is made to travel through a convoluted path before eventually released outside at the front of the suppressor, thus dissipating its energy over a larger area and a longer time. This reduces both the intensity of the blast (thus lower loudness ) and the recoil generated (as for the same impulse , force is inversely proportional to time). For small arms,

5415-454: The gun backwards, but may also cause it to rotate about its center of mass or recoil mount. This is particularly true of older firearms, such as the classic Kentucky rifle , where the butt stock angles down significantly lower than the barrel, providing a pivot point about which the muzzle may rise during recoil. Modern firearms, such as the M16 rifle , employ stock designs that are in direct line with

5510-401: The gun equal to the backward momentum supplied by the recoil force, in order to bring the gun to a halt. There are two special cases of counter recoil force: Free-recoil , in which the time duration of the counter-recoil force is very much larger than the duration of the recoil force, and zero-recoil, in which the counter-recoil force matches the recoil force in magnitude and duration. Except for

5605-431: The gun mount are not exceeded. Modern cannons also employ muzzle brakes very effectively to redirect some of the propellant gasses rearward after projectile exit. This provides a counter-recoiling force to the barrel, allowing the buffering system and gun mount to be more efficiently designed at even lower weight. Propellant gases are even more tapped in recoilless guns , where much of the high pressure gas remaining in

5700-449: The gun mount. To apply this counter-recoiling force, modern mounted guns may employ recoil buffering comprising springs and hydraulic recoil mechanisms , similar to shock-absorbing suspension on automobiles. Early cannons used systems of ropes along with rolling or sliding friction to provide forces to slow the recoiling cannon to a stop. Recoil buffering allows the maximum counter-recoil force to be lowered so that strength limitations of

5795-406: The image, excessive recoil can create serious range safety concerns, if the shooter cannot adequately restrain the firearm in a down-range direction. Perception of recoil is related to the deceleration the body provides against a recoiling gun, deceleration being a force that slows the velocity of the recoiling mass. Force applied over a distance is energy. The force that the body feels, therefore,

SECTION 60

#1732859579123

5890-1047: The initial velocities are known, the final velocities are given by v A 2 = ( m A − m B m A + m B ) v A 1 + ( 2 m B m A + m B ) v B 1 v B 2 = ( m B − m A m A + m B ) v B 1 + ( 2 m A m A + m B ) v A 1 . {\displaystyle {\begin{aligned}v_{A2}&=\left({\frac {m_{A}-m_{B}}{m_{A}+m_{B}}}\right)v_{A1}+\left({\frac {2m_{B}}{m_{A}+m_{B}}}\right)v_{B1}\\v_{B2}&=\left({\frac {m_{B}-m_{A}}{m_{A}+m_{B}}}\right)v_{B1}+\left({\frac {2m_{A}}{m_{A}+m_{B}}}\right)v_{A1}\,.\end{aligned}}} If one body has much greater mass than

5985-481: The jerking motion is almost certain to disturb the alignment of the gun and may result in a miss. The shooter may also be physically injured by firing a weapon generating recoil in excess of what the body can safely absorb or restrain; perhaps getting hit in the eye by the rifle scope, hit in the forehead by a handgun as the elbow bends under the force, or soft tissue damage to the shoulder, wrist and hand; and these results vary for individuals. In addition, as pictured in

6080-425: The maximum forces accelerating the gun during the short time the projectile is in the barrel. To mitigate these large recoil forces, recoil buffering mechanisms spread out the counter-recoiling force over a longer time, typically ten to a hundred times longer than the duration of the forces accelerating the projectile. This results in the required counter-recoiling force being proportionally lower, and easily absorbed by

6175-411: The momentum of one particle after the collision is known, the law can be used to determine the momentum of the other particle. Alternatively if the combined kinetic energy after the collision is known, the law can be used to determine the momentum of each particle after the collision. Kinetic energy is usually not conserved. If it is conserved, the collision is called an elastic collision ; if not, it

6270-426: The mount (or to the ground on which the gun has been placed). In a soft-recoil system , the spring (or air cylinder) that returns the barrel to the forward position starts out in a nearly fully compressed state, then the gun's barrel is released free to fly forward in the moment before firing; the charge is then ignited just as the barrel reaches the fully forward position. Since the barrel is still moving forward when

6365-742: The negative sign indicating that the forces oppose. Equivalently, d d t ( p 1 + p 2 ) = 0. {\displaystyle {\frac {\text{d}}{{\text{d}}t}}\left(p_{1}+p_{2}\right)=0.} If the velocities of the particles are v A1 and v B1 before the interaction, and afterwards they are v A2 and v B2 , then m A v A 1 + m B v B 1 = m A v A 2 + m B v B 2 . {\displaystyle m_{A}v_{A1}+m_{B}v_{B1}=m_{A}v_{A2}+m_{B}v_{B2}.} This law holds no matter how complicated

6460-400: The net force is equal to the mass of the particle times its acceleration . Example : A model airplane of mass 1 kg accelerates from rest to a velocity of 6 m/s due north in 2 s. The net force required to produce this acceleration is 3  newtons due north. The change in momentum is 6 kg⋅m/s due north. The rate of change of momentum is 3 (kg⋅m/s)/s due north which

6555-447: The other, its velocity will be little affected by a collision while the other body will experience a large change. In an inelastic collision, some of the kinetic energy of the colliding bodies is converted into other forms of energy (such as heat or sound ). Examples include traffic collisions , in which the effect of loss of kinetic energy can be seen in the damage to the vehicles; electrons losing some of their energy to atoms (as in

6650-456: The particles are numbered 1 and 2, the second law states that F 1 = ⁠ d p 1 / d t ⁠ and F 2 = ⁠ d p 2 / d t ⁠ . Therefore, d p 1 d t = − d p 2 d t , {\displaystyle {\frac {{\text{d}}p_{1}}{{\text{d}}t}}=-{\frac {{\text{d}}p_{2}}{{\text{d}}t}},} with

6745-421: The particles is moving, the center of mass of the system will generally be moving as well (unless the system is in pure rotation around it). If the total mass of the particles is m {\displaystyle m} , and the center of mass is moving at velocity v cm , the momentum of the system is: p = m v cm . {\displaystyle p=mv_{\text{cm}}.} This

6840-516: The peak force that is conveyed to whatever the gun is mounted on. Old-fashioned cannons without a recoil system roll several meters backwards when fired; systems were used to somewhat limit this movement (ropes, friction including brakes on wheels, slopes so that the recoil would force the gun uphill,...), but utterly preventing any movement would just have resulted in the mount breaking. As a result, guns had to be put back into firing position and carefully aimed again after each shot, dramatically slowing

6935-434: The projectile and affect its flight dynamics ), creating a phenomenon known as the muzzle blast . The forward vector of this blast creates a jet propulsion effect that exerts back upon the barrel, and creates an additional momentum on top of the backward momentum generated by the projectile before it exits the gun . The overall recoil applied to the firearm is equal and opposite to the total forward momentum of not only

7030-423: The projectile exits the muzzle. In hand-held small arms , the shooter will apply this force using their own body, resulting in a noticeable impulse commonly referred to as a "kick". In heavier mounted guns, such as heavy machine guns or artillery pieces , recoil momentum is transferred through the platform on which the weapon is mounted . Practical weight gun mounts are typically not strong enough to withstand

7125-504: The projectile, but also the ejected gas. Likewise, the recoil energy given to the firearm is affected by the ejected gas. By conservation of mass , the mass of the ejected gas will be equal to the original mass of the propellant (assuming complete burning). As a rough approximation, the ejected gas can be considered to have an effective exit velocity of α V 0 {\displaystyle \alpha V_{0}} where V 0 {\displaystyle V_{0}}

7220-399: The rear, and the recoil is taken up by a cylinder which is similar in operation to an automotive gas-charged shock absorber , and is commonly visible as a cylinder shorter and smaller than the barrel mounted parallel to it. The cylinder contains a charge of compressed air that will act as a spring, as well as hydraulic oil; in operation, the barrel's energy is taken up in compressing the air as

7315-538: The rear, balancing the recoil. They are used often as light anti-tank weapons. The Swedish-made Carl Gustav 84mm recoilless gun is such a weapon. In machine guns following Hiram Maxim 's design – e.g. the Vickers machine gun – the recoil of the barrel is used to drive the feed mechanism. Mega- Too Many Requests If you report this error to the Wikimedia System Administrators, please include

7410-451: The recoil force on the firearm: ∫ 0 t r F r ( t ) d t = m f v f = − m p v p {\displaystyle \int _{0}^{t_{\text{r}}}F_{\text{r}}(t)\,dt=m_{\text{f}}v_{\text{f}}=-m_{\text{p}}v_{\text{p}}} where: Assuming the forces are somewhat evenly spread out over their respective durations,

7505-410: The recoil process generally lasts much longer than the time needed to move the ejecta down the barrel. An example of near zero-recoil would be a gun securely clamped to a massive or well-anchored table, or supported from behind by a massive wall. However, employing zero-recoil systems is often neither practical nor safe for the structure of the gun, as the recoil momentum must be absorbed directly through

7600-408: The recoil: imparting momentum to the projectile requires imparting opposite momentum to the gun. A change in momentum of a mass requires applying a force (this is Newton's laws of motion ). In a firearm forces wildly change, so what matters is impulse : the change of momentum is equal to the impulse. The rapid change of velocity ( acceleration ) of the gun is a shock and will countered as if by

7695-402: The relevant laws of physics. Suppose x is a position in an inertial frame of reference. From the point of view of another frame of reference, moving at a constant speed u relative to the other, the position (represented by a primed coordinate) changes with time as x ′ = x − u t . {\displaystyle x'=x-ut\,.} This is called

7790-538: The resulting direction and speed of motion of objects after they collide. Below, the basic properties of momentum are described in one dimension. The vector equations are almost identical to the scalar equations (see multiple dimensions ). The momentum of a particle is conventionally represented by the letter p . It is the product of two quantities, the particle's mass (represented by the letter m ) and its velocity ( v ): p = m v . {\displaystyle p=mv.} The unit of momentum

7885-402: The role of gun mount, and must similarly dissipate the gun's recoiling momentum over a longer period of time than the bullet travel-time in the barrel, in order not to injure the shooter. Hands, arms and shoulders have considerable strength and elasticity for this purpose, up to certain practical limits. Nevertheless, "perceived" recoil limits vary from shooter to shooter, depending on body size,

7980-644: The same speed. Adding the speed of the center of mass to both, we find that the body that was moving is now stopped and the other is moving away at speed v . The bodies have exchanged their velocities. Regardless of the velocities of the bodies, a switch to the center of mass frame leads us to the same conclusion. Therefore, the final velocities are given by v A 2 = v B 1 v B 2 = v A 1 . {\displaystyle {\begin{aligned}v_{A2}&=v_{B1}\\v_{B2}&=v_{A1}\,.\end{aligned}}} In general, when

8075-507: The same. Increasing the firearm weight also lowers recoil, again all else being the same. The following are base examples calculated through the Handloads.com free online calculator, and bullet and firearm data from respective reloading manuals (of medium/common loads) and manufacturer specs: In addition to the overall mass of the gun, reciprocating parts of the gun will affect how the shooter perceives recoil. While these parts are not part of

8170-418: The second reference frame is also an inertial frame and the accelerations are the same: a ′ = d v ′ d t = a . {\displaystyle a'={\frac {{\text{d}}v'}{{\text{d}}t}}=a\,.} Thus, momentum is conserved in both reference frames. Moreover, as long as the force has the same form, in both frames, Newton's second law

8265-400: The sides, increasing the lateral blast intensity (hence louder to the sides) but reducing the thrust from the forward-projection (thus less recoil). Similarly, recoil compensators divert the gas ejecta mostly upwards to counteract the muzzle rise . However, suppressors work on a different principle, not by vectoring the gas expansion laterally but instead by modulating the forward speed of

8360-569: The time period during which it acts will yield the momentum supplied by that force. The counter-recoil force must supply enough momentum to the firearm to bring it to a halt. This means that: ∫ 0 t cr F cr ( t ) d t = − m f v f = m p v p {\displaystyle \int _{0}^{t_{\text{cr}}}F_{\text{cr}}(t)\,dt=-m_{\text{f}}v_{\text{f}}=m_{\text{p}}v_{\text{p}}} where: A similar equation can be written for

8455-430: The total momentum of the system (ammunition, gun and shooter/shooting platform)) equals zero just as it did before the trigger was pulled. From a practical engineering perspective, therefore, through the mathematical application of conservation of momentum, it is possible to calculate a first approximation of a gun's recoil momentum and kinetic energy simply based on estimates of the projectile speed (and mass) coming out

8550-453: The use of recoil padding , individual pain tolerance, the weight of the firearm, and whether recoil buffering systems and muzzle devices ( muzzle brake or suppressor ) are employed. For this reason, establishing recoil safety standards for small arms remains challenging, in spite of the straightforward physics involved. There are two conservation laws at work when a gun is fired: conservation of momentum and conservation of energy . Recoil

8645-592: The usual system. However, the need to reliably achieve ignition at a single precise instant is a major practical difficulty with this system; and unlike the usual hydro-pneumatic system, soft-recoil systems do not easily deal with hangfires or misfires . One of the early guns to use this system was the French 65 mm mle.1906 ; it was also used by the World War II British PIAT man-portable anti-tank weapon. Recoilless rifles and rocket launchers exhaust gas to

8740-414: The vector sum, magnitude and direction, of the momentum of all the bodies involved does not change; that is, momentum of the system is conserved. This conservation of momentum is why gun recoil occurs in the opposite direction of bullet projection—the mass times velocity of the projectile (gas included) in the positive direction equals the mass times velocity of the gun in the negative direction. In summation,

8835-406: The very short time (typically only a few milliseconds) it is travelling inside the barrel. Meanwhile, the same pressures acting on the base of the projectile are acting on the rear face of the gun chamber, accelerating the gun rearward during firing with just the same force it is accelerating the projectile forward. This moves the gun rearward and generates the recoil momentum. This recoil momentum

8930-421: The very small distance of elastic deformation of the materials the gun and mount are made from, perhaps exceeding their strength limits. For example, placing the butt of a large caliber gun directly against a wall and pulling the trigger risks cracking both the gun stock and the surface of the wall. The recoil of a firearm, whether large or small, is a result of the law of conservation of momentum. Assuming that

9025-414: The way in which the shooter perceives the recoil, or kick , can have a significant impact on the shooter's experience and performance. For example, a gun that is said to "kick like a mule " is going to be approached with trepidation, and the shooter may anticipate the recoil and flinch in anticipation as the shot is released. This leads to the shooter jerking the trigger, rather than pulling it smoothly, and

#122877