Why is recoil easier to control on a more massive gun compared to a smaller gun with the same bullet. Presumably the bullet leaves both guns with the same momentum, but the larger gun seems easier to control. Since the momentum you have to control is the same in both cases, why do we perceive less recoil on a bigger gun?
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asked Jul 28, 2015 at 0:52
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4$\begingroup$ For every action there is an equal and opposite reaction. Fire a gun and some momentum is given to the bullet. The same momentum must be given to the gun, going the other direction. A light gun will have to move faster to absorb the same momentum as the heavier gun, and what you feel is that fast movement. (Imagine what it would be like if the gun weighed the same as the bullet.) $\endgroup$– Hot LicksJul 28, 2015 at 1:10
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3 Answers
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Felt recoil is partly a matter of momentum, partly a matter of force.
When a bullet with mass m leaves a gun with a velocity v, the gun must have an equal-but-opposed momentum MV, where M is the mass of the gun and V is the recoil velocity, or $$mv + MV = 0$$. If there are two possible gun sizes, $M_1$ and $M_2$, each will have a recoil velocity $V_1$ and $V_2$. If, for instance, $M_2 = 2M_1$, $$M_1 V_1 = M_2 V_2$$ and $$V_1 = 2V_2$$ Why does this matter? Consider kinetic energy. Let $K_1$ be the kinetic energy of $M_1$, and $K_2$ is that of $M_2$. Then $$\frac {K_1}{K_2} = \frac {\frac{M_1{V_1}^2}{2}}{\frac{M_2{V_2}^2}{2}} = \frac{M_1}{M_2} {(\frac{V_1}{V_2})}^2 = \frac{1}{2} 2^2 = 2$$
So the lighter gun has twice the kinetic energy of the heavier gun and this shows up in two ways. First, since both guns need to stop in about the same distance, the force applied to the lighter gun must be greater than that applied to the heavier. By Newton's First Law, this means that the lighter gun pushes harder on the hand or shoulder of the shooter. Second, the duration of acceleration must be smaller for the lighter gun, since $$S_1 = \frac{a_1{t_1}^2}{2} = \frac{{V_1}t_1}{2} = S_2 = \frac{a_2{t_2}^2}{2} = \frac{{V_2}t_2}{2} $$and $${V_1}{t_1} = {V_2}{t_2}$$ or $$ \frac {t_1}{t_2} = \frac{V_2}{V_1} = \frac{1}{2}$$ So not only is the recoil force greater for the lighter gun, it lasts a shorter time and is therefore "sharper".
answered Jul 28, 2015 at 1:57
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$\begingroup$ +1 By sharper, do you mean a shorter but stronger impulse, which the body detects more acutely than a longer but softer blow? $\endgroup$– user81619Jul 28, 2015 at 2:04
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$\begingroup$ @AcidJazz - You bet. Think of the difference between being hit by a car at 15 miles an hour (the equivalent of running a 4-minute mile and hitting a wall) and 30 miles an hour. Ouch. Peak forces are much greater. $\endgroup$ Jul 28, 2015 at 2:10
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Newtons 3rd Law:
When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.
Momentum is product of mass and velocity. The heavier gun has more mass, so, for the same momentum, it must have less "backwards" velocity, so less felt recoil.
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$\begingroup$ But all of the momentum that goes into the gun ultimately goes to my body, doesn't it? In that case, shouldn't the recoil felt be the same? $\endgroup$ Jul 28, 2015 at 5:08
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$\begingroup$ @Dargscisyhp Momentum, yes. But that's not the only important thing by far. For example, the kinetic energy is proportional to velocity squared - so the same amount of momentum applied to a heavier object will result in less kinetic energy than with a lighter object. But more importantly, recoil is a shock of sorts - you're balancing against a sudden force that last a tiny bit and disappears a moment later. And that's the point - you can easily handle the energy and momentum thrown at you, but the stronger the shock, the harder it is to maintain control of the gun. $\endgroup$– LuaanJul 28, 2015 at 6:55
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1$\begingroup$ @Dargscisyhp In other words, yes, ultimately the recoil is the same - if you define recoil as the momentum the gun transfers to you. That's why no matter how smart, a big enough gun shooting a big enough projectile fast enough will throw you around. But that's not important for rifles and handguns - in a handgun, your hands and arms mostly absorb the blow. Rifles give you extra support from your shoulders, back, better stance... allowing you to easily absorb recoil that would make you lose control of a handgun - you have more mass immediately countering the recoil. $\endgroup$– LuaanJul 28, 2015 at 6:59
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1$\begingroup$ But if you continue the progression, other factors come into play. For instance, a WWII rifle grenade had to be fired from standing, rather than prone. Fired standing up the recoil would knock the firer backwards, but that was recoverable with practice. Firing prone could break the firer's shoulder. $\endgroup$ Jul 28, 2015 at 14:54
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The larger firearm has more mass, and therefore more inertia for the recoil momentum of the bullet to overcome.
Also, small firearms may be more difficult to secure a good grip on.
