Edit *solved*
Thanks for all your responses, I understand now
For clarification:
I’ve heard a lot about the limitations of personal armour, and how it will still result in broken ribs and potential blunt force organ damage even if it stops penetration.
But if it’s spreading the force out over a large area, (arguably larger than the stock of the gun) how is that impact worse than the equal and opposite impact of the rifle recoiling into the shooter’s shoulder?
In: Physics
When the bullet is fired, the explosion sends the bullet out, the gun pushes back, there is a loud noise and some heat. So energy is divided in many directions, while the travelling bullet is very concentrated energy
When the bullet exits the rifle, and the rifle kicks back, the force is controllably distributed to the rifle and the gunner.
But when the bullet hits the target, most of the momentum and the force is concentrated just at the point where the bullet hits, causing a bigger force peak.
There are two things involved here – momentum and energy.
Newton’s laws are all about momentum – momentum = mass × velocity. It tells us that the momentum will effect both people equally. The person firing the shot will be pushed back similarly to the way the person struck by the bullet will be pushed back.
But the other half is energy, and the equation for that is E=½mv^2. That velocity squared makes a big difference. The gun weighs a few kilograms, so the momentum law says its velocity, its speed, will be low. When you square a small number, you get another small number. The bullet weighs a few grams, so its velocity will be very high. When you square this very high number, you get a very, very high number.
It is this energy, deposited into the target, that causes the damage.
The bullet is acellerating in the barrel of the gun for what? 700mm say, then decellerating againt the target for 100mm. The force is concentrated in a tiny point and spread out through the armour, but is transfering that force in a much shorter time and over a much shorter distance
The gun accelerates the bullet over the length of the rifle’s barrel. On the “receiving” end, the bullet is decelerated by body armor in the length of the bullet. Since energy is mostly conserved, dumping the energy in a much shorter distance gives a much larger pulse of force.
When a bullet hit you you like to stop is in a few centimeters but the bullet accelerates for the length of the barrel length.
So put on a steel plate on your chest that has the same mass a the rifle. it might move 10mm when it stops the bullet.
Let’s fire the round from a bolt action rifle so the feed system with springs etc has no effect. A German WWII Karabiner 98k has a 600mm barrel
The bullet needs to stop in 1/60 of the length
Force = mass* acceleration
So if you assume that the acceleration is linear in both cases then
Velocity = acceleration* time
You also have
velocity =distance / time or time =distance/velocity
for linear acc
So for linear acceleration, you get
force= mass*velocity/time = mass*velocity^2/distance
So if time and therefore the distance the force increase. Half the stopping distance is twice the force 1/60 the distance 60x the force
This is why when you drive a car and use the brake to stop the force you feel is higher the faster you stop. A very quick stop like hitting a solid rock then the distance is just the length of the deforming part of the car this will kill you at high speed. Compare that to just pressing the clutch and stop by win resistance. You might travel a 1km and never feel the acceleration that stops you. This shows that distance to stop has a huge effect.
To stop in 1/60 of the length result in 60x higher in the impact than when fired if you hit a metal plate with the same mass
Because the mass of the riles and the plat is assumed identical the force on you will be 60x higher on impact
A rifle butt on the shoulder is supported by muscles and bones that quite tick and to much recoil will move it backward. You could get damaged muscles but that does kill you or even a broken bone. You arm also absorb a lot of force and move with it so even more mass than the rife moves back
Ribs are a lot weaker and they have to ben or move the whole abdomen back. The organs also if they get damages is a lot more of a problem then a muscle. If you would hold a rifle again you ribs with no arm support it would damage you a lot more then a rifle to the shoulder with arms support.
But no to the important part of the answer and that is most body armor is not the hard metal plate but soft kevlar or ceramics that deform. If you have a steel plate that the bullet does not penetrate the damage is minimal on you.
kevlar is effective against pistols and will deform a lot but spread out the force and stoping it from penetrating the skin. Look at [this video](https://www.youtube.com/watch?v=kgBzqzSLdlY) and you can see the deformation. This can break a rib.
The calculation above is not valid for the case because it a mass you hit identical to the rifle. A 9mm pistol like a Beretta 92S has a mass of 950g and 125mm barrel. If the bullet stop when penetrating 50mm into you when it moves, for example, 100g of west and tissue you still take about 20x the force on a more sensitive area of the body
If you use ceramic plate they will deform and the area that they force is concentrated will be similar to a rifle butt. Look at [This video](https://www.youtube.com/watch?v=RVpo4MwZoHM) where they fire on a steel plate and a steel-ceramic hybrid and you will see the effect on a clay backing. The ceramics will deform more then you expect
So the answer is that the force is implied on a sorter time. If you have a plate with the mass of a rife that is not penetrated the effect on you is minimal. But with material that deforms when hit you have smaller area then you expect, less mass to accelerate and shorter distance to do the same amount of acceleration. The result is more damaging impact on your body
The easiest way to explain is to imagine the firearm and operator as a single mass, the projectile as a single mass, and the gunpowder as the force. I’ll skip over the mechanics and physics of how firearms accomplish this, as that’s super complex mechanical and physics manipulation. Just know that they do accomplish this feat.
Now, when the gun powder goes off, it does apply equal force to the projectile and the firearm/operator. Newton’s laws state that an object’s ability to resist change is equally to it’s speed and mass. Speed is irrelevant as the projectile and firearm/operator, atleast initially, are not moving and therefore the same. But the mass of the two objects is not the same. A human male on average weighs 155 lbs, a rifle usually weighs about 15, and a projectile can be between 1-6 six ounces.
So at the point of discharge, just by mass alone, the firearm/operator is something like 27,000% more resistant to the force than the projectile. However, as the bullets accelerates, the force of the hot gas trying to escape starts taking the path of least resistance. The firearm/operator is resisting, but the projectile is not. Atleast not as much. This results in the firearm/operator kinda of “reflecting” the force back towards the projectile, causing it to accelerate even more. If you look at the ratio of energy absorbed by the projectile versus the energy absorbed by the firearm/operator, it’s something between 30-40 times greater on the projectile.
So if you took this step by step it would go something like this: the gun powder applies force, the firearm/operator resists the force, the force is reflected to the projectile, the projectile absorbs the force and accelerates.
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