I've had this discussion on this subreddit countless times and I figure it's worth exploring the physics. The conversation goes something like:
"I’m sure the armour will stop the bullet, but it will leave a puddle of human on the inside."
On the simple level, conservation of momentum tells us that the momentum imparted by the bullet on the person, assuming their armour stops the round and the momentum is spread out over some area, should be no more than the recoil experienced by the person firing the firearm. In reality, modern firearms have various mechanisms to reduce the amount of recoil experienced, so we have to go deeper.
We can sort of put this question to rest with simple physics. The momentum of a bullet, once in the air, is mass * velocity. If the bullet impacts the armour and stops, this is the maximum momentum imparted. (If the bullet impacts but bounces back at the same speed, the maximum momentum is double that of the bullet, but it's unlikely so we'll ignore it for now)
The mass of a 5.56x45mm round is about 3.56 grams (I can't tell if that's the weight of the bullet+cartridge or just bullet – if someone can let me know I can edit this accordingly – let's assume it's the bullet for now for an upper bound). The maximum velocity is around 993 m/s.
So the momentum of the round is 3.535 kg.m/s. Ok, but what does this mean?
We can convert this to the speed at which an 90kg target (human plus body armour) will be moving after impact with the equation momentum/mass(t).
3.535/90 = 0.039 m/s
Not much, and certainly not enough to 'turn someone into a puddle'. This video comes to mind (do NOT try this for the love of Zeus).
What about something bigger, like a .50 cal round?
The mass of a .50 cal round is about 42 grams (again unsure if cartridge+bullet or just bullet). The maximum velocity is around 928 m/s.
So the momentum of a .50 cal round is 38.976 kg.m/s.
The velocity of the 90 kg target is therefore 38.976/90 = 0.433 m/s
Sizeable, but not enough to 'turn someone into a puddle'. Heck, a car going 10 km/hr has ~ 100 times the momentum of a .50 cal round.
Happy for someone to contest the maths/physics I've presented, but no one has presented any physics any time I've had this discussion on the EFT subreddit, they just downvote me and say I'm clearly wrong.
Remember, the main source of a bullets damage is the fact that it's small so its force is imparted over a small area and penetrates the body. This is how body armour is designed to work. From Wikipedia:
Contusions – in simple terms, big bruises. Not the catastrophic damage people seem to think occurs thanks to the influence of movies and popular representation.
*If* it's the case that the force is spread over an area of body armour, I stand my ground that unless it's something heavier/faster than a .50 cal round, they won't 'turn in to a puddle' at all.
If anyone has a reference for how much recoil force is reduced via the various mechanisms in modern firearms I'd love to see it.