How dense would something have to be to stop a bullet?

How big is the bullet? What shape is it? Does the material have a grainy structure or a homogenous gelatin structure?

 

kevlar isn't particularly dense, and it will easily stop a bullet. Why do you think density is a good indicator of a material's ability to stop a bullet?

 

that would be pretty cool. uh... If you know the range of a weapon, then you could work out the mass of air it travelled through. And then you can calculate the equivalent density of air which would give the same mass over a different volume.

But this would probably not work very well, since the air resistance forces are going to be non-linear for sure. Maybe assuming a force which goes as velocity squared would be better.

 

Not compressed at all. Our atmosphere can stop a bullet, it just takes a while. You need more information to answer this kind of thing. Do you want to stop it in a mile? In an inch? In a nanometer?

 

High density material employs conservation of energy and conservation of momentum to spread the impact load. The bullet fragments becomes part of the high density material.

A bullet delivers a high kinetic energy to a small area on the target over a very short period of time. To stop a bullet you must quickly convert that KE to heat, and / or spread the material over a much greater area. Some bullets are designed to focus their energy. Although they will be vaporised on impact there will still be a significant KE in that vapour. The vapour may be toxic.

 

He kinda was...

 

Ahh, my bad.

 

Stopping a bullet requires two things:
1. The energy of the bullet must be expended
2. The momentum of the bullet must be absorbed

A bale of cotton will do both. So will a pile of dirt.

 

The bullet will decelerate but never quite stop. The profile and weight of the bullet will make a big difference. You must redefine “stop” and provide some idea of the distance that is available to decelerate the bullet to a “safe” speed.

 

If it's of interest you could estimate the acceleration the bullet must undertake in order to stop in 3 inches. If we assume constant acceleration (which might not be a valid assumption) then we can use SUVAT equation...

V²= U² + 2as

where

V = final velocity (eg 0)
U = initial velocity (304 meters/sec)
s = distance (eg 3 inches = 0.075 meters)
a = acceleration.

Rearranging the equation gives

a= (V² - U²)/2s

a = -304²/0.150
= -62,000 ms^-2

Roughly 6200g

 

I suspect a bullet would pass through 75mm of liquid nitrogen (at atmospheric pressure but that's not quite what you are asking).