physics question about bullets

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A 2000 lb car is rolling at 10 mph
A 5000 lb car is rolling at 10 mph

Which would be harder to stop?
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The heavier car would because it has more KE. but what if you sped the lighter car up to match the KE of the heavier car , say to 20 mph ( I'm guessing) then which is harder to stop?

In the bullet situation, because velocities are different but the energy is the same I'm starting to think they would penetrate to equal distance.
 
At that velocity, the bee would pass through the locomotive like it was cardboard... It wouldn't slow down the locomotive at all.
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Yep... Defense research (DARPA research) in the 1960's-1970's showed that small projectiles with very high velocity could defeat the best steel armor. Even a small plastic pellet of a few grams could blast a fist size hole through 6 inches of armor plate, if accelerated above 80,000 ft/sec.

Even high velocity water droplets can punch holes through hardened steel... thus, water jet cutters :)

This DARPA research led to better shaped charge designs for HEAT anti-tank weapons, and to better sabot designs for the M1 Abrams.
 
If you have the same amount of (muzzle)energy 500 ft/lbs the outcome will be the same only with bullets that are equal in size/shape and not expansion or deformation. Now since the heavier bullet is longer, there is going to be drag resistent along the longer portion of the bullet.
 
Guys....

Stop talking about kinetic energy.

Kinetic energy is not responsible for penetration.

Momentum is responsible for penetration.

Kinetic energy can be (and is) lost to all sorts of other, non-movement sources. Friction, bullet deformation, heat, etc.

Momentum is a vector quantity and always conserved in both quantity and direction.

Momentum is responsible for penetration, NOT kinetic energy.
 
What? No.

Muzzle energy is kinetic energy.

The bullet also has momentum. Momentum is not "stored" kinetic energy. Momentum is momentum.

Momentum is responsible for penetration. Kinetic energy is not.

If kinetic energy gave penetration, lighter, faster bullets would penetrate more than slower, heavier bullets.

They do not.

Why?

Because momentum is responsible for penetration, kinetic energy is not.
 
OK, so is the 180 grain bullet going to have more momentum than the lighter one, even though they are producing the same KE? The lighter bullet is moving faster.
 
The heavier car would because it has more KE. but what if you sped the lighter car up to match the KE of the heavier car , say to 20 mph ( I'm guessing) then which is harder to stop?

In the bullet situation, because velocities are different but the energy is the same I'm starting to think they would penetrate to equal distance.
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The one harder to stop it the one with the highest momentum.

Kinetic energy is one half mass times velocity squared .5(m*V^2)
Momentum is mass times velocity m*v

I will convert and use SI units...

5000lb=2267.96kg
10mph=4.47m/s

Kinetic energy is 22.65794kJ (I feel the extra significant digits are important due to the conversion of joules to kilojoules)

So we need to match that with the lighter car.

2000lb=907.18kg

22.65794kJ=22657.94J

Lets work backwards...

22657.94*2=45315.88
45315.88/907.18=49.95
square root of 49.95=7.07m/s
7.07m/s=15.81mph

5000lb car at 10mph has the same energy as a 2000lb car moving 15.81mph


Now lets look at the Momentum (we will ignore the vector quantity aspect, as it is unimportant to our purposes at the moment, we only need the magnitude, not the vector.)

momentum of the 5000lb car
2267.96kg*4.47m/s= 10137.78 kg per m/s

momentum of the 2000lb car
907.18kg*7.07m/s=6413.76 kg per m/s


The 5000lb car has a much larger momentum, and therefore a much larger force will be needed to stop it.

A you can see from the calculations... while energy is the same, momentum is not... To equal momentum you will need much more velocity. Of course energy will also be higher...

The 2000lb car would need to be moving 11.18m/s or 25mph to equal the momentum of the 5000lb car. More than double the speed...



Since force is quantified as acceleration or deceleration... it is has a function of time. Example- 5m/s per second

And since Momentum is a vector quantity, time means movement in the direction of the vector. So given a constant decelerating force, the object with the highest momentum will travel farther before stopping.


This is why heavy for caliber bullets penetrate farther than lighter ones, even ones with higher kinetic energy.

It is also why in hollowpoints that they are more consistent. As every shot fired is slightly different, and forces can be different... more momentum helps ensure better more consistent penetration.

Sectional density comes into play when comparing different calibers. As well as frontal surface area and drag characteristics... Fontal area is larger on higher calibers, and that affects the drag and therefore the deceleration force acting on it.

Any friction between a longer heavier bullet and a shorter lighter one in a given caliber will be minor overall. Given solid non expanding projectiles that is.

Sectional density, frontal surface area and drag, also comes into play in expanding bullets because they change as the round expands.

As a round expands, the decelerating force acting upon it also increases.



This all combines into a very complex set of variables that determine final penetration.
 
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I wish I had some 180 fmj bullet for my loading. I would load some 125fmj and 180 so they had the same muzzle energy. Then line up some jugs of water and do a penetration test.

Maybe someone here has already did that.
 
A rather narrow question, which has gotten a lot of broad replies.

10mm bullet, 180 and 155gr, speed adjusted to give the same ME (muzzle energy). Which one will penetrate more?

Inertia (call it momentum) says the heavier one will tend to keep going more than the lighter one.

The shape of the nose will play a part, as well.
Just look at how spitzer vs. round nose rifle bullets shed velocity, when all they are penetrating is the air.

Of course, what you are penetrating makes a difference too.
 
Well... Inertia is an inherent quality of a mass, it has it when moving or still.

Inertia and momentum are kind of related, in that inertia is a bit of the driving factor.


Inertia is not related to velocity, so velocity can decrease and inertia will stay constant. (so long as the mass stays constant)


Loosing mass hurts inertia and that hurts resistance to changes in momentum.
 
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Thank you both marine6680 and 44amp.
What I'm seeing from all the answers is that all else being equal (shape, me/ke, caliber, target media) that the heavier bullet will penetrate farther.
 
You got it.

Of course "all things being equal" is very difficult to achieve in something like a real life bullet fired into real test media.

Bullets do not have perfect repeatability... even with careful loading, there is a velocity spread. Mass can be the same to a high degree of precision, if using a very sensitive measuring device to sort them. (maybe needing a little manipulation and material removal to get a few in line)

Test media may not be uniformly even in density.


But most of those effects are minor overall if care is taken to minimize them. Like making one large batch of test media to make all test blocks, and keeping them all the same temps in testing.


But still, the effects are not as important if media is calibrated properly... even tests between different people in different ambient conditions can be compared. As long as velocities are not too drastically effected by firearm selection and ambient conditions.
 
Of course "all things being equal" is very difficult to achieve in something like a real life bullet fired into real test media
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Understood. It was simply an arbitrary starting point from which to begin looking at bullet performances.
 
Energy is ½mV²
Momentum is simple mV
Although they are related, they are not the same.

At the ordinary bullet regimes we are talking about now:

- " mV " is responsible for penetration
- " ½mV² " is responsible for destruction along the way



Note: In the hypervelocity regimes where plastic pellets blast through heavy armor, the ½mV² factor immediately destroys the armor material upon contact -- turns it into a plasma -- and then continues to convert/direct that plasma itself through the rest of the armor plate. (Note also that the target material itself plays a significant role)
 
We have two indian spears both are identical, but A weighs 4 lbs and B weighs 8 lbs.
Drop them in the mud, both from the same elevation 10 feet and B will have more penenatration-more ft/lb of energy.
Then move spear A to higher elevations until it equals the energy or depth of spear B.

I'm not going to do the math, but A might be 25 ft higher than B. Allowing A to store its energy in form of velocity
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Lift them both to a 10 ft elevation and the 4 lb spear will have 40 ft lb of potential energy and the 8 lb spear will have 80 ft lb of potential energy.
As they fall down, the potential energy is converted into kinetic energy. At elevation zero, the potential energy is zero and the 4 lb spear is traveling at a velocity that gives it 40 ft-lb of kinetic energy and the 8 lb spear is traveling at a velocity that gives it 80 ft-lb of kinetic energy.
To give the 4 lb spear 80 ft-lb of potential energy, you have to raise it 20 ft in the air.
A roller coaster has potential energy from its elevation and kinetic energy from its velocity. Assuming zero friction and no braking or energy input after the initial lift, the kinetic energy + the potential energy is constant. As the track goes up, the kinetic energy is converted into potential energy as the cars slow down and when the track goes down, the potential energy is converted into kinetic energy as the car speeds up, but the sum of the two is constant.
Bernoulli's equations are also based on this principle. Pressure = potential energy, velocity = kinetic energy, as one goes up, the other goes down.
 
I wonder why my 357 mag 125 fmj will go through 3/4" bullet resistant glass and the 180 grn bullet doesn't.

Please explain
 
Kinetic Energy can also be expressed in terms of momentum....

KE=1/2mv^2
p=mv

Solve for v...

v=p/m

Replace "v" in KE formula...

KE= 1/2m(p/m)^2

Simplify...

KE=p^2/2m
 
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