Question about transfer of Energy and Momentum in Ballistics

[Assaltwaffle]

My apologies if the prefix is too high of complexity. I don't know where this would fall, difficulty or academically speaking.

While it may be surprising to some given Hollywood's portrayal of it in movies, if a person in wearing hard bulletproof armor is struck by a projectile, the person is not thrown back or knocked down so long as he is not simply startled into falling over; for many it is said to feel comparable to a punch.
Yet a punch has a massively lower amount of energy compared to pretty much any bullet (about 100 joules in an untrained punch vs 1,800 joules in a round of 5.56x45mm). So, my question is this: why, mechanically, is the person not knocked over despite such a high amount of energy striking the individual relative to a weaker force that can achieve this?
My initial belief was that the plate itself allows the bullet to dissipate energy by striking it through the work of bending the metal or cracking the ceramic, thus "spending" some of the forward kinetic energy before spreading the rest across a larger surface area. However, after an exhaustive long reddit thread on the topic, I am now lead to think that it is not the transfer of energy than knocks someone over from a collision, but rather a transfer of momentum, and therefore the reason for the bullet not knocking someone over is because it has lower momentum than a strong punch or kick despite its high energy.

Could someone help clarify the reasoning behind this? Is the person not thrown back because the plate dissipates the energy by giving the bullet's energy work to do, or is it because only a transfer of momentum can knock someone down, and bullets have low momentum? Any help would be appreciated!

 

[Baluncore]

Is the person not thrown back because the plate dissipates the energy by giving the bullet's energy work to do, or is it because only a transfer of momentum can knock someone down, and bullets have low momentum?

Hollywood and the theatre do not follow the laws of physics.

 

[Assaltwaffle]

Hollywood and the theatre do not follow the laws of physics.

I'm aware. Hence why I asked why they are not thrown down since the amount of energy transferred into a plate by a bullet is quite a lot. As I said, about 18x more than an average punch and even done over a far smaller period of time.
My apologies if you thought I was meaning that Hollywood is real and people do fly back. They very clearly do not and I just want to know the exact mechanics of why, particularly involving an armor plate.

 

[Baluncore]

A punch is low velocity, the recipient can see and react by moving away from or deflect the punch. The target instinctively deflects the incident energy to minimise injury, in a way that removes them from the next punch.
A fast projectile will impart energy that will both deform the armor and deliver the remaining momentum to the armored body.

 

[Assaltwaffle]

A punch is low velocity, the recipient can see and react by moving away from or deflect the punch. The target instinctively deflects the incident energy to minimise injury, in a way that removes them from the next punch.
A fast projectile will impart energy that will both deform the armor and deliver the remaining momentum to the armored body.

I've already stated in the OP that the only reason for someone falling over from a bullet being someone gets startled from not knowing it's coming. That doesn't really answer my question, though.
If someone is punched or kicked hard enough, the person can be physically shoved back even if prepared. With a bullet, if the person is prepared to take the hit, it does not shove the individual back.

So a more concise question would be both: what knocks someone down, energy or momentum, and, also, where does that energy in the bullet go, assuming the plate retains 100% of the bullet? How does a kick or punch of inferior energy knock someone back, but a bullet with superior energy fail to do this? Is it because the plate's deformation uses up the energy, or because it's momentum is inferior?

 

[Baluncore]

The target might be able to stand up to counter the impact, but only if they knew it was coming and had great confidence in their armour.

We do not know how much energy will be lost to bullet and armour deformation and the generation of heat.
I would calculate the final velocity of the target by momentum.

 

[Assaltwaffle]

The target might be able to stand up to counter the impact, but only if they knew it was coming and had great confidence in their armour.

We do not know how much energy will be lost to bullet and armour deformation and the generation of heat.
I would calculate the final velocity of the target by momentum.

So it is, in the end, about the armor's reduction of the transfer of energy and not about momentum being inherently low?

 

[jbriggs444]

So it is, in the end, about the armor's reduction of the transfer of energy and not about momentum being inherently low?

If you are talking about knocking someone down by force then momentum is the relevant quantity. Armor does nothing to reduce the momentum of an impact. It can dissipate the energy and spread the force, but momentum is a conserved quantity. It cannot be disposed of so easily.

 

[Assaltwaffle]

If you are talking about knocking someone down by force then momentum is the relevant quantity. Armor does nothing to reduce the momentum of an impact. It can dissipate the energy and spread the force, but momentum is a conserved quantity. It cannot be disposed of so easily.

So a bullet striking armor doesn't knock someone down because of the bullet's low momentum? Am I getting that right? It's not the energy transfer that moves someone?

 

[jbriggs444]

So a bullet striking armor doesn't knock someone down because of the bullet's low momentum? Am I getting that right? It's not the energy transfer that moves someone?

Bear in mind that you are asking us to explain something that is fictional in origin. But yes, to the extent that the bullet's momentum is not enough to push the [well-prepared] shooter over from recoil, it is not enough to push over the [surprised] target from impact.

Then again, preparation and surprise are very different situations.

Energy by itself has essentially no "push" to it. You can burn a hole in a target with a laser without bowling it over.

Edit: Without a muzzle brake, the momentum transferred by recoil to the shooter will, in general be greater than the momentum transferred to the target by impact. The muzzle gasses add to the recoil but do not contribute to the impact.

 

[sysprog]

Maybe consider that the momentum imparted to the target can't be more than that imparted to the shooter.

 

[Assaltwaffle]

Bear in mind that you are asking us to explain something that is fictional in origin. But yes, to the extent that the bullet's momentum is not enough to push the [well-prepared] shooter over from recoil, it is not enough to push over the [surprised] target from impact.

Then again, preparation and surprise are very different situations.

Energy by itself has essentially no "push" to it. You can burn a hole in a target with a laser without bowling it over.

Edit: Without a muzzle brake, the momentum transferred by recoil to the shooter will, in general be greater than the momentum transferred to the target by impact. The muzzle gasses add to the recoil but do not contribute to the impact.

I suppose I'm just confused about what all the energy "does". If not push, then what does 1,800 joules of energy do when it strikes the target if only momentum can push? The energy has to do and I thought the only work the energy could do is the bend or brake the plate and/or push it.
I know that energy is not necessarily kinetic and doesn't always push, but you're saying that kinetic energy even given a specific direction, such as is found in a bullet does not push, only momentum?

 

[Assaltwaffle]

Maybe consider that the momentum imparted to the target can't be more than that imparted to the shooter.

I think I need more clarification on the differences in what momentum and kinetic energy do in regards to moving or doing anything at all to an object. It seems to be said here that momentum alone can push, not energy. If that's the case then where does that 1,800 joules of energy go? Is not energy the capacity of an object to do work? Work such as moving a target body or shattering a ceramic plate?

 

[jbriggs444]

what all the energy "does". If not push

Crack, bend, crush, melt, sever, tear, shock, spall, penetrate. Yes, there is push, but the amount of push is quantified accurately by momentum, not by energy.

 

[Assaltwaffle]

Crack, bend, crush, melt, sever, tear, shock, spall, penetrate. Yes, there is push, but the amount of push is quantified accurately by momentum, not by energy.

So the energy can push but only proportionately to its momentum and all energy beyond that must do something else, like bend or crush?

 

[sysprog]

Maybe do a search on third law of motion.

 

[Filip Larsen]

It is fairly easy to calculate the momentum of a bullet and compare it with, say, a 1 kg "momentum-equivalent" impact which may be easier to imagine the effect of. For instance, a 9 mm bullet with mass 8 g and muzzle speed of 400 m/s carries an impulse of up to 3.2 kg m/s which momentum-wise is equivalent to a 1 kg object moving at 3.2 m/s. And going further, a 80 kg person standing still on a frictionless floor receiving this impulse will have speed of 4 cm/s after the impact.

 

[Assaltwaffle]

It is fairly easy to calculate the momentum of a bullet and compare it with, say, a 1 kg "momentum-equivalent" impact which may be easier to imagine the effect of. For instance, a 9 mm bullet with mass 8 g and muzzle speed of 400 m/s carries an impulse of up to 3.2 kg m/s which momentum-wise is equivalent to a 1 kg object moving at 3.2 m/s.

I'm not asking how to calculate momentum. That is easily Google-able. I'm confused about what principles govern how one is pushed back by a projectile and what the energy of the projectile does.
It seems to be indicated by some others in the thread that energy can only transfer a pushing force proportionate to the object's momentum. However, the project has vastly more kinetic energy than that. Take your momentum equivalent. Yes, momentum wise it's equal to a 1kg object moving at 3.2 m/s. But its kinetic energy is vastly, vastly higher; the KE of a 9mm projectile of 8 grams and a velocity of 400 m/s carries 480 joules of energy. A 1kg rock moving at 3.2 m/s carries 5 joules.

So what does that energy do, if not push? If only momentum can result in something being pushed, where does all the extra energy being carried by the projectile go? Is there simply a cap on how much energy a projectile can impart based on its energy and, after that cap, it must only bend, break, or tear beyond that?

What principle governs such a cap (if I'm not grievously misunderstanding) and why is only momentum capable of imparting physical motion to an object when it seems like energy is the capacity for work and there is WAY more energy than is used to move the plate and person?

 

[wrobel]

why, mechanically, is the person not knocked over despite such a high amount of energy striking the individual relative to a weaker force that can achieve this?

first the linear momentum is a matter here not the energy
and the second a man is not knocked over by the fist blow ; he loses consciousness and falls down

 

[Assaltwaffle]

first the linear momentum is a matter here not the energy
and the second a man is not knocked over by the fist blow ; he loses consciousness and falls down

I suppose it's easier to imagine a hanging punching bag with an armor plate instead of a person.

So only linear momentum determines how much an object, such as a punching bag, is pushed? If that's the case, then where does the extra energy go/what does the energy do? The projectile is carrying far more energy than what is translated to rearward movement of the bag. It initially seems strange than kinetic energy seems to not matter when moving an object back appears to be work that should be done by energy.

 

[Filip Larsen]

So what does that energy do, if not push?

The more energy the more "potential for work", or in case of bullet impacts, damage to the body. Considering the energy of that 9 mm bullet from before it is energy-wise equivalent to being hit by a 1kg object moving at roughly 36 m/s (while standing up against a wall). It is not hard to imagine that a 1 kg at 36 m/s hitting, say, your head in a soft spot very likely will be lethal.

 

[Assaltwaffle]

The more energy the more "potential for work", or in case of bullet impacts, damage to the body. Considering the energy of that 9 mm bullet from before it is energy-wise equivalent to being hit by a 1kg object moving at roughly 36 m/s (while standing up against a wall). It is not hard to imagine that a 1 kg at 36 m/s hitting, say, your head in a soft spot very likely will be lethal.

That I understand. Is not pushing an object work? Why does the energy from the bullet only transfer a small amount of its energy into a pushing force?
Using the 1kg rock again, a 1kg rock at 36 m/s hitting a punching bag would likely push it back quite a bit, and certainly more than only 9mm of equal energy. Why does the rock, despite having equal energy, push the bag a greater distance than the 9mm?
Is only momentum valid when determining pushing force?

 

[Filip Larsen]

The projectile is carrying far more energy than what is translated to rearward movement of the bag.

In (simple) inelastic collisions some fraction of the kinetic energy is converted into inner energy, e.g. heat. This is the "damage potential" I mentioned. The higher the speed of the impactor is (for the same kinetic energy), the more of that kinetic energy will be lost in a fully inelastic collision.

 

[Assaltwaffle]

In (simple) in-elastic collisions some fraction of the kinetic energy is converted into inner energy, e.g. heat. This is the "damage potential" I mentioned. The higher the speed of the impactor is (for the same kinetic energy), the more of that kinetic energy will be lost in a fully in-elastic collision.

So the higher energy projectile only moves back the object proportionate to its momentum and all excess energy must be used to do different types of work, such as becoming heat or bending/cracking the armor plate or tearing through flesh? So momentum determines the physical shoving force and all energy beyond that must be expended in others ways. Is that a proper understanding?

 

[Filip Larsen]

So momentum determines the physical shoving force and all energy beyond that must be expended in others ways

If you leave out "beyond that" it is close to being correct in a physical sense. That is, momentum determines the "push" (physical name for this is impulse) and energy determines the potential for doing "damage" to whatever is hit.

Note that the mass ratio is rather important. In an inelastic collision the loss of energy is given by $$\Delta E = \frac{M}{m+M} E_0,$$ where $m$ is the impactor (bullet) mass, $M$ is the impacted object originally standing still, and $E_0$ is the energy of the impactor. Since the mass of bullets in general are insignificant compared to the mass of a normal person it means nearly all of the kinetic energy of a bullet that stays in the body ends up being used "for damage".

 

[Assaltwaffle]

If you leave out "beyond that" it is close to being correct in a physical sense. That is, momentum determines the "push" (physical name is "impulse") and energy determines the potential for doing "damage" to whatever is hit.

Note that the mass ratio is rather important. In an inelastic collision the loss of energy is given by $$\Delta E = \frac{M}{m+M} E_0,$$ where $m$ is the impactor (bullet) mass, $M$ is the impacted object originally standing still, and $E_0$ is the energy of the impactor. Since the mass of bullets in general are insignificant compared to the mass of a normal person it means nearly all of the kinetic energy of a bullet that stays in the body ends up being used "for damage".

I think see now. It seems I had a fundamental misunderstanding about what determined the "impulse" and how what work the bullet did is determined.

So, if the round is fully stopped by a plate or fully capture by a body, the full momentum is transferred as an impulse. Energy does this work and transfers energy accordingly to make that impulse, but the rest of the energy of the projectile is lost through doing "damage", such as bending metal, breaking ceramic, or tearing flesh.
If I understand it, it would seem to reason that if the plate failed because it could not resist the projectile and the object passed through the target, that the impulse would be less because the round still had energy and momentum remaining upon exit and did not transfer all of its momentum. Is that correct?

 

[Drakkith]

So the energy can push but only proportionately to its momentum and all energy beyond that must do something else, like bend or crush?

Not really. Let's look at some examples. Note that some quantities have been rounded and may not add exactly up.

First, consider two elastic collisions. The first of which has a 1 kg projectile impacting a 100 kg stationary target at 100 m/s. The projectile rebounds at -98.02 m/s and the target is sent backwards at 1.98 m/s. This means that out of the original 5,000 J of kinetic energy the projectile keeps 4,804 J and the target has 196 J. The initial momentum of the projectile is +100 kg*m/s and -98 after (momentum is a vector quantity). This is a change of 198 kg*m/s, which means that the target must have that amount in order for momentum to be conserved. And in fact it does have exactly that amount.

To summarize:
For a projectile moving at 100 m/s. KE of 5,000 J and momentum of 100 kg*m/s.
Final projectile velocity, KE and momentum: -98.02 m/s, 4804 J and -98 kg*m/s
Final target velocity, KE and momentum: 1.98 m/s, 196 J and 198 kg*m/s.

Let's now increase the velocity to 200 m/s. Initial projectile KE and momentum is 20,000 J and 200 kg*m/s.
Final projectile velocity, KE and momentum: -196 m/s, 19,208 J and -196 kg*m/s
Final target velocity, KE and momentum: 3.96 m/s, 792 J and 396 kg*m/s.

Now for two fully inelastic collisions (no rebound) with the same values.
For a projectile moving at 100 m/s initially. KE is 5,000 J.
Final projectile velocity, KE and momentum: 0.99 m/s, 0.49 J, and 0.99 kg*m/s.
Final target velocity, KE, and momentum: 0.99 m/s, 49 J, and 99 kg*m/s.

For a projectile moving at 200 m/s initially. KE is 20,000 J.
Final projectile velocity, KE and momentum: 1.98 m/s, 1.96 J, and 1.98 kg*m/s.
Final target velocity, KE, and momentum: 1.98 m/s, 196 J, and 198 kg*m/s.

As shown, the more elastic the collisions is, the more the target is shoved backwards. Conceptually you can imagine catching a baseball in space and then having to throw it back. The catch will accelerate you backwards a bit and the throw will do it again. Note that in an elastic collision all of the kinetic energy is retained and shared between the projectile and target, so a perfectly elastic collision can't damage either the target or the projectile since that would require energy. Needless to say, perfectly elastic collisions don't exist in real life.

Now let's look at an inelastic collision involving a 10 kg mass moving at 100 m/s and compare it to our first inelastic collision above.
Initial projectile velocity, KE, and momentum: 100 m/s, 50,000 J, 1,000 kg*m/s.
Final projectile velocity, KE, and momentum: 9.09 m/s, 413 J, 90.9 kg*m/s.
Final target velocity, KE, and momentum: 9.09 m/s, 4,131 J, 909 kg*m/s

As you can see, in the 1 kg collision virtually all of the KE was lost. Only about 1% was retained by the target and projectile. But in the 10 kg collision about 9% was retained.

Let's also look at a 20 kg projectile traveling at 50 m/s since it also has 1,000 kg*m/s of momentum like the example just above. (Inelastic collision again)
Initial projectile velocity, KE, and momentum: 50 m/s, 25,000 J, 1,000 kg*m/s.
Final projectile velocity, KE, and momentum: 8.33 m/s, 693.88 J, 166.6 kg*m/s.
Final target velocity, KE, and momentum: 8.33 m/s, 3,469 J, 833 kg*m/s

Even though the momentum was the same, the final velocity was less than before. This makes sense, as if we continue to increase the mass, the velocity must drop to keep the momentum the same. At the extreme, the initial velocity approaches zero as our mass increases without bound.

Finally, let's compare a projectile with the same kinetic energy as our 10 kg projectile but different momentums. (Inelastic again)
Projectile mass is 20 kg.
Initial projectile velocity, KE, and momentum: 70.75 m/s, 50,000 J, 1415 kg*m/s.
Final projectile velocity, KE, and momentum: 11.79 m/s, 1,390 J, 235.8 kg*m/s.
Final target velocity, KE, and momentum: 11.79 m/s, 6,950 J, 1179 kg*m/s

So momentum determines the physical shoving force and all energy beyond that must be expended in others ways. Is that a proper understanding?

As I've shown, it's a bit more complicated than that. In my final 3 examples, two of the projectiles have identical momentums and yet still don't end up with identical final velocities. Two of them had identical kinetic energies and different final velocities. And two of them had identical masses and different final velocities.

Keep in mind these are idealized scenarios. Real collisions are somewhere in between elastic and inelastic collisions and don't typically happen in a frictionless vacuum.

 

[Filip Larsen]

Energy does this work and transfers energy accordingly to make that impulse

You still seem to mix energy and momentum a bit. You should consider energy and momentum as two independent quantities that "just happens" to be related via mass and velocity. The momentum of the bullet (minus loss during flight) is transferred from the shooter to the target and, independently of this, some part of the energy released by the gun powder is first transferred to the bullet in the barrel and then transferred (except for loss during flight) to the target.

If I understand it, it would seem to reason that if the plate failed because it could not resist the projectile and the object passed through the target, that the impulse would be less because the round still had energy and momentum remaining upon exit and did not transfer all of its momentum. Is that correct?

Yes, that is correct. If a bullet passes through its target while maintaining significant speed the corresponding momentum and energy for this speed can not have been "deposited" in the target.

 

[Assaltwaffle]

Full reply placeholder

Thank you for the in-depth reply! So, in the calculations, the energy that is lost through the inelastic collisions is converted into different methods of work, such as bending, tearing, or heating, correct? Certainly energy cannot simply be erased.
The different comparative examples help me think of how this all relates beyond seeing a singular formula.

 

[Nugatory]

If I understand it, it would seem to reason that if the plate failed because it could not resist the projectile and the object passed through the target, that the impulse would be less because the round still had energy and momentum remaining upon exit and did not transfer all of its momentum. Is that correct?

Yes. Consider the extreme case of a bullet fired through a sheet of paper: the paper doesn’t move noticeably and the bullet doesn’t slow down noticeably.

 

[Assaltwaffle]

You still seem to mix energy and momentum a bit. You should consider energy and momentum as two independent quantities that "just happens" to be related via mass and velocity. The momentum of the bullet (minus loss during flight) is transferred from the shooter to the target and, independently of this, some part of the energy released by the gun powder is first transferred to the bullet in the barrel and then transferred (except for loss during flight) to the target.

I think this was my misconception. It was my understanding that any motion undertaken by an object is a byproduct of energy and thus must be caused by energy. So the transfer of momentum causes the object to move accordingly, not the transfer of kinetic energy, and even though the kinetic energy transfer isn't zero (because kinetic energy and momentum both are a product of mass and velocity), the energy is not the cause of motion. Am I still misunderstanding?
So, then, how does work fit in?

 

[Drakkith]

So the transfer of momentum causes the object to move accordingly, not the transfer of kinetic energy, and even though the kinetic energy transfer isn't zero (because kinetic energy and momentum both are a product of mass and velocity), the energy is not the cause of motion. Am I still misunderstanding?

Movement requires BOTH energy and momentum. Your misunderstanding, in my opinion, centers around the idea that motion is related to one more fundamentally than the other. This is not true. Both are equally fundamental, they just behave and are transferred somewhat differently. Neither is the cause of motion.

 

[jbriggs444]

I think see now. It seems I had a fundamental misunderstanding about what determined the "impulse" and how what work the bullet did is determined.

So, if the round is fully stopped by a plate or fully capture by a body, the full momentum is transferred as an impulse. Energy does this work and transfers energy accordingly to make that impulse, but the rest of the energy of the projectile is lost through doing "damage", such as bending metal, breaking ceramic, or tearing flesh.

I would say "yes", you have this correct. Some small portion of the kinetic energy of the bullet is expended in the kinetic energy corresponding to the bulk motion of the target body. Energy is conserved. The remaining energy must be accounted for otherwise. As you put it, bending metal, breaking ceramic or tearing flesh. Or is retained as the bullet proceeds on and through the target.

If I understand it, it would seem to reason that if the plate failed because it could not resist the projectile and the object passed through the target, that the impulse would be less because the round still had energy and momentum remaining upon exit and did not transfer all of its momentum. Is that correct?

Yes. If the bullet carries on through the target, it retains some of its original momentum. Less has been transferred to the target.

 

[Filip Larsen]

So the transfer of momentum causes the object to move accordingly, not the transfer of kinetic energy

To the extend you mean that change in momentum of an isolated object causes change in its velocity, then you are correct. In classical mechanics Newtons 2nd law states that time rate change in momentum for an object equals the net force on that object, so if you can model what forces an object experiences as a function of its position and velocity then you can calculate how the object moves without directly having to consider the energy of anything.

However, for some dynamical models it can be much easier to derive the dynamics considering the energy flow between the interacting parts, especially so when the interactions can be modeled as frictionless. And in some cases, like for fully elastic collisions, you need to consider (conservation of) both momentum and energy at the same time to arrive at a solution.

Another reason why mass, momentum and energy are important (and independent) concepts in classical mechanics is that the sum of each of these quantities for the parts of an isolated system can be considered a conserved quantity (i.e. constant over time), with energy being conserved as the sum of kinetic energy and internal energy in case of friction.

 

[Mister T]

why, mechanically, is the person not knocked over despite such a high amount of energy striking the individual relative to a weaker force that can achieve this?

You made quite the leap there from energy to force. Just because something has more energy doesn't mean it exerts a larger force. You have to take into account the distance the object moves while it's slowing down.