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Let's assume we have 2 bullets. The first bullet has $450$ $J$ of KE and $3$ $kg.m/s$ of momentum, while the second bullet has $250$ $J$ of KE and $5$ $kg.m/s$. Now if they are both shot at ballistic gelatin, which one is expected to cause more damage if both are stopped by the gelatin? By damage here I mean more penetration, bigger cavity, heat and any other deformation of the gelatin.

In other words, if the velocity of a projectile is doubled, will the amount of damage it causes when it collides with an object double or quadruple?

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Now if they are both shot at ballistic gelatin, which one is expected to cause more damage if both are stopped by the gelatin ? By damage here I mean more penetration, bigger cavity, heat and any other deformation of the gelatin. In other words, if the velocity of a projectile is doubled, will the amount of damage it causes when it collides with an object double or quadruple ?

That was how the formula of KE was proved by experiments:

soon after Leibniz' death, the quadratic relation was confirmed by experiments independently by the Italian Poleni in 1719 and the Dutch Gravesande in 1722, who dropped balls from varying heights onto soft clay and found that balls with twice speed produced and indentation four times deeper

It is energy that matters, it is energy that produces (motion =) momentum, but they cannot be separated, they are two sides of the same coins. Your question asks for an impossible choice. In some cases we consider energy, in other cases , like collision, we consider momentum. In the link quoted above, if you are interested in abstract principles, you can read about the historic dispute between Leibniz and Newton-Descartes on the primacy of energy and momentum

Update:

Hoping to give you a clear insight into the issue: consider a unitary mass travelling at v. Apart from units, momentum = v and energy = $v^2/2$, you can refer at your own choice to any of the values or any of the concepts, but each one implies the other. From a conceptual point of view energy is the cause and motion (momentum) is one of the consequences. That is why Leibniz, a philosopher, advocated the primacy of energy

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    $\begingroup$ "it is energy that produces momentum" Uhg. It is not meaningful or helpful to try to build a hierarchy between energy and momentum, even in Newtonian physics (and less so in Einsteinian physics where they are components of a four-vector are mixed into different proportions depending on the observer). $\endgroup$ Oct 17, 2014 at 15:57
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    $\begingroup$ @dmckee , considering the level of question and the precise reference to double or quadruple this seems the best possible answer. The alternative is not to reply or to say both as in the other answer, and OP wouldn't have had a clue, I suppose. $\endgroup$
    – bobie
    Oct 17, 2014 at 16:20
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    $\begingroup$ By holding the mass fixed, you've effectively ignored the question. $\endgroup$
    – WillO
    Aug 6, 2019 at 15:11
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Here's a more physical argument for why damage is caused by kinetic energy instead of momentum. When I say "caused by" I mean that two projectiles with the same kinetic energy will cause the same amount of damage, whereas two projectiles with the same momentum may not.

When a projectile (a baseball, for example) impacts on an object (a glass window), there can be two results: the object stops or reflects the projectile with no damage, or the projectile penetrates the object, causing damage until it comes to rest inside the object or breaks through the other side. By Newton's Third Law, the force that the window exerts to stop the baseball is equal to the force the baseball exerts on the window. So, even if the window doesn't break, it feels a force and bends in response--the greater the force, the greater the bending. A window can only bend so far before it breaks, so there is a maximum force that can be applied to the window before it breaks. By Newton's Third Law again, there is a maximum force that the window can exert on the ball.

So, as the ball is thrown harder against the window, it takes greater and greater force to stop the ball, causing greater and greater bending in the window. Once the force is great enough, the window breaks and the ball continues on its way at a slower speed due to the force of the window.

Now imagine, instead of a single window pane, there is a large stack of window panes in the way of the baseball. The baseball will break through a certain number of windows until it slows down enough for one window deep in the stack to stop it. What does each window do? When the ball makes contact, the window bends in response until it reaches it's breaking point, applying a force to the ball over a certain distance. This means that each broken window does work on the ball, which means the ball does some amount of work (force over a distance) to break the window. If the windows are identical copies, each will take the same amount of work to break, and the amount of damage is the number of windows broken, which is the amount of work done on the windows, which must be equal to the kinetic energy of the ball.

Momentum doesn't matter because it doesn't matter how much time a ball pushes against a window. You can rest a ball on top of a horizontal pane of glass and nothing will happen, even as the ball applies an unlimited amount of impulse (force times time) to the window. What matters is the distance that the ball deforms the window, which requires a certain amount of force, and exerting a force over a distance is work, i.e., energy.

You can try an experiment for yourself. Get a box of sand and some hunks of clay. You can measure damage by measuring the size of the crater formed when the lumps of clay are dropped onto the sand. You'll need to perform at least 4 drops: 2 where different masses of clay impact with the same kinetic energy and different momenta, and 2 where different masses impact with the same momentum but different kinetic energies. One of these pairs of drops should make similarly sized holes and the other pair should make different sized holes.

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  • $\begingroup$ "When the ball makes contact, the window bends in response until it reaches it's breaking point, applying a force to the ball over a certain distance" How do we know that the force applied is constant? In fact it seems like it shouldn't be, and it seems like your argument relies on the force being constant. if the force is not constant, it can vary precisely in such a way that it doesn't take a fixed amount of work to break the glass, right? $\endgroup$
    – user56834
    Aug 6, 2019 at 4:49
  • $\begingroup$ @user56834 The force is definitely not constant. I would expect the force to be similar to that of a spring: the greater the deformation, the greater the force on the ball. Nothing in my argument requires a constant force between the ball and window. $\endgroup$
    – Mark H
    Aug 6, 2019 at 6:44
  • $\begingroup$ But what if the force applied during the bullet going throught the first glass window is different from that applied during the second glass window, due to the bullet having slowed down? Then the work done on the two glass windows are also different. $\endgroup$
    – user56834
    Aug 6, 2019 at 8:13
  • $\begingroup$ @user56834 If the windows are all identical, then it should take the same force to break each one. Putting the same force on the window should produce the same amount of deformation. Therefore, each window takes the same energy to break, which takes away the same amount of energy from the bullet. $\endgroup$
    – Mark H
    Aug 6, 2019 at 14:47
  • $\begingroup$ @user56834 If the windows are different (in material or thickness or something else), then each window breaking would take away different amounts of energy. This is fine. In my answer, I assumed that all the windows were identical so that more windows breaking = more damage = more energy in the bullet. $\endgroup$
    – Mark H
    Aug 6, 2019 at 14:50
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Momentum: p=mv Kinetic energy: E(k) = 1/2mv^2

So the answer is both. Kinetic energy and momentum are both functions of mass and velocity, and so functions of each other. One goes up, the other goes up:

from momentum: v = p/m

plugging back into Kinetic energy: E(k) = 1/2m[p/m]^2

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Yes, energy and momentum are not separated with each other. In formula, they have such relation: E = P^2/2m. So, two particles with the same energy, the one with larger mass (m) has larger momentum, and vice verse.

To illustrate which determines the damage strength, we could see the following instance in micro-world. Imagine two beams of charged particles: electron and ion (e.g. Argon ion) with the same energy (1000 eV, for instance) shine on some surface (e.g. metal) for a while. The result would be that the one shined with electrons has little change, whereas the top few layers atoms ( a few nm) of the one shined with ions were knocked off. In fact, scientists use high energy ions to sputter metal surfaces to obtain clean surfaces for research purpose. while, they use high-energy electrons as tools to 'see' their materials, like SEM, or TEM, and so on, because high energy electrons has little damage to the shined materials.

Now we could see that momentum has more direct effect in the damage.

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Ok so I thought about this a lot, and realized there's a cool answer...

Momentum (mysterious physics concept) is mathematically the derivative of kinetic energy (KE)!

I have always wondered why KE looks like a multiply-by-1/2-derivative-trick to make the 2 disappear.

$$\mathrm{KE} = \frac{1}{2} \times m \times v ^2$$ $$\ \ \ \ \ \ \ \ = \frac{1}{2} \times m \times \frac{s^2} {t^2}$$

s = distance

t = time

derivative[KE] $= m \times v$ , i.e ,momentum

Now it makes sense why they are both conserved, and related.

Derivative means momentum is equivalent to the rate of change of KE.

If you divide the units:   KE / Momentum   =  m / s   =  velocity
Dividing is analogous to 'y/x' slope type of derivative

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    $\begingroup$ Hi, Jason; welcome to Physics.SE; this is a MathJax-enabled site; do use that facility to better format your equations. For a quick reference of MathJax, check this Meta Math.SE post. $\endgroup$
    – user36790
    Dec 21, 2016 at 5:40
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    $\begingroup$ Momentum is not $d(KE)/dt$. The time derivative of KE has units of power not momentum. $\endgroup$ Apr 18, 2022 at 16:17

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