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If a car hits a wall at a speed of $v$, and an identical car hits a wall at a speed of $2\,v$, with the kinetic energy equation $E_K = \frac{1}{2}mv^2$ (the speed goes from $v$ to $0$) the second car will lose 4 times the energy, and therefore cause 4 times the damage, but with the force equation $F = ma$ (the car hits the wall and decelerates), it will hit with 2 times the force, and cause 2 times the damage.

How can this be explained?

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  • $\begingroup$ There are several parts of your question that seem ambiguous. 1. Why are we justified in assuming that energy (loss) is proportional to damage? 2. Using F=ma sounds like you are assuming that both collisions occur over the same interval of time. (e.g. are you assuming $a=\Delta v/\Delta t$, with identical $\Delta t$ for each collision?) Is this assumption also justified? For a problem like this, I would imagine you would want to begin by looking at the impulse $\vec{J}=\vec{F}_{avg}\Delta t = \Delta \vec{p}$, but I still don't know how you intend to quantify "damage." $\endgroup$ – Bob Jan 9 '18 at 3:56
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it will hit with 2 times the force, and cause 2 times the damage

Let us examine this statement starting with the initial conditions that car $1$, mass $m$, hits the wall at a speed $v$ and the wall exerts a constant force $F$ on it and car $2$ with the same mass as car $1$ with initial speed $2v$ hits the wall and the wall exerts a constant force of $2F$ on it.

Foe car $1$ the work done in stopping the car is $Fs_1$ where $s_1$ distance that car $1$ moves whilst slowing down.

So you have $Fs_1=\frac 12 mv^2$ and for car $2$ you have $2Fs_2 = \frac 12 m (2v)^2$

Combining these two equations gives $2s_1=s_2$.

So car $2$ has twice the force exerted on it and travels twice the distance as compared with car $1$.

Whatever the forces acting it will always take four times the work to stop car $2$ as compared with car $1$.

Your confusion might arise because you have equated "damage" with force and then relating this to the loss of kinetic energy.
It is certainly true that a larger force will increase the chances of permanent deformation (damage) and that is reason for wearing seat belts in a car but in such an example the work done in stopping the occupant of a car from a certain speed will be the same irrespective of the force applied to the occupant whilst stopping.
A smaller force (ie less damage) will mean that the distance traveled before stopping will be larger with the work done being the same as when a largewr force is applied.

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your confusion is that 2v doesn't mean 2f it means 4f.

Assuming the wall has a constant elasticity through it's course of collapsing while stopping the incoming car while it moves back to resist and stop the car, basically like a spring. Then the area of wall's work should be equal to kinetic energy of the car that needs to be stopped.

Area of wall resistance work= (kx^2)/2 (K is spring constant of the wall, x is the recoil length), triangular work area starting from zero!So the wall will set back 2 times.

However the energy dissipated during a collaps is never linearly proportional to the wall set back; it is much more complicated.

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The crux of the question comes down to, "which causes more damage: the energy lost or the force exerted?"

The answer is the former. For example, a building exerts several mega-newtons of force on the Earth's surface, but nothing happens to the surface. On the other hand, a nuclear bomb dropped from a moderate height will exert a much smaller force when it lands, but cause much greater destruction.

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  • $\begingroup$ Assuming, that is, the bomb detonates. If it does not detonate, then it will produce very little damage (or at least, not any more than a rock of similar mass dropped from the same height would.). Granted, it could very well break open and release its radioactives into the atmosphere as noxious poison, and this could have severe biological effects on a great many people (which is ultimately, due to a release of energy as well but at vulnerable places within the microstructure of their bodies, and actually surprisingly not very much energy),(ctd) $\endgroup$ – The_Sympathizer Jan 9 '18 at 8:43
  • $\begingroup$ (ctd) but in terms of raw mechanical damage, it won't do much. The bomb would have to be built to detonate on impact, in which case the "height" is meaningless (even dropping from space, it will yield more by explosion than impact). And when it explodes, that will exert a very great force indeed (albeit a rather indirect one, since what actually happens is the X-rays from the bomb vaporize the surroundings and that in turn generates the actual explosive force, more than the bomb gas itself doing work directly on them.) $\endgroup$ – The_Sympathizer Jan 9 '18 at 8:44
  • $\begingroup$ (FWIW the energy to kill an average human, about 63 kg body mass, is only about 400 J for radiation. This is about as much as a wimpy handgun bullet, but not anywhere close to what, say, it would take in terms of thermal radiation.) $\endgroup$ – The_Sympathizer Jan 9 '18 at 8:47
  • $\begingroup$ So the connection between the two is rather less clear than one might think. (Indeed the detonation pressure in a warhead can reach into the peta-pascal range, which is comparable to the pressures inside stellar cores, and the radiation pressure alone from the X-ray glow vastly exceeds the pressure at the center of the Earth!) A small force can deliver a very large energy yet still not be destructive. Your car accelerates you from maybe 0 to 30 m/s. In doing so it imparts about 28 kJ of kinetic energy to your body, which is enough that were it to hit you in the form of a bullet, it would be $\endgroup$ – The_Sympathizer Jan 9 '18 at 8:51
  • $\begingroup$ gruesomely lethal (like a .50 BMG, actually even more.). (That's if you're 63 kg i.e. average human. If you're more, it will be more.) The reason it don't kill you is the force is gentle. And when you stop the car, the retarding force is similarly gentle. But smash into a wall at that speed, and you're gonna be seriously hurt, because the force generated on impact will be very extreme. $\endgroup$ – The_Sympathizer Jan 9 '18 at 8:52
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The relationship between force, energy, and damage is complex.

To understand it, you should really think of force not as "mass times acceleration" but rather as the "time rate of change of momentum" or the "flux of momentum into/out of an object":

$$\vec{F} = \frac{d\vec{p}}{dt}$$

And arguably, this is the more fundamental way of looking at it (it is what you need to get into relativistic and quantum mechanics), and makes Newton's third law understandable - e.g. when one body crashes into another elastically so as to come to a "stop", there is a sudden and rapid transfer of all its momentum to the other body, and this fast transfer exerts - or actually, is - a great force. And it's these contact forces and their sequelae within the objects themselves, that are what do the damage.

The degree of force will depend then, precisely on the speed at which that momentum is transferred, as evident from the above, and this depends on the distance over which it is exerted. That distance must be large enough to do enough work to make the object come to a stop (or at least its center of mass), if we imagine a fixed barrier and inelastic collision. And that's where the kinetic energy comes in: that determines the distance (since the work done to stop is the kinetic energy, essentially by definition) since

$$W = Fd$$

thus

$$d = \frac{W}{F} = \frac{K.E.}{F}$$

and if the material cannot supply that distance without breaking, then it will break as long and deep as is needed until it has finally been supplied, or it is pushed out of the way or the impactor ricochets, and the collision was not entirely inelastic. Thus also, the precise specifics will depend on the materials involved, and those that can provide a very long distance over which to stop without breaking, will provide a much reduced impact force and much less damage, even with a relatively large kinetic energy.

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protected by Qmechanic Jan 9 '18 at 7:27

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