# What is the physics of a car collision?

Send to a Friend via Email

Recipient's Email

This field is required.

Separate multiple addresses with commas. Limited to 10 recipients. We will not share any of the email addresses on this form with third parties.

**Question:** What is the physics of a car collision?

This topic was presented by, Anton Tuomi, a reader from Sweden. Many thanks to Anton for the very rich situation!

I have the theory that: **if you drive a vehicle into a static, unbreakable wall, you will feel the same G-force and get the same injuries as if you would drive into your exact copy but mirrored (same car, weight, velocity, angle) head to head.**

Everybody [says] there would be more energy transferred to the drivers and more injuries. but I don't agree. I think the effect should be exactly the same when the energy is divided between the bodies no matter if it's a wall or your mirror image/clone.

**Answer:**

There's a lot going on in this example, so I'll be focusing on only a few of the points you bring up -- mainly the distinction (or lack thereof) between force and energy .

### Force - Colliding With a Wall

Consider case A, in which car A collides with static unbreakable wall. The situation begins with car A traveling at a velocity *v* and it ends with a velocity of 0. The force of this situation is defined by Newton's second law of motion. Force equals mass times acceleration. In this case, the acceleration is ( *v* - 0)/ *t*. where *t* is whatever time it takes car A to come to a stop.

The car exerts this force in the direction of the wall, but the wall (which is static and unbreakable) exerts an equal force back on the car, per Newton's third law of motion. It is this equal force which causes cars to accordion up during collisions.

It is important to note that this is an idealized model. In case A, the car slams into the wall and comes to an immediate stop, which is a perfectly inelastic collision. Since the wall doesn't break or move at all, the full force of the car into the wall has to go somewhere. Either the wall is so massive that it accelerates/moves an imperceptible amount or it doesn't move at all, in which case the force of the collision actually acts on the entire planet - which is, obviously, so massive that the effects are negligible.

### Force - Colliding With a Car

In case B, where car A collides with car B, we have some different force considerations. Assuming that car A and car B are complete mirrors of each other (again, this is a highly idealized situation), they would collide with each other going at precisely the same speed (but opposite directions). From conservation of momentum. we know that they must both come to rest. The mass is the same. Therefore, the force experienced by car A and car B are identical and are identical to that acting on the car in case A.

This explains the force of the collision, but there is a second part of Anton's question - the energy considerations of the collision.

### Now For Energy.

Force is a vector quantity while kinetic energy is a scalar quantity, calculated with the formula *K* = 0.5 *mv* 2 .

In each case, therefore, each car has kinetic energy *K* directly before the collision. At the end of the collision, both cars are at rest, and the total kinetic energy of the system is 0.

Since these are inelastic collisions, the kinetic energy is not conserved, but total energy

is *always* conserved, so the kinetic energy "lost" in the collision has to convert into some other form - heat, sound, etc.

In case A, there is only one car moving, so the energy released during the collision is *K*. In case B, however, there are two cars moving, so the total energy released during the collision is 2*K*. So the crash in case B is clearly more energetic than the case A crash, which brings us to Anton's next point.

### From Cars to Particles

Anton's letter goes on to say: A couple of people bring up the particle accelerator and tell me that there's a reason why they accelerate two particles against each other. "It will create more energy, thus damaging the particles more, as with the cars and their drivers."

But I think that only shatters the atoms more, like throwing two glass bottles really hard and they shatter all over more than just throwing a glass at a wall. Cars don't shatter like that so I don't think it applies when the bodies come to a stop.

First, it's important to consider the major differences between the two situations. At the quantum level of particles, energy and matter can basically swap between states. The physics of a car collision will never, no matter how energetic, emit a completely new car!

The car would experience exactly the same force in both cases. The only force that acts on the car is the sudden deceleration from *v* to 0 velocity in a brief period of time, due to the collision with another object.

However, when viewing the total system, the collision in case B releases twice as much energy as the case A collision. It's louder, hotter, and likely messier. In all likelihood, the cars have fused into each other, pieces flying off in random directions.

And this is why colliding two beams of particles are useful, because in particle collisions we don't really care about the force of the particles (which we never even really measure), we care instead about the energy of the particles.

A particle accelerator speeds particles up but does so with a very real speed limitation (dictated by the speed of light barrier from Einstein's theory of relativity ). To squeeze some extra energy out of the collisions, instead of colliding a beam of near-lightspeed particles with a stationary object, it's better to collide it with another beam of near-lightspeed particles going the opposite direction.

**NOTE:** Kinetic energy considerations in the relativistic situation within particle accelerators, where the particles are traveling near the speed of light, are not quite as straightforward as the classical case involving cars. but using relativity and Lorentz transformations. analogous equations come out fairly easily that correct for the relativistic changes. When these are taken into account, colliding opposite beams still yields more energy output than colliding a beam with stationary matter. From the particle's standpoint I don't know that it would so much "shatter more," but definitely when the two particles collide more energy is released. In collisions of particles, this energy can take the form of other particles, and the more energy you pull out of the collision, the more exotic the particles are.

### Conclusion

In answer to Anton's original theory, therefore, I believe that he's basically correct. His hypothetical passenger would not be able to tell any difference whether he was colliding with a static, unbreakable wall or with his exact mirror twin.

His friends are also right that the particle accelerator beams get more energy out of the collision if the particles are going in opposite directions, but they get more energy out of the total system - each individual particle can only give up so much energy, because it only contains so much energy.

Source: physics.about.com

Category: Accident