Suppose a car crashes at a speed $v$ against a wall and comes to a stop. Now if the car crashes at $2v$, does that mean it suffers twice as much destruction, if that can be objectively measured?

If force is what causes the damage and $F = ma$ and $a = \Delta v/\Delta t$, so $F = m \Delta v/\Delta t$. Assuming $\Delta t$ remains the same, then the force acting upon the car in the crash at $2v$ is just twice the force of the crash at $v$.

But by conservation of energy, if the cars come to a stop all the kinetic energy is converted into other forms of energy. So a crash at twice the speed is four times as energetic, since $E = m v^2 / 2$. If that's the case, shouldn't it cause 4 times as much destruction?


2 Answers 2


A car crashing at twice the speed has 4x the energy and will experience much more damage.

The forces involved are more complicated to analyse, a car travelling at twice the speed and coming to rest in the same time will only experience twice the force on a component. But the way a structural component behaves is a complex function of the speed that the force is applied.


OK, for a car, its a bit hard to calculate. Aside from the fact that "destruction" is not something you can quantify (there is no meaning for "twice as much destruction"), a car has just too many complicated parts. You'd need to know Young's moduli and breaking stresses for each part, and have a large computer to do the calculations.

Anyways, here are some points that may make the situation clearer.

1) When you have any sort of collision, there is a rapid deceleration. This implies a large force in a short time. As $F=\frac{dp}{dt}$, the impulse felt by this body is written as $$\Delta p=J=\int Fdt$$. Now, we cannot calculate the value of F from just this equation, as we do not know how the body behaves when deformed. For a simple body which obeys Hooke's law/Young's moduli, F as a function of time(for the duration of the collision with the wall, will be: $$mv\left(\frac{m}{k}\right)^{\frac{3}{2}}\left(sin{\sqrt{\frac{m}{k}}t}\right)$$ Where $k=\frac{YA}{L}$ of the body obeying Young's modulus (Or $k$ can be thought of as the sprong constant). The above equation is easily calculated using SHM equations of a spring-mass system. If you want the equations for two bodies striking each other, replace $k$ with $\frac{k_1k_2}{k_1+k_2}$, and $m$ with $\frac{m_1m_2}{m_1+m_2}$. You will also need to replace $v$ with $v_{rel}$ between the two bodies before collision.

So, if by deformation, you want the amount of force that acts on it, then $F\propto v$.

2)On the other hand, you may only want the amount that the body is crushed; i.e the length that it is shortened. This is easy to calculate for a spring-mass system (one that obeys Hooke's law/Young's modulus). Equating energies, we get $$\frac{1}{2}mv^2=\frac{1}{2}kx^2$$ in standard symbols, thus $$x=v\sqrt{\frac{m}{k}}$$ The same substitutions as above will generalize this for a two-body problem. Again, we got that destruction is proportional to initial velocity.

3)We may also look at destruction as the amount of energy used in destroying the body. In this case, we get that it is proportional to $v^2$, as long as no energy is lost through sound (in car collisions, quite a lot is).

In conclusion:

  1. As destruction is qualitative not quantitative, we cannot say that "destruction increases 4x" until we define what we mean by destruction is a quantitative manner.
  2. Cars collisions are complex, generalizing any of the above formulae for them is not even a good approximation, due to loss of energy by heat and sound, as well as the complexity of parts (overall, a car does NOT obey Hooke's law or Young's modulus; and neither do half of its parts at the magnitudes of collision forces). THe answer will vary from car to car.

So really your question is unanswerable, we can only elaborate on the underlying concept.

If you want, though, you could try to find a automobile research center. They computer-simulate these crashes on their computers, so they'll have a more accurate answer to this question.

  • $\begingroup$ Strange. @DavidZaslavsky Know anything about this? $\endgroup$ Commented Feb 8, 2012 at 18:30

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