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.

I don't think any of the other answers have made the following point clear enough, so I am going to give it a try. Both scenarios are very similar before the collision, but they differ greatly afterwards...

From a stationary reference, you see the cars driving towards each other at 50mph, but of course if you choose a reference frame moving with the first car, then the second will be headed toward it at 100 mph. How is this different from the wall scenario?

Well, from a stationary reference frame, after the crash both cars remain at rest, so the kinetic energy dissipated is $2\times \frac{1}{2}mv^2$.

From the reference frame moving with the first car, the kinetic energy before the crash is $\frac{1}{2}m(2v)^2=4\times\frac{1}{2}mv^2$, but after the crash the cars do not remain at rest, but keep moving in the direction of the second car at half the speed. So of course the kinetic energy after the crash is $2\times\frac{1}{2}mv^2$, and the total kinetic energy lost in the crash is the same as when considering a stationary reference frame.

In the car against a wall, you do have the full dissipation of a kinetic energy of $4\times\frac{1}{2}mv^2$.

## Best Answer

Basically the answer is that yes, the two cars colliding at a closing speed of $2v$ is the same as a single car hitting an immovable wall at a speed of $v$. The argument is that the cars involved lose the same amount of kinetic energy, $\frac{1}{2}mv^2$, during both crashes, and this energy can go into bending the cars by the same amount.

In detail the forces experienced are likely to be different in the two collisions. Unless you can make both cars identical, and have them crash precisely lined up, during the crash the cars will deform each other asymmetrically and the force:time curve will be different from a crash into an unyielding wall. However the end results will be broadly similar.