# Cars crashing head-on or into a wall

Q: *Suppose two identical cars with 2 clone drivers collided with each other head-on, each traveling at a speed of 100 km/h. Would the impact that the two driver feel be equal to the impact of them driving into a solid wall at 200 km/h? This is what I intuitively think, but several driving instructors told me that the impact would be like as if driving into a solid wall at 400 km/h, citing something to do with physics.*

A: Neither: It's like hitting a wall at 100kph. There is a simple answer and a MUCH more complicated answer. Your driving instructors are picking the complicated one and failing to understand the full consequences of that decision. I'll try to explain - but it's gets horribly complicated very fast!

### SIMPLE ANSWER

The total energy that has to be dispersed by crumpling metal is twice as high when two cars collide (if both are travelling at 100kph) than if one car hits a totally immobile brick wall at 100kph - but the crumple distance is also doubled and so is the amount of bent metal at the end (which is where all that kinetic energy went) - so the accelleration is the same when two cars collide at 100km/h versus one car hitting a wall at 100km/h. So no - driving into another car with both going at 100km/h would be the same as driving into a wall at 100km/h. (In fact, no wall is going to perfectly fail to absorb any energy - so you'd probably be fractionally better off hitting the wall in the real world).

### UGLY RELATIVE MOTION ANSWER

The relative motion argument is another perfectly valid way to look at it - but you have to be utterly consistant about that. So from the point of view of a frame of reference that's moving at a uniform speed of 100kph towards you: You are initially rushing towards that frame of reference at 200kph and the other car is stationary (compared to that frame of reference) - then you collide - but then both cars continue on at 100kph - the frame of reference didn't stop moving. So your speed (in that frame or reference) went from 200kph to 100kph - so you lost 100kph and it's no worse than driving into a brick wall at 100kph in a stationary frame of reference.

### DRIVING INSTRUCTOR ANSWER (which is bogus)

I suspect these guys read that kinetic energy is mass times the SQUARE of the velocity (that's true - if you hit a brick wall at twice the speed - you'll take four times the amount of damage - so drive slowly!). So they reason that the kinetic energy at double the approach speed is four times higher. But then they make the mistake of picking the other car's frame of reference, instead of having two cars, both with energy (E) given by E=(Mass x 100kph x 100kph) - they assumed that this was the same thing as one car moving at twice the speed (Mass x 200 x 200)...which is 4E. But that's only allowed if you take the 'UGLY RELATIVE MOTION ANSWER' - and in that case, the car doesn't lose all of it's kinetic energy - it only loses enough to drop its speed from 200kph to 100kph (3E) - and (critically) in this rather strange frame of reference, the other car GAINED kinetic energy because it starts off stationary and winds up moving backwards at 100kph, gaining -E in the process. The resulting energy (2E) is then shared between the two cars - so (as if by magic), each car has to dissipate exactly 1E of damage...just the same as it was if you pick a less mind-bending stationary frame of reference.