Hm...your explanation is a little more understandable to me than Cisco Qid's (not that I am disparaging Cisco's - it was just a little over my head. Thanks to both of you for your answers - clearly both of you know vastly more physics than I). I think I might *almost *understand it.

Case A: Single car hits immovable wall at 50mph = 50mph worth of damage absorbed by one car.

Case B: Single car hits stationary car at 50 mph = 50mph worth of damage absorbed by

*two* cars.

Case C: Single car hits stationary car at 100 mph = 100mph worth of damage absorbed by

*two* cars.

Case D: Two cars hit each other head on at 50mph each (combined closing speed of 100mph) = 100mph worth of damage absorbed by

*two* cars.

Case B will obviously involve less damage to the moving car than A. Case C should be the same damage per car as in A, with twice the momentum but also twice the material (two cars) absorbing the impact. Case D is exactly the same as C, only described from a different frame of reference, so should be the same damage per car as both C and A.

That was my initial reasoning, but this is obviously wrong. Hitting a wall at 50 is far more than twice as bad as hitting the same wall at 25. It's not momentum (m x v) that determines damage, but rather the kinetic energy (m x v^2). So double the speed is a lot more than just double the damage. It's exponential. So Case B is half the damage (per car) as A (same energy, but two cars to absorb it), while case C will be more damage per car than A (two cars absorbing the impact, but much more than twice the energy). And Case D remains the same as Case C.

So I think the answer to your question is neither. The damage per car in the head-on crash of Case D will be twice that of hitting an immovable wall at the same speed (of 50mph), yet still only half the damage of hitting an immovable wall at twice the speed (i.e. 100mph).