Car Crash Question... (Conservation of Momentum?)

Question

Dumb car crash question. I think this is a conservation of momentum problem.

In one scenario, two 4000 lb cars (car A and car B) each traveling at exactly 35 MPH in opposite directions have a perfect head on collision with each other.
In the other scenario, one 4000 lb car (Car A) traveling at 35 MPH hits a steel reinforced concrete bridge abutment (i.e. a huge mass with zero speed.)

Which Car A sustains more 'damage'?

My gut feel is that both experience exactly the same crash damage, but I'm at a loss to describe why.

(Question inspired by a temporary stretch of freeway near my home, with just a painted double line between opposing directions of traffic.. It's quite scary...)

@Neil Funny, I was reading up on the link provided by John Rennie, which says mythbusters says the damage is equal (and not twice like you've mentioned...) I guess I'll search for that episode... — zipzit, Mar 31 '16 at 08:42
Mythbusters youtube video, 6 minutes long.... 50MPH into wall damage = 50MPH Head On Collision damage... — zipzit, Mar 31 '16 at 09:31
Twice as much kinetic energy, twice as many cars to absorb it. — Solomon Slow, Mar 31 '16 at 16:11
@ James large: So yes twice the KE and two damaged cars. Once the KE then one car damaged and one wall damaged.Same? — , Mar 31 '16 at 21:02

score 4 · Accepted Answer · answered Mar 31 '16 at 08:52

4

Your gut feel is correct. Both are exactly the same. Look at the acceleration in both scenarios. 35 mph to 0 in the time it takes for the cars to fold up and stop. Everybody gets this wrong. Good question.

answered Mar 31 '16 at 08:52

Alex Tworkowski

62

I'm trying to do the math 1/2 mv^2... and I keep getting stuck... in first case Mass car a = mass car b, in the second case Mass abutment = infinity, velocity abutment = 0... – zipzit Mar 31 '16 at 08:54
@zipzit, Instead of saying "mass of abutment = infinity", say "mass of abutment as large as you care to make it." – Solomon Slow Mar 31 '16 at 16:13
I recently had a physicist tell me that using KE would imply that there is twice as much KE in the system. He went on to say the head-on would be worse. He was wrong and I told him so. The negative acceleration from speed to zero speed is the same in both systems. Is there a distinction between KE and Momentum. I don't know. Also if KE=1/2 MV^2 then when bowling, get the lightest ball you can and fire it down the lane as fast as you can. It doesn't work. You can't know the pins over with a baseball. I use MV now. Don't know why KE won't work. – Alex Tworkowski Jan 15 '18 at 14:38

score 0 · Answer 2 · answered Mar 31 '16 at 08:39

Remember that it is not the speed that kills you, it’s the sudden stop at the end of the road ;)

Translated into momentum conservation, this means that in your first scenario the two vehicles are approaching each other with a speed of 70 MPH whereas in the second the approach speed is only 35 MPH.

The details of the collisions are of course very different but considering approach speed alone, you would be worse of hitting something that is moving towards you than hitting something that is standing still.

Car Crash Question... (Conservation of Momentum?)

2 Answers2