Re: crash tests

[Huw Powell <audi@mediaone.net>]

> ½mV²
> Ok, for arguments sake, one car has a mass of 2000Kg (we'll call it
> cadillac), and a velocity of 10m/s (36Kph) = 100KiloJoules (I think that's
> the right unit).  And that cadillac at 20m/s has 400KJ.
> That means a head on collision with another cadillac of similar mass and
> velocity has 200KJ Joules to release instead of the 400KJ of one cadillac
> at 20m/s running into a parked cadillac (negating any solid barricade
> argument).  However, it might be less severe to run into a solid object at
> 10m/s than a deformable one (eg: car) at 20m/s.
> 
> To recap, 1+1=2, but (1*2)²+0=4.
> 
> Make sense?  It's been 10 years...what am I forgetting?

Relativity.  

You are measuring V relative to an external frame of reference.  The
subjects in question, the cars, only have energy relative to each
other.  in the 0/60 collision, if it were not for friction bringing the
mangled clump to a halt eventually, the two damaged cars would keep
travelling at about 30 mph (less shrapnel/fragmentation/deformation
energy losses) in the direction the moving car was travelling
indefinitely - this is where the "missing" energy ends up.  Filmed from
one car the accident would be indistinguishable from the 30/30 accident.

This is why it is different than the barrier accident, where the
terminal velocity relative to the observer is always zero.

One problem here I think is that people imagine that the 60/0 collision
takes place at a fixed point in their frame of reference.  It doesn't. 
(unlike the barrier versions where everything stops dead and the 60 mph
collision is twice as bad...)

-- 
Huw Powell