Re: crash tests

[Huw Powell <audi@mediaone.net>]

> Look at the collision as two separate accidents.  One car traveling west
> (let's say) strikes an obstacle and comes to an almost instant stop from 30
> mph to zero mph.  A second identical car traveling east at 30 mph strikes
> an obstacle and almost instantly stops moving.  The change in kinetic
> energy in each case is KE = (mV^2)/2.  Each car "releases" that amount of
> energy.  True, the total energy is double because there are two cars but it
> is spread over two cars, not one.

not one?

> 
> BTW a single car traveling at 60 mph striking a barricade 

the quote I was agreeing with used two cars, no barricade.  The "non
moving" object is another car.  the "barricade" example where the
baricade is immobile is *completely different*.

> releases four
> times as much energy as the same car traveling at 30 mph would release, not
> double.  Remember, the energy change is proportion to the *square* of the
> velocity change.

This I remember.  Now remember your relativity and look at the closed
system.  Each car is going 60 mph relative to the other one and ends up
going 30 mph after it hits it.  

> 
> At 12:00 PM 12/13/1999 -0500, you wrote:
> >>                 ... Be aware that two cars travelling towards each other,
> >> each at 30 MPH is an extremely severe collision, visually, it would LOOK
> >> like driving into a parked car while you were doing 60 MPH, brakes OFF.

-- 
Huw Powell

http://www.humanspeakers.com/audi/

82 Audi Coupe; 84 4kq; 85 Coupe GT; 73 F250

http://people.ne.mediaone.net/audi/thoughts.htm