Re: crash tests, from a Physics Teacher

[Lawrence C Leung <l.leung@juno.com>]

Unless you test the actual collision, it is impossible, given our
limitations, to predict the speed or direction of the combined wrecked
cars. If the collision were to be PERFECTLY elastic, then both cars will
bounce backwards at 60 MPH, no K.E. would be lost. Since cars are
relatively inelastic, they are better approximated by in-elasticity. The
reality is somewhere in-between, the degree is determined by how much the
designers of the car (since we are doing only one brand, model at at
time) wanted it to absorb. If, by the design limits it ended up absorbing
1/2 the energy, than the car would bounce back at 1/4 the collision speed
(the rest of the car's crush zones would have been used up at that
point). Note, 1/2 the energy at 60 MPH would have HANDILY taken ALL of
the energy of a 30 MPH crash, which has only 1/4 the energy (see previous
post). So, without knowing the energy absorption design limits of the
car's in the collision, we end up in idle speculation as to how fast a
car will be moving backwards after a collision, thus determine how much
energy is exchanged to the passengers during the collision. 

SO, where does this leave us? With empirical evidence. The assumption I
made (total in-elastic collision) was based upon what little experience I
have had at 30 MPH head-on crashes, in a sense video of car crash tests
on TV. Most of the crashes do have a small amount of bounce back, but I
don't feel at liberty to judge the rate of the bounce. It certainly does
not appear to be 1/2 or even 1/4 of the original speed. If anyone wishes
to take the time to find any videos and then compare the bounce back
speed to the original crash speeds and report back, I can revise the 30
MPH energy estimates. Obviously, since there are no 60 MPH videos (with
good scientific backing) that I know of and no other info on the energy
absorption from the factories, I can't revise those estimates.  Thus, in
order to make the relative calculations reasonable to understand, AND
from estimates based off of crash videos, I felt a reasonable (note
REASONABLE) approximation was that of an inelastic collision for the
30/30 crash. 

SO, for the 60/0 crash, we will need to take into account the amount of
energy absorbed by the 30/30 car, and apply that to each of the 60/0 cars
(HOW did we get to THIS situation anyway. As I said before, this was ONLY
a visual reference). Now the situation can be calculated the same way as
before, noting directions of the moving car and expected directions of
the crashed cars. You MUST use both equations (the TOTAL momentum and
TOTAL energy equations) to arrive at a solution, solving for the common
velocity in both sets of equations. SO, if you have my last post, I will
leave the proof for you to do, as I DO have to prep for my next lesson. I
will happily check your work and conclusions. For now, I must go. Good
Luck.

Larry - '89 2CTQ, '85 GTi (Solo2)

On Mon, 13 Dec 1999 22:45:04 -0500 Huw Powell <audi@mediaone.net> writes:
>> Okay, you guys need something to do!
>> 
>> EVERYONE MISSED THE POINT!
>
>big snip
>
>> We must look at total energy and total momentum.
>
>big agreement
>
>> Total KE after the crash: Since all is stopped
>
>big disagreement, assumption problem.  All does not stop because of 
>the
>crash - in the 60/0 collision, the big smoking wreck is travelling at 
>30
>mph until it grinds to a halt on the road due to friction.  That 
>energy
>is not dissipated in the crash, which essentially occurs at a speed
>relative to the observer of 30 mph.
>
>-- 
>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