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Re: crash tests, from a Physics Teacher




Okay, you guys need something to do!

EVERYONE MISSED THE POINT! From the drivers point of view, the RELATIVE
VELOCITY of the two cars doing 30 crashing head on would LOOK (see that
word, LOOK) like him crashing into the front (again the point is LOOK
LIKE) of a parked car while doing 60. The energy of the crash is NOT
related. People tend to think that they are surviving more severe crashes
than they've actually experienced. I would bet, none of those testimonals
would not even come close to APPEARING as if they were crashing into
something stationary while they were going head long, no brakes at 60
MPH. They had to have braked or slid (either case scrubs speed remarkably
well) and crashed at a significantly smaller speed. So, the 30 MPH
barrier test is generally more severe than the testimonials, which means
that the NHTSA tests are probably more on target than they  think. (Sorry
Audi, but compared to todays cars, the old ones just didn't cut it. Then
again, few did)


To solve the problem that I apparently started:

The equations are:
K.E. = mV

(momentum) p   = mV

We must look at total energy and total momentum. Due to the absorbtion of
energy in the (generally) in-elastic collisions of cars (very little
bounce-back) we can assume the following.

V initial of car 1 = 30 MPH
V initial of car 2 = - 30 MPH (travelling in opposite direction. Think
number line)

Since the mass of the cars are assumed to be the same, mass can
ultimately be dropped from the picture. Note a 2200 lb car has a mass of
about 1000 kg. 30 MPH translates to roughly 15 m/s

Total KE before crash:   K.E. = mV + mV

= m(V1 + V2)

=  (1000 kg) (15 + (-15) = 225,000 J or 225 kJ

Total KE after the crash: Since all is stopped, V = 0, K.E. = 0 so a
total of 225 kJ must be absorbed, one car absorbs 112.5 kJ.

As for Momentum:

Total p before crash:    p = mV1+ mV2

   = 1000 kg (15 + (-15)) = 0 kg m/s (surprise!)

Total p after crash: Gee, same reason as above p = 0

Change of p = m(V1 - Vend) = 1000 kg (15 - 0) = 15,000 kg m/s

Momentum CHANGE for EACH car is 15,000 kg m/s, since both came to a stop
from 15 m/s. If the car's bounced backwards at, say 2.5 m/s (5 MPH,
actually pretty fast for a bounce) then the change for each car would
INCREASE to 17,500 kg m/s.

ENERGY change for each INDIVIDUAL car in the bounce case would go up by a
BIGGER factor

change in K.E. = m(V-V)
         =  (1000) * (15 - (-2.5))  (remember, directions the -2.5 is
opposite the 15)
         = 153.1 kJ

a significant increase over 112.5 kJ. THIS is why cars are built to
deform, rather than being built to be rigid crashing structure (it's
possible to do, just not too survivable).

So, why the barrier test?

change in K.E. = K.E. = m(V-V)
         =  (1000) * (15 - (0))  (remember car was doing 15 came to
rest)
         = 112.5 kJ

change of p = mV1 final - mV1 before
      = (1000 kg)* (0 - 15)
      = - 15,000 kg m/s or 15k kg m/s is lost by the crashing car

SO, the barrier crash duplicates the effect of a TWO car head on crash of
TWO cars of equal mass travelling towards each other at 30 MPH.

NOW, for the question, of ONE car hitting a barrier at 60 MPH
(approximately 30 m/s)

change in K.E. = K.E. = m(V-V)
         =  (1000) * (30 - (0))  (remember car was doing 15 came to
rest)
         = 450 kJ  (4 x the ENERGY!)

change of p = mV1 final - mV1 before
      = (1000 kg)* (0 - 30)
      = - 30,000 kg m/s or 30k kg m/s is lost by the crashing car

SO the 60 MPH crash is 4 x more severe than ONE car crashing at 30 MPH,
and TWICE as severe as TWO cars crashing into one another head-on EACH
doing 30 MPH. SO, relative velocities be damned when it comes to
determining the energies of the crash.

AGAIN, my only point in expressing the RELATIVE velocities would LOOK
like 60 MPH was to illustrate the point that people have most likely
survived MUCH less severe crashes than they thought.

As for why would I spend the time to write this post? I'm a Physics
Teacher and didn't intend to bolster any misconceptions, I was actually
trying to bust one (about the NHTSA tests being bogus, I think they are
quite valid) which had many people thinking (hoping) that their older
cars are as survivable in a crash as the newer ones. Unfortunately not
so, and I own one such older car, one of the more up to date series. As
for the 2CQ, I believe most of the crash worthiness comes from the Seat
Belt Pretensioners (see the '90 passenger side test for the 5K to 1C,
from * to 5*(s), and yet no airbag), which (hopefully) means 88 - up
80-90 owners get as much benefit over the 4K (too bad, I did love my old
4KQ).  So, hopefully this ends this potential thread. Hope this wasn't
too much.

Any Questions?

Larry - '89 2CTQ (5 *), '85 GTi (Solo2, so no traffic - my guess 1 * ?)


Yes, would you like to settle the 60,0 two car solution.
I agree with all of your statements and I believe the biggest problem is
listening, as I believe most of us agree on almost all points, even you
answered the 60,barrier question but not the 60,0 two car question.  However
as you stated conservation of momentum states that both cars will be
traveling at 30 immediatly after.

  Dave G
3x  84 4kq's (death traps)
  No one should be driving these older quattro's anymore and I will accept
any given to me, just for everyone elses good.