# Re: crash tests

```Sorry, Huw.  Your reasoning is flawed.  Each car in your example goes from
traveling at 30 mph to "traveling" at 0 mph in a very short period of time.
Whether this happens as a result of striking an immovable barricade or
whether it results from one car striking the other is of no consequence.

Imagine a car striking an immovable barricade.  What would be the effect
*on the barricade* of being struck by a car from one side?  It would feel a
force applied from the direction from which the car struck the barricade.
Correct?  The magnitude and duration of that force is calculable but is not
necessary for the moment.

Now imagine that another car strikes the same barricade at the exact same
instant as the first one strikes the barricade.  The two cars are
identical.  The only difference being that one approaches from one side and
the other approaches from the other side of the barricade.  (OK?  With me
so far?)  The "first car" exerts a force on the barricade from one
direction.  The "second car" exerts an exactly equal magnitude force on the
barricade but from exactly the opposite direction.  (Still with me?)

Now, what net force is exerted on the barricade?  The resultant is zero.
The two forces cancel each other out.  There is no net force exerted on the
barricade.  If that is the case then what is the barricade doing during the
collision?  The answer is "it is doing nothing".  Since this is the case,
the barricade is not necessary for the example.  In fact, a suitable
"barricade" could be made from a piece of tissue paper.  (Actually, less
than that.)

To each car, the effect of the head-on collision is exactly the same as the
effect of the car colliding with the proverbial immovable barricade.  In
effect, each car becomes the other car's immovable barricade.

At 01:24 PM 12/13/1999 -0500, Huw Powell wrote:
>> 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
>
>
___
Bob
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