Consider two problems:
A. Two identical cars (same mass, same crumplability, etc,) traveling in
exactly opposite directions have a head-on collision, each traveling (say)
at 35 mph instantaneous speed in the mathematical instant prior to first
contact.
B. One of the above cars approaches an "immovable object" (i.e., an object
whose mass is MUCH larger than the mass of the car, such as a concrete
abutment quite firmly attached to planet Earth). Assume further that this
object has significantly LESS crumplability than does the car (i.e., a
rather rigid object). This car is traveling at 35 mph instantaneous speed
in the mathematical instant prior to first contact. The collision is also
square-on to an extended flat surface of the "immovable object".
Compare and contrast; talk amongst yourselves; consider the momentum and
energy (in all its forms) in the mathematical instant prior to first
contact, versus these quantities some time after first contact, when
everything has stopped moving, the dust has settled, etc.
By symmetry, it is evident that the two cars in A will come to rest with
their crumpled interface (on average) located above the exact point of the
pavement above which the first contact occurred; both cars will now have
zero momentum (because the sum of the vector mementa prior to the collision
was exactly zero). The kinetic energy (two times one-half*m*vee-squared
prior to the collision) will be zero after the collision. Since energy is
also conserved, where did this K.E. go? ANSWER: This is a completely
inelastic collision; most of the energy will have gone into plastic
deformation of the metal of each car (some of that ends up as heat which
then adds to infrared radiation trying to escape the atmosphere); into
acoustic energy -- the noise of the crash (which also ends up as thermal
and infrared); into some kinetic energy of shrapnel (which, after the
fragments fall to the ground, etc., produce more acoustic and thermal and
infrared energy).
In case B, the point of first contact, once again, is the resting point of
the (average)front surface of the now crumpled car.
In both cases, the crumpling of the car would follow the same time-profile,
with the same average forces on the chassis of all cars involved, and the
same total duration of collision, so F times delta-tee will be the same
throughout.
So cases A and B are EQUIVALENT (in terms of average forces on the
crash-test dummies <biological or otherwise> and in terms of damage
sustained by each car. The damage sustained, by the way, is independent of
the frame in which the event is described -- you can't fool Mother Nature
that way!
Clearly, colliding with a mature redwood tree (ten or more feet in
diameter) at an incoming speed of 70 mph is a worse situation, since the
base of the trunk redwood tree makes a pretty good approximation to an
"immovable object" as defined above -- the momentum of the incoming car
which must be dissipated is twice as much (again, this does NOT -- in
Newtonian approximation -- depend on the frame) and the kinetic energy to
be transformed is quadrupled (ditto re: frames). To satisfy both of these
conditions, both the average force on the car during the collision (and
thus the average force on the occupants) will be rather larger, AND the
length of car which is crumpled (this is essentially the distance the
center of mass of the car travels from first contact until coming to rest)
will be larger than in cases A and B above. (Both F times delta-tee and F
times delta-x must increase.) How much each of these increases will depend
on what fraction of the initial K.E. goes into plastic deformation versus
all the other forms mentioned above.
Peter Vajk
St. Joseph Notre Dame High School
1011 Chestnut Street
Alameda, CA 94501