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Re: momentum



I agree with your posts concerning the analysis of this problem. (So do
most of the authors on my shelf.) I must also say that cars don't
bounce. Having also been an EMT for several years, I have seen my share
of MVA's (Motor Vehicle Accidents), and NONE of the head-on collisions
has resulted in a "bounce".

Peter Schoch

Michael Edmiston wrote:

Jack Uretsky says I am incorrect that a collision with a tree (or brick
wall) brings the car to rest at the tree/wall. I think Jack needs to
view a few test-crash videos, or visit a few accident sites. Any
"bounce back" is quite small. The front bumper of the car will be very
close to (or wrapped around) the tree. Because the bumper may be lower
than the center of mass, the car will often have the back end rise off
the ground during the collision. When this comes back down, it may
give one the impression that the car rebounded. But there is actually
very little rebound.

Jack would be correct that there would be some rebound if this were a
partly elastic collision. But this is almost perfectly inelastic. Our
familiar elastic-collision equations are of little value to us here.

Jack says I am incorrect about a moving car hitting a parked car. He
says the parked car will shoot ahead and the first car will come to
rest. He does admit that his analysis assumes hardened cars colliding
elastically. But why would we want to make those assumptions when they
are so obviously incorrect. This is one place where I will acknowledge
that it makes some sense to bring energy analysis into the picture
(along with momentum analysis). Jack also says that my statement that
the time of deceleration increases does not follow. I beg to differ.
Here is the analysis.

(1) Is there damage (crumpling) of either car when a moving car hits a
parked car? Yes. Then some kinetic energy was converted to thermal
energy... perhaps quite a bit.

(2) If kinetic energy is not conserved, then (by definition) this is an
inelastic collision.

(3) ONLY in the case of an elastic collision (between equal mass and
frictionless cars) will the first car stop and the second car assume
the velocity of the first car. In an inelastic collision between equal
mass cars (one parked) the moving car CANNOT stop at the point of
impact; it must travel beyond the point of impact.

(4) If the first car travels beyond the point of impact, it must move
for a longer period of time than the time it would have traveled had it
come to rest at the point of impact. The deceleration is spread over
greater time and greater distance.

(5) If the deceleration of the car is spread over longer time, the
forces on the car are reduced (Newton's second law). Hence, the damage
is less. Incidentally, this is why you "give" with the ball when you
catch a fast baseball.

Michael D. Edmiston, Ph.D. Phone/voice-mail: 419-358-3270
Professor of Chemistry & Physics FAX: 419-358-3323
Chairman, Science Department E-Mail edmiston@bluffton.edu
Bluffton College
280 West College Avenue
Bluffton, OH 45817