Q: (Initially posted April 16, 2004) FROM STUDENT
MEMBER ALEXIS K. IN VA
I learned in driver's education that if a head-on collision
occurs, the car that is
going faster causes the other car to bounce backwards. If the
cars were
going the same speed, their kinetic energies would cancel out
when they
crash into each other. Here's what I would like to know: If
a head-on
between two cars is inevitable, where one car is going faster,
let's say 60
mph, and the other is going slower, 40 mph, would it better
for the slower
car to accelerate a little bit to meet the other car's force
or to brake
(This is assuming the faster car neither brakes nor speeds up)?
The reason I
ask this is because I wonder if it would be better to help cancel
out the
momentum rather than take on the entirety of the force. When
I envision the
situation in my mind, it seems like braking would be worse,
because then the
momentum of the slower car disappears and it's as if it no longer
has any
defense against the force that is about to crush it. Either
way, the outcome
isn't going to be pretty, but I'm wondering if it's at all possible
to
lessen the amount of damage. |
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April 19, 2004
A: FROM MENTOR MOLLY WILLIAMS
IN MI
No, No, No, Alexis! Do NOT speed up into a head-on collision!
The
damage from a collision happens because of the kinetic energy
that must be
dissipated. Kinetic energy is proportional to the square of
the velocity,
so when a car stops from an initial speed of 60 mph it has
more than twice
as much energy to dissipate as there would be if the car were
traveling 40
mph. (60^2 divided by 40^2 = 2.25). That energy has to go
somewhere when
the car stops. In normal driving, much of that energy goes
safely into
heating up the brakes. However, in a collision, the energy
crunches the car
and also injures the passengers. Airbags also serve as cushions
to absorb
energy and minimize passenger injury. The slower the speeds
in a collision,
the less damage occurs overall. The best course of action
(assuming the
collision is not otherwise avoidable) is to brake, reducing
your own speed
as much as you can.
*******************
A: FROM MENTOR LESLIE WAITE
IN CA
Hi Alexis!
When two cars collide, assuming that they are of similar size
and
weight, The total momentum of the crash is split equally between
them. Thus, the slower car will bounce back because half of
the total
crash momentum is more than its original momentum. BOTH CARS
STILL
TAKE THE SAME IMPACT, WHICH IS EQUAL TO HALF THE MOMENTUM
OF THE
CRASH. In the end, both cars go from whatever speed they are
going to
zero in a fraction of a second.
When I was 17, I was in a crash which is the extreme example
of what
you are talking about: I was going about 40 MPH on a street,
and ran
right into a car that was stopped waiting to make a left hand
turn.
Being the faster car in the collision really didn't change
the fact
that I went from 40 MPH to zero MPH in less than a quarter
of a
second. I still flew at the windshield and hit it with great
force, I
still slammed against the steering wheel, and that was WITH
a seat
belt on (although the shoulder restraint did give way, so
I was not
as protected as I could have been). My little car was destroyed,
and
the big car that I hit was slightly damaged. Which brings
up the next
point: collisions won't always involve cars of equal size
and crash
resistance. If you are in a VW beetle, and you slam into an
SUV, you
are going to be worse off, no matter how fast you are going,
and how
slow they are going. It's not just speed, it's momentum.
The other thing to realize is that collisions happen fast.
If you are
going to have a head-on collision, and you have the time to
realize
it and react, that time would be FAR better spent maneuvering
to
avoid the collision or to minimize the impact.
BUT setting aside reality and going with the theoretical aspects
of
your question- as I mention above, occupants inside BOTH cars
take
the impact, not just those in the slower one. In your scenario,
if
one car is going at 60 MPH and the other at 40 MPH (assuming
they are
similar size and weight), BOTH cars will hit with the force
of 100
MPH. The slower car will bounce back, but that is NOT because
it
takes more of the impact, it is because it is absorbing half
the
momentum of the crash, and half the momentum of the crash
is more
than its original momentum (see my math below, then take physics-
this will be clearer). Speeding up to "match" the
other car's speed
will just increase the force of the collision to 120 MPH.
All you are
doing is increasing the total force of the crash for BOTH
cars: both
will absorb the same impact force in the end (equal to half
the total
momentum of the crash). So in the end, speeding up will do
more
damage to BOTH cars.
So the question is: do you want to hit a wall at 100 MPH,
120 MPH, or
have both of you slow down and hit at 80 MPH? The answer is:
wear
your seat belt!
MATH BELOW: let's assume both cars weigh 1 "unit".
Momentum is mass X
velocity, so:
for car 1 going at 60 MPH, it's momentum is 60 X 1 = 60 "momentums"
(my unit for this case)
for car 2, 40 MPH X 1= 40 momentums.
In the collision, total momentum is 60 + 40 momentums=100
momentums total.
This will be split equally between the two cars: 100/2=50
momentums per car.
So for car one, it will absorb 50 momentums, and it may even
move
forward a bit after the crash. BUT- it still absorbs 50 momentums
of
impact.
For car two, it will absorb 50 momentums of impact, but since
it
started at a momentum of 40, it will go backwards after the
crash.
BUT-It will still be absorbing the same amount of impact as
car one.
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