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Comparing Collisions

Date: 05/20/2002 at 23:56:10
From: Steve Ginkel
Subject: kenetic energy

If there are two cars with the same weight going head on at each 
other at exactly 35 mph, will the crash be equivalent to a 70 mph 
crash?

That is, would it be like one car hitting a brick wall at 70 mph? 


Date: 05/21/2002 at 09:42:28
From: Doctor Rick
Subject: Re: kenetic energy

Hi, Steve.

The relative velocity between the cars is the same as the relative 
velocity between the car and the brick wall. Thus the crash happens 
just as fast, and that is significant. But to decide which crash does 
more damage, we need to consider how much energy a car absorbs in the 
collision; that's the best measure we can get of the amount of damage 
sustained.

In the head-on collision, the total momentum is zero. If all the 
kinetic energy is absorbed by the cars, they will end up stationary. 
The total momentum after the collision must be the same as before the 
collision, namely zero, and ending up with both cars stationary is 
consistent with this. They might also bounce off each other, 
retaining some kinetic energy and reducing the energy absorbed, but 
I'm considering the worst case. The amount of kinetic energy absorbed 
in this case is

  (1/2)m(35mi/hr)^2 + (1/2)m(35 mi/hr)^2 = 1225m

where m is the mass of each car. That's in strange units (pound-miles 
squared per hour squared), but since all I want to do is compare two 
numbers, I don't need to convert this to more conventional units.

We can suppose, in the head-on collision, that the cars will absorb equal
amounts of energy, 612.5m units each. This is not necessarily the case,
unless the crash is perfectly symmetrical, but it's a reasonable
assumption for simplicity's sake.

Now we want to consider the car hitting the wall. To get there, 
though, let's first think about a similar case: the car going 70 mph 
smashes into a stationary car of mass M. Only the one car is moving 
at the start, so the initial kinetic energy is

  (1/2)m(70 mi/hr)^2 = 2450m

The momentum going into the collision is

  m*70 mi/hr + M*0 mi/hr

Let's say the cars are welded together by the impact, so that they 
move together afterward, with velocity v. The final momentum is 

  (m+M)*v

The final momentum must equal the initial momentum, so we can solve 
for v:

  v = 70 mi/hr * m/(m+M)

The final kinetic energy, then, is

  (1/2)(m+M)(70*m/(m+M))^2 = (1/2)70^2 * m^2/(m+M)

If the two cars are the same mass m (as in the head-on collision), 
the final kinetic energy is (1/4)70^2 m = 1225m. Thus the total amount of 
energy that must be absorbed in the impact is

  2450m - 1225m = 1225m

That's the same as in the head-on collision. Thus we can say that a 
head-on collision is just as bad as a car going twice as fast hitting 
a stationary *car*. Again it's reasonable to suppose that each car
absorbs half the energy, or 612.5m units.

On the other hand, consider the case in which the second car's mass M 
is much larger than m. Then the denominator (m+M) is not much 
different from M alone, and the final kinetic energy is

  (1/2)70^2 * m^2/M

The larger the mass M, the smaller the final kinetic energy, and the 
more energy must be absorbed in the collision. If we consider the 
wall to be a "car" of essentially infinite mass (because it's 
attached to the earth), the final kinetic energy is zero, and the 
full 2450m units of energy must be absorbed in the collision. Thus a 
head-on collision is only half as bad as hitting a *wall* at twice 
the speed.

I have made assumptions here: for one thing, that the cars stick 
together rather than bouncing away, in which case there would be some 
kinetic energy at the end and not all the energy would go into 
damaging the cars. 

Also, I assumed that the car and the wall absorb the same amount of 
energy -- that is, the collision hurts the wall as much as the car -- 
because that's more or less what will happen when two similar cars 
collide. When it comes to the car hitting the wall, it isn't so easy
to  say how the total energy absorbed in the collision is divided --
which would sustain more damage, a car or a wall.

If the wall "crumples" more easily than the car, thus absorbing
much of the energy, then the car could suffer less damage than my
total-energy calculations indicate. I could imagine a "wall"
(not brick or concrete, but like a big air bag) absorbing *all* the
energy, so the car could drive away from the accident!

On the other hand, if the wall is untouched and the car absorbs all the
energy, then the wall acts like a mirror: the car hitting the wall is
just like a car hitting its reflection -- a car traveling toward it at
the same speed. The car absorbs the entire 2450m units of energy, 4 times
the energy absorbed by each car in the 35 mph head-on.

Those are the extreme possibilities: the energy absorbed by the car when
it hits the wall at 70 mph is somewhere between 0 and 4 times the energy
it would absorb in a 35 mph head-on collision with an identical car.
That's about all I can say without actually doing the experiment!

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Physics/Chemistry

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