On Fri, 17 Mar 2000, K M wrote:
> Steve,
> I am a bit brain dead but are you saying that the force of two cars hitting
> headon at 60 mph would be the same as one car going 60 hitting an immovable
> wall. That is my understanding of the physics but I have to admit that I
> never took that subject in school. By the way, that is what a physics book
> that I have here says. I always thought that it would be 120 but apparently
> that is not true because of the equal mass and the elasticity. Kim
==========================================
"Force" isn't really the operational concept here. If the wall isn't
"hurt" by the collision, then the only other place for the energy to go
(assuming - and it's a big assumption - that the car does not
"bounce" off the wall but comes to a dead stop), then all the energy will
go towards messing up the car.
What I am saying is the final effect (meaning the condition of each car
after the collision) is the same whether it hits an elastic wall (meaning
the wall "reflects" all the energy back) and doesn't bounce, etc., or
whether it hits an oncoming car (head on, no bouncing off, etc.) going at
the same rate of speed. Remember that in the 2-car case there's twice as
much energy to be abosrbed, but there are 2 cars !
And of course all this is based on the assumption that inelastic energy
absorbtion is what trashes the car.
===========================================================
> >Let's assume that (a) the amount of damage caused to a car is directly
> >proportional to the amount of energy it absorbed, and (b) what you want to
> >know is how fast the single Bricklin must be going when it hits the wall
> >to have the same amount of damage as each of two Bricks hitting each
> >other.
> >
> >Two car case: Energy of Car 1 = (1/2)mv^2 m=mass v=velocity
> > Energy of Car 2 = same
> >
> > Making the usual assumptions (they hit perfectly symmetrically,
> >both cars come to a complete stop immediately after the collison), each
> >Car will absorb an equal amount of energy, i.e. (1/2)mv^2 worth and be
> >damaged equally.
> >
> >One car case: Assume the wall is totally elastic (i.e. does not
>absorb any
> >energy), then all the energy of the collision is absorbed by the
> >Car. Then clearly this car must also be going at the same velocity as the
> >cars in the "two car" example.
> >
> >And remember, the above example is hinged on the assumption that absorbed
> >energy is what causes damage.
> >
> >=================================================
> >
> >Now if you assume that in the two car case, one of the cars is like
> >"Kitt" from Knight Rider, then the second car sustains all the
> >damage. Assume Kitt is totally elastic and Car 2 is totally inelastic,
> >and that Kitt has the same mass as Car 2.
> >
> > "Damage" sustained by Car 2 = 2 x (1/2)mv^2
> > "Damage" sustained by Kitt = zippo
> >
> >So now in the 1 car/wall case, this car must be going at SQRT(2) x v, or
> >about 1.414 x v to be trashed as much as Car 2 in the "Kitt" example.
> >
> >Definitons: v = velocity
> > m = mass
> > ^ = "raised to the power"
> > SQRT = square root
> > elastic = does not absorb any energy
> > inelastic = absorbs all the energy
> >
> >
> >------------------------------------------------------------------------
> >I am Pentium of Borg, precision is futile,
> >you will be approximated...
> >
> > _____ . .
> > ' \\ . .
> >|>>
> > O// . . |
> > \_\ . . . |
> > | | . . . . |
> > / | . sbepko@bcpl.net . . . |
> > / .| Baltimore, MD. . ....o |
> >-------------------------------------------------------------------------
> >
>
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------------------------------------------------------------------------
I am Pentium of Borg, precision is futile,
you will be approximated...
_____ . .
' \\ . . |>>
O// . . |
\_\ . . . |
| | . . . . |
/ | . sbepko@bcpl.net . . . |
/ .| Baltimore, MD. . ....o |
-------------------------------------------------------------------------
|