On Thu, 9 Mar 2000, K M wrote:
> Fill in please. If two Bricklins, each going 60 mph, have a head-on
> collision, that is like one Bricklin running into a brick wall at (how many
> mph)? Thanks, it will settle a debate, and it doesn't have to be Bricklins
> but it does have to be two moving, equal masses hitting head to head and the
> wall has to be immovable. Thanks, Kim.
>
>
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
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