Re: crash tests, from a Physics Teacher

[Orin Eman <orin@wolfenet.com>]

> > SO, for the 60/0 crash, we will need to take into account the amount of
> > energy absorbed by the 30/30 car, and apply that to each of the 60/0 cars

>  The momentum of the system becomes
> 1000kg(30m/s) + 1000kg(0m/s)  = 2000kg (15m/s)
> which states that both vehicles (assuming inelastic collision) will be
> traveling at 15m/s (30 mph) immediatly after the collision and the impact
> becomes identical to the 30,-30 collision.
>   What have I missed.

Nothing.

You can pick any frame of reference you want, moving or stationary.
Taking the inside of a moving car as your frame of reference,
you go from 0 to -30 mph in both cases (assuming inelastic collisions).

BTW, the energy numbers do work out in the 'stationary'
(excluding rotation of earth etc. etc.) frame of reference:

In both the following v = 60, m = mass of car.

60 case.

Kinetic Energy initially = mv^2/2.
KE finally = 2m (v/2)^2/2 = mv^2/4
Change in KE = mv^2/4

30/30 case. 

KE initially  = 2m(v/2)^2/2  = mv^2/4
KE finally = 0.
Change in KE = mv^2/4.

For the pedants, the parked car would be worse due
to extra forces... frictional forces (parking brake if on,
driveline friction if out of gear and parking brake off).

Orin.