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Re: crash tests, from a Physics Teacher

You need to solve this as system of simultaneous equations, so you must
simultaneously solve the Energy Equation (1/2 mv^2) with the momentum
equation (mv) for the final velocity. You solved the momentum equation,
now use the info in the energy equation to find the total changes in
energy, which would reflect the damage to the passengers. Note, in the
energy equation you need to subtract the amount of energy absorbed by
both cars, which may be at the maximum that can be absorbed by both cars.

On Tue, 14 Dec 1999 23:28:40 -0800 "David G" <daveglu@hotmail.com>
>> SO, for the 60/0 crash, we will need to take into account the amount 
>> energy absorbed by the 30/30 car, and apply that to each of the 60/0 
> The momentum of the system becomes
>1000kg(30m/s) + 1000kg(0m/s)  = 2000kg (15m/s)
>which states that both vehicles (assuming inelastic collision) will 
>traveling at 15m/s (30 mph) immediatly after the collision and the 
>becomes identical to the 30,-30 collision.
>  What have I missed.  I have no intention of starting a torsen 
>thread, I
>just want to help solve poeples misconceptions especially if they are 
>  Dave G