# Re: crash tests, from a Physics Teacher, 2 cars at 60MPH

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>To solve the question of Dave:
>
>The equations STILL are:
>K.E. = ½mV²
>
>(momentum) p   = mV
>
>We must look at total energy and total momentum. Due to the absorbtion
>of
>energy in the (generally) in-elastic collisions of cars (very little
>bounce-back) we can assume the following.
>
>V initial of car 1 = 60 MPH
>V initial of car 2 = - 60 MPH (travelling in opposite direction. Think
>number line)
>
>Since the mass of the cars are assumed to be the same, mass can
>ultimately be dropped from the picture. Note a 2200 lb car has a mass
>of
>about 1000 kg. 60 MPH translates to roughly 30 m/s
>
>Total KE before crash:   K.E. = ½mV² + ½mV²
>
>= ½m(V²1 + V²2)
>
>= ½ (1000 kg) (30² + (-30²) = 900 kJ (note the energy goes up by 4 X
over the 30MPH crash!)

NOTE THIS: Most car crashes (by design, no less) are relatively
in-elastic, that is, very little bounce back. This minimizes the total
transfer of energy to the passengers (the goal, I think). So, in this
model, there is no bounce back and the final velocity after the head-on
crash is 0 m/s for BOTH cars. (They are stuck together at the point of
collision in one, cold, dead, stationary lump). In reality, there
probably is bounce, I am guessing in the order of maybe 5 - 10 MPH, but
since I don't know, I am not about to speculate. So, as before, we will
use the inelastic model here.
>
>Total KE after the crash: Since all is stopped, V = 0, K.E. = 0 so a
>total of 900 kJ must be absorbed, one car absorbs 450 kJ.
>
>As for Momentum:
>
>Total p before crash:    p = mV1+ mV2
>
>   = 1000 kg (30 + (-30)) = 0 kg m/s (surprise!)
>
>Total p after crash: Gee, same reason as above p = 0
>
>Change of p = m(V1 - Vend) = 1000 kg (30 - 0) = 30,000 kg m/s
>
>Momentum CHANGE for EACH car is 30,000 kg m/s, since both came to a
>stop
>from 30 m/s. If the car's bounced backwards at, say 2.5 m/s (5 MPH,
>actually pretty fast for a bounce) then the change for each car would
>INCREASE to 32,500 kg m/s.
>
>ENERGY change for each INDIVIDUAL car in the bounce case would go up
>by a
>BIGGER factor
>
>change in K.E. = ½m(V-V)²
>         = ½ (1000) * (30 - (-2.5))²  (remember, directions the -2.5
>is
>opposite the 15)
>         = 528.1 kJ !!!! FOUR times as much and nearly 20% more with
bounce-back than without
>
Again, THIS is why cars are built to
>deform, rather than being built to be rigid crashing structure (it's
>possible to do, just not too survivable).
>
>Yes, would you like to settle the 60,0 two car solution.
>I agree with all of your statements and I believe the biggest problem
>is
>listening, as I believe most of us agree on almost all points, even
>you
>answered the 60,barrier question but not the 60,0 two car question.
>However
>as you stated conservation of momentum states that both cars will be
>traveling at 30 immediatly after.  (Not true - I assumed that both cars
will stop at the point of collision, an in-elastic collision. THis is
what crumple s=zones are supposed to do.
>
>  Dave G
>3x  84 4kq's (death traps)
>  No one should be driving these older quattro's anymore and I will
>accept
>any given to me, just for everyone elses good.

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