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*To*: daveglu@hotmail.com, quattro@audifans.com*Subject*: Re: crash tests, from a Physics Teacher, 2 cars at 60MPH*From*: Lawrence C Leung <l.leung@juno.com>*Date*: Tue, 14 Dec 1999 20:20:13 -0500*Sender*: owner-quattro@audifans.com

>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.

**Follow-Ups**:**Re: crash tests, from a Physics Teacher, 2 cars at 60MPH***From:*"David G" <daveglu@hotmail.com>

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