Grand Prix Racing   The Science of Fast Pinewood Cars 
F  Force of collision 
m  Car's mass 
v  Speed of car 
vi  Initial speed of car 
t  Time duration 
ti  Initial time 
O  Collision angle 
n  Median rubbing energy loss per foot 
l  Distance that the median was rubbed by the car 
p  Portion of energy lost in a collision 
Important expressions derived for this model include:
Average force of Impact  F = m(vvi)/(tti) 
Inelastic bump and scrape  v = vi\[1p(1n)^l] 
In an inelastic collision, there is also a loss of energy to the median. The energy given up is kinetic energy. Before the bump, the sideways kinetic energy is m(visinO)2/2. After it is zero. All of the sideways kinetic energy is lost. The car remains traveling forward next to the median, possibly rubbing it. The loss of energy due to the initial impact is,
Let m = 0.01295 ozs2/in, vi = 160 and O = 5.5 deg (the maximum) so sinO = 0.09677
0.01295(160(0.09677))2/2 = 1.55 ozin.
The total kinetic energy before the collision was mvi2/2
0.01295(160)2/2 = 165.76 ozin
So the energy loss, 1.55 ozin, represents 0.94% (=100sin2O) of the linear kinetic energy.

What is the reduction in speed caused by this collision, considering only linear kinetic energy? Use a speed of 160 in/s.
v = vi\[1p] = 160\[10.0094] = 159.25 in/s.
This is a reduction of about 0.7 in/s.
To see the impact at the end of the race, plug the initial speed, vi, into the time equation of the flat track part with the flat trajectory length. Use that time but the lower speed to get the distance traveled at the lower speed on the flat part of the track. The difference between this distance and the trajectory length is the distance the bumped car lost by.
Because the across track speed of the car would be zero after a perfectly inelastic impact, these collisions do not result in zigzag paths. What would the force be? Using the change in momentum approach from the discussion of ellastic dynamics,
Fave = 0.01295(0+160)0.09677/0.01 = 20 oz

But if the across track speed of a car is effectively zero after an impact, that means the car would stay up against that side of the median and continue to scrape it all the way down! So a car that hugs one side of the lane median exhibits perfectly inelastic collision behavior, even if the collision is a little elastic to begin with. You could say that the elasticity is damped out by the force (be it misalignment or uneven axle friction) that is pulling the car to the median.
Inelastic cars are characterized by hugging the lane median all the way down. As determined when we looked at wheel alignment, this car can lose up to 1/2% of its energy every foot it rubs. This suggests a worst case model, where we assume the car has an inelastic collision after accelerating to near top speed on the ramp.
v = vi\[1p(1n)^l]
note: the "^" in X^p is read "X raised to the power of p"
where the final velocity v is the initial velocity, vi, times the square root of the energy loss factors, 1p, due to the initial inelastic impact and 1n, for n the loss per foot, raised to the power of the number of feet, l, traveled while continuing to rub the lane median.
Let's use an initial speed, vi, of 160 inches per second. Say the car bumps the median at the end of the ramp, so it rubs the median for the length of the flat, l = 249.5 inches. Assuming a bad inelastic collision, p = 0.0094, and an energy loss of 1/2 percent every foot, n = 0.005.
v = 160\[10.0094(10.005)^(249.5/12)] = 160\[0.99060.995^20.8] = 160\[0.990.9] = 160(0.3) = 48 in/s
That's about a quarter of the initial speed on the flat. Of course, the actual speed at the finish line would also be reduced by the energy lost to the other forms of friction as well. But this wild estimate gives us a hint of the effect of just the inelastic bumping and scraping alone.
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Grand Prix Racing   The Science of Fast Pinewood Cars 
Copyright © 1997, 2004 by Michael Lastufka, All rights reserved worldwide. 