An 8 g cartridge, a 55 g car, a 20 m lane, about one second. This pack models the whole run. The impulse from the gas, the momentum and kinetic energy the car gains, the work lost to drag and to the bearings, and the energy account at the finish line. The simulator is the same one we use in our own design reviews.
Define impulse as force × time and link it to the change in momentum (J = FΔt = mΔv).
02
Calculate kinetic energy (½mv²) and the work done against a force (W = Fd).
03
Build an energy account for a run: thrust work equals the kinetic energy at the finish plus the drag loss plus the rolling loss.
04
Use the model to explain why lighter cars and lower CdA finish sooner, and why the gain is smaller than you might expect.
02From the Nightmare car
Every Nightmare design review ends the same way. We put the change into the run model and read off the finish time. The physics is short. The cartridge delivers about 1.5 N·s of impulse in roughly a quarter of a second, and that sets the momentum. Drag then takes energy back for the whole lane, hardest near top speed. The bearings take a small, steady slice. Whatever is left at 20 m is the kinetic energy the stop has to absorb. Fig. 5 is that story as one picture.
Fig. 5 Left: the cartridge’s thrust fades quickly, and the area under the curve is the impulse. Right: where the thrust work has gone by the time the baseline car crosses 20 m. The numbers come from the simulator below.
03Core idea
J = F · Δt = Δp = m · Δv Impulse in N·s equals the change in momentum in kg·m/s.
When the force changes with time, the impulse is the area under the force against time graph. The cartridge cannot change how much impulse it gives. So the only way to a bigger Δv is a smaller m. In STEM Racing, mass is speed.
KE = ½ m v² · W = F · d · Wthrust = KEfinish + Wdrag + Wrolling
The last line is the work-energy idea applied to the whole run. Everything the gas gave the car either arrives at the finish as kinetic energy or was spent pushing air and warming the bearings. The simulator works out all four numbers so you can check that the account balances.
i
Why isn’t a lighter car proportionally faster? A lighter car gets a bigger Δv from the same impulse, so it reaches a higher speed sooner. But drag grows with v², so it also loses more energy to the air. The simulator shows the net effect. It is real, but smaller than J/m on its own would suggest.
04Race simulator and the full energy account
Set up the car, then run it. The chart can show speed against time, distance against time, or force against time. The energy bars are the account: thrust work in, and the three places it can go. Try to make the balance fail. You can’t.
Model: F(t) = F₀e−t/τ. Drag = ½ρv²CdA with ρ = 1.20. Rolling = Crr·mg plus 0.02 N of tether friction. Worked out in 0.5 ms steps. The impulse J = F₀τ.
—sFinish time
—N·sImpulse J
—m/sJ ÷ m, with no losses
—m/sReal top speed
Thrust work in
—
KE at the finish
—
Lost to drag
—
Lost to rolling
—
05Worked example
Impulse: the cartridge gives an average thrust of 5 N for 0.30 s.
J = F·Δt = 5 × 0.30 = 1.5 N·s
Ideal speed change with no drag and no rolling, for m = 0.055 kg:
Δv = J / m = 1.5 / 0.055 = 27.3 m/s
KEideal = ½ × 0.055 × 27.3² = 20.5 J
Real finish from the simulator, baseline car: vfinish ≈ 20.9 m/s.
KEfinish = ½ × 0.055 × 20.9² = 12.0 J
Lost = Wthrust − KE = 18.8 − 12.0 = 6.8 J, which is about 6.3 J of drag and 0.5 J of rolling
What it means: a third of the thrust work went into the air. That is why the aero pack (LAB-03) exists, and why the bearings pack (LAB-01) is about the last 3%.
06Student questions Teacher mode is off. Answers are hidden.
Q1 · Impulse[3]
A cartridge gives an average of 4.8 N for 0.32 s. Work out the impulse. What speed would a 0.050 kg car reach if there were no losses?
Answer key
J = 4.8 × 0.32 = 1.54 N·s
Δv = J/m = 1.54 / 0.050 = 30.7 m/s
One mark for the impulse, one for rearranging, one for the answer with a unit.
Q2 · Energy[2]
Find the kinetic energy of a 0.055 kg car at 20 m/s. Then write the drag loss of 6.3 J as a percentage of the thrust work of 18.8 J.
Answer key
KE = ½ × 0.055 × 400 = 11 J
6.3 / 18.8 = 33.5%, about 34%
Q3 · Explain[3]
Use the model to explain why a lighter car finishes sooner with the same cartridge. Then give one reason the gain is smaller than Δv = J/m predicts.
Answer key
Same impulse and a smaller m means a larger Δv (from J = mΔv), so the car accelerates harder and reaches speed sooner (2). But drag grows with v², so the faster car loses a bigger share of its energy to the air. The rules also set a minimum mass, so there is a limit to how light you can go (1).
Q4 · Work[3]
Drag averages 0.20 N over the 20 m lane. How much energy is lost to drag? If the thrust work was 18.8 J, what fraction reaches the finish? Ignore rolling.
Accept any equivalent working. The marks in brackets match the student sheet.
Teacher copy Do not hand out
1. [2] J = FΔt = 6 × 0.25 = 1.5 N·s
2. [2] Δv = J/m = 1.5 / 0.060 = 25 m/s
3. [2] KE = ½ × 0.055 × 21² = 12.1 J
4. [4] Baseline: thrust work 18.8 J (100%), KE 12.0 J (64%), drag 6.3 J (33%), rolling 0.56 J (3%). Accept the student’s own simulator numbers if the percentages are consistent (2). The balance is close to zero because energy is conserved. All the work the gas did is either still in the car as KE or has gone into the air and the bearings as heat (2).
5. [3] Same impulse and a smaller mass gives a larger Δv, so the car accelerates harder and reaches top speed sooner (2). Drag grows with v², so the faster car loses a bigger share to the air, and the minimum-mass rule limits how light you can go (1).