The method and the results are on the terminal velocity lab. Do the analysis yourself first, then check the answers below.
Quick answer: Terminal velocity is the steady speed at which a resistive force balances the weight, so the net force is zero. These answers go with the AP Physics C lab on terminal velocity: the graph, the calculations, the uncertainties and the conclusion explained, for all 7 analysis questions.
Model answers use the sample data above. Your own values will differ a little, but the graph should still be a straight line through the origin with c near a few times 10⁻³ kg/m for cases of this size.
Level: AP Physics C (Mechanics and E&M, calculus-based) · Curriculum: AP PC:M Unit 2
v = 1.49, 2.13, 2.63, 3.03, 3.33 m/s and v² = 2.23, 4.54, 6.92, 9.18, 11.09 m²/s² for n = 1 to 5. Result: v² rises from 2.2 to 11.1 m²/s².
Question 2 Graph
The points lie on a straight line through the origin. Using m = 4.00 g = 4.00 × 10⁻³ kg: gradient = 11.09/(4.00 × 10⁻³) = 2.8 × 10³ m²s⁻²kg⁻¹.
Question 3 Drag constant
Gradient = g/c, so c = 9.81/(2.8 × 10³) = 3.5 × 10⁻³ kg/m.
Question 4 Why the last metre?
The cases start from rest and accelerate before reaching terminal speed. Timing only the last metre, after 1.5 m of fall, means v is close to vt, so the speed is nearly constant over the timed section.
Question 5 Uncertainty
Percentage uncertainty in t = 0.02/0.30 = 6.7 %, so in v it is 6.7 %. Squaring doubles the percentage, so the uncertainty in v² is 13 %.
Question 6 Conclusion
v² ∝ m, so vt ∝ √m. This agrees with quadratic drag, vt = √(mg/c), and rules out linear drag, which would give v ∝ m.
Question 7 Improvements
Use a longer drop so terminal speed is surely reached, and film at a higher frame rate to cut the timing uncertainty. Averaging more repeats also helps. Longer drop, faster video.
