The questions are on the angular momentum test. Take it under real conditions before reading the mark scheme.
Quick answer: Angular momentum is L = Iω, and it stays constant when no net external torque acts, so a spinner who pulls mass inward spins faster. This mark scheme gives the model answers and the marks for the 8 questions of the AP Physics 1 test on angular momentum, from definitions to ball up a slope, with the equivalent answers to accept.
Model answers with the mark scheme. Award method marks for correct conservation equations even if the arithmetic slips, and apply the unit mark in each calculation.
Level: AP Physics 1 (algebra-based) · Curriculum: AP P1 Unit 6
Question 1 Definitions (2 marks)
a. L = Iω.
b. The net external torque on the system must be zero.
Marking: 1 mark each.
Question 2 Flywheel (2 marks)
a. K = ½Iω² = ½ × 0.80 × (25)² = 2.5 × 10² J (250 J).
b. L = Iω = 0.80 × 25 = 20 kg m²/s.
Marking: 1 mark each, with units.
Question 3 Spinning skater (3 marks)
a. Iiωi = Ifωf: ωf = 3.0 × 2.4 ÷ 1.2 = 6.0 rad/s.
b. Ki = ½ × 3.0 × (2.4)² = 8.64 J and Kf = ½ × 1.2 × 36 = 21.6 J, so the kinetic energy increases: she does work pulling her arms inwards.
Marking: (a) 1 mark for the conservation equation, 1 mark for the result; (b) 1 mark for the increase with the reason.
Question 4 Rolling cylinder (3 marks)
a. Ktrans = ½mv² = ½ × 2.0 × 16 = 16 J.
b. Krot = ½ × ½MR² × (v/R)² = ¼Mv² = 8.0 J.
c. Ktotal = 16 + 8.0 = 24 J.
Marking: 1 mark each; allow error carried forward in (c).
Question 5 Racing down a ramp (3 marks)
a. The solid disk reaches the bottom first.
b. Both lose the same gravitational potential energy mgh. A hoop has I = MR², so a larger share of its energy is kept as rotation and its centre moves slower; the disk has I = ½MR², so more energy becomes translation and it reaches a higher speed (√(4gh/3) compared with √(gh)).
Marking: (a) 1 mark; (b) 1 mark for equal energy available, 1 mark for the greater share of rotational energy in the hoop.
Question 6 Clay lump on a turntable (3 marks)
a. If = 0.20 + 0.050 = 0.25 kg m². ωf = 0.20 × 6.0 ÷ 0.25 = 4.8 rad/s.
b. Ki = ½ × 0.20 × 36 = 3.6 J; Kf = ½ × 0.25 × (4.8)² = 2.88 J. Lost = 0.72 J.
Marking: (a) 1 mark for the conservation equation, 1 mark for the result; (b) 1 mark.
Question 7 Work by a torque (2 marks)
a. W = τΔθ = 3.0 × 8.0 = 24 J.
b. ½Iω² = 24 J, so ω = √(2 × 24 ÷ 0.30) = √160 = 13 rad/s (12.6).
Marking: 1 mark each.
Question 8 Ball up a slope (2 marks)
Ktotal = ½mv² + (1/5)mv² = (7/10)mv². Setting this equal to mgh gives h = 0.70 v²/g = 0.70 × 12.25 ÷ 9.81 = 0.87 m.
Marking: 1 mark for total kinetic energy in terms of v; 1 mark for the height.
