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IB Physics SL Paper 2 Structured Questions Practice Paper | 6 questions, 50 marks, 1 hour 15 minutes
IB Physics SL Paper 2
IB Physics SL Paper 2 practice paper: 6 questions, 50 marks, 75 minutes; Section A data-based questions on free fall, circuits, calorimetry and waves, Section B on circular motion and decay, with full solutions.
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Paper details Level: IB SL | Subject: Physics | Time: 1 hour 15 minutes | Marks: 50 | Questions: 6
Section A: data-based questions (30 marks)
Structured questions: show your working and give correct units; answer every part of each question. Total 50 marks.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A small metal ball is released from rest and its speed is recorded by a data logger. Time / s: 0, 0.2, 0.4, 0.6, 0.8; speed / m s⁻¹: 0.0, 1.9, 3.9, 5.9, 7.8. (a) State what physical quantity the gradient of the straight line through these data represents. (1 mark) (b) Determine that quantity from the data and calculate the percentage difference from g = 9.81 m s⁻². (2 marks) (c) Calculate the distance fallen between 0 and 0.8 s. (2 marks) (d) Explain why the acceleration found in (b) is slightly smaller than g. (2 marks) |
Answers: (a) The gradient represents acceleration. (b) a = (7.8 − 0.0) m s⁻¹ ÷ (0.8 − 0.0) s = 9.75 m s⁻²; percentage difference = |9.75 − 9.81| ÷ 9.81 × 100 = 0.61%. (c) Distance = area under the speed–time graph, summed in strips: 0.2 × [(0.0 + 1.9) + (1.9 + 3.9) + (3.9 + 5.9) + (5.9 + 7.8)] ÷ 2 = 3.12 m. (d) Air resistance acts upwards, so the resultant force is smaller than the weight and the acceleration is below g; drag grows with speed, so the average acceleration over the whole fall is less than 9.81 m s⁻². |
| 2 | A battery has electromotive force 6.0 V. When a 10.0 Ω fixed resistor is connected across it, the current is 0.55 A and the potential difference across the resistor is 5.5 V. (a) Calculate the internal resistance of the battery. (3 marks) (b) Calculate the energy dissipated inside the battery each second. (2 marks) (c) Calculate the efficiency of this circuit. (1 mark) (d) State whether the output power is greatest when the external resistance equals the internal resistance, and calculate that maximum output power. (2 marks) |
Answers: (a) r = (E − V) ÷ I = (6.0 − 5.5) V ÷ 0.55 A = 0.91 Ω. (b) Power dissipated inside = I²r = (0.55)² × 0.909 = 0.28 W, i.e. 0.28 J each second. (c) Efficiency = V ÷ E × 100 = 5.5 ÷ 6.0 × 100 = 91.7%. (d) Yes: output power is greatest when the external resistance equals the internal resistance; P = E² ÷ (4r) = 36 ÷ (4 × 0.909) = 9.9 W. |
| 3 | A 1500 W immersion heater raises the temperature of 0.50 kg of water from 18.0 °C to 62.0 °C in 68 s. The specific heat capacity of water is 4180 J kg⁻¹ K⁻¹. (a) Calculate the energy absorbed by the water. (3 marks) (b) Calculate the energy supplied by the heater in this time. (1 mark) (c) Calculate the efficiency of the heating process. (2 marks) (d) State one way in which energy is lost and how that loss can be reduced. (2 marks) |
Answers: (a) Q = mcΔT = 0.50 × 4180 × 44.0 = 91960 J. (b) E = Pt = 1500 × 68 = 102000 J. (c) Efficiency = Q ÷ E × 100 = 91960 ÷ 102000 × 100 = 90.2%; 10040 J is lost. (d) Energy is lost by conduction, convection and radiation through the walls of the container; adding insulation (for example a foam lid) or reducing the exposed surface area reduces the loss. |
| 4 | A string of length 1.20 m is fixed at both ends and vibrates at 180 Hz in a standing wave with three antinodes. Its linear mass density is 2.0 × 10⁻³ kg m⁻¹. (a) Calculate the wavelength of the standing wave. (2 marks) (b) Calculate the speed of the wave on the string. (2 marks) (c) Calculate the tension in the string. (2 marks) (d) State the number of nodes on the string. (1 mark) |
Answers: (a) Three antinodes means the third harmonic: L = 3λ ÷ 2, so λ = 2L ÷ 3 = 2 × 1.20 ÷ 3 = 0.80 m. (b) v = fλ = 180 × 0.80 = 144 m s⁻¹. (c) v = √(T ÷ μ), so T = μv² = 2.0 × 10⁻³ × 144² = 41.5 N. (d) Nodes = antinodes + 1 = 4. |
Section B: extended questions (20 marks)
Structured questions: show your working and give correct units; answer every part of each question. Total 50 marks.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | A ball of mass 0.25 kg on a light string of length 0.80 m moves in a horizontal circle at 2.5 revolutions per second. (a) Calculate the angular speed of the ball. (2 marks) (b) Calculate the linear speed of the ball. (2 marks) (c) Calculate the centripetal acceleration. (2 marks) (d) Calculate the tension in the string. (2 marks) (e) Calculate the kinetic energy of the ball. (2 marks) |
Answers: (a) ω = 2πf = 2π × 2.5 = 15.71 rad s⁻¹. (b) v = ωr = 15.71 × 0.80 = 12.6 m s⁻¹. (c) a = ω²r = 15.71² × 0.80 = 197.4 m s⁻² (about 20.1 g). (d) The tension provides the centripetal force: T = ma = 0.25 × 197.4 = 49.3 N. (e) KE = ½mv² = 0.5 × 0.25 × 12.6² = 19.7 J. |
| 6 | A radioactive source has an activity of 4800 Bq and a half-life of 8.0 days. (a) Calculate the activity after 24 days. (3 marks) (b) State how activity depends on the number of undecayed nuclei, and give the fraction of nuclei remaining after 24 days. (2 marks) The mass defect of a helium-4 nucleus is 0.0304 u, and 1 u is equivalent to 931.5 MeV. (c) Calculate the binding energy of the helium-4 nucleus in MeV. (3 marks) (d) Calculate the binding energy per nucleon. (2 marks) |
Answers: (a) 24 days is 3 half-lives, so activity = 4800 ÷ 2³ = 600 Bq. (b) Activity is proportional to the number of undecayed nuclei (A = λN); the fraction remaining after 24 days is 1 ÷ 2³ = 0.125, i.e. 12.5%. (c) Binding energy = 0.0304 × 931.5 = 28.3 MeV. (d) A helium-4 nucleus has 4 nucleons, so the binding energy per nucleon is 28.3 ÷ 4 = 7.08 MeV. |
A TutorZone practice paper; original questions and passages, not publisher material.
