IB Physics SL Paper 2 Practice Paper | 10 Questions, 100 Marks, 75 Minutes

2026-11-01

IB Physics SL

IB Physics SL Paper 2 practice paper: electromagnetic induction, radioactive decay and efficiency; 10 questions, 100 marks, 75 minutes, with full answers and marking notes.

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Section A: Kinematics and mechanics (Questions 1-4, 40 marks)

Answer all questions. A calculator and the data booklet may be used. Show all working; answers without working score no marks.

#	Question	Answer and marking notes
1	(a) A car accelerates uniformly from 12 m/s to 30 m/s in 6.0 s. Find its acceleration. (3 marks) (b) Find the distance travelled during these 6.0 s. (3 marks) (c) The car then decelerates uniformly at 2.5 m/s² until it stops. Find the time taken. (4 marks)	(a) a = (30 – 12) / 6.0 = 3.0 m/s². (3 marks) (b) s = (12 + 30) / 2 × 6.0 = 126 m. (3 marks) (c) t = 30 / 2.5 = 12 s. (4 marks)
2	(a) A ball is thrown horizontally at 15 m/s from a height of 20 m (g = 9.81 m/s²). Find the time of flight. (3 marks) (b) Find the horizontal range. (3 marks) (c) Find the speed of the ball on landing. (4 marks)	(a) 20 = ½ × 9.81 × t², so t = √(2 × 20 / 9.81) = 2.02 s. (3 marks) (b) range = 15 × 2.02 = 30.3 m. (3 marks) (c) vertical velocity = 9.81 × 2.02 = 19.8 m/s; speed = √(15² + 19.8²) = 24.8 m/s. (4 marks)
3	(a) A block of mass 4.0 kg slides down a slope inclined at 30° to the horizontal. Find the component of weight along the slope. (3 marks) (b) The friction force is 6.0 N. Find the work done by the resultant force while the block moves 5.0 m from rest. (4 marks) (c) Find the speed of the block at the bottom of the slope. (3 marks)	(a) component = 4.0 × 9.81 × sin 30° = 19.62 N. (3 marks) (b) resultant = 19.62 – 6.0 = 13.62 N; work = 13.62 × 5.0 = 68.1 J. (4 marks) (c) ½ × 4.0 × v² = 68.1, so v = 5.84 m/s. (3 marks)
4	(a) A trolley of mass 2.0 kg moving at 3.0 m/s collides with a stationary 3.0 kg trolley and they stick together. Find their common speed after the collision. (3 marks) (b) Find the kinetic energy lost in the collision. (4 marks) (c) Find the impulse acting on the 3.0 kg trolley during the collision. (3 marks)	(a) 2.0 × 3.0 = 5.0 × v, so v = 1.2 m/s. (3 marks) (b) initial kinetic energy = ½ × 2.0 × 3.0² = 9.0 J; final = ½ × 5.0 × 1.2² = 3.6 J; loss = 5.4 J. (4 marks) (c) impulse = 3.0 × 1.2 = 3.6 N s, in the direction of the original motion. (3 marks)

Section B: Circular motion, oscillations and waves (Questions 5-6, 20 marks)

Answer all questions. A calculator and the data booklet may be used. Show all working; answers without working score no marks.

#	Question	Answer and marking notes
5	(a) A car of mass 1200 kg travels at 18 m/s round a horizontal bend of radius 45 m. Find the centripetal force required. (4 marks) (b) The maximum coefficient of static friction between the tyres and the road is 0.65. Determine whether the car can take the bend safely at 18 m/s, and find the maximum safe speed. (3 marks) (c) Find the angle of banking that would allow the bend to be taken at 18 m/s without friction. (3 marks)	(a) F = m v² / r = 1200 × 18² / 45 = 8640 N. (4 marks) (b) maximum friction = 0.65 × 1200 × 9.81 = 7652 N, which is less than 8640 N, so 18 m/s is not safe; the maximum safe speed is v = √(0.65 × 9.81 × 45) = 16.9 m/s. (3 marks) (c) tan θ = v² / (r g) = 18² / (45 × 9.81) = 0.734, so θ = 36.3°. (3 marks)
6	(a) A simple pendulum has length 0.80 m. Find its period (g = 9.81 m/s²). (4 marks) (b) A sound wave has frequency 425 Hz and speed 340 m/s. Find its wavelength. (3 marks) (c) A string of length 0.90 m vibrates in its third harmonic with a wave speed of 120 m/s. Find the frequency. (3 marks)	(a) T = 2π √(L / g) = 2π √(0.80 / 9.81) = 1.79 s. (4 marks) (b) λ = v / f = 340 / 425 = 0.80 m. (3 marks) (c) third harmonic: λ = 2L/3 = 0.60 m, so f = 120 / 0.60 = 200 Hz. (3 marks)

Section C: Thermal physics and electricity (Questions 7-8, 20 marks)

Answer all questions. A calculator and the data booklet may be used. Show all working; answers without working score no marks.

#	Question	Answer and marking notes
7	(a) 2.0 mol of an ideal gas occupies 0.045 m³ at 300 K. Find the pressure (R = 8.314 J mol⁻¹ K⁻¹). (4 marks) (b) Find the energy needed to heat 0.50 kg of water from 20 °C to 60 °C (specific heat capacity 4180 J kg⁻¹ K⁻¹). (3 marks) (c) 0.10 kg of ice melts completely on absorbing 3.34 × 10⁴ J. Find the specific latent heat of fusion of ice. (3 marks)	(a) p = n R T / V = 2.0 × 8.314 × 300 / 0.045 = 110853 Pa, about 1.1 × 10⁵ Pa. (4 marks) (b) E = m c ΔT = 0.50 × 4180 × 40 = 83600 J. (3 marks) (c) L = E / m = 3.34 × 10⁴ / 0.10 = 334000 J/kg. (3 marks)
8	(a) A battery of e.m.f. 12 V and internal resistance 0.5 Ω is connected in series with 4.0 Ω and 6.0 Ω resistors. Find the current. (4 marks) (b) Find the terminal potential difference of the battery. (3 marks) (c) Find the power dissipated in the 4.0 Ω resistor. (3 marks)	(a) I = 12 / (4.0 + 6.0 + 0.5) = 1.14 A. (4 marks) (b) V = 12 – 1.14 × 0.5 = 11.43 V. (3 marks) (c) P = I² R = 1.14² × 4.0 = 5.22 W. (3 marks)

Section D: Induction, nuclear physics and energy (Questions 9-10, 20 marks)

Answer all questions. A calculator and the data booklet may be used. Show all working; answers without working score no marks.

#	Question	Answer and marking notes
9	(a) A coil has 200 turns and a total area of 0.015 m². The magnetic flux density through the coil rises from 0.20 T to 0.60 T in 0.10 s. Find the induced e.m.f. (4 marks) (b) Find the flux linkage when the flux density reaches 0.60 T. (3 marks) (c) The flux density then falls from 0.60 T to zero in 0.05 s. Find the induced e.m.f. (3 marks)	(a) ΔΦ = (0.60 – 0.20) × 0.015 = 0.006 Wb; ε = N ΔΦ / Δt = 200 × 0.006 / 0.10 = 12 V. (4 marks) (b) flux linkage = N B A = 200 × 0.60 × 0.015 = 1.8 Wb turns. (3 marks) (c) ε = 200 × 0.60 × 0.015 / 0.05 = 36 V, in the opposite direction to part (a). (3 marks)
10	(a) A radioactive nuclide has a half-life of 5.27 years. Find the fraction of nuclei remaining after 15.8 years. (4 marks) (b) A nuclear reaction has a mass defect of 0.0025 u. Find the energy released (1 u = 931.5 MeV). (3 marks) (c) A power station has a thermal input of 1500 MW and an electrical output of 500 MW. Find its efficiency. (3 marks)	(a) 15.8 / 5.27 = 3 half-lives, so the fraction is (1/2)³ = 0.125, i.e. 12.5%. (4 marks) (b) E = 0.0025 × 931.5 = 2.33 MeV, about 2.3 MeV. (3 marks) (c) efficiency = 500 / 1500 × 100% = 33.3%. (3 marks)

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