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HKDSE Physics Paper 1B Structured Practice Paper | 10 questions, 100 marks, 2 hours 30 minutes
HKDSE Physics Paper 1B
HKDSE Physics Paper 1B practice paper: 10 questions, 100 marks, 2.5 hours, covering kinematics, forces, energy, momentum, waves, refraction, circuits, gas laws and half-life, with full worked answers.
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Paper details Level: HKDSE | Subject: Physics | Time: 2 hours 30 minutes | Marks: 100 | Questions: 10
Section A: mechanics and waves (50 marks)
Structured questions: write down the formulae and your working, give units and use appropriate significant figures, and answer every part. Total 100 marks.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | (Uniform acceleration) A car starts from rest and accelerates uniformly, reaching 20.0 m s⁻¹ after 8.0 s. It then travels at constant speed. (a) Find its acceleration. (b) Find the distance travelled in the first 8.0 s. (c) If it then moves at constant speed for a further 6.0 s, find the average speed for the whole journey. (10 marks) | Answer: (a) a = (v − u) ÷ t = (20.0 − 0) ÷ 8.0 = 2.50 m s⁻². (b) s = (u + v) ÷ 2 × t = (0 + 20.0) ÷ 2 × 8.0 = 80.0 m. (c) Distance at constant speed = 20.0 × 6.0 = 120.0 m, total distance = 200.0 m, total time = 14.0 s, so average speed = 200.0 ÷ 14.0 = 14.29 m s⁻¹. (10 marks) |
| 2 | (Forces on a slope) A 5.0 kg block slides from rest down a smooth (frictionless) slope inclined at 30° to the horizontal. Take g = 9.81 m s⁻². (a) Find its acceleration along the slope. (b) Find the normal reaction from the slope. (c) If the slope is 2.0 m long, find the speed at the bottom. (10 marks) | Answer: (a) Only the component of weight mg sin θ acts along the slope, so a = g sin θ = 9.81 × sin 30° = 4.90 m s⁻². (b) Perpendicular to the slope the forces balance, so N = mg cos θ = 5.0 × 9.81 × cos 30° = 42.5 N. (c) v² = 2as = 2 × 4.90 × 2.0 = 19.6, so v = 4.43 m s⁻¹. (10 marks) |
| 3 | (Conservation of energy) A 0.50 kg ball is released from rest 1.8 m above the ground and rebounds to a height of 1.2 m. Take g = 9.81 m s⁻² and ignore air resistance. (a) Find its speed just before hitting the ground. (b) Find its kinetic energy just before impact. (c) Find the mechanical energy lost during the bounce, and the percentage lost. (10 marks) | Answer: (a) mgh = ½mv², so v = √(2gh) = √(2 × 9.81 × 1.8) = 5.94 m s⁻¹. (b) KE = ½mv² = 0.5 × 0.50 × 5.94² = 8.829 J, the same value as mgh. (c) Energy lost = mg(h₁ − h₂) = 0.50 × 9.81 × 0.6 = 2.943 J; percentage lost = 2.943 ÷ 8.829 × 100% = 33.3%. The rebound speed is 4.85 m s⁻¹. (10 marks) |
| 4 | (Conservation of momentum) A 0.40 kg trolley moving at 1.5 m s⁻¹ collides with a stationary 0.60 kg trolley and they move off together. (a) Find their common velocity after the collision. (b) Find the total kinetic energy before and after the collision, and the percentage lost. (c) Explain why conservation of momentum alone cannot give the kinetic energy after the collision. (10 marks) | Answer: (a) By conservation of momentum, 0.40 × 1.5 = (0.40 + 0.60) × v, so v = 0.60 ÷ 1.00 = 0.60 m s⁻¹. (b) KE before = ½ × 0.40 × 1.5² = 0.450 J; KE after = ½ × 1.00 × 0.60² = 0.180 J; energy lost = 0.270 J, i.e. 60.0%. (c) Momentum is a vector quantity whereas kinetic energy depends on the square of speed; deformation and heating remove energy in the collision, so momentum conservation alone cannot determine how the energy is shared. (10 marks) |
| 5 | (Circular motion) A 0.20 kg ball on a light string of length 0.8 m moves at a constant speed of 3.0 m s⁻¹ in a circle. (a) Find the tension in the string. (b) Find the time for one complete revolution. (c) If the string can withstand a maximum tension of 5.0 N, find the maximum allowable speed. (10 marks) | Answer: (a) The tension provides the centripetal force, so T = mv² ÷ r = 0.20 × 3.0² ÷ 0.80 = 2.25 N. (b) Circumference = 2πr = 2π × 0.8 = 5.027 m, so the period = 5.027 ÷ 3.0 = 1.68 s. (c) v = √(T_max r ÷ m) = √(5.0 × 0.8 ÷ 0.20) = 4.47 m s⁻¹. (10 marks) |
Section B: light, electricity and atomic physics (50 marks)
Structured questions: write down the formulae and your working, give units and use appropriate significant figures, and answer every part. Total 100 marks.
| # | Question | Answer and marking notes |
|---|---|---|
| 6 | (Stationary waves) A string of length 1.20 m fixed at both ends vibrates at 120 Hz with 3 antinodes. Its mass per unit length is 2.0e-03 kg m⁻¹. (a) Find the wavelength. (b) Find the wave speed. (c) Find the tension in the string. (d) If the string is to vibrate with 2 antinodes at the same wave speed, what frequency is needed? (10 marks) | Answer: (a) With both ends fixed and 3 antinodes, L = 3 × λ ÷ 2, so λ = 2L ÷ 3 = 2 × 1.20 ÷ 3 = 0.80 m. (b) v = f λ = 120 × 0.80 = 96 m s⁻¹. (c) v = √(T ÷ μ), so T = μ v² = 2.0e-03 × 96² = 18.43 N. (d) With 2 antinodes λ = 2L ÷ 2 = 1.20 m and f = v ÷ λ = 96 ÷ 1.20 = 80 Hz. (10 marks) |
| 7 | (Refraction and total internal reflection) A ray of light passes from glass of refractive index 1.50 into air. (a) Find the critical angle at the glass-air boundary. (b) If light enters the glass from air at an angle of incidence of 30.0°, find the angle of refraction. (c) Explain how an optical fibre carries light over long distances with little loss. (10 marks) | Answer: (a) sin C = 1 ÷ n = 1 ÷ 1.50 = 0.6667, so C = 41.8°. (b) n = sin i ÷ sin r, so sin r = n sin i = 1.50 × sin 30.0° = 0.750 and r = 48.6°. (c) Inside the core the light strikes the core-cladding boundary at angles greater than the critical angle, so total internal reflection occurs; almost no energy leaves the core and the light travels long distances by repeated reflection. (10 marks) |
| 8 | (Circuits) A battery of e.m.f. 12 V and negligible internal resistance is connected to a parallel combination of 6 Ω and 3 Ω resistors, in series with a 4 Ω resistor. (a) Find the equivalent resistance. (b) Find the current and the total power delivered by the battery. (c) Find the voltage across the parallel section and the current through the 6 Ω resistor. (10 marks) | Answer: (a) For the parallel part 1/R = 1/6 + 1/3, giving R_p = 2.0 Ω; the equivalent resistance is R = 2.0 + 4.0 = 6.0 Ω. (b) I = E ÷ R = 12 ÷ 6.0 = 2.0 A and P = EI = 12 × 2.0 = 24 W. (c) The voltage across the parallel section is V = I R_p = 2.0 × 2.0 = 4.0 V; the current through the 6 Ω resistor is 4.0 ÷ 6 = 0.67 A, leaving 1.33 A in the other branch. (10 marks) |
| 9 | (Gas laws) An ideal gas in a fixed-volume container has a pressure of 2.0e+05 Pa at 300 K. (a) Find the pressure when the temperature rises to 450 K. (b) Explain, in terms of molecular motion, why the pressure rises at constant volume. (c) If a little gas leaks out while the container is heated, would the actual pressure be higher or lower than your answer in (a)? Explain. (10 marks) | Answer: (a) At constant volume P ÷ T is constant, so P₂ = P₁ × T₂ ÷ T₁ = 2.0e+05 × 450 ÷ 300 = 3.0e+05 Pa. (b) Raising the temperature increases the average kinetic energy and speed of the molecules, so they strike the walls more often and each collision involves a larger change of momentum; the average force per unit area, and hence the pressure, increases. (c) The pressure would be lower, because fewer molecules strike the walls than assumed in (a) when gas escapes. (10 marks) |
| 10 | (Radioactive half-life) A radioisotope has a half-life of 5.0 days and its initial activity is 800 Bq. (a) Find the activity after 20 days. (b) Find the percentage of the original atoms that remain undecayed after 20 days. (c) Explain why activity gives no information about the chemical properties of the isotope. (10 marks) | Answer: (a) 20 days is 4 half-lives, so A = 800 × (1/2)^4 = 50 Bq. (b) The fraction remaining is (1/2)^4 = 1/16 = 6.25%. (c) Radioactive decay is a nuclear process independent of the electron arrangement; chemical properties are determined by the electrons, so the activity does not change the chemical behaviour of the element. (10 marks) |
A TutorZone practice paper; original questions and passages, not publisher material.
