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HKDSE Physics Paper 1B Calculations and Data Practice Paper (10 Questions, 100 Marks)
HKDSE Physics Paper 1B Calculations
HKDSE Physics Paper 1B calculations and data practice paper: 10 original questions worth 100 marks in 60 minutes, covering uniform deceleration and reaction distance.
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Paper details Level: S6 (DSE) | Subject: Physics | Time: 60 minutes | Marks: 100 | Questions: 10
Section A Mechanics, Energy and Matter (50 marks)
Questions 1-5
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A car travels at 25 m/s. The driver sees a red light and brakes with a uniform deceleration of 2.5 m/s². (a) Find the time taken from the start of braking to a complete stop. (4 marks) (b) Find the distance travelled during braking. (3 marks) (c) If the reaction time is 0.8 s, find the total distance from seeing the red light to stopping. (3 marks) |
Answer: (a) t = u ÷ a = 25 ÷ 2.5 = 10 s. (b) s = u² ÷ (2a) = 625 ÷ 5 = 125 m. (c) Reaction distance = 25 × 0.8 = 20 m; total distance = 20 + 125 = 145 m. Marking notes: treat deceleration as positive; reaction and braking distances are computed separately then added. |
| 2 | A ball is projected at 40 m/s at 30° above the horizontal. Air resistance is negligible and g = 10 m/s². (a) Find the horizontal and vertical components of the initial velocity. (4 marks) (b) Find the time of flight and the horizontal range. (3 marks) (c) Find the maximum height reached. (3 marks) |
Answer: (a) Horizontal component = 40 × cos30° = 34.64 m/s; vertical component = 40 × sin30° = 20 m/s. (b) Time of flight = 2 × 20 ÷ 10 = 4 s; range = 34.64 × 4 = 138.6 m. (c) Maximum height = 20² ÷ (2 × 10) = 20 m. Marking notes: the range uses the horizontal component times the total flight time; the peak uses the vertical component only. |
| 3 | A 5 kg block slides down a rough plane inclined at 30° to the horizontal. The coefficient of kinetic friction is 0.20 and g = 10 m/s². (a) Find the component of weight along the plane and the normal reaction. (4 marks) (b) Find the magnitude of the friction force. (3 marks) (c) Find the acceleration of the block down the plane. (3 marks) |
Answer: (a) Component along plane = 5 × 10 × sin30° = 25 N; normal reaction = 5 × 10 × cos30° = 43.3 N. (b) Friction = 0.20 × 43.3 = 8.66 N. (c) Net force = 25 − 8.66 = 16.34 N; acceleration = 16.34 ÷ 5 = 3.27 m/s². Marking notes: normal reaction uses cos and the along-plane component uses sin; friction opposes motion. |
| 4 | A 1,200 kg car accelerates from rest to 20 m/s while climbing 50 m up a slope in 20 s. Take g = 10 m/s². (a) Find the gain in kinetic energy and in gravitational potential energy. (4 marks) (b) Find the average output power of the engine. (3 marks) (c) If the engine efficiency is 80%, find the average input power. (3 marks) |
Answer: (a) Kinetic energy = ½ × 1,200 × 20² = 240,000 J; potential energy = 1,200 × 10 × 50 = 600,000 J. (b) Total work = 840,000 J; power = 840,000 ÷ 20 = 42,000 W. (c) Input power = 42,000 ÷ 0.80 = 52,500 W. Marking notes: total work must include both kinetic and potential energy; efficiency is used as a decimal so input exceeds output. |
| 5 | A 0.5 kg ball hits a wall perpendicularly at 8 m/s and rebounds at 6 m/s. The contact time is 0.02 s. (a) Find the change in momentum of the ball, taking the direction towards the wall as positive. (4 marks) (b) Find the average force exerted by the wall on the ball. (3 marks) (c) In another test, a 2 kg trolley moving at 3 m/s hits a stationary 1 kg trolley and they stick together; find the common velocity and the kinetic energy lost. (3 marks) |
Answer: (a) Δp = 0.5 × (8 + 6) = 7 N s, directed away from the wall. (b) F = Δp ÷ t = 7 ÷ 0.02 = 350 N. (c) By conservation of momentum: v = (2 × 3) ÷ (2 + 1) = 2 m/s; energy lost = 9 − 6 = 3 J. Marking notes: the rebound reverses direction so the momentum changes add; a perfectly inelastic collision conserves momentum only. |
Section B Electricity, Waves and Radioactivity (50 marks)
Questions 6-10
| # | Question | Answer and marking notes |
|---|---|---|
| 6 | A sealed 2.0 × 10⁻³ m³ container holds an ideal gas at 300 K and 1.0 × 10⁵ Pa. Take R = 8.31 J mol⁻¹ K⁻¹. (a) The volume is fixed while the gas is heated to 450 K; find the new pressure. (4 marks) (b) Find the number of moles of gas in the container. (3 marks) (c) Explain, in microscopic terms, why the pressure rises when the gas is heated at constant volume. (3 marks) |
Answer: (a) p₂ = p₁T₂ ÷ T₁ = 1.0 × 10⁵ × 450 ÷ 300 = 1.5 × 10⁵ Pa. (b) n = pV ÷ RT = (1.0 × 10⁵ × 2.0 × 10⁻³) ÷ (8.31 × 300) = 0.0802 mol. (c) Heating raises the mean kinetic energy and speed of the molecules, so they strike the walls more often and each impact changes momentum more, raising the pressure. Marking notes: at constant volume p/T is constant; the microscopic answer needs both collision frequency and impact strength. |
| 7 | A 6 Ω and a 3 Ω resistor are connected in parallel across a battery of e.m.f. 12 V and internal resistance 0.5 Ω. (a) Ignoring internal resistance, find the equivalent resistance of the parallel pair and the current in each resistor. (4 marks) (b) Including the internal resistance, find the total current and the terminal voltage. (3 marks) (c) Find the power dissipated in the internal resistance and explain why the terminal voltage is below the e.m.f. (3 marks) |
Answer: (a) Equivalent resistance = (6 × 3) ÷ (6 + 3) = 2 Ω; I₆ = 12 ÷ 6 = 2 A and I₃ = 12 ÷ 3 = 4 A. (b) Total current = 12 ÷ (2 + 0.5) = 4.8 A; terminal voltage = 12 − 4.8 × 0.5 = 9.6 V. (c) Power in internal resistance = 4.8² × 0.5 = 11.5 W; part of the e.m.f. is dropped across the internal resistance, so the terminal voltage is lower. Marking notes: a parallel combination is always smaller than its smallest branch; the terminal voltage is the e.m.f. minus the internal drop. |
| 8 | A 200-turn coil of area 0.015 m² lies perpendicular to a uniform magnetic field that decreases steadily from 0.40 T to 0.10 T in 0.25 s. (a) Find the change in magnetic flux through one turn. (4 marks) (b) Find the magnitude of the induced e.m.f. in the coil. (3 marks) (c) If the coil resistance is 1.2 Ω, find the induced current and state how its direction relates to the decreasing flux. (3 marks) |
Answer: (a) ΔΦ = 0.015 × (0.40 − 0.10) = 0.0045 Wb. (b) ε = NΔΦ ÷ Δt = 200 × 0.0045 ÷ 0.25 = 3.6 V. (c) I = 3.6 ÷ 1.2 = 3 A; the induced current produces a field in the same direction as the original field so as to oppose the decrease (Lenz law). Marking notes: Faraday law must include the number of turns; Lenz law opposes the change, not the original field. |
| 9 | A ray of light passes from air (refractive index 1.00) into glass of refractive index 1.50 at an angle of incidence of 30°. Take the speed of light in vacuum as 3.0 × 10⁸ m/s. (a) Find the angle of refraction in the glass. (4 marks) (b) Find the speed of light in the glass. (3 marks) (c) Find the critical angle for light travelling from the glass into air. (3 marks) |
Answer: (a) sin r = sin30° ÷ 1.50 = 0.3333, so r = 19.5°. (b) v = c ÷ n = 3.0 × 10⁸ ÷ 1.50 = 2.0 × 10⁸ m/s. (c) sin c = 1 ÷ 1.50 = 0.6667, so c = 41.8°. Marking notes: the denser medium has the slower speed and the smaller angle; a critical angle exists only for light going from denser to less dense medium. |
| 10 | A radioactive isotope has a half-life of 8 days. A sample has an initial activity of 640 Bq. (a) Find the activity of the sample after 24 days. (4 marks) (b) Find the time needed for the activity to fall to 40 Bq. (3 marks) (c) Find the fraction of nuclei that have decayed after 16 days. (3 marks) |
Answer: (a) 24 days is 3 half-lives, so activity = 640 ÷ 2³ = 80 Bq. (b) 640 → 40 is 4 half-lives, so the time = 4 × 8 = 32 days. (c) 16 days is 2 half-lives, so the fraction remaining is 1 ÷ 4 = 0.25 and the fraction decayed is 0.75 (75%). Marking notes: activity falls by powers of two; the decayed fraction is one minus the remaining fraction. |
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