A-Level OCR Physics: Unit Test Practice Paper | A-Level OCR 物理：单元测试卷

📚 A-Level OCR Physics: Unit Test Practice Paper | A-Level OCR 物理：单元测试卷

This article presents a comprehensive unit test practice paper for A-Level OCR Physics, covering core topics such as mechanics, electricity, waves, thermal physics, and nuclear physics. Each question is followed by a full solution with step‑by‑step reasoning. The paper is designed to mirror actual OCR examination style, helping you consolidate knowledge and improve problem‑solving skills.

本文提供一套全面的 A‑Level OCR 物理单元测试练习卷，涵盖力学、电学、波动、热物理和核物理等核心主题。每道题都配有完整解答和逐步推理过程。试卷仿照 OCR 真题风格设计，帮助你巩固知识、提升解题能力。

1. Kinematics of a Projectile | 抛体运动学

A ball is kicked from ground level with a speed of 25 m s⁻¹ at an angle of 40° to the horizontal. Calculate the horizontal range of the ball. Ignore air resistance. (Take g = 9.81 m s⁻²)

一个足球从地面以 25 m s⁻¹ 的初速度、与水平方向成 40° 角踢出。忽略空气阻力，计算球的水平射程。（取 g = 9.81 m s⁻²）

Solution | 解答

Resolve the initial velocity: uₓ = u cosθ = 25 × cos40° ≈ 19.15 m s⁻¹; uᵧ = u sinθ = 25 × sin40° ≈ 16.07 m s⁻¹. Time of flight: T = 2uᵧ / g = 2 × 16.07 / 9.81 ≈ 3.276 s. Range = uₓ × T = 19.15 × 3.276 ≈ 62.7 m.

分解初速度：水平分量 uₓ = u cosθ = 25 × cos40° ≈ 19.15 m s⁻¹；竖直分量 uᵧ = u sinθ = 25 × sin40° ≈ 16.07 m s⁻¹。飞行时间 T = 2uᵧ / g = 2 × 16.07 / 9.81 ≈ 3.276 s。射程 = uₓ × T = 19.15 × 3.276 ≈ 62.7 m。

2. Forces in Equilibrium | 力的平衡

A 5.0 kg mass is suspended by two light inextensible strings. One string is horizontal, the other makes an angle of 30° with the vertical. Find the tension in each string.

一个 5.0 kg 的重物由两根轻质不可伸长的绳悬挂。一根绳水平，另一根绳与竖直方向成 30° 角。求每根绳的张力。

Solution | 解答

Weight W = mg = 5.0 × 9.81 = 49.05 N. Let T₁ be tension in the horizontal string, T₂ in the angled string. Vertical equilibrium: T₂ cos30° = W → T₂ = 49.05 / cos30° ≈ 56.6 N. Horizontal equilibrium: T₁ = T₂ sin30° = 56.6 × 0.5 = 28.3 N.

重力 W = mg = 5.0 × 9.81 = 49.05 N。设水平绳张力为 T₁，斜绳张力为 T₂。竖直方向平衡：T₂ cos30° = W → T₂ = 49.05 / cos30° ≈ 56.6 N。水平方向平衡：T₁ = T₂ sin30° = 56.6 × 0.5 = 28.3 N。

3. Work, Energy and Power | 功、能与功率

A car of mass 1200 kg accelerates uniformly from rest to 20 m s⁻¹ in 8.0 s along a level road. Calculate the average power developed by the engine, assuming no resistive forces.

一辆质量为 1200 kg 的汽车在水路面上从静止匀加速到 20 m s⁻¹ 用时 8.0 s。假设没有阻力，求发动机输出的平均功率。

Solution | 解答

Work done = gain in kinetic energy = ½ m v² = 0.5 × 1200 × (20)² = 240 000 J. Time = 8.0 s. Average power = work done / time = 240 000 / 8.0 = 30 000 W = 30 kW.

做功 = 增加的动能 = ½ m v² = 0.5 × 1200 × (20)² = 240 000 J。时间 8.0 s。平均功率 = 做功 / 时间 = 240 000 / 8.0 = 30 000 W = 30 kW。

4. Ohm’s Law and Resistivity | 欧姆定律与电阻率

A wire of length 2.5 m and cross‑sectional area 3.0 × 10⁻⁷ m² has a resistance of 4.8 Ω. Calculate the resistivity of the material. When a potential difference of 12 V is applied across the wire, what is the current?

一根长 2.5 m、横截面积 3.0 × 10⁻⁷ m² 的导线电阻为 4.8 Ω。计算材料的电阻率。如果在导线两端施加 12 V 电压，电流是多少？

Solution | 解答

R = ρ L / A → ρ = R A / L = 4.8 × (3.0×10⁻⁷) / 2.5 = 5.76×10⁻⁷ Ω m. Current I = V / R = 12 / 4.8 = 2.5 A.

R = ρ L / A → ρ = R A / L = 4.8 × (3.0×10⁻⁷) / 2.5 = 5.76×10⁻⁷ Ω m。电流 I = V / R = 12 / 4.8 = 2.5 A。

5. Internal Resistance and EMF | 内阻与电动势

A cell of EMF 1.5 V is connected to a 5.0 Ω resistor. The terminal voltage is measured as 1.3 V. Determine the internal resistance of the cell.

一个电动势为 1.5 V 的电池连接到一个 5.0 Ω 的电阻上，测得端电压为 1.3 V。求电池的内阻。

Solution | 解答

Circuit current I = V_R / R = 1.3 / 5.0 = 0.26 A. Using ε = V + I r → 1.5 = 1.3 + 0.26 r → r = (1.5 – 1.3) / 0.26 = 0.20 / 0.26 ≈ 0.769 Ω.

电路电流 I = V_R / R = 1.3 / 5.0 = 0.26 A。由 ε = V + I r 得 1.5 = 1.3 + 0.26 r → r = (1.5 – 1.3) / 0.26 = 0.20 / 0.26 ≈ 0.769 Ω。

6. Wave Interference and Path Difference | 波的干涉与程差

Two coherent sound sources emit in phase at a frequency of 680 Hz. A point P is located 4.0 m from one source and 4.75 m from the other. The speed of sound is 340 m s⁻¹. Determine whether constructive or destructive interference occurs at P.

两个同相位相干声源发出频率为 680 Hz 的声波。点 P 距离一个声源 4.0 m，距离另一个声源 4.75 m。声速为 340 m s⁻¹。判断 P 点发生加强干涉还是减弱干涉。

Solution | 解答

Wavelength λ = v / f = 340 / 680 = 0.50 m. Path difference = 4.75 – 4.0 = 0.75 m = 1.5 λ. Since the path difference is an odd multiple of half‑wavelengths (1.5 = 3×½), destructive interference occurs (minimum).

波长 λ = v / f = 340 / 680 = 0.50 m。路程差 = 4.75 – 4.0 = 0.75 m = 1.5 λ。由于路程差是半波长的奇数倍 (1.5 = 3×½)，发生减弱干涉（极小值）。

7. Photoelectric Effect and Work Function | 光电效应与功函数

Ultraviolet light of wavelength 200 nm is incident on a metal surface. The maximum kinetic energy of emitted electrons is 2.10 eV. Calculate the work function of the metal in electronvolts. (Planck constant h = 6.63 × 10⁻³⁴ J s, 1 eV = 1.60 × 10⁻¹⁹ J, c = 3.00 × 10⁸ m s⁻¹)

波长为 200 nm 的紫外光照射到金属表面，发射电子的最大动能为 2.10 eV。计算该金属的功函数（以 eV 为单位）。（普朗克常量 h = 6.63 × 10⁻³⁴ J s，1 eV = 1.60 × 10⁻¹⁹ J，c = 3.00 × 10⁸ m s⁻¹）

Solution | 解答

Photon energy E = h f = h c / λ = (6.63×10⁻³⁴ × 3.00×10⁸) / (200×10⁻⁹) = 9.945×10⁻¹⁹ J. Convert to eV: 9.945×10⁻¹⁹ / 1.60×10⁻¹⁹ = 6.216 eV. Einstein’s equation: E = φ + K_max → φ = E – K_max = 6.216 – 2.10 = 4.12 eV (≈ 4.1 eV).

光子能量 E = h f = h c / λ = (6.63×10⁻³⁴ × 3.00×10⁸) / (200×10⁻⁹) = 9.945×10⁻¹⁹ J。换算为 eV：9.945×10⁻¹⁹ / 1.60×10⁻¹⁹ = 6.216 eV。爱因斯坦方程：E = φ + K_max → φ = E – K_max = 6.216 – 2.10 = 4.12 eV（≈ 4.1 eV）。

8. Radioactive Decay and Half‑life | 放射性衰变与半衰期

A sample of pure ¹³¹I has an initial activity of 8.0 × 10⁶ Bq. The half‑life of ¹³¹I is 8 days. Calculate its activity after 32 days. How many ¹³¹I nuclei decay in the first 16 days? (1 Bq = 1 decay per second)

一份纯 ¹³¹I 样品的初始活度为 8.0 × 10⁶ Bq。¹³¹I 的半衰期为 8 天。计算 32 天后的活度。前 16 天有多少个 ¹³¹I 原子核发生衰变？（1 Bq = 每秒 1 次衰变）

Solution | 解答

After n half‑lives: A = A₀ / 2ⁿ. n = 32 / 8 = 4. A = 8.0×10⁶ / 2⁴ = 8.0×10⁶ / 16 = 5.0×10⁵ Bq. Decay constant λ = ln2 / T₁/₂ = 0.693 / (8×24×3600) ≈ 1.0×10⁻⁶ s⁻¹. Initial number of nuclei N₀ = A₀ / λ = 8.0×10⁶ / 1.0×10⁻⁶ = 8.0×10¹². Number remaining after 16 days (two half‑lives): N = N₀ / 4 = 2.0×10¹². Number decayed = N₀ – N = 6.0×10¹².

经过 n 个半衰期：A = A₀ / 2ⁿ。n = 32 / 8 = 4。A = 8.0×10⁶ / 2⁴ = 8.0×10⁶ / 16 = 5.0×10⁵ Bq。衰变常数 λ = ln2 / T₁/₂ = 0.693 / (8×24×3600) ≈ 1.0×10⁻⁶ s⁻¹。初始核数目 N₀ = A₀ / λ = 8.0×10⁶ / 1.0×10⁻⁶ = 8.0×10¹²。16 天后（两个半衰期）剩余核数 N = N₀ / 4 = 2.0×10¹²。已衰变数 = N₀ – N = 6.0×10¹²。

9. Ideal Gases and Kinetic Theory | 理想气体与分子动理论

A cylinder contains 0.25 mol of an ideal gas at 27 °C. The gas is compressed isothermally to half its original volume. Calculate the work done on the gas during this compression. (Universal gas constant R = 8.31 J mol⁻¹ K⁻¹)

一个气缸内装有 0.25 mol 的理想气体，温度为 27 °C。该气体被等温压缩到原体积的一半。计算压缩过程中对气体做的功。（普适气体常量 R = 8.31 J mol⁻¹ K⁻¹）

Solution | 解答

Isothermal work done on gas W = nRT ln(V₁/V₂). Here V₂ = V₁/2, so V₁/V₂ = 2. T = 27 + 273 = 300 K. W = 0.25 × 8.31 × 300 × ln(2) = 0.25 × 8.31 × 300 × 0.693 ≈ 432 J.

等温压缩对气体做的功 W = nRT ln(V₁/V₂)。已知 V₂ = V₁/2，故 V₁/V₂ = 2。T = 27 + 273 = 300 K。W = 0.25 × 8.31 × 300 × ln(2) = 0.25 × 8.31 × 300 × 0.693 ≈ 432 J。

10. Simple Harmonic Motion (SHM) | 简谐运动

A mass of 0.50 kg attached to a spring oscillates with an amplitude of 4.0 cm and a period of 0.60 s. Calculate the maximum acceleration of the mass and the total energy of the system. (spring constant k = m ω²)

一个 0.50 kg 的物体连接在弹簧上做简谐运动，振幅为 4.0 cm，周期为 0.60 s。计算物体的最大加速度和系统的总能量。（弹簧劲度系数 k = m ω²）

Solution | 解答

Angular frequency ω = 2π / T = 2π / 0.60 ≈ 10.47 rad s⁻¹. Amplitude A = 0.040 m. Maximum acceleration a_max = ω² A = (10.47)² × 0.040 ≈ 4.38 m s⁻². Total energy E = ½ m ω² A² = 0.5 × 0.50 × (10.47)² × (0.040)² = 0.0438 J ≈ 4.4×10⁻² J.

角频率 ω = 2π / T = 2π / 0.60 ≈ 10.47 rad s⁻¹。振幅 A = 0.040 m。最大加速度 a_max = ω² A = (10.47)² × 0.040 ≈ 4.38 m s⁻²。总能量 E = ½ m ω² A² = 0.5 × 0.50 × (10.47)² × (0.040)² = 0.0438 J ≈ 4.4×10⁻² J。

11. Gravitational Field Strength | 引力场强

A planet has a mass of 6.4 × 10²³ kg and a radius of 3.4 × 10⁶ m. Find the gravitational field strength at its surface. At what height above the planet’s surface does the field strength become half of its surface value? (G = 6.67 × 10⁻¹¹ N m² kg⁻²)

一个行星质量为 6.4 × 10²³ kg，半径为 3.4 × 10⁶ m。求它表面的引力场强。在行星表面上方多高处场强变为表面值的一半？(G = 6.67 × 10⁻¹¹ N m² kg⁻²)

Solution | 解答

Surface g = G M / R² = (6.67×10⁻¹¹ × 6.4×10²³) / (3.4×10⁶)² = (4.27×10¹³) / (1.156×10¹³) ≈ 3.69 N kg⁻¹. Let half g at distance r from planet centre: ½ g = G M / r² → r² = 2 G M / g = 2 R² → r = R √2. Height h = r – R = R(√2 – 1) = 3.4×10⁶ × (1.414 – 1) ≈ 1.41×10⁶ m.

表面 g = G M / R² = (6.67×10⁻¹¹ × 6.4×10²³) / (3.4×10⁶)² ≈ 3.69 N kg⁻¹。设场强为一半处距行星中心 r：½ g = G M / r² → r² = 2 R² → r = R √2。高度 h = r – R = R(√2 – 1) = 3.4×10⁶ × 0.414 ≈ 1.41×10⁶ m。

12. Capacitor Discharge | 电容器放电

A 100 μF capacitor is charged to 12 V and then discharged through a 47 kΩ resistor. Find the time constant and the voltage across the capacitor after 3.0 seconds.

一个 100 μF 的电容器被充电至 12 V，然后通过 47 kΩ 的电阻放电。求时间常数和 3.0 秒后电容器两端的电压。

Solution | 解答

Time constant τ = R C = 47×10³ × 100×10⁻⁶ = 4.7 s. Discharge equation: V = V₀ e^(−t/τ). After 3.0 s: V = 12 × e^(−3.0 / 4.7) = 12 × e^(−0.638) ≈ 12 × 0.528 = 6.34 V.

时间常数 τ = R C = 47×10³ × 100×10⁻⁶ = 4.7 s。放电方程：V = V₀ e^(−t/τ)。3.0 s 后：V = 12 × e^(−3.0 / 4.7) ≈ 12 × 0.528 = 6.34 V。

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A-Level OCR Physics: Unit Test Practice Paper | A-Level OCR 物理：单元测试卷