📚 A-Level OCR Physics: Worked Examples Explained | A-Level OCR 物理：典型例题详解
Mastering A-Level OCR Physics requires more than memorising formulas – it demands a deep understanding of how to apply concepts to unfamiliar problems. This article walks you through ten carefully chosen worked examples, each targeting a key topic from the OCR specification, including mechanics, electricity, waves, thermal physics, nuclear physics, and fields. Every solution is broken down step by step with clear reasoning, and all mathematical expressions use standard notation. Follow these demonstrations to sharpen your problem-solving skills and build confidence for the exam.
精通 A-Level OCR 物理不仅需要记住公式，更需要深刻理解如何将概念应用于陌生问题。本文精选了十个典型例题，涵盖 OCR 考试大纲中的核心主题，包括力学、电路、波、热物理、核物理以及场。每道题的解答都逐步拆解，逻辑清晰，所有数学表达式均使用标准符号。通过这些演示，你可以磨练解题技巧，为考试建立信心。
1. Kinematics with Constant Acceleration | 匀加速运动学
A car accelerates uniformly from rest at 2.5 m s⁻² for 8.0 s on a straight horizontal road. It then maintains a constant velocity for a further 12.0 s before decelerating uniformly to rest in 10.0 s. Determine the total distance travelled by the car.
一辆汽车从静止开始以 2.5 m s⁻² 的加速度在平直水平道路上匀加速 8.0 s，随后保持匀速行驶 12.0 s，再以匀减速运动在 10.0 s 内停下。求汽车行驶的总距离。
We break the motion into three phases. Phase 1: acceleration, u = 0, a = 2.5 m s⁻², t = 8.0 s. Final velocity v = u + at = 0 + 2.5 × 8.0 = 20 m s⁻¹. Distance s₁ = ut + ½at² = 0 + 0.5 × 2.5 × (8.0)² = 80 m. Phase 2: constant velocity 20 m s⁻¹ for 12.0 s, so s₂ = v × t = 20 × 12.0 = 240 m. Phase 3: deceleration to rest, initial velocity 20 m s⁻¹, final velocity 0, t = 10.0 s. Acceleration a = (v – u) / t = (0 – 20) / 10.0 = -2.0 m s⁻². Distance s₃ = ut + ½at² = 20 × 10.0 + 0.5 × (-2.0) × (10.0)² = 200 – 100 = 100 m. Total distance = 80 + 240 + 100 = 420 m.
我们将运动分为三个阶段。第一阶段：加速，u = 0，a = 2.5 m s⁻²，t = 8.0 s。末速度 v = u + at = 0 + 2.5 × 8.0 = 20 m s⁻¹。位移 s₁ = ut + ½at² = 0 + 0.5 × 2.5 × (8.0)² = 80 m。第二阶段：以 20 m s⁻¹ 匀速运动 12.0 s，故 s₂ = 20 × 12.0 = 240 m。第三阶段：匀减速至停下，初速 20 m s⁻¹，末速 0，时间 10.0 s。加速度 a = (0 – 20) / 10.0 = -2.0 m s⁻²。位移 s₃ = 20 × 10.0 + 0.5 × (-2.0) × (10.0)² = 200 – 100 = 100 m。总距离 = 80 + 240 + 100 = 420 m。
2. Projectile Motion and SUVAT | 抛体运动与 SUVAT 方程
A stone is thrown horizontally from the top of a vertical cliff 45 m high with a speed of 15 m s⁻¹. Calculate the time taken for the stone to reach the sea and the horizontal distance from the base of the cliff where it lands. Ignore air resistance.
一块石头从高 45 m 的竖直悬崖顶部以 15 m s⁻¹ 的水平速度抛出。忽略空气阻力，求石头落到海面所用的时间以及落点距悬崖底部的水平距离。
Vertical motion: initial vertical velocity u_y = 0, displacement s_y = -45 m (taking downward as negative, but we can also use magnitude with g = 9.81 m s⁻² downward). Using s = ut + ½at² with downward positive: s = 45 m, u = 0, a = g = 9.81 m s⁻². Then 45 = 0 × t + ½ × 9.81 × t² → t² = (2 × 45) / 9.81 = 90 / 9.81 ≈ 9.17 → t = √9.17 ≈ 3.03 s. Horizontal motion: constant velocity 15 m s⁻¹. Distance = speed × time = 15 × 3.03 ≈ 45.5 m. Thus the stone takes 3.03 s and lands 45.5 m from the cliff base.
竖直方向：初始竖直速度 u_y = 0，位移 s_y = 45 m 向下。取向下为正，用 s = ut + ½at²：45 = 0 + ½ × 9.81 × t²，得 t² = 90 / 9.81 ≈ 9.17，t ≈ 3.03 s。水平方向：匀速 15 m s⁻¹。水平距离 = 15 × 3.03 ≈ 45.5 m。因此石头经过 3.03 s 落到海里，距悬崖底部的水平距离约 45.5 m。
3. Newton’s Laws and Connected Particles | 牛顿定律与连接体
Two blocks of mass 3.0 kg and 5.0 kg are connected by a light inextensible string that passes over a smooth pulley. The 5.0 kg block rests on a rough horizontal table with coefficient of friction μ = 0.25, while the 3.0 kg block hangs freely. The system is released from rest. Find the acceleration of the blocks and the tension in the string.
两个质量分别为 3.0 kg 和 5.0 kg 的物块通过轻质不可伸长的绳子相连，绳子跨过一个光滑滑轮。5.0 kg 的物块放在粗糙水平桌面上，摩擦系数 μ = 0.25，3.0 kg 的物块自由悬挂。系统由静止释放。求物块的加速度和绳中的张力。
Let tension be T. For the 5.0 kg mass on the table, friction force F_f = μ × R = 0.25 × (5.0 × 9.81) = 0.25 × 49.05 ≈ 12.26 N. Equation of motion: T – F_f = 5.0 × a (1). For the hanging 3.0 kg mass: 3.0g – T = 3.0a → 29.43 – T = 3.0a (2). Add (1) and (2): (T – 12.26) + (29.43 – T) = 8.0a → 17.17 = 8.0a → a = 17.17 / 8.0 ≈ 2.15 m s⁻². Substitute into (2): T = 29.43 – 3.0 × 2.15 = 29.43 – 6.45 = 22.98 N ≈ 23.0 N. Acceleration is 2.15 m s⁻².
设张力为 T。对桌面上 5.0 kg 物块，摩擦力 F_f = μ × R = 0.25 × (5.0 × 9.81) = 12.26 N。运动方程：T – 12.26 = 5.0a (1)。对悬挂的 3.0 kg 物块：3.0g – T = 3.0a → 29.43 – T = 3.0a (2)。(1)+(2) 得 17.17 = 8.0a，a ≈ 2.15 m s⁻²。代入 (2) 得 T ≈ 23.0 N。所以加速度为 2.15 m s⁻²，张力约 23.0 N。
4. Work, Energy and Power | 功、能与功率
A cyclist of mass 75 kg cycles up a slope inclined at 6.0° to the horizontal at a constant speed of 4.0 m s⁻¹. The total resistive force acting on the cyclist and bicycle is 28 N. Calculate the useful power output the cyclist must produce. (Assume g = 9.81 m s⁻²)
一个质量为 75 kg 的自行车手沿着与水平面成 6.0° 的斜坡以 4.0 m s⁻¹ 的恒定速度向上骑行。作用在车手和自行车上的总阻力为 28 N。计算车手必须产生的有效功率。（设 g = 9.81 m s⁻²）
At constant speed, the driving force F_d equals the sum of resistive forces plus the component of weight down the slope. Weight component = mg sin θ = 75 × 9.81 × sin 6.0°. sin 6.0° ≈ 0.1045, so weight component ≈ 75 × 9.81 × 0.1045 ≈ 76.9 N. Total opposing force = 76.9 + 28 = 104.9 N. Hence F_d = 104.9 N. Power P = F_d × v = 104.9 × 4.0 = 419.6 W ≈ 420 W.
匀速运动时，驱动力 F_d 等于阻力之和加上重力沿斜坡向下的分力。重力分力 = mg sin θ = 75 × 9.81 × sin 6.0°。sin 6.0° ≈ 0.1045，故重力分力 ≈ 76.9 N。总反向力 = 76.9 + 28 = 104.9 N。因此 F_d = 104.9 N。功率 P = F_d v = 104.9 × 4.0 ≈ 420 W，即约 420 W。
5. Momentum and Impulse | 动量与冲量
A ball of mass 0.150 kg strikes a vertical wall at 12 m s⁻¹ and rebounds at 8.0 m s⁻¹. The contact time with the wall is 0.040 s. Calculate the magnitude of the average force exerted by the wall on the ball.
一个质量为 0.150 kg 的球以 12 m s⁻¹ 的速度撞击竖直墙壁，并以 8.0 m s⁻¹ 的速度反弹。与墙的接触时间为 0.040 s。求墙壁对球施加的平均力的大小。
Take the direction towards the wall as positive. Initial velocity u = +12 m s⁻¹, final velocity v = -8.0 m s⁻¹ (rebounds). Change in momentum Δp = m(v – u) = 0.150 × (-8.0 – 12) = 0.150 × (-20) = -3.0 kg m s⁻¹. Magnitude of impulse is 3.0 N s. Average force F = Δp / Δt = 3.0 / 0.040 = 75 N. So the average force is 75 N in the direction away from the wall.
取向墙为正方向。初速度 u = +12 m s⁻¹，末速度 v = -8.0 m s⁻¹。动量变化 Δp = m(v – u) = 0.150 × (-8.0 – 12) = -3.0 kg m s⁻¹。冲量大小为 3.0 N s。平均力 F = Δp / Δt = 3.0 / 0.040 = 75 N。因此墙对球的平均力大小为 75 N，方向远离墙壁。
6. Electrical Circuits and Internal Resistance | 电路与内阻
A battery of emf 9.0 V and internal resistance 0.80 Ω is connected to a variable resistor. When the current is 1.5 A, calculate the terminal potential difference and the power dissipated in the external resistor.
一个电动势为 9.0 V、内阻为 0.80 Ω 的电池与一个可变电阻连接。当电流为 1.5 A 时，计算路端电压和外电阻消耗的功率。
Terminal pd V = ε – Ir = 9.0 – (1.5 × 0.80) = 9.0 – 1.2 = 7.8 V. The external resistance R = V / I = 7.8 / 1.5 = 5.2 Ω. Power dissipated P = I²R = (1.5)² × 5.2 = 2.25 × 5.2 = 11.7 W. Alternatively, P = V × I = 7.8 × 1.5 = 11.7 W. So terminal pd is 7.8 V and power is 11.7 W.
路端电压 V = ε – Ir = 9.0 – (1.5 × 0.80) = 7.8 V。外电阻 R = V / I = 7.8 / 1.5 = 5.2 Ω。功率 P = I²R = (1.5)² × 5.2 = 11.7 W，或 P = V × I = 7.8 × 1.5 = 11.7 W。因此路端电压为 7.8 V，功率为 11.7 W。
7. Wave Interference and Double Slit | 波的干涉与双缝
In a Young’s double-slit experiment, monochromatic light of wavelength 589 nm illuminates two slits separated by 0.50 mm. The interference pattern is observed on a screen 1.80 m away. Determine the fringe separation on the screen.
在杨氏双缝实验中，波长为 589 nm 的单色光照射间距为 0.50 mm 的双缝。观察屏与双缝距离为 1.80 m。求屏上相邻亮条纹的间距。
Fringe spacing Δy = λD / d, where λ = 589 nm = 589 × 10⁻⁹ m, D = 1.80 m, d = 0.50 mm = 0.50 × 10⁻³ m. So Δy = (589 × 10⁻⁹ × 1.80) / (0.50 × 10⁻³) = (1060.2 × 10⁻⁹) / (0.50 × 10⁻³) = 2.1204 × 10⁻³ m = 2.12 mm. Fringe separation is 2.12 mm.
条纹间距 Δy = λD / d，其中 λ = 589 nm = 5.89×10⁻⁷ m，D = 1.80 m，d = 0.50 mm = 5.0×10⁻⁴ m。Δy = (5.89×10⁻⁷ × 1.80) / (5.0×10⁻⁴) = 2.12×10⁻³ m = 2.12 mm。条纹间距为 2.12 mm。
8. Thermal Physics and Ideal Gases | 热物理与理想气体
A cylinder contains 2.0 mol of an ideal gas at a temperature of 300 K and pressure 1.5 × 10⁵ Pa. The gas is heated at constant volume until the pressure doubles. Calculate the final temperature and the energy supplied to the gas if its molar heat capacity at constant volume is 12.5 J K⁻¹ mol⁻¹.
一个容器装有 2.0 mol 的理想气体，初始温度为 300 K，压强为 1.5 × 10⁵ Pa。气体在体积不变的条件下被加热，直至压强翻倍。计算最终温度以及供给气体的能量，已知该气体的摩尔定容热容为 12.5 J K⁻¹ mol⁻¹。
At constant volume, P₁/T₁ = P₂/T₂. P₂ = 2P₁ → T₂ = 2T₁ = 600 K. The change in temperature ΔT = 300 K. Heat supplied Q = n C_v ΔT = 2.0 × 12.5 × 300 = 2.0 × 3750 = 7500 J. So final temperature is 600 K and energy supplied is 7.50 × 10³ J.
定容条件下，P₁/T₁ = P₂/T₂。P₂ = 2P₁，所以 T₂ = 2T₁ = 600 K。温度变化 ΔT = 300 K。供给的热量 Q = n C_v ΔT = 2.0 × 12.5 × 300 = 7500 J。因此最终温度为 600 K，吸收能量为 7.50 × 10³ J。
9. Radioactive Decay and Half-life | 放射性衰变与半衰期
A sample of a radioactive isotope has an initial activity of 2400 Bq. After 36 hours its activity has fallen to 150 Bq. Determine the decay constant and half-life of the isotope.
某放射性同位素样品的初始活度为 2400 Bq，36 小时后活度降至 150 Bq。求该同位素的衰变常数和半衰期。
Activity A = A₀ e^{-λt}. So 150 = 2400 e^{-λ × 36 hours}. Divide: 150 / 2400 = 0.0625 = e^{-36λ}. Take natural log: ln(0.0625) = -36λ → λ = -ln(0.0625) / 36. ln(0.0625) = -2.7726 (since 0.0625 = 1/16). So λ = 2.7726 / 36 = 0.0770 h⁻¹. Half-life t½ = ln 2 / λ = 0.693 / 0.0770 ≈ 9.0 hours. Thus λ = 0.077 h⁻¹ and t½ = 9.0 h.
活度 A = A₀ e^{-λt}。150 = 2400 e^{-λ × 36}。比值 150/2400 = 0.0625 = e^{-36λ}。取自然对数：ln(0.0625) = -36λ，ln(0.0625) = ln(1/16) = -2.7726，故 λ = 2.7726 / 36 = 0.0770 h⁻¹。半衰期 t½ = ln 2 / λ = 0.693 / 0.0770 ≈ 9.0 h。衰变常数为 0.077 h⁻¹，半衰期为 9.0 小时。
10. Gravitational Fields and Orbital Motion | 引力场与轨道运动
A satellite orbits the Earth in a circular orbit of radius 7.0 × 10⁶ m. The Earth’s mass is 5.97 × 10²⁴ kg and the gravitational constant G = 6.67 × 10⁻¹¹ N m² kg⁻². Calculate the orbital speed of the satellite and its period.
一颗人造卫星在半径为 7.0 × 10⁶ m 的圆形轨道上绕地球运行。地球质量为 5.97 × 10²⁴ kg，引力常量 G = 6.67 × 10⁻¹¹ N m² kg⁻²。求卫星的轨道速率和运行周期。
For circular orbit, centripetal force is provided by gravity: mv²/r = GMm/r² → v = √(GM / r). v = √[(6.67 × 10⁻¹¹ × 5.97 × 10²⁴) / (7.0 × 10⁶)] = √[(39.8 × 10¹³) / (7.0 × 10⁶)] = √(5.686 × 10⁷) ≈ 7.54 × 10³ m s⁻¹ = 7.54 km s⁻¹. Period T = 2πr / v = (2π × 7.0 × 10⁶) / (7.54 × 10³) = (4.398 × 10⁷) / (7.54 × 10³) ≈ 5.83 × 10³ s ≈ 97 minutes (1 hour 37 minutes).
圆轨道运动中，引力提供向心力：mv²/r = GMm/r² → v = √(GM / r)。v = √[(6.67×10⁻¹¹ × 5.97×10²⁴) / (7.0×10⁶)] = √(3.98×10¹⁴ / 7.0×10⁶) = √(5.686×10⁷) ≈ 7.54×10³ m s⁻¹ = 7.54 km s⁻¹。周期 T = 2πr / v = (2π × 7.0×10⁶) / 7.54×10³ ≈ 5.83×10³ s，约 97 分钟。
11. Simple Harmonic Motion | 简谐运动
A mass of 0.50 kg attached to a spring oscillates with an amplitude of 4.0 cm and frequency 2.0 Hz. Find the maximum acceleration of the mass and the maximum kinetic energy. (Neglect damping)
一质量为 0.50 kg 的物体连在弹簧上，以振幅 4.0 cm、频率 2.0 Hz 作简谐运动。忽略阻尼，求最大加速度和最大动能。
Angular frequency ω = 2πf = 2π × 2.0 = 4π rad s⁻¹ ≈ 12.57 rad s⁻¹. Maximum acceleration a_max = ω²A = (12.57)² × 0.040 = 157.9 × 0.040 ≈ 6.32 m s⁻². Maximum speed v_max = ωA = 12.57 × 0.040 = 0.503 m s⁻¹. Max kinetic energy KE_max = ½ m v_max² = 0.5 × 0.50 × (0.503)² = 0.25 × 0.253 = 0.06325 J. So a_max = 6.3 m s⁻², KE_max = 6.33 × 10⁻² J.
角频率 ω = 2πf = 2π×2.0 = 4π rad s⁻¹ ≈ 12.57 rad s⁻¹。最大加速度 a_max = ω²A = (12.57)² × 0.040 ≈ 6.32 m s⁻²。最大速度 v_max = ωA = 12.57 × 0.040 = 0.503 m s⁻¹。最大动能 KE_max = ½ m v_max² = 0.5×0.50×(0.503)² ≈ 0.0633 J。故最大加速度约 6.3 m s⁻²，最大动能约 6.33×10⁻² J。
12. Capacitors and Exponential Decay | 电容器与指数衰减
A 470 µF capacitor is charged to 12 V and then discharged through a 10 kΩ resistor. Calculate the time constant of the circuit and the voltage across the capacitor after 4.7 s.
一个 470 µF 的电容器充电至 12 V，然后通过一个 10 kΩ 的电阻放电。计算电路的时间常数以及 4.7 s 后电容器两端的电压。
Time constant τ = RC = (10 × 10³ Ω) × (470 × 10⁻⁶ F) = 4.7 s. Voltage during discharge: V = V₀ e^{-t/RC}. At t = 4.7 s, t/RC = 4.7 / 4.7 = 1. So V = 12 × e⁻¹ = 12 × 0.3679 ≈ 4.41 V. The time constant is 4.7 s and the voltage after one time constant is 4.41 V.
时间常数 τ = RC = (10×10³ Ω) × (470×10⁻⁶ F) = 4.7 s。放电电压 V = V₀ e^{-t/RC}。当 t = 4.7 s 时，t/RC = 1，因此 V = 12 × e⁻¹ ≈ 12 × 0.3679 = 4.41 V。时间常数为 4.7 s，经过一个时间常数后电压为 4.41 V。
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