A-Level CIE Physics: Detailed Worked Examples | A-Level CIE 物理：典型例题详解

📚 A-Level CIE Physics: Detailed Worked Examples | A-Level CIE 物理：典型例题详解

Mastering A-Level Physics requires more than memorizing formulas; it demands the ability to apply concepts to novel problems. This article walks you through carefully selected typical questions from the CIE Physics syllabus, with fully worked solutions explained in both English and Chinese. Each example targets a core topic, helping you build the analytical skills and exam confidence needed for top grades.

攻克 A-Level 物理不能只靠死记硬背公式，必须学会将概念灵活运用到陌生题目中。本文为您精选了 CIE 物理考纲中极具代表性的典型例题，并配以中英双语详解。每个例题覆盖一个核心知识点，旨在帮助您建立解题思维，提升应考信心，冲刺高分。

1. Kinematics: Projectile Motion | 运动学：抛体运动

Question: A ball is projected from ground level with an initial speed of 20 m s⁻¹ at an angle of 30° to the horizontal. Ignoring air resistance, calculate its time of flight, maximum height, and horizontal range. (Take g = 9.81 m s⁻²)

题目：一个小球从地面以 20 m s⁻¹ 的初速率、与水平面成 30° 角抛出。忽略空气阻力，计算其飞行时间、最大高度和水平射程。（取 g = 9.81 m s⁻²）

Resolve the initial velocity into components: horizontal uₓ = 20 cos 30° ≈ 17.32 m s⁻¹, vertical u_y = 20 sin 30° = 10.0 m s⁻¹.

将初速度分解为分量：水平分量 uₓ = 20 cos 30° ≈ 17.32 m s⁻¹，竖直分量 u_y = 20 sin 30° = 10.0 m s⁻¹。

Time of flight depends only on vertical motion. Using s = u_y t − ½ g t² and setting s = 0 for landing, we get t_total = 2 u_y / g = (2 × 10.0) / 9.81 ≈ 2.04 s.

飞行时间仅由竖直运动决定。由 s = u_y t − ½ g t²，落地时 s = 0，得到 t_total = 2 u_y / g = (2 × 10.0) / 9.81 ≈ 2.04 s。

Maximum height occurs when vertical velocity is zero: v_y = u_y − g t = 0 ⇒ t_peak = u_y / g = 10.0 / 9.81 ≈ 1.02 s. Then H = u_y t_peak − ½ g t_peak² = (10.0)(1.02) − ½ × 9.81 × (1.02)² ≈ 5.10 m.

最大高度出现在竖直速度为零时：v_y = u_y − g t = 0 ⇒ t_peak = u_y / g ≈ 1.02 s。于是 H = u_y t_peak − ½ g t_peak² ≈ 5.10 m。

Horizontal range is R = uₓ × t_total = 17.32 × 2.04 ≈ 35.3 m. Alternatively, using the formula R = (u² sin 2θ)/g = (20² sin 60°)/9.81 gives the same result.

水平射程为 R = uₓ × t_total = 17.32 × 2.04 ≈ 35.3 m。也可用公式 R = (u² sin 2θ)/g = (20² sin 60°)/9.81 得到相同结果。

2. Newton’s Laws: Connected Particles | 牛顿定律：连接体问题

Question: A block of mass 4.0 kg rests on a smooth plane inclined at 30° to the horizontal. It is connected by a light inextensible string passing over a frictionless pulley to a 2.0 kg mass hanging freely. Find the acceleration of the system and the tension in the string. (g = 9.81 m s⁻²)

题目：质量为 4.0 kg 的木块放在倾角 30° 的光滑斜面上，通过轻绳和光滑定滑轮与一个 2.0 kg 的悬挂重物相连。求系统的加速度和绳中张力。（g = 9.81 m s⁻²）

For the hanging mass m₂ = 2.0 kg, the equation is m₂ g − T = m₂ a. For the block on the incline m₁ = 4.0 kg, the component of weight down the slope is m₁ g sin 30°, so T − m₁ g sin 30° = m₁ a.

对悬挂重物 m₂ = 2.0 kg，运动方程为 m₂ g − T = m₂ a。对斜面上的木块 m₁ = 4.0 kg，重力沿斜面分量为 m₁ g sin 30°，故 T − m₁ g sin 30° = m₁ a。

Adding the two equations eliminates T: m₂ g − m₁ g sin 30° = (m₁ + m₂) a. Substitute values: (2.0 × 9.81) − (4.0 × 9.81 × 0.5) = (4.0 + 2.0) a → 19.62 − 19.62 = 6.0 a → a = 0 m s⁻². This is a special balanced case, giving zero acceleration, so tension is simply T = m₂ g = 19.62 N.

将两式相加消去 T：m₂ g − m₁ g sin 30° = (m₁ + m₂) a。代入数值：(2.0 × 9.81) − (4.0 × 9.81 × 0.5) = (4.0 + 2.0) a → 19.62 − 19.62 = 6.0 a → a = 0 m s⁻²。这是一个平衡的特例，加速度为零，因此张力即为 T = m₂ g = 19.62 N。

If the masses were different, say m₂ = 3.0 kg, then net force would be 29.43 − 19.62 = 6.0 a → a = 1.64 m s⁻². The method remains the same: write separate equations and solve simultaneously.

若质量不同，比如 m₂ = 3.0 kg，则合力为 29.43 − 19.62 = 6.0a → a = 1.64 m s⁻²。解题方法不变：分别写方程并联立求解。

3. Circular Motion: Banked Track | 圆周运动：倾斜弯道

Question: A car rounds a bend of radius 120 m on a road banked at 20° to the horizontal. Assuming no lateral friction, calculate the safe speed for the car. (g = 9.81 m s⁻²)

题目：一辆汽车在半径 120 m、倾角 20° 的倾斜弯道上行驶。假设没有侧向摩擦力，求汽车的安全速度。（g = 9.81 m s⁻²）

The normal reaction N has two components: N cos 20° balances weight mg, and N sin 20° provides the centripetal force mv²/r.

支持力 N 有两个分量：N cos 20° 与重力 mg 平衡，N sin 20° 提供向心力 mv²/r。

Dividing these gives tan 20° = v² / (r g). Rearranging: v = √(r g tan 20°).

两式相除得到 tan 20° = v² / (r g)。整理得 v = √(r g tan 20°)。

Substituting: v = √(120 × 9.81 × tan 20°) = √(120 × 9.81 × 0.3640) ≈ √428.6 ≈ 20.7 m s⁻¹. In km/h, that is 20.7 × 3.6 ≈ 74.5 km/h.

代入数值：v = √(120 × 9.81 × tan 20°) = √(120 × 9.81 × 0.3640) ≈ √428.6 ≈ 20.7 m s⁻¹。换算成 km/h 为 20.7 × 3.6 ≈ 74.5 km/h。

This design speed depends only on the radius, bank angle, and g, not on the mass of the vehicle.

该设计速度仅取决于半径、倾角和重力加速度，与车辆质量无关。

4. Work, Energy and Power: Spring and Energy Conservation | 功与能：弹簧与能量守恒

Question: A block of mass 0.60 kg is pushed against a horizontal spring of force constant k = 480 N m⁻¹, compressing it by 0.15 m. The block is released from rest on a smooth surface. Find its speed when it leaves the spring.

题目：质量为 0.60 kg 的物块压缩一个劲度系数 k = 480 N m⁻¹ 的水平弹簧，压缩量为 0.15 m。物块从静止释放，表面光滑。求它离开弹簧时的速度。

The elastic potential energy stored in the spring is converted entirely into kinetic energy of the block: ½ k x² = ½ m v².

弹簧储存的弹性势能完全转化为物块的动能：½ k x² = ½ m v²。

Cancel ½ and solve for v: v = √(k x² / m) = √(480 × 0.15² / 0.60).

消去 ½ 并求解 v：v = √(k x² / m) = √(480 × 0.15² / 0.60)。

Compute stepwise: 0.15² = 0.0225, k x² = 480 × 0.0225 = 10.8 J. Then v = √(10.8 / 0.60) = √18 ≈ 4.24 m s⁻¹.

分步计算：0.15² = 0.0225，k x² = 480 × 0.0225 = 10.8 J。于是 v = √(10.8 / 0.60) = √18 ≈ 4.24 m s⁻¹。

This is a straightforward energy conservation problem, assuming no friction or air resistance.

这是一道直接运用能量守恒的题目，假设没有摩擦和空气阻力。

5. Electric Fields: Point Charges | 电场：点电荷

Question: Two point charges, Q₁ = +2.0 μC at x = 0 and Q₂ = +3.0 μC at x = 0.40 m, are fixed on the x-axis. Determine the electric field strength and potential at point P, x = 0.20 m. (k = 8.99 × 10⁹ N m² C⁻²)

题目：两个点电荷 Q₁ = +2.0 μC（位于 x = 0）和 Q₂ = +3.0 μC（位于 x = 0.40 m）固定在 x 轴上。求在 P 点（x = 0.20 m）处的电场强度和电势。（k = 8.99 × 10⁹ N m² C⁻²）

At P, distance from Q₁ is 0.20 m, and from Q₂ is 0.20 m as well. E₁ points away from Q₁ (to the right), magnitude E₁ = k Q₁ / r₁² = (8.99×10⁹ × 2.0×10⁻⁶) / 0.20² = 18.0×10³ / 0.04 = 4.50×10⁵ N C⁻¹ to the right.

P 点到 Q₁ 的距离为 0.20 m，到 Q₂ 的距离也为 0.20 m。E₁ 方向远离 Q₁（向右），大小 E₁ = k Q₁ / r₁² = (8.99×10⁹ × 2.0×10⁻⁶) / 0.20² = 4.50×10⁵ N C⁻¹ 向右。

E₂ points away from Q₂, so at P it points to the left. Magnitude E₂ = k Q₂ / r₂² = (8.99×10⁹ × 3.0×10⁻⁶) / 0.20² = 6.74×10⁵ N C⁻¹ to the left.

E₂ 方向远离 Q₂，因此在 P 点指向左。大小 E₂ = k Q₂ / r₂² = (8.99×10⁹ × 3.0×10⁻⁶) / 0.20² = 6.74×10⁵ N C⁻¹ 向左。

Net field is E = E₂ − E₁ = (6.74 − 4.50)×10⁵ = 2.24×10⁵ N C⁻¹, directed to the left.

合场强 E = E₂ − E₁ = (6.74 − 4.50)×10⁵ = 2.24×10⁵ N C⁻¹，方向向左。

Electric potential is scalar: V = k Q₁ / r₁ + k Q₂ / r₂ = (8.99×10⁹ × 2.0×10⁻⁶)/0.20 + (8.99×10⁹ × 3.0×10⁻⁶)/0.20 = 8.99×10⁴ + 1.35×10⁵ = 2.25×10⁵ V.

电势为标量：V = k Q₁ / r₁ + k Q₂ / r₂ = 8.99×10⁴ + 1.35×10⁵ = 2.25×10⁵ V。

6. DC Circuits: Kirchhoff’s Laws and Internal Resistance | 直流电路：基尔霍夫定律与内阻

Question: A 12 V battery of internal resistance 1.0 Ω is connected in a circuit with two external resistors: R₁ = 4.0 Ω and R₂ = 6.0 Ω in parallel. Find the terminal voltage of the battery and the current through each resistor.

题目：一个电动势 12 V、内阻 1.0 Ω 的电池，外接两个并联电阻 R₁ = 4.0 Ω 和 R₂ = 6.0 Ω。求电池的端电压和流过每个电阻的电流。

First find the equivalent external resistance: 1/R_ext = 1/4.0 + 1/6.0 = 3/12 + 2/12 = 5/12, so R_ext = 12/5 = 2.4 Ω.

先求外电路等效电阻：1/R_ext = 1/4.0 + 1/6.0 = 5/12，故 R_ext = 12/5 = 2.4 Ω。

Total circuit resistance is R_total = R_ext + r = 2.4 + 1.0 = 3.4 Ω. Total current I = ε / R_total = 12 / 3.4 ≈ 3.53 A.

总电阻 R_total = R_ext + r = 2.4 + 1.0 = 3.4 Ω。总电流 I = ε / R_total = 12 / 3.4 ≈ 3.53 A。

Terminal voltage V = ε − I r = 12 − 3.53 × 1.0 = 8.47 V. This is the voltage across the parallel combination.

端电压 V = ε − I r = 12 − 3.53 × 1.0 = 8.47 V。这就是并联部分两端的电压。

Current through R₁: I₁ = V / R₁ = 8.47 / 4.0 ≈ 2.12 A. Current through R₂: I₂ = V / R₂ = 8.47 / 6.0 ≈ 1.41 A. Check: I₁ + I₂ ≈ 3.53 A, matching the total current.

流过 R₁ 的电流：I₁ = V / R₁ = 8.47 / 4.0 ≈ 2.12 A。流过 R₂ 的电流：I₂ = V / R₂ = 8.47 / 6.0 ≈ 1.41 A。验证：I₁ + I₂ ≈ 3.53 A，与总电流一致。

7. Magnetic Fields: Force on a Current-Carrying Conductor | 磁场：载流导线受力

Question: A straight wire of length 0.25 m carries a current of 4.0 A and is placed in a uniform magnetic field of 0.60 T. The wire makes an angle of 30° with the field lines. Determine the magnitude of the magnetic force and state the direction relative to both current and field.

题目：一根长 0.25 m 的直导线通有 4.0 A 电流，置于均匀磁场中，磁感应强度为 0.60 T。导线与磁场线夹角为 30°。求磁力的大小，并说明其相对于电流和磁场的方向。

The magnetic force on a current-carrying wire is given by F = B I L sin θ, where θ is the angle between the current direction and the magnetic field.

载流导线在磁场中所受磁力由 F = B I L sin θ 给出，θ 为电流方向与磁场的夹角。

Substitute the values: F = 0.60 × 4.0 × 0.25 × sin 30° = 0.60 × 4.0 × 0.25 × 0.5 = 0.30 N.

代入数值：F = 0.60 × 4.0 × 0.25 × sin 30° = 0.60 × 4.0 × 0.25 × 0.5 = 0.30 N。

The direction of the force is given by Fleming’s left-hand rule: thumb (force), forefinger (field), middle finger (current). The force is perpendicular to both the current and the field, out of the plane defined by them.

力的方向由弗莱明左手定则确定：拇指为力，食指为磁场，中指为电流。力垂直于电流和磁场所决定的平面。

8. Waves: Young’s Double-Slit Interference | 波动：杨氏双缝干涉

Question: A double-slit experiment uses light of wavelength 520 nm. The slits are separated by 0.40 mm and the screen is 1.5 m away. Find the fringe separation. How far from the central maximum is the fourth bright fringe?

题目：双缝干涉实验使用波长 520 nm 的光，缝距为 0.40 mm，屏幕距双缝 1.5 m。求条纹间距。第四明纹离中央明纹多远？

The fringe spacing Δy is given by Δy = λ D / d. Ensure consistent units: λ = 520 × 10⁻⁹ m, D = 1.5 m, d = 0.40 × 10⁻³ m.

条纹间距 Δy = λ D / d。注意统一单位：λ = 520 × 10⁻⁹ m，D = 1.5 m，d = 0.40 × 10⁻³ m。

Δy = (520 × 10⁻⁹ × 1.5) / (0.40 × 10⁻³) = (7.80 × 10⁻⁷) / (4.0 × 10⁻⁴) = 1.95 × 10⁻³ m = 1.95 mm.

Δy = (520 × 10⁻⁹ × 1.5) / (0.40 × 10⁻³) = (7.80 × 10⁻⁷) / (4.0 × 10⁻⁴) = 1.95 × 10⁻³ m = 1.95 mm。

The fourth bright fringe corresponds to n = 4 (or order m = 4). Its distance from the central maximum is yₙ = n Δy = 4 × 1.95 mm = 7.80 mm.

第四明纹对应 n = 4。它与中央明纹的距离为 yₙ = n Δy = 4 × 1.95 mm = 7.80 mm。

This assumes small-angle approximation, which is valid here because y ≪ D.

此处小角度近似成立，因为 y ≪ D。

9. Thermal Physics: Ideal Gas and the First Law | 热学：理想气体与热力学第一定律

Question: Two moles of an ideal monatomic gas are held in a cylinder at an initial temperature of 300 K. The gas is heated at constant pressure of 1.0 × 10⁵ Pa until the temperature reaches 400 K. Calculate the work done by the gas, the increase in internal energy, and the heat supplied. (C_{v,m} = 12.5 J mol⁻¹ K⁻¹, R = 8.31 J mol⁻¹ K⁻¹)

题目：2 摩尔单原子理想气体装在气缸中，初始温度为 300 K。在恒定压强 1.0 × 10⁵ Pa 下加热至 400 K。求气体做功、内能增量及吸收的热量。（C_{v,m} = 12.5 J mol⁻¹ K⁻¹，R = 8.31 J mol⁻¹ K⁻¹）

Work done at constant pressure: W = p ΔV. For an ideal gas, p ΔV = n R ΔT, so W = n R (T₂ − T₁) = 2 × 8.31 × (400 − 300) = 2 × 8.31 × 100 = 1662 J.

等压过程中气体做功：W = p ΔV。对理想气体，p ΔV = n R ΔT，因此 W = n R (T₂ − T₁) = 2 × 8.31 × 100 = 1662 J。

Change in internal energy: ΔU = n C_{v,m} ΔT = 2 × 12.5 × 100 = 2500 J.

内能变化：ΔU = n C_{v,m} ΔT = 2 × 12.5 × 100 = 2500 J。

By the first law, ΔU = Q − W, so heat supplied Q = ΔU + W = 2500 + 1662 = 4162 J.

由热力学第一定律 ΔU = Q − W，得吸收的热量 Q = ΔU + W = 2500 + 1662 = 4162 J。

Notice that for a monatomic gas C_{v,m} = (3/2)R, here given as 12.5, consistent with 1.5 × 8.31 ≈ 12.5.

注意单原子气体 C_{v,m} = (3/2)R，此处给出 12.5，与 1.5 × 8.31 ≈ 12.5 吻合。

10. Quantum Physics: Photoelectric Effect and de Bro

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A-Level CIE Physics: Detailed Worked Examples | A-Level CIE 物理：典型例题详解