Edexcel Physics: Detailed Worked Examples | Edexcel 物理：典型例题详解

📚 Edexcel Physics: Detailed Worked Examples | Edexcel 物理：典型例题详解

This article presents a selection of typical Edexcel A Level Physics problems with step-by-step solutions. Each example is chosen to reinforce key concepts across mechanics, electricity, waves, quantum physics and nuclear physics. The worked solutions demonstrate effective problem-solving strategies and clear application of formulae, helping you master the skills required for the exam.

本文精选了 Edexcel A Level 物理中的典型例题，并给出逐步详解。每个例子都旨在巩固力学、电学、波动、量子物理和核物理的核心概念。这些解题过程展示了高效的问题解决策略和公式的清晰应用，帮助同学们掌握考试必备的技能。

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1. Kinematics: Two-Stage Motion Problem | 运动学：两阶段运动问题

A car starts from rest, accelerates uniformly for 10 s and then travels at constant speed for a further 20 s. The total distance covered is 400 m. Calculate the acceleration of the car.

一辆汽车从静止开始，先匀加速运动 10 秒，然后以恒定速度行驶 20 秒。总路程为 400 m。求汽车的加速度。

Let the acceleration be a. In the first phase, u = 0, t₁ = 10 s. Distance s₁ = ut₁ + ½at₁² = ½a(10)² = 50a. The final velocity after acceleration is v = u + at₁ = 10a.

设加速度为 a。在第一阶段，初速 u = 0，时间 t₁ = 10 s。位移 s₁ = ut₁ + ½at₁² = ½a(10)² = 50a。加速后的末速度 v = u + at₁ = 10a。

During the second phase, the car moves at constant velocity v for t₂ = 20 s. Distance s₂ = v t₂ = (10a) × 20 = 200a.

在第二阶段，汽车以恒定速度 v 运动 t₂ = 20 s。位移 s₂ = v t₂ = (10a) × 20 = 200a。

Total distance s = s₁ + s₂ = 50a + 200a = 250a. Set equal to 400 m: 250a = 400, therefore a = 400 / 250 = 1.6 m s⁻².

总位移 s = s₁ + s₂ = 50a + 200a = 250a。等于 400 m：250a = 400，因此加速度 a = 400 / 250 = 1.6 m s⁻²。

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2. Newton’s Laws: Force at an Angle with Friction | 牛顿定律：斜向拉力与摩擦力

A block of mass 5.0 kg rests on a rough horizontal surface. A force of 30 N is applied at 30° above the horizontal. The coefficient of kinetic friction between the block and the surface is 0.40. Find the acceleration of the block.

一个质量为 5.0 kg 的木块放在粗糙水平面上。一个大小为 30 N 的力以与水平方向成 30° 向上拉木块。木块与地面间的动摩擦因数为 0.40。求木块的加速度。

Resolve the applied force: horizontal component Fₓ = 30 cos 30° = 30 × (√3/2) ≈ 25.98 N. Vertical component Fᵧ = 30 sin 30° = 15 N upwards.

分解拉力：水平分量 Fₓ = 30 cos 30° = 30 × (√3/2) ≈ 25.98 N。竖直分量 Fᵧ = 30 sin 30° = 15 N（向上）。

Weight of block W = mg = 5.0 × 9.81 = 49.05 N downward. The normal reaction N = W – Fᵧ = 49.05 – 15 = 34.05 N.

木块重力 W = mg = 5.0 × 9.81 = 49.05 N（向下）。支持力 N = W – Fᵧ = 49.05 – 15 = 34.05 N。

Kinetic friction f = μ N = 0.40 × 34.05 = 13.62 N opposing motion.

动摩擦力 f = μ N = 0.40 × 34.05 = 13.62 N，方向与运动相反。

Net horizontal force F_net = Fₓ – f = 25.98 – 13.62 = 12.36 N. Acceleration a = F_net / m = 12.36 / 5.0 ≈ 2.47 m s⁻².

水平方向合力 F_net = Fₓ – f = 25.98 – 13.62 = 12.36 N。加速度 a = F_net / m = 12.36 / 5.0 ≈ 2.47 m s⁻²。

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3. Energy Conservation: Incline and Spring | 能量守恒：斜面与弹簧

A 2.0 kg mass slides from rest down a frictionless incline of height 5.0 m. At the bottom, it hits a horizontal spring with spring constant k = 200 N m⁻¹. Find the maximum compression of the spring.

一个 2.0 kg 的物体从静止沿光滑斜面下滑，斜面高度 5.0 m。在底端物体撞击一个水平放置的弹簧，弹簧劲度系数 k = 200 N m⁻¹。求弹簧的最大压缩量。

Since there is no friction, mechanical energy is conserved. Loss in gravitational potential energy = mgh = 2.0 × 9.81 × 5.0 = 98.1 J.

由于无摩擦，机械能守恒。减少的重力势能 = mgh = 2.0 × 9.81 × 5.0 = 98.1 J。

This energy is converted entirely into elastic potential energy of the spring: ½ k x_max² = 98.1 J.

这部分能量全部转化为弹簧的弹性势能：½ k x_max² = 98.1 J。

Rearrange: x_max² = (2 × 98.1) / 200 = 196.2 / 200 = 0.981. Therefore x_max = √0.981 ≈ 0.990 m (about 99 cm).

整理得：x_max² = (2 × 98.1) / 200 = 196.2 / 200 = 0.981。因此最大压缩量 x_max = √0.981 ≈ 0.990 m（约 99 cm）。

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4. Elastic Collision in One Dimension | 一维弹性碰撞

A 0.50 kg ball moving at 4.0 m s⁻¹ collides elastically with a stationary 0.30 kg ball on a frictionless track. Determine the velocities of both balls after the collision.

一个 0.50 kg 的小球以 4.0 m s⁻¹ 的速度在光滑轨道上运动，与一个静止的 0.30 kg 小球发生弹性碰撞。求碰撞后两球的速度。

For a perfectly elastic head-on collision, both momentum and kinetic energy are conserved. Using the standard formulae for one stationary object:

对于完全弹性正碰，动量和动能均守恒。使用经典公式（一球静止）：

v₁’ = (m₁ – m₂) / (m₁ + m₂) * u₁, v₂’ = (2m₁) / (m₁ + m₂) * u₁.

v₁’ = (m₁ – m₂) / (m₁ + m₂) · u₁, v₂’ = (2m₁) / (m₁ + m₂) · u₁。

Substitute m₁ = 0.50 kg, m₂ = 0.30 kg, u₁ = 4.0 m s⁻¹:

代入 m₁ = 0.50 kg，m₂ = 0.30 kg，u₁ = 4.0 m s⁻¹：

v₁’ = (0.50 – 0.30) / (0.50 + 0.30) × 4.0 = (0.20 / 0.80) × 4.0 = 0.25 × 4.0 = 1.0 m s⁻¹.

v₁’ = (0.50 – 0.30) / (0.50 + 0.30) × 4.0 = (0.20 / 0.80) × 4.0 = 0.25 × 4.0 = 1.0 m s⁻¹。

v₂’ = (2 × 0.50) / (0.80) × 4.0 = (1.0 / 0.80) × 4.0 = 1.25 × 4.0 = 5.0 m s⁻¹.

v₂’ = (2 × 0.50) / 0.80 × 4.0 = (1.0 / 0.80) × 4.0 = 1.25 × 4.0 = 5.0 m s⁻¹。

The 0.50 kg ball continues forward at 1.0 m s⁻¹; the 0.30 kg ball moves forward at 5.0 m s⁻¹.

0.50 kg 的球以 1.0 m s⁻¹ 向前运动；0.30 kg 的球以 5.0 m s⁻¹ 向前运动。

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5. DC Circuits: Kirchhoff’s Laws | 直流电路：基尔霍夫定律

In the circuit below, E₁ = 10 V, E₂ = 5 V, R₁ = 2 Ω, R₂ = 3 Ω, R₃ = 4 Ω. Use Kirchhoff’s laws to find the current through each resistor. (Assume the circuit forms two loops: the left loop contains E₁, R₁ and R₂; the right loop contains R₂, R₃ and E₂, with R₂ shared.)

在下图电路中，E₁ = 10 V，E₂ = 5 V，R₁ = 2 Ω，R₂ = 3 Ω，R₃ = 4 Ω。用基尔霍夫定律求通过每个电阻的电流。（假设电路有两个回路：左回路包含 E₁、R₁ 和 R₂；右回路包含 R₂、R₃ 和 E₂，R₂ 为共用电阻。）

Assign loop currents: I₁ flows clockwise in the left loop, I₂ flows clockwise in the right loop. Then current through R₂ is I₁ – I₂ (downward).

设定回路电流：左回路顺时针方向电流 I₁，右回路顺时针方向电流 I₂。则通过 R₂ 的电流（向下）为 I₁ – I₂。

Apply Kirchhoff’s Voltage Law to left loop: E₁ – I₁R₁ – (I₁ – I₂)R₂ = 0 → 10 – 2I₁ – 3(I₁ – I₂) = 0 → 10 – 5I₁ + 3I₂ = 0. (1)

对左回路应用基尔霍夫电压定律：E₁ – I₁R₁ – (I₁ – I₂)R₂ = 0 → 10 – 2I₁ – 3(I₁ – I₂) = 0 → 10 – 5I₁ + 3I₂ = 0。 (1)

Right loop: – (I₂ – I₁)R₂ – I₂R₃ – E₂ = 0? Be careful with polarity: going clockwise, the voltage drop across R₂ due to I₂ is I₂R₂, but direction of I₁ also contributes. Write as: -I₂R₃ – E₂ + (I₁ – I₂)R₂ = 0? Standard: Starting from negative terminal of E₂, go clockwise: rise +E₂, then drop across R₃: -I₂R₃, then drop across R₂: -I₂R₂, but current through R₂ is I₁ – I₂ downward, so the drop in the clockwise direction is +(I₁ – I₂)R₂? Let’s derive systematically: the right loop goes through E₂ (from – to +) gives +5 V, then through R₃ with current I₂, drop = -4I₂, then through R₂ with current I₂ relative to this loop? The correct KVL for right loop (clockwise) is: -E₂ + I₂R₃ + (I₂ – I₁)R₂ = 0 if we define same current direction. Many textbooks: For loop with E₂, resistor R₃ and shared R₂, the sum: -5 + 4I₂ + 3(I₂ – I₁) = 0 → 7I₂ – 3I₁ = 5. Let’s use that.

右回路：顺时针绕行，先经过 E₂ 从负极到正极电位升 +5 V，再经过 R₃ 电位降 -4I₂，然后经过 R₂：由于通过 R₂ 的电流为 I₁ – I₂（向下），顺时针方向经过 R₂ 的电位降为 +3(I₂ – I₁)。方程：-5 + 4I₂ + 3(I₂ – I₁) = 0 → 7I₂ – 3I₁ = 5。 (2)

Solve equations (1): 5I₁ – 3I₂ = 10 and (2): -3I₁ + 7I₂ = 5.

解方程组：(1) 5I₁ – 3I₂ = 10，(2) -3I₁ + 7I₂ = 5。

Multiply (1) by 3 and (2) by 5: 15I₁ – 9I₂ = 30; -15I₁ + 35I₂ = 25. Add: 26I₂ = 55 → I₂ = 55/26 ≈ 2.115 A. Then from (1): 5I₁ = 10 + 3I₂ = 10 + 6.346 = 16.346 → I₁ = 3.269 A. Current through R₂ = I₁ – I₂ ≈ 1.154 A.

(1)×3: 15I₁ – 9I₂ = 30；(2)×5: -15I₁ + 35I₂ = 25。相加得 26I₂ = 55，I₂ ≈ 2.115 A。代入 (1)：5I₁ = 10 + 3I₂ = 16.346，I₁ ≈ 3.269 A。R₂ 中的电流 = I₁ – I₂ ≈ 1.154 A。

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6. Young’s Double-Slit Interference | 杨氏双缝干涉

In a Young’s double-slit experiment, the slit separation is 0.50 mm and the screen is placed 2.0 m from the slits. The fringe separation on the screen is found to be 2.0 mm. Calculate the wavelength of the light used.

在杨氏双缝实验中，双缝间距为 0.50 mm，屏幕距双缝 2.0 m。测得条纹间距为 2.0 mm。计算所用光的波长。

The formula for fringe spacing Δy is Δy = λD / d, where λ is wavelength, D is distance to screen, d is slit separation.

条纹间距公式为 Δy = λD / d，其中 λ 是波长，D 是屏距，d 是双缝间距。

Rearrange: λ = Δy d / D. Convert all lengths to metres: Δy = 2.0 × 10⁻³ m, d = 0.50 × 10⁻³ = 5.0 × 10⁻⁴ m, D = 2.0 m.

整理得：λ = Δy d / D。统一单位：Δy = 2.0 × 10⁻³ m，d = 5.0 × 10⁻⁴ m，D = 2.0 m。

λ = (2.0 × 10⁻³) × (5.0 × 10⁻⁴) / 2.0 = (10.0 × 10⁻⁷) / 2.0 = 5.0 × 10⁻⁷ m = 500 nm.

λ = (2.0 × 10⁻³) × (5.0 × 10⁻⁴) / 2.0 = (10.0 × 10⁻⁷) / 2.0 = 5.0 × 10⁻⁷ m = 500 nm。

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7. Photoelectric Effect: Kinetic Energy Calculation | 光电效应：动能计算

Sodium has a work function of 2.3 eV. Light of wavelength 300 nm is incident on a sodium surface. Determine the maximum kinetic energy of the emitted photoelectrons.

钠的功函数为 2.3 eV。波长为 300 nm 的光照射到钠表面。求发射出的光电子的最大动能。

Photon energy E_photon = hc / λ. Using hc = 1240 eV·nm (approximation), E_photon = 1240 eV·nm / 300 nm = 4.13 eV.

光子能量 E_photon = hc / λ。利用近似关系 hc = 1240 eV·nm，得 E_photon = 1240 eV·nm / 300 nm = 4.13 eV。

Einstein’s photoelectric equation: K_max = E_photon – Φ = 4.13 eV – 2.3 eV = 1.83 eV.

爱因斯坦光电方程：K_max = E_photon – Φ = 4.13 eV – 2.3 eV = 1.83 eV。

If required in joules: 1 eV = 1.60 × 10⁻¹⁹ J, so K_max = 1.83 × 1.60 × 10⁻¹⁹ J ≈ 2.93 × 10⁻¹⁹ J.

如需以焦耳表示：1 eV = 1.60 × 10⁻¹⁹ J，故 K_max = 1.83 × 1.60 × 10⁻¹⁹ J ≈ 2.93 × 10⁻¹⁹ J。

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8. Mass-Energy Equivalence in Nuclear Reactions | 核反应中的质能等价

A nuclear reaction has a mass defect of 0.10 u. Calculate the energy released in MeV. (1 u = 931.5 MeV/c²)

某核反应的质量亏损为 0.10 u。计算释放的能量（以 MeV 为单位）。(1 u = 931.5 MeV/c²)

The energy released is given by ΔE = Δm c². With Δm in atomic mass units, ΔE = 0.10 u × 931.5 MeV/u = 93.15 MeV.

释放能量用质能方程 ΔE = Δm c² 计算。用原子质量单位，ΔE = 0.10 u × 931.5 MeV/u = 93.15 MeV。

The reaction releases approximately 93 MeV of energy, which is typical for nuclear processes and much larger than chemical energy changes.

该反应释放约 93 MeV 的能量，这是核过程的典型值，远大于化学反应的能量变化。

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