📚 Typical Worked Examples in A-Level CCEA Physics | A-Level CCEA 物理：典型例题详解
This article presents a collection of typical worked examples drawn from the CCEA A-Level Physics specification. Each example illustrates key principles and problem-solving techniques required for success in examinations. Work through each solution carefully to deepen your understanding of core topics.
本文精选了一系列来自 CCEA A-Level 物理大纲的典型例题，逐一进行详细解析。每个例题都展示了考试中必备的核心原理与解题方法。请仔细阅读每道题的解答过程，以加深对知识点的理解。
1. Uniformly Accelerated Motion | 匀加速直线运动
A car accelerates uniformly from rest at 2.0 m s⁻² for 5.0 s. Calculate the displacement and the final velocity.
一辆汽车从静止开始以 2.0 m s⁻² 的加速度匀加速运动 5.0 s。求位移和末速度。
Step 1 – List known quantities. Initial velocity u = 0, acceleration a = 2.0 m s⁻², time t = 5.0 s.
步骤 1 – 列出已知量。 初速度 u = 0，加速度 a = 2.0 m s⁻²，时间 t = 5.0 s。
Step 2 – Select the appropriate equation for displacement. Use s = ut + ½at².
步骤 2 – 选择适当的位移公式。 使用 s = ut + ½at²。
s = (0)(5.0) + ½ × 2.0 × (5.0)² = 0 + 0.5 × 2.0 × 25 = 25 m
Substituting the values gives a displacement of 25 m.
代入数值后，得到位移为 25 m。
Step 3 – Calculate final velocity using v = u + at.
步骤 3 – 使用 v = u + at 计算末速度。
v = 0 + 2.0 × 5.0 = 10 m s⁻¹
The final velocity is 10 m s⁻¹.
末速度为 10 m s⁻¹。
2. Newton’s Second Law and Friction | 牛顿第二定律与摩擦力
A 5.0 kg box rests on a horizontal surface. A horizontal force of 20 N is applied. The coefficient of kinetic friction between the box and the surface is 0.25. Calculate the acceleration of the box. (Use g = 9.8 m s⁻².)
一个 5.0 kg 的箱子放在水平面上，受到 20 N 的水平拉力。箱子与平面之间的动摩擦系数为 0.25。求箱子的加速度。（取 g = 9.8 m s⁻²）
Step 1 – Determine the normal reaction force. On a horizontal surface, N = mg.
步骤 1 – 计算支持力。 在水平面上，N = mg。
N = 5.0 × 9.8 = 49 N
Step 2 – Calculate the frictional force. F_f = μ N.
步骤 2 – 计算摩擦力。 F_f = μ N。
F_f = 0.25 × 49 = 12.25 N
Step 3 – Find the net force and apply Newton’s second law. F_net = applied force − friction = ma.
步骤 3 – 求合力并应用牛顿第二定律。 F_net = 拉力 − 摩擦力 = ma。
F_net = 20 − 12.25 = 7.75 N
a = F_net / m = 7.75 / 5.0 = 1.55 m s⁻²
The acceleration of the box is 1.55 m s⁻².
箱子的加速度为 1.55 m s⁻²。
3. Resistor Networks | 电阻网络
A 6.0 Ω resistor and a 3.0 Ω resistor are connected in parallel; this combination is then connected in series with a 4.0 Ω resistor. A 12 V battery is connected across the whole network. Calculate the total current from the battery and the potential difference across the 4.0 Ω resistor.
一个 6.0 Ω 和一个 3.0 Ω 的电阻并联，然后与一个 4.0 Ω 的电阻串联。整个网络接入 12 V 电源。求电池输出的总电流以及 4.0 Ω 电阻两端的电压。
Step 1 – Find the equivalent resistance of the parallel pair. 1/R_p = 1/6.0 + 1/3.0.
步骤 1 – 计算并联部分的等效电阻。 1/R_p = 1/6.0 + 1/3.0。
1/R_p = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 0.5 ⇒ R_p = 2.0 Ω
Step 2 – Calculate total resistance. R_total = R_p + 4.0 = 2.0 + 4.0 = 6.0 Ω.
步骤 2 – 计算总电阻。 R_total = R_p + 4.0 = 2.0 + 4.0 = 6.0 Ω。
Step 3 – Use Ohm’s law for the whole circuit to find total current. I = V / R_total.
步骤 3 – 用欧姆定律求总电流。 I = V / R_total。
I = 12 / 6.0 = 2.0 A
Step 4 – The current through the 4.0 Ω resistor is the total current. Calculate its p.d. V = I R = 2.0 × 4.0 = 8.0 V.
步骤 4 – 流过 4.0 Ω 电阻的电流就是总电流。计算其电压。 V = I R = 2.0 × 4.0 = 8.0 V。
The potential difference across the 4.0 Ω resistor is 8.0 V (and the remaining 4.0 V appears across the parallel pair).
4.0 Ω 电阻两端电压为 8.0 V（剩余的 4.0 V 加在并联部分）。
4. Young’s Double-Slit Interference | 杨氏双缝干涉
In a double-slit experiment, the slit separation is 0.50 mm and the screen is placed 1.50 m from the slits. The third bright fringe (m = 3) is found to be 24 mm from the central maximum. Determine the wavelength of the light used.
在双缝实验中，缝间距为 0.50 mm，屏幕距双缝 1.50 m。观察到第 3 级亮纹（m = 3）距中央明纹 24 mm。求所用光的波长。
Step 1 – Convert all quantities to SI units. d = 0.50 mm = 5.0 × 10⁻⁴ m, D = 1.50 m, y = 24 mm = 2.4 × 10⁻² m, m = 3.
步骤 1 – 将所有量转换为国际单位。 d = 0.50 mm = 5.0 × 10⁻⁴ m, D = 1.50 m, y = 24 mm = 2.4 × 10⁻² m, m = 3。
Step 2 – Apply the fringe-spacing formula for bright fringes. d sinθ ≈ d (y/D) = mλ, for small angles.
步骤 2 – 使用亮纹公式（小角度近似）。 d sinθ ≈ d (y/D) = mλ。
λ = y d / (m D)
Step 3 – Substitute and calculate.
步骤 3 – 代入数值计算。
λ = (2.4 × 10⁻²) × (5.0 × 10⁻⁴) / (3 × 1.50) = (1.2 × 10⁻⁵) / 4.5 = 2.67 × 10⁻⁷ m
Wavelength λ ≈ 2.7 × 10⁻⁷ m, which is 270 nm (ultraviolet).
波长约为 2.7 × 10⁻⁷ m，即 270 nm（紫外光）。
5. Photoelectric Effect | 光电效应
Light of wavelength 250 nm is incident on a metal surface whose work function is 2.3 eV. Calculate (a) the maximum kinetic energy of the emitted photoelectrons in joules and eV, and (b) the stopping potential. (h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹, e = 1.60 × 10⁻¹⁹ C.)
波长为 250 nm 的光照射到功函数为 2.3 eV 的金属表面。计算 (a) 逸出光电子的最大动能（以 J 和 eV 表示），(b) 截止电压。（h = 6.63 × 10⁻³⁴ J s，c = 3.00 × 10⁸ m s⁻¹，e = 1.60 × 10⁻¹⁹ C。）
Step 1 – Calculate the photon energy in joules. E_photon = hf = hc / λ.
步骤 1 – 计算光子能量（单位 J）。 E_photon = hf = hc / λ。
E_photon = (6.63 × 10⁻³⁴) × (3.00 × 10⁸) / (250 × 10⁻⁹) = 1.989 × 10⁻²⁵ / 2.50 × 10⁻⁷ = 7.96 × 10⁻¹⁹ J
Step 2 – Convert the work function to joules. Φ = 2.3 eV × 1.60 × 10⁻¹⁹ J eV⁻¹ = 3.68 × 10⁻¹⁹ J.
步骤 2 – 将功函数转换为焦耳。 Φ = 2.3 eV × 1.60 × 10⁻¹⁹ = 3.68 × 10⁻¹⁹ J。
Step 3 – Apply Einstein’s photoelectric equation. E_k max = hf − Φ.
步骤 3 – 应用爱因斯坦光电方程。 E_k max = hf − Φ。
E_k max = (7.96 − 3.68) × 10⁻¹⁹ = 4.28 × 10⁻¹⁹ J
In eV: E_k max = 4.28 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 2.68 eV.
以 eV 为单位：E_k max = 2.68 eV。
Step 4 – The stopping potential V_s is related to maximum kinetic energy by E_k max = eV_s.
步骤 4 – 截止电压 V_s 满足 E_k max = eV_s。
V_s = E_k max / e = 2.68 V
The stopping potential is 2.68 V.
截止电压为 2.68 V。
6. Circular Motion | 圆周运动
A car travels at a constant speed of 20 m s⁻¹ around a bend of radius 50 m. Calculate (a) the centripetal acceleration, and (b) the minimum coefficient of friction required between the tyres and the road to prevent skidding.
一辆汽车以 20 m s⁻¹ 的恒定速率沿半径 50 m 的弯道行驶。计算 (a) 向心加速度，以及 (b) 防止侧滑所需的轮胎与路面之间的最小摩擦系数。
Step 1 – Determine the centripetal acceleration. a_c = v² / r.
步骤 1 – 求向心加速度。 a_c = v² / r。
a_c = (20)² / 50 = 400 / 50 = 8.0 m s⁻²
Step 2 – The centripetal force is provided by friction. F_f = m a_c and F_f = μ N, where N = mg on a horizontal road.
步骤 2 – 向心力由摩擦力提供。 F_f = m a_c，而 F_f = μ N，水平路面上 N = mg。
μ m g = m a_c ⇒ μ = a_c / g
Step 3 – Substitute values (taking g = 9.8 m s⁻²).
步骤 3 – 代入数值（取 g = 9.8 m s⁻²）。
μ = 8.0 / 9.8 ≈ 0.82
A coefficient of friction of at least 0.82 is required.
摩擦系数至少需要 0.82。
7. Capacitor Discharge | 电容放电
A 100 μF capacitor is charged to 6.0 V and then discharged through a 10 kΩ resistor. Calculate (a) the time constant, (b) the initial discharge current, and (c) the time taken for the voltage to fall to 1.5 V.
一个 100 μF 的电容充电至 6.0 V，然后通过 10 kΩ 的电阻放电。计算 (a) 时间常数，(b) 初始放电电流，(c) 电压降至 1.5 V 所需的时间。
Step 1 – Time constant τ = RC.
步骤 1 – 时间常数 τ = RC。
τ = (10 × 10³) × (100 × 10⁻⁶) = 1.0 s
Step 2 – Initial current I₀ = V₀ / R, because the capacitor initially behaves like a source of emf V₀.
步骤 2 – 初始电流 I₀ = V₀ / R，因为起始时刻电容相当于一个电压为 V₀ 的电源。
I₀ = 6.0 / (10 × 10³) = 6.0 × 10⁻⁴ A = 0.60 mA
Step 3 – For a discharging capacitor, V = V₀ e^(-t/RC). Rearrange to solve for t.
步骤 3 – 电容放电公式 V = V₀ e^(-t/RC)。变形求时间 t。
V / V₀ = e^(-t/τ) ⇒ ln(V₀/V) = t / τ
t = τ ln(V₀ / V) = 1.0 × ln(6.0 / 1.5) = ln 4 = 1.39 s
The time needed for the voltage to drop to 1.5 V is about 1.4 s.
电压降至 1.5 V 约需 1.4 s。
8. Radioactive Decay | 放射性衰变
A sample of a radioactive isotope has an initial activity of 1.6 × 10⁵ Bq. Its half-life is 8.0 days. Calculate (a) the decay constant λ, and (b) the activity after 16 days.
某放射性同位素样品的初始活度为 1.6 × 10⁵ Bq，其半衰期为 8.0 天。计算 (a) 衰变常数 λ，以及 (b) 16 天后的活度。
Step 1 – Determine the decay constant using λ = ln 2 / t½.
步骤 1 – 利用 λ = ln 2 / t½ 求衰变常数。
λ = 0.693 / 8.0 = 0.0866 day⁻¹
Step 2 – Use the exponential decay law A = A₀ e^(-λt) to find the activity after 16 days.
步骤 2 – 使用指数衰变公式 A = A₀ e^(-λt) 计算 16 天后的活度。
t = 16 days, λt = 0.0866 × 16 = 1.386
A = 1.6 × 10⁵ × e^(-1.386) = 1.6 × 10⁵ × 0.25 = 4.0 × 10⁴ Bq
Alternatively, after two half-lives the activity is (½)² of the original: 1.6 × 10⁵ / 4 = 4.0 × 10⁴ Bq, which confirms the result.
另一种方法：经过两个半衰期后，活度变为原来的 (½)²，即 1.6 × 10⁵ /
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