📚 Mastering Calculation Skills for WJEC A-Level Physics | A-Level WJEC 物理计算题专项训练
WJEC A-Level Physics exams demand confident and accurate numerical problem-solving. Whether it's mechanics, electricity, fields, or nuclear physics, a large proportion of the marks depend on your ability to select the correct equation, convert units properly, substitute values carefully, and present your answer with the appropriate significant figures. This guide is designed as a targeted calculation workout, covering the most common quantitative topics across the WJEC specification. Each section highlights key formulae, typical pitfalls, and a worked example to consolidate your technique.
WJEC A-Level 物理考试对计算题的准确性和熟练度要求很高。从力学、电学到场和核物理，大量分数都取决于你能否正确选择公式、规范地进行单位换算、准确代入数据并用恰当的有效数字呈现答案。本指南是一份针对性的计算题专项训练，覆盖 WJEC 考纲中最常见的定量计算主题。每个小节都突出核心公式、常见易错点，并通过典型例题来强化你的解题技能。
1. Kinematics: The Four Equations of Motion | 运动学：匀加速运动四大公式
In uniformly accelerated linear motion, four equations link initial velocity u, final velocity v, acceleration a, displacement s, and time t. Always define a positive direction and use consistent units (e.g. metres, seconds).
在匀加速直线运动中，有四个方程将初速度 u、末速度 v、加速度 a、位移 s 和时间 t 联系起来。你必须定义一个正方向，并统一单位（例如米、秒）。
v = u + at
s = ut + ½at²
s = ½(u+v)t
v² = u² + 2as
A common mistake is to use the wrong sign for acceleration when an object is slowing down. If the chosen positive direction is the initial motion, deceleration is negative.
常见错误是物体减速时加速度的符号用错。如果选定初始运动方向为正，减速时加速度应取负值。
15 = 0 + a×3.0, so a = 5.0 m s⁻². Then s = ½(u+v)t = ½(0+15)×3.0 = 22.5 m ≈ 23 m (2 s.f.).
例题：一辆汽车从静止加速到 15 m s⁻¹，用时 3.0 s。求加速度以及此过程中的位移。由 v = u + at 得 15 = 0 + a×3.0，故 a = 5.0 m s⁻²。再由 s = ½(u+v)t = ½(0+15)×3.0 = 22.5 m ≈ 23 m（两位有效数字）。
2. Forces and Newton’s Laws: From Free-Body Diagrams to Friction | 力与牛顿定律：从受力分析到摩擦力
The resultant force on an object equals its mass multiplied by acceleration: Fnet = ma. In problems involving slopes or pulleys, drawing a clear free-body diagram is essential. The frictional force is often given by f = μR, where R is the normal reaction and μ is the coefficient of friction.
物体所受的合力等于质量乘以加速度：F合 = ma。在涉及斜面或滑轮的问题中，画出清晰的受力分析图至关重要。摩擦力通常由 f = μR 给出，其中 R 为法向反作用力，μ 为摩擦系数。
iting; if so, f = μR. For the WJEC exam, be careful to convert grams to kilograms and cm to m.
在倾角为 θ 的斜面上，重力沿斜面的分量为 mg sinθ，法向反作用力为 mg cosθ。务必检查摩擦力是否达到极限；若是，则 f = μR。WJEC 考试中要特别注意将克换算为千克、厘米换算为米。
. Find the acceleration. Resolve forces: vertical component reduces normal reaction R = mg – 20 sin30°. Then f = μR, and horizontal net force = 20 cos30° – f. Finally, a = Fnet/m.
例题：质量 5.0 kg 的物块在粗糙水平面上被拖动，拉力为 20 N、方向与水平面成 30° 向上，μ = 0.25。求加速度。分解力：竖直分量减小法向反力 R = mg – 20 sin30°。然后 f = μR，水平合力 = 20 cos30° – f。最后 a = F合/m。
3. Momentum and Impulse: Conservation in Collisions | 动量与冲量：碰撞中的守恒
Linear momentum is defined as p = mv and is conserved in all isolated systems. Impulse is the change in momentum: FΔt = Δp = mv – mu. The area under a force–time graph also represents impulse.
线动量定义为 p = mv，在孤立系统中守恒。冲量等于动量的变化量：FΔt = Δp = mv – mu。力–时间图像下方的面积也表示冲量。
For collisions, distinguish between perfectly elastic (kinetic energy conserved) and inelastic collisions. In WJEC questions you will often combine conservation of momentum with energy calculations to find velocities or energy lost.
对于碰撞问题，要区分完全弹性碰撞（动能守恒）和非弹性碰撞。WJEC 的题目常将动量守恒与能量计算结合，以求解速度或损失的能量。
collision and the kinetic energy lost. By conservation: (2.0×4.0) = (2.0+3.0)v, v = 1.6 m s⁻¹. Initial KE = ½×2×16 = 16 J; final KE = ½×5×2.56 = 6.4 J; loss = 9.6 J.
例题：质量 2.0 kg 的小车以 4.0 m s⁻¹ 的速度运动，与静止的质量为 3.0 kg 的小车相撞后粘在一起。求碰撞后的速度及动能损失。由动量守恒：(2.0×4.0) = (2.0+3.0)v，v = 1.6 m s⁻¹。初始动能 = ½×2×16 = 16 J；末动能 = ½×5×2.56 = 6.4 J；损失 = 9.6 J。
4. Circular Motion: Centripetal Force and Angular Velocity | 圆周运动：向心力与角速度
For an object moving in a circle of radius r at constant speed v, the centripetal acceleration is a = v²/r or a = ω²r, where angular velocity ω = 2π/T = 2πf. The net inward force causing this acceleration is F = mv²/r = mω²r.
对于以恒定速率 v 在半径为 r 的圆周上运动的物体，向心加速度为 a = v²/r 或 a = ω²r，其中角速度 ω = 2π/T = 2πf。产生该加速度的向内的合力为 F = mv²/r = mω²r。
Common sources of centripetal force include tension (string), friction (car on a bend), and gravitational force (orbits). Always identify which force provides the centripetal component and set it equal to mv²/r. Be careful with unit conversions: radius in metres, period in seconds.
常见的向心力来源有绳的张力、摩擦（汽车转弯）和万有引力（轨道运动）。务必要明确是哪个力提供向心力分量，并令其等于 mv²/r。注意单位换算：半径用米，周期用秒。
ce. First find ω = 2π×2.0 = 12.57 rad s⁻¹. Then F = mω²r = 0.80 × (12.57)² × 1.5 ≈ 190 N (2 s.f.).
例题：质量 0.80 kg 的石块在水平圆周上旋转，半径 1.5 m，每秒 2.0 转。求向心力。先求 ω = 2π×2.0 = 12.57 rad s⁻¹。然后 F = mω²r = 0.80 × (12.57)² × 1.5 ≈ 190 N（两位有效数字）。
5. Electric Fields: Coulomb’s Law and Uniform Fields | 电场：库仑定律与匀强电场
Electric field strength E can be defined as E = F/q. For a point charge, the field strength at a distance r is E = kQ/r² (k = 1/(4πε₀) ≈ 8.99×10⁹ N m² C⁻²). In a uniform field between parallel plates, E = V/d.
电场强度 E 可以定义为 E = F/q。对于点电荷，距离 r 处的场强为 E = kQ/r²（k = 1/(4πε₀) ≈ 8.99×10⁹ N m² C⁻²）。在平行板间的匀强电场中，E = V/d。
done W = qΔV = qEd. Remember that the direction of the field is from positive to negative.
WJEC 试题经常要求你将 E = V/d 与带电粒子的受力 F = qE 结合，或求解做功 W = qΔV = qEd。记住电场方向是由正电荷指向负电荷。
Worked example: An electron (charge – .60×10⁻¹⁹ × 5.0×10³ = 8.0×10⁻¹⁶ N. Then a = F/m = 8.0×10⁻¹⁶ / 9.11×10⁻³¹ ≈ 8.8×10¹⁴ m s⁻².
例题：一个电子（电荷量 –1.60×10⁻¹⁹ C）置于场强为 5.0×10³ V m⁻¹ 的匀强电场中。求所受电场力的大小及其加速度（mₑ = 9.11×10⁻³¹ kg）。F = qE = 1.60×10⁻¹⁹ × 5.0×10³ = 8.0×10⁻¹⁶ N。然后 a = F/m = 8.0×10⁻¹⁶ / 9.11×10⁻³¹ ≈ 8.8×10¹⁴ m s⁻²。
6. DC Circuits: Kirchhoff’s Rules and Internal Resistance | 直流电路：基尔霍夫定律与内阻
Ohm's law V = IR applies to individual resistors, but for whole circuits you need Kirchhoff's laws: current junction rule (ΣIin = ΣIout) and voltage loop rule (Σε = ΣIR). Power can be calculated as P = IV = I²R = V²/R.
欧姆定律 V = IR 适用于单个电阻，但对于整个电路则需要基尔霍夫定律：电流节点定律（ΣI入 = ΣI出）和电压回路定律（Σε = ΣIR）。功率可表示为 P = IV = I²R = V²/R。
A real cell has internal resistance r, so the terminal p.d. is V = ε – Ir. When calculating current, the total resistance in the circuit is Rtotal = Rexternal + r. Always express currents in amperes and emf in volts.
实际电池具有内阻 r，因此端电压为 V = ε – Ir。计算电流时，电路中的总电阻为 R总 = R外 + r。电流务必用安培，电动势用伏特。
n the external resistor. I = ε / (R+r) = 12.0 / (3.5+0.5) = 3.0 A. Power P = I²R = (3.0)² × 3.5 = 31.5 W ≈ 32 W (2 s.f.).
例题：电动势为 12.0 V、内阻为 0.50 Ω 的电池接在一个 3.5 Ω 的电阻上。求电流以及外电阻上耗散的功率。I = ε / (R+r) = 12.0 / (3.5+0.5) = 3.0 A。功率 P = I²R = (3.0)² × 3.5 = 31.5 W ≈ 32 W（两位有效数字）。
7. Capacitors: Time Constants and Exponential Decay | 电容器：时间常数与指数衰减
Capacitance is C = Q/V. The energy stored is E = ½CV² = ½QV. For a discharging capacitor through a resistor, the p.d. and charge fall exponentially: V = V₀ e–t/RC, and the time constant is τ = RC.
电容的定义为 C = Q/V。储存的能量为 E = ½CV² = ½QV。对于通过电阻放电的电容器，电压和电荷量按指数规律下降：V = V₀ e–t/RC，时间常数 τ = RC。
One time constant is the time for the voltage to fall to 37% of its initial value. The half-life is related by t½ = RC ln 2. In WJEC practical-based questions you may analyse log-linear graphs of ln V against t.
经过一个时间常数，电压降至初始值的 37%。半衰期与时间常数的关系为 t½ = RC ln 2。在 WJEC 与实验相关的题目中，你可能需要分析 ln V 对 t 的对数–线性图。
and the time for the p.d. to halve. τ = RC = 220×10⁻⁶ × 47×10³ = 10.34 s. For half-life, t½ = τ ln2 ≈ 10.34 × 0.693 = 7.2 s.
例题：一个 220 μF 的电容器通过 47 kΩ 的电阻放电。求时间常数以及电压减半所需的时间。τ = RC = 220×10⁻⁶ × 47×10³ = 10.34 s。半衰期 t½ = τ ln2 ≈ 10.34 × 0.693 = 7.2 s。
8. Radioactive Decay: Half-life and Activity Calculations | 放射性衰变：半衰期与活度计算
Activity A is the number of decays per second and is given by A = λN, where λ is the decay constant. The exponential decay law is N = N₀ e–λt and the half-life T½ = ln2 / λ.
活度 A 是每秒的衰变次数，由 A = λN 给出，其中 λ 为衰变常量。指数衰变规律为 N = N₀ e–λt，半衰期 T½ = ln2 / λ。
You must be able to convert between mass, number of nuclei, and activity using the Avogadro constant NA NA. Often, WJEC questions require you to find the age of a sample using the decay equation rearranged to t = (1/λ) ln(N₀/N).
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