A2 Physics: Calculation Drills Mastery | A2物理计算题专项训练

📚 A2 Physics: Calculation Drills Mastery | A2物理计算题专项训练

Calculation problems form the backbone of A2 Physics examinations, testing your ability to apply theoretical principles to quantitative scenarios. This article provides a structured drill across major topics, equipping you with step-by-step strategies, typical worked examples, and common pitfalls to avoid. Strengthen your numerical fluency and approach every calculation with confidence.

计算题是A2物理考试的核心，考查你将理论知识应用于定量情境的能力。本文围绕主要专题提供结构化训练，传授分步解题策略、典型范例与常见误区，帮助你提升数字运算的流畅度，从容应对每一道计算。

1. Mechanics & Energy | 力学与能量计算

Mechanics questions often demand a systematic approach: sketch a free-body diagram, resolve forces, apply Newton’s second law, and choose between kinematic equations or energy conservation. Always convert quantities to SI units (kg, m, s, N) before substituting into formulas.

力学题通常需要系统的方法：画出受力图，分解力，应用牛顿第二定律，并在运动学方程与能量守恒之间做出选择。代入公式前务必把物理量转换为国际单位（kg, m, s, N）。

Example: A 1500 kg car accelerates uniformly from rest to 25 m/s over a distance of 200 m. Calculate the net force acting on the car.

例题：一辆1500 kg的汽车从静止开始均匀加速，经过200 m 后速度达到25 m/s。计算汽车所受的净力。

Solution steps: (1) Use v² = u² + 2as to find acceleration: (25)² = 0 + 2a(200) → a = 1.5625 m/s². (2) Apply F = ma: F = 1500 × 1.5625 = 2343.75 N ≈ 2340 N (3 s.f.). Alternatively, calculate kinetic energy gain (½mv²) and use work–energy theorem: W = ΔEₖ = ½ × 1500 × (25)² = 468 750 J, then F = W/d = 468 750/200 = 2343.75 N. Both routes give the same result.

解题步骤：（1）用 v² = u² + 2as 求加速度：(25)² = 0 + 2a(200) → a = 1.5625 m/s²。（2）应用 F = ma：F = 1500 × 1.5625 = 2343.75 N ≈ 2340 N（三位有效数字）。或者，计算动能增量（½mv²）并利用功能定理：W = ΔEₖ = ½ × 1500 × (25)² = 468 750 J，然后 F = W/d = 468 750/200 = 2343.75 N。两种方法结果一致。

Common errors include forgetting to square the velocity, using the wrong mass, or mixing kinematics with forces without linking through acceleration. Always check if resistive forces are negligible; if not, net force = driving force − resistance.

常见错误包括忘记速度平方、用错质量、或在未经加速度连接的情况下混淆运动学与力。始终检查阻力是否可忽略；若不可忽略，净力 = 驱动力 − 阻力。

2. Circular Motion & Gravitation | 圆周运动与万有引力

For an object moving in a circle at constant speed, the centripetal force is F = mv²/r = mrω². In gravitational contexts, the centripetal force is provided by gravity: GMm/r² = mv²/r or GMm/r² = mrω². This yields the orbital speed v = √(GM/r) and Kepler’s third law T² ∝ r³.

对于匀速圆周运动，向心力为 F = mv²/r = mrω²。在引力情境中，向心力由引力提供：GMm/r² = mv²/r 或 GMm/r² = mrω²，得出轨道速度 v = √(GM/r) 以及开普勒第三定律 T² ∝ r³。

Example: A satellite orbits Earth at an altitude of 400 km. Given Earth’s mass 5.97 × 10²⁴ kg and radius 6370 km, find the orbital period.

例题：一颗卫星在离地400 km的轨道上运行。已知地球质量为5.97 × 10²⁴ kg，半径为6370 km，求轨道周期。

Steps: (1) Orbital radius r = (6370 + 400) km = 6.77 × 10⁶ m. (2) Using T² = (4π²/GM) r³, with G = 6.67 × 10⁻¹¹ N m² kg⁻². Compute r³ = (6.77×10⁶)³ = 3.10 × 10²⁰ m³. (3) T² = [4π²/(6.67×10⁻¹¹ × 5.97×10²⁴)] × 3.10×10²⁰ = (39.48 / 3.98×10¹⁴) × 3.10×10²⁰ ≈ 9.91×10⁻¹⁴ × 3.10×10²⁰ = 3.07×10⁷ s². (4) T = √(3.07×10⁷) ≈ 5540 s ≈ 92.3 min. Always convert km to m.

步骤：（1）轨道半径 r = (6370 + 400) km = 6.77 × 10⁶ m。（2）用 T² = (4π²/GM) r³，取 G = 6.67 × 10⁻¹¹ N m² kg⁻²。计算 r³ = (6.77×10⁶)³ = 3.10 × 10²⁰ m³。（3）T² = [4π²/(6.67×10⁻¹¹ × 5.97×10²⁴)] × 3.10×10²⁰ = (39.48 / 3.98×10¹⁴) × 3.10×10²⁰ ≈ 9.91×10⁻¹⁴ × 3.10×10²⁰ = 3.07×10⁷ s²。（4）T = √(3.07×10⁷) ≈ 5540 s ≈ 92.3 min。务必把 km 换算为 m。

A frequent mistake is using Earth’s radius as the orbital radius instead of adding the altitude. Also ensure to square π correctly and use the value of G in SI. If a geostationary orbit is asked, the period must be 24 hours (86 400 s) – equate and solve for r.

常见错误是把地球半径当作轨道半径而忘了加上高度。还要确保正确平方 π 并使用 SI 制下的 G 值。若要求同步轨道，周期必须为24小时（86 400 s），列方程求解 r。

3. Simple Harmonic Motion | 简谐运动

SHM is defined by a = −ω²x. The displacement x = A cos(ωt) or A sin(ωt), velocity v = ±ω√(A² − x²), maximum speed v_max = ωA, maximum acceleration a_max = ω²A. The period of a mass–spring system T = 2π√(m/k) and for a simple pendulum T = 2π√(L/g).

简谐运动由 a = −ω²x 定义。位移 x = A cos(ωt) 或 A sin(ωt)，速度 v = ±ω√(A² − x²)，最大速度 v_max = ωA，最大加速度 a_max = ω²A。弹簧振子的周期 T = 2π√(m/k)，单摆周期 T = 2π√(L/g)。

Example: A 0.500 kg mass on a spring oscillates with amplitude 4.0 cm and period 0.80 s. Determine the maximum kinetic energy and the spring constant.

例题：一0.500 kg的物体连接在弹簧上，振幅为4.0 cm，周期为0.80 s。求最大动能和劲度系数。

Steps: (1) ω = 2π/T = 2π/0.80 = 7.854 rad/s. (2) v_max = ωA = 7.854 × 0.040 = 0.314 m/s. (3) Eₖ_max = ½ m v_max² = 0.5 × 0.500 × (0.314)² = 0.0247 J. (4) Using T = 2π√(m/k) → k = 4π²m/T² = (4π² × 0.500) / 0.64 = 19.63 / 0.64 ≈ 30.7 N/m. Alternatively, k = mω² = 0.500 × (7.854)² = 30.8 N/m. Always convert cm to m.

步骤：（1）ω = 2π/T = 2π/0.80 = 7.854 rad/s。（2）v_max = ωA = 7.854 × 0.040 = 0.314 m/s。（3）Eₖ_max = ½ m v_max² = 0.5 × 0.500 × (0.314)² = 0.0247 J。（4）由 T = 2π√(m/k) 得 k = 4π²m/T² = (4π² × 0.500) / 0.64 = 19.63 / 0.64 ≈ 30.7 N/m。或者 k = mω² = 0.500 × (7.854)² = 30.8 N/m。始终将 cm 转换为 m。

Watch out for mixing amplitude in cm without conversion, and forgetting that total energy = ½kA² = maximum kinetic energy. Also note that the mass in the pendulum formula is not needed, but for spring systems it is crucial.

注意不能直接将 cm 代入而忘记换算，总能量 = ½kA² = 最大动能。单摆周期公式与质量无关，但弹簧振子中质量至关重要。

4. Electric Fields & Potential | 电场与电势

For point charges, electric field strength E = kQ/r² (where k = 1/(4π ε₀) = 8.99×10⁹ N m² C⁻²), force F = qE, and electric potential V = kQ/r. The work done in moving a charge q through a potential difference ΔV is W = qΔV. Uniform fields follow E = V/d.

对于点电荷，电场强度 E = kQ/r²（其中 k = 1/(4π ε₀) = 8.99×10⁹ N m² C⁻²），作用力 F = qE，电势 V = kQ/r。移动电荷 q 经过电势差 ΔV 做的功为 W = qΔV。匀强电场满足 E = V/d。

Example: Two point charges, +2.0 μC and −4.0 μC, are separated by 30 cm. Find the electric field strength at the midpoint between them.

例题：两点电荷，+2.0 μC 和 −4.0 μC，相距30 cm。求它们连线中点的电场强度。

Steps: (1) Distance from each charge to midpoint r = 0.15 m. (2) E₁ due to +2.0 μC = k × 2.0×10⁻⁶ / (0.15)² = (8.99×10⁹ × 2.0×10⁻⁶) / 0.0225 = 17 980 / 0.0225 ≈ 7.99×10⁵ N/C (away from positive charge). (3) E₂ due to −4.0 μC = k × 4.0×10⁻⁶ / (0.15)² = 2 × 7.99×10⁵ = 1.598×10⁶ N/C (towards negative charge). Since both fields point in the same direction (towards the negative charge), resultant E = 7.99×10⁵ + 1.598×10⁶ = 2.397×10⁶ N/C ≈ 2.4×10⁶ N/C towards the negative charge.

步骤：（1）各电荷到中点的距离 r = 0.15 m。（2）由 +2.0 μC 产生的 E₁ = k × 2.0×10⁻⁶ / (0.15)² = (8.99×10⁹ × 2.0×10⁻⁶) / 0.0225 = 17 980 / 0.0225 ≈ 7.99×10⁵ N/C（背离正电荷）。（3）由 −4.0 μC 产生的 E₂ = k × 4.0×10⁻⁶ / (0.15)² = 2 × 7.99×10⁵ = 1.598×10⁶ N/C（指向负电荷）。由于两者方向相同（均指向负电荷），合场强 E = 7.99×10⁵ + 1.598×10⁶ = 2.397×10⁶ N/C ≈ 2.4×10⁶ N/C，方向指向负电荷。

Beware of sign conventions: field is a vector; determine direction carefully. For potential energy, use scalar addition with signs. The midpoint potential V = k(+2.0×10⁻⁶)/0.15 + k(−4.0×10⁻⁶)/0.15 = (k/0.15)(−2.0×10⁻⁶) = −1.2×10⁵ V.

注意符号惯例：电场是矢量，需仔细判断方向。电势能使用标量叠加并保留正负号。中点电势 V = k(+2.0×10⁻⁶)/0.15 + k(−4.0×10⁻⁶)/0.15 = (k/0.15)(−2.0×10⁻⁶) = −1.2×10⁵ V。

5. Capacitance | 电容器计算

Key equations: Q = CV, capacitance of a parallel-plate capacitor C = ε₀A/d, energy stored E = ½CV² = ½QV = ½Q²/C. For capacitors in series, 1/C_total = Σ(1/C_i); in parallel, C_total = ΣC_i. Time constant for RC discharge: τ = RC, and the discharge equation: V = V₀ e^(−t/RC).

核心公式：Q = CV，平行板电容器的电容 C = ε₀A/d，储能 E = ½CV² = ½QV = ½Q²/C。串联电容：1/C_total = Σ(1/C_i)；并联电容：C_total = ΣC_i。RC 放电时间常数 τ = RC，放电方程：V = V₀ e^(−t/RC)。

Example: A 100 μF capacitor is charged to 12 V and then discharged through a 50 kΩ resistor. Calculate the charge after 3.0 s and the time taken for the voltage to drop to 5.0 V.

例题：一100 μF 的电容器充电至12 V，然后通过50 kΩ 电阻放电。计算3.0 s后的电荷量以及电压降至5.0 V所需的时间。

Steps: (1) Time constant τ = RC = 50×10³ × 100×10⁻⁶ = 5.0 s. (2) Initial charge Q₀ = CV = 100×10⁻⁶ × 12 = 1.2×10⁻³ C. (3) Q = Q₀ e^(−t/τ) = 1.2×10⁻³ × e^(−3.0/5.0) = 1.2×10⁻³ × e^(−0.6) ≈ 1.2×10⁻³ × 0.549 = 6.59×10⁻⁴ C. (4) For voltage, V = V₀ e^(−t/τ) → 5.0 = 12 e^(−t/5.0) → e^(−t/5.0) = 5/12 → −t/5.0 = ln(5/12) = −0.875 → t = 4.38 s. Always keep units consistent: convert μF to F (10⁻⁶) and kΩ to Ω (10³).

步骤：（1）时间常数 τ = RC = 50×10³ × 100×10⁻⁶ = 5.0 s。（2）起始电荷 Q₀ = CV = 100×10⁻⁶ × 12 = 1.2×10⁻³ C。（3）Q = Q₀ e^(−t/τ) = 1.2×10⁻³ × e^(−3.0/5.0) = 1.2×10⁻³ × e^(−0.6) ≈ 1.2×10⁻³ × 0.549 = 6.59×10⁻⁴ C。（4）电压：V = V₀ e^(−t/τ) → 5.0 = 12 e^(−t/5.0) → e^(−t/5.0) = 5/12 → −t/5.0 = ln(5/12) = −0.875 → t = 4.38 s。始终统一单位：μF 转换为 F (10⁻⁶)，kΩ 转换为 Ω (10³)。

Common errors include using milli-, micro-, or kilo- prefixes incorrectly in calculations, and forgetting that the exponential uses the ratio t/τ. Also, for charging, the formula is V = V₀(1 − e^(−t/RC)). Practice plotting and interpreting logarithmic graphs for exponential decay.

常见错误包括在计算中混用毫、微、千等词头，以及忘记指数中使用 t/τ 的比值。另外，充电公式为 V = V₀(1 − e^(−t/RC))。建议练习绘制和解读指数衰减的对数图像。

6. Magnetic Fields & Forces | 磁场与力

For a charged particle moving in a magnetic field, the force is F = Bqv sinθ. If the motion is perpendicular to the field, it follows a circular path with radius r = mv/(Bq). The force on a current-carrying wire is F = BIL sinθ. Magnetic flux density is measured in tesla (T).

带电粒子在磁场中运动时，受力 F = Bqv sinθ。若运动方向与磁场垂直，粒子将做圆周运动，半径 r = mv/(Bq)。载流导线在磁场中受力 F = BIL sinθ。磁通量密度的单位为特斯拉 (T)。

Example: An electron (mass = 9.11×10⁻³¹ kg, charge = −1.60×10⁻¹⁹ C) moves at 2.0×10⁷ m/s perpendicular to a uniform magnetic field of 0.50 T. Calculate the radius of its path and the time for one complete revolution.

例题：一电子（质量 9.11×10⁻³¹ kg，电量 −1.60×10⁻¹⁹ C）以2.0×10⁷ m/s的速度垂直进入0.50 T的匀强磁场。求其运动轨迹半径和旋转一周的时间。

Steps: (1) r = mv/(Bq) = (9.11×10⁻³¹ × 2.0×10⁷) / (0.50 × 1.60×10⁻¹⁹) = (1.822×10⁻²³) / (8.0×10⁻²⁰) = 2.28×10⁻⁴ m. (2) Period T = 2πr/v = 2π × 2.28×10⁻⁴ / 2.0×10⁷ ≈ 7.16×10⁻¹¹ s. Alternatively, T = 2πm/(Bq) = (2π × 9.11×10⁻³¹) / (0.50 × 1.60×10⁻¹⁹) = 5.72×10⁻³¹ / 8.0×10⁻²⁰ = 7.15×10⁻¹¹ s. Note that the period is independent of speed.

步骤：（1）r = mv/(Bq) = (9.11×10⁻³¹ × 2.0×10⁷) / (0.50 × 1.60×10⁻¹⁹) = (1.822×10⁻²³) / (8.0×10⁻²⁰) = 2.28×10⁻⁴ m。（2）周期 T = 2πr/v = 2π × 2.28×10⁻⁴ / 2.0×10⁷ ≈ 7.16×10⁻¹¹ s。或用 T = 2πm/(Bq) = (2π × 9.11×10⁻³¹) / (0.50 × 1.60×10⁻¹⁹) = 5.72×10⁻³¹ / 8.0×10⁻²⁰ = 7.15×10⁻¹¹ s。注意周期与速度无关。

Make sure to use the magnitude of charge for circular motion calculations. The direction of force is given by Fleming’s left-hand rule for conventional current (or right-hand rule for negative charges). In the wire force equation, angle θ is between current and magnetic field.

计算圆周运动时要使用电荷量的大小。力的方向由左手定则判断（对正电荷或常规电流），负电荷则用右手。在导线受力公式中，θ 是电流与磁场方向的夹角。

7. Electromagnetic Induction | 电磁感应

Faraday’s law: induced emf ε = −N Δφ/Δt, where magnetic flux φ = BA cosθ. Lenz’s law gives the direction. For a conductor of length L moving perpendicularly across a field B at speed v, ε = BLv. In an AC generator, ε = NBAω sin(ωt).

法拉第定律：感应电动势 ε = −N Δφ/Δt，磁通量 φ = BA cosθ。楞次定律决定感应电流方向。长为L的导体以速度 v 垂直切割磁感线时，ε = BLv。在交流发电机中，ε = NBAω sin(ωt)。

Example: A coil of 200 turns, each of area 0.025 m², is in a magnetic field of 0.40 T. The field is reduced uniformly to zero in 0.50 s. Calculate the average induced emf. If the coil is then rotated at 50 Hz, find the peak emf.

例题：一200匝的线圈，每匝面积为0.025 m²，置于0.40 T的磁场中。磁场在0.50 s内均匀减小到零。计算平均感应电动势。若线圈随后以50 Hz旋转，求峰值电动势。

Steps: (1) Initial flux per turn φ = BA = 0.40 × 0.025 = 0.010 Wb. (2) Δφ = 0 − 0.010 = −0.010 Wb (change magnitude 0.010 Wb). (3) Average emf magnitude ε = N|Δφ/Δt| = 200 × 0.010 / 0.50 = 40 / 0.50 = 80 V. (4) For rotation: peak emf ε₀ = NBAω. ω = 2πf = 100π = 314 rad/s. ε₀ = 200 × 0.40 × 0.025 × 314 = 200 × 0.010 × 314 = 2.0 × 314 = 628 V. Ensure the flux change uses the correct sign, but magnitude is usually sufficient for average emf.

步骤：（1）每匝的初始磁通量 φ = BA = 0.40 × 0.025 = 0.010 Wb。（2）Δφ = 0 − 0.010 = −0.010 Wb（变化量大小为0.010 Wb）。（3）平均电动势大小 ε = N|Δφ/Δt| = 200 × 0.010 / 0.50 = 40 / 0.50 = 80 V。（4）旋转时：峰值电动势 ε₀ = NBAω。ω = 2πf = 100π = 314 rad/s。ε₀ = 200 × 0.40 × 0.025 × 314 = 200 × 0.010 × 314 = 2.0 × 314 = 628 V。磁通量变化需用正确符号，但求平均电动势大小通常取绝对值。

Students often confuse area with cross-sectional area or use the angle incorrectly. The angle in φ = BA cosθ is between the field and the normal to the coil. In rotation, the flux linkage varies as NBA cos(ωt), leading to emf = NBAω sin(ωt).

学生常混淆面积与横截面积，或使用角度出错。φ = BA cosθ 中的角度是磁场与线圈法线之间的夹角。旋转时磁链为 NBA cos(ωt)，电动势为 NBAω sin(ωt)。

8. Alternating Current | 交流电

AC calculations involve peak (I₀, V₀) and root-mean-square (rms) values: I_rms = I₀/√2, V_rms = V₀/√2. Power in a resistive load P = I_rms V_rms = I_rms² R. Transformers follow V_s/V_p = N_s/N_p and for an ideal transformer P_p = P_s, hence V_p I_p = V_s I_s.

交流电计算涉及峰值（I₀, V₀）和有效值：I_rms = I₀/√2，V_rms = V₀/√2。电阻性负载的功率 P = I_rms V_rms = I_rms² R。变压器遵循 V_s/V_p = N_s/N_p，理想变压器满足 P_p = P_s，即 V_p I_p = V_s I_s。

Example: A mains supply of 230 V rms (50 Hz) is connected to a 100 Ω resistor. Calculate the peak current and the power dissipated. If a transformer steps this down to 12 V rms for a device drawing 2.0 A, find the primary current assuming 100% efficiency.

例题：230 V 有效值（50 Hz）的市电连接一个100 Ω 电阻。计算峰值电流和耗散功率。若用变压器降至12 V 有效值，供一设备消耗2.0 A电流，假设效率100%，求初级电流。

Steps: (1) V₀ = V_rms × √2 = 230 × 1.414 = 325 V. (2) I₀ = V₀/R = 325/100 = 3.25 A, I_rms = 230/100 = 2.30 A. (3) P = I_rms² R = (2.30)² × 100 = 529 W (or P = V_rms I_rms = 230 × 2.30 = 529 W). (4) For transformer: V_p = 230 V, V_s = 12 V, I_s = 2.0 A. Ideal transformer: V_p I_p = V_s I_s → I_p = (V_s I_s)/V_p = (12 × 2.0)/230 = 0.104 A ≈ 0.10 A. Always check whether the given voltage is rms or peak.

步骤：（1）V₀ = V_rms × √2 = 230 × 1

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A2 Physics: Calculation Drills Mastery | A2物理计算题专项训练