📚 Year 13 AQA Physics Mock Test Breakdown | AQA 物理模拟卷解析
Mock tests are a vital part of preparing for AQA Year 13 Physics, helping you master advanced mechanics, fields, and nuclear physics. This article provides a detailed walkthrough of a typical unit test paper, offering step‑by‑step solutions, key techniques, and common mistakes to avoid. By engaging with these questions, you will sharpen your understanding and boost your confidence for the final examinations.
模拟测试是备考 AQA Year 13 物理的重要环节，有助于掌握进阶力学、场论和核物理。本文对一份典型单元测试卷进行详细解析，提供逐步解答、关键技巧和易犯错误。通过练习这些题目，你将加深理解，增强考试信心。
1. Circular Motion & Centripetal Force | 圆周运动与向心力
A car of mass 1200 kg drives at a constant speed of 15 m s⁻¹ around a roundabout of radius 25 m. Calculate the centripetal acceleration and the frictional force that provides the centripetal force.
一辆质量为 1200 kg 的汽车以 15 m s⁻¹ 的恒定速率绕半径为 25 m 的环岛行驶。计算向心加速度以及提供向心力的摩擦力。
We start with the equation for centripetal acceleration, a = v² / r. Substituting the numbers gives a = (15 m s⁻¹)² / 25 m. The squared velocity is 225 m² s⁻², so a = 225 / 25 = 9.0 m s⁻². The centripetal force is then F = m a = 1200 kg × 9.0 m s⁻² = 1.08 × 10⁴ N. For a car on a horizontal roundabout, static friction between the tyres and the road supplies this force, pointing towards the centre of the circle.
我们首先使用向心加速度公式 a = v² / r。代入数值计算：a = (15 m s⁻¹)² / 25 m，速度平方为 225 m² s⁻²，因此 a = 225 / 25 = 9.0 m s⁻²。向心力 F = m a = 1200 kg × 9.0 m s⁻² = 1.08 × 10⁴ N。对于水平环岛上的汽车，轮胎与路面间的静摩擦力提供该力，方向指向圆心。
a = v² / r = 9.0 m s⁻² F = m v² / r = 1.08 × 10⁴ N
A common error is forgetting that the centripetal force is not an extra ‘new’ force but the resultant of real forces, here friction. Also check that speed is constant for uniform circular motion; any change in speed would imply a tangential acceleration as well.
常见错误是忘记向心力并非额外的“新”力，而是真实力（此处为摩擦力）的合力。同时注意匀速圆周运动中速率恒定；若速率变化，还会存在切向加速度。
2. Simple Harmonic Motion (SHM) | 简谐运动
A mass‑spring system oscillates with an amplitude of 0.050 m and a frequency of 2.0 Hz. Determine the maximum speed, the maximum acceleration, and write an expression for the displacement x as a function of time t, assuming zero phase constant.
一个弹簧–质量系统以振幅 0.050 m、频率 2.0 Hz 振动。求最大速度、最大加速度，并写出位移 x 随时间 t 变化的表达式，设初相为零。
First calculate the angular frequency: ω = 2πf = 2π × 2.0 Hz = 4π rad s⁻¹ (≈ 12.57 rad s⁻¹). The maximum speed in SHM is vₘₐₓ = ωA = 4π × 0.050 m = 0.628 m s⁻¹. The maximum acceleration is given by aₘₐₓ = ω²A = (4π)² × 0.050 = 16π² × 0.050 ≈ 7.90 m s⁻². With zero phase constant, the displacement can be written as x = A cos(ωt) = 0.050 cos(4π t) m, or equivalently x = 0.050 sin(4π t + π/2) m.
首先计算角频率：ω = 2πf = 2π × 2.0 Hz = 4π rad s⁻¹（约 12.57 rad s⁻¹）。SHM 中的最大速度 vₘₐₓ = ωA = 4π × 0.050 m = 0.628 m s⁻¹。最大加速度 aₘₐₓ = ω²A = (4π)² × 0.050 = 16π² × 0.050 ≈ 7.90 m s⁻²。设初相为零，位移可写为 x = A cos(ωt) = 0.050 cos(4π t) m，或等效为 x = 0.050 sin(4π t + π/2) m。
vₘₐₓ = ωA = 0.628 m s⁻¹ aₘₐₓ = ω²A = 7.90 m s⁻²
Be careful with the relationship between frequency and angular frequency; a common mistake is to use f directly instead of 2πf. Also remember that the maximum acceleration occurs at the points of maximum displacement, where the velocity is zero.
注意频率与角频率的关系；常见错误是直接用 f 代替 2πf。同时记住最大加速度发生在最大位移处，此时速度为零。
3. Gravitational Fields & Satellite Orbits | 引力场与卫星轨道
A satellite orbits the Earth at an altitude of 400 km. Given: Earth’s mass M = 5.97 × 10²⁴ kg, radius R = 6370 km, G = 6.67 × 10⁻¹¹ N m² kg⁻². Find the orbital speed and the period of the satellite. Take the orbital radius r = R + h = 6770 km.
一颗卫星在距地面 400 km 的高度上绕地球运行。已知：地球质量 M = 5.97 × 10²⁴ kg，半径 R = 6370 km，G = 6.67 × 10⁻¹¹ N m² kg⁻²。求卫星的轨道速率和周期。取轨道半径 r = R + h = 6770 km。
First convert the orbital radius to metres: r = 6770 km = 6.77 × 10⁶ m. For a circular orbit, centripetal force is provided by gravity: G M m / r² = m v² / r, giving v = √(G M / r). Substitute the values: v = √[(6.67 × 10⁻¹¹ × 5.97 × 10²⁴) / (6.77 × 10⁶)] = √(3.98 × 10¹⁴ / 6.77 × 10⁶) = √(5.88 × 10⁷) ≈ 7.67 × 10³ m s⁻¹. The period T = 2π r / v = (2π × 6.77 × 10⁶) / (7.67 × 10³) ≈ 5.55 × 10³ s (about 92.5 minutes).
首先将轨道半径转换为米：r = 6770 km = 6.77 × 10⁶ m。对圆周轨道，向心力由引力提供：G M m / r² = m v² / r，得 v = √(G M / r)。代入数值：v = √[(6.67 × 10⁻¹¹ × 5.97 × 10²⁴) / (6.77 × 10⁶)] = √(3.98 × 10¹⁴ / 6.77 × 10⁶) = √(5.88 × 10⁷) ≈ 7.67 × 10³ m s⁻¹。周期 T = 2π r / v = (2π × 6.77 × 10⁶) / (7.67 × 10³) ≈ 5.55 × 10³ s（约 92.5 分钟）。
v = √(G M / r) = 7.67 × 10³ m s⁻¹ T ≈ 5.55 × 10³ s
A frequent error is using the wrong radius – remember r is measured from the centre of the Earth, not the surface. Also keep all quantities in SI units to avoid magnitude mistakes.
常见错误是使用错误的半径——记住 r 从地心测量，而非地表。同时统一使用国际单位，避免数量级错误。
4. Electric Fields & Superposition | 电场与叠加
Two point charges, +2.0 μC and –4.0 μC, are placed 0.30 m apart in a vacuum. Determine the electric field strength (magnitude and direction) at the midpoint between them. Use k = 1/(4πε₀) = 8.99 × 10⁹ N m² C⁻².
两点电荷 +2.0 μC 和 –4.0 μC 在真空中相距 0.30 m。求它们连线中点处的电场强度（大小和方向）。使用 k = 1/(4πε₀) = 8.99 × 10⁹ N m² C⁻²。
The midpoint is 0.15 m from each charge. The field due to the positive charge points away from it, and the field due to the negative charge points towards it. At the midpoint, both fields point in the same direction – from the positive towards the negative charge. Compute each field: E₁ = k Q₁ / r² = 8.99×10⁹ × 2.0×10⁻⁶ / (0.15)² ≈ 7.99×10⁵ N C⁻¹. E₂ = k |Q₂| / r² = 8.99×10⁹ × 4.0×10⁻⁶ / (0.15)² ≈ 1.60×10⁶ N C⁻¹. The resultant field is E_total = E₁ + E₂ ≈ 2.40×10⁶ N C⁻¹, directed towards the negative charge.
中点距每个电荷 0.15 m。正电荷的场背离正电荷，负电荷的场指向负电荷。在中点，两个场方向相同——从正电荷指向负电荷。分别计算：E₁ = k Q₁ / r² = 8.99×10⁹ × 2.0×10⁻⁶ / (0.15)² ≈ 7.99×10⁵ N C⁻¹。E₂ = k |Q₂| / r² = 8.99×10⁹ × 4.0×10⁻⁶ / (0.15)² ≈ 1.60×10⁶ N C⁻¹。合场强 E_total = E₁ + E₂ ≈ 2.40×10⁶ N C⁻¹，方向指向负电荷。
E = k |Q| / r² E_total = 2.40 × 10⁶ N C⁻¹ (towards –4.0 μC)
Students often confuse the sign convention for field direction: a test positive charge would be repelled by the +2.0 μC and attracted by the –4.0 μC, so the net force is indeed towards the negative charge. Vector addition must account for direction, not just magnitudes.
学生常混淆电场方向的正负号约定：正的试探电荷会被 +2.0 μC 排斥，被 –4.0 μC 吸引，因此合力的确指向负电荷。矢量叠加必须考虑方向，而非仅大小。
5. Capacitor Charging & Time Constant | 电容器充电与时间常数
A 470 μF capacitor is connected in series with a 22 kΩ resistor and a 9.0 V battery. Calculate the time constant τ, the initial charging current, and the time taken for the capacitor voltage to reach 6.0 V. Assume V = V₀ (1 – e^{-t/τ}).
一只 470 μF 的电容器与 22 kΩ 的电阻和 9.0 V 的电池串联。计算时间常数 τ、初始充电电流，以及电容器两端电压达到 6.0 V 所需的时间。设 V = V₀ (1 – e^{-t/τ})。
The time constant is τ = R C = 22 × 10³ Ω × 470 × 10⁻⁶ F = 10.34 s. At t = 0 the capacitor is uncharged, so the initial current is I₀ = V₀ / R = 9.0 V / 22000 Ω = 4.09 × 10⁻⁴ A = 0.409 mA. To find when V = 6.0 V, set 6.0 = 9.0 (1 – e^{-t/τ}). Rearranging: 1 – e^{-t/τ} = 2/3, so e^{-t/τ} = 1/3. Taking natural logs: –t/τ = ln(1/3) = –ln 3, hence t = τ ln 3 ≈ 10.34 × 1.099 = 11.4 s.
时间常数 τ = R C = 22 × 10³ Ω × 470 × 10⁻⁶ F = 10.34 s。在 t = 0 时电容器未充电，初始电流 I₀ = V₀ / R = 9.0 V / 22000 Ω = 4.09 × 10⁻⁴ A = 0.409 mA。求 V = 6.0 V 的时间：6.0 = 9.0 (1 – e^{-t/τ})。整理得 1 – e^{-t/τ} = 2/3，即 e^{-t/τ} = 1/3。取自然对数：–t/τ = ln(1/3) = –ln 3，因此 t = τ ln 3 ≈ 10.34 × 1.099 = 11.4 s。
τ = R C = 10.3 s I₀ = 0.409 mA t ≈ 11.4 s
A typical mistake is confusing the charging equation V = V₀(1 – e^{-t/τ}) with the discharging equation. Also ensure that when using logs, you keep the sign consistent. The time constant gives the time to reach about 63% of full charge, which is a useful check.
典型错误是混淆充电公式 V = V₀(1 – e^{-t/τ}) 与放电公式。同时注意使用对数时保持符号一致。时间常数对应达到满充电约 63% 的时间，可作为快速检验。
6. Charged Particle in a Magnetic Field | 带电粒子在磁场中的运动
An electron (mass m = 9.11 × 10⁻³¹ kg, charge magnitude e = 1.60 × 10⁻¹⁹ C) enters a uniform magnetic field of 0.50 T at a speed of 3.0 × 10⁶ m s⁻¹, perpendicular to the field. Find the radius of the resulting circular path and the period of one complete revolution.
电子（质量 m = 9.11 × 10⁻³¹ kg，电荷量 e = 1.60 × 10⁻¹⁹ C）以 3.0 × 10⁶ m s⁻¹ 的速度垂直进入 0.50 T 的匀强磁场。求其圆周轨迹的半径和旋转周期。
The magnetic force provides the centripetal force: B e v = m v² / r. Solving for radius: r = m v / (B e). Substitute: r = (9.11 × 10⁻³¹ × 3.0 × 10⁶) / (0.50 × 1.60 × 10⁻¹⁹) = (2.733 × 10⁻²⁴) / (8.0 × 10⁻²⁰) = 3.416 × 10⁻⁵ m, but let’s recalculate correctly: 9.11e-31 * 3e6 = 2.733e-24, divided by (0.5*1.6e-19=8e-20) gives about 3.42e-5 m, which seems too small. Let me re-evaluate: 2.733e-24 / 8e-20 = 3.416e-5 m = 0.0342 mm. However, typical electron radius in 0.5 T is indeed small. Let’s verify: mv = 9.11e-31 * 3e6 = 2.733e-24. Be = 0.5
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