📚 Year 12 AQA Physics: Unit Test Mock Paper Walkthrough | Year 12 AQA 物理：单元测试模拟卷解析
This article provides a detailed, step-by-step walkthrough of a mock unit test for Year 12 AQA Physics. It covers key topics from mechanics, materials, waves, electricity, and quantum phenomena. Each question is broken down with clear reasoning, relevant equations, and numerical solutions, helping you consolidate your understanding and exam technique.
本文为 Year 12 AQA 物理单元测试模拟卷提供逐步解析，涵盖力学、材料、波、电学和量子现象等核心内容。每道题目均配有清晰的思路、相关方程和数值解答，帮助巩固理解与应试技巧。
1. Uniformly Accelerated Motion (SUVAT) | 匀加速直线运动 (SUVAT)
A car accelerates uniformly from rest at 2.5 m/s² for 8.0 s. Determine its final velocity and the distance travelled during this time.
一辆汽车从静止以 2.5 m/s² 匀加速运动 8.0 s。求末速度和这段时间内的位移。
Use v = u + at with u = 0, a = 2.5 m/s², t = 8.0 s.
使用 v = u + at，其中 u = 0, a = 2.5 m/s², t = 8.0 s。
v = 0 + 2.5 × 8.0 = 20 m/s
For distance s, use s = ut + ½ at².
求位移 s，使用 s = ut + ½ at²。
s = 0 + ½ × 2.5 × (8.0)² = 1.25 × 64 = 80 m
Hence the final velocity is 20 m/s and the distance covered is 80 m.
因此，末速度为 20 m/s，行驶距离为 80 m。
2. Projectile Motion | 抛体运动
A ball is projected horizontally from a cliff 20 m high with an initial speed of 15 m/s. Take g = 9.81 m/s². Find the time of flight and the horizontal range.
一球从 20 m 高的悬崖以 15 m/s 的初速度水平抛出。取 g = 9.81 m/s²。求飞行时间和水平射程。
Vertical motion: h = ½ g t², solve for t.
竖直方向：h = ½ g t²，解出 t。
t = √(2h/g) = √(2 × 20 / 9.81) = √4.077 ≈ 2.02 s
Horizontal range R = vₓ × t = 15 × 2.02 ≈ 30.3 m.
水平射程 R = vₓ × t = 15 × 2.02 ≈ 30.3 m。
The time of flight is 2.02 s and the horizontal range is 30.3 m.
飞行时间为 2.02 s，水平射程为 30.3 m。
3. Newton’s Second Law with Friction | 牛顿第二定律与摩擦力
A block of mass 5.0 kg is pulled along a rough horizontal surface by a force of 30 N applied at 30° above the horizontal. The coefficient of kinetic friction μₖ = 0.20. Calculate the acceleration of the block.
一个 5.0 kg 的木块在粗糙水平面上受到 30 N 的拉力，拉力方向与水平成 30° 角。动摩擦因数 μₖ = 0.20。求木块的加速度。
Resolve the pulling force: horizontal component Fₓ = 30 cos30° ≈ 26.0 N, vertical component Fᵧ = 30 sin30° = 15 N.
分解拉力：水平分量 Fₓ = 30 cos30° ≈ 26.0 N，竖直分量 Fᵧ = 30 sin30° = 15 N。
Normal reaction N = mg – Fᵧ = 5.0×9.81 – 15 ≈ 49.05 – 15 = 34.05 N.
法向反力 N = mg – Fᵧ = 5.0×9.81 – 15 ≈ 49.05 – 15 = 34.05 N。
Frictional force f = μₖ N = 0.20 × 34.05 ≈ 6.81 N.
摩擦力 f = μₖ N = 0.20 × 34.05 ≈ 6.81 N。
Net horizontal force F_net = Fₓ – f = 26.0 – 6.81 = 19.19 N.
净水平力 F_net = Fₓ – f = 26.0 – 6.81 = 19.19 N。
Acceleration a = F_net / m = 19.19 / 5.0 ≈ 3.84 m/s².
加速度 a = F_net / m = 19.19 / 5.0 ≈ 3.84 m/s²。
The block accelerates at about 3.8 m/s².
木块的加速度约为 3.8 m/s²。
4. Work, Energy and Power | 功、能量和功率
A crane lifts a 200 kg load vertically at a constant speed of 0.50 m/s. Calculate the work done in 10 s and the power output of the crane. Take g = 9.81 m/s².
一台起重机以 0.50 m/s 的恒定速度竖直提升 200 kg 的重物。计算 10 s 内做的功和起重机的输出功率。取 g = 9.81 m/s²。
The lifting force equals the weight: F = mg = 200 × 9.81 = 1962 N.
提升力等于重力：F = mg = 200 × 9.81 = 1962 N。
Distance moved in 10 s: s = v × t = 0.50 × 10 = 5.0 m.
10 s 内移动的距离：s = v × t = 0.50 × 10 = 5.0 m。
Work done W = F × s = 1962 × 5.0 = 9810 J (≈ 9.8 kJ).
做的功 W = F × s = 1962 × 5.0 = 9810 J (≈ 9.8 kJ)。
Power P = W / t = 9810 / 10 = 981 W, or P = F × v = 1962 × 0.50 = 981 W.
功率 P = W / t = 9810 / 10 = 981 W，或 P = F × v = 1962 × 0.50 = 981 W。
Thus, 9810 J of work is done and the power output is 981 W.
因此，做功 9810 J，输出功率为 981 W。
5. Stress, Strain and Young Modulus | 应力、应变与杨氏模量
A metal wire of length 2.00 m and cross-sectional area 1.5 × 10⁻⁶ m² extends by 1.2 mm under a load of 50 N. Calculate the stress, strain and Young modulus of the material.
一根长 2.00 m、横截面积 1.5 × 10⁻⁶ m² 的金属丝在 50 N 荷载下伸长 1.2 mm。求应力、应变和材料的杨氏模量。
Stress σ = Force / Area = 50 / (1.5×10⁻⁶) = 3.33 × 10⁷ Pa.
应力 σ = 力 / 面积 = 50 / (1.5×10⁻⁶) = 3.33 × 10⁷ Pa。
Strain ε = extension / original length = (1.2×10⁻³ m) / 2.00 = 6.0 × 10⁻⁴ (no units).
应变 ε = 伸长量 / 原长 = (1.2×10⁻³ m) / 2.00 = 6.0 × 10⁻⁴（无单位）。
Young modulus E = σ / ε = (3.33×10⁷) / (6.0×10⁻⁴) = 5.56 × 10¹⁰ Pa.
杨氏模量 E = σ / ε = (3.33×10⁷) / (6.0×10⁻⁴) = 5.56 × 10¹⁰ Pa。
The material’s Young modulus is about 5.6 × 10¹⁰ Pa, typical for metals like copper.
该材料的杨氏模量约为 5.6 × 10¹⁰ Pa，与铜等金属的典型值相符。
6. Wave Properties | 波的性质
A progressive wave of amplitude 2.0 mm travels with a frequency of 250 Hz and speed 340 m/s. Determine the wavelength and write an equation of the form y = A sin(ωt – kx).
一列振幅为 2.0 mm 的行波以 250 Hz 的频率和 340 m/s 的速度传播。求波长，并写出形式为 y = A sin(ωt – kx) 的波动方程。
Wavelength λ = v / f = 340 / 250 = 1.36 m.
波长 λ = v / f = 340 / 250 = 1.36 m。
Angular frequency ω = 2πf = 2π × 250 ≈ 1571 rad/s.
角频率 ω = 2πf = 2π × 250 ≈ 1571 rad/s。
Wave number k = 2π / λ = 2π / 1.36 ≈ 4.62 rad/m.
波数 k = 2π / λ = 2π / 1.36 ≈ 4.62 rad/m。
Amplitude A = 2.0 × 10⁻³ m. Hence the wave equation is:
振幅 A = 2.0 × 10⁻³ m。因此波动方程为：
y = 2.0×10⁻³ sin(1571 t – 4.62 x)
with y and x in metres, t in seconds.
其中 y 和 x 以米为单位，t 以秒为单位。
7. Stationary Waves on a String | 弦上的驻波
A string of length 0.80 m has a mass of 4.0 g. It is stretched under a tension of 100 N. Find the fundamental frequency of vibration.
一根长 0.80 m、质量为 4.0 g 的弦在 100 N 张力下拉伸。求振动的基频。
Linear density μ = mass / length = (4.0×10⁻³ kg) / 0.80 m = 5.0 × 10⁻³ kg/m.
线密度 μ = 质量 / 长度 = (4.0×10⁻³ kg) / 0.80 m = 5.0 × 10⁻³ kg/m。
Fundamental frequency f₁ = (1 / 2L) √(T / μ).
基频 f₁ = (1 / 2L) √(T / μ)。
f₁ = 1/(2×0.80) × √(100 / 5.0×10⁻³) = 1/1.60 × √(2.0×10⁴)
f₁ = 0.625 × 141.4 ≈ 88.4 Hz
The fundamental frequency is approximately 88 Hz.
基频约为 88 Hz。
8. DC Circuit Analysis | 直流电路分析
Three resistors R₁ = 4 Ω, R₂ = 6 Ω are connected in series, and this combination is connected in parallel with R₃ = 12 Ω. The network is supplied by a 12 V battery with negligible internal resistance. Determine the total resistance and the current drawn from the battery.
三个电阻 R₁ = 4 Ω、R₂ = 6 Ω 串联，然后与 R₃ = 12 Ω 并联。该网络由内阻可忽略的 12 V 电池供电。求总电阻和电池输出的电流。
Series combination R_series = 4 + 6 = 10 Ω.
串联组合 R_series = 4 + 6 = 10 Ω。
Parallel with 12 Ω: 1/R_total = 1/10 + 1/12 = (6 + 5) / 60 = 11/60.
与 12 Ω 并联：1/R_total = 1/10 + 1/12 = (6 + 5) / 60 = 11/60。
R_total = 60/11 ≈ 5.45 Ω
Current I = V / R_total = 12 / 5.45 ≈ 2.20 A.
电流 I = V / R_total = 12 / 5.45 ≈ 2.20 A。
The circuit draws about 2.2 A from the battery.
该电路从电池获取约 2.2 A 电流。
9. Resistivity | 电阻率
A metal wire is 15 m long and has a diameter of 0.50 mm. Its resistance is measured as 2.0 Ω. Calculate the resistivity of the metal.
一根金属丝长 15 m，直径为 0.50 mm，测得电阻为 2.0 Ω。求该金属的电阻率。
Cross-sectional area A = π (d/2)². Diameter d = 0.50 mm = 5.0×10⁻⁴ m, radius = 2.5×10⁻⁴ m.
横截面积 A = π (d/2)²。直径 d = 0.50 mm = 5.0×10⁻⁴ m，半径 = 2.5×10⁻⁴ m。
A = π × (2.5×10⁻⁴)² = π × 6.25×10⁻⁸ ≈ 1.96×10⁻⁷ m²
Using R = ρ L / A → ρ = R A / L.
根据 R = ρ L / A → ρ = R A / L。
ρ = 2.0 × 1.96×10⁻⁷ / 15 ≈ 2.61×10⁻⁸ Ω m
The resistivity is about 2.6 × 10⁻⁸ Ω m, typical for a metal such as aluminium.
电阻率约为 2.6 × 10⁻⁸ Ω m，与铝等金属的典型值相符。
10. Photoelectric Effect | 光电效应
Ultraviolet light of frequency 1.20 × 10¹⁵ Hz is incident on a metal surface with a work function of 2.30 eV. Calculate the maximum kinetic energy of the emitted photoelectrons in joules and the stopping potential. (h =
Published by TutorHao | Year 12 Physics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply