Pre-U CIE Physics Unit Test Mock Paper Analysis | Pre-U CIE 物理：单元测试模拟卷解析

📚 Pre-U CIE Physics Unit Test Mock Paper Analysis | Pre-U CIE 物理：单元测试模拟卷解析

This mock paper analysis is designed to help Pre-U CIE Physics students consolidate their understanding of core topics through a structured walkthrough of a representative unit test. Each section presents a question followed by a detailed solution, highlighting key concepts, common pitfalls, and the reasoning expected in high-scoring answers. The questions span mechanics, waves, electricity, particle physics, and quantum phenomena, mirroring the breadth and depth of the CIE Pre-U syllabus.

这份模拟卷解析旨在帮助 Pre-U CIE 物理学生通过系统的讲解来巩固对核心主题的理解。每个部分先给出一道典型试题，然后提供详细的解答，突出关键概念、常见错误以及高分答案应有的推理过程。题目覆盖力学、波动、电学、粒子物理和量子现象，反映了 CIE Pre-U 教学大纲的广度和深度。

1. Kinematics and Projectile Motion | 运动学与抛体运动

A stone is projected from a cliff top 45.0 m above the sea with a speed of 24.0 m s⁻¹ at an angle of 35° above the horizontal. Calculate the total time of flight before the stone hits the water, ignoring air resistance.

一块石头从高出海面45.0米的悬崖顶以24.0米/秒的初速度、与水平方向成35°角向上抛出。忽略空气阻力，计算石头落到水面之前的总飞行时间。

We resolve the initial velocity vertically: u_y s positive, the vertical displacement is +45.0 m, acceleration is +9.81 m s⁻². Using s = u_y t + ½ a t², but careful with sign convention: if upward is positive, then initial u_y t − 4.905 t². Rearranging: 4.905 t² − 13.77 t − 45.0 = 0. Solving quadratic: t = [13.77 ± √13.77² + 4×4.905×45.0] / (2×4.905). Discriminant √189.6 + 882.9 = √1072.5 ≈ 32.75. Positive root: t = (13.77 + 32.75) / 9.81 ≈ 4.74 s.

我们将初速度沿竖直方向分解：u_y = 24.0 sin 35° ≈ 13.77 m s⁻¹。取向下为正方向，则竖直位移为 +45.0 m，加速度为 +9.81 m s⁻²。利用 s = u_y t + ½ a t²，但要注意符号约定：若取向上为正，则 u_y = +13.77 m s⁻¹，位移 s = −45.0 m，加速度 a = −9.81 m s⁻²。于是 −45.0 = 13.77 t − 4.905 t²，整理得 4.905 t² − 13.77 t − 45.0 = 0。解二次方程：t = [13.77 ± √13.77² + 4×4.905×45.0] / (2×4.905)，判别式 √189.6 + 882.9 = √1072.5 ≈ 32.75，取正根得 t = (13.77 + 32.75) / 9.81 ≈ 4.74 s。

---

2. Newton’s Laws and Friction | 牛顿定律与摩擦力

A block of mass 5.0 kg rests on a rough slope inclined at 25° to the horizontal. The coefficient of static friction is 0.62. Determine whether the block remains at rest when released.

一个质量为5.0千克的物块静止在倾角为25°的粗糙斜面上。静摩擦系数为0.62。判断释放后物块是否会保持静止。

.81 × cos 25° ≈ 0.62 × 49.05 × 0.9063 ≈ 27.6 N. Since maximum friction (27.6 N) exceeds the downhill component (20.7 N), static friction is sufficient to hold the block. The block remains at rest.

沿斜面向下的重力分力为 mg sinθ = 5.0 × 9.81 × sin 25° ≈ 20.7 N。最大静摩擦力 = μ_s × 法向反作用力 = μ_s mg cosθ = 0.62 × 5.0 × 9.81 × cos 25° ≈ 0.62 × 49.05 × 0.9063 ≈ 27.6 N。由于最大静摩擦力（27.6 N）大于下滑分力（20.7 N），静摩擦力足以使物块保持静止。物块将保持静止。

---

3. Work, Energy and Power | 功、能量与功率

A car of mass 1200 kg accelerates uniformly from rest to 30 m s⁻¹ in 8.0 s on a horizontal road. Assuming constant resistive forces of 600 N, calculate the average power developed by the engine during the acceleration.

一辆质量为1200千克的汽车在水平路面上由静止均匀加速，8.0秒内达到30米/秒。设阻力恒为600牛，计算加速过程中发动机输出的平均功率。

Acceleration a = (v−u)/t = 30/8.0 = 3.75 m s⁻². Net force from engine thrust F_net = ma = 1200 × 3.75 = 4500 N. Engine thrust must overcome resistance, so total thrust F = F_net e power = F × average velocity = 5100 × 15 = 76 500 W ≈ 76.5 kW. Alternatively, using work– tance = average velocity × time = 15 × 8 = 120 m. So work against resistance = 600 × 120 = 72 000 J. Total work = 612 000 J. Average power = 612 000 / 8.0 = 76 500 W.

加速度 a = (v−u)/t = 30/8.0 = 3.75 m s⁻²。发动机提供的净力 F_net = ma = 1200 × 3.75 = 4500 N。发动机推力需克服阻力，因此总推力 F = F_net + 阻力 = 4500 + 600 = 5100 N。加速过程中的平均速度 = (0+30)/2 = 15 m s⁻¹。平均功率 = F × 平均速度 = 5100 × 15 = 76 500 W ≈ 76.5 kW。也可用功能关系：发动机做功 = 动能增量 + 克服阻力做功。动能增量 = ½ × 1200 × 30² = 540 000 J。克服阻力做功 = 600 N × 位移，位移 = 平均速度 × 时间 = 15 × 8 = 120 m，因此阻力做功 = 600 × 120 = 72 000 J。总功 = 612 000 J，平均功率 = 612 000 / 8.0 = 76 500 W。

---

4. Momentum and Impulse | 动量与冲量

A tennis ball of mass 0.058 kg strikes a wall perpendicularly at 42 m s⁻¹ and rebounds at 38 m s⁻¹. The contact time is 0.026 s. Calculate the average force exerted by the wall on the ball.

一个质量为0.058千克的网球以42米/秒的速度垂直撞击墙壁，并以38米/秒的速度反弹。接触时间为0.026秒。计算墙壁对网球的平均作用力。

s⁻¹. Change in momentum Δp = final − initial = −2.204 − 2.436 = −4.640 kg m s⁻¹. Impulse F_avg Δt = Δp, so F_avg s force is opposite to initial direction. Magnitude of average force is about 1.8 × 10² N.

取初速度方向为正。初动量 = +0.058 × 42 = +2.436 kg m s⁻¹。末动量 = −0.058 × 38 = −2.204 kg m s⁻¹。动量变化 Δp = 末动量 − 初动量 = −2.204 − 2.436 = −4.640 kg m s⁻¹。冲量 F_avg Δt = Δp，故 F_avg = Δp / Δt = −4.640 / 0.026 ≈ −178.5 N。负号表示力与初速度方向相反。平均作用力的大小约为 1.8 × 10² N。

---

5. Circular Motion and Gravitation | 圆周运动与万有引力

97 × 10²⁴ kg and G = 6.67 × 10⁻¹¹ N m² kg⁻², find the satellite’s orbital period in hours.

一颗卫星在半径为4.23 × 10⁷米的圆形轨道上绕地球飞行。已知地球质量 M = 5.97 × 10²⁴ 千克，万有引力常量 G = 6.67 × 10⁻¹¹ N m² kg⁻²，求卫星的轨道周期（以小时为单位）。

gravity: centripetal force mv²/r = GMm/r², cancelling m gives v² = GM/r. Orbital speed v = √GM/r = √[(6.67×10⁻¹¹ × 5.97×10²⁴) / (4.23×10⁷)] = √3.98×10¹⁴ / 4.23×10⁷ = √9.41×10⁶ 640 / 3600 ≈ 24.07 h. This matches the geostationary orbit period, confirming the radius.

对于仅受引力作用的圆周运动：向心力 mv²/r = GMm/r²，消去 m 得 v² = GM/r。轨道速率 v = √GM/r = √[(6.67×10⁻¹¹ × 5.97×10²⁴) / (4.23×10⁷)] = √3.98×10¹⁴ / 4.23×10⁷ = √9.41×10⁶ ≈ 3.068×10³ m s⁻¹。周期 T = 2πr / v = 2 × 3.1416 × 4.23×10⁷ / 3068 ≈ 2.658×10⁸ / 3068 ≈ 86 640 s。换算为小时：86 640 / 3600 ≈ 24.07 小时。这与地球同步轨道周期一致，验证了轨道半径。

---

6. Simple Harmonic Motion | 简谐运动

A mass-spring system oscillates with amplitude 0.12 m and period 0.80 s. Determine the maximum acceleration and the displacement when the speed is half its maximum value.

一弹簧振子以振幅0.12米、周期0.80秒做简谐运动。求最大加速度及速度为最大值一半时的位移。

Angular frequency ω = 2π/T = 2 × 3.1416 / 0.80 ≈ 7.854 rad s⁻¹. Maximum acceleration a_max = ω² A = (7.854)² × 0.12 ≈ 61.7 × 0.12 = 7.40 m s⁻². Maximum speed v_max = ωA = 7.854 × 0.12 = 0.9425 m s⁻¹. When v = v_max/2 = 0.4712 m s⁻¹. Using v = ω √A² − x² 20 / 61.69 ≈ 0.00360. Hence x² = 0.0144 − 0.00360 = 0.01080, so x ≈ 0.104 m (or −0.104 m).

角频率 ω = 2π/T = 2 × 3.1416 / 0.80 ≈ 7.854 rad s⁻¹。最大加速度 a_max = ω² A = (7.854)² × 0.12 ≈ 61.7 × 0.12 = 7.40 m s⁻²。最大速率 v_max = ωA = 7.854 × 0.12 = 0.9425 m s⁻¹。当 v = v_max/2 = 0.4712 m s⁻¹ 时，利用 v = ω √A² − x²，得 (0.4712)² = (7.854)² (0.12² − x²)，即 0.2220 = 61.69 (0.0144 − x²)。于是 0.0144 − x² = 0.2220 / 61.69 ≈ 0.00360，故 x² = 0.01080，x ≈ 0.104 米（或 −0.104 米）。

---

广告 / Sponsored

7. Wave Phenomena and Superposition | 波动现象与叠加

Two coherent sound sources 0.85 m apart emit waves of frequency 680 Hz in phase. The speed of sound is 340 m s⁻¹. An observer walks along a line parallel to the line joining the sources, at a distance of 6.0 m. Calculate the distance between adjacent maxima (fringe spacing) observed along this line.

两个相干声源相距0.85米，同相发射频率为680赫兹的声波。声速为340米/秒。观察者沿着一条与两源连线平行且相距6.0米的直线行走。计算沿此线相邻极大值之间的距离（条纹间距）。

om sources to observation line (6.0 m), d is separation of sources (0.85 m). So Δy = 0.50 × 6.0 / 0.85 ≈ 3.0 / 0.85 ≈ 3.53 m. However, this formula is exact only for small angles; here D ≫ d, so approximation holds.

波长 λ = v/f = 340/680 = 0.50 米。这类似于双缝干涉。条纹间距 Δy = λD / d，其中 D 为声源到观测线的垂直距离（6.0 米），d 为声源间距（0.85 米）。因此 Δy = 0.50 × 6.0 / 0.85 ≈ 3.0 / 0.85 ≈ 3.53 米。注意此公式仅在小角度近似下严格成立，此处 D ≫ d 满足近似条件。

---

8. Electric Fields and Potential | 电场与电势

Two point charges, +8.0 nC and −4.0 nC, are placed 0.30 m apart in a vacuum. Find the electric potential at the midpoint between them and the electric field strength at that point.

两个点电荷 +8.0 nC 和 −4.0 nC 在真空中相距0.30米。求两者连线中点处的电势和该点的电场强度。

Midpoint distance from each charge r = 0.15 m. Potential V = k Σ(q_i / r_i c field: E is vector sum. Due to +8.0 nC, field directed away: E₁ = kq/r² = 8.99×10⁹ × 8.0×10⁻⁹ / 0.15² = 71.92 / 0.0225 ≈ 3196 N C⁻¹ to the right (assuming + charge on left). Due to −4.0 nC, field directed towards it: E₂ = 8.99×10⁹ × 4.0×10⁻⁹ / 0.15² = 35.96 / 0.0225 ≈ 1598 N C⁻¹ also to the right. Both point in same direction, so total E = 3196 + 1598 ≈ 4794 N C⁻¹, directed towards the negative charge.

中点到每个电荷的距离 r = 0.15 m。电势 V = k Σ(q_i / r_i)，k = 8.99×10⁹ N m² C⁻²。 )/0.15 + (−4.0×10⁻⁹)/0.15 ] = 8.99×10⁹ × (4.0×10⁻⁹ / 0.15) = 8.99×10⁹ × 2.667×10⁻⁸ ≈ 240 V。电场强度：E 为矢量叠加。+8.0 nC 产生背离电荷的电场：E₁ = kq/r² = 8.99×10⁹ × 8.0×10⁻⁹ / 0.15² ≈ 3196 N C⁻¹ 方向向右（设正电荷在左）。−4.0 nC 产生指向电荷的电场：E₂ = 8.99×10⁹ × 4.0×10⁻⁹ / 0.15² ≈ 1598 N C⁻¹ 也向右。两者同向，合场强 E = 3196 + 1598 ≈ 4794 N C⁻¹，方向指向负电荷。

---

9. Capacitance and Energy Storage | 电容与储能

A 470 μF capacitor is charged to a p.d. of 12 V. It is then disconnected and connected in parallel to an uncharged 220 μF capacitor. Calculate the final energy stored in the system and compare it with the initial energy.

一个470微法的电容器充电至12伏电势差。然后断开电源，与一个未充电的220微法电容器并联。计算系统中最终储存的能量，并与初始能量比较。

total charge Q = C₁ V = 470×10⁻⁶ × 12 = 5.64×10⁻³ C is conserved. Equivalent capacitance C_total = 470 + 220 = 690 μF. Final p.d. V_f = Q / C_total = 5.64×10⁻³ / 690×10⁻⁶ = 8.174 V. Final energy E_final = ½ C_total V_f gy lost = 0.03384 − 0.02305 = 0.01079 J, dissipated as heat/radiation in connecting wires.

初始能量 E_initial = ½ C₁ V² = 0.5 × 470×10⁻⁶ × 12² = 0.03384 J。并联后，总电荷 Q = C₁ V = 470×10⁻⁶ × 12 = 5.64×10⁻³ C 守恒。等效电容 C_total = 470 + 220 = 690 μF。最终电压 V_f = Q / C_total = 5.64×10⁻³ / 690×10⁻⁶ ≈ 8.174 V。最终能量 E_final = ½ C_total V_f² = 0.5 × 690×10⁻⁶ × (8.174)² ≈ 0.02305 J。能量损失为 0.01079 J，以热量或电磁辐射形式耗散在连接导线中。

---

10. Photoelectric Effect and Photons | 光电效应与光子

Ultraviolet light of wavelength 220 nm is incident on a metal surface with work function 4.5 eV. Determine the maximum kinetic energy of emitted electrons in joules and their stopping potential.

波长为220纳米的紫外光照射在功函数为4.5电子伏特的金属表面。求发射电子的最大动能（以焦耳为单位）及其遏止电势。

hc = 1240 eV·nm (useful constant). E = 1240 / 220 ≈ 5.636 eV. Work function Φ = 4.5 eV. K_max .636 − 4.5 = 1.136 eV. In joules: 1.136 × 1.602×10⁻¹⁹ ≈ 1.82×10⁻¹⁹ J. Stopping potential V_s = K_max / e = 1.136 V (since kinetic energy in eV equals stopping potential in volts).

光子能量 E = hc/λ，利用常数 hc = 1240 eV·nm，得 E = 1240 / 220 ≈ 5.636 eV。功函数 Φ = 4.5 eV，最大动能 K_max = E − Φ = 5.636 − 4.5 = 1.136 eV。换算为焦耳：1.136 × 1.602×10⁻¹⁹ ≈ 1.82×10⁻¹⁹ J。遏止电势 V_s = K_max / e = 1.136 V（以 eV 为单位的动能数值等于遏止电压的伏特数）。

---

11. Nuclear Physics and Mass–Energy Equivalence | 核物理与质能等价

nergy. Calculate the total mass defect in atomic mass units (u). Given 1 u = 931.5 MeV/c².

在一个核反应中，一个铀-235核俘获一个中子并发生裂变，释放出 3.2 × 10⁻¹¹ 焦的能量。计算总的质量亏损，以原子质量单位（u）表示。已知 1 u = 931.5 MeV/c²。

= energy (MeV) / 931.5 = 199.75 / 931.5 ≈ 0.2145 u. Alternatively, directly Δm = E/c², but using the stated conversion factor is straightforward.

首先把能量转换为 MeV：3.2 × 10⁻¹¹ J ÷ 1.602×10⁻¹³ J/MeV = 199.75 MeV。质量亏损 Δm = ΔE / c²。利用原子质量单位换算：Δm (u) = 能量 (MeV) / 931.5 = 199.75 / 931.5 ≈ 0.2145 u。也可直接使用质能公式，但利用给定换因子更直接。

---

12. Alternating Current and Rectification | 交流电与整流

A sinusoidal a.c. supply of peak voltage 17.0 V is connected to a single diode half-wave rectifier with a 50 Ω load resistor. Calculate the average power dissipated in the resistor, assuming an ideal diode.

峰值为17.0伏的正弦交流电源连接到一个由单个二极管构成的半波整流器，负载电阻为50欧姆。假设二极管为理想元件，计算电阻上消耗的平均功率。

For a half-wave rectified sine wave, the r.m.s. voltage across the load is V_rms = V_peak / 2 = 17.0 / 2 = 8.50 V. (Note: for half-wave, r.m.s. = V_peak/2, not V_peak/√2). Average power P_avg = V_rms² / R = (8.50)² / 50 = 72.25 / 50 = 1.445 W. Alternatively, using average value: P_avg = (V_peak² / R) × (1/4) = (17² / 50) × 0.25 = (289/50)×0.25 = 5.78 × 0.25 = 1.445 W.

对于半波整流后的正弦波形，负载上的均方根电压 V_rms = V_peak / 2 = 17.0 / 2 = 8.50 V。（注意：半波整流的 r.m.s. 值为峰值的一半，而非峰值除以 √2）。平均功率 P_avg = V_rms² / R = (8.50)² / 50 = 72.25 / 50 = 1.445 W。也可用平均值法：P_avg = (V_peak² / R) × (1/4) = (17² / 50) × 0.25 = 1.445 W。

---

Published by TutorHao | Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）