📚 Year 10 Edexcel Physics: Unit Test Mock Paper Walkthrough | 爱德思十年级物理：单元测试模拟卷解析
This walkthrough takes you through a typical Year 10 Edexcel Physics unit test paper, covering the core topics of motion, forces, energy, waves and electricity. Each question is solved step by step with clear explanations of the underlying principles and common pitfalls. By working through these model answers, you will strengthen your understanding and build the problem‑solving skills needed to excel in your assessments.
本解析将带你回顾一份典型的爱德思十年级物理单元测试卷，涵盖运动、力、能量、波及电学等核心主题。每个问题都逐步求解，并清楚说明背后的物理原理和常见易错点。通过认真研读这些示范解答，你能够加深理解并锻炼解题能力，为单元测试做好充分准备。
1. Speed, Distance and Time | 速度、距离与时间
A runner completes a circular track of 400 m in exactly 50 seconds. Calculate the runner’s average speed in metres per second. Express your final answer also in kilometres per hour.
一位运动员沿 400 米圆形跑道用 50 秒跑完一圈。计算该运动员的平均速度（以米/秒为单位），并将结果转换为千米/小时。
Solution approach: Average speed is total distance divided by time. No direction is considered, so this is a scalar quantity. Convert 400 m to km and 50 s to hours to obtain the answer in km/h. Watch out for zero errors on stopwatches — here we assume an ideal measurement.
解题思路： 平均速度等于总路程除以时间。因不涉及方向，这里计算的是标量。进行单位换算：将 400 米换算为千米，50 秒换算为小时，即可得到以 km/h 为单位的结果。注意假设测量理想，停表没有零位误差。
average speed = total distance / time
so speed = 0.40 ÷ 0.01389 ≈ 28.8 km/h. Alternatively multiply 8.0 by 3.6 to get 28.8 km/h.
代入数值：速度 = 400 m ÷ 50 s = 8.0 m/s。转换为 km/h：路程 = 0.40 km，时间 = 50/3600 h ≈ 0.01389 h，速度 = 0.40 ÷ 0.01389 ≈ 28.8 km/h。也可以直接将 8.0 乘以 3.6 得到 28.8 km/h。
2. Acceleration Calculation | 加速度计算
A car accelerates uniformly from 5.0 m/s to 25.0 m/s in 4.0 seconds. Calculate the acceleration and the distance travelled during these 4.0 seconds.
一辆汽车从 5.0 m/s 匀加速到 25.0 m/s，历时 4.0 秒。求出加速度的大小以及这 4.0 秒内行驶的距离。
Method: Acceleration is the rate of change of velocity. Use a = (v − u) / t. Since the motion is uniform, you can find distance with average velocity: s = (u + v) t / 2. Always check that units are consistent: velocities in m/s, time in s.
方法： 加速度是速度的变化率，使用 a = (v − u) / t。由于是匀加速运动，可利用平均速度求距离：s = (u + v) t / 2。注意单位统一：速度用 m/s，时间用 s。
a = (25.0 − 5.0) / 4.0 = 5.0 m/s²
s = ((5.0 + 25.0) × 4.0) / 2 = 60 m
Common error: forgetting that the ‘t²’ term appears only in s = ut + ½at². Using average speed avoids this and works perfectly here.
常见错误：忘记 s = ut + ½at² 中含有 t² 项。本题直接使用平均速度可避免此错误，且完全正确。
3. Newton's Second Law | 牛顿第二定律
A resultant force of 15 N is applied to a trolley of mass 3.0 kg on a friction‑free horizontal surface. Determine the acceleration of the trolley. Describe and explain what happens to the acceleration if the mass is doubled while the force remains unchanged.
在无摩擦的水平面上，对一辆质量为 3.0 kg 的小车施加 15 N 的合力。求小车的加速度。若保持力不变，质量加倍，试描述并解释加速度会如何变化。
Core equation:
F = ma
.0 kg, new acceleration becomes 15 / 6.0 = 2.5 m/s². This shows inverse proportionality: a ∝ 1/m for a constant force. The explanation refers to inertia — a larger mass resists changes in motion more strongly.
变形：a = F / m = 15 N / 3.0 kg = 5.0 m/s²。当质量加倍至 6.0 kg 时，新的加速度变为 15 / 6.0 = 2.5 m/s²。这表明加速度与质量成反比（合力一定时）。原理在于惯性：质量越大，运动状态越难改变。
4. Kinetic and Potential Energy | 动能与势能
ssuming all kinetic energy is converted to gravitational potential energy, determine the vertical height gained. (Use g = 9.8 m/s²)
一辆质量为 1200 kg 的汽车以 15 m/s 的速度行驶。计算其动能。随后汽车冲上一个无摩擦的斜坡，直至瞬间停下。假设动能全部转化为重力势能，求汽车上升的垂直高度。（取 g = 9.8 m/s²）
KE = ½mv² = ½ × 1200 × (15)² = 135 000 J
1.5 m. The mass cancels if you set ½mv² = mgh, giving h = v²/(2g) = 225/19.6 ≈ 11.5 m. Both methods yield the same result.
重力势能增加量等于初始动能：mgh = 135 000 J。所以高度 h = 135 000 / (1200 × 9.8) = 135 000 / 11 760 ≈ 11.5 m。若直接由 ½mv² = mgh 消去质量，得 h = v²/(2g) = 225/19.6 ≈ 11.5 m，两种方法结果一致。
5. Energy Efficiency | 能量效率
ful work done on the mass, and hence determine the efficiency of the motor. (g = 9.8 m/s²)
一台电动机将 10 kg 的重物竖直提升 5.0 m。提供给电动机的电能为 600 J。计算对重物做的有用功，并求出电动机的效率。（g = 9.8 m/s²）
Useful work output = increase in GPE = mgh = 10 × 9.8 × 5.0 = 490 J. Energy input = 600 J.
Efficiency = (useful output / total input) × 100%
Efficiency = (490 / 600) × 100% ≈ 81.7%. The remaining 18.3% d friction. Stating the formula and showing the substitution are essential for full marks.
有用功 = 重力势能增加量 = mgh = 10 × 9.8 × 5.0 = 490 J。输入总能量 = 600 J。效率 = (490 / 600) × 100% ≈ 81.7%。大约 18.3% 的能量以热能等形式耗散在电机绕组和摩擦中。写出公式并代入数据是拿到满分的关键。
6. Wave Equation | 波动方程
A water wave has a frequency of 8.0 Hz and a wavelength of 0.25 m. Calculate the wave speed and the period of the wave.
一个水波的频率为 8.0 Hz，波长为 0.25 m。计算波速和波的周期。
v = fλ
of frequency: T = 1/f = 1/8.0 = 0.125 s. When solving wave problems, keep an eye on units — hertz is equivalent to s⁻¹, so multiplying by metres gives m/s automatically.
波速 v = 8.0 Hz × 0.25 m = 2.0 m/s。周期 T 是频率的倒数：T = 1/f = 1/8.0 = 0.125 s。解题时注意单位：赫兹即 s⁻¹，乘以米后自然得到 m/s。
7. Refraction and Snell's Law | 折射与斯涅尔定律
A ray of light passes from air into a glass block. The angle of incidence in air is 30°. The refractive index of the glass is 1.5. Calculate the angle of refraction inside the glass.
一束光从空气射入玻璃块。空气中的入射角为 30°。玻璃的折射率为 1.5。计算玻璃内的折射角。
n₁ sin θ₁ = n₂ sin θ₂
ends towards the normal because it enters an optically denser medium. Rounding to appropriate significant figures (3 s.f.) gives 19.5°.
已知 n₁ = 1.0（空气），θ₁ = 30°，n₂ = 1.5。代入：1.0 × sin 30° = 1.5 × sin θ₂ → sin θ₂ = 0.5 / 1.5 = 0.3333，所以 θ₂ = sin⁻¹(0.3333) ≈ 19.5°。由于光进入光密介质，折射光线向法线靠拢。结果保留三位有效数字得 19.5°。
8. Ohm's Law and Resistance | 欧姆定律与电阻
A resistor is connected to a 12 V supply and a current of 0.50 A flows through it. Calculate the resistance of the resistor and the power dissipated in it.
一只电阻器接在 12 V 电源上，通过它的电流为 0.50 A。计算该电阻器的电阻值及其消耗的功率。
V = IR
use P = I²R = (0.50)² × 24 = 6.0 W. Both forms are acceptable; remember to include units.
变形得 R = V / I = 12 V / 0.50 A = 24 Ω。消耗功率可用 P = IV = 0.50 A × 12 V = 6.0 W 或 P = I²R = (0.50)² × 24 = 6.0 W 求得。两种公式均可，务必带上单位。
9. Electrical Power and Energy Cost | 电功率与电能费用
A 2000 W electric heater is used for 3.0 hours each day. Electricity costs 15 pence per kilowatt‑hour (kWh). Calculate the daily energy consumption in kWh and the daily cost of running the heater.
一台 2000 W 的电暖器每天使用 3.0 小时。电费为每千瓦时 15 便士（pence）。计算每天的用电量（kWh）以及每天的电费。
u convert watts to kilowatts before multiplying by hours; a common mistake is to use 2000 instead of 2.0, giving an answer 1000 times too large.
功率单位换算：2000 W = 2.0 kW。电能 = 功率 × 时间 = 2.0 kW × 3.0 h = 6.0 kWh。费用 = 用电量 × 单价 = 6.0 kWh × 15 便士/kWh = 90 便士。一定要先将瓦特转换为千瓦再乘以小时，常见错误是直接用 2000 计算，导致结果夸大 1000 倍。
10. Domestic Wiring Safety | 家庭电路安全
Explain the roles of the live, neutral and earth wires in a three‑pin plug. Why is a fuse connected to the live wire, and how does it protect the appliance?
解释三脚插头中火线、零线和地线的作用。为什么保险丝要接在火线上？它是如何保护电器的？
Key points: The live wire carries the alternating potential difference (230 V in the UK) to the appliance. The neutral wire completes the circuit by providing a return path at approximately 0 V. The earth wire is a safety feature connected to the metal casing; if a fault causes the casing to become live, a large current flows to earth, which blows the fuse and disconnects the appliance. The fuse must be in the live wire so that when it melts, the appliance is completely isolated from the high potential, preventing electric shock.
要点： 火线将 230 V 交流电输送到电器；零线为电流提供接近 0 V 的返回路径，构成完整回路；地线则是安全保护线，连接到金属外壳。一旦发生故障使外壳带电，将有较大电流经地线流入大地，使保险丝熔断并切断电路。保险丝必须串联在火线上，这样熔断后电器才能彻底脱离高电位，以防触电事故发生。
Published by TutorHao | Physics Revision Series | aleveler.com
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