📚 Year 11 SQA Physics: Unit Test Mock Paper Analysis | SQA物理单元测试模拟卷解析
This article walks you through a typical SQA National 5 Physics unit test mock paper, providing full solutions, detailed annotations, and exam-focused commentary. Mastering these worked examples will help you tackle real assessments with confidence, whether you are revising dynamics, electricity, waves, or radioactivity.
本文带你完整解析一套典型的SQA国家5物理单元测试模拟卷，提供详细解答、批注及考试要点点评。无论你正在复习动力学、电学、波还是辐射，吃透这些例题都能让你在真实考试中更加从容。
1. Overview of the Mock Paper | 模拟试卷概览
This mock unit test mirrors the structure and difficulty of a live SQA National 5 Physics unit assessment. It covers key areas: kinematics, forces, energy, electricity, waves, and nuclear radiation. The paper contains two sections — Section A (multiple-choice) and Section B (short-answer and extended-response). Total marks: 25. Time allowed: 45 minutes. No data booklet is required, but students are expected to recall and apply standard equations.
本模拟单元测试仿照正式的SQA国家5物理单元评估结构与难度，涵盖运动学、力、能量、电学、波与核辐射等核心领域。试卷分为A部分（选择题）和B部分（简答与拓展题），满分25分，限时45分钟。考试中不提供数据手册，要求考生记忆并运用标准公式。
2. Multiple-Choice: Scalar and Vector Quantities | 选择题：标量与矢量
Question: Which of the following is a vector quantity?
A. Speed
B. Distance
C. Velocity
D. Mass
Vectors have both magnitude and direction. Velocity is defined as the rate of change of displacement and acts in a specific direction. Speed, distance, and mass are all scalar quantities — they have magnitude only. Therefore, the correct answer is C. Velocity.
题目：以下哪一个是矢量？
A. 速率
B. 路程
C. 速度
D. 质量
矢量既有大小也有方向。速度定义为位移的变化率，并具有特定方向。速率、路程和质量都是标量——只有大小。因此正确答案是C. 速度。
3. Short Answer: Average Speed and Velocity | 简答题：平均速率与平均速度
Question: A cyclist travels 300 m due north in 50 s, then 100 m due south in 20 s. For the whole journey, calculate:
(a) the average speed;
(b) the average velocity.
= 70 s. Net displacement = 300 m north - 100 m south = 200 m north. Using the definitions:
average speed = total distance ÷ total time
= 400 m ÷ 70 s ≈ 5.71 m/s
average velocity = displacement ÷ total time
= 200 m north ÷ 70 s ≈ 2.86 m/s north
Note that speed is a scalar and carries no direction, while velocity must state the direction.
题目：一名自行车手先向北骑行300 m，用时50 s，再向南骑行100 m，用时20 s。针对整个行程，计算：
(a) 平均速率；
(b) 平均速度。
总路程 = 300 m + 100 m = 400 m。总时间 = 50 s + 20 s = 70 s。合位移 = 向北200 m。根据定义：
平均速率 = 总路程 ÷ 总时间
= 400 m ÷ 70 s ≈ 5.71 m/s
平均速度 = 位移 ÷ 总时间
= 200 m 向北 ÷ 70 s ≈ 2.86 m/s 向北
注意速率是标量，无需标明方向；而速度必须标明方向。
4. Short Answer: Newton's First Law and Inertia | 简答题：牛顿第一定律与惯性
Question: A passenger standing on a bus lurches forward when the bus brakes suddenly. Use Newton's first law to explain this observation.
Newton's first law states that an object will remain at rest or move with constant velocity unless acted upon by an unbalanced force. The passenger's body was moving forward at the same speed as the bus. When the bus decelerates, the lower body (in contact with the bus) slows down, but the upper body continues moving forward due to inertia until an unbalanced force (e.g., from a handrail) acts. This gives the sensation of lurching forward.
题目：公交车突然刹车时，站立的乘客会向前倾倒。请用牛顿第一定律解释这一现象。
牛顿第一定律指出，任何物体都将保持静止或匀速直线运动状态，除非受到不平衡力的作用。乘客的身体原本与车同速前进。当车减速时，与车接触的下半身随之减速，而上半身由于惯性继续保持原来的运动状态，直到受到不平衡力（如扶手）作用。这就产生了向前倾倒的感觉。
5. Short Answer: Newton's Second Law (F = ma) | 简答题：牛顿第二定律 (F = ma)
Question: A car of mass 1200 kg accelerates at 2.5 m/s². A constant frictional force of 500 N opposes the motion.
(a) Calculate the resultant force acting on the car.
(b) Determine the driving force the engine must produce.
Resultant force is given by F = ma. Thus:
F_resultant = 1200 kg × 2.5 m/s² = 3000 N
The driving force must overcome friction and provide the resultant force:
F_driving = F_resultant + friction = 3000 N + 500 N = 3500 N
Always draw a free-body diagram to check the direction of forces; this avoids sign errors.
题目：一辆质量为1200 kg的汽车以2.5 m/s²的加速度行驶，运动中受到恒定的500 N摩擦力。
(a) 计算作用在汽车上的合力。
(b) 求发动机必须提供的驱动力。
合力由 F = ma 得出：
合力 = 1200 kg × 2.5 m/s² = 3000 N
驱动力既要克服摩擦力，又要提供合力：
驱动力 = 合力 + 摩擦力 = 3000 N + 500 N = 3500 N
务必画受力示意图检查力的方向，这样能避免符号错误。
6. Short Answer: Gravitational Potential Energy and Kinetic Energy | 简答题：重力势能与动能
Question: A ball of mass 0.50 kg is dropped from rest at a height of 10 m. Take g = 9.8 m/s².
(a) Calculate the initial gravitational potential energy (GPE) of the ball.
(b) Assuming no air resistance, determine the velocity of the ball just before it hits the ground.
GPE = m g h:
GPE = 0.50 × 9.8 × 10 = 49 J
tic energy (KE) at the ground: KE = ½ m v². Setting m g h = ½ m v² and cancelling m gives:
v = √2 g h = √2 × 9.8 × 10 = √196 = 14 m/s
The mass cancels; in the absence of air resistance all objects fall at the same rate regardless of mass.
题目：一个质量为0.50 kg的小球从10 m高处由静止释放。取 g = 9.8 m/s²。
(a) 计算小球最初的引力势能。
(b) 假设没有空气阻力，求小球落地前一瞬间的速度。
引力势能 = m g h：
GPE = 0.50 × 9.8 × 10 = 49 J
落地时势能全部转化为动能：½ m v²。令 m g h = ½ m v²，消去 m 得：
v = √2 × 9.8 × 10 = √196 = 14 m/s
质量被消去；在没有空气阻力的情况下，所有物体不论质量大小都会以相同的加速度下落。
7. Short Answer: Series and Parallel Circuits | 简答题：串联与并联电路
Question: Two resistors, 4 Ω and 6 Ω, are connected in series to a 12 V battery.
(a) Calculate the total resistance.
(b) Determine the current in the circuit.
(c) Find the voltage across each resistor.
In series, total resistance Rtotal = R₁ + R₂ = 4 Ω + 6 Ω = 10 Ω. Using Ohm's law:
I = V ÷ Rtotal = 12 V ÷ 10 Ω = 1.2 A
is the same at all points in a series circuit. Voltage across each resistor is V = I × R:
V₁ = 1.2 A × 4 Ω = 4.8 V
V₂ = 1.2 A × 6 Ω = 7.2 V
Check: 4.8 V + 7.2 V = 12 V — the sum of p.d.s equals the supply voltage, which confirms the calculation.
题目：两个电阻（4 Ω 和 6 Ω）串联后接到12 V电池上。
(a) 计算总电阻。
(b) 计算电路中的电流。
(c) 求每个电阻两端的电压。
串联时总电阻 Rtotal = 4 Ω + 6 Ω = 10 Ω。由欧姆定律：
电流 I = 12 V ÷ 10 Ω = 1.2 A
串联电路中各处电流相等。每个电阻的电压为 V = I R：
V₁ = 1.2 A × 4 Ω = 4.8 V
V₂ = 1.2 A × 6 Ω = 7.2 V
验证：4.8 V + 7.2 V = 12 V ——各电阻两端电压之和等于电源电压，说明计算正确。
8. Short Answer: Ohm's Law and Resistive Components | 简答题：欧姆定律与电阻元件
Question: An ohmic conductor carries a current of 0.50 A when a voltage of 3.0 V is applied.
(a) Calculate its resistance.
(b) Predict the current if the voltage is doubled to 6.0 V, assuming the conductor remains ohmic.
. For an ohmic conductor, resistance is constant, so doubling the voltage doubles the current:
Inew = 6.0 V ÷ 6.0 Ω = 1.0 A
A straight-line V-I graph passing through the origin confirms ohmic behaviour. If the component were non-ohmic (e.g., a filament lamp), resistance would increase with current due to heating.
题目：某欧姆导体在3.0 V电压下通过的电流为0.50 A。
(a) 计算其电阻。
(b) 若电压加倍至6.0 V，假设导体仍保持欧姆特性，预测此时的电流。
电阻 R = V / I = 3.0 V ÷ 0.50 A = 6.0 Ω。欧姆导体的电阻保持不变，因此电压加倍，电流也加倍：
新电流 I = 6.0 V ÷ 6.0 Ω = 1.0 A
一条通过原点的V-I直线可以证明欧姆特性。如果是非欧姆元件（如灯丝），由于发热，电阻会随电流增大而增加。
9. Short Answer: Wave Equation (v = f λ) | 简答题：波速公式 (v = f λ)
Question: A water wave has a frequency of 5.0 Hz and a wavelength of 0.34 m.
(a) Calculate the wave speed.
(b) Determine the period of the wave.
The wave equation is v = f × λ:
v = 5.0 Hz × 0.34 m = 1.7 m/s
Period T is the reciprocal of frequency:
T = 1 / f = 1 ÷ 5.0 Hz = 0.20 s
In SQA questions, always check unit consistency. Frequency must be in hertz (Hz) and wavelength in metres (m) for speed in m/s.
题目：水波的频率为5.0 Hz，波长为0.34 m。
(a) 计算波速。
(b) 求该波的周期。
波速公式 v = f × λ：
v = 5.0 Hz × 0.34 m = 1.7 m/s
周期 T 是频率的倒数：
T = 1 / 5.0 Hz = 0.20 s
SQA考题中要始终检查单位的一致性。频率必须用赫兹，波长用米，速度才得到 m/s。
10. Short Answer: Radioactive Half-life | 简答题：放射性半衰期
Question: A sample of a radioactive isotope has an initial activity of 800 Bq. Its half-life is 2.0 hours. Calculate the activity after 6.0 hours.
time ÷ half-life = 6.0 h ÷ 2.0 h = 3 half-lives. After each half-life the activity halves:
After 1 half-life: 800 → 400 Bq
After 2 half-lives: 400 → 200 Bq
After 3 half-lives: 200 → 100 Bq
100 Bq. The mass of the undecayed isotope also follows the same exponential decay pattern.
题目：某种放射性同位素样品的初始活度为800 Bq，半衰期为2.0小时。计算6.0小时后的活度。
经过的半衰期个数 = 总时间 ÷ 半衰期 = 6.0 h ÷ 2.0 h = 3个。每经过一个半衰期，活度减半：
1个半衰期后：800 → 400 Bq
2个半衰期后：400 → 200 Bq
3个半衰期后：200 → 100 Bq
或使用公式：活度 = 初始活度 × (½)ⁿ，n为半衰期个数。因此 A = 800 × (½)³ = 100 Bq。未衰变的同位素质量也遵循相同的指数衰减规律。
11. Extended Response: Designing an Experiment for Acceleration | 拓展题：设计加速度实验
Question: Describe an experiment to determine the acceleration of a trolley rolling down a slope. Include a labelled diagram (description only), measurements taken, and how the acceleration is calculated from the data.
A typical setup uses a dynamics trolley on a slightly inclined runway, with a light gate or motion sensor. A card of known length (e.g., 0.100 m) is attached to the trolley. The time taken for the card to pass through a light gate gives the instantaneous speed at that point. Using two light gates at different positions, the initial speed u at the first gate and final speed v at the second gate are recorded. The time t between the gates is measured with a stopwatch or the data-logger. Acceleration is found from:
a = (v – u) ÷ t
Alternatively, a single light gate and a double card can be used to measure the acceleration directly if the software does the calculation. Students must repeat measurements for different slopes or release distances, plot a speed–time graph, and determine the gradient, which equals acceleration. Key control variables: angle of slope, mass of trolley, and starting point. Safety: ensure a catcher at the bottom. Common error: not accounting for friction or the card's finite width causing timing uncertainty.
题目：描述一个测定小车沿斜面下滑加速度的实验。要求包含示意图（文字描述即可）、需要测量的量，以及如何从数据计算出加速度。
典型装置使用动力小车、微倾斜轨道和光门或运动传感器。在小车上固定已知宽度的遮光片（如0.100 m）。遮光片通过光门的时间可求得该点的瞬时速度。在轨道不同位置设置两个光门，记录第一个光门处的初速度u和第二个光门处的末速度v，并用秒表或数据采集器测量两光门间的时间t。加速度由以下公式求得：
a = (v – u) ÷ t
也可使用单个光门和双遮光片，由软件直接计算加速度。学生应改变斜面倾角或释放距离重复实验，绘制速度-时间图，其斜率即为加速度。需控制的变量：斜面倾角、小车质量、起始位置。安全措施：在底端放置缓冲物。常见错误：未考虑摩擦影响，或遮光片宽度引起的计时不确定度。
12. Exam Tips and Common Pitfalls | 答题技巧与常见错误
Always write down the correct formula before substituting numbers. This earns marks for method even if the arithmetic slips. In SQA unit assessments, final answers should usually be rounded to two or three significant figures and must include correct units. Never mix scalar and vector quantities in equations — e.g., do not substitute speed into an equation that requires displacement. Check that your answer makes physical sense; for example, a single resistor's p.d. cannot exceed the supply voltage. In extended-response questions, structure your answer with clear bullet points or numbered steps: aim, apparatus, method, measurements, risk assessment, and data analysis. Finally, practise reading values from graphs carefully: the gradient and intercept are common sources of marks.
做题前先把公式写出来，再代入数字。即使计算有误，正确的方法也能得分。在SQA单元评估中，最终答案通常取两到三位有效数字，并必须带上正确的单位。切勿在方程中混用标量和矢量——例如不能把速率代入需要位移的公式。检查结果的物理合理性：比如单个电阻两端的电压不能超过电源电压。答拓展题时，结构要清晰，用分点或编号步骤写出：目的、器材、方法、测量量、风险评估和数据分析。最后，学会仔细读取图线数值：斜率和截距是常见的得分点。
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