Year 12 SQA Physics: Unit Test Mock Paper Walkthrough | SQA 物理单元测试模拟卷解析

📚 Year 12 SQA Physics: Unit Test Mock Paper Walkthrough | SQA 物理单元测试模拟卷解析

This article provides a full walkthrough of a mock Unit 1 test for SQA Higher Physics (Our Dynamic Universe). Each question is explained step by step, with emphasis on command words, common pitfalls, and application of relationships from the SQA data booklet. Use this as a revision tool to check your understanding of kinematics, dynamics, momentum, projectile motion, gravitation, circular motion, and energy.

本文为 SQA Higher 物理单元一（我们的动态宇宙）模拟测试卷提供详细解析。每一道题目都会逐步拆解，重点关注指令词、常见错误以及数据手册中关系式的运用。你可以将其作为复习工具，检验自己对运动学、动力学、动量、抛体运动、万有引力、圆周运动和能量等知识的掌握情况。

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1. Mock Paper Overview | 模拟卷概览

The mock paper contains nine questions totalling 50 marks, designed to be completed in 60 minutes. It mimics the style of an SQA end‑of‑unit test, with multiple‑choice style calculations, structured problems, and an uncertainty question. The questions primarily cover the 'Our Dynamic Universe' unit, but some require cross‑topic data‑handling skills.

本模拟卷共九道题，满分 50 分，建议用时 60 分钟。试卷模仿 SQA 单元结束测验的风格，包含类似选择题的计算题、结构化问题以及一道不确定度分析题。题目主要覆盖“我们的动态宇宙”单元，但部分题目需要跨主题的数据处理能力。

A thorough understanding of equations from the SQA Higher Physics data booklet is expected. You are advised to show all working and pay attention to significant figures and units, as these are explicitly marked in many SQA questions.

你需要熟练掌握 SQA Higher 物理数据手册中的公式。建议写出完整的解题步骤，并注意有效数字和单位——许多 SQA 考题会对这些细节进行评分。

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2. Q1: Constant Acceleration | 问题一：匀加速运动

Question (English): A cyclist accelerates from rest at 1.5 m/s² for 8.0 s. Calculate the final speed and the distance travelled during this time.

题目（中文）：一位自行车运动员从静止开始以 1.5 m/s² 的加速度加速 8.0 s。计算末速度与在此时间内行驶的距离。

Solution (English): ½at². Since u = 0, s = ½ × 1.5 × (8.0)² = ½ × 1.5 × 64 = 48 m. Always include units and check significant figures (t has 2 s.f., but a has 2 s.f., so answers are best given as 12 m/s and 48 m).

解析（中文）：写出已知量：u = 0, a = 1.5 m/s², t = 8.0 s。求末速度用 v = u + at，代入得 v = 12 m/s。求位移用 s = ut + ½at²，因为初速为零，得 s = 48 m。注意单位并检查有效数字：最终答案保留两位有效数字为 12 m/s 和 48 m。

v = u + at   s = ut + ½at²

Common mistake: Forgetting to square the time in the displacement equation or using the wrong sign for acceleration. Also, many students confuse the two equations.

常见错误：在位移公式中忘记对时间平方，或者加速度符号用错。此外，很多学生将两个运动学公式混淆。

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3. Q2: Newton’s Second Law with Friction | 问题二：牛顿第二定律与摩擦力

Question: A block of mass 4.0 kg is pulled along a horizontal surface by a force of 25 N applied horizontally. The block experiences a constant frictional force of 5.0 N. Calculate the acceleration of the block.

题目：一个质量为 4.0 kg 的木块在水平面上受到一个 25 N 的水平拉力。木块受到恒定的摩擦力 5.0 N。计算木块的加速度。

Solution: F/m = 20 / 4.0 = 5.0 m/s². The direction of acceleration is the same as the applied force.

解析：求合力：F_合 = 拉力 − 摩擦力 = 25 − 5.0 = 20 N。应用牛顿第二定律 F_合 = ma，得 a = 5.0 m/s²。加速度的方向与拉力方向相同。

F_unbalanced = ma

Key point: d : Always resolve forces and find the net force before using F = ma. SQA expects you to show the subtraction step clearly.

关键点：务必先求出合力，再使用 F = ma。SQA 要求清楚地展示求合力的步骤。

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4. Q3: Conservation of Momentum | 问题三：动量守恒

Question: Trolley A of mass 2.0 kg moves at 3.0 m/s to the right. It collides and sticks to stationary trolley B of mass 1.0 kg. Find the velocity of the combined trolleys after the collision.

题目：小车 A 质量为 2.0 kg，以 3.0 m/s 的速度向右运动，与静止的小车 B（质量 1.0 kg）碰撞并粘在一起。求碰撞后合体小车的速度。

Solution: Total momentum before = m_A u_A + m_B u_B 2.0 m/s to the right. This is an inelastic collision, so kinetic energy is not conserved.

解析：碰撞前总动量 = 2.0 × 3.0 + 1.0 × 0 = 6.0 kg m/s。碰撞后总质量 3.0 kg，由动量守恒得 v = 2.0 m/s 向右。这是非弹性碰撞，动能不守恒。

m₁u₁ + m₂u₂ = (m₁ + m₂)v

Common pitfall: Forgetting that momentum is a vector. Here it is one‑dimensional, but always assign a direction. Negative signs are crucial in explosion or rebound problems.

常见陷阱：忘记动量是矢量。本题虽然是一维问题，但始终要规定正方向。在反冲或反弹问题中，负号至关重要。

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5. Q4: Projectile Motion | 问题四：抛体运动

Question: A ball is kicked from ground level with an initial speed of 20 m/s at an angle of 30° to the horizontal. Calculate the maximum height reached and the horizontal range. Ignore air resistance.

题目：一个足球从地面以初速度 20 m/s、与水平方向成 30° 角踢出。计算能到达的最大高度和水平射程。忽略空气阻力。

Solution: Resolve velocity: u_x = 20 cos30° = 17.3 m/s; u_y = 20 sin30° = 10 m/s. At maximum height, vertical velocity = 0. Use v_y² = u_y² − 2gH. With g = 9.8 m/s², H = (10²) / (2 × 9.8) = 100/19.6 ≈ 5.1 m. Time of flight: t_total = 2u_y/g = 20/9.8 ≈ 2.04 s. Range = u_x × t_total = 17.3 × 2.04 ≈ 35.3 m.

解析：分解速度：水平分量 u_x = 20 cos30° = 17.3 m/s，竖直分量 u_y = 20 sin30° = 10 m/s。最高点竖直速度为零。用 v_y² = u_y² − 2gH，取 g = 9.8 m/s²，得 H ≈ 5.1 m。总时间 t = 2u_y/g ≈ 2.04 s。水平射程 = u_x × t ≈ 35.3 m。

u_x = u cosθ   u_y = u sinθ   v² = u² − 2gh   Range = u_x × (2u_y/g)

Advice: Always list the data for horizontal and vertical motions separately. Remember that horizontal acceleration is zero and that time connects both components.

建议：始终将水平和竖直运动的已知量分开罗列。牢记水平加速度为零，时间是联系两个方向的桥梁。

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6. Q5: Gravitation and Circular Motion | 问题五：万有引力与圆周运动

Question: ting gravitational force to centripetal force, show that the orbital speed is given by v = √GM/r and calculate its value.

题目：一颗卫星在轨道半径 7.0 × 10⁶ m 处绕地球运行。地球质量为 5.97 × 10²⁴ kg。通过万有引力等于向心力，证明轨道速率公式为 v = √GM/r 并计算其数值。

Solution: nal force F = GMm/r² provides centripetal force mv²/r. Cancel m and one r: GM/r = v² → v = √GM/r. Using G = 6.67 × 10⁻¹¹ N m²/kg², v = √[(6.67×10⁻¹¹ × 5.97×10²⁴) / (7.0×10⁶)] ≈ √5.69×10⁷ ≈ 7.54×10³ m/s.

解析：万有引力 F = GMm/r² 提供向心力 mv²/r。消去 m 和一个 r 得 v = √GM/r。代入数据得 v ≈ 7.54×10³ m/s。

GMm / r² = mv² / r   →   v = √GM / r

Extended: found using T = 2πr/v. This often appears in SQA questions about geostationary satellites.

拓展：随后可用 T = 2πr/v 求周期。这类推导经常出现在 SQA 关于地球同步卫星的试题中。

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7. Q6: Energy and Work Done on a Slope | 问题六：斜面上的功和能

Question: A crate of mass 50 kg is pulled at constant speed 2.0 m up a smooth slope inclined at 20° to the horizontal. The pulling force acts parallel to the slope. Calculate the work done by the pulling force and the gain in gravitational potential energy.

题目：一个 50 kg 的板条箱以恒定速度沿光滑斜面向上拉动 2.0 m，斜面倾角 20°。拉力平行于斜面。计算拉力做的功以及重力势能的增加量。

Solution: ne = F × d = 167.6 × 2.0 ≈ 335 J. Vertical height gained = 2.0 sin20° ≈ 0.684 m. Gain in E_p = mgh = 50 × 9.8 × 0.684 ≈ 335 J. The work done equals the energy gained (no friction).

解析：匀速运动，合力为零，因此拉力 F = 沿斜面下滑的重力分量 mg sinθ ≈ 167.6 N。做功 = F × d ≈ 335 J。上升的竖直高度 = 2.0 sin20° ≈ 0.684 m。重力势能增加 = mgh ≈ 335 J。拉力做的功等于获得的势能（无摩擦）。

W = Fd   ΔE_p = mgΔh   Δh = d sinθ

Note: If friction were present, work done against friction would increase the pulling force and the total work input.

注意：如果存在摩擦力，克服摩擦力做功将增大所需拉力和总输入功。

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8. Q7: Impulse from a Force–Time Graph | 问题七：力-时间图像中的冲量

Question: A footballer kicks a stationary ball of mass 0.43 kg. The force applied by the foot varies as shown: it rises linearly from 0 to 150 N in 0.010 s, stays constant at 150 N for 0.005 s, then falls to 0 in 0.010 s. Estimate the speed of the ball after the kick.

题目：一名足球运动员踢一个静止的 0.43 kg 的足球。脚施加的力如下变化：0 到 0.010 s 从 0 线性增至 150 N，之后保持 150 N 恒定 0.005 s，然后在 0.010 s 内降为 0。估算球被踢后的速度。

Solution: Impulse = area under F– ectangle (0.005 s), triangle (0.010 s fall). Area = ½×0.010×150 + 0.005×150 + ½×0.010×150 = 0.75 + 0.75 + 0.75 = 2.25 N s. Impulse = change in momentum = mv − 0 → v = 2.25 / 0.43 ≈ 5.2 m/s.

解析：冲量 = F–t 图线下的面积。图形可视为梯形，总面积 = 2.25 N s。冲量 = 动量变化量 = 0.43v，解得 v ≈ 5.2 m/s。

Impulse = Area = Δp = m(v − u)

Approach: When a graph has curved sections, SQA may ask for an estimate using counting squares. Here the linear segments allow exact calculation.

方法：当图线包含曲线段时，SQA 可能要求通过数方格进行估算。本题中线性段允许精确计算。

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9. Q8: Uncertainty Analysis | 问题八：不确定度分析

Question: rmine the absolute uncertainty in the calculated acceleration due to gravity, g = 4π²L/T².

题目：一名学生测得摆长 L = 0.800 ± 0.005 m，周期 T = 1.78 ± 0.05 s。首先计算百分比不确定度，然后确定重力加速度 g = 4π²L/T² 的绝对不确定度。

Solution: %U_L = (0.005/0.800)×100 = 0.625%. %U_T = (0.05/1.78)×100 ≈ 2.81%. For g, T has power 2, so total %U_g = %U_L + 2×%U_T = 0.625 + 5.62 = 6.25%. Nominal g = 4π²×0.800/(1.78²) ≈ 3.95×0.800 / 3.1684? 4 ≈ 9.97 m/s². Absolute uncertainty = 0.0625×9.97 ≈ 0.62 m/s². Hence g = 9.97 ± 0.62 m/s².

解析：百分比不确定度：长度 0.625%，周期 2.81%。由于 T 的指数为 2，g 的百分比不确定度 = %U_L + 2×%U_T = 6.25%。计算 g ≈ 9.97 m/s²，绝对不确定度 ≈ 0.62 m/s²，结果表示为 g = 9.97 ± 0.62 m/s²。

%U_g = %U_L + 2%U_T

Note: Remember to multiply the fractional uncertainty of T by its power. SQA accepts using either fractional or percentage methods, but the final absolute uncertainty should be given to 1 or 2 significant figures.

注意：记得将周期的不确定度乘以其指数。SQA 接受分数法或百分比法，但最终绝对不确定度一般保留 1 至 2 位有效数字。

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10. Q9: Geostationary Satellite Orbit Radius | 问题九：同步卫星轨道半径

Question: f planetary motion, then calculate its value given T = 24 h and Earth mass = 5.97×10²⁴ kg.

题目：利用开普勒第三定律推导地球同步卫星的轨道半径表达式，并根据 T = 24 h 和地球质量 5.97×10²⁴ kg 计算其数值。

Solution: 7×10²⁴ × (86400)²) / (4π²). Compute numerator: 6.67×10⁻¹¹ × 5.97×10²⁴ = 3.98×10¹⁴, × (7.46×10⁹) ≈ 2.97×10²⁴. Divide by 39.478 ≈ 7.53×10²². Cube root gives r ≈ 4.22×10⁷ m (about 42,200 km from centre).

解析：由万有引力提供向心力：GMm/r² = mrω²，且 ω = 2π/T。整理得 r³ = GMT²/4π²。将 T = 86400 s 代入计算得 r ≈ 4.22×10⁷ m（距地心约 42200 km）。

r³ = (GMT²) / (4π²)

Common confusion: synchronous orbits. SQA questions often ask you to explain why a geostationary satellite must be above the equator.

常见混淆：区分地球同步轨道（T = 24 h，固定在赤道上空）和对地静止轨道。SQA 经常要求解释为何同步卫星必须在赤道平面上方。

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11. Unit Test Tips | 单元测试技巧

Before the test, ensure you can recall all relationships from the 'Our Dynamic Universe' page of the data booklet. Practice by writing them out from memory and checking conditions (e.g., constant acceleration, conservation of momentum in closed systems).

考试前，务必能够默写数据手册“我们的动态宇宙”部分的所有关系式。通过默写并检查适用条件（如匀加速度、封闭系统动量守恒）进行练习。

Pay close attention to command words: 'Show that…' requires an algebraic derivation; 'Calculate' expects a numerical answer with correct units; 'Explain' demands a justifiable scientific reason. Always underline key physics terms in your explanation.

密切关注指令词：“证明……”需要代数推导；“计算”要求给出带正确单位的数值答案；“解释”需要合理解释，并加上科学依据。在解释中划出关键的物理术语。

Manage your time: mark allocation indicates roughly the number of minutes to spend. If stuck, move on and return later. A blank often scores zero, but a partial attempt with correct equations can pick up method marks.

合理分配时间：分值大致对应所需的分钟数。如果卡住，先做下一题，回头再补。留白只得零分，而写出正确的公式通常能得到方法分。

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12. Conclusion and Further Practice | 总结与强化练习

Working through this mock paper has highlighted the importance of a systematic problem‑solving approach: list knowns, write the appropriate relationship, substitute values, compute, and finally check units and significant figures. This method helps reduce unnecessary errors and improves clarity for markers.

通过这份模拟卷的演练，可以看出一套系统化解题方法的重要性：罗列已知量，选择正确关系式，代入数据，计算，最后检查单位和有效数字。这种方法能减少无谓的失误，并让阅卷人更清晰地看到解题思路。

Regular practice with past‑paper questions under timed conditions will build confidence and speed. Use the SQA marking principles to self‑assess: be strict about 'no unit' penalties and missing method steps. For further support, revisit your class notes on deriving the equations of motion from graphs and the vector nature of momentum and impulse.

定期在计时条件下练习真题能够增强信心和解题速度。使用 SQA 评分原则进行自我评估：严格对待缺单位的扣分以及漏掉的解题步骤。如需进一步巩固，可复习课堂笔记中关于用图像推导运动方程以及动量与冲量的矢量特性等内容。

Published by TutorHao | Physics Revision Series | aleveler.com

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