SQA Higher Physics Mock Unit Test Analysis | SQA 高等物理模拟单元测试解析

📚 SQA Higher Physics Mock Unit Test Analysis | SQA 高等物理模拟单元测试解析

The SQA Higher Physics course challenges students with in-depth concepts in mechanics, dynamics, relativity, and astrophysics. This mock unit test focuses on the ‘Our Dynamic Universe’ unit, which covers kinematics, forces, energy, momentum, circular motion, gravitation, special relativity, and the expanding universe. Each question is designed to mirror the style of a typical SQA unit assessment, combining problem‑solving with the application of fundamental physical principles. In this article, we provide a complete walk‑through of eight representative questions, showing step‑by‑step solutions and key reasoning.

SQA 高等物理课程深入考察学生在力学、动力学、相对论及天体物理等领域的理解。本模拟单元测试聚焦于“我们的动态宇宙”单元，涵盖运动学、力、能量、动量、圆周运动、万有引力、狭义相对论和宇宙膨胀。每道题目均仿照 SQA 单元评估的典型风格，将问题求解与基本原理的应用相结合。在本文中，我们将完整解析八道代表性题目，逐步展示解题过程与关键思路。

1. Uniform Acceleration | 匀加速直线运动

Question: A car travelling at 15 m s⁻¹ applies its brakes and experiences a constant deceleration of 3.0 m s⁻². Calculate the distance it takes to come to a complete stop.

问题：一辆汽车以 15 m s⁻¹ 的速度行驶，刹车时产生恒定的 3.0 m s⁻² 减速度。求汽车完全停下所需要的距离。

We list the known quantities: initial speed u = 15 m s⁻¹, final speed v = 0, acceleration a = -3.0 m s⁻² (since it is opposite to motion). The unknown is displacement s. The appropriate kinematic equation without time is v² = u² + 2as.

列出已知量：初速度 u = 15 m s⁻¹，末速度 v = 0，加速度 a = -3.0 m s⁻²（因其与运动方向相反）。未知量为位移 s。不涉及时间的运动学方程为 v² = u² + 2as。

Substitute the values: (0)² = (15)² + 2 × (-3.0) × s → 0 = 225 – 6.0 s → 6.0 s = 225 → s = 37.5 m.

代入数值：(0)² = (15)² + 2 × (-3.0) × s → 0 = 225 – 6.0 s → 6.0 s = 225 → s = 37.5 m。

Therefore, the braking distance is 37.5 m. This demonstrates the importance of correctly assigning signs to vector quantities in one‑dimensional motion.

因此，制动距离为 37.5 m。这显示了在一维运动中正确处理矢量符号的重要性。

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2. Conservation of Momentum | 动量守恒

Question: A 2.0 kg trolley moves at 4.0 m s⁻¹ on a frictionless track and collides with a stationary 3.0 kg trolley. After the collision, they couple together. Determine the common velocity and comment on the type of collision.

问题：质量为 2.0 kg 的小车以 4.0 m s⁻¹ 的速度在光滑轨道上运动，与静止的质量为 3.0 kg 的小车发生碰撞，碰撞后二者连在一起。求共同速度，并判断碰撞类型。

Since no external horizontal force acts, momentum is conserved. Before collision: total momentum = m₁u₁ + m₂u₂ = (2.0 × 4.0) + (3.0 × 0) = 8.0 kg m s⁻¹. After collision the combined mass is (2.0 + 3.0) = 5.0 kg, with common velocity v.

由于无水平外力作用，动量守恒。碰撞前总动量：m₁u₁ + m₂u₂ = (2.0 × 4.0) + (3.0 × 0) = 8.0 kg m s⁻¹。碰撞后总质量为 (2.0 + 3.0) = 5.0 kg，共同速度为 v。

Applying conservation: 8.0 = 5.0 × v → v = 1.6 m s⁻¹ in the original direction of the 2.0 kg trolley.

应用守恒律：8.0 = 5.0 × v → v = 1.6 m s⁻¹，方向与 2.0 kg 小车初速度方向相同。

Total kinetic energy before collision: Eₖ = ½ × 2.0 × (4.0)² + 0 = 0.5 × 2.0 × 16 = 16 J. After collision: Eₖ = ½ × 5.0 × (1.6)² = 2.5 × 2.56 = 6.4 J. The kinetic energy is not conserved (16 J → 6.4 J); it has decreased. Hence the collision is inelastic, specifically perfectly inelastic because the trolleys stick together.

碰撞前总动能：Eₖ = ½ × 2.0 × (4.0)² + 0 = 16 J。碰撞后：Eₖ = ½ × 5.0 × (1.6)² = 6.4 J。动能减少，不守恒，因此该碰撞为非弹性碰撞，且因碰后粘连，属于完全非弹性碰撞。

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3. Conservation of Energy | 能量守恒

Question: A ball of mass 0.50 kg is dropped from rest at a height of 10.0 m above the ground. Assuming air resistance is negligible, calculate the speed just before impact and the kinetic energy at that instant. (Use g = 9.8 m s⁻²)

问题：质量为 0.50 kg 的小球从距地面 10.0 m 高处由静止释放，忽略空气阻力，求小球撞击地面瞬间的速度及此时的动能。（取 g = 9.8 m s⁻²）

The principle of conservation of mechanical energy applies: the loss in gravitational potential energy equals the gain in kinetic energy. ΔEₚ = mgh = 0.50 × 9.8 × 10.0 = 49 J.

机械能守恒：重力势能的减少等于动能的增加。ΔEₚ = mgh = 0.50 × 9.8 × 10.0 = 49 J。

Thus, the kinetic energy just before impact is 49 J. To find the speed, use Eₖ = ½mv² → 49 = 0.5 × 0.50 × v² → v² = 49 / 0.25 = 196 → v = √196 = 14 m s⁻¹.

因此，碰地前瞬间的动能为 49 J。由 Eₖ = ½mv² 求速度：49 = 0.5 × 0.50 × v² → v² = 196 → v = 14 m s⁻¹。

Alternatively, the kinematic equation v² = u² + 2as with a = g gives v² = 0 + 2 × 9.8 × 10 → v = √196 = 14 m s⁻¹. Both methods agree.

也可用运动学方程 v² = u² + 2as（a = g）求解：v² = 0 + 2 × 9.8 × 10 = 196 → v = 14 m s⁻¹。两者结果一致。

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4. Circular Motion | 圆周运动

Question: A 0.20 kg rubber stopper is whirled in a horizontal circle of radius 0.50 m at a frequency of 2.0 revolutions per second. Determine (a) the angular velocity, (b) the linear speed, and (c) the centripetal force acting on the stopper.

问题：一个 0.20 kg 的橡胶塞在半径 0.50 m 的水平圆周上以每秒 2.0 转的频率旋转。求 (a) 角速度，(b) 线速度，(c) 作用在塞子上的向心力。

Angular velocity ω is related to frequency f by ω = 2πf. Thus ω = 2π × 2.0 = 4π rad s⁻¹ ≈ 12.6 rad s⁻¹.

角速度 ω 与频率 f 的关系为 ω = 2πf，因此 ω = 2π × 2.0 = 4π rad s⁻¹ ≈ 12.6 rad s⁻¹。

Linear speed v = rω = 0.50 × 4π = 2π ≈ 6.28 m s⁻¹.

线速度 v = rω = 0.50 × 4π = 2π ≈ 6.28 m s⁻¹。

Centripetal force F = mrω² = 0.20 × 0.50 × (4π)² = 0.10 × 16π² ≈ 15.8 N. The calculation can also be performed using F = mv²/r, yielding the same result.

向心力 F = mrω² = 0.20 × 0.50 × (4π)² = 0.10 × 16π² ≈ 15.8 N。也可用 F = mv²/r 计算，结果相同。

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5. Gravitational Fields | 万有引力场

Question: Given the Earth’s mass M = 5.97 × 10²⁴ kg, radius R = 6.38 × 10⁶ m, and gravitational constant G = 6.67 × 10⁻¹¹ N m² kg⁻², (a) calculate the gravitational field strength g at the Earth’s surface, and (b) estimate the orbital speed required for a satellite in low Earth orbit (r ≈ R).

问题：已知地球质量 M = 5.97 × 10²⁴ kg，半径 R = 6.38 × 10⁶ m，引力常量 G = 6.67 × 10⁻¹¹ N m² kg⁻²，(a) 计算地球表面的引力场强度 g，(b) 估算近地轨道（r ≈ R）卫星的轨道速度。

(a) g = GM / R² = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴) / (6.38 × 10⁶)². Numerator: 3.98 × 10¹⁴; denominator: 4.07 × 10¹³. Thus g ≈ 9.78 m s⁻², close to the accepted 9.8 m s⁻².

(a) g = GM / R² = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴) / (6.38 × 10⁶)²。分子约 3.98 × 10¹⁴，分母约 4.07 × 10¹³，故 g ≈ 9.78 m s⁻²，与常用值 9.8 m s⁻² 相近。

(b) For a circular orbit, gravitational force provides the centripetal force: GMm/r² = mv²/r. Cancelling m and one r gives v = √(GM/r). Substituting r ≈ R: v = √(6.67 × 10⁻¹¹ × 5.97 × 10²⁴ / 6.38 × 10⁶) ≈ √(6.24 × 10⁷) ≈ 7.90 × 10³ m s⁻¹. This is the approximate first cosmic velocity, about 7.9 km s⁻¹.

(b) 对圆轨道，万有引力提供向心力：GMm/r² = mv²/r。消去 m 和一个 r 得 v = √(GM/r)。代入 r ≈ R：v = √(6.67 × 10⁻¹¹ × 5.97 × 10²⁴ / 6.38 × 10⁶) ≈ √(6.24 × 10⁷) ≈ 7.90 × 10³ m s⁻¹。此即第一宇宙速度，约 7.9 km s⁻¹。

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6. Special Relativity – Time Dilation | 狭义相对论 – 时间膨胀

Question: A muon has a rest lifetime of 2.2 × 10⁻⁶ s. Suppose a muon is created in the upper atmosphere travelling at 0.99c relative to the Earth. According to an Earth‑based observer, what distance does the muon travel before decaying? (c = 3.00 × 10⁸ m s⁻¹)

问题：μ子的静止寿命为 2.2 × 10⁻⁶ s。若 μ 子在大气层顶部产生并以相对地面 0.99c 的速度运动，从地面参考系来看，μ 子在衰变前能飞行多远？（c = 3.00 × 10⁸ m s⁻¹）

The Earth observer sees the muon’s internal clock running slow. The dilated lifetime t’ = t₀ / √(1 – v²/c²), where t₀ is the proper lifetime (2.2 × 10⁻⁶ s). First compute the Lorentz factor γ = 1/√(1 – (0.99)²) = 1/√(1 – 0.9801) = 1/√(0.0199) ≈ 1/0.1411 ≈ 7.09.

地面观察者认为 μ 子的内部时钟变慢。膨胀后的寿命 t’ = t₀ / √(1 – v²/c²)，其中 t₀ 为固有寿命（2.2 × 10⁻⁶ s）。先计算洛伦兹因子 γ = 1/√(1 – 0.99²) = 1/√(0.0199) ≈ 1/0.1411 ≈ 7.09。

Hence, t’ = γ t₀ = 7.09 × 2.2 × 10⁻⁶ ≈ 1.56 × 10⁻⁵ s.

于是 t’ = γ t₀ = 7.09 × 2.2 × 10⁻⁶ ≈ 1.56 × 10⁻⁵ s。

Distance travelled in the Earth frame = v × t’ = 0.99 × 3.00 × 10⁸ × 1.56 × 10⁻⁵ = 0.99 × 3.00 × 10⁸ × 1.56 × 10⁻⁵ ≈ 4.63 × 10³ m (about 4.6 km). Without time dilation, the distance would be only about 660 m, confirming the necessity of relativistic effects.

在地球参考系中飞行距离 = v × t’ = 0.99 × 3.00 × 10⁸ × 1.56 × 10⁻⁵ ≈ 4.63 × 10³ m（约 4.6 km）。若不计时间膨胀，距离仅约 660 m，这证实了相对论效应的必要性。

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7. The Doppler Effect | 多普勒效应

Question: A spectral line normally observed at 500.0 nm in the laboratory is observed at 520.0 nm in the light from a distant galaxy. Assuming the redshift is due to the galaxy’s motion along the line of sight, calculate the recession speed of the galaxy. (c = 3.00 × 10⁸ m s⁻¹)

问题：实验室中通常观测到的一条谱线波长为 500.0 nm，在来自遥远星系的光中却被观测到为 520.0 nm。设此红移因星系沿视线方向的运动所致，求星系的退行速度。（c = 3.00 × 10⁸ m s⁻¹）

For speeds much smaller than c, the redshift relation is Δλ/λ₀ ≈ v/c. Here λ₀ = 500.0 nm, λ_observed = 520.0 nm, so Δλ = 20.0 nm. Therefore, v/c ≈ 20.0 / 500.0 = 0.040, giving v ≈ 0.040 × 3.00 × 10⁸ = 1.20 × 10⁷ m s⁻¹.

对远小于光速的速度，红移关系为 Δλ/λ₀ ≈ v/c。此处 λ₀ = 500.0 nm，观测波长 = 520.0 nm，故 Δλ = 20.0 nm。于是 v/c ≈ 20.0/500.0 = 0.040，v ≈ 0.040 × 3.00 × 10⁸ = 1.20 × 10⁷ m s⁻¹。

It is good practice to confirm that the relativistic formula gives a similar value at this speed. The relativistic Doppler formula gives a comparable result, but the approximation is acceptable for v << c. The galaxy is receding at 1.2 × 10⁷ m s⁻¹, which is about 12 000 km s⁻¹.

为严谨起见，可使用相对论多普勒公式验证，但在 v << c 情况下近似结果可以接受。该星系以 1.2 × 10⁷ m s⁻¹（约 12 000 km s⁻¹）的速度退行。

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8. Hubble’s Law | 哈勃定律

Question: Using the recession speed of 1.20 × 10⁷ m s⁻¹ obtained in Question 7, and taking the Hubble constant H₀ = 70 km s⁻¹ Mpc⁻¹, estimate the distance to the galaxy. Express your answer in both Mpc and light‑years. (1 pc ≈ 3.26 light‑years)

问题：利用第 7 题中求得的退行速度 1.20 × 10⁷ m s⁻¹ 和哈勃常数 H₀ = 70 km s⁻¹ Mpc⁻¹，估算该星系的距离，结果分别用 Mpc 和光年表示。（1 pc ≈ 3.26 光年）

First convert the velocity to km s⁻¹: v = 1.20 × 10⁷ m s⁻¹ = 1.20 × 10⁴ km s⁻¹ = 12 000 km s⁻¹. Hubble’s law states v = H₀ d, so d = v / H₀ = 12 000 / 70 ≈ 171.4 Mpc.

首先将速度单位转换为 km s⁻¹：v = 1.20 × 10⁷ m s⁻¹ = 1.20 × 10⁴ km s⁻¹ = 12 000 km s⁻¹。哈勃定律 v = H₀ d，因此 d = v / H₀ = 12 000 / 70 ≈ 171.4 Mpc。

To express this in light‑years: d = 171.4 × 10⁶ pc × 3.26 ly pc⁻¹ = 5.59 × 10⁸ ly (approximately 560 million light‑years).

换算为光年：d = 171.4 × 10⁶ pc × 3.26 ly pc⁻¹ = 5.59 × 10⁸ 光年（约 5.6 亿光年）。

This distance illustrates the vast scale of the Universe and supports the Big Bang theory, which predicts a proportional relationship between redshift‑derived recession speeds and cosmic distances.

这一距离展示了宇宙的巨大尺度，并支持大爆炸理论，该理论预言了红移得出的退行速度与宇宙距离之间的正比关系。

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9. Graphical Analysis of Motion | 运动图像分析

Question: A trolley’s motion is recorded, producing the following velocity–time data. Plot a velocity–time graph and use it to find (a) the acceleration during the first 4 seconds, (b) the total displacement in the first 6 seconds. Data: at t=0 s, v=0 m s⁻¹; t=2 s, v=6 m s⁻¹; t=4 s, v=12 m s⁻¹; t=6 s, v=12 m s⁻¹ (constant speed).

问题：记录了某小车的运动，得到如下速度-时间数据。

Published by TutorHao | Year 12 Physics Revision Series | aleveler.com

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