Year 8 Cambridge Physics: Mock Unit Test Walkthrough & Solutions | Year 8 剑桥物理单元测试模拟卷解析

📚 Year 8 Cambridge Physics: Mock Unit Test Walkthrough & Solutions | Year 8 剑桥物理单元测试模拟卷解析

This article provides a complete walkthrough of a mock unit test for Year 8 Cambridge Physics, covering key topics such as forces, motion, energy, density, pressure, electricity, light, sound, and the particle model. Each question is explained step by step to help you revise and build confidence for your exams.

本文提供了针对 Year 8 剑桥物理单元测试的模拟卷完整解析，涵盖了力、运动、能量、密度、压强、电学、光、声音和粒子模型等关键主题。每道题都逐步讲解，帮助你复习并为考试树立信心。

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1. Balanced Forces and Constant Speed | 平衡力与匀速运动

Question: A car travels along a straight horizontal road at a constant speed of 20 m/s. Name all the horizontal and vertical forces acting on the car and explain why it does not accelerate even though the engine is providing a forward driving force.

问题：一辆汽车以 20 m/s 的恒定速度沿笔直的水平道路行驶。请说出作用在汽车上的所有水平与竖直方向的力，并解释为什么尽管发动机提供了向前的驱动力，汽车却没有加速。

Solution: In the horizontal direction, the forward driving force (thrust) is balanced by the backward frictional force (air resistance and drag). In the vertical direction, the weight of the car pulling downwards is balanced by the upward normal contact force from the road. Because all forces are balanced, the resultant force is zero. According to Newton's First Law, an object with zero resultant force will maintain its state of motion — so the car continues moving at a constant velocity without acceleration.

解答：水平方向上，向前的驱动力（推力）与向后的摩擦力（空气阻力和阻力）平衡。竖直方向上，向下的重力与路面向上的法向接触力平衡。由于所有力都平衡，合力为零。根据牛顿第一定律，合力为零的物体会保持其运动状态——因此汽车继续以恒定速度行驶，不会加速。

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2. Calculating Average Speed | 计算平均速度

Question: A cyclist rides 15 kilometres in 30 minutes. Calculate her average speed in metres per second (m/s).

问题：一名自行车手在 30 分钟内骑行了 15 公里。请以米每秒 (m/s) 为单位计算她的平均速度。

. Time: 30 min = 30 × 60 = 1 800 s. Average speed is total distance divided by total time.

解答：首先，把单位转换成标准国际单位。距离：15 km = 15 000 m。时间：30 min = 30 × 60 = 1 800 s。平均速度等于总距离除以总时间。

v = d / t = 15 000 m / 1 800 s = 8.33 m/s

Therefore, the cyclist's average speed is 8.3 m/s (to two significant figures). Remember always to check units and convert to metres and seconds when calculating speed in physics.

因此，自行车手的平均速度为 8.3 m/s（保留两位有效数字）。计算物理中的速度时，请务必检查单位并转换为米和秒。

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3. Weight and Gravitational Field Strength | 重量与重力场强

.6 N/kg. (a) Calculate his weight on Earth. (b) Calculate his weight on the Moon. (c) Explain why his mass stays the same but his weight changes.

问题：一名宇航员的质量为 70 kg。地球上的重力场强度 g = 10 N/kg。月球上 g ≈ 1.6 N/kg。(a) 计算他在地球上的重量。(b) 计算他在月球上的重量。(c) 解释为什么他的质量不变而重量却改变了。

Solution: (a) Weight = mass × gravitational field strength.

解答：(a) 重量 = 质量 × 重力场强度。

W_earth = m × g_earth = 70 kg × 10 N/kg = 700 N

(b) On the Moon,

W_moon = m × g_moon = 70 kg × 1.6 N/kg = 112 N

(c) Mass is the amount of matter in an object; it does not depend on location. Weight is the force of gravity acting on this mass, so it changes with the strength of gravity. The Moon's weaker gravity pulls less on the same mass, resulting in a lower weight.

(c) 质量是物体所含物质的多少，与位置无关。重量是作用在该质量上的重力，因此它会随重力强度变化。月球引力较弱，对相同质量产生的拉力较小，因此重量更轻。

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4. Pressure Exerted by a Solid | 固体施加的压强

.6 m by 0.4 m. Calculate the pressure exerted by the box on the table. (Take g = 10 N/kg.)

问题：一个矩形箱子的质量为 24 kg，静置在桌面上。箱子的底面大小为 0.6 m × 0.4 m。计算箱子对桌面施加的压强。（取 g = 10 N/kg。）

Solution: The force exerted on the table is the weight of the box.

解答：箱子施加在桌面上的力就是箱子的重量。

F = m × g = 24 kg × 10 N/kg = 240 N

The area of the base in contact with the table is:

与桌面接触的底面面积为：

A = length × width = 0.6 m × 0.4 m = 0.24 m²

Pressure is force per unit area:

压强是单位面积上承受的压力：

p = F / A = 240 N / 0.24 m² = 1 000 Pa (or 1 kPa)

So the pressure is 1 000 pascals. Ensure you use square metres for area when calculating pressure in pascals.

因此压强为 1 000 帕斯卡。计算压强（单位为帕斯卡）时，务必使用平方米作为面积单位。

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5. Density and Material Identification | 密度与材料识别

Use the data below to suggest which metal the cube is made of: aluminium = 2.7 g/cm³, iron = 7.9 g/cm³, lead = 11.3 g/cm³.

问题：一个金属立方体质量为 540 g，体积为 200 cm³。(a) 以 g/cm³ 为单位计算该金属的密度。(b) 将此密度转换为 kg/m³。(c) 利用以下数据，判断该立方体是由哪种金属制成的：铝 = 2.7 g/cm³，铁 = 7.9 g/cm³，铅 = 11.3 g/cm³。

Solution: (a) Density = mass / volume.

解答：(a) 密度 = 质量 / 体积。

ρ = m / V = 540 g / 200 cm³ = 2.7 g/cm³

To convert to kg/m³, note that 1 g/cm³ = 1 000 kg/m³. Therefore, 2.7 g/cm³ = 2 700 kg/m³.

(b) 要转换为 kg/m³，请注意 1 g/cm³ = 1 000 kg/m³。因此，2.7 g/cm³ = 2 700 kg/m³。

(c) The calculated density of 2.7 g/cm³ exactly matches the density of aluminium. Therefore, the cube is most likely made of aluminium.

(c) 计算出的密度 2.7 g/cm³ 与铝的密度完全吻合。因此，该立方体很可能由铝制成。

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6. Energy Transfers in a Roller Coaster | 过山车中的能量转换

Question: A roller coaster car is pulled to the top of the first hill, which is the highest point. Describe the energy changes that take place as the car travels from the top of the first hill to the bottom of the hill. Ignore friction and air resistance.

问题：一辆过山车被拉到第一个山坡的最高点。请描述过山车从第一个山坡的顶部行驶到底部时所发生的能量变化。忽略摩擦和空气阻力。

Solution: At the top of the hill, the car has maximum gravitational potential energy (GPE) because it is at the greatest height. Its kinetic energy (KE) is low, possibly zero if starting from rest. As the car rolls down the hill, its height decreases, so GPE is transferred into KE. The car speeds up, and kinetic energy increases. At the bottom of the hill, GPE is at a minimum (assuming the bottom is the reference height) and kinetic energy reaches a maximum. Throughout the motion, the total mechanical energy (GPE + KE) remains constant if we ignore friction and air resistance, obeying the principle of conservation of energy.

解答：在山顶，过山车处于最高点，因此具有最大重力势能 (GPE)。它的动能 (KE) 较低，如果从静止出发则为零。当过山车向下行驶时，高度降低，GPE 转化为 KE。车子加速，动能增加。到达坡底时，GPE 最小（假设底部为参考高度），动能达到最大。在整个运动过程中，如果忽略摩擦和空气阻力，总机械能（GPE + KE）保持不变，遵循能量守恒原理。

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7. Series and Parallel Circuits | 串联与并联电路

Question: A series circuit is built with a 6 V battery and two identical lamps. A parallel circuit uses the same battery and same two lamps, with each lamp on its own branch. (a) In the series circuit, if the current is 0.5 A, what is the current through each lamp and the voltage across each lamp? (b) In the parallel circuit, what is the voltage across each lamp and how does the current behave?

问题：用一个 6 V 电池和两个相同的灯泡分别组成一个串联电路和一个并联电路（每个灯泡在独立支路上）。(a) 在串联电路中，如果电流为 0.5 A，通过每个灯泡的电流和每个灯泡两端的电压分别是多少？(b) 在并联电路中，每个灯泡两端的电压是多少？电流又是如何分配的？

Solution: (a) In a series circuit, the current is the same at all points. So the current through each lamp is 0.5 A. The voltage of the battery is shared across components. Since the lamps are identical, the voltage divides equally: each lamp receives 3 V.

解答：(a) 串联电路中，各处电流相等。因此通过每个灯泡的电流为 0.5 A。电池电压被各元件分担。因为两灯泡相同，电压平均分配：每个灯泡两端电压为 3 V。

(b) In a parallel circuit, each branch receives the full battery voltage, so each lamp has 6 V across it. The total current from the battery splits between the branches. With identical lamps, the current divides equally; if the total current is 1.0 A, each branch would carry 0.5 A. The key rules: in series, current constant, voltage shared; in parallel, voltage constant, current shared.

(b) 并联电路中，每个支路都获得全部电源电压，因此每个灯泡两端电压为 6 V。电池提供的总电流在各支路之间分配。使用相同灯泡时，电流平均分配；若总电流为 1.0 A，则每支路电流为 0.5 A。关键规律：串联时电流处处相等、电压相加；并联时电压处处相等、电流相加。

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8. Law of Reflection | 反射定律

Question: A ray of light strikes a plane mirror with an angle of incidence of 35°. State the law of reflection and determine the angle of reflection. In your answer, describe how the incident ray, reflected ray, and the normal are positioned relative to the mirror.

问题：一束光线以 35° 的入射角照射到平面镜上。陈述反射定律，并确定反射角的大小。在你的回答中，描述入射光线、反射光线和法线相对于镜面的位置。

Solution: The law of reflection states that the angle of incidence equals the angle of reflection. Both angles are measured from the normal, which is an imaginary line perpendicular to the mirror surface at the point of incidence. Therefore, if the angle of incidence is 35°, the angle of reflection is also 35°. The incident ray, reflected ray, and the normal all lie in the same plane. A ray diagram would show the normal as a dashed line at 90° to the mirror, with the incident and reflected rays on opposite sides of the normal forming equal angles.

解答：反射定律指出，入射角等于反射角。两个角都从法线量起，法线是在入射点处垂直于镜面的一条假想线。因此，如果入射角为 35°，反射角也为 35°。入射光线、反射光线和法线三者位于同一平面内。光线图会用法线（虚线，与镜面成 90°）以及位于法线两侧、形成相等角度的入射和反射光线来表示。

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9. Sound Echo and Distance Calculation | 声音回声与距离计算

Question: A person stands a certain distance from a vertical cliff and shouts. She hears the echo 3.0 seconds later. Given the speed of sound in air is 340 m/s, calculate the distance between the person and the cliff. Explain why you must divide the total distance by two.

问题：一个人站在离峭壁某段距离处大声呼喊，3.0 秒后听到回声

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