📚 Case Study Practice: Mastering Physics for Year 9 Cambridge | 案例分析实战演练：精通 Year 9 剑桥物理
This article presents a series of real-world physics case studies designed to sharpen your problem-solving skills for the Year 9 Cambridge Physics syllabus. Each case focuses on a core concept — from speed and acceleration to moments and energy — with step-by-step explanations and calculations. You will learn how to apply formulas correctly, interpret data, and explain physical phenomena in both English and Chinese. Work through these examples, and you will build the confidence needed to tackle any exam question.
本文提供一系列真实世界中的物理案例分析，旨在提升你解决 Year 9 剑桥物理考题的能力。每个案例聚焦一个核心概念——从速度、加速度到力矩和能量——并配有逐步解释和计算。你将学会如何正确应用公式、解读数据，并用中英文解释物理现象。认真完成这些演练，你将建立起应对任何考试问题的信心。
1. Average Speed of a Runner | 跑步者的平均速度
e. Remember that speed is the distance travelled per unit time. First, convert the time into seconds: 5 min = 5 × 60 = 300 s. Then apply the formula.
在学校运动会上，一名学生在5分钟内跑完1200米。为了求她的平均速度，我们使用距离与时间的关系。记住速度是单位时间内通过的距离。首先将时间换算为秒：5分钟 = 5 × 60 = 300秒。然后代入公式。
v = d ÷ t
Substituting the values: v = 1200 m ÷ 300 s = 4.0 m/s. The runner' s 4 metres every second on average. In an exam, always check that the units are consistent — metres for distance and seconds for time — to obtain speed in m/s.
代入数值：v = 1200 m ÷ 300 s = 4.0 m/s。这位跑步者的平均速度是每秒4.0米。这意味着她平均每秒跑4米。在考试中，务必检查单位是否一致——距离用米、时间用秒——这样才能得到以 m/s 为单位的速度。
2. Acceleration of a Car | 汽车的加速度
the initial velocity u = 0 m/s, and the final velocity v = 20 m/s. The time taken t = 8 s.
一辆汽车从静止（0 m/s）加速到20 m/s，用时8秒。加速度的定义是速度的变化率。因为汽车从静止起步，初速度 u = 0 m/s，末速度 v = 20 m/s。所用时间 t = 8 s。
a = (v - u) ÷ t
Plug in the numbers: a = (20 m/s - 0 m/s) ÷ 8 s = 20 ÷ 8 = 2.5 m/s². The car' 2.5 m/s every second. The SI unit for acceleration is m/s², which you must always include.
代入数字：a = (20 m/s - 0 m/s) ÷ 8 s = 20 ÷ 8 = 2.5 m/s²。汽车的加速度是2.5米每二次方秒。这告诉我们速度每秒增加2.5 m/s。加速度的国际单位是 m/s²，你必须始终带上它。
3. Newton's Second Law in Action | 牛顿第二定律的应用
A trolley of mass 15 kg is pushed with a resultant force of 45 N. Newton' elationship is given by F = m × a, where F is force in newtons, m is mass in kilograms, and a is acceleration in m/s².
一个质量为15 kg的小车受到45 N的合力推动。牛顿第二定律指出，物体的加速度与所受合力成正比，与其质量成反比。关系式为 F = m × a，其中 F 是力（牛顿），m 是质量（千克），a 是加速度（m/s²）。
F = m × a
ccelerates at 3.0 m/s². Always remember that the force used must be the resultant (unbalanced) force. If friction or other forces are present, subtract them to find the net force first.
重新排列求加速度：a = F ÷ m = 45 N ÷ 15 kg = 3.0 m/s²。小车以3.0 m/s²加速。始终记住，所用的力必须是合力（非平衡力）。如果存在摩擦力或其他力，需要先减去它们来得到净力。
4. Weight on Different Planets | 不同行星上的重量
gth. On Earth, g ≈ 9.8 N/kg (often 10 N/kg for simplicity). On the Moon, g = 1.6 N/kg. Let's compare her weight on Earth and on the Moon.
一位宇航员的质量为80 kg。重量是作用在物体质量上的重力，用公式 W = m × g 计算，其中 g 是引力场强度。在地球上，g ≈ 9.8 N/kg（通常简化为10 N/kg）。在月球上，g = 1.6 N/kg。我们来比较她在地球和月球上的重量。
W = m × g
kg): W = 80 kg × 10 N/kg = 800 N. On the Moon: W = 80 kg × 1.6 N/kg = 128 N. The astronaut' ains unchanged. Mass is a measure of the amount of matter and does not depend on location.
在地球上（取 g = 10 N/kg）：W = 80 kg × 10 N/kg = 800 N。在月球上：W = 80 kg × 1.6 N/kg = 128 N。宇航员在月球上的重量小得多，因为 g 更小，但她的质量保持不变。质量是物质多少的量度，不随位置改变。
5. Pressure Exerted by a Brick | 砖块施加的压强
ting per unit area. The formula is p = F ÷ A, where p is pressure in pascals (Pa), F is the force (weight, here acting perpendicular), and A is the contact area.
一块长方体砖放在桌子上。它的重量为24 N，与桌面接触的面面积为0.03 m²。压强定义为单位面积上所受的力。公式为 p = F ÷ A，其中 p 是压强（帕斯卡，Pa），F 是力（这里是垂直作用的重力），A 是接触面积。
p = F ÷ A
e exerted by the brick is 800 pascals. If the brick were placed on a smaller face, the contact area would decrease and the pressure would increase for the same weight. This explains why sharp objects penetrate more easily — they have a tiny contact area.
p = 24 N ÷ 0.03 m² = 800 Pa。因此砖块施加的压强为800帕斯卡。如果将砖块以更小的面放置，接触面积减小，同样重量下压强会增大。这就解释了为什么尖锐物体更容易刺入——它们的接触面积非常小。
6. Work Done Lifting a Box | 提升箱子所做的功
First, find the weight of the box (force needed to lift it steadily). Taking g = 10 N/kg, weight = 12 kg × 10 N/kg = 120 N.
仓库工人将一个12 kg的箱子竖直向上提升2.5 m。当一个力使物体沿力的方向移动时，就做了功。首先求出箱子的重量（匀速提起所需的力）。取 g = 10 N/kg，重量 = 12 kg × 10 N/kg = 120 N。
W = F × d
The worker does 300 joules of work. Note that the distance must be in metres. If the box is moved horizontally at constant speed, no work is done against gravity, although work may be done against friction.
所做的功 = 力 × 沿力的方向移动的距离 = 120 N × 2.5 m = 300 J。工人做了300焦耳的功。注意距离必须以米为单位。如果箱子是水平匀速移动的，则没有克服重力做功，但可能克服摩擦力做功。
7. Moments and Equilibrium | 力矩与平衡
A uniform metre rule is pivoted at its centre (50 cm mark). A 2.0 N weight is hung at the 20 cm mark. We wish to balance the rule by hanging a weight W at the 80 cm mark. The moment of a force about a pivot is the product of the force and the perpendicular distance from the pivot to the line of action.
一根均匀的米尺在其中点（50 cm刻度处）支起。在20 cm刻度处悬挂一个2.0 N的砝码。我们想在80 cm刻度处悬挂一个重量 W 来平衡米尺。力对支点的力矩等于力的大小与从支点到力的作用线的垂直距离的乘积。
moment = F × d
0 N × 0.30 m = 0.60 N m. The unknown weight W is 30 cm (0.30 m) from the pivot on the other side: clockwise moment = W × 0.30 m. Setting them equal gives W × 0.30 = 0.60, so W = 2.0 N. The rule balances with equal weights at equal distances.
为达到平衡（不转动），顺时针力矩 = 逆时针力矩。2.0 N 的砝码距离支点30 cm（0.30 m）：逆时针力矩 = 2.0 N × 0.30 m = 0.60 N m。未知重量 W 在另一侧距支点30 cm（0.30 m）：顺时针力矩 = W × 0.30 m。令两者相等得到 W × 0.30 = 0.60，因此 W = 2.0 N。等距离悬挂等重砝码，米尺平衡。
8. Density of an Irregular Object | 不规则物体的密度
An irregular stone has a mass of 360 g measured on a balance. To find its volume, we use a displacement method: a measuring cylinder contains 50 cm³ of water; after gently lowering the stone, the water level rises to 80 cm³. Density is mass per unit volume.
一块不规则石头的质量为360克（用天平测得）。为测出其体积，我们采用排水法：量筒中装有50 cm³的水；轻轻放入石头后，水面上升到80 cm³。密度是单位体积的质量。
ρ = m ÷ V
m³, which is denser than most common rocks, suggesting it might contain metal ores. Remember that 1 g/cm³ = 1000 kg/m³.
石头体积 = 最终体积 - 初始体积 = 80 cm³ - 50 cm³ = 30 cm³。质量 = 360 g。因此密度 ρ = 360 g ÷ 30 cm³ = 12 g/cm³。这相当于12000 kg/m³，比大多数常见岩石密度大，可能含有金属矿物。记住 1 g/cm³ = 1000 kg/m³。
9. Interpreting a Speed-Time Graph | 解读速度-时间图
A cyclist's journey is recorded on a speed-time graph. From 0 to 10 s, the line slopes upward from 0 m/s to 8 m/s, showing constant acceleration. From 10 s to 25 s, the line is horizontal at 8 m/s — constant speed. From 25 s to 30 s, the line slopes down to 0 m/s, showing deceleration. A speed-time graph provides acceleration from the gradient and distance from the area under the graph.
一位自行车手的行程以速度-时间图记录。从0到10秒，线条从0 m/s上升到8 m/s，表示匀加速。从10秒到25秒，线条保持在8 m/s水平——匀速。从25秒到30秒，线条下降至0 m/s，表示减速。速度-时间图的斜率给出加速度，线下面积给出行驶距离。
a = gradient; distance = area under graph
tance = 8 m/s × 15 s = 120 m. Deceleration distance = area of second triangle = ½ × 5 s × 8 m/s = 20 m. Total distance = 180 m. This analysis is typical of Cambridge exam graph questions.
加速阶段：a = (8 - 0) ÷ 10 = 0.8 m/s²。加速段距离 = 三角形面积 = ½ × 底 × 高 = ½ × 10 s × 8 m/s = 40 m。匀速段距离 = 8 m/s × 15 s = 120 m。减速段距离 = 第二个三角形面积 = ½ × 5 s × 8 m/s = 20 m。总距离 = 180 m。这种分析是剑桥考试图形题的典型内容。
10. Energy Transformations in a Roller Coaster | 过山车中的能量转换
energy (GPE) into kinetic energy (KE). Ignoring friction, mechanical energy is conserved. GPE = m × g × h, and KE = ½ × m × v².
一辆质量为500 kg的过山车被拉到第一个坡顶，离地30 m。然后它向下滑动，将重力势能（GPE）转化为动能（KE）。忽略摩擦，机械能守恒。GPE = m × g × h，KE = ½ × m × v²。
GPE = m g h; KE = ½ m v²
(150 000 × 2) ÷ 500 = 600, so v ≈ 24.5 m/s. In reality, some energy is lost as heat and sound, so the actual speed is lower. This conservation principle is fundamental in physics.
在顶端，GPE = 500 kg × 10 N/kg × 30 m = 150 000 J。在底部，所有 GPE 转化为 KE（假设无能量损失）：150 000 J = ½ × 500 kg × v²。求解：v² = (150 000 × 2) ÷ 500 = 600，所以 v ≈ 24.5 m/s。现实中部分能量转化为热能和声能，所以实际速度更低。这一守恒原理是物理学的基础。
11. Terminal Velocity of a Skydiver | 跳伞者的终极速度
A skydiver jumps from a plane. Initially, gravity is the dominant force, causing acceleration. As speed increases, air resistance (drag) builds up. Eventually, the upward drag equals the downward weight. At this point, the resultant force is zero, and the skydiver falls at a constant maximum speed called terminal velocity. This is a classic example of balanced forces in a fluid.
一名跳伞者从飞机上跳下。起初，重力占主导，导致加速。随着速度增加，空气阻力（拖曳力）逐渐增大。最终，向上的阻力等于向下的重量。此时合力为零，跳伞者以恒定的最大速度下落，即终极速度。这是流体中受力平衡的经典例子。
Weight = Drag → zero resultant force → terminal velocity
When the parachute opens, the surface area increases dramatically, so drag becomes much larger than weight. The skydiver decelerates rapidly until a new, much lower terminal velocity is reached for a safe landing. A velocity-time graph would show an initial curved acceleration, then a horizontal line at terminal velocity, then a sharp deceleration after parachute deployment, and finally a lower constant speed.
当降落伞打开时，表面积急剧增大，因此阻力远大于重量。跳伞者迅速减速，直至达到一个新的、低得多的终极速度，以便安全着陆。速度-时间图会显示起初的曲线加速，然后是终极速度的水平线，接着是开伞后的急剧减速，最后是较低的恒定速度。
12. Combined Forces and Motion | 合力与运动
. To find the acceleration, we must first calculate the resultant force. Resultant force = driving force - resistive force = 3000 N - 800 N = 2200 N forward.
一辆1000 kg的汽车以3000 N的驱动力向前行驶。同时有800 N的总阻力（摩擦和空气阻力）向后作用。为了求加速度，我们必须先计算合力。合力 = 驱动力 - 阻力 = 3000 N - 800 N = 2200 N，方向向前。
F_resultant = F_driving - F_resistive
Now apply Newton' .2 m/s². The car accelerates at 2.2 m/s². If the driving force were equal to the resistive force, the resultant force would be zero, and the car would move at constant velocity. Always draw a free-body diagram to identify all forces before starting calculations.
现在应用牛顿第二定律：a = F ÷ m = 2200 N ÷ 1000 kg = 2.2 m/s²。汽车以2.2 m/s²加速。如果驱动力等于阻力，合力将为零，汽车将匀速运动。在开始计算前，一定要画受力图以识别所有作用力。
Published by TutorHao | Physics Revision Series | aleveler.com
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