Case Study Practical Exercises | 案例分析实战演练

📚 Case Study Practical Exercises | 案例分析实战演练

Case studies provide a powerful way to link physics concepts to real-world situations. In this revision guide, you will work through ten practical scenarios that test your understanding of motion, forces, energy, electricity, waves and matter. Each case challenges you to analyse data, apply equations and explain observations – exactly the skills needed for your KS3 Edexcel physics assessments.

案例分析是把物理概念与现实情境联系起来的有效方法。在这份复习指南中，你将通过十个实战场景来检验对运动、力、能量、电学、波动和物质的理解。每个案例都要求你分析数据、应用公式并解释观察结果，这些正是 KS3 Edexcel 物理考试所需要的技能。

1. Analysing a Sprinter’s Race | 分析短跑运动员的比赛

A sprinter’s 100 m race is split into three segments by the coach. The recorded distances and times are shown in the table below.

一位短跑运动员的 100 米赛跑被教练分成三段。记录的距离和时间如下表所示。

Segment	Distance (m)	Time (s)
Start to 20 m	20	3.0
20 m to 50 m	30	2.5
50 m to 100 m	50	4.5

We can calculate the average speed for each segment using speed = distance ÷ time.

我们可以用 速度 = 距离 ÷ 时间 计算每一段的平均速度。

average speed = distance moved ÷ time taken

For the first segment: speed = 20 m ÷ 3.0 s = 6.67 m/s. The second segment gives 30 m ÷ 2.5 s = 12.0 m/s, and the final segment yields 50 m ÷ 4.5 s ≈ 11.1 m/s.

第一段：速度 = 20 m ÷ 3.0 s = 6.67 m/s。第二段：30 m ÷ 2.5 s = 12.0 m/s，最后一段：50 m ÷ 4.5 s ≈ 11.1 m/s。

The sprinter was fastest during the middle part of the race. This suggests a strong acceleration phase followed by a slight drop in speed as fatigue sets in.

这名运动员在比赛的中段速度最快。这表明有一个较强的加速阶段，之后随着疲劳出现，速度略有下降。

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2. Investigating Density of an Irregular Object | 探究不规则物体的密度

A metal nugget of mass 300 g is lowered into a measuring cylinder containing 50 cm³ of water. The water level rises to 80 cm³.

一块质量为 300 g 的金属块被放入装有 50 cm³ 水的量筒中。水面上升到 80 cm³。

The volume of the metal is the difference in water level: 80 cm³ – 50 cm³ = 30 cm³.

金属的体积就是水面变化的差值：80 cm³ – 50 cm³ = 30 cm³。

density = mass ÷ volume

Substituting the values: density = 300 g ÷ 30 cm³ = 10 g/cm³.

代入数值：密度 = 300 g ÷ 30 cm³ = 10 g/cm³。

A density of 10 g/cm³ is close to that of silver (about 10.5 g/cm³) or lead (11.3 g/cm³). Further tests, like checking magnetic properties or colour, would be needed to identify it precisely.

10 g/cm³ 的密度接近银 (约 10.5 g/cm³) 或铅 (11.3 g/cm³)。要进一步精确鉴定，还需测试磁性或颜色等性质。

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3. Understanding Series and Parallel Circuits | 理解串联和并联电路

Two circuits are built: one with a single lamp in series, and another with two identical lamps connected in parallel. The current is measured at several points.

搭建了两个电路：一个只有一盏灯串联，另一个有两盏相同的灯并联。在多个点测量了电流。

Circuit type	Location	Current (A)
Series	Any point in the loop	0.30
Parallel	Main branch (near cell)	0.50
Parallel	Through one lamp	0.25

In the series circuit, the current is the same everywhere because there is only one loop for the charges to flow through.

在串联电路中，电流处处相等，因为电荷只有一条回路可以流动。

In the parallel circuit, the main current (0.50 A) splits equally between the two branches (0.25 A each). The sum of the branch currents equals the current from the cell.

在并联电路中，干路电流 (0.50 A) 平均分配到两条支路中 (各 0.25 A)。各支路电流之和等于流出电池的总电流。

Voltage measurements would show the cell voltage is shared in series but remains the same across parallel branches. This explains why lamps in parallel shine with equal brightness.

电压测量则会显示串联时电池电压被各灯分担，而并联时各支路两端电压相同。这就是为什么并联的灯泡亮度一样。

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4. Energy Transfers in a Roller Coaster | 过山车中的能量转移

A roller coaster car of mass 60 kg descends from point A (height 50 m) to the lowest point B (height 5 m). We assume friction and air resistance are negligible, and take gravitational field strength as 10 N/kg.

一辆质量为 60 kg 的过山车从 A 点 (高度 50 m) 下降到最低点 B (高度 5 m)。假设摩擦和空气阻力可忽略，重力场强度取 10 N/kg。

The loss in gravitational potential energy (GPE) is calculated as:

重力势能 (GPE) 的减少量为：

ΔGPE = m × g × Δh = 60 kg × 10 N/kg × (50 m – 5 m) = 27 000 J

This energy is converted into kinetic energy (KE) at point B. Using the formula:

这些能量在 B 点全部转化为动能 (KE)。使用公式：

KE = ½ × m × v²

Setting GPE lost = KE gained gives 27 000 J = ½ × 60 kg × v². Solving for v yields v² = 900, so v = 30 m/s.

令 GPE 减少量 = KE 增加量，得 27 000 J = ½ × 60 kg × v²。解得 v² = 900，因此 v = 30 m/s。

The coaster therefore reaches a speed of 30 m/s at the bottom. In reality, some energy is always transferred as heat, so the actual speed would be slightly lower.

过山车在底部达到 30 m/s 的速度。实际上，总有一部分能量转化为热量，所以实际速度会略低。

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5. The Bouncing Ball Experiment | 弹跳球实验

A rubber ball of mass 0.10 kg is dropped from a height of 1.00 m above a hard floor. It bounces back to a maximum height of 0.70 m.

一个质量为 0.10 kg 的橡胶球从离硬地板 1.00 m 高处落下。它反弹到的最大高度为 0.70 m。

We can compare the GPE before and after the bounce to find the energy dissipated.

我们可以比较碰撞前后的重力势能，求出耗散的能量。

Initial GPE = m × g × h₁ = 0.10 kg × 10 N/kg × 1.00 m = 1.0 J

Final GPE = m × g × h₂ = 0.10 kg × 10 N/kg × 0.70 m = 0.7 J

The energy lost is 1.0 J – 0.7 J = 0.3 J, which is 30% of the original energy.

损失的能量为 1.0 J – 0.7 J = 0.3 J，占原有能量的 30%。

This ‘lost’ energy has not disappeared but has been transferred into thermal energy in the ball and the floor, as well as sound waves. You can often hear the bounce and feel a slight warming after several bounces.

这些“损失”的能量并没有消失，而是转化为了球和地板中的热能，以及声波。多次弹跳后你常能听到声音并感到轻微发热。

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6. Cooling Curve of Stearic Acid | 硬脂酸的冷却曲线

Stearic acid is heated until it becomes a clear liquid at about 80 °C. It is then allowed to cool while the temperature is recorded every 30 seconds.

将硬脂酸加热，直到约 80 °C 时变成澄清液体。然后让其自然冷却，每 30 秒记录一次温度。

Time (min)	0.0	1.0	2.0	3.0	4.0	5.0	6.0
Temp (°C)	80	72	64	56	54	54	54
Time (min)	7.0	8.0	9.0	10.0	11.0	12.0	13.0
Temp (°C)	53	50	46	41	37	34	32

From the data, the temperature falls steadily until it reaches about 54 °C. It then stays constant at 54 °C for about three minutes before decreasing again.

从数据可以看出，温度稳定下降，直到约 54 °C。接着在 54 °C 保持恒定约三分钟，然后再次下降。

The flat region represents the freezing point where liquid stearic acid turns into solid. During this change of state, energy is being released to the surroundings without a change in temperature.

这一段平台代表了凝固点，此时液态硬脂酸变为固态。在物态变化过程中，能量向外传递而温度保持不变。

This horizontal plateau is a characteristic of pure substances. Impurities would make the freezing point less sharp and usually lower.

这种水平平台是纯净物质的特征。杂质会使凝固点变得不分明，并通常使其降低。

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7. Moments and Levers in Daily Life | 日常生活中的力矩与杠杆

A mechanic uses a spanner to tighten a bolt. She applies a force of 50 N at a perpendicular distance of 0.30 m from the pivot (the bolt).

一位机修工使用扳手拧紧螺栓。她在距支点（螺栓）垂直距离 0.30 m 处施加了 50 N 的力。

moment of a force = force × perpendicular distance from pivot

The turning effect is: moment = 50 N × 0.30 m = 15 N m.

转动力矩为：力矩 = 50 N × 0.30 m = 15 N m。

If she grips the spanner closer to the bolt, at 0.20 m, the moment becomes 50 N × 0.20 m = 10 N m. To produce the same 15 N m torque from a shorter distance, a larger force would be needed.

如果她握住离螺栓更近的位置，比如 0.20 m，力矩就变为 50 N × 0.20 m = 10 N m。要在较短距离处产生同样 15 N m 的力矩，就需要更大的力。

This explains why longer spanners make it easier to loosen tight bolts – the same force creates a greater turning effect, or a smaller force can be used.

这就解释了为何长扳手更容易拧松螺栓——相同的力能产生更大的转动效果，或者可以用更小的力。

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8. Investigating Friction on Different Surfaces | 探究不同表面的摩擦力

A wooden block is pulled at a steady speed across a smooth table, a sheet of sandpaper and a carpet. The force needed to keep it moving is measured with a newton meter.

用一个木块分别在光滑桌面、砂纸和地毯上匀速拉动。用弹簧测力计测量维持运动所需的力。

Surface	Pulling force (N)
Smooth table	1.5
Sandpaper	3.2
Carpet	4.8

At constant speed, the pulling force exactly balances the friction force. Therefore, the friction force is 1.5 N on the smooth table, 3.2 N on sandpaper and 4.8 N on carpet.

匀速运动时，拉力与摩擦力恰好平衡。因此，摩擦力在光滑桌面为 1.5 N，砂纸上为 3.2 N，地毯上为 4.8 N。

Rougher surfaces have higher friction because the microscopic bumps interlock more, making it harder to slide. Lubricants or smoother materials reduce

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