📚 Year 7 Cambridge Physics: Case Study Practice | 剑桥七年级物理：案例分析实战演练
Physics comes alive when you link theory to real-world situations. Working through case studies sharpens your analytical thinking and prepares you for exam-style questions. The following eight case studies below cover the key topics of from the Year 7 Cambridge Physics syllabus: motion, forces, energy, electricity, waves and magnetism. For each one, read the scenario carefully, apply your physics knowledge step by step, and then check your answer against the worked solution.
当你把理论与现实情境联系起来时，物理就变得生动起来。通过案例研究可以锻炼你的分析思维，并为应对考试题型做好准备。下面八个案例分析涵盖了剑桥七年级物理课程的核心主题：运动、力、能量、电、波以及磁。对每一个案例，请先仔细阅读情景，一步步运用你的物理知识，然后再与分步解答进行核对。
1. How to Use These Case Studies | 如何使用这些案例
Every case study in this article follows the same three-step routine: identify the known values, choose the correct equation, and substitute with units. This is exactly the routine that earns marks in a Cambridge checkpoint or end-of-topic test.
本文中的每一个案例都遵循同样的三步流程：找出已知量、选择正确的公式、带单位代入计算。这正是在剑桥阶段性测评或单元测试中拿分的标准流程。
Step 1 - Read and highlight. Underline every number and every unit in the scenario, and note what the question is actually asking for.
第一步——阅读并标注。把情景中的每一个数字和单位都画出来，并弄清楚题目到底在求什么。
Step 2 - Write the equation first. Write the equation in symbols before you put any numbers in. A correct equation on the page can still earn a method mark even if the arithmetic slips.
第二步——先写公式。在代入任何数字之前，先用符号写出公式。即使运算出错，写在卷面上的正确公式仍可能拿到方法分。
Step 3 - Substitute, solve, and check the unit. Always carry units through the calculation and ask yourself whether the final answer is sensible.
第三步——代入、求解并检查单位。计算过程中始终带着单位，并问自己最后的答案是否合理。
- Write down: known values, unknown value, equation. | 写下：已知量、未知量、公式。
- Check: does the unit of the answer match what was asked? | 检查：答案的单位是否与题目要求一致？
- Review: state the physics reason in words, not just the number. | 回顾：用文字说出物理原因，而不只是给出数字。
2. Case Study 1: The Speedy Cyclist | 案例一：飞驰的自行车手
A cyclist completes a 15 km ride in 30 minutes. You need to determine his average speed in kilometres per hour and predict how far he can travel in 2 hours at the same speed.
一名自行车手在30分钟内完成了15公里的骑行。你需要求出他以公里/小时为单位的平均速度，并预测以同样的速度在2小时内他能行驶多远。
Question: (a) Calculate the average speed. (b) How far will the cyclist travel in 2 hours if the speed stays the same?
题目：（a）计算平均速度。（b）如果速度保持不变，该自行车手2小时内能骑行多远？
first. 30 minutes = 0.5 hours. Then use speed = distance ÷ time = 15 km ÷ 0.5 h = 30 km/h.
第（a）问：时间以分钟给出，所以先换算。30分钟 = 0.5小时。然后利用速度 = 路程 ÷ 时间 = 15 km ÷ 0.5 h = 30 km/h。
v = s ÷ t
s cancel nothing here, so the answer stays in kilometres, which is exactly what was asked.
第（b）问：把同一个公式变形即可。路程 = 速度 × 时间 = 30 km/h × 2 h = 60 km。注意这里"小时"并没有被消掉，所以答案的单位仍是千米，正好符合题目要求。
m ÷ 3600 s ≈ 8.3 m/s. The conversion factor between km/h and m/s is always divide by 3.6.
拓展：剑桥考官常常还会要求用米每秒表示速度。30 km/h = 30 × 1000 m ÷ 3600 s ≈ 8.3 m/s。km/h 与 m/s 之间的换算永远是除以 3.6。
Key idea: average speed is the total distance divided by the total time. It tells you nothing about how fast the cyclist was going at any single instant, because he may have slowed down at a junction or sped up downhill.
核心概念：平均速度等于总路程除以总时间。它并不能说明自行车手在某一瞬间到底骑得多快，因为他可能在路口减速，也可能在下坡时加速。
3. Case Study 2: The Skydiver's Fall | 案例二：跳伞运动员的下落
ces acting on her at three moments during the fall and work out her acceleration each time.
一名质量为70 kg的跳伞运动员从飞机上跳下。取 g = 10 N/kg。描述她在下落过程中三个时刻所受的力，并分别求出她的加速度。
Weight = mass × gravitational field strength = 70 kg × 10 N/kg = 700 N, acting downwards.
第一步——求重力。重力 = 质量 × 重力场强度 = 70 kg × 10 N/kg = 700 N，方向竖直向下。
W = m × g F = m × a
ation = resultant force ÷ mass = 700 N ÷ 70 kg = 10 m/s². She is effectively in free fall.
时刻一——刚跳出机舱。此时几乎没有空气阻力，所以合力为向下的 700 N。加速度 = 合力 ÷ 质量 = 700 N ÷ 70 kg = 10 m/s²。她实际上处于自由落体状态。
wards. Acceleration = 300 N ÷ 70 kg ≈ 4.3 m/s². She is still speeding up, but more slowly.
时刻二——下落途中。空气阻力增大到向上的 400 N。此时合力为 700 N − 400 N = 300 N，方向向下。加速度 = 300 N ÷ 70 kg ≈ 4.3 m/s²。她仍在加速，但加速得越来越慢。
Moment 3 - Terminal velocity. Air resistance has risen to 700 N upwards, exactly balancing her weight. The resultant force is zero, so the acceleration is zero and her speed stays constant. This steady speed is called terminal velocity.
时刻三——收尾速度。空气阻力增大到向上的 700 N，恰好与她的重力平衡。合力为零，因此加速度为零，速度保持不变。这个稳定的速度就叫做收尾速度（终端速度）。
Extra twist: when the parachute opens, air resistance may jump to about 1400 N. The resultant force is then 700 N upwards, so she decelerates at roughly 10 m/s² until the forces balance again at a much lower terminal velocity.
额外变化：当降落伞打开时，空气阻力可能骤增到约 1400 N。此时合力为向上的 700 N，于是她以约 10 m/s² 减速，直到受力再次平衡，并在一个低得多的收尾速度上稳定下来。
- Balanced forces ⇌ constant velocity (which can be zero). | 受力平衡 ⇌ 速度不变（速度也可以为零）。
- Unbalanced forces → acceleration in the direction of the resultant. | 受力不平衡 → 沿合力方向产生加速度。
- Air resistance increases with speed, so it is a self-limiting force. | 空气阻力随速度增大而增大，因此它是一种"自我限制"的力。
4. Case Study 3: The Roller Coaster Energy Audit | 案例三：过山车的能量核算
g. Ignoring friction, find the gravitational potential energy stored and the speed of the car at the bottom of the hill.
一辆质量为500 kg的过山车静止在20 m高的坡顶。取 g = 10 N/kg。忽略摩擦，求它储存的重力势能以及到达坡底时的速度。
itational potential energy. Eₚ = m × g × h = 500 kg × 10 N/kg × 20 m = 100 000 J = 100 kJ.
第一步——重力势能。Eₚ = m × g × h = 500 kg × 10 N/kg × 20 m = 100 000 J = 100 kJ。
Eₚ = m g h Eₖ = ½ m v²
all 100 000 J of potential energy becomes kinetic energy at the bottom. So Eₖ = 100 000 J.
第二步——能量守恒。如果没有能量转移到周围环境，那么 100 000 J 的势能在坡底将全部转化为动能。因此 Eₖ = 100 000 J。
speed. ½ m v² = 100 000 J, so v² = 2 × 100 000 J ÷ 500 kg = 400 m²/s², giving v = 20 m/s.
第三步——变形求速度。½ m v² = 100 000 J，所以 v² = 2 × 100 000 J ÷ 500 kg = 400 m²/s²，得 v = 20 m/s。
Real-world check: in a real theme park the answer is always slightly less than 20 m/s, because friction and air resistance dissipate some energy as thermal energy. Examiners often ask you to explain this difference using the words "energy is transferred to the thermal store of the surroundings".
现实检验：在真正的主题公园里，答案总会略小于 20 m/s，因为摩擦和空气阻力会把一部分能量以热能形式耗散掉。考官常要求你用"能量被转移到周围环境的热store中"这样的表述来解释这一差异。
- At the top: maximum Eₚ, minimum Eₖ. | 在坡顶：Eₚ 最大，Eₖ 最小。
- At the bottom: maximum Eₖ, minimum Eₚ. | 在坡底：Eₖ 最大，Eₚ 最小。
- The total energy stays the same - it only changes form. | 总能量保持不变——只是形式发生了变化。
5. Case Study 4: Inside a Torch | 案例四：手电筒内部
A torch uses two 1.5 V cells connected in series and a bulb of resistance 6 Ω. Work out the current through the bulb, the power it transfers, and the charge that flows in one minute.
一支手电筒使用两节串联的1.5 V电池和一个电阻为6 Ω的小灯泡。求通过灯泡的电流、它传输的功率，以及一分钟内流过的电荷量。
s: 1.5 V + 1.5 V = 3 V. Cells in parallel would keep the voltage at 1.5 V but last longer.
第一步——总电压。串联电池的电压相加：1.5 V + 1.5 V = 3 V。若并联，电压仍为 1.5 V，但使用时间更长。
V = I × R P = V × I Q = I × t
Step 2 - Current. Rearrange V = I × R to get I = V ÷ R = 3 V ÷ 6 Ω = 0.5 A.
第二步——电流。将 V = I × R 变形得 I = V ÷ R = 3 V ÷ 6 Ω = 0.5 A。
× 0.5 A = 1.5 W. You could also use P = I² R = 0.5² × 6 = 1.5 W, which is a useful check.
第三步——功率。P = V × I = 3 V × 0.5 A = 1.5 W。你也可以用 P = I² R = 0.5² × 6 = 1.5 W，这是一个很好的验证方法。
- Charge. Convert time to seconds: 1 minute = 60 s. Then Q = I × t = 0.5 A × 60 s = 30 C.
第四步——电荷量。先把时间换算成秒：1分钟 = 60 s。然后 Q = I × t = 0.5 A × 60 s = 30 C。
Circuit skills to remember: an ammeter goes in series with the component, a voltmeter goes in parallel across it, and every circuit needs a complete loop back to the cell before any current can flow.
需要记住的电路技能：电流表与被测元件串联，电压表与被测元件并联，而且任何电流流动之前，电路必须形成回到电池的完整回路。
6. Case Study 5: The Cliff Echo | 案例五：悬崖回声
A student claps her hands in front of a cliff and hears the echo 1.2 s later. The speed of sound in air is 340 m/s. How far away is the cliff, and what is the wavelength of a 170 Hz note?
一名学生在悬崖前拍手，1.2 s 后听到回声。空气中的声速为 340 m/s。悬崖有多远？一个 170 Hz 的音符波长是多少？
Step 1 - Recognise the double journey. The sound travels to the cliff and back again, so the total distance is twice the distance to the cliff. This is the single most common trap in echo questions.
第一步——识别"双程"。声音先传到悬崖再返回，所以总路程是到悬崖距离的两倍。这是回声题中最常见的陷阱。
Step 2 - Total distance. Distance = speed × time = 340 m/s × 1.2 s = 408 m.
第二步——总路程。路程 = 速度 × 时间 = 340 m/s × 1.2 s = 408 m。
Step 3 - Halve it. Distance to the cliff = 408 m ÷ 2 = 204 m.
第三步——取一半。到悬崖的距离 = 408 m ÷ 2 = 204 m。
v = f × λ
Rearrange v = f × λ to get λ = v ÷ f = 340 m/s ÷ 170 Hz = 2 m. The wavelength is 2 metres.
第四步——波长。将 v = f × λ 变形得 λ = v ÷ f = 340 m/s ÷ 170 Hz = 2 m。波长为2米。
ho would return in only 408 m ÷ 1500 m/s ≈ 0.27 s. This is why sonar works so well at sea.
拓展：声音在水中的传播速度要快得多，约为 1500 m/s。如果同样的拍手发生在水下，回声只需 408 m ÷ 1500 m/s ≈ 0.27 s 就会返回。这就是声呐在海上如此有效的原因。
- Sound needs a medium - it cannot travel through a vacuum. | 声音需要介质——它无法在真空中传播。
- Higher frequency = higher pitch; bigger amplitude = louder sound. | 频率越高音调越高；振幅越大声音越响。
- Echoes are reflections of sound waves from a hard surface. | 回声是声波从坚硬表面反射形成的。
7. Case Study 6: The Laser and the Mirror | 案例六：激光与平面镜
A laser beam strikes a plane mirror at 35° to the normal. Find the angle of reflection and the angle between the incident and reflected rays. Then explain what happens if the mirror is rotated by 10°.
一束激光以与法线成35°的角射到平面镜上。求反射角以及入射光线与反射光线之间的夹角。然后解释如果把镜子旋转10°会发生什么。
Step 1 - Apply the law of reflection. The angle of incidence equals the angle of reflection, both measured from the normal, which is the dashed line perpendicular to the mirror surface.
第一步——应用反射定律。入射角等于反射角，两者都从法线量起，法线就是垂直于镜面的那条虚线。
i = r
Step 2 - Angle of reflection. Since i = 35°, then r = 35° as well.
第二步——反射角。因为 i = 35°，所以 r 也等于 35°。
wo rays lie on opposite sides of the normal, so the angle between them is 35° + 35° = 70°.
第三步——两条光线之间的夹角。两条光线位于法线两侧，所以它们的夹角为 35° + 35° = 70°。
Step 4 - Rotating the mirror. If the mirror turns through 10°, the normal also turns through 10°, and the reflected ray turns through twice that amount, 20°. This doubling effect is why small adjustments to a mirror in a periscope or a laser experiment cause large movements of the spot.
第四步——旋转镜子。如果镜子转过 10°，法线也转过 10°，而反射光线会转过两倍的角度，即 20°。这种"倍增效应"正解释了为什么潜望镜或激光实验中对镜子做微小调整，光斑却会移动很大距离。
Refraction link: when light passes from air into glass it slows down and bends towards the normal; going back out into air it speeds up and bends away from the normal. Refraction is what makes a pencil look bent in a glass of water.
与折射的联系：光从空气进入玻璃时速度变慢，并向法线偏折；从玻璃射回空气时速度变快，并偏离法线。正是折射使铅笔在装水的玻璃杯中看起来像是折断的。
8. Case Study 7: The Electromagnet Crane | 案例七：电磁铁起重机
A student winds 50 turns of insulated wire around an iron nail and connects it to a 6 V battery. The current is 0.4 A. Explain how the strength of the electromagnet can be changed and why a scrap-yard crane uses an electromagnet rather than a permanent magnet.
一名学生把50匝绝缘导线绕在一根铁钉上，并接到6 V电池上。电流为0.4 A。请解释如何改变电磁铁的强度，以及废料场起重机为什么使用电磁铁而不是永磁体。
Point 1 - More turns, stronger magnet. Doubling the turns to 100 would roughly double the magnetic effect, because more coils mean more aligned magnetic fields.
要点一——匝数越多，磁性越强。把匝数增加到100匝，磁效应大约会翻倍，因为更多的线圈意味着更多方向一致的磁场。
Point 2 - Bigger current, stronger magnet. Increasing the current by raising the voltage increases the magnetic field strength. If the current doubles to 0.8 A, the field strength roughly doubles too.
要点二——电流越大，磁性越强。通过升高电压来增大电流，可以增强磁场强度。如果电流增大到 0.8 A，磁场强度也大约翻倍。
Point 3 - An iron core concentrates the field. Soft iron is used because it is easily magnetised and easily demagnetised
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图