📚 Year 11 OCR Physics: Case Study Practice Drill | 案例分析实战演练
In the OCR GCSE Physics exam, applying concepts to real-world scenarios is key. This article presents nine case studies covering forces, energy, electricity, waves, radioactivity, and more, guiding you through step-by-step analysis and calculations.
在OCR GCSE物理考试中，将概念应用到现实世界情境至关重要。本文通过九个案例，涵盖力、能量、电路、波、放射性等内容，带领你进行逐步分析和计算。
1. Stopping a Car on a Wet Road | 湿滑路面上的汽车刹车
A car of mass 1200 kg is travelling at 20 m/s on a wet road. The driver brakes hard, and the tyres skid. The average friction force acting on the car is 4800 N. We need to find the braking distance.
一辆质量为1200 kg的汽车在湿滑路面上以20 m/s行驶。司机猛踩刹车，轮胎打滑。作用在汽车上的平均摩擦力为4800 N。我们需要求出刹车距离。
The car's kinetic energy is converted into work done against friction. Initial kinetic energy:
Eₖ = ½ m v² = ½ × 1200 × (20)² = 240 000 J
汽车的动能转化为克服摩擦力所做的功。初始动能：Eₖ = ½ m v² = ½ × 1200 × (20)² = 240 000 J。
= W. Therefore d = Eₖ / F = 240 000 J / 4800 N = 50 m. The braking distance is 50 metres.
摩擦力做功：W = F × d。由于所有动能都用于做功，Eₖ = W。因此 d = Eₖ / F = 240 000 J / 4800 N = 50 m。刹车距离为50米。
Notice that braking distance is proportional to v². If the speed doubles to 40 m/s, kinetic energy becomes 4 times larger, and the stopping distance becomes 200 m. This huge increase explains why speed limits are lower on wet roads.
注意刹车距离与v²成正比。如果车速加倍至40 m/s，动能增大为原来的4倍，刹车距离变为200 m。这种剧烈增长解释了为什么湿滑路面限速更低。
| Speed (m/s) | Kinetic Energy (J) | Braking Distance (m) |
| 20 | 240 000 | 50 |
| 40 | 960 000 | 200 |
2. Fault in a Parallel Lighting Circuit | 并联照明电路故障
Three identical lamps are connected in parallel to a 230 V mains supply. Each lamp has a resistance of 920 Ω. One lamp's filament breaks. Determine the current drawn from the supply before and after the fault.
三盏相同的灯泡并联连接到230 V市电电源。每个灯泡的电阻为920 Ω。其中一盏灯泡的灯丝烧断。求故障前后从电源汲取的电流。
urrent I = V / R = 230 V / 920 Ω = 0.25 A. The total current from the supply is the sum: Itotal = 3 × 0.25 A = 0.75 A. The total resistance can also be found: 1/Rtotal = 1/920 + 1/920 + 1/920 = 3/920, so Rtotal = 920/3 ≈ 307 Ω.
故障前，每盏灯泡流过的电流 I = V / R = 230 V / 920 Ω = 0.25 A。电源总电流为三者之和：Itotal = 3 × 0.25 A = 0.75 A。也可求出总电阻：1/Rtotal = 1/920 + 1/920 + 1/920 = 3/920，Rtotal = 920/3 ≈ 307 Ω。
urrent is 2 × 0.25 A = 0.50 A. The failed lamp does not affect the brightness of the others because they are in parallel.
当一盏灯泡烧坏，该支路变为断路。剩余两盏灯泡两端电压仍为230 V，因此每盏仍流过0.25 A。新的总电流为 2 × 0.25 A = 0.50 A。损坏的灯泡不会影响其他灯泡的亮度，因为它们并联连接。
If a 3 A fuse protects the circuit, it will not blow in either case. However, if the lamps were mistakenly wired in series, a single failure would break the entire circuit and all lamps would go out.
如果电路使用3 A保险丝，两种情况下保险丝都不会熔断。然而，如果灯泡被错误地串联在一起，一处损坏就会使整个电路断开，所有灯泡都会熄灭。
3. Water Waves Approaching a Shore | 水波靠近海岸
Water waves in deep ocean travel at 4.0 m/s with a frequency of 2.0 Hz. As they approach the shore, they enter shallow water where the speed drops to 2.4 m/s. The frequency remains unchanged. Calculate the wavelengths in deep and shallow water, and explain the change in direction.
深海水波以4.0 m/s的速度传播，频率为2.0 Hz。当它们靠近海岸时，进入浅水区，波速降至2.4 m/s。频率保持不变。计算深水和浅水中的波长，并解释传播方向的变化。
Using the wave equation v = f λ. In deep water: λ_deep = vdeep In shallow water: λ_shallow = v_shallow / f = 2.4 / 2.0 = 1.2 m. The wavelength decreases.
使用波动方程 v = f λ。深水中：λ_deep = vdeep / f = 4.0 / 2.0 = 2.0 m。浅水中：λ_shallow = v_shallow / f = 2.4 / 2.0 = 1.2 m。波长减小。
Because the wave speed changes at the boundary, the wavefronts bend – this is refraction. If the waves approach at an angle, the change in speed causes the wave direction to move closer to the normal in shallow water, just like light entering a denser medium.
由于波速在界面处改变，波阵面会发生弯曲——这就是折射。当波浪以一定角度入射时，波速的变化会导致波在浅水中的传播方向更靠近法线，就像光进入光密介质一样。
4. Carbon-14 Dating of an Ancient Bone | 碳-14测定古代骨头年代
Living organisms contain carbon-14 (¹⁴C) which decays with a half-life of 5730 years. The activity of a fresh bone is 0.25 Bq per gram of carbon. An ancient bone sample has an activity of 0.03125 Bq per gram. How old is the bone?
活体生物含有碳-14 (¹⁴C)，其半衰期为5730年。新鲜骨骼的活度为每克碳0.25 Bq。一个古代骨骼样本的活度为每克0.03125 Bq。该骨骼的年龄是多少？
be written as (½)³, so three half-lives have elapsed. Age = 3 × 5730 years = 17 190 years.
首先，求出¹⁴C的剩余比例：0.03125 / 0.25 = 0.125。这可以写成 (½)³，因此经过了三个半衰期。年龄 = 3 × 5730 年 = 17 190 年。
An alternative method uses the formula: A = A₀ (½)t/T, where T is the half-life. Taking logs gives the same result. This dating method is reliable only up to about 50 000 years due to the very low activity in older samples.
另一种方法是使用公式：A = A₀ (½)t/T，其中T为半衰期。取对数可得相同结果。由于较古老样本的活度极低，这种年代测定方法仅在大约5万年以内有效。
| Half-lives passed | Fraction remaining | Age (years) |
| 1 | 0.5 | 5 730 |
| 2 | 0.25 | 11 460 |
| 3 | 0.125 | 17 190 |
5. Spring-Powered Toy Car Launch | 弹簧动力玩具车发射
A toy car of mass 0.20 kg uses a compressed spring to launch itself forward. The spring constant is 800 N/m, and the spring is compressed by 0.050 m before release. Calculate the car's speed if all elastic potential energy transfers to kinetic energy.
一辆质量为0.20 kg的玩具车利用压缩弹簧向前弹射。弹簧劲度系数为800 N/m，释放前弹簧被压缩0.050 m。若所有弹性势能全部转化为动能，计算车的速度。
gained: Eₖ = ½ m v², so 1.0 = ½ × 0.20 × v². Solving gives v² = 1.0 / 0.10 = 10, thus v = √10 ≈ 3.16 m/s.
弹性势能：Eₑ = ½ k x² = ½ × 800 × (0.050)² = ½ × 800 × 0.0025 = 1.0 J。获得的动能：Eₖ = ½ m v²，因此 1.0 = ½ × 0.20 × v²。求解得 v² = 1.0 / 0.10 = 10，故 v = √10 ≈ 3.16 m/s。
In reality, some energy is lost to friction and sound. If the energy transfer is only 80% efficient, the useful kinetic energy is 0.80 J. Then v = √2 × 0.80 / 0.20 = √8 ≈ 2.83 m/s. This highlights the importance of efficiency in energy transfers.
实际上，部分能量会因摩擦和声音而损失。如果能量转化效率仅为80%，则有用动能为0.80 J。此时 v = √2 × 0.80 / 0.20 = √8 ≈ 2.83 m/s。这凸显了能量传递中效率的重要性。
6. Torque on a DC Motor Coil | 直流电动机线圈的力矩
A DC motor has a rectangular coil of 50 turns, with sides of length 0.050 m and width 0.030 m, placed in a uniform magnetic field of 0.20 T. A current of 0.50 A flows through the coil. Calculate the force on one side and the maximum torque produced.
一个直流电动机有一个50匝的矩形线圈，边长0.050 m、宽0.030 m，置于0.20 T的匀强磁场中。线圈中通有0.50 A的电流。计算单边所受的力以及产生的最大力矩。
0015 m². Maximum torque when the coil plane is parallel to the field: τ = n B I A = 50 × 0.20 × 0.50 × 0.0015 =
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