Case Study Practical Drills for Year 11 Edexcel Physics | Year 11 Edexcel 物理：案例分析实战演练

📚 Case Study Practical Drills for Year 11 Edexcel Physics | Year 11 Edexcel 物理：案例分析实战演练

Welcome to your focused revision on applying physics to real-world scenarios. This article guides you through a series of case studies – from motion and forces to electricity and waves – each carefully built around the Edexcel Year 11 specification. You will learn how to break down a problem, identify relevant principles, and solve it step by step.

欢迎来到物理实际应用专题复习。本文将通过一系列案例分析——从运动与力到电学与波动——每个案例都紧扣 Edexcel 十一年级考纲设计。你将学会如何拆解问题、识别相关原理并逐步求解。

1. Stopping Distance of a Car | 汽车的制动距离

A car is travelling at 20 m/s on a dry road. The driver reacts to an obstacle in 0.7 s. The braking deceleration is 6.0 m/s². Calculate the total stopping distance.

一辆汽车在干燥路面上以 20 m/s 行驶。司机对前方障碍做出反应需 0.7 秒。刹车的减速度为 6.0 m/s²。计算总停车距离。

First, find the thinking distance: s_t × (−6.0) × s. Solve: s = 400 / 12 = 33.3 m. Total stopping distance = 14 + 33.3 = 47.3 m.

首先，计算反应距离：s_t = v × t = 20 × 0.7 = 14 m。然后对刹车过程使用 v² = u² + 2as，末速度 v = 0：0 = 20² + 2 × (−6.0) × s。解得 s = 400 / 12 = 33.3 m。总停车距离 = 14 + 33.3 = 47.3 m。

---

2. Lifting a Load with a Pulley System | 滑轮组提升重物

A worker uses a pulley system to lift a 250 kg mass at constant speed. The system has three rope sections supporting the load. Calculate the effort force needed.

一位工人使用滑轮组以恒定速度提升 250 kg 的重物。该系统有三段绳索支撑负载。计算所需的动力大小。

For a three‑rope pulley, the tension in each rope supports one‑ weight (ignoring friction). Weight W = mg = 250 × 9.8 = 2450 N. Effort F = W / 3 ≈ 817 N.

对于三段绳索的滑轮组，每段绳的张力承担重量的三分之一（忽略摩擦）。重量 W = mg = 250 × 9.8 = 2450 N。动力 F = W / 3 ≈ 817 N。

Explain why in practice the effort is slightly higher: friction at the pulley axle and the weight of the ropes themselves increase the required force.

请解释实际中所需动力为何略高：滑轮轴的摩擦以及绳索自身的重量都会增大所需力。

---

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

A roller coaster car of mass 500 kg is at the top of a 30 m hill. Neglecting friction, find its speed at the bottom.

一个质量为 500 kg 的过山车位于 30 m 高的坡顶。忽略摩擦，求其到达坡底时的速度。

Gravitational potential energy lost = mgh = 500 × 9.8 × 30 = 147 000 J. This becomes kinetic energy: ½ mv² = 147 000. v = √2 × 147 000 / 500 = √588 ≈ 24.2 m/s.

减少的重力势能 = mgh = 500 × 9.8 × 30 = 147 000 J。这些能量转化为动能：½ mv² = 147 000。v = √2 × 147 000 / 500 = √588 ≈ 24.2 m/s。

400 N acting over a track length of 80 m, the work done against friction is 400 × 80 = 32 000 J. The new kinetic energy becomes 147 000 − 32 000 = 115 000 J, giving a lower speed of about 21.4 m/s.

若存在平均 400 N 的摩擦力，作用长度为 80 m，则克服摩擦做功为 400 × 80 = 32 000 J。新的动能变为 147 000 − 32 000 = 115 000 J，得到更低的速度约 21.4 m/s。

---

4. Series and Parallel Resistors in a Circuit | 电路中的串联与并联电阻

A 12 V battery is connected to two resistors: R₁ = 4 Ω and R₂ power dissipated in each resistor when they are connected (a) in series, (b) in parallel.

一个 12 V 电池连接两个电阻：R₁ = 4 Ω，R₂ = 6 Ω。求电阻 (a) 串联和 (b) 并联时通过每个电阻的电流及其消耗的功率。

(a) Series: total R = 10 Ω. Current I = 12/10 = 1.2 A. P₁ = I²R₁ = 1.44 × 4 = 5.76 W; P₂ = 1.44 × 6 = 8.64 W. (b) Parallel: each gets full 12 V. I₁ = 12/4 = 3 A, P₁ = 12 × 3 = 36 W. I₂ = 12/6 = 2 A, P₂ = 24 W.

(a) 串联：总电阻 = 10 Ω。电流 I = 12/10 = 1.2 A。P₁ = I²R₁ = 1.44 × 4 = 5.76 W；P₂ = 1.44 × 6 = 8.64 W。(b) 并联：每个电阻获得完整的 12 V 电压。I₁ = 12/4 = 3 A，P₁ = 12 × 3 = 36 W。I₂ = 12/6 = 2 A，P₂ = 24 W。

Notice how parallel connection draws much more total current (5 A) compared to series (1.2 A). This affects battery drain and safety.

注意并联时总电流（5 A）远大于串联（1.2 A）。这会影响电池消耗和安全性。

---

5. Wave Speed on a Spring | 弹簧上的波速

A physics student generates a transverse wave on a slinky spring. The wave has a frequency of 2.5 Hz and a wavelength of 0.80 m. Calculate the wave speed and the time taken for the wave to travel 6.0 m.

一位物理学生在螺旋弹簧上产生横波。波的频率为 2.5 Hz，波长为 0.80 m。计算波速及波传播 6.0 m 所需的时间。

ave speed v = f × λ = 2.5 × 0.80 = 2.0 m/s. Time t = distance / speed = 6.0 / 2.0 = 3.0 s.

波速 v = f × λ = 2.5 × 0.80 = 2.0 m/s。时间 t = 距离 / 速率 = 6.0 / 2.0 = 3.0 s。

If the tension in the spring is increased, wave speed increases. For the same frequency, the wavelength would become longer.

若增大弹簧张力，波速增大。在频率不变的情况下，波长将变长。

---

6. Refraction of Light through a Glass Block | 光通过玻璃块折射

A ray of light strikes a rectangular glass block at an angle of incidence of 40°. The refractive index of glass is 1.5. Calculate the angle of refraction inside the glass.

一束光线以 40° 的入射角射向矩形玻璃块。玻璃的折射率为 1.5。计算玻璃内部的折射角。

Using Snell's law: n_air sin i = n_glass 0° = 1.5 × sin r. sin r = sin 40° / 1.5 ≈ 0.643 / 1.5 = 0.4287. r = sin⁻¹(0.4287) ≈ 25.4°.

使用斯涅尔定律：n_空气 sin i = n_玻璃 sin r。1.0 × sin 40° = 1.5 × sin r。sin r = sin 40° / 1.5 ≈ 0.643 / 1.5 = 0.4287。r = sin⁻¹(0.4287) ≈ 25.4°。

Describe what happens to the ray as it leaves the opposite side of the block parallel to the table. The emergent ray is parallel to the incident ray because the two opposite faces are parallel – only a lateral shift occurs.

描述光线从玻璃块另一侧射出时的情况：由于两个相对面平行，出射光线与入射光线平行——只产生侧向位移。

---

广告 / Sponsored

7. Radioactive Decay and Half‑Life | 放射性衰变与半衰期

A sample of iodine‑131 has an initial activity of 800 Bq. Its half‑life is 8 days. Calculate the activity after 24 days and the fraction of nuclei remaining.

一份碘‑131 样品的初始活度为 800 Bq，半衰期为 8 天。计算 24 天后的活度及剩余原子核的比例。

Number of half‑lives = 24 / 8 = 3. Activity after 3 half‑lives = 800 × (½)³ = 800 × 1/8 = 100 Bq. The fraction remaining = 1/8 = 0.125 (or 12.5%).

经过的半衰期个数 = 24 / 8 = 3。3 个半衰期后的活度 = 800 × (½)³ = 800 × 1/8 = 100 Bq。剩余比例 = 1/8 = 0.125（即 12.5%）。

Explain why the count rate measured by a detector would be lower than the true activity because of background radiation and detector efficiency.

解释为什么探测器测得的计数率会低于真实活度，原因包括本底辐射和探测器效率。

---

8. Electromagnetic Induction | 电磁感应

A student moves a bar magnet into a coil connected to a sensitive ammeter. The needle deflects to the right, then returns to zero when the magnet stops moving. Explain what happens when the magnet is pulled out quickly.

一位学生将条形磁铁插入连接灵敏电流计的线圈中。指针向右偏转，磁铁停止运动时指针归零。解释当磁铁快速抽出时会发生什么。

When the magnet moves into the coil, the magnetic flux through the coil increases, inducing an emf that drives current in one direction (deflection right). When pulled out, flux decreases, inducing an emf in the opposite direction, so the needle deflects to the left. The faster the motion, the larger the deflection.

磁铁插入线圈时，穿过线圈的磁通量增加，产生感应电动势并驱动电流沿某一方向流动（指针右偏）。抽出时磁通量减小，感应出相反方向的电动势，指针因此向左偏转。运动越快，指针偏转越大。

This demonstrates Faraday’s law: the induced emf is proportional to the rate of change of magnetic flux. Lenz’s law determines the direction.

这验证了法拉第定律：感应电动势与磁通量变化率成正比。楞次定律决定其方向。

---

9. Pressure in a Hydraulic System | 液压系统中的压强

area 0.050 m². A force of 150 N is applied to the small piston. Calculate the maximum load the large piston can lift.

一台液压千斤顶的小活塞面积为 2.0 × 10⁻³ m²，大活塞面积为 0.050 m²。对小活塞施加 150 N 的力。计算大活塞能举起的最大载荷。

Pressure is transmitted equally: P = F₁/A₁ = 150 / (2.0 × 10⁻³) = 75 000 Pa. Force on large piston F₂ = P × A₂ = 75 000 × 0.050 = 3750 N. This corresponds to a mass of about 3750 / 9.8 ≈ 383 kg.

压强等量传递：P = F₁/A₁ = 150 / (2.0 × 10⁻³) = 75 000 Pa。大活塞上的力 F₂ = P × A₂ = 75 000 × 0.050 = 3750 N。相应质量约为 3750 / 9.8 ≈ 383 kg。

aller than that of the small piston, conserving work (Fd = constant, neglecting friction).

注意大活塞移动的距离远小于小活塞，这符合功的守恒（Fd = 常数，忽略摩擦）。

---

10. Momentum Conservation in an Explosion | 爆炸中的动量守恒

Two stationary ice skaters push apart. Skater A (mass 65 kg) moves to the left at 1.2 m/s. Skater B (mass 50 kg) moves to the right. Calculate her velocity.

两位静止的溜冰者互相推开。溜冰者 A（质量 65 kg）以 1.2 m/s 向左运动。溜冰者 B（质量 50 kg）向右运动。求她的速度。

Initial total momentum = 0. After push: m_Av_A + m_Bv_B = 0. Taking right as positive: 65 × (−1.2) + 50 × v_B = 0. −78 + 50 v_B = 0 → v_B = 78 / 50 = 1.56 m/s to the right.

初始总动量 = 0。推后：m_Av_A + m_Bv_B = 0。以向右为正方向：65 × (−1.2) + 50 × v_B = 0。−78 + 50 v_B = 0 → v_B = 78 / 50 = 1.56 m/s，向右。

Thus momentum is conserved in the isolated system, and the more massive skater moves slower.

因此动量在这个孤立系统中守恒，且质量较大的溜冰者运动较慢。

Published by TutorHao | Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）