GCSE Physics: Typical Example Problems Explained | GCSE 物理：典型例题详解

📚 GCSE Physics: Typical Example Problems Explained | GCSE 物理：典型例题详解

Mastering GCSE Physics requires more than memorising facts — it demands the ability to apply concepts to unfamiliar situations. In this article, we work through ten carefully selected example problems that mirror the style and difficulty of GCSE exam questions. Each solution is broken down step by step, with paired English and Chinese explanations to strengthen your understanding and boost your confidence.

掌握 GCSE 物理不仅需要记忆知识点，更要求将概念应用到陌生的情境中。本文精心挑选了十道典型例题，难度和风格贴近真实考试。每道题的解答都配有详细步骤，并以中英对照的方式呈现，帮助你加深理解、提升信心。

1. Kinematics: Uniform Acceleration | 运动学：匀加速运动

A car accelerates uniformly from rest to 20 m/s in 10 s. Calculate the acceleration and the distance covered during this time.

一辆汽车从静止开始匀加速，10 秒后速度达到 20 m/s。求加速度和在此时间内行驶的距离。

Use the equation v = u + at. With u = 0, v = 20 m/s, t = 10 s, the acceleration a = (v – u) / t = 2 m/s².

使用公式 v = u + at。代入 u = 0, v = 20 m/s, t = 10 s，加速度 a = (v – u) / t = 2 m/s²。

The distance s = ut + ½at² = 0 + ½ × 2 × 10² = 100 m. You can also use average velocity: s = (u+v)/2 × t = 10 × 10 = 100 m.

距离 s = ut + ½at² = 0 + ½ × 2 × 10² = 100 m。也可用平均速度： s = (u+v)/2 × t = 10 × 10 = 100 m。

a = 2 m/s², s = 100 m

加速度 2 m/s²，距离 100 m

2. Newton’s Second Law | 牛顿第二定律

A 1200 kg car experiences a resultant force of 3600 N forwards. Find its acceleration. If it starts from rest, how fast will it be moving after 5 s?

一辆 1200 kg 的汽车受到 3600 N 的向前的合力。求它的加速度。如果它从静止开始，5 秒后速度有多大？

Applying F = ma gives a = F / m = 3600 N / 1200 kg = 3 m/s².

应用 F = ma 得 a = F / m = 3600 N / 1200 kg = 3 m/s²。

Using v = u + at with u = 0, t = 5 s, we find v = 0 + 3 × 5 = 15 m/s.

用 v = u + at，u = 0, t = 5 s，得 v = 0 + 3 × 5 = 15 m/s。

a = 3 m/s², v = 15 m/s

加速度 3 m/s²，末速度 15 m/s

3. Energy Efficiency | 能量效率

An electric heater transfers 50 000 J of electrical energy to the surroundings. Only 38 000 J is transferred as useful thermal energy to the room. Calculate the efficiency of the heater.

一台电暖器向周围传递了 50 000 J 的电能，其中只有 38 000 J 以有用的热能传递到房间里。求该暖器的效率。

Efficiency = (useful energy output / total energy input) × 100%. So efficiency = (38 000 J / 50 000 J) × 100% = 76%.

效率 = (有用的输出能量 / 总输入能量) × 100%。因此效率 = (38 000 J / 50 000 J) × 100% = 76%。

The remaining 24% is dissipated mostly as sound and heat in the heater’s own components. This is why no device is 100% efficient.

剩下的 24% 主要耗散为声音和暖器自身元件的热量。这就是为什么任何设备都不会 100% 效率。

4. Specific Heat Capacity | 比热容

A student heats 1.5 kg of water from 25 °C to 85 °C. The specific heat capacity of water is 4200 J/(kg °C). Determine the energy transferred to the water.

一名学生加热 1.5 kg 的水，温度从 25 °C 上升到 85 °C。水的比热容为 4200 J/(kg °C)。求传递给水的能量。

The energy Q = m c Δθ. Here m = 1.5 kg, c = 4200 J/(kg °C), and Δθ = 85 °C – 25 °C = 60 °C.

能量 Q = m c Δθ。这里 m = 1.5 kg，c = 4200 J/(kg °C)，Δθ = 85 °C – 25 °C = 60 °C。

Q = 1.5 × 4200 × 60 = 378 000 J, or 378 kJ.

Q = 1.5 × 4200 × 60 = 378 000 J，即 378 kJ。

Δθ = 60 °C, Q = 378 kJ

温度变化 60 °C，能量 378 kJ

5. Ohm’s Law and Series Circuits | 欧姆定律与串联电路

Two resistors, 6 Ω and 12 Ω, are connected in series to a 9 V battery. Find: (a) the total resistance, (b) the current in the circuit, and (c) the potential difference across each resistor.

两个电阻，6 Ω 和 12 Ω，串联后接在 9 V 电池上。求：(a) 总电阻，(b) 电路中的电流，(c) 每个电阻两端的电压。

For series, R_total = R₁ + R₂ = 6 + 12 = 18 Ω.

串联时，总电阻 R_total = R₁ + R₂ = 6 + 12 = 18 Ω。

Using Ohm’s law I = V / R_total = 9 V / 18 Ω = 0.5 A.

根据欧姆定律 I = V / R_total = 9 V / 18 Ω = 0.5 A。

V₁ = I × R₁ = 0.5 × 6 = 3 V, and V₂ = I × R₂ = 0.5 × 12 = 6 V. Notice how the supply voltage divides in proportion to the resistances.

V₁ = I × R₁ = 0.5 × 6 = 3 V，V₂ = I × R₂ = 0.5 × 12 = 6 V。注意供电电压按照电阻值比例分配。

Resistor	Resistance (Ω)	Voltage (V)
R₁	6	3
R₂	12	6

I = 0.5 A, V₁ = 3 V, V₂ = 6 V

电流 0.5 A，R₁ 电压 3 V，R₂ 电压 6 V

6. Electrical Power | 电功率

A 230 V mains lamp is rated at 36 W. Calculate the current flowing through it and its resistance. How much energy does it consume in 5 minutes?

一盏额定 230 V、36 W 的灯泡。求通过它的电流和电阻。它在 5 分钟内消耗多少能量？

Using P = I V, I = P / V = 36 W / 230 V ≈ 0.157 A.

由 P = I V 得 I = P / V = 36 W / 230 V ≈ 0.157 A。

Resistance can be found from R = V / I = 230 / 0.157 ≈ 1468 Ω, or directly from P = V² / R, so R = V² / P = 230² / 36 ≈ 1469 Ω.

电阻可由 R = V / I = 230 / 0.157 ≈ 1468 Ω，或直接用 P = V² / R，R = V² / P = 230² / 36 ≈ 1469 Ω。

Energy E = P × t = 36 W × (5 × 60 s) = 36 × 300 = 10 800 J.

能量 E = P × t = 36 W × (5 × 60 s) = 36 × 300 = 10 800 J。

I ≈ 0.16 A, R ≈ 1.47 kΩ, E = 10.8 kJ

电流约 0.16 A，电阻约 1.47 kΩ，能量 10.8 kJ

7. Wave Speed Equation | 波速公式

A water wave has a wavelength of 2.5 m and a frequency of 0.8 Hz. Calculate its speed. If the frequency is doubled while the speed remains constant, what happens to the wavelength?

一个水波的波长为 2.5 m，频率为 0.8 Hz。求波速。如果频率加倍而波速不变，波长会如何变化？

The wave speed v = f λ. With f = 0.8 Hz, λ = 2.5 m, v = 0.8 × 2.5 = 2 m/s.

波速 v = f λ。代入 f = 0.8 Hz, λ = 2.5 m，得 v = 0.8 × 2.5 = 2 m/s。

If speed stays at 2 m/s and frequency becomes 1.6 Hz, then λ = v / f = 2 / 1.6 = 1.25 m. So the wavelength halves when frequency doubles.

如果波速保持 2 m/s，频率变为 1.6 Hz，则 λ = v / f = 2 / 1.6 = 1.25 m。可见频率加倍时波长减半。

v = f λ

波速 2 m/s；频率翻倍，波长减半

8. Half-Life Calculations | 半衰期计算

A radioactive isotope has a half-life of 4 hours. The initial count rate from a sample is 640 counts per minute. Determine the count rate after 16 hours.

某种放射性同位素半衰期为 4 小时。某样品初始计数率为 640 次/分钟。求 16 小时后的计数率。

Number of half-lives n = total time / half-life = 16 h / 4 h = 4.

半衰期个数 n = 总时间 / 半衰期 = 16 h / 4 h = 4。

After each half-life, the count rate is halved. Thus, after 4 half-lives, activity = initial × (½)⁴ = 640 × (½)⁴.

每经过一个半衰期，计数率减半。因此，经过 4 个半衰期后，活性 = 初始值 × (½)⁴ = 640 × (½)⁴。

(½)⁴ = 1/16, so count rate = 640 / 16 = 40 counts per minute.

(½)⁴ = 1/16，所以计数率 = 640 / 16 = 40 次/分钟。

n = 4, final count rate = 40 min⁻¹

4 个半衰期，最终计数率 40 次/分钟

9. Hooke’s Law | 胡克定律

A spring stretches 3.0 cm when a 6.0 N load is attached. Determine the spring constant k. How much would the spring extend under a 10.0 N load, assuming the elastic limit is not exceeded?

一根弹簧在 6.0 N 的负载下伸长 3.0 cm。求弹簧常数 k。假设未超过弹性限度，在 10.0 N 的负载下弹簧会伸长多少？

Hooke’s law: F = k x, where x must be in metres. 3.0 cm = 0.030 m. So k = F / x = 6.0 N / 0.030 m = 200 N/m.

胡克定律：F = k x，其中 x 必须以米为单位。3.0 cm = 0.030 m。因此 k = F / x = 6.0 N / 0.030 m = 200 N/m。

For F = 10.0 N, extension x = F / k = 10.0 / 200 = 0.050 m = 5.0 cm.

当 F = 10.0 N 时，伸长量 x = F / k = 10.0 / 200 = 0.050 m = 5.0 cm。

k = 200 N/m, extension = 5.0 cm

弹簧常数 200 N/m，伸长量 5.0 cm

10. Momentum Conservation | 动量守恒

A 3.0 kg cart moves at 4.0 m/s to the right on a frictionless track. It collides with a stationary 2.0 kg cart, and they couple together. Calculate the velocity of the combined carts after the collision.

一辆 3.0 kg 的小车以 4.0 m/s 的速度向右在无摩擦轨道上运动。它与一辆静止的 2.0 kg 小车碰撞并连在一起。求碰撞后两车共同的速度。

Total momentum before = m₁ u₁ + m₂ u₂ = (3.0 × 4.0) + (2.0 × 0) = 12 kg m/s.

碰撞前总动量 = m₁ u₁ + m₂ u₂ = (3.0 × 4.0) + (2.0 × 0) = 12 kg m/s。

After collision, total mass m = 3.0 + 2.0 = 5.0 kg. By conservation of momentum, (m₁+m₂) v = 12, so v = 12 / 5.0 = 2.4 m/s to the right.

碰撞后总质量 m = 3.0 + 2.0 = 5.0 kg。根据动量守恒，(m₁+m₂) v = 12，所以 v = 12 / 5.0 = 2.4 m/s，方向向右。

Notice that the velocity decreases because mass increases — momentum is conserved, not velocity.

注意速度减小了，因为质量增加了——守恒的是动量，而不是速度。

v = 2.4 m/s

碰撞后速度 2.4 m/s

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GCSE Physics: Typical Example Problems Explained | GCSE 物理：典型例题详解