Year 11 AQA Physics Unit Test Mock Paper Walkthrough | Year 11 AQA 物理单元测试模拟卷解析

📚 Year 11 AQA Physics Unit Test Mock Paper Walkthrough | Year 11 AQA 物理单元测试模拟卷解析

This article presents a full walkthrough of a mock unit test for Year 11 AQA GCSE Physics, covering key topics from Paper 1 including energy, electricity, particle model of matter and atomic structure. Each question is broken down with step-by-step reasoning, key equations and exam tips to help you master the required practical skills and calculation techniques.

本文对一份 Year 11 AQA GCSE 物理单元模拟试卷进行全程解析，涵盖试卷一的核心主题：能量、电学、物质粒子模型和原子结构。每道题都配有逐步推导、关键公式和应试技巧，帮助同学们扎实掌握必备的实践技能与计算方法。

---

1. Kinetic Energy Calculation | 动能计算

Question: A car of mass 1200 kg accelerates from rest to a speed of 25 m/s. Calculate its kinetic energy. [2 marks]

题目：一辆质量为 1200 kg 的汽车从静止加速到 25 m/s，求它的动能。[2 分]

Kinetic energy is the energy stored in a moving object. The relevant equation is E_k = ½ m v².

动能是运动物体储存的能量，对应公式为 Eₖ = ½ m v²。

Eₖ = ½ m v²

Substitute the values: m = 1200 kg, v = 25 m/s. Then E_k = 0.5 × 1200 × (25)².

代入数值：质量 m = 1200 kg，速度 v = 25 m/s。计算 Eₖ = 0.5 × 1200 × (25)²。

5² = 625), then multiply: 0.5 × 1200 × 625 = 600 × 625 = 375 000 J.

先算平方 (25² = 625)，再相乘：0.5 × 1200 × 625 = 600 × 625 = 375 000 J。

The kinetic energy of the car is 375 kJ. Always include the unit joules (J) and convert to kilojoules when appropriate to match significant figures.

汽车的动能为 375 kJ。务必写上单位焦耳 (J)，并根据有效数字要求转换为千焦。

---

2. Gravitational Potential Energy | 重力势能

of 12 m. Calculate the change in gravitational potential energy. (g = 9.8 N/kg) [2 marks]

题目：起重机将 500 kg 的重物竖直提升 12 m。计算重力势能的变化量。(g = 9.8 N/kg) [2 分]

Gravitational potential energy (GPE) depends on mass, gravitational field strength and height. The formula is E_p = m g h.

重力势能取决于质量、重力场强度和高度，公式为 Eₚ = m g h。

Eₚ = m g h

Plug in the numbers: m = 500 kg, g = 9.8 N/kg, h = 12 m. E_p = 500 × 9.8 × 12.

代入数据：m = 500 kg, g = 9.8 N/kg, h = 12 m，Eₚ = 500 × 9.8 × 12。

hen 4900 × 12 = 58 800 J. The load gains 58 800 J of gravitational potential energy.

先算 500 × 9.8 = 4900，再乘 12 得 58 800 J。重物获得了 58 800 J 的重力势能。

AQA sometimes uses g = 9.8 N/kg; do not accidentally use 10 unless the question states it.

注意 AQA 常使用 g = 9.8 N/kg，除非题目明确说明，否则不要擅自使用 10。

---

3. Power and Efficiency | 功率与效率

Question: A motor does 48 000 J of useful work in 15 seconds. The total electrical energy supplied is 60 000 J. Calculate the power output and the efficiency of the motor. [4 marks]

题目：一台电动机在 15 秒内做有用功 48 000 J，电源供给的总能量为 60 000 J。计算电动机的输出功率和效率。[4 分]

Power is the rate of doing work. P = W / t, where W is work done and t is time.

功率是做功的快慢，P = W / t，W 为做功，t 为时间。

P = W ÷ t

Power output = 48 000 J ÷ 15 s = 3200 W (or 3.2 kW).

输出功率 = 48 000 J ÷ 15 s = 3200 W (3.2 kW)。

Efficiency = (useful energy output ÷ total energy input) × 100%.

效率 = (有效输出能量 ÷ 总输入能量) × 100%。

Efficiency = (E_useful ÷ E_input) × 100%

Substitute: (48 000 ÷ 60 000) × 100% = 0.80 × 100% = 80%. The motor is 80% efficient; the remaining 20% is wasted as thermal energy and sound.

代入：(48 000 ÷ 60 000) × 100% = 0.80 × 100% = 80%。电动机效率为 80%，剩余 20% 以内能（热）和声能等形式耗散。

---

4. Specific Heat Capacity | 比热容

Question: 0.5 kg of water is heated, and its temperature rises from 20 °C to 40 °C. The specific heat capacity of water is 4200 J/(kg °C). Calculate the energy transferred to the water. [2 marks]

题目：加热 0.5 kg 水，水温由 20°C 升至 40°C。水的比热容为 4200 J/(kg °C)。计算传递给水的能量。[2 分]

The energy change during heating is given by ΔE = m c Δθ.

温度变化时热量的计算公式为 ΔE = m c Δθ。

ΔE = m c Δθ

Here m = 0.5 kg, c = 4200 J/(kg °C), Δθ = 40 - 20 = 20 °C.

本题中 m = 0.5 kg，c = 4200 J/(kg °C)，温差 Δθ = 40 - 20 = 20 °C。

imes; 4200 × 20 = 0.5 × 84 000 = 42 000 J. The water absorbs 42 000 J (42 kJ).

ΔE = 0.5 × 4200 × 20 = 0.5 × 84 000 = 42 000 J。水吸收了 42 000 J (42 kJ) 的能量。

---

5. Ohm's Law and Resistance | 欧姆定律与电阻

Question: A resistor has a resistance of 10 Ω. A current of 0.5 A passes through it. Calculate the potential difference across the resistor. [1 mark]

题目：一个 10 Ω 的电阻中流过 0.5 A 的电流，求电阻两端的电势差（电压）。[1 分]

Ohm's law links potential difference (V), current (I) and resistance (R): V = I R.

欧姆定律描述了电压 V、电流 I 和电阻 R 的关系：V = I R。

V = I R

Put in the numbers: V = 0.5 A × 10 Ω = 5.0 V. The p.d. is 5 volts.

代入计算：V = 0.5 A × 10 Ω = 5.0 V。电压为 5 伏特。

---

广告 / Sponsored

6. Series Circuit Analysis | 串联电路分析

Question: Two resistors, 10 Ω and 15 Ω, are connected in series to a 12 V battery. Calculate the total resistance, the circuit current, and the potential difference across each resistor. [4 marks]

题目：10 Ω 和 15 Ω 的两个电阻串联后接到 12 V 电池上。求总电阻、电路中的电流以及每个电阻两端的电压。[4 分]

For series resistors, total resistance R_total = R₁ + R₂.

串联电阻的总电阻为 R_总 = R₁ + R₂。

R_total = 10 Ω + 15 Ω = 25 Ω.

总电阻 R_总 = 10 Ω + 15 Ω = 25 Ω。

Use Ohm's law to find the current: I = V ÷ R_total = 12 V ÷ 25 Ω = 0.48 A.

利用欧姆定律求电流：I = V ÷ R_总 = 12 V ÷ 25 Ω = 0.48 A。

The same current flows through each resistor in series. Then V₁ = I × R₁ = 0.48 A × 10 Ω = 4.8 V; V₂ = I × R₂ imes; 15 Ω = 7.2 V. Check: 4.8 V + 7.2 V = 12 V, which satisfies the supply voltage.

串联电路中各处电流相等。因此 V₁ = I × R₁ = 0.48 × 10 = 4.8 V；V₂ = 0.48 × 15 = 7.2 V。验证：4.8 V + 7.2 V = 12 V，与电源电压一致。

---

7. Electrical Energy and Cost | 电能与电费

Question: A 2000 W electric heater is used for 3 hours. Calculate the energy transferred in kilowatt-hours (kWh) and the cost of running the heater if electricity costs 15 pence per kWh. [4 marks]

题目：一个 2000 W 的电暖器使用了 3 小时。计算消耗的电能（以千瓦时为单位）以及电费，假设电价为每千瓦时 15 便士。[4 分]

convert the power to kilowatts: 2000 W = 2 kW. Energy (kWh) = power (kW) × time (h).

先把功率换算为千瓦：2000 W = 2 kW。电能 (千瓦时) = 功率 (kW) × 时间 (h)。

E = P t

Energy used = 2 kW × 3 h = 6 kWh.

消耗电能 = 2 kW × 3 h = 6 kWh。

Cost = energy (kWh) × price per kWh = 6 × 15 p = 90 p.

费用 = 电能 (kWh) × 单价 = 6 × 15 便士 = 90 便士 (0.90 英镑)。

---

8. Density Calculation | 密度计算

Question: A metal block has a mass of 540 g and a volume of 200 cm³. Calculate the density of the metal. Give your answer in g/cm³ and in kg/m³. [3 marks]

题目：一块金属的质量为 540 g，体积为 200 cm³。计算该金属的密度，分别用 g/cm³ 和 kg/m³ 表示。[3 分]

Density is mass per unit volume: ρ = m ÷ V.

密度等于质量除以体积：ρ = m ÷ V。

ρ = m ÷ V

In g/cm³: ρ = 540 g ÷ 200 cm³ = 2.7 g/cm³;.

以 g/cm³ 表示：ρ = 540 g ÷ 200 cm³ = 2.7 g/cm³;。

ply by 1000: 2.7 g/cm³ × 1000 = 2700 kg/m³;. This is typical for aluminium.

换算为 kg/m³，乘以 1000：2.7 g/cm³ × 1000 = 2700 kg/m³;。该密度与铝的密度接近。

---

9. Specific Latent Heat | 比潜热

Question: Calculate the energy required to melt 0.2 kg of ice at 0 °C. The specific latent heat of fusion of water is 334 000 J/kg. [2 marks]

题目：计算将 0.2 kg 0°C 的冰完全熔化为水所需要的能量。冰的熔化比潜热为 334 000 J/kg。[2 分]

s not change. The energy needed is E = m L, where L is the specific latent heat of fusion.

熔化过程中温度不变。所需能量 E = m L，其中 L 为熔化比潜热。

E = m L

2 kg × 334 000 J/kg = 66 800 J. So 66.8 kJ of energy must be transferred to the ice.

E = 0.2 kg × 334 000 J/kg = 66 800 J。必须向冰传递 66.8 kJ 的能量。

---

10. Radioactive Decay and Half-life | 放射性衰变与半衰期

Question: A radioactive sample has an initial count rate of 800 counts per second. The half-life of the isotope is 2 days. Calculate the count rate after 6 days. [2 marks]

题目：某放射性样品的初始计数率为每秒 800 次，该同位素的半衰期为 2 天。求 6 天后的计数率。[2 分]

ed: time elapsed = 6 days, half-life = 2 days, so number of half-lives = 6 ÷ 2 = 3.

首先确定经过的半衰期个数：经过时间为 6 天，半衰期为 2 天，所以半衰期个数 = 6 ÷ 2 = 3。

After each half-life the count rate halves. Start with 800, after 1 half-life → 400, after 2 → 200, after 3 → 100 counts/s.

每经过一个半衰期，计数率减半。从 800 开始，1 个半衰期后 → 400，2 个后 → 200，3 个后 → 100 次/秒。

Alternatively, use final count rate = initial count rate × (½)ⁿ, where n = 3: 800 × (1/8) = 100 counts/s.

也可用公式：最终计数率 = 初始计数率 × (½)ⁿ，其中 n = 3，即 800 × (1/8) = 100 次/秒。

---

Published by TutorHao | Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）