Year 11 Cambridge Physics Unit Test Mock Paper with Solutions | 剑桥物理单元测试模拟卷解析

📚 Year 11 Cambridge Physics Unit Test Mock Paper with Solutions | 剑桥物理单元测试模拟卷解析

This article presents a mock unit test designed for Year 11 Cambridge IGCSE Physics, covering key topics from mechanics, energy, waves, electricity, and thermal physics. Each question is followed by a detailed bilingual solution to strengthen understanding and exam technique.

本文为11年级剑桥IGCSE物理单元测试模拟卷，涵盖力学、能量、波、电学和热物理核心知识点。每道题目后配有详细的中英双语解析，帮助巩固概念、提升解题技巧。

1. Question 1: Kinematics | 运动学

A car accelerates uniformly from rest to 20 m/s in 8.0 seconds. Calculate its acceleration.

一辆汽车从静止开始匀加速，8.0秒后速度达到20 m/s。计算其加速度。

Solution / 解析:

Use the equation a = (v – u) / t. Here u = 0, v = 20 m/s, t = 8.0 s. So a = (20 – 0) / 8.0 = 2.5 m/s².

使用公式 a = (v – u) / t。这里初速度 u = 0，末速度 v = 20 m/s，时间 t = 8.0 s。因此 a = (20 – 0) / 8.0 = 2.5 m/s²。

Always remember to include the correct unit for acceleration: metres per second squared (m/s²).

务必加上正确的加速度单位：米每二次方秒 (m/s²)。

---

2. Question 2: Resultant Force and Newton’s Second Law | 合力与牛顿第二定律

A 5.0 kg box is pushed across a frictionless surface with a constant force of 15 N. What is the acceleration of the box?

一个质量为5.0 kg的箱子在无摩擦表面上受到15 N的恒力推动。箱子的加速度是多少？

Solution / 解析:

Newton's second law: F = m × a. Rearranged: a = F / m = 15 N / 5.0 kg = 3.0 m/s².

牛顿第二定律：F = m × a。变形得 a = F / m = 15 N / 5.0 kg = 3.0 m/s²。

Note that 'N' is equivalent to kg·m/s², so the unit cancels correctly.

注意 N 等于 kg·m/s²，因此单位约分后得到正确结果。

---

3. Question 3: Work and Gravitational Potential Energy | 功与重力势能

a 200 kg load vertically by 12 m. Calculate the work done by the crane. (Take g = 10 m/s²)

一台起重机将200 kg的重物竖直吊起12 m。计算起重机所做的功。（取 g = 10 m/s²）

Solution / 解析:

e weight of the load: weight = m × g = 200 × 10 = 2000 N. Work = 2000 N × 12 m = 24 000 J.

功 = 力 × 沿力方向移动的距离。所需力等于重物的重力：重力 = m × g = 200 × 10 = 2000 N。功 = 2000 N × 12 m = 24 000 J。

ional potential energy (GPE = mgh). Always check if the question expects answer in joules.

这也等于物体增加的重力势能 (GPE = mgh)。务必注意题目是否要求以焦耳为单位作答。

---

4. Question 4: Density and Buoyancy | 密度与浮力

side length of 0.10 m. Calculate the density of the material. Will the cube float in water? (Density of water = 1000 kg/m³)

一个实心立方体质量为2.4 kg，边长为0.10 m。计算该材料的密度。该立方体在水中会漂浮吗？（水的密度 = 1000 kg/m³）

Solution / 解析:

f cube = side³ = (0.10)³ = 0.001 m³. Density ρ = mass / volume = 2.4 / 0.001 = 2400 kg/m³.

立方体体积 = 边长³ = (0.10)³ = 0.001 m³。密度 ρ = 质量 / 体积 = 2.4 / 0.001 = 2400 kg/m³。

Since 2400 kg/m³ > 1000 kg/m³, the cube will sink. An object floats only if its density is less than the fluid.

由于 2400 kg/m³ > 1000 kg/m³，该立方体将会下沉。物体只有当密度小于流体时才会漂浮。

---

5. Question 5: Ohm’s Law and Resistance | 欧姆定律与电阻

A resistor is connected to a 6.0 V battery. The current flowing through it is 0.40 A. Calculate the resistance. What current would flow if the voltage is increased to 9.0 V?

一个电阻器连接到6.0 V的电池上，流过的电流为0.40 A。计算其电阻。如果电压增至9.0 V，电流会变为多少？

Solution / 解析:

Using Ohm' R = V / I = 6.0 V / 0.40 A = 15 Ω. For the new voltage: I = V / R = 9.0 V / 15 Ω = 0.60 A.

使用欧姆定律：R = V / I = 6.0 V / 0.40 A = 15 Ω。对于新电压：I = V / R = 9.0 V / 15 Ω = 0.60 A。

Assuming the resistor obeys Ohm’s law (constant resistance at constant temperature).

假设该电阻遵循欧姆定律（温度恒定下电阻不变）。

---

6. Question 6: Wave Speed, Frequency and Wavelength | 波速、频率和波长

A sound wave has a frequency of 500 Hz and a wavelength of 0.68 m. Calculate its speed. What would be the wavelength of the same wave if the frequency is increased to 1000 Hz? (Assume the speed remains constant.)

一个声波的频率为500 Hz，波长为0.68 m。计算其波速。若频率增加到1000 Hz，波长变为多少？（假设波速不变）

Solution / 解析:

= f × λ. Speed v = 500 Hz × 0.68 m = 340 m/s. At 1000 Hz, λ = v / f = 340 / 1000 = 0.34 m.

波动方程：v = f × λ。波速 v = 500 Hz × 0.68 m = 340 m/s。频率为1000 Hz时，λ = v / f = 340 / 1000 = 0.34 m。

Notice that wavelength halves when frequency doubles, a key relationship for a given wave speed.

注意在波速一定的情况下，频率翻倍时波长减半，这是非常重要的关系。

---

7. Question 7: Specific Heat Capacity | 比热容

How much energy is needed to raise the temperature of 2.0 kg of water from 20°C to 80°C? (Specific heat capacity of water = 4200 J/(kg·°C))

将2.0 kg的水从20°C加热到80°C需要多少能量？（水的比热容 = 4200 J/(kg·°C)）

Solution / 解析:

Energy = m × c × Δθ, where Δθ = 80 – 20 = 60°C. So E = 2.0 × 4200 × 60 = 504 000 J (or 504 kJ).

能量 = m × c × Δθ，其中 Δθ = 80 – 20 = 60°C。因此 E = 2.0 × 4200 × 60 = 504 000 J（即 504 kJ）。

Always convert final answer to kilojoules if the number is large, but check the question's requirement.

如果数字较大，可将答案转换为千焦，但要检查题目是否有特定要求。

---

广告 / Sponsored

8. Question 8: Half-life and Radioactive Decay | 半衰期与放射性衰变

A radioactive sample has a half-life of 6 hours. If the initial activity is 800 counts per minute, what will be the activity after 24 hours?

某放射性样品的半衰期为6小时。若初始活性为每分钟800计数，24小时后的活性是多少？

Solution / 解析:

s each half-life: after 1: 400; after 2: 200; after 3: 100; after 4: 50 counts per minute.

半衰期数 = 总时间 / 半衰期 = 24 h / 6 h = 4。每过一个半衰期活性减半：1次后：400；2次后：200；3次后：100；4次后：50 计数/分钟。

You can also use formula A = A₀ (1/2)ⁿ, where n = number of half-lives.

也可用公式 A = A₀ (1/2)ⁿ，n 为半衰期个数。

---

9. Question 9: Moments and Equilibrium | 力矩与平衡

A uniform metre rule is pivoted at its 50 cm mark. A 2.0 N weight is hung at the 20 cm mark. At which point must a 4.0 N weight be hung to balance the rule?

一根均匀的米尺在50 cm刻度处被支起。在20 cm处悬挂一个2.0 N的重物。若要使尺子平衡，应在何处悬挂一个4.0 N的重物？

Solution / 解析:

moments. Take pivot at 50 cm. The 2.0 N at 20 cm gives anticlockwise moment: distance = 50 – .15 m = 15 cm from pivot. Since it must be on opposite side, hang at 50 + 15 = 65 cm mark.

力矩原理：顺时针力矩之和 = 逆时针力矩之和。支点在50 cm处。2.0 N在20 cm处产生逆时针力矩：距离 = 50 – 20 = 30 cm = 0.30 m。力矩 = 2.0 N × 0.30 m = 0.60 N·m。为平衡，顺时针力矩也需为0.60 N·m。所以 4.0 N × d = 0.60，d = 0.15 m = 15 cm。由于需在支点另一侧，悬挂位置为50 + 15 = 65 cm刻度处。

Always check the direction of moments; units must be consistent (convert cm to m for standard units, but comparing cm is acceptable as long as consistent).

务必检查力矩方向；单位要保持一致（转换为米为标准单位，但若都用厘米且相等则无妨）。

---

10. Question 10: Pressure and Gas Laws | 压强与气体定律

a and volume is 400 cm³. The piston is pushed in until the volume becomes 100 cm³, keeping temperature constant. What is the new pressure?

气体被活塞封闭在气缸内。初始压强为1.0 × 10⁵ Pa，体积为400 cm³。保持温度不变，将活塞推入使体积变为100 cm³。新的压强是多少？

Solution / 解析:

Pa, V₁ = 400 cm³, V₂ = 100 cm³. So p₂ = (p₁V₁) / V₂ = (1.0×10⁵ × 400) / 100 = 4.0×10⁵ Pa.

波义耳定律：p₁V₁ = p₂V₂（温度恒定）。p₁ = 1.0×10⁵ Pa，V₁ = 400 cm³，V₂ = 100 cm³。因此 p₂ = (p₁V₁) / V₂ = (1.0×10⁵ × 400) / 100 = 4.0×10⁵ Pa。

Notice that volume units cancel if they are the same; no need to convert to m³ here.

注意如果体积单位相同，可以直接约去，无需转换为立方米。

---

11. Question 11: Series and Parallel Circuits | 串联与并联电路

Two resistors, 10 Ω and 15 Ω, are connected in series across a 12 V battery. Calculate the total resistance and the current. If they are reconnected in parallel, what is the total resistance?

两个电阻10 Ω和15 Ω串联连接在12 V电池上。计算总电阻和电流。若改为并联，总电阻为多少？

Solution / 解析:

Series / 串联	Rtotal = R₁ + R₂ = 10 + 15 = 25 Ω
Current / 电流	I = V / Rtotal = 12 / 25 = 0.48 A
Parallel / 并联	1/Rtotal = 1/10 + 1/15 = 3/30 + 2/30 = 5/30 → Rtotal = 30/5 = 6 Ω

In parallel circuits, total resistance is less than the smallest individual resistance. This is a common exam trick.

在并联电路中，总电阻小于任何一个单独的电阻值，这是考试中常见的陷阱点。

---

12. Exam Tips and Common Mistakes | 考试技巧与常见错误

Always include units in your final answers; missing units can lose marks. Show all working steps clearly – even if the final answer is wrong, method marks can be awarded. Be careful with prefixes like centi (10⁻²) and kilo (10³); convert to base units when using standard formulas. In electricity, distinguish between series and parallel current/voltage rules.

最终答案务必包含单位；缺少单位会丢分。清晰写出所有解题步骤——即使最终答案有误，也可能获得方法分。注意厘清厘(10⁻²)和千(10³)等前缀；在使用标准公式时转换为基本单位。在电学中，注意区分串联和并联的电流/电压规律。

Finally, manage your time: do not spend too long on one question. If a question seems too difficult, move on and return to it later.

最后，管理好时间：不要在一道题上花太多时间。若某题太困难，先跳过去，最后再回来处理。

Published by TutorHao | Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）