📚 Year 11 OCR Physics Unit Test Mock Paper: Waves in Matter | Year 11 OCR 物理单元测试模拟卷：物质中的波
This mock paper focuses on the 'Waves in Matter' topic from the OCR GCSE Physics specification. It is designed to test your understanding of wave properties, behaviour, and applications. Each question is followed by detailed bilingual solutions to help you revise effectively and achieve your best in the unit test.
这套模拟卷紧扣 OCR GCSE 物理大纲中“物质中的波”主题，旨在考查你对波的属性、行为及应用的掌握。每道题后都配有详细的双语解析，助你高效备考并在单元测试中冲刺高分。
1. Question 1: Wave Speed and Period | 问题1：波速与周期
A water wave has a frequency of 12 Hz and a wavelength of 0.25 m. Calculate (a) the wave speed and (b) the period of the wave.
一个水波的频率为 12 Hz，波长为 0.25 m。计算 (a) 波速，(b) 波的周期。
is found using the wave equation v = f × λ. Substituting in gives v = 12 × 0.25 = 3.0 m/s.
使用波速公式 v = f × λ，代入得 v = 12 × 0.25 = 3.0 m/s。
al of frequency: T = 1/f = 1/12 ≈ 0.0833 s. This is the time for one complete oscillation.
周期 T 是频率的倒数：T = 1/f = 1/12 ≈ 0.0833 s。这是完成一次完整振动所需的时间。
Always remember: wave speed is constant in a given medium, so if frequency increases, wavelength must decrease to keep the speed the same.
切记：在给定介质中波速恒定，因此若频率增大，波长必须减小以保持波速不变。
2. Question 2: Transverse and Longitudinal Waves | 问题2：横波与纵波
Describe the difference between transverse and longitudinal waves. Give one example of each. Explain how particles in a sound wave move.
描述横波与纵波的区别，并各举一例。解释声波中粒子的运动方式。
In a transverse wave, the oscillations are perpendicular to the direction of energy transfer. Examples include light waves, water ripples, and waves on a string. The particles move up and down as the wave travels horizontally.
横波中，振动方向与能量传播方向垂直。例如光波、水波涟漪和绳波。当波水平传播时，粒子上下振动。
In a longitudinal wave, the oscillations are parallel to the direction of energy transfer. Sound waves in air are longitudinal; the particles vibrate back and forth, creating compressions (high pressure) and rarefactions (low pressure).
纵波中，振动方向与能量传播方向平行。空气中的声波是纵波；粒子前后振动，形成压缩（高压）和稀疏（低压）。
For a sound wave travelling through air, air molecules oscillate from side to side in the same line as the wave direction, producing alternating regions of high and low density.
声波在空气中传播时，空气分子沿波传播的同一直线左右振动，产生疏密相间的区域。
3. Question 3: Law of Reflection | 问题3：反射定律
A ray of light strikes a plane mirror. The angle between the incident ray and the mirror surface is 60°. Determine the angle of incidence and state the angle of reflection.
一束光射向平面镜，入射光线与镜面之间的夹角为 60°。求入射角，并说明反射角。
he mirror surface. Since the ray makes 60° with the mirror, the angle to the normal is 90° – 60° = 30°. Therefore, the angle of incidence is 30°.
入射角是入射光线与法线之间的夹角，而非与镜面的夹角。光线与镜面成 60°，则与法线夹角为 90° – 60° = 30°，因此入射角为 30°。
According to the law of reflection, the angle of reflection equals the angle of incidence: 30°. The incident ray, reflected ray and normal all lie in the same plane.
根据反射定律，反射角等于入射角：30°。入射光线、反射光线和法线位于同一平面内。
Draw the mirror as a straight line; the normal is perpendicular to it. The ray approaches at 30° to the normal and reflects symmetrically on the other side.
绘制平面镜为一直线；法线垂直于镜面。光线以与法线 30° 的方向入射，在另一侧对称反射。
4. Question 4: Refraction and Wave Speed | 问题4：折射与波速
A light ray passes from air into glass at an angle. It bends towards the normal. Explain why this happens in terms of wave speed and give a diagram description.
一束光从空气斜射入玻璃，光线向法线偏折。从波速的角度解释原因，并描述示意图。
When light travels from air (faster medium) to glass (slower medium), its speed decreases. The wavefronts arrive at the boundary at an angle; the part of the wavefront entering the glass first slows down, while the rest continues at a higher speed. This causes the wave to change direction.
光从空气（较快介质）进入玻璃（较慢介质）时速度变慢。波前以一定角度到达界面；先进入玻璃的那部分波前先减速，其余部分仍以较高速度行进，导致波的方向改变。
As a result, the ray bends towards the normal. The frequency of the wave remains constant, so the wavelength decreases in the denser medium.
因此，光线向法线偏折。波的频率保持不变，所以在较密介质中波长变短。
The diagram should show the boundary, incident ray in air, refracted ray in glass closer to the normal, and the normal line. Wavefronts should be drawn as parallel lines that are closer together in the glass.
示意图应显示界面、空气中的入射光线、玻璃中更靠近法线的折射光线以及法线。波前应画成平行线，在玻璃中间距更小。
5. Question 5: Electromagnetic Spectrum | 问题5：电磁波谱
List the main regions of the electromagnetic spectrum in order of increasing frequency. State one practical use for microwaves and one for X-rays.
按频率递增的顺序列出电磁波谱的主要区域，并分别说明微波和 X 射线的一种实际用途。
The spectrum from lowest to highest frequency is: radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays. A good memory aid is 'Rabbits Mate In Very Unusual Xciting Gardens'.
频率从低到高依次为：无线电波、微波、红外线、可见光、紫外线、X 射线、伽马射线。记忆口诀为“无线微红见紫 X 伽”。
| Radiation / 辐射类型 | Frequency order / 频率顺序 | Typical use / 典型应用 |
| Radio waves / 无线电波 | Lowest / 最低 | Broadcasting / 广播 |
| Microwaves / 微波 | Cooking, satellite communication / 烹饪、卫星通信 | |
| Infrared / 红外线 | Thermal imaging, remote controls / 热成像、遥控器 | |
| Visible light / 可见光 | Seeing, photography / 视觉、摄影 | |
| Ultraviolet / 紫外线 | Security marking, tanning / 防伪标记、美黑 | |
| X-rays / X 射线 | Medical imaging, airport security / 医学成像、机场安检 | |
| Gamma rays / 伽马射线 | Highest / 最高 | Cancer treatment, sterilisation / 癌症治疗、灭菌 |
Microwaves are used for cooking because they are absorbed by water molecules, causing them to vibrate and heat the food. X-rays penetrate soft tissue but are absorbed by bone, making them ideal for imaging fractures.
微波用于烹饪，因为它们被水分子吸收，使其振动加热食物。X 射线能穿透软组织但被骨骼吸收，因此非常适合骨折成像。
6. Question 6: Ultrasound and Distance Measurement | 问题6：超声波与距离测量
A ship uses sonar to detect the seafloor. An ultrasound pulse is emitted and the echo is received after 0.4 s. The speed of sound in seawater is 1500 m/s. Calculate the depth of the sea.
一艘船使用声呐探测海底，发射超声波脉冲后 0.4 s 收到回波。海水中声速为 1500 m/s。计算海深。
d × time = 1500 × 0.4 = 600 m. This is the round trip. The depth is half of this distance.
脉冲传播的总距离 = 速度 × 时间 = 1500 × 0.4 = 600 m。这是往返路程，海深为此距离的一半。
Depth d = (v × t) / 2 = (1500 m/s × 0.4 s) / 2 = 300 m
海深 d = (v × t) / 2 = (1500 m/s × 0.4 s) / 2 = 300 m
Ultrasound has a frequency above 20 kHz, beyond human hearing. It is widely used in sonar, medical scans, and industrial flaw detection.
超声波的频率高于 20 kHz，超出人耳听觉范围，广泛用于声呐、医学扫描和工业探伤。
7. Question 7: Seismic Waves and Earth's Structure | 问题7：地震波与地球结构
Explain why S-waves cannot travel through the Earth's outer core. What does this observation tell us about the nature of the outer core?
解释为什么 S 波不能穿过地球的外核。这一观测结果告诉我们外核的什么性质？
S-waves (secondary waves) are transverse waves. They require a solid medium to propagate because transverse motion needs a restoring force that liquids and gases cannot supply. When S-waves encounter the liquid outer core, they are absorbed and do not pass through.
S 波（次波）是横波，需要固体介质才能传播，因为横向运动需要液体和气体无法提供的恢复力。当 S 波遇到液态外核时，被吸收而无法穿过。
This creates an S-wave shadow zone on the Earth's surface opposite the earthquake epicentre. Scientists infer that the outer core is liquid, while the inner core is solid because P-waves can travel through both but S-waves only through the solid.
这在地震震中相对的地表形成 S 波影区。科学家由此推断外核是液态的，而内核是固态的，因为 P 波两种介质均可穿过，而 S 波只能穿过固体。
This evidence supports the model of a layered Earth with a liquid outer core composed mainly of molten iron and nickel.
这一证据支持地球分层模型，外核主要由液态铁镍合金组成。
8. Question 8: Converging Lens (Higher Tier) | 问题8：凸透镜（高阶内容）
e image distance v. Describe the image (real/virtual, magnified/diminished, upright/inverted).
一物体置于焦距 10 cm 的凸透镜前 30 cm 处。使用透镜公式 1/f = 1/u + 1/v 求像距 v，并描述像的性质（实/虚、放大/缩小、正立/倒立）。
Given f = 10 cm, u = 30 cm. Rearranging: 1/v = 1/f – 1/u = 1/10 – 1/30 = (3 – 1)/30 = 2/30 = 1/15. Therefore v = 15 cm.
已知 f = 10 cm，u = 30 cm。整理得：1/v = 1/f – 1/u = 1/10 – 1/30 = (3-1)/30 = 2/30 = 1/15。因此 v = 15 cm。
e image is half the height of the object, so it is diminished. Since v is positive, the image is on the opposite side of the lens and is real. A real image is always inverted.
放大率 m = v/u = 15/30 = 0.5。像高为物体的一半，因此是缩小的。v 为正，像在透镜对侧，是实像。实像总是倒立的。
The image is real, inverted, and diminished. This is the same configuration used in a camera.
像为倒立、缩小的实像，与相机成像原理相同。
9. Question 9: Black Body Radiation | 问题9：黑体辐射
A metal bar is heated until it glows red. Explain how the intensity and peak wavelength of the emitted radiation change as the temperature increases further. What colour would the bar glow at extremely high temperatures?
一金属棒加热至发红光。解释温度进一步升高时，辐射强度和峰值波长如何变化。温度极高时棒会发什么颜色的光？
As temperature increases, the total intensity of emitted radiation increases dramatically (it is proportional to the fourth power of absolute temperature, Stefan‑Boltzmann law). The peak wavelength shifts to shorter values according to Wien's law, moving from red towards blue.
温度升高时，辐射总强度急剧增大（与绝对温度的四次方成正比，斯特藩-玻尔兹曼定律）。根据维恩定律，峰值波长向短波方向移动，从红光移向蓝光。
At lower temperatures the metal glows red; as it gets hotter, it becomes orange, yellow, and eventually white-hot. At extremely high temperatures, it would appear blue-white because the peak emission enters the visible short-wavelength region.
较低温度时金属发红光；温度更高时，变为橙色、黄色，最终达到白热。温度极高时，会呈现蓝白色，因为峰值辐射进入可见光短波区域。
This behaviour is characteristic of an ideal black body and is the reason stars have different colours based on their surface temperature.
这种特性是理想黑体的标志，也是恒星的表面温度决定其颜色的原因。
10. Question 10: Wave Energy and Amplitude | 问题10：波的能量与振幅
A loudspeaker produces a sound wave. Describe the relationship between amplitude and energy. If the amplitude of the wave is doubled, by what factor does the energy transmitted increase? Explain your reasoning.
扬声器产生声波。描述振幅与能量的关系。如果将振幅加倍，传输的能量增加为原来的多少倍？解释你的推理。
For mechanical waves like sound, the energy carried by the wave is proportional to the square of the amplitude. This is because both the displacement and the restoring force are proportional to amplitude, resulting in energy ∝ (amplitude)².
对于声波等机械波，波所携带的能量与振幅的平方成正比。这是因为位移和回复力均与振幅成正比，导致能量正比于振幅的平方。
so the energy increases by a factor of 4. This explains why loud sounds (large amplitude)
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