📚 Year 13 CCEA Physics: Unit Test Mock Paper Analysis | Year 13 CCEA 物理：单元测试模拟卷解析
This mock paper analysis is designed to help Year 13 students consolidate key concepts from CCEA AS Physics Units 1 and 2. Each question has been carefully selected to reflect the style and demand of actual unit tests. Through step-by-step bilingual explanations, you will learn how to approach problems logically, apply relevant equations, and avoid common pitfalls. Focus on the scientific reasoning behind each solution and make sure you understand how marks are allocated for structured questions.
这份单元测试模拟卷解析旨在帮助 Year 13 学生巩固 CCEA AS 物理单元一和单元二的核心内容。每道题目都经过精心挑选，力求贴近真实考试的风格与难度。通过逐步的中英双语解析，你将学会如何有逻辑地处理问题、运用正确的公式并避开常见错误。请重点关注每个解答背后的科学推理，并确保你理解结构化问题中的得分要点。
1. Projectile Motion and Vector Resolution | 抛体运动与矢量分解
It releases a supply package. Assume air resistance is negligible and take g = 9.8 m s⁻².
一架救援飞机以 60 m s⁻¹ 的恒定速度水平飞行，飞行高度为 490 m。飞机投下一个物资包裹。假设空气阻力可忽略不计，取 g = 9.8 m s⁻²。
d a = g. This gives 490 = ½ × 9.8 × t², leading to t = √( (490 × 2) / 9.8 ) = √100 = 10 s.
（a）计算包裹落到地面所需的时间。由于竖直初速度为零，运用公式 s = uᵧ t + ½ a t²，其中 uᵧ = 0，s = 490 m，a = g。由此得到 490 = ½ × 9.8 × t²，解得 t = 10 s。
re is no horizontal acceleration. Distance = horizontal velocity × time = 60 × 10 = 600 m.
（b）求包裹从释放点起的水平飞行距离。水平方向不受力，运动为匀速直线运动。水平距离 = 水平速度 × 时间 = 60 × 10 = 600 m。
A common error is to combine horizontal and vertical velocities too early. Always treat the two components independently before using time to link them.
常见错误是过早地将水平与竖直速度合成。务必先将两个分量独立处理，再通过时间将它们联系起来。
2. Newton's Laws and the Inclined Plane | 牛顿定律与斜面
friction between the block and the surface is μₖ = 0.40. The block is released from rest.
一质量为 5.0 kg 的木块放置在与水平面成 30° 的粗糙斜面上。木块与斜面间的动摩擦因数 μₖ = 0.40。木块由静止释放。
ting down the slope: W_parallel = mg sinθ = 5.0 × 9.8 × sin30° = 5.0 × 9.8 × 0.5 = 24.5 N.
（a）计算重力沿斜面向下的分量：W_平行 = mg sinθ = 5.0 × 9.8 × sin30° = 5.0 × 9.8 × 0.5 = 24.5 N。
866 = 42.4 N. Then kinetic friction f = μₖ R = 0.40 × 42.4 ≈ 17.0 N (acting up the slope).
（b）求摩擦力的大小。首先计算法向反作用力 R = mg cosθ = 5.0 × 9.8 × cos30° ≈ 42.4 N。动摩擦力 f = μₖ R = 0.40 × 42.4 ≈ 17.0 N（方向沿斜面向上）。
the slope = 24.5 N - 17.0 N = 7.5 N. Using F = ma, acceleration a = 7.5 / 5.0 = 1.5 m s⁻².
（c）求木块的加速度。沿斜面向下的合力 = 24.5 N - 17.0 N = 7.5 N。利用牛顿第二定律 F = ma，加速度 a = 7.5 / 5.0 = 1.5 m s⁻²。
Always check that the frictional force does not exceed the maximum possible static friction if the object was initially at rest. Here kinetic friction is used because the block starts moving.
务必检查如果物体初始静止，摩擦力是否超过最大静摩擦力。本题中木块一旦释放便开始滑动，因此使用动摩擦力。
3. Work, Energy and Power | 做功、能量与功率
A skateboarder of mass 65 kg starts from rest at the top of a smooth quarter-pipe ramp of vertical height 4.0 m. The ramp is frictionless.
一位质量为 65 kg 的滑板运动员从竖直高度为 4.0 m 的光滑四分之一管道顶部由静止出发。管道无摩擦。
he speed of the skateboarder at the bottom using energy conservation: mgh = ½ m v², so v = √2gh = √2 × 9.8 × 4.0 = √78.4 ≈ 8.85 m s⁻¹.
（a）利用能量守恒计算运动员到达底部时的速度：mgh = ½ m v²，因此 v = √2gh = √2 × 9.8 × 4.0 = √78.4 ≈ 8.85 m s⁻¹。
= f × d = 50 × 6.0 = 300 J. Initial GPE = mgh = 65 × 9.8 × 4.0 = 2548 J. KE at bottom = 2548 - 300 = 2248 J. v = √2 × 2248 / 65 ≈ √69.17 ≈ 8.32 m s⁻¹.
若沿 6.0 m 长的弧面存在 50 N 的恒定摩擦力，求运动员到达底部时的新速度。克服摩擦力做功 = f × d = 50 × 6.0 = 300 J。初始重力势能 GPE = mgh = 65 × 9.8 × 4.0 = 2548 J。底部动能 = 2548 - 300 = 2248 J。速度 v = √2 × 2248 / 65 ≈ 8.32 m s⁻¹。
Notice that the work–energy principle is especially useful when the path is curved but the friction is constant along it.
注意，当路径为曲线但摩擦力恒定时，功能原理尤其方便。
4. DC Circuits and Internal Resistance | 直流电路与内阻
A battery of e.m.f. 12.0 V and internal resistance r is connected to an external resistor of 8.0 Ω. A voltmeter reads the terminal potential difference as 10.0 V.
一个电动势为 12.0 V、内阻为 r 的电池连接到一个 8.0 Ω 的外电阻。电压表测得路端电压为 10.0 V。
ircuit is 1.25 A. Using V = IR for the external resistor: I = V / R = 10.0 / 8.0 = 1.25 A.
（a）证明电路中的电流为 1.25 A。对外电阻使用 V = IR：I = 10.0 / 8.0 = 1.25 A。
so 10.0 = 12.0 - 1.25 r. Rearranging gives r = (12.0 - 10.0) / 1.25 = 2.0 / 1.25 = 1.6 Ω.
（b）计算内阻 r。路端电压 V = ε - I r，代入得 10.0 = 12.0 - 1.25 r，解得 r = 1.6 Ω。
R = (1.25)² × 8.0 = 1.5625 × 8.0 = 12.5 W. Alternatively, P = V I = 10.0 × 1.25 = 12.5 W.
（c）求外电阻消耗的功率。P = I² R = (1.25)² × 8.0 = 12.5 W，或 P = V I = 10.0 × 1.25 = 12.5 W。
Always distinguish between e.m.f. and terminal p.d., especially when the internal resistance is not negligible.
务必区分电动势和路端电压，尤其在内阻不可忽略的情况下。
5. Young's Double-Slit Interference | 杨氏双缝干涉
In a double-slit experiment, monochromatic light passes through two slits separated by 0.50 mm. The interference pattern is observed on a screen 2.0 m away. The distance between ten successive bright fringes (10 fringe spacing) is 24.0 mm.
在双缝实验中，单色光通过相距 0.50 mm 的双缝，在 2.0 m 远的屏幕上观察到干涉图样。10 个连续亮纹之间的距离为 24.0 mm。
m = 2.67 × 10⁻³ m. The formula is Δy = λ D / a, so λ = Δy × a / D. Here a = 0.50 mm = 5.0 × 10⁻⁴ m, D = 2.0 m. λ = (2.67 × 10⁻³ × 5.0 × 10⁻⁴) / 2.0 = 6.68 × 10⁻⁷ m = 668 nm.
计算光的波长。首先计算条纹间距 Δy：10 个亮纹对应 9 个间隔，因此条纹间距 Δy = 24.0 mm / 9 = 2.67 mm = 2.67 × 10⁻³ m。公式为 Δy = λ D / a，变形得 λ = Δy × a / D。其中 a = 0.50 mm = 5.0 × 10⁻⁴ m，D = 2.0 m。λ = (2.67 × 10⁻³ × 5.0 × 10⁻⁴) / 2.0 = 6.68 × 10⁻⁷ m = 668 nm。
State the effect on the fringe pattern if the slit separation is halved while other conditions remain the same. Since Δy ∝ 1/a, halving a doubles the fringe spacing, making the pattern more spread out but dimmer because less light reaches the screen.
若将双缝间距减半而其他条件不变，说明对图样的影响。由于 Δy 与 1/a 成正比，间距减半会使条纹间距加倍，图样更加分散，但同时因为通光量减小，亮度降低。
6. Photoelectric Effect | 光电效应
Light of wavelength 450 nm is incident on a clean caesium surface. The work function ϕ caesium is 2.1 eV. Take h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹, 1 eV = 1.60 × 10⁻¹⁹ J.
波长为 450 nm 的光照射清洁的铯表面。铯的逸出功 ϕ = 2.1 eV。取 h = 6.63 × 10⁻³⁴ J s，c = 3.00 × 10⁸ m s⁻¹，1 eV = 1.60 × 10⁻¹⁹ J。
× 10⁸) / (450 × 10⁻⁹) = 4.42 × 10⁻¹⁹ J. In eV: E = 4.42 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ ≈ 2.76 eV.
（a）计算光子能量（以焦耳和电子伏特表示）：E = hf = hc / λ = 4.42 × 10⁻¹⁹ J，换算为 2.76 eV。
(b) Determine the maximum kinetic energy of emitted photoelectrons: Kmax = E - ϕ = 2.76 eV - 2.1 eV = 0.66 eV, or 1.06 × 10⁻¹⁹ J.
（b）求逸出光电子的最大动能：Kmax = E - ϕ = 2.76 eV - 2.1 eV = 0.66 eV（即 1.06 × 10⁻¹⁹ J）。
(c) Calculate the stopping potential: Vₛ = Kmax stopping potential in volts equals maximum kinetic energy in eV for an electron charge e.
（c）计算遏止电压：Vₛ = Kmax / e = 0.66 V，因为对电子而言遏止电压的数值等于以 eV 为单位的最大动能。
Explain why the maximum kinetic energy of photoelectrons does not depend on light intensity. Each photon interacts with a single electron; increasing intensity only increases the number of photons (and thus the photocurrent), but the energy per photon stays the same.
解释为何光电子的最大动能不依赖于光强。每个光子与单个电子相互作用；增加光强仅增加光子数目（从而增大光电流），但每个光子的能量不变。
7. Standing Waves on a String | 弦上的驻波
der a tension of 80 N. The string is fixed at both ends and its vibrating length is 1.2 m.
一根拉紧的弦线质量线密度 μ = 0.0050 kg m⁻¹，所受张力为 80 N。弦两端固定，振动长度为 1.2 m。
(a) Calculate the speed of transverse waves on the string: v = √T / μ = √80 / 0.0050 = √16000 = 126.5 m s⁻¹ (approximately 127 m s⁻¹).
（a）计算弦上横波的波速：v = √T / μ = √80 / 0.0050 = √16000 ≈ 126.5 m s⁻¹。
oth ends, the fundamental wavelength λ₁ = 2L = 2.4 m. f₁ = v / λ₁ = 126.5 / 2.4 ≈ 52.7 Hz.
（b）求基频（第一谐波）。对于两端固定的弦，基波波长 λ₁ = 2L = 2.4 m，f₁ = v / λ₁ ≈ 52.7 Hz。
ency f₃ = 3 f₁ ≈ 158 Hz. The pattern consists of three loops with nodes at the ends and two interior nodes.
（c）求第三谐波的频率并画出驻波图样。第三谐波有三个波腹，λ₃ = 2L / 3 = 0.80 m，f₃ = 3 f₁ ≈ 158 Hz。该波形包含三个波节（两端和两个内部节点），形成三个完整波段。
8. Hubble's Law and Redshift | 哈勃定律与红移
ift can be approximated by z ≈ v / c. Take H₀ = 70 km s⁻¹ Mpc⁻¹ and c = 3.00 × 10⁵ km s⁻¹.
室女座星系团中一个星系观测到的红移 z = 0.050。对低速运动，红移可近似为 z ≈ v / c。取 H₀ = 70 km s⁻¹ Mpc⁻¹，c = 3.00 × 10⁵ km s⁻¹。
e the recessional velocity of the galaxy: v ≈ z c = 0.050 × 3.00 × 10⁵ = 1.5 × 10⁴ km s⁻¹.
（a）估算该星系的退行速度：v ≈ z c = 0.050 × 3.00 × 10⁵ = 1.5 × 10⁴ km s⁻¹。
(b) Use Hubble' = (1.5 × 10⁴ km s⁻¹) / (70 km s⁻¹ Mpc⁻¹) ≈ 214 Mpc. This is roughly 7.0 × 10⁸ light-years.
（b）利用哈勃定律求星系的距离：d = v / H₀ ≈ 214 Mpc，约合 7.0 × 10⁸ 光年。
(c) Explain what a redshift of 0.050 tells us about the early Universe. The observed redshift is due to the expansion of space; the further the galaxy, the faster it recedes, supporting the Big Bang model. This provides evidence that the Universe is expanding uniformly.
（c）解释红移 0.050 对早期宇宙的启示。观测到的红移源于空间膨胀；星系越远退行越快，这支持了大爆炸宇宙模型，为宇宙的均匀膨胀提供了证据。
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