CCEA Year 12 Physics Unit Test Mock Paper Analysis | Year 12 CCEA 物理单元测试模拟卷解析

📚 CCEA Year 12 Physics Unit Test Mock Paper Analysis | Year 12 CCEA 物理单元测试模拟卷解析

This article walks you through a full mock unit test paper for Year 12 CCEA Physics, covering core topics from AS Unit 1 (Forces, Energy and Electricity) and AS Unit 2 (Waves, Photons and Astronomy). Each question is worked out step by step, with the physics reasoning clearly explained and common exam traps highlighted.

本文逐步解析一套 Year 12 CCEA 物理单元测试模拟卷，涵盖 AS 第1单元（力、能量和电学）与第2单元（波、光子和天文学）的核心内容。每一道题都按步骤推演，清晰地解释物理思路，并指出常见的考试失分点。

1. Constant Acceleration | 匀加速直线运动

A cyclist accelerates uniformly from rest with an acceleration of 1.5 m/s² for 8.0 seconds. Use the equations of motion to find the final velocity and the distance travelled.

骑自行车的人从静止以 1.5 m/s² 的加速度匀加速 8.0 秒。用运动学方程求末速度和行驶距离。

th u = 0, a = 1.5 m/s², t = 8.0 s: v = 0 + (1.5 × 8.0) = 12 m/s, giving a speed of 12 m/s.

用 v = u + at，初速 u = 0，加速度 a = 1.5 m/s²，时间 t = 8.0 s，得 v = 1.5 × 8.0 = 12 m/s，即速度为 12 m/s。

= 0 + ½ × 1.5 × (8.0)² = ½ × 1.5 × 64 = 48 m. The distance covered is 48 m.

利用 s = ut + ½at² = ½ × 1.5 × (8.0)² = ½ × 1.5 × 64 = 48 m，位移为 48 m。

You can also verify this with average velocity: s = (u+v)/2 × t = (0+12)/2 × 8 = 48 m.

也可用平均速度验证：s = (0+12)/2 × 8 = 48 m。

2. Projectile Motion | 抛体运动

al. Calculate the maximum height reached and the total time of flight. Take g = 9.81 m/s².

一球以初速 25 m/s、与水平方向成 40° 抛出。计算最大高度和总飞行时间。取 g = 9.81 m/s²。

Resolve the initial velocity: vertical component v_y ≈ 25 × 0.643 = 16.1 m/s. (Horizontal component is not needed for height and flight time.)

分解初速度：竖直分量 v_y = 25 sin40° ≈ 16.1 m/s（最大高度和飞行时间不需要水平分量）。

At the maximum height, the vertical velocity becomes zero. Using v_y² = u_y² - 2gH, with v_y = 0, gives H = u_y²/(2g) = (16.1)² / (2 × 9.81) ≈ 259 / 19.62 ≈ 13.2 m.

在最大高度处竖直速度为零。用 v_y² = u_y² - 2gH，令 v_y = 0，得 H = u_y²/(2g) = (16.1)² / (2×9.81) ≈ 13.2 m。

The time to reach the highest point is t_up = u_y/g = 16.1 / 9.81 ≈ 1.64 s. By symmetry, the total time of flight T = 2 × t_up ≈ 3.28 s.

上升到最高点的时间 t_up = u_y/g ≈ 1.64 s。由于对称性，总飞行时间 T = 2 × 1.64 = 3.28 s。

3. Newton's Laws and Friction | 牛顿定律与摩擦力

A box of mass 10 kg is pulled across a horizontal floor by a rope exerting a force of 40 N at an angle of 30° above the horizontal. The coefficient of kinetic friction between the box and the floor is 0.25. Determine the acceleration of the box.

一个质量为 10 kg 的箱子被绳子以 40 N 的力、与水平方向成 30° 角拉动。箱子与地面之间的动摩擦因数为 0.25。求箱子的加速度。

Resolve the pulling force: horizontal component F_x = 40 cos30° ≈ 34.6 N; vertical component F_y = 40 sin30° = 20 N (upwards).

分解拉力：水平分量 Fₓ = 40 cos30° ≈ 34.6 N；竖直分量 F_y = 40 sin30° = 20 N（向上）。

The upward component reduces the normal contact force: N = mg - F_y = (10 × 9.81) - 20 = 98.1 - 20 = 78.1 N.

向上的分量减轻了正压力：N = mg - F_y = 98.1 - 20 = 78.1 N。

Kinetic friction fₖ = μₖ N = 0.25 × 78.1 ≈ 19.5 N (opposing motion).

动摩擦力 fₖ = μₖ N = 0.25 × 78.1 ≈ 19.5 N（方向与运动相反）。

The net horizontal force is F_net = F_x - fₖ = 34.6 - 19.5 = 15.1 N. Using F = m a, acceleration a = F_net / m = 15.1 / 10 = 1.51 m/s² (approximately 1.5 m/s²).

水平方向合力 F_net = Fₓ - fₖ = 34.6 - 19.5 = 15.1 N。由 F = ma，加速度 a = 15.1/10 = 1.51 m/s²（约 1.5 m/s²）。

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4. Conservation of Momentum | 动量守恒

A trolley of mass 2.0 kg moves at 3.0 m/s on a frictionless track and collides with a stationary trolley of mass 1.0 kg. The two trolleys stick together. Find their common velocity just after the collision and the kinetic energy lost.

质量为 2.0 kg 的小车以 3.0 m/s 在光滑轨道上运动，与静止的 1.0 kg 小车碰撞后粘在一起。求碰撞后的共同速度和损失的动能。

momentum after. m₁u₁ + m₂u₂ = (m₁+m₂)v. Take direction of the moving trolley as positive.

根据动量守恒：碰撞前总动量等于碰撞后总动量。m₁u₁ + m₂u₂ = (m₁+m₂)v，设运动方向为正。

2.0 × 3.0 + 1.0 × 0 = (2.0+1.0) v → 6.0 = 3.0 v → v = 2.0 m/s in the original direction.

2.0×3.0 + 1.0×0 = (2.0+1.0) v → 6.0 = 3.0 v → v = 2.0 m/s，方向不变。

Initial kinetic energy: KEᵢ = ½ m₁ u₁² = 0.5 × 2.0 × (3.0)² = 9.0 J.

初动能：KEᵢ = ½ × 2.0 × 3.0² = 9.0 J。

(2.0)² = 6.0 J. Kinetic energy lost = 9.0 - 6.0 = 3.0 J (transformed into heat and sound).

末动能：KE_f = ½ × 3.0 × 2.0² = 6.0 J。损失的动能 = 9.0 - 6.0 = 3.0 J（转化为热与声）。

5. DC Circuits: Series and Parallel | 直流电路：串联与并联

A 12 V battery is connected to a 4.0 Ω resistor in series with a parallel combination of a 6.0 Ω and a 3.0 Ω resistor. Calculate the total current drawn from the battery and the voltage across the 6.0 Ω resistor.

一个 12 V 电池与一个 4.0 Ω 电阻串联，然后与由 6.0 Ω 和 3.0 Ω 电阻并联构成的组合相连。求电池输出的总电流和 6.0 Ω 电阻两端的电压。

First find the equivalent resistance of the parallel branch: 1/R_par = 1/6.0 + 1/3.0 = (1+

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