📚 Year 12 CIE Science: Unit Test Mock Paper Analysis | CIE 12年级科学：单元测试模拟卷解析
This resource provides a detailed walkthrough of a Unit Test mock paper for Year 12 CIE Science, focusing on Mechanics. Each question is analysed with step-by-step reasoning to reinforce key principles and exam technique.
本资源为 CIE 12年级科学力学单元测试模拟卷提供详细解析。每道题目均通过逐步推理进行分析，以巩固核心原理和考试技巧。
1. Kinematics Graph Analysis | 运动学图像分析
A velocity-time graph shows an object accelerating uniformly from rest to 20 m s⁻¹ in 5 s, then moving at constant velocity for 10 s, and finally decelerating uniformly to rest in 4 s. Calculate the total displacement.
一速度-时间图像显示物体从静止均匀加速到 20 m s⁻¹ 用时 5 s，然后匀速运动 10 s，最后均匀减速到静止用时 4 s。计算总位移。
The area under a velocity-time graph represents displacement. Divide the motion into three segments: triangle (acceleration), rectangle (constant velocity), and triangle (deceleration).
速度-时间图像下的面积表示位移。将运动分为三段：三角形（加速）、矩形（匀速）和三角形（减速）。
Acceleration phase area: (1/2) × base × height = (1/2) × 5 s × 20 m s⁻¹ = 50 m. Constant velocity area: 10 s × 20 m s⁻¹ = 200 m. Deceleration phase area: (1/2) × 4 s × 20 m s⁻¹ = 40 m. Total displacement = 50 + 200 + 40 = 290 m.
加速阶段面积：(1/2) × 底 × 高 = (1/2) × 5 s × 20 m s⁻¹ = 50 m。匀速阶段面积：10 s × 20 m s⁻¹ = 200 m。减速阶段面积：(1/2) × 4 s × 20 m s⁻¹ = 40 m。总位移 = 50 + 200 + 40 = 290 m。
2. Newton’s Second Law Application | 牛顿第二定律应用
A block of mass 5.0 kg is pulled along a rough horizontal surface by a force of 30 N applied at an angle of 30° above the horizontal. The coefficient of kinetic friction is 0.25. Determine the acceleration of the block.
一个质量为 5.0 kg 的木块在粗糙水平面上被一个与水平方向成 30° 向上的 30 N 力拉动。动摩擦因数为 0.25。求木块的加速度。
Resolve the pulling force into horizontal and vertical components. Horizontal: Fx = 30 cos30° ≈ 25.98 N. Vertical: Fy = 30 sin30° = 15.0 N upward. The normal reaction force N = mg − Fy = (5.0 kg × 9.81 m s⁻²) − 15.0 N = 49.05 N − 15.0 N = 34.05 N.
将拉力分解为水平和竖直分量。水平：Fₓ = 30 cos30° ≈ 25.98 N。竖直：Fᵧ = 30 sin30° = 15.0 N 向上。法向反作用力 N = mg − Fᵧ = (5.0 kg × 9.81 m s⁻²) − 15.0 N = 49.05 N − 15.0 N = 34.05 N。
Frictional force f = μN = 0.25 × 34.05 N ≈ 8.51 N. Net horizontal force Fnet = Fx − f = 25.98 N − 8.51 N = 17.47 N. Then acceleration a = Fnet / m = 17.47 N / 5.0 kg ≈ 3.49 m s⁻².
摩擦力 f = μN = 0.25 × 34.05 N ≈ 8.51 N。净水平力 F_net = Fₓ − f = 25.98 N − 8.51 N = 17.47 N。则加速度 a = F_net / m = 17.47 N / 5.0 kg ≈ 3.49 m s⁻²。
3. Conservation of Energy in a Pendulum | 单摆中的能量守恒
A pendulum bob of mass 0.200 kg is released from rest at point A where the string makes an angle of 60° with the vertical. The string length is 0.80 m. Find the speed of the bob at the lowest point B. Ignore air resistance.
一个质量为 0.200 kg 的单摆摆球从静止释放，释放点 A 处绳与竖直方向夹角为 60°。绳长 0.80 m。求摆球在最低点 B 的速度。忽略空气阻力。
Use conservation of mechanical energy. The vertical height lost from A to B is Δh = L − L cosθ = 0.80 m × (1 − cos60°) = 0.80 m × (1 − 0.5) = 0.40 m. Loss in gravitational potential energy = mgΔh = 0.200 kg × 9.81 m s⁻² × 0.40 m = 0.7848 J.
使用机械能守恒。从 A 到 B 下降的竖直高度 Δh = L − L cosθ = 0.80 m × (1 − cos60°) = 0.80 m × (1 − 0.5) = 0.40 m。重力势能减少量 = mgΔh = 0.200 kg × 9.81 m s⁻² × 0.40 m = 0.7848 J。
This energy is converted into kinetic energy at B: (1/2)mv² = 0.7848 J. Solving for v: v = √(2 × 0.7848 / 0.200) = √(7.848) ≈ 2.80 m s⁻¹.
此能量在 B 点转化为动能：(1/2)mv² = 0.7848 J。解得 v = √(2 × 0.7848 / 0.200) = √(7.848) ≈ 2.80 m s⁻¹。
4. Momentum Conservation in Collisions | 碰撞中的动量守恒
Trolley X (mass 1.5 kg, velocity 4.0 m s⁻¹ to the right) collides with stationary trolley Y (mass 2.5 kg). After the collision, X moves at 1.2 m s⁻¹ to the left. Find the velocity of Y after the collision and verify whether the collision is elastic.
小车 X（质量 1.5 kg，速度 4.0 m s⁻¹ 向右）与静止的小车 Y（质量 2.5 kg）碰撞。碰撞后 X 以 1.2 m s⁻¹ 向左运动。求碰撞后 Y 的速度，并验证碰撞是否为弹性碰撞。
Choose right as positive. Initial momentum: pi = mXuX + mYuY = (1.5 × 4.0) + (2.5 × 0) = 6.0 kg m s⁻¹. After collision, X velocity vX = −1.2 m s⁻¹. Let vY be Y’s velocity. Conservation of momentum: 6.0 = 1.5 × (−1.2) + 2.5 vY ⇒ 6.0 = −1.8 + 2.5 vY ⇒ 2.5 vY = 7.8 ⇒ vY = 3.12 m s⁻¹ to the right.
取向右为正。初始动量：p_i = mₓuₓ + mᵧuᵧ = (1.5 × 4.0) + (2.5 × 0) = 6.0 kg m s⁻¹。碰撞后 X 速度 vₓ = −1.2 m s⁻¹。设 vᵧ 为 Y 的速度。动量守恒：6.0 = 1.5 × (−1.2) + 2.5 vᵧ ⇒ 6.0 = −1.8 + 2.5 vᵧ ⇒ 2.5 vᵧ = 7.8 ⇒ vᵧ = 3.12 m s⁻¹ 向右。
Check kinetic energy: initial KE = 0.5 × 1.5 × 4.0² = 12.0 J. Final KE = 0.5 × 1.5 × 1.2² + 0.5 × 2.5 × 3.12² = 1.08 + 12.168 = 13.248 J. The total KE has increased, indicating an inelastic collision — energy appears to have been added, possibly due to an error in data, but in a standard analysis this would be classified as super-elastic or a calculation check is needed. However, using typical exam values, if final KE equals initial KE, it’s elastic; here it does not, so it is inelastic.
检验动能：初始动能 = 0.5 × 1.5 × 4.0² = 12.0 J。末动能 = 0.5 × 1.5 × 1.2² + 0.5 × 2.5 × 3.12² = 1.08 + 12.168 = 13.248 J。总动能增加，说明是非弹性碰撞 — 可能数据有误，但在标准分析中由于动能不守恒，判定为非弹性碰撞。
5. Equilibrium of Forces and Torque | 力的平衡与力矩
A uniform beam of length 4.0 m and weight 200 N is pivoted at one end. A cable attached 3.0 m from the pivot exerts a tension at 40° above the horizontal to keep the beam horizontal. Find the tension in the cable.
一根长 4.0 m、重 200 N 的均匀梁一端铰接。一根缆绳在距铰接点 3.0 m 处施加拉力，方向与水平方向成 40° 向上，以保持梁水平。求缆绳中的张力。
Take moments about the pivot. Weight acts at the centre (2.0 m from pivot), producing clockwise moment = 200 N × 2.0 m = 400 Nm. Tension T has a vertical component T sin40° at 3.0 m, producing anticlockwise moment = T sin40° × 3.0 m. For equilibrium, clockwise moment = anticlockwise moment: 400 = 3T sin40°. Solve: T = 400 / (3 sin40°) = 400 / (3 × 0.6428) ≈ 400 / 1.9284 ≈ 207.4 N.
对铰接点取矩。重力作用于中心（距铰接点 2.0 m），产生顺时针力矩 = 200 N × 2.0 m = 400 Nm。张力 T 在 3.0 m 处有竖直分量 T sin40°，产生逆时针力矩 = T sin40° × 3.0 m。平衡时顺时针力矩等于逆时针力矩：400 = 3T sin40°。解得 T = 400 / (3 sin40°) = 400 / (3 × 0.6428) ≈ 400 / 1.9284 ≈ 207.4 N。
6. Stress, Strain and Young Modulus | 应力、应变与杨氏模量
A steel wire of diameter 0.50 mm and original length 2.00 m stretches by 4.0 mm under a load of 80 N. Calculate the stress, strain, and Young modulus of the steel.
一根直径 0.50 mm、原长 2.00 m 的钢丝在 80 N 的载荷下伸长 4.0 mm。计算应力、应变和钢的杨氏模量。
Cross-sectional area A = π(d/2)² = π × (0.25 × 10⁻³ m)² ≈ 1.9635 × 10⁻⁷ m². Stress σ = F / A = 80 N / 1.9635×10⁻⁷ m² ≈ 4.07 × 10⁸ Pa. Strain ε = extension / original length = (4.0 × 10⁻³ m) / 2.00 m = 2.0 × 10⁻³ (no units). Young modulus E = σ / ε = (4.07 × 10⁸) / (2.0 × 10⁻³) ≈ 2.04 × 10¹¹ Pa.
横截面积 A = π(d/2)² = π × (0.25 × 10⁻³ m)² ≈ 1.9635 × 10⁻⁷ m²。应力 σ = F / A = 80 N / 1.9635×10⁻⁷ m² ≈ 4.07 × 10⁸ Pa。应变 ε = 伸长量 / 原长 = (4.0 × 10⁻³ m) / 2.00 m = 2.0 × 10⁻³（无单位）。杨氏模量 E = σ / ε = (4.07 × 10⁸) / (2.0 × 10⁻³) ≈ 2.04 × 10¹¹ Pa。
7. Simple Harmonic Motion: Pendulum Period | 简谐运动：单摆周期
A simple pendulum has a period of 2.00 s on Earth where g = 9.81 m s⁻². What is its length? On a planet where the acceleration due to gravity is one-quarter of Earth’s, what would its period be?
一个单摆在地球（g = 9.81 m s⁻²）上的周期为 2.00 s。其摆长是多少？在一个重力加速度为地球四分之一的星球上，它的周期会是多少？
Period of a simple pendulum: T = 2π√(L/g). Rearrange: L = gT² / (4π²). With T = 2.00 s, L = (9.81 × 4) / (4π²) = 9.81 / π² ≈ 9.81 / 9.8696 ≈ 0.994 m. On the new planet, g’ = g/4 = 2.4525 m s⁻². New period T’ = 2π√(L / g’) = 2π√(0.994 / 2.4525) = 2π√(0.4053) ≈ 2π × 0.6367 ≈ 4.00 s. Alternatively, since T ∝ 1/√g, halving g doubles T, so T’ = 4.00 s.
单摆周期公式：T = 2π√(L/g)。变形得：L = gT² / (4π²)。代入 T = 2.00 s，L = (9.81 × 4) / (4π²) = 9.81 / π² ≈ 9.81 / 9.8696 ≈ 0.994 m。在新星球上，g’ = g/4 = 2.4525 m s⁻²。新周期 T’ = 2π√(L / g’) = 2π√(0.994 / 2.4525) = 2π√(0.4053) ≈ 2π × 0.6367 ≈ 4.00 s。或者，由于 T ∝ 1/√g，g 减半则 T 加倍，所以 T’ = 4.00 s。
8. Wave Properties: Speed, Frequency, Wavelength | 波的特性：波速、频率、波长
A transverse wave on a string is described by y = 0.050 sin(6.0x − 240t) where x and y are in metres and t in seconds. Determine the amplitude, wavelength, frequency, and wave speed.
一列弦上的横波由 y = 0.050 sin(6.0x − 240t) 描述，其中 x 和 y 以米为单位，t 以秒为单位。求振幅、波长、频率和波速。
General wave equation: y = A sin(kx − ωt). Amplitude A = 0.050 m. Wave number k = 6.0 rad m⁻¹, so wavelength λ = 2π/k = 2π/6.0 ≈ 1.047 m. Angular frequency ω = 240 rad s⁻¹, so frequency f = ω / 2π = 240 / 2π ≈ 38.2 Hz. Wave speed v = fλ = (ω/k) = 240/6.0 = 40 m s⁻¹.
一般波动方程：y = A sin(kx − ωt)。振幅 A = 0.050 m。波数 k = 6.0 rad m⁻¹，因此波长 λ = 2π/k = 2π/6.0 ≈ 1.047 m。角频率 ω = 240 rad s⁻¹，因此频率 f = ω / 2π = 240 / 2π ≈ 38.2 Hz。波速 v = fλ = (ω/k) = 240/6.0 = 40 m s⁻¹。
9. Superposition and Stationary Waves | 叠加与驻波
Two identical waves travelling in opposite directions on a string interfere to form a stationary wave. The distance between two consecutive nodes is 0.30 m, and the frequency is 250 Hz. Calculate the speed of each progressive wave.
两列相同的波在弦上相向传播，干涉形成驻波。相邻两个波节之间的距离为 0.30 m，频率为 250 Hz。计算每列行波的速度。
For a stationary wave, distance between consecutive nodes = λ/2. Thus λ = 2 × 0.30 m = 0.60 m. Using v = fλ, speed of the progressive waves = 250 Hz × 0.60 m = 150 m s⁻¹.
对于驻波，相邻波节间距 = λ/2。因此 λ = 2 × 0.30 m = 0.60 m。利用 v = fλ，行波速度 = 250 Hz × 0.60 m = 150 m s⁻¹。
10. Projectile Motion Analysis | 抛体运动分析
A ball is kicked horizontally from a cliff 45 m high with a speed of 15 m s⁻¹. Determine the time of flight, the horizontal range, and the velocity on impact (magnitude and direction).
一个足球从 45 m 高的悬崖上以 15 m s⁻¹ 的水平速度踢出。求飞行时间、水平射程以及撞击时的速度（大小和方向）。
Vertical motion: uy = 0, a = 9.81 m s⁻², s = 45 m downward. Use s = uyt + ½at² → 45 = 0 + ½ × 9.81 × t² → t² = 90/9.81 ≈ 9.174 → t ≈ 3.03 s. Horizontal motion: constant velocity vx = 15 m s⁻¹, so range = vx × t = 15 × 3.03 ≈ 45.5 m. Final vertical velocity vy = uy + at = 0 + 9.81 × 3.03 ≈ 29.7 m s⁻¹ downwards. Impact speed = √(vx² + vy²) = √(15² + 29.7²) ≈ √(225 + 882.1) ≈ √1107.1 ≈ 33.3 m s⁻¹. Direction θ = tan⁻¹(vy/vx) = tan⁻¹(29.7/15) ≈ 63.2° below the horizontal.
竖直方向：u_y = 0，a = 9.81 m s⁻²，位移 s = 45 m 向下。利用 s = u_y t + ½at² → 45 = 0 + ½ × 9.81 × t² → t² = 90/9.81 ≈ 9.174 → t ≈ 3.03 s。水平方向：匀速 v_x = 15 m s⁻¹，因此水平射程 = v_x × t = 15 × 3.03 ≈ 45.5 m。末竖直速度 v_y = u_y + at = 0 + 9.81 × 3.03 ≈ 29.7 m s⁻¹ 向下。撞击速度大小 = √(v_x² + v_y²) = √(15² + 29.7²) ≈ √(225 + 882.1) ≈ √1107.1 ≈ 33.3 m s⁻¹。方向 θ = tan⁻¹(v_y/v_x) = tan⁻¹(29.7/15) ≈ 63.2° 在水平线下方。
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