📚 IGCSE CCEA Physics: Unit Test Paper | IGCSE CCEA 物理：单元测试卷
This practice unit test paper has been carefully compiled for IGCSE CCEA Physics students. It addresses core concepts across mechanics, energy, waves and thermal physics, mirroring the style and depth of official assessments. Each question is paired with a fully worked solution, allowing you to identify any gaps and refine your problem‑solving technique. Complete all questions under timed conditions, then review the explanations to strengthen your understanding.
这套练习单元测试卷专为 IGCSE CCEA 物理考生精心编撰，覆盖力学、能量、波和热物理中的核心概念，难度与深度贴近官方测评。每道题目均配有完整解答，帮助你查找薄弱环节、优化解题方法。请在限时条件下完成所有试题，再对照解析巩固理解。
1. Acceleration & Distance | 加速度与距离
A car accelerates uniformly from rest to 20 m/s in 10 seconds. Calculate its acceleration and the total distance travelled during this time.
一辆汽车从静止匀加速至20 m/s，用时10秒。计算其加速度及在此过程中行驶的总距离。
Solution: Use the definition of acceleration a = (v − u) / t, where u = 0 m/s, v = 20 m/s and t = 10 s.
解答：利用加速度定义式 a = (v − u) / t，其中 u = 0 m/s，v = 20 m/s，t = 10 s。
a = (20 − 0) / 10 = 2 m/s²
For distance s, apply the kinematic equation s = ut + ½at².
求距离 s 可用运动学方程 s = ut + ½at²。
s = 0 × 10 + ½ × 2 × (10)² = 100 m
An alternative approach uses average velocity: vav = (u + v) / 2 = (0 + 20) / 2 = 10 m/s, so s = vav × t = 10 × 10 = 100 m. Both methods give an acceleration of 2 m/s² and a distance of 100 m.
也可用平均速度法：vav = (u + v) / 2 = 10 m/s，故 s = vav × t = 100 m。两种方法均得出加速度2 m/s²，距离100 m。
2. Newton’s First Law | 牛顿第一定律
State Newton’s first law of motion and give one real‑world example that demonstrates it.
陈述牛顿第一运动定律，并给出一个展示该定律的实际例子。
Solution: Newton’s first law (the law of inertia) states that a body will remain at rest or continue to move with constant velocity unless acted upon by a resultant external force.
解答：牛顿第一定律（惯性定律）指出，除非受到合外力的作用，否则物体将保持静止或匀速直线运动状态。
A classic example occurs when a bus brakes sharply: standing passengers lurch forward. Their bodies try to maintain the original forward motion due to inertia, while the bus floor decelerates beneath them. Seat belts provide the unbalanced force needed to change the passengers’ motion.
典型例子是公交车急刹车：站立的乘客向前倾倒。由于惯性，他们的身体试图保持原有的向前运动，而车底板已减速。安全带提供了改变乘客运动所需的非平衡力。
3. Change in Momentum | 动量变化
A ball of mass 0.5 kg moves horizontally at 4 m/s. It strikes a wall and rebounds at 3 m/s in the opposite direction. Taking the initial direction as positive, calculate the change in momentum of the ball.
一个质量为0.5 kg的小球以4 m/s的水平速度运动。它撞墙后以3 m/s的速度沿相反方向反弹。设初速度方向为正，计算小球动量的变化。
Solution: Initial momentum pi = m × u = 0.5 × 4 = 2 kg m/s. Final velocity v = −3 m/s, so final momentum pf = 0.5 × (−3) = −1.5 kg m/s.
解答：初动量 pi = m × u = 0.5 × 4 = 2 kg m/s。末速度 v = −3 m/s，故末动量 pf = 0.5 × (−3) = −1.5 kg m/s。
Δp = pf − pi = −1.5 − 2 = −3.5 kg m/s
The negative sign indicates the change in momentum is 3.5 kg m/s in the opposite direction to the initial motion. The magnitude of the momentum change is 3.5 kg m/s.
负号表示动量变化大小为3.5 kg m/s，方向与初运动方向相反。动量变化量的绝对值为3.5 kg m/s。
4. Work Done & Power | 功与功率
A horizontal force of 50 N pushes a box 3 m across a smooth floor. Calculate the work done on the box. If the box was moved in 2 seconds, find the power developed.
一个50 N的水平力将一个箱子沿光滑地面推行了3 m。计算对箱子所做的功。若箱子在2秒内被移动，求产生的功率。
Solution: Work done W = force × distance moved in the direction of the force, so W = 50 N × 3 m = 150 J.
解答：功 W = 力 × 沿力方向移动的距离，因此 W = 50 N × 3 m = 150 J。
Power is the rate of doing work: P = W / t = 150 J / 2 s = 75 W. Thus, 75 joules of energy are transferred every second.
功率是做功的速率：P = W / t = 150 J / 2 s = 75 W。即每秒传递75焦耳的能量。
5. Free‑fall Velocity | 自由落体速度
An object is dropped from a height of 20 m above the ground. Assuming g = 10 m/s² and negligible air resistance, calculate its velocity just before it hits the ground.
一个物体从距地面20 m高处自由释放。设 g = 10 m/s² 且空气阻力可忽略，计算它即将撞击地面前的瞬时速度。
Solution: Use the equation v² = u² + 2as, where u = 0, a = g = 10 m/s², and s = 20 m.
解答：使用方程 v² = u² + 2as，其中 u = 0，a = g = 10 m/s²，s = 20 m。
v² = 0 + 2 × 10 × 20 = 400 → v = √400 = 20 m/s
The object hits the ground at 20 m/s. Notice that the mass of the object does not affect the final velocity in free‑fall, demonstrating that gravitational acceleration is independent of mass.
物体以20 m/s的速度撞击地面。注意在自由落体中物体的质量不影响末速度，这体现了重力加速度与质量无关。
6. Terminal Velocity | 终端速度
Explain why a skydiver reaches a terminal velocity during free‑fall before opening the parachute.
解释为什么跳伞者在打开降落伞前的自由下落过程中会达到终端速度。
Solution: Immediately after jumping, the skydiver’s weight is much larger than air resistance, causing a downward acceleration. As speed builds, air resistance increases until it equals the weight. The resultant force becomes zero, so acceleration ceases and the skydiver continues at a constant maximum speed — the terminal velocity.
解答：刚跳下时，重力远大于空气阻力，产生向下的加速度。随着速度增大，空气阻力不断增加，直至与重力相等。此时合外力为零，加速度消失，跳伞者以恒定的最大速度运动，即终端速度。
Factors such as body shape, cross‑sectional area and air density affect the magnitude of air resistance and therefore the terminal velocity. Before the parachute opens, a typical skydiver reaches about 55 m/s in a spread‑eagle position.
身体姿态、横截面积和空气密度等因素会影响空气阻力的大小，从而影响终端速度。在开伞前，采用四肢伸展姿势的典型终端速度约为55 m/s。
7. Wave Speed | 波速
A sound wave has a frequency of 50 Hz and a wavelength of 0.4 m. Calculate its speed.
一列声波的频率为50 Hz，波长为0.4 m。计算其波速。
Solution: The wave equation is v = f × λ. Substitute the given values:
解答：波动方程为 v = f × λ。代入已知数值：
v = 50 × 0.4 = 20 m/s
The wave travels at 20 m/s. This relationship holds for all types of waves — including light, water waves and sound — provided the medium is unchanged.
该波以20 m/s的速度传播。只要介质不变，这一
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