📚 Year 10 Cambridge Physics Unit Test Mock Exam Analysis | 剑桥物理十年级单元测试模拟卷解析
This article analyses a mock unit test designed for Year 10 Cambridge IGCSE Physics students. It covers key topics including motion, forces, energy, pressure, thermal physics, waves, optics, and electricity. Each question is followed by a step-by-step solution in both English and Chinese to reinforce understanding.
本文解析一份专为十年级剑桥IGCSE物理设计的单元模拟测试卷，涵盖运动、力、能量、压强、热物理、波动、光学和电学等核心主题。每道题目均附有中英文分步解答，以加深理解。
1. Speed and Acceleration | 速度和加速度
A car accelerates uniformly from rest to 20 m/s in 5 seconds. Calculate the acceleration and the distance travelled during this time.
一辆汽车从静止开始匀加速，5秒后速度达到20 m/s。求加速度以及这段时间内行驶的距离。
ce, apply the equation s = ut + ½ a t². Substituting: s = 0 × 5 + ½ × 4 × (5)² = 0 + ½ × 4 × 25 = 2 × 25 = 50 m. Alternatively, use average velocity: s = (u + v)/2 × t = (0 + 20)/2 × 5 = 10 × 5 = 50 m. Always check units.
根据加速度定义：a = (v - u) / t。初速度 u = 0，末速度 v = 20 m/s，时间 t = 5 s。所以 a = (20 - 0)/5 = 4 m/s²。求距离用公式 s = ut + ½ a t²。代入：s = 0×5 + ½×4×(5)² = 2×25 = 50 m。也可用平均速度：s = (u+v)/2 × t = (0+20)/2×5 = 50 m。注意检查单位。
2. Newton's Second Law | 牛顿第二定律
A force of 10 N acts on a mass of 5 kg initially at rest on a frictionless surface. Determine the acceleration and the velocity after 2 seconds.
一个10 N的力作用在5 kg的物体上，物体初始静止，表面无摩擦。求加速度和2秒后的速度。
Newton' arting from rest, u = 0, so v = 0 + 2 m/s² × 2 s = 4 m/s. The direction of velocity is the same as the applied force.
根据牛顿第二定律：F = m a，得 a = F/m = 10 N / 5 kg = 2 m/s²。由于力恒定，加速度恒定。速度公式 v = u + a t。初速度 u = 0，所以 v = 0 + 2×2 = 4 m/s。速度方向与力的方向相同。
3. Conservation of Energy | 能量守恒
ation of energy to calculate its speed just before hitting the ground. (Take g = 9.8 m/s²)
一个0.2 kg的小球从10 m高处自由落下。不计空气阻力，利用能量守恒求它刚撞击地面前的速度。（取 g = 9.8 m/s²）
servation: m g h = ½ m v². Mass m cancels out: g h = ½ v². Rearranging: v² = 2 g h, so v = √2 g h. Substitute: v = √2 × 9.8 × 10 = √196 ≈ 14 m/s. Notice mass does not affect the final speed.
在最高点，小球具有重力势能 GPE = m g h。在最低点，所有重力势能转化为动能 KE = ½ m v²。根据能量守恒：m g h = ½ m v²。质量 m 抵消，得 g h = ½ v²，整理得 v² = 2 g h，所以 v = √2 g h。代入数值：v = √2×9.8×10 = √196 ≈ 14 m/s。注意最终速度与质量无关。
4. Work and Power | 功与功率
A person pushes a box with a constant horizontal force of 50 N over a distance of 4 m in 2 seconds. Calculate the work done on the box and the power output.
一个人用50 N的恒定水平力推动箱子移动了4 m，用时2 s。计算对箱子做的功和输出的功率。
ive: P = F × v, where average velocity v = d / t = 4 m / 2 s = 2 m/s, so P = 50 N × 2 m/s = 100 W.
功 W = 力 × 在力的方向上移动的距离：W = F d = 50 N × 4 m = 200 J。功率是做功的速率：P = W / t = 200 J / 2 s = 100 W。也可用 P = F v，平均速度 v = d/t = 2 m/s，所以 P = 50 N × 2 m/s = 100 W。
5. Pressure | 压强
box has an area of 0.4 m². Calculate the pressure exerted on the floor. (Use g = 10 m/s²)
一个质量为20 kg的箱子放在地板上，箱子的底面积为0.4 m²。求箱子对地板的压强。（取 g = 10 m/s²）
r N/m²). Ensure area is in square metres. If the box were placed on a smaller area, pressure would be greater.
压强 P = 压力 / 面积。压力等于箱子的重量，F = m g = 20 kg × 10 m/s² = 200 N。所以 P = 200 N / 0.4 m
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