📚 Year 11 Edexcel Physics Mock Unit Test Walkthrough | Edexcel 物理单元测试模拟卷解析
This article provides a step-by-step walkthrough of a mock unit test designed for Year 11 Edexcel Physics. The questions cover key topics including motion, forces, energy, momentum, waves, electricity and radioactivity. Each solution is presented with a clear bilingual explanation to help you master exam technique and deepen conceptual understanding.
本文为 Year 11 Edexcel 物理单元测试模拟卷提供逐步解析。题目涵盖运动、力、能量、动量、波动、电学和放射性等核心知识点。每道题的解答均配有清晰的中英双语解释，帮助同学掌握解题技巧并深化概念理解。
1. Average Speed Calculation | 平均速度计算
A runner completes a 400 m lap in 50 seconds. Calculate the average speed.
一名跑步运动员在 50 秒内跑完 400 米一圈。计算平均速度。
Recall that average speed is defined as total distance divided by time.
平均速度的定义是总距离除以时间。
average speed = distance ÷ time
Substitute the values: distance = 400 m, time = 50 s.
代入数值：距离 = 400 m，时间 = 50 s。
average speed = 400 m ÷ 50 s = 8 m/s
The runner's average speed is 8 metres per second. Always check that your answer has the correct unit – in this case m/s.
跑步者的平均速度是 8 米每秒。务必检查答案是否有正确单位——本例为 m/s。
2. Acceleration and Distance Travelled | 加速度与运动距离
A car accelerates uniformly from rest to 25 m/s in 10 seconds. Calculate (a) the acceleration and (b) the distance travelled during this time.
一辆汽车从静止开始匀加速，10 秒后速度达到 25 m/s。计算 (a) 加速度和 (b) 此期间行驶的距离。
equation for uniform acceleration: a = (v − u) / t, where u = 0 m/s, v = 25 m/s, t = 10 s.
第 (a) 部分使用匀加速度定义式：a = (v − u) / t，其中 u = 0 m/s，v = 25 m/s，t = 10 s。
a = (25 m/s − 0 m/s) ÷ 10 s = 2.5 m/s²
The acceleration is 2.5 metres per second squared.
加速度为 2.5 米每二次方秒。
ar starts from rest, we can use s = ut + ½ a t². With u = 0 this simplifies to s = ½ a t².
第 (b) 部分，因为汽车从静止开始，可使用 s = ut + ½ a t²。由于 u = 0，简化为 s = ½ a t²。
s = ½ × 2.5 m/s² × (10 s)² = 0.5 × 2.5 × 100 = 125 m
The car travels 125 metres while accelerating. Remember to square the time correctly.
汽车在加速过程中行驶了 125 米。注意正确计算时间的平方。
3. Newton's Second Law | 牛顿第二定律
A resultant force of 50 N acts on a block of mass 20 kg. Calculate the acceleration of the block and state the law you have used.
一个大小为 50 N 的合力作用在质量为 20 kg 的木块上。计算木块的加速度并说明你使用的定律。
Newton' ultant force acting on it and inversely proportional to its mass. Mathematically: F = m a.
牛顿第二定律指出：物体的加速度与作用在它上面的合外力成正比，与其质量成反比。数学表达式为 F = m a。
a = F ÷ m = 50 N ÷ 20 kg = 2.5 m/s²
The acceleration is 2.5 m/s² in the direction of the resultant force. Always ensure force is in newtons and mass in kilograms.
加速度为 2.5 m/s²，方向与合力方向相同。务必确保力以牛顿为单位，质量以千克为单位。
4. Work Done and Gravitational Potential Energy | 做功与重力势能
/kg. Calculate the work done against gravity and the gain in gravitational potential energy (GPE).
一个 10 kg 的物体被竖直向上提升 2 m。取 g = 10 N/kg。计算克服重力所做的功以及重力势能 (GPE) 的增加量。
gravity = force needed to lift × vertical distance. The force needed equals the weight: Wweight = m × g = 10 kg × 10 N/kg = 100 N.
克服重力所做的功 = 提升所需的力 × 竖直距离。所需力等于重力：Wweight = m × g = 10 kg × 10 N/kg = 100 N。
work done = 100 N × 2 m = 200 J
The work done against gravity is 200 joules. For a lift at constant speed, this is exactly equal to the gain in GPE.
克服重力做功为 200 焦耳。匀速提升时，该值恰好等于重力势能的增加量。
GPE gained = m g h = 10 × 10 × 2 = 200 J. Thus GPE increases by 200 J.
重力势能增加量 = m g h = 10 × 10 × 2 = 200 J。因此 GPE 增加 200 J。
5. Momentum Change | 动量变化
A 0.5 kg ball is travelling horizontally at 12 m/s when it hits a vertical wall. It rebounds straight back at 8 m/s. Calculate the change in momentum of the ball.
一个 0.5 kg 的球以 12 m/s 的速度水平运动并撞向竖直墙壁。它沿原路以 8 m/s 的速度反弹。求球的动量变化。
v₁ = +12 m/s and final velocity v₂ = −8 m/s (the minus indicates the opposite direction).
选取初始方向为正方向。则初速度 v₁ = +12 m/s，末速度 v₂ = −8 m/s（负号表示方向相反）。
Momentum change Δp = final momentum − initial momentum = m v₂ − m v₁ = m (v₂ − v₁).
动量变化 Δp = 末动量 − 初动量 = m v₂ − m v₁ = m (v₂ − v₁)。
Δp = 0.5 kg × (−8 m/s − 12 m/s) = 0.5 × (−20) = −10 kg m/s
The change in momentum is −10 kg m/s; the magnitude of the change is 10 kg m/s, and the negative sign tells us the change is in the reverse direction.
动量变化为 −10 kg m/s；变化量的大小为 10 kg m/s，负号表明变化方向与初速度方向相反。
6. Wave Speed | 波速
A water wave has a frequency of 5 Hz and a wavelength of 2 m. Calculate the speed of the wave.
一个水波的频率为 5 Hz，波长为 2 m。计算波速。
Wave speed is given by the equation v = f λ, where f is frequency and λ is wavelength.
波速由公式 v = f λ 给出，其中 f 为频率，λ 为波长。
v = 5 Hz × 2 m = 10 m/s
The wave travels at 10 metres per second. This is typical for water ripples.
该波以 10 米每秒的速度传播，属于水波常见的数值。
7. Ohm's Law and Power Dissipation | 欧姆定律与功率
A resistor of 10 Ω carries a current of 2 A. Calculate the potential difference (p.d.) across the resistor and the power dissipated by it.
一个 10 Ω 的电阻中流过的电流为 2 A。计算电阻两端的电势差（p.d.）及其消耗的功率。
Ohm's law: V = I R, where V is p.d., I is current and R is resistance.
欧姆定律：V = I R，V 为电势差，I 为电流，R 为电阻。
V = 2 A × 10 Ω = 20 V
The p.d. across the resistor is 20 volts.
电阻两端的电势差为 20 伏特。
Power dissipated can be found from P = I V or P = I² R. Using P = I V:
消耗的功率可通过 P = I V 或 P = I² R 求得。使用 P = I V：
P = 2 A × 20 V = 40 W
The resistor dissipates 40 watts of power as heat.
该电阻以热能形式消耗 40 瓦特的功率。
8. Resistors in Parallel | 并联电阻
Two identical 6 Ω resistors are connected in parallel. Calculate the total effective resistance.
两个相同的 6 Ω 电阻器并联连接。计算总有效电阻。
For resistors in parallel, the total resistance Rtotal is found from:
对于并联电阻，总电阻 Rtotal 由下式求得：
1 / Rtotal = 1 / R₁ + 1 / R₂
Substituting R₁ = 6 Ω and R₂ = 6 Ω:
代入 R₁ = 6 Ω、R₂ = 6 Ω：
1 / Rtotal = 1/6 + 1/6 = 2/6 = 1/3
Therefore Rtotal Ω. When identical resistors are in parallel, the total resistance is half of one resistor's value.
因此 Rtotal = 3 Ω。当相同电阻并联时，总电阻为单个电阻值的一半。
9. Refractive Index | 折射率
is 19°. (sin 30° = 0.500, sin 19° = 0.3256). Calculate the refractive index of the glass.
一束光线从空气射入玻璃。入射角为 30°，折射角为 19°。（sin 30° = 0.500，sin 19° = 0.3256）。计算玻璃的折射率。
Snell' aw: n = sin i / sin r, where i is the angle of incidence and r is the angle of refraction.
斯涅尔定律：n = sin i / sin r，i 为入射角，r 为折射角。
n = sin 30° ÷ sin 19° = 0.500 / 0.3256 ≈ 1.54
The refractive index of the glass is approximately 1.54. This value is slightly larger than typical crown glass, making it a plausible result for a dense glass.
该玻璃的折射率约为 1.54。该值略大于典型的冕玻璃，符合致密玻璃的合理结果。
10. Half-life Calculation | 半衰期计算
A radioactive isotope has a half-life of 3 days. The initial count rate is 800 counts per second. What will the count rate be after 9 days?
某种放射性同位素的半衰期为 3 天。初始计数率为每秒 800 次。9 天后的计数率是多少？
e the number of half-lives that have passed: 9 days ÷ 3 days per half-life = 3 half-lives.
首先确定经过的半衰期个数：9 天 ÷ 每 3 天一个半衰期 = 3 个半衰期。
Each half-life reduces the count rate by a factor of 2. After 3 half-lives the count rate is halved three times.
每经过一个半衰期，计数率减半。经过 3 个半衰期后计数率减半三次。
Count rate after 1 half-life: 800 ÷ 2 = 400 counts/s
After 2 half-lives: 400 ÷ 2 = 200 counts/s
After 3 half-lives: 200 ÷ 2 = 100 counts/s
Alternatively, use the formula: final count rate = initial × (1/2)n = 800 × (1/2)³ = 800 × 1/8 = 100 counts/s.
另一种方法是使用公式：最终计数率 = 初始值 × (1/2)ⁿ = 800 × (1/2)³ = 800 × 1/8 = 100 counts/s。
After 9 days the count rate is 100 counts per second.
9 天后计数率为每秒 100 次。
Published by TutorHao | Physics Revision Series | alevel
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