📚 Year 10 Eduqas Physics Unit Test Mock Paper Walkthrough | Year 10 Eduqas 物理：单元测试模拟卷解析
Welcome to this step-by-step walkthrough of a Year 10 Eduqas Physics unit test mock paper. This resource breaks down typical questions on motion, forces, energy, waves, electricity and radioactivity, showing you exactly how to reach the correct answer and where marks are gained. Use it to sharpen your exam technique and deepen your understanding of the key concepts.
欢迎阅读这份 Year 10 Eduqas 物理单元测试模拟卷的逐步解析。本文拆解了运动、力、能量、波、电学和放射性等典型题目，清晰展示如何得出正确答案以及如何获得分数。用它来打磨你的考试技巧并加深对核心概念的理解。
1. Displacement–Time Graph Analysis | 位移-时间图分析
The question presents a displacement−time graph with a straight line rising from the origin to 30 m in 6 s.
题目给出了一条从原点出发，在 6 s 内上升到 30 m 的直线型位移−时间图。
A straight line on a displacement–time graph means constant velocity, because the gradient does not change.
位移−时间图上的一条直线代表速度恒定，因为斜率不变。
Velocity = gradient = Δs / Δt = 30 m / 6 s = 5 m/s.
速度 = 斜率 = Δs / Δt = 30 m ÷ 6 s = 5 m/s。
Always include the unit, and remember that the gradient of a displacement–time graph gives velocity; a curved line would indicate acceleration.
务必带上单位，记住位移−时间图的斜率给出速度；曲线则表明存在加速度。
2. Uniform Acceleration Calculations | 匀加速度计算
A car accelerates uniformly from rest and travels 100 m in 8 s. Find the acceleration and final velocity.
一辆汽车从静止开始匀加速，8 s 内行驶了 100 m。求加速度和末速度。
s = ut + ½at². Since u = 0, s = ½ a t² → 100 = ½ × a × 64 → a = 100 × 2 / 64 = 3.125 m/s².
使用公式 s = ut + ½at²。因为 u = 0，得 s = ½ a t² → 100 = ½ × a × 64 → a = 100 × 2 ÷ 64 = 3.125 m/s²。
Final velocity: v = u + at = 0 + 3.125 × 8 = 25 m/s.
末速度：v = u + at = 0 + 3.125 × 8 = 25 m/s。
u² + 2as; substituting gives v² = 0 + 2 × 3.125 × 100 = 625, so v = 25 m/s, which matches.
也可以用 v² = u² + 2as 验证；代入得 v² = 0 + 2 × 3.125 × 100 = 625，因此 v = 25 m/s，结果一致。
3. Newton’s Second Law | 牛顿第二定律
A 1200 kg car experiences a resultant force of 3600 N. Calculate its acceleration and explain how the acceleration changes if the mass is increased.
一辆 1200 kg 的汽车受到 3600 N 的合外力。计算其加速度并解释若质量增加，加速度如何变化。
F = m a → a = F / m = 3600 N / 1200 kg = 3.0 m/s².
F = m a → a = F / m = 3600 N ÷ 1200 kg = 3.0 m/s²。
If the mass increases while the resultant force stays the same, acceleration decreases because a is inversely proportional to m.
若质量增加而合外力不变，加速度减小，因为 a 与 m 成反比。
Always quote the relationship: for a constant force, doubling the mass halves the acceleration.
始终点明关系：力恒定时，质量加倍则加速度减半。
4. Kinetic Energy and Energy Conservation | 动能与能量守恒
A cyclist of total mass 80 kg is moving at 10 m/s. She applies brakes and the bike stops without skidding. Explain the energy transfers using the principle of conservation of energy.
一个总质量为 80 kg 的自行车手以 10 m/s 行驶。她捏下刹车，车子不打滑地停下来。用能量守恒原理解释能量转化。
Initial kinetic energy Eₖ = ½ m v² = ½ × 80 × 10² = 4000 J.
初始动能 Eₖ = ½ m v² = ½ × 80 × 10² = 4000 J。
During braking, this kinetic energy is transferred mechanically into thermal energy in the brake pads and rims, and ultimately into the surroundings as heat. No energy is destroyed; 4000 J is dissipated as thermal energy.
刹车过程中，这些动能通过机械方式转化为刹车片和轮圈的热能，最终以热的形式散失到周围环境中。能量没有消失；4000 J 以热能形式耗散。
Conservative of energy means total energy before equals total energy after, just in different forms.
能量守恒意味着总能量在转化前后相等，只是形式不同。
5. Wave Speed, Frequency and Wavelength | 波速、频率与波长
An oscilloscope shows a sound wave with a period of 0.004 s. If the wavelength is 1.36 m, calculate the speed.
示波器显示一个周期为 0.004 s 的声波。若波长为 1.36 m，计算波速。
Frequency f = 1 / T = 1 / 0.004 = 250 Hz.
频率 f = 1 / T = 1 ÷ 0.004 = 250 Hz。
Wave speed v = f λ = 250 × 1.36 = 340 m/s.
波速 v = f λ = 250 × 1.36 = 340 m/s。
This is the typical speed of sound in air at room temperature, which confirms our working.
这是室温下空气中声音传播的典型速度，验证了我们的计算。
6. Radioactive Half-Life | 放射性半衰期
A sample of iodine-131 has an initial activity of 3200 Bq and a half-life of 8 days. Calculate the activity remaining after 24 days and state the fraction undecayed.
一份碘-131 样品的初始活度为 3200 Bq，半衰期为 8 天。计算 24 天后的剩余活度，并说明未衰变的比例。
Number of half-lives = total time / half-life = 24 / 8 = 3.
半衰期数 = 总时间 ÷ 半衰期 = 24 ÷ 8 = 3。
After each half-life the activity halves: 3200 → 1600 → 800 → 400 Bq.
每过一个半衰期活度减半：3200 → 1600 → 800 → 400 Bq。
Fraction remaining = (½)³ = 1/8. So 400 Bq is 1/8 of the original activity.
剩余比例 = (½)³ = 1/8。因此 400 Bq 是原来活度的 1/8。
7. Circuit Calculations with Ohm’s Law | 欧姆定律电路计算
A 12 Ω resistor is connected across a 6 V battery. Calculate the current and the charge flowing in 5 minutes.
一个 12 Ω 的电阻器连接到 6 V 的电池上。计算电流和 5 分钟内流过的电荷量。
I = V / R = 6 V / 12 Ω = 0.5 A.
I = V ÷ R = 6 V ÷ 12 Ω = 0.5 A。
Charge Q = I t. t = 5 min = 300 s. Q = 0.5 A × 300 s = 150 C.
电荷量 Q = I t。t = 5 min = 300 s。Q = 0.5 A × 300 s = 150 C。
If the resistor is replaced with a 24 Ω resistor, the current halves, so the charge in the same time also halves, illustrating the inverse relationship.
若将电阻器换成 24 Ω，电流减半，相同时间内的电荷量也减半，说明了反比关系。
8. Nuclear Equations | 核反应方程
ete the alpha decay equation: ²²⁶₈₈Ra → X + ⁴₂He. Determine X and write the full equation.
完成下面的 α 衰变方程：²²⁶₈₈Ra → X + ⁴₂He。确定 X 并写出完整方程。
the atomic number by 2. So the daughter nuclide X has A = 226 − 4 = 222, Z = 88 − 2 = 86.
α 衰变中质量数减少 4，原子序数减少 2。因此子体核素 X 的 A = 226 − 4 = 222，Z = 88 − 2 = 86。
Element with Z = 86 is radon (Rn). So X is radon-222: ²²²₈₆Rn.
原子序数为 86 的元素是氡 (Rn)。因此 X 是氡-222：²²²₈₆Rn。
Full equation: ²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂He.
完整方程：²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂He。
Alpha particles are helium nuclei, and the equation balances on both mass and charge.
α 粒子是氦原子核，该方程在质量和电荷上均守恒。
9. Momentum in Collisions | 碰撞中的动量
A 2 kg trolley travelling at 4 m/s to the right collides with a stationary 1 kg trolley. After the collision they lock together. Find the combined velocity.
一辆 2 kg 的小推车以 4 m/s 向右运动，撞上一辆静止的 1 kg 小推车。碰撞后两者锁在一起。求共同速度。
Total momentum before = m₁u₁ + m₂u₂ = 2 × 4 + 1 × 0 = 8 kg m/s.
碰撞前总动量 = m₁u₁ + m₂u₂ = 2 × 4 + 1 × 0 = 8 kg m/s。
m, total momentum after = (m₁ + m₂) v = 3 v. So 3 v = 8 → v = 8/3 ≈ 2.67 m/s to the right.
根据动量守恒，碰撞后总动量 = (m₁ + m₂) v = 3 v。因此 3 v = 8 → v = 8/3 ≈ 2.67 m/s 向右。
Even though kinetic energy is not conserved in this inelastic collision, momentum is always conserved in a closed system.
尽管在这种非弹性碰撞中动能不守恒，但在封闭系统中动量始终守恒。
10. Required Practical: Measuring g | 规定实验：测量重力加速度
Describe a valid method to determine the acceleration due to gravity g using a simple pendulum or a light gate and a falling object. Identify the measurements and the graph you would plot.
描述一个用单摆或光门与下落物体测定重力加速度 g 的合理方法。指出需要测量的量以及你绘制的图像。
ht gate (or the time of fall between two light gates). Use s = ut + ½at² for free fall from rest, giving h = ½ g t².
使用光门和一个从静止释放的钢球：用米尺测量下落高度 h，记录通过光门的时间 t（或两个光门之间的下落时间）。对自由落体应用 s = ut + ½at²，得 h = ½ g t²。
or various heights. Plot a graph of h against t². The gradient = ½ g, so g = 2 × gradient.
对不同高度重复实验。绘制 h 对 t² 的图像。斜率 = ½ g，因此 g = 2 × 斜率。
o g = 4π² / gradient. State precautions like small oscillation angles and reliable timing.
若使用单摆：测量不同摆长 L 的周期 T，绘制 T² 对 L 的图像，斜率 = 4π² / g，因此 g = 4π² / 斜率。说明注意事项，如小摆角和高精度计时。
This experimental method is a common practical skills question; always link the graph gradient to the physics formula.
该实验方法是常见的实验技能考题；务必将图像斜率与物理公式联系起来。
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