📚 Year 10 SQA Physics Case Study Practice | 苏格兰SQA物理案例分析实战演练
Mastering physics is not just about memorising formulas—it is about applying them to real-world situations. This article presents a series of case study exercises designed for Year 10 pupils following the SQA Physics curriculum. Each case focuses on a key topic, from motion and energy to electricity and waves, guiding you through the logical steps of problem-solving. By working through these examples, you will strengthen your understanding and build the confidence needed for assessments.
掌握物理不只是记忆公式——更重要的是将公式应用到真实情境中。本文为学习SQA物理课程的十年级学生设计了一系列案例分析练习。每个案例聚焦一个核心主题，从运动和能量到电学与波动，引导你一步步完成逻辑解题过程。通过练习这些实例，你将加深理解并树立应对考核的信心。
1. Motion with Constant Acceleration | 匀加速运动案例分析
A car accelerates from rest at a constant rate of 3.0 m/s² along a straight road. Calculate its velocity after 8.0 seconds and the distance it covers in that time.
一辆汽车从静止开始以3.0 m/s²的恒定加速度沿直线道路行驶。请计算8.0秒后的速度以及在这段时间内行驶的距离。
city: v = u + at, where u = 0, a = 3.0 m/s², t = 8.0 s. Thus v = 0 + (3.0 × 8.0) = 24 m/s.
我们使用运动学方程。求速度：v = u + at，其中u = 0，a = 3.0 m/s²，t = 8.0 s。因此 v = 0 + (3.0 × 8.0) = 24 m/s。
½ × 3.0 × (8.0)² = 0.5 × 3.0 × 64 = 96 m. The car travels 96 metres while reaching 24 m/s.
求距离：s = ut + ½at²。同样u = 0，所以 s = ½ × 3.0 × (8.0)² = 0.5 × 3.0 × 64 = 96 m。汽车达到24 m/s时驶过了96米。
2. Newton’s Second Law in Action | 牛顿第二定律的应用
A trolley of mass 4.5 kg is pulled along a frictionless surface by a horizontal force of 13.5 N. Determine the acceleration of the trolley and the velocity it gains after 5.0 s if it starts from rest.
一辆质量为4.5 kg的小车在无摩擦表面上受到13.5 N的水平拉力。求小车的加速度，以及若从静止起步5.0秒后获得的速度。
s second law: F = m a, so a = F / m = 13.5 / 4.5 = 3.0 m/s². The acceleration is 3.0 m/s².
牛顿第二定律：F = m a，所以 a = F / m = 13.5 / 4.5 = 3.0 m/s²。加速度为3.0 m/s²。
Velocity after 5.0 s: v = u + at = 0 + 3.0 × 5.0 = 15.0 m/s. The trolley moves at 15 m/s.
5.0秒后的速度：v = u + at = 0 + 3.0 × 5.0 = 15.0 m/s。小车的速度为15 m/s。
3. Work and Kinetic Energy | 功与动能的转换
A constant force of 20 N pushes a 2.0 kg block over a distance of 3.0 m along a smooth table. The block was initially at rest. Find the work done by the force and the final kinetic energy of the block. Hence determine its final speed.
一颗20 N的恒力在光滑桌面上将一块2.0 kg的物块推动了3.0 m。物块起初静止。求力所做的功以及物块的末动能，并由此计算末速度。
20 × 3.0 = 60 J. Since the surface is smooth, all work converts into kinetic energy. So Ek = 60 J.
功 W = F × d = 20 × 3.0 = 60 J。因为表面光滑，所有功转化为动能。因此 Ek = 60 J。
Kinetic energy formula: Ek = ½ m v². Rearranging: v = √2E_k / m = √2 × 60 / 2.0 = √120 / 2.0 = √60 ≈ 7.75 m/s. The block reaches about 7.7 m/s.
动能公式：Ek = ½ m v²。变形得 v = √2E_k / m = √2 × 60 / 2.0 = √120 / 2.0 = √60 ≈ 7.75 m/s。物块的末速度约为7.7 m/s。
4. Gravitational Potential Energy | 重力势能的计算
.8 N/kg.) If the backpack falls freely from the shelf, at what speed will it hit the floor? (Ignore air resistance.)
一名学生把一个5.0 kg的书包从地板举到1.8 m高的架子上。计算增加的重力势能。（取g = 9.8 N/kg。）如果书包从架子自由下落，忽略空气阻力，它撞击地板时的速度是多少？
Gravitational potential energy: Ep = m g h = 5.0 × 9.8 × 1.8 = 88.2 J. The backpack gains 88.2 J of potential energy.
重力势能：Ep = m g h = 5.0 × 9.8 × 1.8 = 88.2 J。书包获得了88.2 J的势能。
If it falls, all Ep converts to kinetic energy just before impact. So ½ m v² = 88.2 J. Then v = √2 × 88.2 / 5.0 = √176.4 / 5.0 = √35.28 ≈ 5.94 m/s. Impact speed is about 5.9 m/s.
如果下落，落地前所有势能转化为动能。因此 ½ m v² = 88.2 J。于是 v = √2 × 88.2 / 5.0 = √176.4 / 5.0 = √35.28 ≈ 5.94 m/s。撞击速度约为5.9 m/s。
5. Conservation of Momentum | 动量守恒分析
A 2.0 kg trolley moving at 3.0 m/s collides with a stationary 1.0 kg trolley. They stick together after the collision. Find the velocity of the combined trolleys just after the collision.
一辆2.0 kg的小车以3.0 m/s的速度与一辆静止的1.0 kg小车碰撞，碰后粘连在一起。求碰撞后瞬间组合小车的速度。
, moving at velocity v. So 3.0 × v = 6.0, giving v = 2.0 m/s. The joined trolleys move at 2.0 m/s in the original direction.
根据动量守恒：碰前总动量等于碰后总动量。碰前：p = (2.0 × 3.0) + (1.0 × 0) = 6.0 kg m/s。碰后：总质量为3.0 kg，以速度v运动。于是3.0 × v = 6.0，解得v = 2.0 m/s。粘连小车以2.0 m/s沿原方向运动。
6. Ohm's Law and Simple Circuits | 欧姆定律与简单电路
A resistor of 15 Ω is connected to a 6.0 V battery. Calculate the current flowing through the resistor and the charge that passes through it in 2 minutes.
一个15 Ω的电阻器连接到6.0 V电池上。计算通过电阻器的电流以及2分钟内通过它的电荷量。
Ohm’s law: V = I R, so I = V / R = 6.0 / 15 = 0.40 A. The current is 0.40 A.
欧姆定律：V = I R，因此 I = V / R = 6.0 / 15 = 0.40 A。电流为0.40 A。
2 min = 120 s. Q = 0.40 × 120 = 48 C. A charge of 48 coulombs passes through the resistor.
电荷量 Q = I × t。时间 t = 2 min = 120 s。Q = 0.40 × 120 = 48 C。有48库仑的电荷通过电阻器。
7. Electrical Power and Energy | 电功率与电能消耗
An electric heater operates at 230 V and draws a current of 4.0 A. Determine its power rating and the energy consumed in 30 minutes, giving the answer in joules and in kilowatt-hours.
一个电暖器工作电压为230 V，通过电流为4.0 A。求其额定功率，以及30分钟内消耗的能量，分别以焦耳和千瓦时为单位作答。
Power P = V × I = 230 × 4.0 = 920 W (0.92 kW).
功率 P = V × I = 230 × 4.0 = 920 W (0.92 kW)。
kilowatt-hours: E = 0.92 kW × 0.5 h = 0.46 kWh. The heater uses 0.46 units of electricity.
焦耳计的能量：E = P × t = 920 × (30 × 60) = 920 × 1800 = 1,656,000 J = 1.656 MJ。以千瓦时计：E = 0.92 kW × 0.5 h = 0.46 kWh。暖器消耗了0.46度电。
8. Gas Laws and the Kelvin Scale | 气体定律与开尔文温标
A fixed mass of gas has a volume of 250 cm³ at 27 °C. Its pressure remains constant while it is heated. At what Celsius temperature will its volume reach 300 cm³?
某固定质量的气体在27 °C时体积为250 cm³。保持压强不变进行加热。当其体积达到300 cm³时，摄氏温度是多少？
00 × 300) / 250 = 90000 / 250 = 360 K. Convert back to Celsius: 360 - 273 = 87 °C. The gas must be heated to 87 °C.
查理定律：V₁/T₁ = V₂/T₂，温度需用开尔文。将27 °C换算为开尔文：T₁ = 27 + 273 = 300 K。V₁ = 250 cm³，V₂ = 300 cm³。因此 T₂ = (V₂ × T₁) / V₁ = (300 × 300) / 250 = 90000 / 250 = 360 K。换回摄氏度：360 - 273 = 87 °C。气体需要加热到87 °C。
9. Wave Properties and the Wave Equation | 波的性质与波动方程
A water wave has a frequency of 5.0 Hz and a wavelength of 0.40 m. Calculate the wave speed. If the frequency doubles while the speed remains the same, what is the new wavelength?
一列水波的频率为5.0 Hz，波长为0.40 m。计算波速。如果频率加倍而波速不变，新的波长是多少？
Wave equation: v = f λ. So v = 5.0 × 0.40 = 2.0 m/s.
波动方程：v = f λ。因此 v = 5.0 × 0.40 = 2.0 m/s。
to 10 Hz while v stays 2.0 m/s, then λ = v / f = 2.0 / 10 = 0.20 m. The wavelength halves.
如果频率加倍到10 Hz而v保持2.0 m/s，那么 λ = v / f = 2.0 / 10 = 0.20 m。波长减半。
10. Refraction and Snell’s Law | 折射与斯涅尔定律
Light travels from air into a glass block. The angle of incidence in air is 45°, and the angle of refraction in the glass is 28°. Calculate the refractive index of the glass. (Assume the refractive index of air is 1.00.)
光线从空气进入玻璃块。空气中的入射角为45°，玻璃中的折射角为28°。求玻璃的折射率。（设空气的折射率为1.00。）
5°) / (sin 28°) = 0.7071 / 0.4695 ≈ 1.51. The refractive index of the glass is about 1.51.
斯涅尔定律：n₁ sin θ₁ = n₂ sin θ₂。此处 n₁ = 1.00，θ₁ = 45°，θ₂ = 28°。因此 n₂ = (sin 45°) / (sin 28°) = 0.7071 / 0.4695 ≈ 1.51。玻璃的折射率约为1.51。
11. Radioactive Decay and Half-Life | 放射性衰变与半衰期
A sample of a radioactive isotope has an initial activity of 800 Bq. Its half-life is 3.0 hours. What will its activity be after 9.0 hours? How much time is needed for the activity to fall below 100 Bq?
一块放射性同位素样品的初始活度为800 Bq，半衰期为3.0小时。9.0小时后的活度是多少？活度降至100 Bq以下需要多长时间？
r of half-lives in 9.0 hours: n = 9.0 / 3.0 = 3. Activity after n half-lives: A = A₀ × (½)n = 800 × (½)3 = 800 × 1/8 = 100 Bq.
9.0小时包含的半衰期个数：n = 9.0 / 3.0 = 3。n个半衰期后的活度：A = A₀ × (½)n = 800 × (½)3 = 800 × 1/8 = 100 Bq。
es (9.0 h) activity is exactly 100 Bq, so after 3.5 half-lives it would be about 800 × (½)3.5 ≈ 70.7 Bq. Thus the activity drops below 100 Bq just after 9.0 hours—trictly, any time beyond 9.0 hours achieves this. The practical answer is after 9.0 hours.
要降至100 Bq以下，样品需要超过3个半衰期。3个半衰期（9.0小时）后活度恰好为100 Bq，所以3.5个半衰期后活度约为800 × (½)3.5 ≈ 70.7 Bq。因此活度在9.0小时之后立刻降至100 Bq以下。实际答案是9.0小时后即可。
12. The Cost of Electricity | 电力成本计算
A 2.5 kW kettle is used for 4.0 minutes to boil water. If electricity costs 15 pence per kilowatt-hour, calculate the cost of this operation. Also, how much energy is transferred in joules?
一个2.5 kW的电热水壶工作4.0分钟烧水。如果每千瓦时电费为15便士，计算此次烧水的费用，并求出以焦耳为单位的能量转移量。
Wh: E = P × t = 2.5 kW × 0.06667 h = 0.1667 kWh. Cost = 0.1667 kWh × 15 p/kWh = 2.5 pence.
时间以小时计：4.0 min = 4.0 / 60 = 0.06667 h。以千瓦时计的能量：E = P × t = 2.5 kW × 0.06667 h = 0.1667 kWh。费用 = 0.1667 kWh × 15 p/kWh = 2.5便士。
67 kWh = 0.1667 × 3,600,000 J = 600,000 J (or 600 kJ). So 600 kJ of energy is transferred.
以焦耳计的能量：0.1667 kWh = 0.1667 × 3,600,000 J = 600,000 J（即600 kJ）。因此转移了600 kJ的能量。
Published by TutorHao | Physics Revision Series | aleveler.com
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