📚 Year 7 OCR Physics: Case Study Practical Exercises | 七年级OCR物理：案例分析实战演练
In Year 7 OCR Physics, mastering concepts is all about seeing how ideas work in real situations. This article takes you through ten practical case studies, from bouncing balls to electric circuits, to sharpen your analytical skills and build confidence for assessments. Each case is broken down with clear physical principles, steps to investigate, and calculations where needed, so you can treat these as your own practice scenarios.
在七年级OCR物理中，掌握概念的关键在于观察它们在真实情境中如何起作用。本文通过十个实践案例——从弹跳球到电路——来提升你的分析能力，并为评估考试建立信心。每个案例都分解为清晰的物理原理、探究步骤和必要的计算，你可以把它们当作自己的练习场景。
1. Measuring Length and Time: The Bouncing Ball Experiment | 测量长度与时间：弹跳球实验
A rubber ball is dropped from a fixed height of 100 cm. We want to find the average speed of the ball during the upward part of its first bounce. The distance travelled upwards is the bounce height, which we measure with a metre ruler. We also use a stopwatch to record the time from the moment the ball leaves the floor until it reaches the highest point.
一只橡皮球从固定高度100厘米落下。我们希望求出球在第一次反弹上升过程中的平均速度。向上移动的距离就是反弹高度，我们用米尺测量。我们还使用秒表记录从球离开地面瞬间至到达最高点的时间。
A typical result shows a bounce height of 62 cm (0.62 m) and a time of 0.31 s. Using the speed equation:
average speed = distance travelled ÷ time taken
一次典型的结果显示反弹高度为62厘米（0.62米），时间为0.31秒。使用速度公式：
平均速度 = 移动距离 ÷ 所花时间
Plugging in the values: average speed = 0.62 m ÷ 0.31 s = 2.0 m/s. To improve reliability, we repeat the drop five times, discard any obviously anomalous readings, and calculate the mean bounce height and mean time before using the formula. This reduces the effect of random errors such as delayed stopwatch presses.
代入数值：平均速度 = 0.62米 ÷ 0.31秒 = 2.0米/秒。为了提高可靠性，我们重复实验五次，剔除任何明显异常读数，再计算平均反弹高度和平均时间，然后代入公式。这可以减少随机误差的影响，例如按下秒表延迟。
2. Balanced and Unbalanced Forces: The Tug-of-War | 平衡力与非平衡力：拔河比赛
Two teams pull on opposite ends of a rope. Team A exerts a force of 450 N to the left, and Team B exerts 450 N to the right. The forces are equal in size but opposite in direction. When we add them as vectors, the resultant force is 0 N. The rope remains stationary or, if already moving, continues at a constant speed in a straight line.
两支队伍在绳子的两端拉扯。A队向左施加450牛的力，B队向右施加450牛的力。这两个力大小相等、方向相反。当我们用矢量加法时，合力为0牛。绳子保持静止，或者如果已经在运动，则沿直线保持匀速运动。
Imagine Team A suddenly increases its pull to 520 N while Team B stays at 450 N. Now the net force is 520 N − 450 N = 70 N towards Team A. This unbalanced force causes the rope (and Team B) to accelerate to the left. The acceleration depends on the total mass being pulled. A free-body diagram with labelled arrows helps us visualise the forces and predict motion.
想象A队突然将拉力增加到520牛，而B队仍保持450牛。现在合力为520牛 − 450牛 = 70牛，方向朝向A队。这个非平衡力导致绳子和B队向左加速。加速度取决于被拉动的总质量。带有标注箭头的隔离体图可以帮助我们直观看到各个力并预测运动。
3. Gravity and Weight: Astronaut on Different Planets | 重力与重量：不同行星上的宇航员
An astronaut has a mass of 70 kg. Mass is the amount of matter and does not change with location. Weight is the force of gravity acting on that mass, calculated by:
weight = mass × gravitational field strength
一位宇航员的质量为70千克。质量是物体所含物质的多少，不随位置改变。重量是作用在该质量上的重力，可以通过以下公式计算：
重量 = 质量 × 重力场强度
On Earth, gravitational field strength g ≈ 10 N/kg, so weight = 70 × 10 = 700 N. On the Moon, g = 1.6 N/kg, giving weight = 70 × 1.6 = 112 N. On Jupiter, with g = 25 N/kg, weight would be 1750 N. The astronaut’s mass stays 70 kg everywhere. This table summarises the comparison:
在地球上，重力场强度g约为10牛/千克，因此重量 = 70 × 10 = 700牛。在月球上，g = 1.6牛/千克，重量 = 70 × 1.6 = 112牛。在木星上，g = 25牛/千克，重量将达到1750牛。宇航员的质量始终是70千克。下表总结了这一对比：
| Location | Mass (kg) | g (N/kg) | Weight (N) |
|---|---|---|---|
| Earth | 70 | 10 | 700 |
| Moon | 70 | 1.6 | 112 |
| Jupiter | 70 | 25 | 1750 |
We can use a newton meter to measure weight directly on Earth. To investigate g on another planet, we would divide the measured weight by the known mass.
我们可以用弹簧测力计直接测量地球上的重量。若要探究另一星球上的g值，可以用测得重量除以已知质量。
4. Friction and Air Resistance: The Parachute Design | 摩擦力与空气阻力：降落伞设计
When you design a model parachute, air resistance is the key force that slows the fall. A larger canopy catches more air particles, increasing the upward drag force. We can test this by making parachutes of different areas from a lightweight material and attaching the same plasticene ‘jumper’ to each.
当你设计一个降落伞模型时，空气阻力是减缓下落的关键力。更大的伞布能捕获更多的空气粒子，增大向上的阻力。我们可以用轻质材料制作不同面积的降落伞，并在每个下面悬挂相同的橡皮泥“跳伞者”来进行验证。
Dropping them from a fixed height and timing the descent shows that the largest canopy takes the longest to reach the ground. At first, the weight is greater than air resistance, so the parachute accelerates downwards. As speed increases, air resistance builds up. When air resistance equals the weight, the resultant force becomes zero and the parachute falls at a constant terminal velocity. A bigger canopy reaches terminal velocity sooner, at a lower speed.
从固定高度投放它们并计时下降过程，结果显示出最大伞布到达地面所需时间最长。开始时，重量大于空气阻力，降落伞向下加速。随着速度增加，空气阻力逐渐增大。当空气阻力等于重量时，合力为零，降落伞以恒定的终端速度下落。更大的伞布会更早在更低速度下达到终端速度。
You can plot a bar chart of average descent time against canopy area. This investigation helps explain why real parachutes are large and why a skydiver spreads their body to increase drag before opening the main chute.
你可以绘制平均下降时间与伞布面积的条形图。这个探究有助于解释为什么真正的降落伞很大，也解释了跳伞者为什么在打开主伞前会展开身体以增大阻力。
5. Density Calculations: Identifying Unknown Materials | 密度计算：识别未知物质
You are given a silvery metal block and asked to determine whether it is aluminium, iron, or zinc. Mass is found using a digital balance: 178 g. Volume is found by measuring the block’s dimensions: length 4.0 cm, width 3.0 cm, height 2.5 cm. Volume = 4.0 × 3.0 × 2.5 = 30 cm³.
给你一个银色金属块，要求判断它是铝、铁还是锌。用数字天平称得质量为178克。通过测量块的尺寸得到体积：长4.0厘米，宽3.0厘米，高2.5厘米。体积 = 4.0 × 3.0 × 2.5 = 30立方厘米。
Now calculate density:
density = mass ÷ volume
现在计算密度：
密度 = 质量 ÷ 体积
So density = 178 g ÷ 30 cm³ ≈ 5.93 g/cm³. Comparing with known values (aluminium ≈ 2.7 g/cm³, zinc ≈ 7.1 g/cm³, iron ≈ 7.9 g/cm³), the closest match is zinc. The slight difference could be due to measurement uncertainty or impurities. If the object had an irregular shape, we would use the displacement method with a measuring cylinder to find its volume.
因此密度 = 178克 ÷ 30立方厘米 ≈ 5.93克/立方厘米。与已知值（铝约2.7克/立方厘米，锌约7.1克/立方厘米，铁约7.9克/立方厘米）比较，最接近的是锌。微小的差异可能是由测量不确定性或杂质引起的。如果物体形状不规则，我们会使用排水法用量筒测量其体积。
6. Energy Transfers: Roller Coaster Energy | 能量转换：过山车能量
A roller coaster car of mass 500 kg is at rest at the top of a hill 20 m above the lowest point. The store of gravitational potential energy (GPE) can be calculated:
GPE = mass × g × height
一辆过山车质量为500千克，静止在距离最低点上方20米的山顶。其储存的重力势能（GPE）可以计算如下：
重力势能 = 质量 × g × 高度
Using g = 10 N/kg, GPE = 500 × 10 × 20 = 100,000 J. As the car descends, GPE is converted into kinetic energy (KE). Ignoring friction for a moment, all 100,000 J of GPE become KE at the very bottom, meaning the car reaches maximum speed there. In reality, some energy is transferred to thermal energy because of friction between wheels and track, and air resistance. This explains why the next hill must be slightly lower – the total mechanical energy has decreased.
取g = 10牛/千克，重力势能 = 500 × 10 × 20 = 100,000焦。当车子下坡时，重力势能转化为动能（KE）。暂时忽略摩擦，所有100,000焦重力势能在底部全部转化为动能，意味着车子在最低点达到最大速度。实际上，由于车轮与轨道间的摩擦以及空气阻力，部分能量转化为热能。这就解释了为什么下一个山坡必须略低——总机械能减少了。
We can draw an energy transfer diagram: chemical energy (from motors lifting the car) → GPE → KE + thermal energy + sound. A Sankey diagram would show the useful and wasted energy paths.
我们可以画一个能量转移图：化学能（来自提升车厢的电机）→ 重力势能 → 动能 + 热能 + 声能。桑基图能显示有用能量和浪费能量的路径。
7. Electric Circuits: The Christmas Tree Lights Problem | 电路：圣诞树彩灯问题
One bulb in a string of 50 Christmas lights goes out, and the whole string goes dark. This tells us the bulbs are connected in series. In a series circuit, there is only one loop for current. If one bulb fails (open circuit), current stops everywhere. To diagnose, a student uses a spare bulb and tests each bulb socket by bridging it – if the rest of the lights turn on, that is the broken one.
一串50个圣诞灯中的一个灯泡熄灭了，然后整串灯都灭了。这告诉我们这些灯泡是串联的。在串联电路中，只有一条电流回路。如果一个灯泡坏了（断路），所有地方的电流都会停止。为了排查故障，一位学生用一个备用灯泡逐个插座旁路测试——如果其余灯泡亮起，那个就是坏的。
Modern Christmas lights often use parallel branches. In a parallel circuit, each bulb has its own loop. If one bulb blows, the other loops remain complete, so the rest stay lit. A circuit diagram with symbols for cells, bulbs and switches helps analyse the paths. We can also predict the brightness: in a series circuit, adding more bulbs makes each dimmer because voltage is shared; in parallel, each bulb gets the full battery voltage and stays bright.
现代圣诞灯通常使用并联支路。在并联电路中，每个灯泡有自己的回路。如果一个灯泡烧坏，其它回路保持完整，因此其余灯泡依然亮着。带有电池、灯泡和开关符号的电路图有助于分析电流路径。我们还可以预测亮度：在串联电路中，增加灯泡会导致每个灯泡变暗，因为电压被分配；在并联电路中，每个灯泡都能获得电池的全部电压，保持明亮。
8. Magnetism: Making a Compass | 磁铁：制作指南针
A simple compass can be built by stroking a steel sewing needle along the same pole of a bar magnet about 20 times in one direction. This magnetises the needle. When the needle is floated on a small piece of cork in a bowl of still water, it gradually aligns itself roughly north–south. The end pointing north is the needle’s north-seeking pole (geographic north).
用一块条形磁铁的同一极沿一个方向摩擦一根钢制的缝衣针约20次，就能制作一个简单的指南针。这样就把针磁化了。当磁化后的针被放在一碗静水中一块小软木片上漂浮时，它会逐渐指向南北向。指向北方的一端就是针
Published by TutorHao | Year 7 Physics Revision Series | aleveler.com
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