📚 AS-Level Physics Unit 1 Question Paper June 2019: Concept Analysis | AS 物理 Unit 1 2019年6月考卷概念解析
The June 2019 Edexcel AS Physics Unit 1 paper assessed core mechanics and materials topics through practical scenarios, from skydiving to stretching wires and falling spheres. This article revisits the fundamental concepts behind each question, helping you solidify your understanding and avoid common misconceptions.
2019年6月Edexcel AS物理第一单元试卷通过跳伞、拉伸金属丝和小球下落等实际场景，考查了力学和材料的核心主题。本文重温每道题目背后的基本概念，帮助你巩固理解并避开常见误区。
1. Interpreting Velocity-Time Graphs for a Skydiver | 解析跳伞运动员的速度-时间图像
A skydiver’s motion provides a rich context for reading velocity-time graphs. The graph typically shows an initial curved increase in speed as the skydiver accelerates downwards under gravity. The slope of the tangent at any point gives the instantaneous acceleration. As speed rises, air resistance builds up, reducing the net force, so acceleration decreases until terminal velocity is reached, where the graph becomes horizontal.
跳伞运动员的运动为解读速度-时间图像提供了丰富的背景。图像通常先呈现速度向下加速的曲线上升段。任意点切线的斜率表示瞬时加速度。随着速度增加，空气阻力增大，合力减小，因而加速度逐渐降低，直至达到终端速度，此时图像变为水平线。
When the parachute opens, a sudden, large drag force causes a sharp deceleration—shown as a steep negative slope. The skydiver then slows to a new, lower terminal velocity. The total displacement during the fall is the area under the v-t graph. Students must be careful to interpret areas above and below the time axis if the motion changes direction, but for a downward fall the area is straightforward.
当降落伞打开时，突然产生巨大的阻力，导致急剧减速，表现为陡峭的负斜率。随后跳伞者减速至一个更低的新终端速度。下落的总位移是速度-时间图像下的面积。虽然本题中运动方向不变，但审题时仍须注意面积的正负规则。
2. Applying SUVAT Equations to Vertical Projection | 应用匀加速方程处理竖直上抛运动
One question involved a ball projected vertically upward. For motion under constant acceleration due to gravity, the SUVAT equations apply. Choosing a sign convention is critical—usually taking upwards as positive makes acceleration a = -9.81 m s⁻². The equation v = u + at helps find the time to reach the highest point, where v = 0.
试卷中有一道涉及小球竖直上抛的题目。对于重力加速度恒定的运动，匀加速运动方程均适用。选取正方向至关重要——通常以向上为正，则加速度 a = -9.81 m s⁻²。利用 v = u + at 可求出到达最高点的时间，该点 v = 0。
To find the maximum height, v² = u² + 2as is convenient. Substituting v = 0 gives s = u²/(2g). The total time of flight is twice the time to the peak if the launch and landing points are at the same height. Students often forget that displacement, velocity and acceleration are vectors, so signs must be consistent.
计算最大高度时，使用 v² = u² + 2as 很方便。代入 v = 0 得到 s = u²/(2g)。若抛出点与落地点在同一高度，则总飞行时间是到达最高点时间的两倍。学生常忘记位移、速度和加速度是矢量，符号须保持一致。
3. Resolving Forces on an Inclined Plane | 斜面上力的分解与平衡
A typical exam scenario places a block on a slope, requiring resolution of weight into components parallel and perpendicular to the plane. The component down the slope is mg sin θ, and the component into the plane is mg cos θ, where θ is the angle of inclination. If the block is in equilibrium, the net force in every direction is zero.
经典考题常把一个物块放在斜面上，要求将重力分解为沿斜面和垂直于斜面的分力。沿斜面向下的分力为 mg sin θ，垂直压向斜面的分力为 mg cos θ，其中 θ 是倾角。物块若处于平衡态，各个方向的合力均为零。
Friction acts up the slope to oppose motion, with a maximum value of μR, where R is the normal reaction. For a stationary block, the frictional force adjusts to match the component of weight down the slope up to the limit of μR. Newton’s third law reminds us that the normal reaction is not simply ‘mg’ on an incline; it equals mg cos θ.
摩擦力沿斜面向上阻碍运动，其最大值为 μR，R 为法向反作用力。对静止的物块，摩擦力会自行调整以平衡下滑分力，直至达到 μR 的极限。牛顿第三定律提醒我们，斜面上的法向反力并不简单等于 mg，而是等于 mg cos θ。
4. Momentum, Impulse and Force-Time Graphs | 动量、冲量与力-时间图像
Momentum p = mv is a vector quantity conserved in all collisions if no external resultant force acts. The June 2019 paper examined a collision where students had to calculate velocity after impact using conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂.
动量 p = mv 是矢量，若无合外力作用，碰撞过程中动量守恒。2019年6月试卷考查了一次碰撞，要求学生用动量守恒式计算撞击后的速度：m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂。
Impulse is the change in momentum, also equal to the area under a force-time graph. For a varying force, this area gives FΔt = Δp. The concept appeared in a question showing a force sensor trace during impact. Students had to estimate impulse by counting squares or using average force.
冲量是动量的变化量，也等于力-时间图像下的面积。对于变力，此面积代表 FΔt = Δp。这一概念在有一题中通过碰撞力传感器图像体现，学生需通过数格或平均力来估算冲量。
5. Work, Energy and Power on a Slope | 斜面上的功、能与功率
A ski-slope problem combined energy conservation with work done against friction. The loss in gravitational potential energy (mgh) converts into kinetic energy (½mv²) and work against friction (Fd). Equating energy transfers gives a neat way to find final speed or frictional force.
一道滑雪坡问题将能量守恒与克服摩擦力做功结合起来。重力势能的减少 mgh 转换为动能 ½mv² 和克服摩擦力做的功 Fd。将能量转化列等式是求解末速度或摩擦力的一种简捷方法。
Power is the rate of doing work, P = Fv for a constant force acting in the direction of velocity. In part of the question, students had to calculate the power output of a skier moving at constant speed down the slope, which required resolving weight and using P = mg sin θ × v.
功率是做功的快慢，对沿速度方向的恒力有 P = Fv。在该题一部分中，学生需计算滑雪者匀速下滑时的输出功率，这需要分解重力并使用 P = mg sin θ × v。
6. Determining the Young Modulus of a Wire | 测量金属丝的杨氏模量
The Young modulus E = stress/strain = (F/A) / (ΔL/L) describes the stiffness of a material. In the exam, a copper wire was stretched by hanging known masses. The extension ΔL was measured for each load, while original length L and cross-sectional area A were recorded using a metre rule and a micrometer screw gauge.
杨氏模量 E = 应力/应变 = (F/A) / (ΔL/L) 描述材料的刚度。试卷中，通过悬挂已知质量来拉伸一根铜丝，测量每次加载的伸长量 ΔL，同时用米尺和千分尺记录原长 L 和截面积 A。
Plotting force F against extension ΔL yields a straight line through the origin for the elastic region. The gradient k = F/ΔL relates to Young modulus by E = kL/A. Therefore, E can be calculated from the gradient, provided all quantities are in SI units: pascals for stress.
在弹性区内绘制力 F 对伸长 ΔL 的图线为一条过原点的直线。斜率 k = F/ΔL 与杨氏模量的关系是 E = kL/A。因此，由斜率可计算出 E，前提是所有物理量使用国际单位制：应力单位为帕斯卡。
7. Stress-Strain Curves: Elastic to Plastic | 应力-应变曲线：从弹性到塑性
A stress-strain graph for a ductile material like copper shows a linear elastic region obeying Hooke’s law, followed by a curved plastic region. The limit of proportionality marks the end of linearity; beyond it, strain is no longer proportional to stress.
像铜这样的韧性材料的应力-应变曲线呈现出符合胡克定律的线性弹性区，随后是弯曲的塑性区。比例极限标志着线性关系的终点；超出此点，应变不再与应力成正比。
The elastic limit is the point beyond which permanent deformation occurs. Further, the yield point may show a drop in stress before plastic flow, and the ultimate tensile stress is the maximum stress before necking and fracture. The area under the curve represents the work done per unit volume to break the material.
弹性极限是发生永久变形的临界点。随后，屈服点可能在塑性流动前显示应力下降，而极限抗拉强度是颈缩和断裂前的最大应力。曲线下的面积代表使材料断裂所需的单位体积功。
8. Fluid Flow and Viscosity Concepts | 流体流动与粘性概念
The paper touched on the behaviour of a sphere falling through oil, connecting to fluid dynamics. In laminar flow, layers of fluid slide smoothly past each other; in turbulent flow, eddies and mixing occur. The Reynolds number Re = ρvd/η predicts flow regime, with low Re indicating laminar conditions.
试卷涉及小球在油中下落的行为，关联到流体动力学。层流中，流体层平滑地相对滑动；湍流中会出现涡流和混合。雷诺数 Re = ρvd/η 可预测流态，低 Re 值对应层流条件。
Viscosity η is a measure of a fluid’s internal friction. The viscous drag force on a sphere in laminar flow is given by Stokes’ law: F = 6πηrv, where r is the sphere radius and v its velocity. This force increases linearly with velocity.
粘度 η 是流体内摩擦的量度。层流中作用于球体的粘滞阻力由斯托克斯定律给出：F = 6πηrv，其中 r 为球体半径，v 为速度。该力随速度线性增大。
9. Stokes’ Law and Falling Sphere Experiment | 斯托克斯定律与小球下落实验
A classic experiment drops a small steel ball into a cylinder of oil. After an initial acceleration, the ball reaches terminal velocity vₜ when the upward viscous drag plus buoyancy equal the weight: 6πηrvₜ + (4/3)πr³ρₗg = (4/3)πr³ρₛg.
经典实验将小钢球投入油柱中。经过短暂加速后，当向上的粘滞阻力与浮力之和等于重力时，球体达到终端速度 vₜ：6πηrvₜ + (4/3)πr³ρₗg = (4/3)πr³ρₛg。
Solving for viscosity gives η = 2r²g(ρₛ − ρₗ) / (9vₜ). To ensure laminar flow and accurate results, the ball must be small, the oil viscous, and terminal velocity measured over a known distance marked by timing rings on the cylinder. The temperature must be kept constant because viscosity is highly temperature-dependent.
求解粘度可得 η = 2r²g(ρₛ − ρₗ) / (9vₜ)。为保证层流和结果准确，球体要小、油液要粘，并通过油柱上的记时环测量已知距离内的终端速度。由于粘度对温度高度敏感，实验时温度必须保持恒定。
10. Handling Experimental Uncertainties | 实验不确定度的处理
The exam required calculating percentage uncertainty. For a measured quantity like diameter d with absolute uncertainty ±Δd, percentage uncertainty is (Δd/d) × 100%. When quantities are multiplied or divided, their percentage uncertainties add together.
试卷要求计算百分比不确定度。对于直径 d 这样的测量量，其绝对不确定度为 ±Δd，百分比不确定度为 (Δd/d) × 100%。当物理量作乘除运算时，其百分比不确定度相加。
In the Young modulus experiment, the quantity r² appears, so the percentage uncertainty in area A is twice that in the radius. Combining uncertainties in force, extension and length then gives the total uncertainty in E. Comparison with a reference value uses percentage difference: |E_measured − E_reference| / E_reference × 100%.
在杨氏模量实验中，由于 r² 的出现，面积 A 的百分比不确定度是半径的两倍。综合力、伸长量和长度的不确定度即可求得 E 的总不确定度。与参考值比较时使用百分差：|E_测量 − E_参考| / E_参考 × 100%。
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