Unit 1: Mechanics and Materials
Pearson Edexcel · International A-Level · Physics トピック 1 22:35 英語ナレーション・英語＋中文字幕 burning-in
章
Transcript
Start with the physical situation, not a formula search.
从物理情境出发，而不是先找公式。
Unit one connects descriptions of motion with causes, then uses momentum and energy to explain interactions.
第一单元把运动的描述与运动的原因联系起来，再用动量和能量解释相互作用。
Materials extend these ideas to fluids and solids.
材料部分把这些概念延伸到流体和固体。
Use the handout as your reference.
把讲义作为参考资料。
Work through the matching skill sheet before its curated past-paper set.
先完成对应的技能练习，再做精选真题。
Each calculation begins by stating the model and its conditions.
每次计算都要先说明所用的模型及其适用条件。
The reviewed Issue three specification sets ninety minutes and eighty marks, with every question compulsory.
已核对的第三版考试大纲规定，考试时间为九十分钟，满分八十分，所有题目必答。
At least thirty-two marks assess Level-two mathematics.
至少三十二分考查二级数学技能。
The formula list supplies equations, but you must recognise the model, identify the quantities and explain a conclusion.
公式表提供方程，但你需要辨认模型、确定各个物理量，并解释结论。
Use nine point eight one for gravitational acceleration.
重力加速度使用九点八一。
These qualification facts come from the specification, not an inferred teaching pattern.
这些考试信息来自大纲，并非根据教学安排推测。
Imagine walking to a shop and returning home.
想象你走到商店再回家。
You travelled a positive distance, but your displacement is zero because your final position equals your initial position.
你走过了正的路程，但位移为零，因为终点与起点相同。
Distance and speed are scalars.
路程和速率是标量。
Displacement and velocity include direction.
位移和速度包含方向。
Choose a positive direction for one-dimensional calculations and keep that convention throughout the working.
在一维计算中，先选定正方向，并在整个解题过程中保持这个约定。
Pause at the diagram.
先停下来观察图。
Read the labelled axes and locate the highlighted regions.
读清坐标轴的名称，找出突出显示的区域。
The curve, values and colours match the handout; this projection version enlarges its lettering.
曲线、数值和颜色与讲义一致；这个投影版本只放大了文字。
Identify what the initial straight gradient and highlighted area mean before advancing to the explanation.
继续看解释之前，先判断起始直线段的斜率和突出显示的面积各代表什么。
The same-looking line means different things on different axes.
看起来相同的线，在不同坐标轴下意义不同。
A displacement-time gradient gives velocity.
位移—时间图的斜率是速度。
A velocity-time gradient gives acceleration, while its signed area gives displacement.
速度—时间图的斜率是加速度，而带正负号的面积是位移。
Area below the time axis represents motion in the negative direction.
时间轴下方的面积表示沿负方向运动。
For distance, add the magnitudes rather than cancelling them.
求路程时，要把面积的大小相加，而不是让它们相互抵消。
On acceleration-time axes, area gives the change in velocity.
在加速度—时间图中，面积表示速度的变化量。
Choose two points on the straight section, not a point and the origin unless the line actually passes through the origin.
在直线段上选两点。 除非直线确实经过原点，否则不要随意把某点与原点配成两点。
The velocity changes by three point two metres per second over twelve seconds.
十二秒内速度变化了每秒三点二米。
Dividing gives zero point two six seven metres per second squared, usually reported as zero point two seven.
相除得到每二次方秒零点二六七米，通常写成零点二七。
The positive answer agrees with the line rising.
结果为正，与直线向上倾斜一致。
Suvat equations are conditional models.
匀加速运动公式是有条件的模型。
They apply to one-dimensional motion with constant acceleration.
它们适用于加速度恒定的一维运动。
If time is absent and you need displacement, use the equation involving the squares of the two velocities.
如果没有时间且需要求位移，就使用包含初、末速度平方的方程。
Do not conclude that every curved graph means changing acceleration: constant acceleration produces a curved displacement-time graph.
不要认为所有曲线图都表示加速度在变：恒定加速度也会产生弯曲的位移—时间图。
It is a curved velocity-time graph that shows the acceleration changing.
弯曲的速度—时间图才说明加速度发生变化。
Attempt this before advancing.
继续之前先自己尝试。
A ball starts one point seven metres above the floor, rises at five point one metres per second, and enters a hoop moving down at two point one.
球从离地一点七米处，以每秒五点一米的速度向上运动，进入篮筐时以每秒二点一米的速度下降。
Upwards is positive, so the final velocity and acceleration are negative.
取向上为正，因此末速度和加速度都为负。
Calculate the rise without time, then add the launch height.
先用不含时间的方程求上升高度，再加上出发高度。
Finish by deciding whether the hoop was three metres high.
最后判断篮筐是否有三米高。
Rearrange the equation symbolically first.
先用符号整理方程。
The squared final velocity minus the squared initial velocity is negative, as is twice the acceleration, so the rise is positive. It is about one point one metres.
末速度平方减初速度平方为负，两倍加速度也为负，所以高度增加量为正，约为一点一米。
Add the initial height to obtain two point eight metres above the floor.
加上初始高度，篮筐离地约二点八米。
Deduce requires the final comparison: this is lower than the proposed three metres.
题目要求推断，因此还要作最后的比较：这个高度低于所提出的三米。
Measure distances and fall times using a light gate or video with a known frame rate.
用光电门，或已知帧率的视频，测量距离和下落时间。
Release from rest without adding a push.
从静止释放，不要额外推一下。
Repeat measurements to estimate random variation.
重复测量，估计随机变化。
Plot distance against time squared and double the gradient to obtain g.
作距离对时间平方的图，把斜率乘二便得到重力加速度。
A fixed distance offset can move the intercept.
固定的距离偏差可能改变截距。
A timing delay can distort the relationship, so automatic timing is preferable to a hand-operated stopwatch.
计时延迟可能使这种关系失真，所以自动计时优于手动秒表。
In a strobe photograph, a shorter flash reduces the distance travelled during the flash, sharpening the image and reducing position uncertainty.
在频闪照片中，闪光越短，物体在闪光期间走过的距离越小，图像越清晰，位置的不确定度也越小。
Choose sine or cosine from the actual angle in your diagram.
根据图中实际标出的角度选择正弦或余弦。
The component adjacent to that angle uses cosine.
与该角相邻的分量使用余弦。
For a weight on a slope, geometry makes the downslope component weight times sine of the slope angle. The perpendicular component is weight times cosine.
对斜面上的重力，几何关系给出沿斜面分量等于重力乘斜面角的正弦，垂直斜面分量等于重力乘余弦。
When recombining perpendicular components, use Pythagoras for magnitude and trigonometry for direction.
合成两个垂直分量时，用勾股定理求大小，用三角关系求方向。
A vector triangle is a calculation by scale drawing.
矢量三角形是按比例作图的计算方法。
Draw each arrow nose to tail, preserving its direction and using the same length scale.
把箭头首尾相接，保持各自的方向，并使用统一的长度比例尺。
If the object is in equilibrium, the final arrow returns to the starting point.
如果物体处于平衡，最后一个箭头应回到起点。
Zero resultant does not mean no forces. It means that the vector sum of the forces acting on the chosen object is zero.
合力为零不表示没有力，而是作用在所选物体上的力的矢量和为零。
Resolve the launch velocity first.
先分解发射速度。
With air resistance neglected, horizontal velocity stays constant while vertical velocity changes under gravity.
忽略空气阻力时，水平速度不变，竖直速度在重力作用下变化。
The two directions share the same elapsed time.
两个方向经历的时间相同。
At the highest point the vertical component is zero, but the horizontal component remains, so the projectile is still moving.
最高点处竖直速度分量为零，但水平分量仍存在，所以物体仍在运动。
Apply the constant-acceleration equations separately to each direction.
分别对两个方向使用恒加速度方程。
The handout uses January twenty twenty-five question nineteen as a taught example.
讲义把二零二五年一月第十九题作为教学例题。
It is therefore exposed and must not be called an unseen test question.
因此，这道题已向学生展示，不能再称为未见过的测试题。
Find time from horizontal distance divided by horizontal velocity.
用水平距离除以水平速度求时间。
Use that same time in the vertical displacement equation and add the initial height.
把同一时间代入竖直位移方程，再加上初始高度。
Keeping unrounded values gives about thirty-two point four seconds and two point nine three kilometres above sea level.
保留未舍入的数值，得到约三十二点四秒，以及海拔约二点九三千米。
Use the time found from horizontal motion in the vertical position equation.
把水平运动求得的时间代入竖直位置方程。
Start from two thousand metres, add the displacement from the initial vertical velocity, and subtract half g t squared.
从海拔两千米出发，加上初始竖直速度对应的位移，再减去二分之一 g t 平方。
Retaining guard digits gives a height of about two point nine three kilometres above sea level. This is above the ship at the stated horizontal position.
保留中间计算位数，得到海拔约二点九三千米，因此在这个水平位置，抛体位于船的上方。
Draw a boundary around the body you are analysing.
先圈定你正在分析的物体。
Include forces acting on that body and exclude forces it applies to others.
只画作用在这个物体上的力，不画它施加给其他物体的力。
Choose a positive direction, then add the force components with signs.
选定正方向，再带正负号相加各力的分量。
Newton’s second law uses the resultant force.
牛顿第二定律使用合力。
When the resultant is zero, acceleration is zero; the body may be stationary or move with constant velocity.
合力为零时，加速度为零；物体可以静止，也可以做匀速直线运动。
Newton’s third law connects two bodies.
牛顿第三定律联系的是两个物体。
The forces are equal and opposite, but they cannot cancel in the free-body diagram of one object because they act on different objects.
两个力大小相等、方向相反，但不能在单个物体的受力图中相互抵消，因为它们作用在不同物体上。
A book’s weight and the table’s normal force can balance, but are not a third-law pair.
书的重力和桌面的支持力可以平衡，却不是第三定律的一对力。
For a falling object at terminal velocity, drag and upthrust balance its weight, so acceleration is zero.
物体以终端速度下落时，阻力与浮力之和平衡重力，因此加速度为零。
Pause and attempt this practice task before the solution.
暂停并计算。
A resultant force of 4800 N accelerates a boat at 0.31 m/s².
合力四千八百牛使船产生每二次方秒零点三一米的加速度。
Use g = 9.81 N/kg to calculate its mass and then its weight.
使用每千克九点八一牛的重力场强度，先求质量，再求重量。
Show the two symbolic equations and units.
写出两个符号方程并保留单位。
The stated four thousand eight hundred newtons is the resultant force, so divide it by zero point three one metres per second squared to determine mass.
题中给出的四千八百牛是合力，因此用它除以每二次方秒零点三一米，先求质量。
The question asks for weight, so one more equation is necessary: multiply the mass by g.
题目要求重力，所以还需要另一个方程：把质量乘以重力加速度。
Keep the unrounded mass in the calculator, then report about one point five times ten to the five newtons.
在计算器中保留未舍入的质量，最后写出约一点五乘十的五次方牛。
Choose the system before invoking momentum conservation.
使用动量守恒之前先选系统。
For a short collision, external forces may have negligible impulse even when they are not literally absent.
短暂碰撞中，外力即使并非严格为零，其冲量也可能可以忽略。
Give opposite velocities opposite signs.
方向相反的速度应有相反的正负号。
Internal forces produce equal and opposite changes of momentum.
内力产生大小相等、方向相反的动量变化。
Kinetic energy is a different quantity and can be transferred into internal energy when objects deform or stick together.
动能是另一个物理量；物体变形或粘在一起时，动能可以转化成内能。
Pause and attempt this practice task before the solution.
暂停并尝试。
A train of mass 3m moves at +v; a truck of mass m moves at −2v.
质量为三 m 的火车以正 v 运动，质量为 m 的货车以负二 v 运动。
They couple with negligible external impulse.
两车连接，外力冲量可忽略。
Find their final velocity and explain its sign.
求共同速度，并解释正负号。
Take the train direction as positive.
取列车方向为正。
Its momentum is three m v, while the truck contributes negative two m v.
列车的动量是三倍的质量参数乘速度，而货车的动量是负二倍的质量参数乘速度。
After coupling, the total mass is four m.
连接后的总质量是该质量参数的四倍。
Conservation gives a final velocity of v divided by four.
由动量守恒，末速度为原速度的四分之一。
The answer is positive and small because the initial momenta nearly cancel.
结果为正且较小，因为最初的动量几乎抵消。
Adding both speeds as positive would describe a different physical event.
把两个速度都当作正值相加，描述的就是另一种物理情境。
Force balance and moment balance answer different questions.
力的平衡和力矩的平衡回答不同问题。
Zero resultant force prevents the centre of mass from accelerating.
合力为零可避免质心加速。
Zero resultant moment prevents angular acceleration about the chosen pivot.
对所选支点，合力矩为零可避免角加速度。
The distance in a moment is perpendicular to the force’s line of action.
力矩中的距离必须垂直于力的作用线。
You can instead resolve a force perpendicular to the lever, but do not apply both corrections to the same moment.
也可以先取垂直于力臂的力分量，但不要对同一个力矩同时作两次这样的修正。
Read the original force diagram before calculating.
计算前先读原题受力图。
The forty-two degree angle is measured from the horizontal.
四十二度是钢轨与水平方向的夹角。
The weight acts at the midpoint of the uniform rail, and the student’s force is perpendicular to the rail.
均匀钢轨的重力作用在中点，学生施加的力垂直于钢轨。
Identify each perpendicular lever arm before taking moments.
求力矩前，先确定每个力对应的垂直力臂。
For the rail in the handout, calculate each moment using its own geometry.
对讲义中的栏杆，按各自的几何关系计算力矩。
The piston supplies about eighteen point three newton metres. The weight supplies about forty-nine point four in the opposing direction.
活塞提供约十八点三牛米，重力在相反方向提供约四十九点四牛米。
A perpendicular handle force must supply the difference over one point four metres.
垂直于把手的力，需要通过一点四米的力臂补足这两者的差。
A vertical handle pull has a shorter perpendicular distance, so a larger force supplies the same moment.
竖直拉把手时，垂直力臂更短，所以产生同样力矩需要更大的力。
Decide what is inside your system and what enters or leaves it.
先确定系统内部包含什么，以及哪些能量进入或离开系统。
Kinetic energy depends on speed squared.
动能取决于速率的平方。
Gravitational potential energy depends on height change.
重力势能取决于高度变化。
Only the force component along displacement does work.
只有沿位移方向的力分量做功。
Efficiency compares useful output with total input for the same process.
效率比较同一过程的有用输出与总输入。
A formula without a defined energy boundary can produce a plausible but wrong denominator.
如果没有明确能量边界，公式可能给出看似合理却分母错误的结果。
Pause and attempt this practice task before the solution.
暂停并计算。
An 1800 kg car moving at 14 m/s brakes to rest while rising 0.76 m. Its battery gains 45000 J.
一千八百千克的汽车以每秒十四米运动，制动至静止时升高零点七六米，电池增加四万五千焦能量。
Calculate braking efficiency using g = 9.81 N/kg, and explain the energy input you choose.
使用每千克九点八一牛，计算制动效率，并说明所选输入能量。
The car begins with one hundred seventy-six thousand four hundred joules of kinetic energy.
汽车最初有十七万六千四百焦的动能。
It ends higher, retaining about thirteen thousand four hundred joules as gravitational potential energy.
最终位置较高，其中约一万三千四百焦保留为重力势能。
The brakes remove the difference, which is the relevant input for energy recovery.
刹车移除的是这两者之差，这才是能量回收过程的相关输入。
Divide the battery gain of forty-five thousand joules by that difference.
把电池增加的四万五千焦除以这个差值。
The efficiency is about twenty-eight percent, less than one hundred percent as required.
效率约为百分之二十八，按物理要求小于百分之百。
Upthrust is the weight of displaced fluid, not its mass or volume.
浮力等于排开流体的重力，不是其质量或体积。
For a small sphere moving slowly through a liquid in laminar flow, Stokes’ law links drag with viscosity, radius and speed.
小球在液体中以低速、层流状态运动时，斯托克斯定律把阻力与黏度、半径和速率联系起来。
At terminal speed the resultant force is zero.
达到终端速率时合力为零。
Include upthrust as well as drag when balancing the forces; neglect it only with a justified approximation.
列力平衡时，除阻力外也要计入浮力；只有经过合理的近似论证，才可忽略浮力。
The time measurement must occur after the ball reaches terminal speed.
必须在小球达到终端速率后再计时。
Measure a suitable distance away from the release point and keep the sphere away from the cylinder walls.
在远离释放点的位置选取合适的测量距离，并使小球远离容器壁。
Find speed from distance divided by time.
用距离除以时间求速率。
Combine weight, upthrust and Stokes’ drag to obtain the stated viscosity equation.
联立重力、浮力与斯托克斯阻力，得到给出的黏度方程。
Control temperature and repeat timings.
控制温度并重复计时。
Radius is squared here, so a radius error has a substantial effect.
这个方程中的半径要平方，所以半径误差影响明显。
This is the reasoning chain in January question eighteen b, checked against the official scheme.
这是依据官方评分方案核对过的一月第十八题第二问的推理链。
At steady speed the resultant force is zero. Tension and upthrust balance weight.
匀速时合力为零，张力与浮力之和平衡重力。
As the submarine emerges, it displaces less water, so upthrust decreases.
潜艇逐渐露出水面时，排开的水减少，浮力随之减小。
Weight remains approximately constant, therefore tension increases.
重力近似不变，因此张力增大。
Once clear of the water, tension equals weight.
完全离开水面后，张力等于重力。
Linked causes matter as well as individual facts.
除了单个事实，因果关系的连接也很重要。
Force and extension depend on the sample geometry.
力和伸长量取决于试样的几何尺寸。
Stress accounts for cross-sectional area and strain accounts for original length.
应力考虑横截面积，应变考虑原长。
Young modulus compares these two quantities within the linear elastic region, describing material response.
杨氏模量在成线性关系的弹性区间内比较这两个量，描述材料的响应。
Use the original length and the cross-sectional area perpendicular to the load.
使用原长，以及垂直于载荷方向的横截面积。
Do not confuse a radius with a diameter, or a total length with an extension.
不要混淆半径与直径，也不要把总长度当作伸长量。
Pause at the diagram.
先停下来观察图。
Read the labelled axes and locate the highlighted regions.
读清坐标轴的名称，找出突出显示的区域。
The curve, values and colours match the handout; this projection version enlarges its lettering.
曲线、数值和颜色与讲义一致；这个投影版本只放大了文字。
Identify what the initial straight gradient and highlighted area mean before advancing to the explanation.
继续看解释之前，先判断起始直线段的斜率和突出显示的面积各代表什么。
The proportional limit and elastic limit answer different questions.
比例极限和弹性极限回答不同问题。
One concerns a linear relationship; the other concerns whether the sample returns to its original dimensions on unloading.
前者关乎是否为线性关系，后者关乎卸载后试样能否恢复原来的尺寸。
Some ductile materials have a distinct yield region, while others yield gradually.
有些延性材料有明显的屈服区，有些则逐渐屈服。
Do not teach a flat yield plateau as universal.
不要把水平屈服平台教成普遍规律。
The initial gradient gives Young modulus, and breaking stress is the stress at fracture.
起始段的斜率是杨氏模量，断裂应力是断裂时的应力。
Pause and attempt this practice task before the solution.
暂停并尝试。
A 0.750 m string stretches to 0.752 m under 36 N. Its radius is 0.85 mm.
原长零点七五零米的弦在三十六牛拉力下伸长到零点七五二米，半径零点八五毫米。
Calculate cross-sectional area, stress, strain and Young modulus. Show each denominator and unit.
依次求截面积、应力、应变和杨氏模量，写清各个分母及单位。
First convert the radius into metres and calculate the cross-sectional area.
先把半径换算成米，再计算横截面积。
Divide force by area to get stress.
用力除以面积求应力。
Calculate extension by subtracting original length from final length, then divide that extension by original length to get strain.
用末长减原长得到伸长量，再用伸长量除以原长求应变。
Finally divide stress by strain.
最后用应力除以应变。
The answer is about six gigapascals.
结果约为六吉帕。
A millimetre conversion or diameter error can change the result by orders of magnitude.
毫米换算错误或把直径当半径，都可能使答案相差几个数量级。
Use a long wire so the extension is measurable.
用较长的金属丝，使伸长量容易测量。
Measure diameter at several positions and orientations, then calculate cross-sectional area.
在多个位置和方向测量直径，再计算横截面积。
Add known loads in steps and measure extension carefully, correcting any initial zero offset.
分步加上已知载荷，仔细测量伸长，并修正初始零点偏差。
Plot stress against strain and take the gradient of the proportional region.
作应力—应变图，取成比例区间的斜率。
Unloading checks whether deformation is recoverable.
卸载可检验形变是否能够恢复。
Secure heavy weights and protect people from a breaking wire.
固定好重物，并防止金属丝断裂伤人。
Area under a force-extension graph is work done.
力—伸长图下的面积是所做的功。
For a linear spring the triangle has area one half force times extension.
线性弹簧对应三角形面积，即力乘伸长量再除以二。
Area under a stress-strain graph has units of pascals, equivalent to joules per cubic metre. It gives energy per unit volume.
应力—应变图下面积的单位是帕，等价于每立方米焦，表示单位体积的能量。
Multiply by volume to obtain total energy.
再乘体积便得到总能量。
The distinction follows directly from dividing force by area and extension by original length.
这一区别直接来自力除以面积，以及伸长量除以原长。
The loading area is work done; it is fully recoverable strain energy only within the elastic region.
加载曲线下的面积是输入的功；只有在弹性区间内，它才全部是可恢复的应变能。
Match your response to the command word.
根据指令词组织答案。
A calculation needs enough working to check the method and a result with a unit.
计算题要写出足够的步骤，使方法可以核查，并给出带单位的结果。
Deduce adds a comparison or conclusion drawn from that result.
推断题还要根据结果作比较或得出结论。
Explain requires linked physical reasons.
解释题需要相互连接的物理原因。
Keep unrounded intermediate values in the calculator and round only the final result.
在计算器中保留未舍入的中间值，只在最终结果处舍入。
Specific marking decisions come from the question’s official scheme, not a universal mark-loss rule.
具体评分依据该题的官方评分方案，而不是所谓普遍适用的扣分规则。
Pause and answer these without consulting the handout.
先停下来，不看讲义回答这些问题。
Name what the velocity-time gradient and signed area represent.
说出速度—时间图的斜率和带正负号面积各代表什么。
Explain why two balancing forces on one body are not automatically a third-law pair.
解释为什么同一物体上两个平衡的力，不一定构成第三定律的一对力。
Distinguish the units and meanings of the two material-graph areas.
区分两种材料图下面积的单位与含义。
Use your answers to decide which skill sheet to revisit before attempting the authentic past-paper set.
根据自己的答案，决定做精选真题前需要回顾哪张技能练习。
Check your answers now.
现在核对答案。
The velocity-time gradient gives acceleration and its signed area gives displacement.
速度—时间图的斜率是加速度，带正负号的面积是位移。
Third-law partners act on different bodies, whereas a force balance is written for one chosen body.
第三定律的一对力作用在不同物体上，而力的平衡针对一个选定物体列式。
Force-extension area gives joules; stress-strain area gives joules per cubic metre.
力—伸长面积的单位是焦；应力—应变面积的单位是每立方米焦。
Correct the reason as well as the final answer, then continue with the matching practice sheet and curated set.
既要修正答案，也要修正理由，然后继续对应的练习和精选真题。