A-Level Physics

A physical quantity 物理量 has two parts: a number (its magnitude 大小) and a unit 单位. The number on its own tells you nothing. You must also say what is measured and in which unit.

Example: "the length is 1.5" is not complete. "The length is 1.5 m" is a physical quantity.

Making estimates

You should be able to estimate 估算 the size of the physical quantities in this syllabus. Learn these rough values:

• mass 质量 of an adult human: $\sim 70\ \text{kg}$
• weight 重力 of an adult human: $\sim 700\ \text{N}$
• height of an adult human: $\sim 1.7\ \text{m}$
• mass of an apple: $\sim 0.1\ \text{kg}$ (so its weight is about $1\ \text{N}$)
• speed of sound in air: $\sim 340\ \text{m s}^{-1}$
• speed of light in a vacuum 真空: $3.0 \times 10^{8}\ \text{m s}^{-1}$
• acceleration 加速度 of free fall 自由落体: $g \approx 9.81\ \text{m s}^{-2}$

A good estimate has the right order of magnitude 数量级 (the right power of ten). For a human, 70 kg is a good guess; 7 kg is not.

Base units

The SI system has seven base quantities 基本量; five of them are used in this topic. You must know these five and their units:

• mass — kilogram, $\text{kg}$
• length 长度 — metre, $\text{m}$
• time — second, $\text{s}$
• current 电流 — ampere 安培, $\text{A}$
• temperature 温度 — kelvin 开尔文, $\text{K}$

Every other unit in this syllabus is built from these five.

Derived units

A derived unit 导出单位 is made by multiplying or dividing base units. You should be able to write any quantity in this syllabus in base units.

Build a derived unit from the equation that defines it:

• speed 速率 = distance / time, so its unit is $\text{m s}^{-1}$
• acceleration = change in velocity 速度 / time, so its unit is $\text{m s}^{-2}$
• force 力 = mass × acceleration, so its unit is $\text{kg m s}^{-2}$. The newton 牛顿 is $1\ \text{N} = 1\ \text{kg m s}^{-2}$.
• work 功 and energy 能量 = force × distance, so the unit is $\text{kg m}^{2}\ \text{s}^{-2}$. The joule 焦耳 is $1\ \text{J} = 1\ \text{kg m}^{2}\ \text{s}^{-2}$.
• power 功率 = energy / time, so the unit is $\text{kg m}^{2}\ \text{s}^{-3}$. The watt 瓦特 is $1\ \text{W} = 1\ \text{kg m}^{2}\ \text{s}^{-3}$.
• pressure 压强 and stress 应力 = force / area, so the unit is $\text{kg m}^{-1}\ \text{s}^{-2}$. The pascal 帕斯卡 is $1\ \text{Pa} = 1\ \text{kg m}^{-1}\ \text{s}^{-2}$.

When a question asks for the SI base units of a quantity, replace each named unit with its base units, then simplify. Example: the SI base units of the watt are $\text{kg m}^{2}\ \text{s}^{-3}$.

Checking that the units match

An equation is homogeneous 量纲一致 when both sides have the same base units. In plain words: the units on both sides match.

Write each side in base units and compare. Take the equation $v^{2} = u^{2} + 2as$:

• left side: $(\text{m s}^{-1})^{2} = \text{m}^{2}\ \text{s}^{-2}$
• right side, first term: $(\text{m s}^{-1})^{2} = \text{m}^{2}\ \text{s}^{-2}$
• right side, second term: $\text{m s}^{-2} \cdot \text{m} = \text{m}^{2}\ \text{s}^{-2}$

Both sides give $\text{m}^{2}\ \text{s}^{-2}$, so the units match.

Be careful: matching units do not prove the whole equation is correct. It could still have a wrong number, or a missing factor of 2. But if the units do not match, the equation is wrong for sure.

Prefixes

A prefix 词头 is a letter put in front of a unit to make it bigger or smaller by powers of ten. You must know these:

To change a prefixed unit into base units, replace the prefix with its factor, then simplify. Example: change $0.25\ \text{kN mm}^{-2}$ into $\text{N m}^{-2}$:

Take special care with squared units like $\text{mm}^{2}$: you must square the factor too.

Every measurement 测量 has some uncertainty 不确定度 — we are never fully sure of the value. A good experimenter knows where the uncertainty comes from, makes a fair estimate of it, and carries it through to the final answer.

Systematic and random errors

A systematic error 系统误差 changes every reading by the same amount, in the same direction. You cannot find it by repeating the measurement. Common causes:

• a zero error 零点误差 (the scale does not read zero when the true value is zero)
• a calibration 校准 error (the scale itself is wrong)
• parallax 视差 (your eye is always to one side of the scale)

A systematic error makes the accuracy 准确度 worse, but it does not change the precision 精密度.

A random error 随机误差 makes readings jump above and below the true value, with no pattern. Causes include how carefully you read the scale, changing conditions, and the smallest step the instrument 仪器 can show. If you repeat the measurement many times and take the mean 平均值 (the average), random errors partly cancel out.

A random error makes the precision worse. But with enough repeats, the mean can still be accurate.

Precision and accuracy

Precision is how close repeated readings are to each other. Precise readings are grouped very close together.

Accuracy is how close a reading (or the mean of several readings) is to the true value.

A set of readings can be:

• precise and accurate — close together and near the true value
• precise but not accurate — close together, but away from the true value (a systematic error)
• accurate but not precise — spread out, but the mean is near the true value
• neither — spread out and away from the true value

When a question gives a table of repeated readings, look at the spread (precision) and the mean (accuracy) separately.

Uncertainty in a derived quantity

A measurement is often written as $x \pm \Delta x$. Here $\Delta x$ is the absolute uncertainty 绝对不确定度. The percentage uncertainty 百分比不确定度 is

A derived quantity 导出量 is one you calculate from measured values. Its uncertainty is found by simple rules:

• Adding or subtracting — add the absolute uncertainties. If $y = a + b$ or $y = a - b$, then $\Delta y = \Delta a + \Delta b$.
• Multiplying or dividing — add the percentage uncertainties. If $y = \dfrac{a \cdot b}{c}$, then
  $$\frac{\Delta y}{|y|} = \frac{\Delta a}{|a|} + \frac{\Delta b}{|b|} + \frac{\Delta c}{|c|}.$$
• Powers — multiply the percentage uncertainty by the power. If $y = a^{n}$, then $\dfrac{\Delta y}{|y|} = |n| \cdot \dfrac{\Delta a}{|a|}$.

Worked example. A ball's diameter 直径 is measured as $d = (5.26 \pm 0.02)\ \text{cm}$. The volume 体积 of a sphere is $V = \tfrac{4}{3}\pi r^{3} = \tfrac{4}{3}\pi (d/2)^{3}$, so $V \propto d^{3}$ ($V$ depends on $d$ cubed).

The percentage uncertainty in $d$ is

Because $V \propto d^{3}$, the percentage uncertainty in $V$ is three times this, about $1.14\%$. The volume is $\tfrac{4}{3}\pi(2.63)^{3} \approx 76.2\ \text{cm}^{3}$. So the absolute uncertainty is $0.0114 \times 76.2 \approx 0.87\ \text{cm}^{3}$. The final answer is $V = (76.2 \pm 0.9)\ \text{cm}^{3}$.

Significant figures

When you write a calculated quantity, give it the same number of significant figures 有效数字 as the least precise measurement you used — usually two or three in this syllabus. Too many significant figures makes the answer look more exact than it really is. Too few loses useful information.

A scalar 标量 has size only. A vector 矢量 has both size and direction.

Examples from the syllabus:

• scalars: mass, time, temperature, energy, work, power, distance, speed, pressure, density 密度, electric charge 电荷
• vectors: displacement 位移, velocity, acceleration, force (including weight), momentum 动量

Quick test: if it makes sense to ask "in which direction?", the quantity is a vector. You cannot ask "in which direction is the temperature?", so temperature is a scalar. You can ask "in which direction is the velocity?", so velocity is a vector.

Adding and subtracting vectors

A vector is drawn as an arrow: the direction of the arrow gives the direction of the quantity, and the length of the arrow (drawn to scale) gives the magnitude.

To add two coplanar 共面 vectors (vectors in the same flat plane), draw them tip to tail. The resultant 合矢量 goes from the tail of the first arrow to the tip of the second.

To find $\vec{X} - \vec{Y}$, add the reverse of $\vec{Y}$: $\vec{X} + (-\vec{Y})$. The reverse of $\vec{Y}$ has the same size as $\vec{Y}$ but points the opposite way.

If the two vectors are at right angles (90°), the size of the resultant is

and its direction comes from $\tan\theta = Y / X$.

Worked example. A swimmer heads north at $1.2\ \text{m s}^{-1}$ across a river that flows east at $0.5\ \text{m s}^{-1}$. Find the size and direction of the resultant velocity.

The two velocities are perpendicular, so

at an angle $\tan\theta = 0.5/1.2$, giving $\theta \approx 23°$ east of north.

If two vectors have the same size $F$ with an angle $2\alpha$ between them, the resultant has size $2F\cos\alpha$ and lies along the line that cuts the angle in half.

Splitting a vector into perpendicular parts

Any vector can be split into two perpendicular 垂直 (at right angles) components 分量. Usually these are horizontal 水平 and vertical 竖直, or along and across a surface. For a vector $\vec{v}$ at angle $\theta$ to the horizontal:

Choose the directions that make the problem easiest. On a slope (an inclined plane 斜面), split the weight into one part along the slope and one part at right angles to it:

where $\theta$ is the angle of the slope to the horizontal.

You split a vector into components whenever you need to know how much of it acts in one direction. For example: the part of a force that acts along a slope, or the horizontal and vertical parts of a ball's velocity after it is thrown.

• Give every answer a unit, and check homogeneity — both sides of an equation must have the same base units.
• Distinguish random error (reduce by repeating and averaging) from systematic error (a zero or calibration error that repeats do not remove).
• Combine uncertainties: add absolute uncertainties when adding/subtracting, add percentage uncertainties when multiplying/dividing (and multiply the % by any power).
• Distinguish precision (small spread) from accuracy (close to the true value).
• Resolve a vector into perpendicular components ($F\cos\theta$, $F\sin\theta$); add vectors tip-to-tail or by components.

These five quantities come up in almost every kinematics 运动学 question. Learn the exact words — the examiner gives marks for precise wording.

• distance 距离 — the total length of the path travelled. A scalar 标量.
• displacement 位移 — the straight-line distance from the start to the end, with a direction. A vector 矢量.
• speed 速率 — the rate of change of distance with time. A scalar.
• velocity 速度 — the rate of change of displacement with time. A vector.
• acceleration 加速度 — the rate of change of velocity with time. A vector.

The unit of speed and velocity is $\text{m s}^{-1}$; the unit of acceleration is $\text{m s}^{-2}$.

A common mistake: deceleration 减速度 just means acceleration in the opposite direction to the velocity. It is not a separate quantity.

Many marks come from reading or drawing motion graphs. Two graphs matter.

Displacement–time graph

The gradient 斜率 (steepness) of a displacement–time graph at a point gives the velocity at that moment.

• flat line → the object is at rest.
• straight sloping line → constant velocity (gradient = velocity).
• curved line → changing velocity. Draw a tangent 切线 at the point and find its gradient.

Velocity–time graph

The gradient of a velocity–time graph gives the acceleration at that moment.

The area between the line and the time axis gives the displacement in that time.

• flat line → constant velocity (zero acceleration).
• straight sloping line → uniform acceleration 匀加速 (constant acceleration).
• curved line → changing acceleration.
• area above the time axis is positive displacement; area below is negative (the object moved backwards).

To find the displacement, split the area into triangles and rectangles, or count grid squares. Area of a triangle is $\tfrac{1}{2} \times \text{base} \times \text{height}$; area of a rectangle is $\text{base} \times \text{height}$.

For motion in a straight line with uniform acceleration, we use five symbols: starting velocity $u$, final velocity $v$, acceleration $a$, displacement $s$, and time $t$. Four equations link them:

Each equation uses four of the five symbols. To pick the right one: write down what you know and what you want, then choose the equation with exactly those four.

Worked example. A car accelerates uniformly from $8\ \text{m s}^{-1}$ to $20\ \text{m s}^{-1}$ over a distance of $56\ \text{m}$. Find its acceleration.

We know $u$, $v$ and $s$ and want $a$, so use $v^{2} = u^{2} + 2as$:

Where the SUVAT equations come from

You should be able to get these from the definitions of velocity and acceleration:

• $v = u + at$ comes from $a = (v - u)/t$, the gradient of the line.
• $s = \tfrac{1}{2}(u + v) t$ is the area under the line — a trapezium 梯形 with parallel sides $u$ and $v$ and width $t$.
• $s = ut + \tfrac{1}{2}at^{2}$ comes from putting $v = u + at$ into the area.
• $v^{2} = u^{2} + 2as$ comes from removing $t$ from the first two.

If a question asks "which equation can be found using only the gradient of a velocity–time graph?", the answer is $v = u + at$ (the gradient is the acceleration).

Choosing a positive direction

Pick a positive direction at the start and keep it. Anything pointing the other way gets a minus sign. For a ball thrown straight up, if "up" is positive: $u$ is positive, $a = -g$ (gravity 重力 pulls down), and at the highest point the displacement is positive but the velocity is zero.

When air resistance 空气阻力 can be ignored, an object in free fall 自由落体 has a constant acceleration $g \approx 9.81\ \text{m s}^{-2}$ downwards. This is the same for every mass.

For a ball dropped from rest and falling a distance $h$:

For a ball thrown straight up with speed $u$:

• greatest height: put $v = 0$ in $v^{2} = u^{2} - 2gh$, giving $h = u^{2}/(2g)$.
• time to reach the top: put $v = 0$ in $v = u - gt$, giving $t = u/g$.
• total time to fall back to the start height: $2u/g$ (the motion is symmetric 对称).

Worked example. A ball is thrown straight up at $20\ \text{m s}^{-1}$. Find the greatest height it reaches (take $g = 9.81\ \text{m s}^{-2}$).

At the highest point $v = 0$, so from $h = u^{2}/(2g)$:

Experiment to find $g$

A common method: drop an object from rest, then measure the distance $h$ it falls and the time $t$ it takes. Then

Repeat for several heights and plot $h$ against $t^{2}$. The gradient of the best straight line is $g/2$, so $g$ is twice the gradient. Repeating reduces random error 随机误差. An electronic timer — using light gates 光电门, or a switch the ball hits — removes reaction-time 反应时间 error.

When an object moves at constant velocity in one direction (say horizontal 水平) and speeds up in a direction at right angles to it (say vertical 竖直, under gravity), the two motions do not affect each other. Treat each direction on its own, with its own SUVAT equation.

Horizontal throw

An object thrown horizontally with speed $u_{\text{H}}$ from height $h$, with air resistance ignored:

• horizontal: constant velocity $u_{\text{H}}$. After time $t$, the horizontal distance is $x = u_{\text{H}} t$.
• vertical: starts from rest and speeds up downwards at $g$. After time $t$, it has fallen $y = \tfrac{1}{2} g t^{2}$ and has vertical velocity $v_{\text{V}} = g t$.

The time to reach the ground depends only on the height $h$, not on $u_{\text{H}}$. Solve $h = \tfrac{1}{2} g t^{2}$ for $t$; then the horizontal range 射程 is $u_{\text{H}} t$.

Worked example. A ball is thrown horizontally at $15\ \text{m s}^{-1}$ from the top of a cliff $20\ \text{m}$ high. Find the time it takes to land and how far from the base it lands (take $g = 9.81\ \text{m s}^{-2}$).

Vertical motion gives the time: from $h = \tfrac{1}{2}g t^{2}$,

Horizontal motion then gives the range: $x = u_{\text{H}} t = 15 \times 2.0 \approx 30\ \text{m}.$

The horizontal-velocity graph is a flat line at $u_{\text{H}}$. The vertical-velocity graph is a straight line from the origin with gradient $g$.

Projectile at an angle

A projectile 抛体 thrown at speed $u$ at angle $\theta$ above the horizontal:

• horizontal component 分量 of the starting velocity: $u_{\text{H}} = u \cos\theta$ (stays constant during the flight).
• vertical component of the starting velocity: $u_{\text{V}} = u \sin\theta$ (gets smaller, becomes zero at the top, then grows downwards).

At the highest point, $v_{\text{V}} = 0$, but $v_{\text{H}}$ is still $u\cos\theta$. The time to the top is $t_{\text{up}} = u\sin\theta / g$; the total flight time (back to the start height) is $2t_{\text{up}}$.

Bouncing ball

When a ball bounces, its velocity–time graph is a set of straight sloping lines (constant $g$) with a sudden jump at each bounce (the velocity flips direction, and gets smaller if some energy 能量 is lost). Add up the times and the distances across the bounces.

When two objects move along the same line in different ways, write a displacement equation for each. Use the same start time and the same positive direction. Then set the two displacements equal (or set their difference to a given gap).

For a goods train at constant velocity $u_{\text{G}}$ and an express train starting from rest with acceleration $a$, both passing the same point at $t = 0$:

They are level again when $s_{\text{G}} = s_{\text{E}}$, giving $t = 2 u_{\text{G}} / a$.

1. Draw a diagram and mark the positive direction.
2. List the SUVAT symbols with their known and unknown values, including signs.
3. Choose the SUVAT equation with exactly the four symbols you have, plus the one you want.
4. For projectile motion, split into horizontal and vertical SUVAT problems, linked only by the time $t$.
5. Always check the units of your answer, and that its size is sensible.

• Use the SUVAT equations only for constant acceleration; list $s, u, v, a, t$ and pick the equation missing your unknown.
• Choose one direction as positive and keep signs consistent (usually $g = -9.81\ \text{m s}^{-2}$).
• On a velocity-time graph, gradient $=$ acceleration and area $=$ displacement.
• For projectiles, treat horizontal (constant velocity) and vertical ($a = g$) motion separately, linked by the same time.

Mass

Mass 质量 tells you how hard it is to change an object's motion. The larger the mass, the larger the force 力 needed to give it a certain acceleration 加速度. Mass is measured in kilograms ($\text{kg}$) and is a scalar 标量.

Momentum

Linear momentum 动量 is the product of mass and velocity:

Momentum is a vector 矢量 — it points the same way as the velocity 速度. Its unit is $\text{kg m s}^{-1}$, which is the same as $\text{N s}$.

Force as the rate of change of momentum

Newton's second law, in its general form: the resultant force 合力 on an object equals the rate of change of its momentum.

When the mass is constant this becomes $F = ma$, because $\Delta p = m\,\Delta v$ and $\Delta v / \Delta t = a$. Cambridge questions often want you to use $F = \Delta p / \Delta t$ directly for a collision 碰撞 or an impulse 冲量: the average force equals the change in momentum divided by the contact time.

A ball hits a wall with momentum $p_{1}$ and bounces back with momentum $p_{2}$ in the opposite direction. The change in momentum is $\Delta p = p_{2} - p_{1}$ (give each direction the correct sign). The average force is $\Delta p / \Delta t$, where $\Delta t$ is the contact time.

Worked example. A $0.20\ \text{kg}$ ball hits a wall at $8.0\ \text{m s}^{-1}$ and bounces straight back at $6.0\ \text{m s}^{-1}$. The contact lasts $0.050\ \text{s}$. Find the average force on the ball.

Take the rebound direction as positive, so $u = -8.0\ \text{m s}^{-1}$ and $v = +6.0\ \text{m s}^{-1}$:

When you know the momentum but not the speed, the change in kinetic energy 动能 is

This comes from $E_{\text{k}} = \tfrac{1}{2} m v^{2} = p^{2} / (2m)$.

First law

An object stays at rest, or keeps moving at constant velocity in a straight line, unless a resultant external force 外力 acts on it. In short: zero resultant force means zero acceleration.

Second law

The resultant force on an object equals its rate of change of momentum, and acts in the same direction as that change. In SI units,

Acceleration and resultant force always point the same way.

Third law

When body A pushes on body B, body B pushes back on body A with an equal and opposite force. The two forces:

• act on different objects,
• are of the same type (both gravitational, both contact, both electrostatic 静电, and so on),
• have the same size and opposite direction.

A common trap: the weight 重力 of a block on a table and the normal contact force 支持力 from the table are not a third-law pair (they act on the same object and are different types). The third-law partner of the block's weight is the pull the block makes on the Earth. The third-law partner of the table's contact force is the push the block makes on the table.

For a rocket: the thrust 推力 on the rocket and the force on the gases are a third-law pair (the engine pushes the gas down, the gas pushes the engine up). Weight and air resistance are not part of this pair.

Weight is the force on an object from a gravitational field 重力场. Near the Earth's surface,

where $g \approx 9.81\ \text{m s}^{-2}$ is the acceleration of free fall. Weight is a vector that points towards the centre of the Earth. Do not mix it up with mass: mass is the same everywhere, but weight changes with place.

Friction and drag forces

A friction 摩擦力 force between two solid surfaces acts along the surface and opposes the sliding. A viscous 黏性 or drag 阻力 force is the resistive force from a fluid 流体 (a liquid or gas) on an object moving through it; air resistance is the case for air. You do not need to use any coefficient 系数 of friction or viscosity.

A simple model: the drag gets bigger as the speed gets bigger. At zero speed, the drag is zero.

An object falling through air

For an object dropped from rest and falling through air:

1. At first, only weight acts, so the object speeds up downwards at $g$.
2. As the speed grows, the upward drag grows. The resultant force gets smaller, so the acceleration gets smaller.
3. In the end, the drag equals the weight. The resultant force is zero, the acceleration is zero, and the speed stays constant — the terminal velocity 收尾速度.

On a velocity–time graph, the line starts straight with gradient $g$, then bends and flattens at the terminal velocity. This shape (fast start, then slowing acceleration, then constant speed) is how "falling with air resistance" differs from "free fall in a vacuum".

Energy during a terminal-velocity fall

At terminal velocity, a parachutist 跳伞者 has constant kinetic energy. But the gravitational potential energy 重力势能 keeps falling as they go down. Where does it go? Almost all of it turns into thermal energy 热能 of the air around them. It does not become kinetic energy of the parachutist — that stays constant.

Cyclist or car at constant speed

A vehicle at constant speed on a flat road has zero resultant force. The forward driving force is equal and opposite to the total resistive force (friction, air resistance, and rolling resistance). At higher speed the drag is larger, so the driving force must be larger too — and so the power 功率 must be larger.

The principle

For a system with no resultant external force, the total momentum stays constant. This is conservation of momentum 动量守恒.

It always holds when there is no outside resultant force — in collisions, explosions 爆炸, and recoil 反冲. In two dimensions, momentum is conserved along each direction on its own.

Worked example. A $2.0\ \text{kg}$ trolley and a $3.0\ \text{kg}$ trolley are held together against a compressed spring, then released from rest. The $2.0\ \text{kg}$ trolley flies off at $6.0\ \text{m s}^{-1}$. Find the speed of the other trolley.

The total momentum stays zero (it started at rest), so

in the opposite direction.

Elastic and inelastic collisions

In every collision, momentum is conserved (if there is no outside resultant force).

An elastic collision 弹性碰撞 is one where the total kinetic energy is also conserved. A quick test: in an elastic collision, the relative speed 相对速率 of approach equals the relative speed of separation.

In an inelastic collision 非弹性碰撞, momentum is conserved but kinetic energy goes down — some becomes thermal, sound, or deformation 形变 energy. If the two objects stick together, the collision is perfectly inelastic.

Solving collision problems (one dimension)

For two objects with masses $m_{1}, m_{2}$ and starting velocities $u_{1}, u_{2}$ that hit head-on 正面, write

Use signed velocities (positive in one chosen direction). If the collision is elastic, add the relative-speed equation

or, the same thing, $\tfrac{1}{2} m_{1} u_{1}^{2} + \tfrac{1}{2} m_{2} u_{2}^{2} = \tfrac{1}{2} m_{1} v_{1}^{2} + \tfrac{1}{2} m_{2} v_{2}^{2}$. That gives two equations for two unknowns.

Worked example. A $1500\ \text{kg}$ car moving at $12\ \text{m s}^{-1}$ runs into a stationary $1000\ \text{kg}$ car and they lock together. Find their common velocity just after the collision.

Momentum is conserved (the cars stick, so $v_{1} = v_{2} = v$):

A useful result for a head-on elastic collision of mass $m$ with a stationary 静止 mass $M$:

Collisions in two dimensions

If the objects move in two dimensions, split the velocities into perpendicular 垂直 components 分量 and apply conservation of momentum along each direction on its own. For a collision where the objects hit at an angle, choose one axis along the first object's motion and one across it. The total momentum is conserved along each axis.

Rocket / pushing out mass

A rocket pushes out gas at velocity $u$ (relative to itself) at a mass-flow rate 质量流率 $\dot m$ (kg per second). It feels a thrust

which comes from $F = \Delta p / \Delta t$. The momentum given to the gas each second equals the thrust on the rocket (Newton's third law: the rocket pushes the gas one way, the gas pushes the rocket the other way).

• Newton's second law is $F = \Delta p / \Delta t$ (rate of change of momentum); $F = ma$ is the special case for constant mass.
• Identify third-law pairs correctly: the same type of force, acting on two different bodies — not the balanced forces on one body.
• Momentum is conserved in every collision; kinetic energy is conserved only in an elastic collision.
• For terminal velocity, explain that drag rises with speed until drag $=$ weight, so the acceleration becomes zero.

Centre of gravity

The weight 重力 of a large object can be treated as acting at one single point, called the centre of gravity 重心 (the same as the centre of mass 质心 in a uniform gravitational field). For a uniform, regular shape — a rectangle, a sphere, a uniform rod — the centre of gravity is at the middle.

When you draw a free-body diagram 受力图, always put the weight arrow at the centre of gravity.

Moment of a force

The moment 力矩 of a force about a point is

where $F$ is the size of the force 力 and $d$ is the perpendicular distance 垂直距离 from the point to the line of action 作用线 of the force. Unit: $\text{N m}$.

If the force acts at angle $\theta$ to a lever arm 力臂 of length $r$ from the pivot 支点, then $d = r\sin\theta$, so $M = F r\sin\theta$. Only the part of the force perpendicular 垂直 to the lever arm makes it turn.

Worked example. A force of $50\ \text{N}$ is applied to the end of a spanner $0.40\ \text{m}$ long, at $30°$ to the spanner. Find the moment of the force about the nut.

A moment is either clockwise 顺时针 or anticlockwise 逆时针 about the chosen point.

Couple and torque

A couple 力偶 is a pair of forces that are:

• equal in size,
• opposite in direction,
• with their lines of action a perpendicular distance apart.

A couple makes the body turn only — its resultant force is zero, so it gives no straight-line acceleration.

The torque 力偶矩 of a couple is the turning effect it makes:

where $F$ is the size of one force and $d$ is the perpendicular distance between the two lines of action. Unit: $\text{N m}$. The torque is the same about any point — a special property of couples.

A common multiple-choice trap: two equal forces in the same direction are not a couple (they have a resultant force and cause translation 平动). A couple needs equal size and opposite direction.

Conditions for equilibrium

A body is in equilibrium 平衡 when:

1. the resultant force 合力 is zero (no straight-line acceleration), AND
2. the resultant moment about any point is zero (no angular acceleration 角加速度).

Both must hold. A body with no resultant force can still be turning; a body with no resultant moment can still be moving in a straight line.

Principle of moments

For a body that is not turning, the total clockwise moment about any point equals the total anticlockwise moment about the same point. This is the principle of moments 力矩原理.

To solve a balance problem:

1. Choose a pivot — usually where an unknown force acts, so that force drops out (its distance is zero).
2. List every force and its perpendicular distance from the pivot.
3. Set $\sum M_{\text{clockwise}} = \sum M_{\text{anticlockwise}}$.
4. Use $\sum F = 0$ if you need a second equation.

A ruler balanced on a pivot with masses on each side is solved this way. For a heavy uniform rod, remember to include its weight acting at its centre of gravity.

Worked example. A uniform beam of length $4.0\ \text{m}$ is pivoted at its centre. A $40\ \text{N}$ weight hangs $1.5\ \text{m}$ from the pivot on one side. How far from the pivot, on the other side, must a $30\ \text{N}$ weight hang to balance the beam?

The beam's own weight acts at the centre (the pivot), so it has no moment. Setting clockwise $=$ anticlockwise moments:

Vector triangle

Three forces in the same plane that are in equilibrium can be drawn as a closed vector triangle 矢量三角形 — drawn tip-to-tail, the three arrows come back to the start. This is a drawing method instead of splitting into components 分量.

Use the sine rule 正弦定理 or the cosine rule 余弦定理 on the triangle to find unknown sizes or directions, or draw the triangle to scale on graph paper.

You can also split each force into horizontal 水平 and vertical 竖直 components and set $\sum F_{x} = 0$ and $\sum F_{y} = 0$.

Density 密度 is the mass per unit volume:

Unit: $\text{kg m}^{-3}$ (or $\text{g cm}^{-3}$; $1\ \text{g cm}^{-3} = 1000\ \text{kg m}^{-3}$). Density is a scalar 标量.

Some useful densities to know:

• water: $1000\ \text{kg m}^{-3}$
• air at room conditions: $\sim 1.2\ \text{kg m}^{-3}$
• iron / steel: $\sim 7800\ \text{kg m}^{-3}$

Pressure 压强 is the force per unit area, where the force acts at right angles to the surface:

Unit: $\text{Pa} = \text{N m}^{-2}$. Pressure is a scalar.

Hydrostatic pressure

Take a column of fluid 流体 with density $\rho$, cross-sectional area 横截面积 $A$ and height $\Delta h$. Its weight is

This weight presses down on the area $A$ at the bottom, so the extra pressure at the bottom compared with the top is

This is the hydrostatic pressure 流体静压强 equation. It depends only on the density of the fluid and the depth 深度 — the shape of the container does not matter.

Worked example. Find the extra pressure due to the water at the bottom of a swimming pool $2.5\ \text{m}$ deep. (Water density $1000\ \text{kg m}^{-3}$, $g = 9.81\ \text{m s}^{-2}$.)

For a submarine at depth $h$ below the surface, the pressure from the water is $\rho_{\text{seawater}}\, g\, h$. For the total pressure, add the atmospheric pressure 大气压强 at the surface (about $1.0 \times 10^{5}\ \text{Pa}$).

Upthrust and Archimedes' principle

When an object is submerged 浸没 in a fluid, the pressure at the bottom of the object is greater than the pressure at the top (by $\rho g \Delta h$, where $\Delta h$ is the object's height). This difference gives a net upward force called the upthrust 浮力.

For an object of volume $V$ (the volume of fluid displaced 排开), the upthrust is

This is Archimedes' principle 阿基米德原理: the upthrust on a body in a fluid equals the weight of the fluid it pushes aside.

Worked example. A metal block of volume $2.0 \times 10^{-3}\ \text{m}^{3}$ is fully submerged in water. Find the upthrust on it. (Water density $1000\ \text{kg m}^{-3}$, $g = 9.81\ \text{m s}^{-2}$.)

For a fully submerged object, $V$ is its full volume. For a floating 漂浮 object, $V$ is only the volume below the surface — the object floats when the upthrust on the part below the surface equals its weight.

Force balance with upthrust

A block held under water by a string tied to the bottom of the container is in equilibrium under three vertical forces: weight (down), tension 张力 (down), upthrust (up). Set $F_{\text{upthrust}} = W + T$ to find the tension.

A submerged block hanging from a newton meter 弹簧测力计 reads less than its weight in air, because of the upthrust: reading $= W - F_{\text{upthrust}}$.

The upthrust depends on the fluid density and the displaced volume, not on the object's material or depth (for an incompressible 不可压缩 fluid). On a planet with smaller $g$, the upthrust is smaller in the same ratio as the weight, so a floating object still floats with the same fraction below the surface.

• Take moments about a chosen pivot; for equilibrium, total clockwise $=$ total anticlockwise moments and the resultant force is zero.
• A body in equilibrium under three forces gives a closed triangle of forces.
• Fluid pressure $= \rho g h$; upthrust $=$ weight of fluid displaced (Archimedes).
• Distinguish mass, weight and density, and give the base unit each time.

Work done by a force

Work 功 is done when a force moves its point of contact along the line of the force. The work done by a constant force $F$ that causes a displacement 位移 $s$ is

where $\theta$ is the angle between the force and the displacement. Only the component 分量 of the force along the displacement does work.

Worked example. A child pulls a sledge $5.0\ \text{m}$ across the snow with a rope, using a force of $20\ \text{N}$ at $60°$ to the ground. Find the work done by the rope.

Unit: $\text{J} = \text{N m}$. Work is a scalar 标量.

Special cases:

• force in the same direction as the motion ($\theta = 0$): $W = Fs$, positive work, energy 能量 given to the object.
• force at right angles to the motion ($\theta = 90°$): $W = 0$. The normal contact force 支持力 on a car on a flat road does no work.
• force opposite to the motion ($\theta = 180°$): $W = -Fs$, negative work, energy taken from the object (for example friction 摩擦力).

For an object moving up a slope at angle $\alpha$ to the horizontal 水平, the work done against gravity in rising a height $h$ is $mgh$, while the work done by a horizontal push over the slope length $L$ uses $\cos\alpha$.

Conservation of energy

Energy is never made or destroyed — it only changes from one form to another, or moves from one object to another. In a closed system 封闭系统, the total energy stays constant. This is conservation of energy 能量守恒.

When you write an energy equation, list every form the energy starts as and ends as. Common forms in this syllabus: kinetic, gravitational potential, elastic potential energy 弹性势能, electrical 电能, thermal, sound, chemical energy 化学能.

A ball rolling down a frictionless 无摩擦 ramp 斜坡 turns gravitational potential energy 重力势能 into kinetic energy 动能: $mgh = \tfrac{1}{2} m v^{2}$, so $v = \sqrt{2gh}$. With friction, some of this energy becomes thermal energy 热能 of the ramp and the air.

Worked example. A ball is released from rest at the top of a smooth ramp $1.2\ \text{m}$ high. Find its speed at the bottom (take $g = 9.81\ \text{m s}^{-2}$).

All the gravitational potential energy becomes kinetic energy, so $v = \sqrt{2gh}$ (the mass cancels):

Efficiency

The efficiency 效率 of a system is

The same idea with power:

Efficiency is always less than 100% in a real system, because some input energy becomes "useless" forms — usually thermal energy.

For an electric motor lifting a load with efficiency $\eta$ at voltage 电压 $V$ and current 电流 $I$, the useful output power is $\eta V I$. From this you can find a force, a lifting speed, or a tension 张力.

Worked example. An electric motor lifts a $50\ \text{kg}$ load at a steady $0.40\ \text{m s}^{-1}$ while drawing $250\ \text{W}$ of electrical power. Find its efficiency (take $g = 9.81\ \text{m s}^{-2}$).

The useful output power is $P = mgv = 50 \times 9.81 \times 0.40 = 196\ \text{W}$, so

Power 功率 is the rate of doing work, or the rate of transferring energy:

Unit: $\text{W} = \text{J s}^{-1}$. Power is a scalar.

Power, force and velocity

For an object moving at velocity 速度 $v$ with a force $F$ along the direction of motion, in a short time $\Delta t$ the displacement is $v\,\Delta t$ and the work done is $F v\,\Delta t$. Dividing by $\Delta t$:

This is one of the most useful results in mechanics.

Worked example. A car travels at a steady $25\ \text{m s}^{-1}$ against a total resistive force of $600\ \text{N}$. Find the output power of its engine.

At constant speed the driving force equals the resistive force, so

• For a car at constant velocity $v$ on a flat road, the engine power must balance the total resistive force: $P = F_{\text{resist}} \cdot v$. If the drag 阻力 grows with $v^{2}$, doubling the speed roughly quadruples the power needed.
• For lifting a weight 重力 $mg$ straight up at constant speed $v$, the useful output power is $P = mg \cdot v$.
• For an aircraft hovering at a fixed height, the lift force equals the weight, and a large power is needed because air must be pushed downwards all the time.

In a uniform gravitational field (close to a planet's surface), the change in gravitational potential energy of mass 质量 $m$ rising or falling through a height $\Delta h$ is

Where it comes from

The work done against gravity to raise a mass $m$ slowly (no change in kinetic energy) through height $\Delta h$ equals the gravitational potential energy gained:

• the gravitational force on the mass is $mg$ downwards,
• the force needed to lift it slowly is $mg$ upwards,
• the work done by this force is $W = F \cdot s = mg \cdot \Delta h$,
• this work becomes $\Delta E_{\text{P}}$.

So $\Delta E_{\text{P}} = mg \Delta h$. To use it you need $m$ and $\Delta h$ (and $g$). You do not need speed or time.

The kinetic energy of an object of mass $m$ moving at speed $v$ is

Where it comes from

Apply a resultant force $F$ to a mass $m$ that starts at rest. It speeds up evenly from $0$ to $v$ over a displacement $s$. From $v^{2} = u^{2} + 2as$ with $u = 0$,

The work done on the mass is

All this work becomes kinetic energy, so $E_{\text{K}} = \tfrac{1}{2} m v^{2}$.

Kinetic energy and momentum

Combining $p = mv$ and $E_{\text{K}} = \tfrac{1}{2} m v^{2}$:

This is handy when the momentum 动量 is given but not the velocity. For a momentum change from $p_{1}$ to $p_{2}$ at constant mass, the change in kinetic energy is $(p_{2}^{2} - p_{1}^{2}) / (2m)$.

A useful plan for problems that mix forces and energy:

1. Find the start and end states. Write the kinetic and potential energies in each.
2. List any work done by outside forces (friction, a push). Friction usually takes energy out; a push can add it.
3. Conservation of energy: $E_{\text{start}} + W_{\text{in}} = E_{\text{end}} + W_{\text{lost as heat etc.}}$.

Examples:

• A box pushed at constant velocity up a ramp of length $L$ rising by $h$: $E_{\text{K}}$ does not change, so the work done by the push goes into $\Delta E_{\text{P}}$ plus the work done against friction.
• A block sliding into a spring 弹簧 with kinetic energy $E_{\text{K}}$ on a frictionless surface: at greatest compression 压缩 $x$, all the kinetic energy has become elastic potential energy $\tfrac{1}{2} k x^{2}$ (where $k$ is the spring constant 劲度系数).
• A ball dropped from height $h_{1}$ that bounces to height $h_{2}$: the ratio $h_{2}/h_{1}$ is the fraction of mechanical energy kept, $h_{2}/h_{1} = (v_{\text{up}}/v_{\text{down}})^{2}$.
• A projectile 抛体 thrown to the same height at different angles: the final speed is the same (only the height matters); use components to get its direction.

• Work $=$ force $\times$ distance moved in the direction of the force — use $Fs\cos\theta$ when they are at an angle.
• Use conservation of energy: loss in GPE $=$ gain in KE ($+$ work done against resistance).
• Power $=$ work $/$ time $= Fv$; efficiency $=$ useful output $\div$ total input.
• GPE uses the vertical height gained, not the distance along a slope.

When a force acts on a solid along its length, the object changes shape (deformation 形变). Two cases (treated as one-dimensional here):

• a tensile 拉伸 force stretches the object — it makes an extension 伸长量 $x$,
• a compressive force squeezes the object — it makes a compression 压缩, treated as a negative extension.

The applied force is the load 负载. The change from the natural length is the extension (or compression).

For many materials at small extensions, the extension is proportional to the load — this is Hooke's law 胡克定律. The constant that links them is the spring constant 劲度系数 $k$:

Unit of $k$: $\text{N m}^{-1}$.

Worked example. A spring stretches by $4.0\ \text{cm}$ when a $2.0\ \text{N}$ load is hung from it. Find its spring constant.

Converting the extension to metres ($4.0\ \text{cm} = 0.040\ \text{m}$):

Reading a graph:

• A force–extension ($F$ against $x$) graph has gradient $k$ in the Hooke's-law region.
• An extension–force ($x$ against $F$) graph has gradient $1/k$ in the Hooke's-law region.

A common trap: if a graph plots length $L$ against force, you can still find the spring constant from the gradient (since $L = L_{0} + F/k$, the gradient is $1/k$ — read it off carefully).

Limit of proportionality

Hooke's law only holds up to the limit of proportionality 比例极限. Past this point the $F$ against $x$ line curves and is no longer straight. The material may still be elastic 弹性 (it returns to its first length when you remove the load) a little further, then it becomes plastic 塑性.

Springs in series and parallel

You may need to combine spring constants:

• Series 串联 (one spring hangs from another): the same load passes through both, the total extension is the sum, so $\dfrac{1}{k_{\text{total}}} = \dfrac{1}{k_{1}} + \dfrac{1}{k_{2}}$.
• Parallel 并联 (two springs side by side holding the same load): each takes half the load (if they are identical), the extensions are equal, so $k_{\text{total}} = k_{1} + k_{2}$.

For a wire of uniform cross-section under a tensile load:

• Stress 应力 $\sigma = \dfrac{F}{A}$, where $F$ is the load and $A$ is the cross-sectional area 横截面积. Unit: $\text{Pa}$.
• Strain 应变 $\varepsilon = \dfrac{x}{L_{0}}$, where $x$ is the extension and $L_{0}$ is the original length. Strain has no unit (it is a ratio of lengths).

The Young modulus 杨氏模量 is the ratio of stress to strain in the Hooke's-law region:

Unit: $\text{Pa}$ (about $10^{11}$ for metals; e.g. steel $\approx 2.0 \times 10^{11}$ Pa).

Worked example. A steel wire of length $2.0\ \text{m}$ and cross-sectional area $1.5 \times 10^{-7}\ \text{m}^{2}$ stretches by $1.0\ \text{mm}$ under a load of $15\ \text{N}$. Find the Young modulus.

The Young modulus is a property of the material — it does not depend on the wire's shape or size. The spring constant $k$ depends on both the material and the size: $k = EA/L_{0}$.

Experiment to find the Young modulus of a metal wire

A standard setup:

1. Clamp one end of a long, thin wire to a fixed support. Pass the wire over a pulley 滑轮 at the edge of the bench so it hangs straight down.
2. Measure the original length $L_{0}$ between the clamp and a marker near the pulley, using a metre rule.
3. Measure the diameter 直径 $d$ of the wire at several places with a micrometer 螺旋测微器 and take the average. Work out $A = \pi d^{2}/4$.
4. Hang weights one at a time. Record the load $F$ and the extension $x$ (how far the marker moves against a fixed scale).
5. Plot $F$ against $x$. In the straight region the gradient is $EA/L_{0}$, so $E = \text{gradient} \times L_{0}/A$.

Why a long, thin wire? To make the extension big enough to measure well. Why repeat readings and measure $d$ at several places? To reduce random error 随机误差 and check the wire is uniform.

As the load grows:

1. Elastic and straight (Hooke obeyed) — up to the limit of proportionality. Removing the load returns the object to its first length.
2. Elastic but curved — between the limit of proportionality and the elastic limit 弹性极限. The extension is no longer straight in the load, but on unloading the object still returns to its first length.
3. Plastic — past the elastic limit. On unloading, the object does not return to its first length; a permanent extension stays.

Hooke's law only holds in the straight, elastic region.

On a force–extension graph for a material taken into the plastic region and then unloaded, the loading line and the unloading line are different. The unloading line is parallel to the first Hooke line but shifted to the right (the permanent extension left when the load reaches zero). The area between the loading and unloading lines is the energy turned into thermal energy 热能 in the material.

The work done in stretching a material from $0$ to extension $x$, as the load grows from $0$ to $F$, is the area under the force–extension graph.

Hooke's-law material

When Hooke's law holds, the $F$ against $x$ graph is a straight line through the origin. The area under it from $0$ to $x$ is a triangle:

This is the elastic potential energy 弹性势能 stored in a spring or wire stretched within its limit of proportionality. An equal form:

Worked example. A spring of spring constant $50\ \text{N m}^{-1}$ is stretched by $0.20\ \text{m}$, within its limit of proportionality. Find the elastic potential energy stored.

Non-Hooke material

For a graph that is not a straight line (a stretched rubber band, or a spring past its limit of proportionality), find the area by counting grid squares or by using trapezia 梯形. The same idea holds: the area under the force–extension graph is the work done on the material.

Comparing stored energy

A common multiple-choice case: two materials are stretched by the same force, or by the same extension. Using $E_{\text{P}} = \tfrac{1}{2} F x$:

• same $F$, smaller $k$ (less stiff) → larger $x$ → more energy stored.
• same $x$, larger $k$ (stiffer) → larger $F$ → more energy stored.

When a stretched spring is released onto a mass, the elastic potential energy becomes kinetic energy 动能 (and gravitational potential energy if the mass rises). Set $\tfrac{1}{2} k x^{2}$ equal to $\tfrac{1}{2} m v^{2}$ (plus any $mgh$) to find the speed or height.

• Hooke's law ($F = kx$) holds only up to the limit of proportionality.
• Stress $= F/A$, strain $= x/L$, Young modulus $=$ stress$/$strain (gradient of the straight part of the stress-strain graph) — watch the units (Pa).
• Energy stored $=$ area under the force-extension graph $= \frac{1}{2}Fx$ in the elastic region.
• Distinguish elastic (returns to shape) from plastic (permanent) deformation.

A wave 波 carries energy 能量 from one place to another without moving matter overall. The particles of the medium 介质 oscillate 振动 about fixed rest positions; only the disturbance (and its energy) propagates 传播. Examples: a transverse wave 横波 on a rope, a longitudinal wave 纵波 on a slinky spring, ripples on water, and sound in air. A wave that travels and carries energy is a progressive wave 行波.

Key terms

• displacement 位移 $y$ — how far a particle has moved from its rest position at a moment. A vector 矢量.
• amplitude 振幅 $A$ — the largest displacement from the rest position.
• wavelength 波长 $\lambda$ — the shortest distance along the wave between two points that move in phase 同相 (for example, two next-door crests 波峰).
• period 周期 $T$ — the time for one full oscillation of a particle.
• frequency 频率 $f$ — the number of full oscillations per second; $f = 1/T$. Unit: hertz 赫兹, $\text{Hz}$.
• speed 速率 $v$ — how fast a crest travels along the medium.
• phase difference 相位差 — the fraction of a cycle by which one oscillation leads or lags another. Given in radians 弧度 (a full cycle is $2\pi$) or degrees (a full cycle is $360°$).

Two points one wavelength apart are in phase (phase difference 0 or $2\pi$). Two points half a wavelength apart are exactly out of phase (phase difference $\pi$).

The wave equation

In one period $T$, the wave moves forward by one wavelength $\lambda$. So speed $= \text{distance} / \text{time} = \lambda / T = \lambda f$:

This comes straight from the definitions of speed, frequency and wavelength, and works for every progressive wave.

Reading a CRO trace

A cathode-ray oscilloscope 示波器 (CRO) draws a voltage signal — for sound, the output of a microphone — against time. Two controls matter:

• time-base 时基 (seconds per division across): turns horizontal distance on the screen into time. Read the period $T$ as the distance between two next-door peaks, then $f = 1/T$.
• y-gain 垂直增益 (volts per division up): turns vertical distance into voltage. The amplitude in volts is the peak height from the centre line.

If the time-base is $5\ \text{ms}/\text{div}$ and one full cycle takes $4$ divisions, then $T = 4 \times 5\ \text{ms} = 20\ \text{ms}$ and $f = 50\ \text{Hz}$.

Intensity of a wave

A wave carries energy. The intensity 强度 at a point is the power 功率 passing through unit area at right angles to the direction of travel:

Unit: $\text{W m}^{-2}$.

Intensity is proportional to the square of the amplitude:

For a point source 点源 sending out energy equally in all directions, the wavefronts 波前 are spheres; the surface area at distance $r$ is $4\pi r^{2}$, so

Doubling the distance cuts the intensity to a quarter, which means the amplitude is halved (since $I \propto A^{2}$).

Worked example. A lamp emits $60\ \text{W}$ of light equally in all directions. Find the intensity of the light $2.0\ \text{m}$ away.

Transverse waves

The particles oscillate perpendicular 垂直 to the direction the energy travels. A wave on a rope, all electromagnetic waves, and S-waves in the Earth are transverse.

Longitudinal waves

The particles oscillate parallel to the direction the energy travels. Sound in any medium, P-waves in the Earth, and the squashes on a slinky are longitudinal. The wave is made of compressions 压缩 (higher pressure, particles close together) and rarefactions 稀疏 (lower pressure, particles spread out).

Graphs of waves

A graph of particle displacement against position at one moment looks like a sine curve 正弦曲线 for both kinds of wave. The difference: for a transverse wave the displacement axis is the real sideways displacement; for a longitudinal wave it is the small back-and-forth displacement along the direction of travel (positive one way, negative the other).

A graph of particle displacement against time at one point in space is also a sine curve for both kinds. Read the period $T$ from this graph.

When the source 波源 of a sound moves relative to a stationary 静止 observer 观察者, the heard frequency is different from the source frequency. This is the Doppler effect 多普勒效应.

• source moving towards the observer: the wavefronts in front are squashed, so the wavelength is shorter and the heard frequency is higher.
• source moving away from the observer: the wavefronts behind are spread out, so the wavelength is longer and the heard frequency is lower.

The formula (source moving at speed $v_{\text{s}}$ along the line to the observer; wave speed $v$, source frequency $f_{\text{s}}$, heard frequency $f_{\text{o}}$):

Choose the sign to match the physics:

• minus sign on the bottom when the source moves towards the observer ($f_{\text{o}} > f_{\text{s}}$),
• plus sign when the source moves away ($f_{\text{o}} < f_{\text{s}}$).

You only need the case of a stationary observer.

Worked example. A car horn at $f_{\text{s}} = 800\ \text{Hz}$ moves at $30\ \text{m s}^{-1}$ towards a still listener. Speed of sound $v = 340\ \text{m s}^{-1}$:

All electromagnetic waves 电磁波 (EM waves) are transverse and travel in a vacuum 真空 at the same speed:

The electromagnetic spectrum 电磁波谱 includes radio waves, microwaves 微波, infrared 红外线, visible light, ultraviolet 紫外线, X-rays X射线 and $\gamma$-rays γ射线.

Approximate wavelength ranges in free space (learn the orders of magnitude):

• radio waves: $> 10^{-1}\ \text{m}$ (up to many km).
• microwaves: $10^{-3}\ \text{m}$ to $10^{-1}\ \text{m}$.
• infrared: $\sim 7 \times 10^{-7}\ \text{m}$ to $10^{-3}\ \text{m}$.
• visible light: $400\ \text{nm}$ (violet) to $700\ \text{nm}$ (red), i.e. $4 \times 10^{-7}\ \text{m}$ to $7 \times 10^{-7}\ \text{m}$.
• ultraviolet: $\sim 10^{-8}\ \text{m}$ to $4 \times 10^{-7}\ \text{m}$.
• X-rays: $\sim 10^{-11}\ \text{m}$ to $10^{-8}\ \text{m}$.
• $\gamma$-rays: $< 10^{-11}\ \text{m}$.

The boundaries between regions are not sharp. Use $c = f\lambda$ to change between wavelength and frequency. Only light with wavelengths $400$–$700\ \text{nm}$ can be seen.

Polarisation 偏振 means making a transverse wave oscillate in one plane only.

• Only transverse waves can be polarised — the oscillation is perpendicular to the direction of travel, so different perpendicular planes are real choices.
• Longitudinal waves (sound) cannot be polarised — the oscillation is along the direction of travel, so there is no other plane.

So polarisation is a test: if a wave can be polarised, it must be transverse.

Malus's law

Plane-polarised 平面偏振 light of intensity $I_{0}$ passes through a polarising filter 偏振片 whose transmission axis 透光轴 is at angle $\theta$ to the plane of polarisation. The transmitted intensity is given by Malus's law 马吕斯定律:

• $\theta = 0°$: filter lined up with the polarisation, $I = I_{0}$, all passes through.
• $\theta = 90°$: filter at right angles, $I = 0$, all blocked.
• $\theta = 60°$: $I = I_{0} \cos^{2} 60° = I_{0} \cdot 0.25 = I_{0}/4$.

Worked example. Plane-polarised light of intensity $12\ \text{W m}^{-2}$ meets a polarising filter whose axis is at $30°$ to the plane of polarisation. Find the transmitted intensity.

For two filters in a row, use Malus's law twice with the angle between each pair. Be careful with the angle each time — after the first filter the polarisation is along that filter's axis, and the second filter's angle is measured from there.

(You do not need to work out the effect of a polarising filter on an unpolarised wave.)

• Define terms precisely (displacement, amplitude, wavelength, period, frequency) and use $v = f\lambda$.
• Distinguish transverse (vibration perpendicular to travel) from longitudinal (parallel); only transverse waves can be polarised.
• For the Doppler effect, the observed frequency rises as the source approaches and falls as it recedes.
• Learn the electromagnetic spectrum order; all its waves travel at $c$ in a vacuum.

When two or more waves 波 overlap at a point, the displacement 位移 there is the vector sum 矢量和 of the displacements each wave would make on its own. This is the principle of superposition 叠加.

The waves pass through each other and come out unchanged. Superposition is the base of everything in this topic.

If two waves of amplitude 振幅 $A_{1}$ and $A_{2}$ meet:

• in phase 同相 (crest 波峰 meets crest): the amplitude is $A_{1} + A_{2}$ (constructive interference 相长干涉).
• exactly out of phase (crest meets trough 波谷, phase difference 相位差 $\pi$): the amplitude is $|A_{1} - A_{2}|$ (destructive interference 相消干涉).
• any other phase difference $\phi$: the amplitude is somewhere between these two.

For intensity 强度, $I \propto A^{2}$. Two equal waves meeting in phase give intensity $(2A)^{2} = 4A^{2}$ — four times the intensity of one wave alone.

When two identical progressive waves 行波 travel in opposite directions and overlap, they make a stationary wave 驻波. Examples: a wave on a string reflected 反射 from a fixed end overlapping the incoming wave; sound in an air column reflected from a closed end; microwaves between an emitter and a metal sheet.

Nodes and antinodes

In a stationary wave:

• node 波节 — a point that is always at zero displacement (the two waves always cancel). The distance between next-door nodes is $\lambda/2$.
• antinode 波腹 — a point of largest amplitude (the two waves always add). The distance between next-door antinodes is $\lambda/2$.
• a node and the next antinode are $\lambda/4$ apart.

Particles between two nodes oscillate in phase with each other, but with different amplitudes (largest at the antinode, zero at the nodes). Particles on opposite sides of a node oscillate in antiphase 反相 (phase difference $\pi$).

Worked example. A string of length $0.80\ \text{m}$ is fixed at both ends and vibrates in its fundamental mode, where the wave speed is $240\ \text{m s}^{-1}$. Find the fundamental frequency.

One loop fits the string, so $\lambda = 2L = 1.60\ \text{m}$. Then

How a stationary wave differs from a progressive wave: it does not carry energy 能量 along its length, the pattern does not move along, and the nodes stay fixed; a progressive wave has the same amplitude everywhere and carries energy.

Measuring wavelength from node spacing

Drive a string with a vibrator at frequency $f$ until a stationary pattern appears. Measure the distance between two well-separated nodes and divide by the number of half-wavelengths 波长 between them. Then $\lambda$ is known, and $v = f\lambda$ gives the wave speed.

For a tube closed at one end and open at the other (a resonance tube 共鸣管), the closed end is a displacement node and the open end is a displacement antinode. The fundamental 基频 has $L = \lambda/4$; the next resonance is at $L = 3\lambda/4$; and so on. For a tube open at both ends, both ends are antinodes; the fundamental is at $L = \lambda/2$.

Diffraction 衍射 is the spreading of a wave after it passes through a gap or around an obstacle 障碍物. All waves diffract — water, sound, light, microwaves.

The amount of spreading depends on the ratio of wavelength to gap width:

• gap much wider than $\lambda$: very little spreading; the wave goes nearly straight through.
• gap about the size of $\lambda$: a lot of spreading; the wave fans out.
• gap smaller than $\lambda$: very strong spreading; the gap acts almost like a point source.

Show this with water waves in a ripple tank 水波槽: straight waves meet a barrier with a gap, and the waves curve more as the gap is made narrower. The same idea is why you can hear someone around a corner (speech has $\lambda$ near 1 m, close to the gap size) but cannot see them (visible light has $\lambda \sim 500\ \text{nm}$, far smaller than the gap).

Interference 干涉 is the superposition of two coherent 相干 waves to give a steady pattern of high-amplitude regions (constructive) and low-amplitude regions (destructive).

Coherence

Two sources are coherent when they emit waves with a constant phase difference (which also needs the same frequency). Two separate lamps are not coherent — their phase changes randomly, so any pattern flickers too fast to see and you get only an average.

To make coherent light from one source, pass it through two slits 狭缝 in a double-slit 双缝 setup. Both slits are lit by the same wavefront, so the two beams keep a fixed phase relationship.

Conditions for a clear pattern

To see two-source fringes you need:

1. two coherent sources (constant phase difference).
2. roughly equal amplitudes (or the dark regions are not very dark).
3. the waves overlap where you look.
4. for light (a transverse wave), the same plane of polarisation 偏振.

Path difference

For two coherent sources, what happens at a point depends on the path difference 路程差 $\Delta x$ between the two waves arriving there:

• constructive: $\Delta x = n\lambda$ (for whole numbers $n = 0, 1, 2, \ldots$).
• destructive: $\Delta x = (n + \tfrac{1}{2})\lambda$.

Double-slit (Young's) experiment

For two slits a distance $a$ apart, with a screen a distance $D$ away (assume $D \gg a$), light of wavelength $\lambda$ makes fringes on the screen.

The fringe spacing 条纹间距 $x$ (one fringe 条纹 to the next) is

Bright fringes (maximum 极大) are where the path difference is a whole number of $\lambda$; dark fringes (minimum 极小) where it is $(n + \tfrac{1}{2})\lambda$. The fringes are equally spaced.

To make the fringe spacing smaller: increase $a$ (slits further apart), reduce $D$ (screen closer), or use a shorter $\lambda$ (bluer light).

Worked example. In a double-slit experiment the slits are $0.50\ \text{mm}$ apart and lit by light of wavelength $600\ \text{nm}$. The screen is $2.0\ \text{m}$ away. Find the fringe spacing.

A diffraction grating 衍射光栅 has many equally spaced slits — often hundreds or thousands per millimetre. Each slit is a coherent source. A maximum is seen at angle $\theta$ from the normal 法线 to the grating when

where $d$ is the slit spacing 缝间距 (centre to centre), $n = 0, \pm 1, \pm 2, \ldots$ is the order 级次, and $\lambda$ is the wavelength.

Worked example. A diffraction grating has $500$ lines per mm. Light of wavelength $600\ \text{nm}$ is shone normally on it. Find the angle of the first-order ($n = 1$) maximum.

The slit spacing is $d = \dfrac{1}{500}\ \text{mm} = 2.0 \times 10^{-6}\ \text{m}$, so

Compared with the double slit, a grating gives much sharper maxima, because more slits add together — every other direction is cancelled by many slits.

Slit spacing from "lines per mm"

If a grating has $N$ lines per millimetre, then $d = 1/N$ millimetres $= 10^{-3}/N$ metres. For $450$ lines per mm, $d = 1/450\ \text{mm} \approx 2.22\ \mu\text{m}$.

Highest order

For a given grating and wavelength, $\sin\theta = n\lambda/d$ cannot be more than $1$, so the highest order seen is

If $d/\lambda = 3.27$, orders up to $n = 3$ exist; $n = 4$ would need $\sin\theta > 1$ and is not seen.

Finding $\lambda$ with a grating

Shine parallel light of unknown wavelength straight at the grating. Measure the angle $\theta_{1}$ of the first-order maximum from the centre. Then $\lambda = d \sin\theta_{1}$. Repeating for higher orders and averaging reduces error.

• Two-source interference: constructive when path difference $= n\lambda$, destructive when $= (n + \tfrac{1}{2})\lambda$; the sources must be coherent.
• Double slit: $\lambda = ax/D$; diffraction grating: $d\sin\theta = n\lambda$ — know every symbol.
• On a stationary wave mark nodes and antinodes; adjacent nodes are $\lambda/2$ apart; it stores energy but does not transfer it.
• A stationary wave needs two waves of the same frequency travelling in opposite directions.

An electric current 电流 is a flow of charge carriers 载流子. In a metal the carriers are negative conduction electrons 电子; in an electrolyte 电解质 they are positive and negative ions 离子; in a semiconductor 半导体 they may be electrons or "holes" 空穴. The conventional current 常规电流 direction is the way positive charge would flow — opposite to the real flow of electrons in a wire.

Charge

Charge is quantised 量子化: the smallest free unit of charge is the elementary charge 基本电荷

Every free charge in this syllabus is a whole-number multiple of $e$. The unit of charge is the coulomb 库仑, $\text{C}$.

Current as the rate of flow of charge

If charge $Q$ passes a point in time $t$, the current is

Unit of current: ampere, $\text{A}$ ($= \text{C s}^{-1}$). For a changing current, the charge that has flowed in a time is the area under an $I$–$t$ graph.

Worked example. A current of $0.50\ \text{A}$ flows for $2.0$ minutes. Find the charge that passes, and how many electrons this represents. ($e = 1.60 \times 10^{-19}\ \text{C}$.)

Drift velocity equation

For a uniform conductor of cross-section area $A$, with $n$ charge carriers per unit volume (the number density 数密度), each carrying charge $q$, moving with average drift velocity 漂移速度 $v$:

Worked example. A copper wire of cross-sectional area $1.0 \times 10^{-6}\ \text{m}^{2}$ carries a current of $5.0\ \text{A}$. Copper has $n = 8.5 \times 10^{28}$ free electrons per $\text{m}^{3}$. Find the drift velocity. ($e = 1.60 \times 10^{-19}\ \text{C}$.)

Rearranging $I = Anvq$ gives $v = \dfrac{I}{Anq}$:

The electrons drift very slowly — less than a millimetre per second.

Use this to compare currents:

• a thinner wire (smaller $A$) at the same $I$ needs a faster drift $v$.
• a semiconductor has far fewer free carriers than a metal (smaller $n$), so for the same $I$ the drift velocity is much larger.
• in series 串联 components, $I$ is the same everywhere, so if $A$ stays the same but the material changes, $nv$ changes the other way.

The potential difference 电势差 (p.d.) across a component is the energy 能量 transferred per unit charge as that charge passes through it:

Unit: volt 伏特, $\text{V}$ ($= \text{J C}^{-1}$).

If $1\ \text{J}$ of electrical energy changes into other forms (thermal, light, kinetic, …) when $1\ \text{C}$ of charge passes through a component, the p.d. across it is $1\ \text{V}$.

The electromotive force 电动势 (e.m.f.) of a source is the energy given per unit charge by the source. The formula is the same as for p.d.; the difference is direction: e.m.f. is energy given to the charge by the source; p.d. is energy given up by the charge to the component.

Combining $V = W/Q$ and $I = Q/t$:

Using Ohm's law $V = IR$:

Pick the form with the quantities you know. Examples:

• two heaters of equal resistance — the one with the larger current gives more power 功率 ($P = I^{2}R$).
• two resistors in parallel 并联 across the same voltage 电压 — the one with smaller $R$ gives more power ($P = V^{2}/R$).
• a kettle marked "$2.4\ \text{kW}, 240\ \text{V}$" draws $I = P/V = 10\ \text{A}$ and has resistance $R = V^{2}/P = 24\ \Omega$.

Energy transferred in time $t$ is $E = P t$.

The resistance 电阻 $R$ of a component is

Unit: ohm 欧姆, $\Omega$ ($= \text{V A}^{-1}$). Resistance depends on the conditions (such as temperature) when it is measured.

Ohm's law

A conductor obeys Ohm's law 欧姆定律 when the current through it is proportional to the p.d. across it, as long as the conditions (especially temperature) stay constant. For such a conductor $R$ is constant and the $I$–$V$ graph is a straight line through the origin.

Ohm's law is an experimental result, not a definition. The definition $R = V/I$ works for any component; only ohmic ones have constant $R$.

$I$–$V$ characteristics

You should be able to sketch these:

• metal wire at constant temperature — a straight line through the origin (constant $R$). Reversing the p.d. drives the current the other way, giving a straight line in both directions.
• filament lamp 灯丝灯泡 — through the origin, steep at first, then flatter as $V$ (and $I$) grow. Reason: more current heats the filament, so its resistance rises and the gradient $1/R$ falls.
• semiconductor diode 二极管 — almost no current for negative $V$ or small positive $V$. Above a "switch-on" voltage (about $0.7\ \text{V}$ for silicon), the current rises sharply.

Resistivity

For a uniform conductor of length $L$ and cross-section area $A$,

$\rho$ is the resistivity 电阻率, a property of the material, with unit $\Omega\ \text{m}$. Doubling the length doubles $R$; doubling the area halves it; halving the diameter quarters the area and so makes $R$ four times bigger.

Worked example. A copper wire of length $2.0\ \text{m}$ and cross-sectional area $1.7 \times 10^{-7}\ \text{m}^{2}$ has resistivity $1.7 \times 10^{-8}\ \Omega\ \text{m}$. Find its resistance.

Typical values: copper at room temperature $\rho \sim 1.7 \times 10^{-8}\ \Omega\ \text{m}$; an insulator 绝缘体 $\rho \sim 10^{15}\ \Omega\ \text{m}$ or more.

The resistivity of a metal rises with temperature (more lattice vibration 晶格振动 scatters 散射 the electrons), which is why the filament lamp's $I$–$V$ line curves.

A light-dependent resistor 光敏电阻 (LDR) is a semiconductor whose resistance falls as the light intensity rises. In bright light $R$ may be a few hundred $\Omega$; in the dark it can be in the megaohms. LDRs are used in light-sensing circuits (street lamps, camera light meters). Here the light intensity 光强 controls the resistance.

In this syllabus a thermistor 热敏电阻 has a negative temperature coefficient 负温度系数: its resistance falls as its temperature rises. This is useful for sensing temperature — put it in a potential divider 分压器 and the output voltage changes with temperature.

This is the opposite of a metal: in a semiconductor, more thermal energy frees more charge carriers, and this matters more than the extra scattering.

• Use $I = Q/t$, $V = W/Q$ (energy per unit charge) and $P = VI = I^2 R = V^2/R$.
• Ohm's law ($V = IR$) applies only to an ohmic conductor at constant temperature — a filament lamp is non-ohmic.
• Sketch and interpret the $I$-$V$ characteristics of a resistor, filament lamp and diode.
• An LDR's resistance falls with light; a thermistor's falls as temperature rises.

e.m.f. and p.d.

The electromotive force 电动势 (e.m.f.) $\varepsilon$ of a source is the energy 能量 given to each unit of charge by the source as it drives the charge around a full circuit. Unit: volt.

The potential difference 电势差 (p.d.) across a component is the energy changed from electrical to other forms by each unit of charge as it passes through that component.

Both are in volts; they differ in energy direction:

• e.m.f. — energy put into the circuit by the source (chemical → electrical in a battery, mechanical → electrical in a generator).
• p.d. — energy taken out of the electrical form (electrical → thermal in a resistor, → light in a lamp, → kinetic in a motor).

In the lab you often build a circuit on a breadboard 面包板 (a board with rows of holes that connect components without soldering) and measure currents and p.d.s with a multimeter 万用表.

Internal resistance

A real source has some internal resistance 内阻 $r$ — usually the resistance of the electrolyte 电解质 in a cell 电池, or the wire windings in a generator. When current $I$ flows, an internal p.d. of $Ir$ is "lost" inside the source, so the terminal p.d. 端电压 across the outside circuit is

So:

• no current (open circuit 开路, $I = 0$): the terminal p.d. equals the e.m.f.
• larger current: the terminal p.d. falls.
• short circuit 短路 ($R_{\text{external}} \to 0$): $I = \varepsilon / r$, a large current, with all the energy turned to heat inside the source.

To measure $r$, change the outside resistance and plot $V_{\text{terminal}}$ against $I$: the line has $y$-intercept $\varepsilon$ and gradient $-r$.

Worked example. A cell of e.m.f. $1.5\ \text{V}$ and internal resistance $0.50\ \Omega$ is connected to a $2.5\ \Omega$ resistor. Find the current and the terminal p.d.

The e.m.f. drives the current through both resistances: $I = \dfrac{\varepsilon}{R + r} = \dfrac{1.5}{2.5 + 0.50} = 0.50\ \text{A}$. Then