A-Level Physics · ⁨Physics A-Level⁩

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. Paper 1 usually opens with a question like this. Learn these rough values:

• mass 质量 of an adult human: $\sim 70\ \text{kg}$; of a car: $\sim 1000\ \text{kg}$; of an apple: $\sim 0.1\ \text{kg}$
• weight 重力 of an adult human: $\sim 700\ \text{N}$; of an apple: $\sim 1\ \text{N}$
• height of an adult human: $\sim 1.7\ \text{m}$; height of a room: $\sim 3\ \text{m}$
• walking speed: $\sim 1.5\ \text{m s}^{-1}$; a car on a fast road: $\sim 30\ \text{m s}^{-1}$
• 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: $g \approx 9.81\ \text{m s}^{-2}$
• density 密度 of water: $1000\ \text{kg m}^{-3}$; of air: $\sim 1.2\ \text{kg m}^{-3}$; of steel: $\sim 8000\ \text{kg m}^{-3}$
• atmospheric pressure 大气压强: $\sim 1.0 \times 10^{5}\ \text{Pa}$
• room temperature: $\sim 20\ ^{\circ}\text{C} \approx 293\ \text{K}$
• power of a kettle: $\sim 2\ \text{kW}$; of a person climbing stairs: $\sim 300\ \text{W}$
• wavelength of visible light: $\sim 5 \times 10^{-7}\ \text{m}$; diameter of an atom: $\sim 10^{-10}\ \text{m}$; of a nucleus: $\sim 10^{-15}\ \text{m}$

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.

To estimate a quantity that is not in the list, build it from ones that are. The kinetic energy of a moving car is $\tfrac{1}{2}mv^{2} \approx \tfrac{1}{2} \times 1000 \times 30^{2} \approx 5 \times 10^{5}\ \text{J}$. The pressure under a standing person is $700\ \text{N} / 0.02\ \text{m}^{2} \approx 4 \times 10^{4}\ \text{Pa}$. Always check that the power of ten looks sensible before you move on.

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.

Worked example. The drag force on a falling ball is given by $F = kv^{2}$, where $v$ is the speed. Find the SI base units of the constant $k$.

Rearrange: $k = F / v^{2}$. In base units, $F$ is $\text{kg m s}^{-2}$ and $v^{2}$ is $\text{m}^{2}\ \text{s}^{-2}$, so

A multiple-choice question often asks "which equation could be correct?". Check the base units of each option; only a homogeneous equation can be correct. A pure number (like 2, $\pi$ or $\tfrac{1}{2}$) has no unit, so it never changes the check.

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.

On a graph the two errors look different. A systematic error moves every point by the same amount, so the line of best fit keeps its gradient but no longer passes through the origin: the intercept 截距 changes. Random errors scatter the points above and below the line; a best-fit line drawn through the middle of the scatter still gives a reliable gradient.

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.

Estimating the uncertainty in a reading

Before you combine uncertainties, you need a sensible uncertainty for each raw reading:

• A single reading on an analogue scale: half the smallest division 分度 (a metre rule reads to $1\ \text{mm}$, so one reading is $\pm 0.5\ \text{mm}$). A length needs two readings, one at each end, so its uncertainty is $\pm 1\ \text{mm}$.
• A digital meter: $\pm 1$ in the last digit shown (a stopwatch showing $2.47\ \text{s}$ is $\pm 0.01\ \text{s}$).
• A hand-timed measurement: your reaction time 反应时间 matters more than the display. Allow about $\pm 0.1$ to $0.2\ \text{s}$, and time many oscillations instead of one, so the same uncertainty is shared between all of them.
• Repeated readings: the uncertainty is half the range 极差的一半. Three timings of $2.42$, $2.48$ and $2.45\ \text{s}$ have a mean of $2.45\ \text{s}$ and a range of $0.06\ \text{s}$, so $t = (2.45 \pm 0.03)\ \text{s}$.

Be honest. If the edge of a shadow is hard to see, the uncertainty in its position is several millimetres, however fine the scale. Examiners do not accept "the smallest division" as the uncertainty of a difficult reading.

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 $V = (76.2 \pm 0.9)\ \text{cm}^{3}$.

Do two values agree? Practical questions often ask whether two calculated values of a constant support a suggested relationship. Do not just say "they are close". Work out the percentage difference 百分比差异 between them,

and compare it with the percentage uncertainty in $k$ (or with the criterion the question gives, often 10%). If the difference is smaller, the two values agree within the uncertainty and the relationship is supported. If it is larger, they do not.

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.

Practical papers ask you to justify the number you chose. Name the raw readings: "$a$ is given to 2 significant figures because $L$ and $t$ were each measured to 2 significant figures." The phrase "because of the raw data" on its own does not earn the mark. In a table of results, judge each row from the least precise raw reading in that row; do not force every row to the same number of figures.

Uncertainties on a graph

Paper 5 asks you to carry uncertainties through a graph:

• Plot each point with an error bar 误差棒 whose length shows the absolute uncertainty in that value.
• Draw the line of best fit 最佳拟合直线 through the points, then the worst acceptable line 最差可接受直线: the steepest (or shallowest) straight line that still passes through every error bar.
• The uncertainty in the gradient is the difference between the two gradients, $\Delta m = |m_{\text{best}} - m_{\text{worst}}|$. The uncertainty in the intercept is found the same way.
• When you plot a logarithm, the absolute uncertainty in $\ln x$ is $\Delta x / x$ (and in $\lg x$ it is $0.434\,\Delta x / x$). Work it out for the largest and smallest values of $x$ separately.

A percentage uncertainty in the gradient then flows into any quantity you calculate from it, using the rules above.

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, 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.

Worked example. Two forces of $6.0\ \text{N}$ act on an object with $60°$ between them. Find the resultant.

Here $2\alpha = 60°$, so $\alpha = 30°$ and $R = 2 \times 6.0 \times \cos 30° = 10.4\ \text{N}$, along the line halfway between the two forces. Resolving gives the same answer: each force has a component $6.0\cos 30°$ along that line, and their components at right angles to it cancel.

For two vectors at any other angle, either make a scale drawing 按比例作图 of the vector triangle and measure the resultant, or resolve both vectors into perpendicular components, add the components, and recombine with $\sqrt{X^{2} + Y^{2}}$.

Worked example. A ball moving east at $5.0\ \text{m s}^{-1}$ is hit so that it moves north at $5.0\ \text{m s}^{-1}$. Find its change in velocity.

A change in velocity is a vector subtraction: $\Delta\vec{v} = \vec{v}_{\text{final}} - \vec{v}_{\text{initial}}$. Draw $5.0\ \text{m s}^{-1}$ north, then add the reverse of the initial velocity, $5.0\ \text{m s}^{-1}$ west. The two are perpendicular, so $|\Delta\vec{v}| = \sqrt{5.0^{2} + 5.0^{2}} = 7.1\ \text{m s}^{-1}$, pointing north-west. The speed did not change, but the velocity did. That is why a force must have acted on the ball (topic 3).

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.

A definition question is marked against fixed wording. Learn these exactly, and give one answer only.

• 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.

Common mistakes

• Substituting 5 cm as 5, or 20 g as 20. Convert the prefix first: 0.05 m, 0.020 kg.
• Giving the percentage uncertainty in $T^{2}$ as the same as in $T$. Squaring doubles it; a square root halves it.
• Treating the percentage uncertainty and the absolute uncertainty as the same number.
• Quoting "the smallest division" as the uncertainty of a reading that was hard to take.
• Offering two answers to a definition. Commit to one.

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 $\text{m s}^{-2}$.

Two more phrases the examiner uses: uniform acceleration 匀加速 means constant acceleration (a straight line on a velocity–time graph), and the acceleration of free fall 自由落体加速度 $g$ is the acceleration of an object falling freely in a uniform gravitational field 匀强重力场 when air resistance is negligible 可忽略的 (small enough to ignore).

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

Speeds are sometimes given in $\text{km h}^{-1}$. Convert before you calculate: $85\ \text{km h}^{-1} = 85 \times 1000 / 3600 = 23.6\ \text{m s}^{-1}$.

Many marks come from reading or drawing motion graphs, the graphical methods 图像法 the syllabus names. 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 (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 $\text{base} \times \text{height}$.

Worked example. The graph shows the velocity of a lift. Find the acceleration in each stage and the total distance travelled.

• Stage A: gradient $= \dfrac{3.0 - 0}{4.0} = 0.75\ \text{m s}^{-2}$.
• Stage B: the line is flat, so the acceleration is zero.
• Stage C: gradient $= \dfrac{0 - 3.0}{2.0} = -1.5\ \text{m s}^{-2}$ (a deceleration of $1.5\ \text{m s}^{-2}$).
• Distance: triangle A $= \tfrac{1}{2} \times 4.0 \times 3.0 = 6.0\ \text{m}$, rectangle B $= 6.0 \times 3.0 = 18\ \text{m}$, triangle C $= \tfrac{1}{2} \times 2.0 \times 3.0 = 3.0\ \text{m}$; total $27\ \text{m}$.

When the graph is a curve, the area is still the displacement: count the squares under the curve (half a square or more counts as one) and multiply by the value of one square.

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. Think of the velocity–time graph of the motion: a straight line from $u$ at $t = 0$ to $v$ at time $t$.

• $v = u + at$: acceleration is the gradient, $a = (v - u)/t$, so $v = u + at$.
• $s = \tfrac{1}{2}(u + v) t$: displacement is the area under the line, a trapezium 梯形 with parallel sides $u$ and $v$ and width $t$, so $s = \tfrac{1}{2}(u + v)t$.
• $s = ut + \tfrac{1}{2}at^{2}$: put $v = u + at$ into the area: $s = \tfrac{1}{2}(u + u + at)t = ut + \tfrac{1}{2}at^{2}$.
• $v^{2} = u^{2} + 2as$: from the first equation $t = (v - u)/a$; put this into $s = \tfrac{1}{2}(u + v)t$ to get $2as = (v + u)(v - u) = v^{2} - u^{2}$.

A "derive" question wants exactly these steps, each starting from a definition or an equation already established, with no numbers.

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).

"Show that" questions. When the question gives the answer ("show that the height is 3.2 km"), the marks are for the method. State the equation, substitute every value with its unit, and give the result to one more significant figure than the value quoted ($3.24\ \text{km}$, which rounds to $3.2\ \text{km}$). Reaching the quoted value proves nothing on its own.

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 $h = u^{2}/(2g)$:

When air resistance is not negligible

The equations above assume the only force is the weight. With air resistance, the resultant force on a falling object is smaller than its weight, so the acceleration is less than $g$; and because air resistance grows with speed, the acceleration keeps decreasing as the object speeds up. On a velocity–time graph the line curves, its gradient falling towards zero as the object approaches terminal velocity 收尾速度 (topic 3). For a projectile, air resistance shortens the range and lowers the maximum height, and the path is no longer a symmetrical parabola 抛物线: the object comes down more steeply than it went up. A "state and explain" question wants the force first, then its effect on the acceleration, then the effect on the motion.

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 $g$.

Comparing two objects. If object A is dropped and object B is thrown horizontally from the same height at the same moment, both reach the ground at the same time: their vertical motions are identical. If instead B is thrown with an upward component, it takes longer, because it must first rise and come back to the start height before it falls the same distance. Its speed on landing is still found from energy: the kinetic energy gained equals the potential energy lost, whatever the direction of the throw, so a ball launched at any angle with the same speed from the same height lands at the same speed (topic 5).

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 $2t_{\text{up}}$.

Sketching the velocity graphs. Examiners often ask you to sketch $v_{\text{H}}$ and $v_{\text{V}}$ against time on the same axes, taking upwards as positive. The horizontal component is a flat line at $u\cos\theta$ for the whole flight. The vertical component is a straight line of gradient $-g$: it starts at $+u\sin\theta$, crosses zero at the top of the flight, and reaches $-u\sin\theta$ on landing at the start height. Label each line, mark the times where the lines start and stop, and make the crossing sit exactly at $t_{\text{up}}$.

Worked example. A ball leaves the ground at $18\ \text{m s}^{-1}$ at $60°$ to the horizontal. Show that it reaches its maximum height at $t = 1.6\ \text{s}$.

$u_{\text{V}} = 18 \sin 60° = 15.6\ \text{m s}^{-1}$. At the top $v_{\text{V}} = 0$, so from $v = u + at$ with $a = -9.81\ \text{m s}^{-2}$: $t = 15.6 / 9.81 = 1.59\ \text{s} \approx 1.6\ \text{s}$.

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.

Read such a graph carefully: the gradient of every sloping line is the same $g$; the ball is at its highest point wherever the line crosses the time axis; and the height of each bounce comes from the area of the triangle above (or below) the axis for that flight.

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.

A definition question is marked against fixed wording. Learn these exactly, and give one answer only.

• 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.

Common mistakes

• Saying the heavier ball hits the ground first, or faster, when air resistance is ignored. Both fall with the same acceleration and land at the same speed.
• Taking the gradient of a $v$–$t$ graph when the question wants the area, or the area when it wants the gradient. Gradient is acceleration; area is displacement.
• Drawing a curve on a $v$–$t$ graph for a body under constant acceleration. Constant acceleration is a straight line; only the sign of $v$ changes at a bounce.
• Using $g = +9.81$ for a ball thrown upwards with "up" as positive. Once up is positive, $a = -9.81\ \text{m s}^{-2}$ for the whole flight, on the way down too.
• Using a SUVAT equation when the acceleration is changing (air resistance, a curved graph). Then only the graph methods work.

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

Momentum–time graphs

Because $F = \Delta p / \Delta t$, the gradient of a momentum–time graph is the resultant force, just as the gradient of a velocity–time graph is the acceleration. A straight line means a constant resultant force; a horizontal line means zero resultant force; a line sloping down means a force acting against the motion.

Worked example. The graph shows the momentum of a motorcycle. Find the resultant force on it during the first $10\ \text{s}$.

Gradient $= \Delta p / \Delta t = 5000 / 10 = 500\ \text{N}$. Between $10\ \text{s}$ and $16\ \text{s}$ the momentum is constant, so the resultant force is zero: the driving force just balances the resistive forces. From $16\ \text{s}$ to $20\ \text{s}$ the momentum falls by $2000\ \text{kg m s}^{-1}$, so the resultant force is $-500\ \text{N}$, acting backwards.

If a question tells you the speed is changing while the resultant force stays constant, that force cannot be air resistance: drag depends on speed, so it would change as the speed changes, and drag always acts against the motion.

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.

To name a third-law partner, swap the two bodies in the sentence. "The back wheel of a bicycle pushes backwards on the road" pairs with "the road pushes forwards on the back wheel", and that forward push is what drives the bicycle. "The Earth pulls the Moon" pairs with "the Moon pulls the Earth", with the same size of force.

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.

On the Moon $g$ is about $1.6\ \text{m s}^{-2}$, so an astronaut's mass is unchanged but their weight is about one sixth of its value on Earth. When a multiple-choice question asks which statement describes weight, the answer is the one that says "the force on the object due to a gravitational field", not "mass times acceleration" and not "the resultant force".

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. Drag also grows with the cross-sectional area 横截面积 of the object and with the density of the fluid; a parachute works by making the area, and so the drag, much larger. These frictional forces and drag forces are all resistive forces: they act against the direction of motion and transfer kinetic energy to thermal energy.

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" in a uniform gravitational field 匀强重力场. On an acceleration–time graph the acceleration starts at $g$ and decreases, curving down to zero.

Two spheres of the same size but different density have different terminal velocities: the denser sphere has the larger weight, so it needs a larger drag to balance it, and drag only reaches that value at a higher speed. That is also why a parachute changes everything: when it opens, the drag suddenly exceeds the weight, the resultant force is upwards, and the skydiver decelerates (the acceleration is upwards while the velocity is still downwards) until the drag has fallen back to equal the weight at a new, much lower terminal velocity.

Falling through a liquid: three forces

A ball falling through a liquid has three forces on it: its weight downwards, the upthrust 浮力 upwards, and the viscous drag upwards. The upthrust comes from hydrostatic pressure 流体静压强: the pressure in a liquid increases with depth, so the liquid pushes harder on the bottom of the ball than on the top, and the resultant of these pushes is upwards. Its size is the weight of the liquid the ball displaces, $U = \rho_{\text{liquid}} V g$ (topic 4). The drag grows with speed. When weight $=$ upthrust $+$ drag, the resultant force is zero and the ball falls at its terminal speed. A "draw labelled arrows" question wants exactly these three, with the upthrust and drag both up and the weight down.

Worked example. A steel ball of radius $2.0\ \text{mm}$ and weight $2.56 \times 10^{-3}\ \text{N}$ falls through oil of density $850\ \text{kg m}^{-3}$. The viscous drag on it is $F = 6\pi \eta r v$, where $\eta$ is the viscosity 黏度 of the oil and $v$ the speed. (a) Find the SI base units of $\eta$. (b) Show that the upthrust on the ball is $2.8 \times 10^{-4}\ \text{N}$. (c) Taking $\eta = 0.20$ in SI units, find the terminal speed.

(a) $\eta = F / (6\pi r v)$, so its units are $\text{N} / (\text{m} \cdot \text{m s}^{-1}) = \text{kg m s}^{-2} / (\text{m}^{2}\ \text{s}^{-1}) = \text{kg m}^{-1}\ \text{s}^{-1}$.

(b) $V = \tfrac{4}{3}\pi r^{3} = \tfrac{4}{3}\pi (2.0 \times 10^{-3})^{3} = 3.35 \times 10^{-8}\ \text{m}^{3}$, so $U = 850 \times 3.35 \times 10^{-8} \times 9.81 = 2.79 \times 10^{-4}\ \text{N} \approx 2.8 \times 10^{-4}\ \text{N}$.

(c) At the terminal speed, drag $= W - U = 2.56 \times 10^{-3} - 2.8 \times 10^{-4} = 2.28 \times 10^{-3}\ \text{N}$. So $v = F / (6\pi \eta r) = 2.28 \times 10^{-3} / (6\pi \times 0.20 \times 2.0 \times 10^{-3}) = 0.30\ \text{m s}^{-1}$.

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 驱动力 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.

On a slope the weight has a component along the road. Going up at constant speed, the driving force must balance the resistive force and the part of the weight that pulls down the slope, $W\sin\theta$; going down, that same component helps the motion.

Worked example. A car of mass $1500\ \text{kg}$ climbs a road inclined at $12°$ to the horizontal at a constant $30\ \text{m s}^{-1}$. The total resistive force is $1600\ \text{N}$. Find the driving force and the useful output power of the engine.

Constant speed means zero resultant force along the slope: $D = 1600 + 1500 \times 9.81 \times \sin 12° = 1600 + 3060 = 4660\ \text{N} \approx 4700\ \text{N}$. Power is force times velocity (topic 5): $P = Dv = 4660 \times 30 = 1.4 \times 10^{5}\ \text{W}$.

The principle

For a system with no resultant external force, the total momentum stays constant. This is conservation of momentum 动量守恒. Stated for the two marks the examiner gives: the total momentum of a system of objects remains constant provided no resultant external force acts on the system. Both halves are needed: "total momentum is constant" alone scores one.

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 $v_{1} = v_{2} = v$):

Questions often go on to ask what fraction of the kinetic energy is transferred to other forms. Before: $E_{\text{k}} = \tfrac{1}{2} \times 1500 \times 12^{2} = 1.08 \times 10^{5}\ \text{J}$. After: $\tfrac{1}{2} \times 2500 \times 7.2^{2} = 6.48 \times 10^{4}\ \text{J}$. So $4.3 \times 10^{4}\ \text{J}$, which is $40\%$ of the original kinetic energy, becomes thermal energy, sound and deformation of the cars. Momentum is conserved; kinetic energy is not.

Worked example. A stationary nucleus of mass $222\,u$ decays by emitting an alpha particle α粒子 of mass $4\,u$ at $1.6 \times 10^{7}\ \text{m s}^{-1}$. Find the speed of the remaining nucleus.

Before the decay the total momentum is zero, so afterwards the two momenta are equal and opposite: $218\,u \times v = 4\,u \times 1.6 \times 10^{7}$, giving $v = 2.9 \times 10^{5}\ \text{m s}^{-1}$ in the opposite direction to the alpha particle. The unit $u$ cancels, so its value is never needed. In a decay the momentum, the charge (proton number), the nucleon number and the total mass–energy are all conserved (topic 11).

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.

Worked example. On a frictionless surface, ball A has momentum $4.0\ \text{N s}$ due east and ball B has momentum $3.0\ \text{N s}$ due north. They collide and stick together. Find the momentum of the combined object.

Momentum is conserved along each axis: $4.0\ \text{N s}$ east and $3.0\ \text{N s}$ north after the collision, exactly as before. The total momentum is the vector sum, $\sqrt{4.0^{2} + 3.0^{2}} = 5.0\ \text{N s}$ at $\tan^{-1}(3.0 / 4.0) = 37°$ north of east; the combined object moves that way at $5.0\ \text{N s}$ divided by the total mass. A multiple-choice diagram of two momentum arrows is asking exactly this: add the arrows tip to tail.

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). An engine ejecting $90\ \text{kg}$ of gas per second at $190\ \text{m s}^{-1}$ produces a thrust of $90 \times 190 = 1.7 \times 10^{4}\ \text{N}$.

Worked example. A rocket of weight $W$ leaves the ground with an initial acceleration $a$. What thrust does its engine produce?

Take upwards as positive. The resultant force is thrust minus weight, so $T - W = ma$. With $m = W / g$, $T = W + Wa / g = W(1 + a / g)$. A rocket of weight $2.0 \times 10^{6}\ \text{N}$ accelerating at $4.0\ \text{m s}^{-2}$ needs a thrust of $2.0 \times 10^{6} \times (1 + 4.0 / 9.81) = 2.8 \times 10^{6}\ \text{N}$. The thrust must exceed the weight before the rocket moves at all.

A definition question is marked against fixed wording. Learn these exactly, and give one answer only.

• 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.

Common mistakes

• Writing "the forces are balanced" or "the forces cancel out". Say "the resultant force on the object is zero".
• Adding both speeds in a head-on collision because momentum is "total". Choose a positive direction, give the opposing velocity a minus sign, then add.
• Calculating kinetic energy from a component of the velocity. Kinetic energy uses the full speed of each body.
• Calling the weight and the normal contact force on one object a third-law pair. A pair acts on two different objects and is one type of force.
• Stating conservation of momentum without the condition. It holds only when no resultant external force acts on the system.

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).

Both are measured from the natural, unstretched length, never from the loaded length: on a diagram of a spring under a box, the distance the spring has shortened is its compression. A spring's own mass is usually said to be negligible 可忽略的, so the only forces on it are the load and the tension it provides.

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 ($4.0\ \text{cm} = 0.040\ \text{m}$):

For the one-mark "state Hooke's law", write it in words: the extension is directly proportional to the applied force (load), provided the limit of proportionality is not exceeded. The spring constant is the force per unit extension; the multiple-choice distractors are "extension per unit force" (that is $1/k$) and $\tfrac{1}{2}Fx$ (the energy stored).

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).

Worked example. A spring has length $90\ \text{mm}$ under a force of $0.80\ \text{N}$ and $115\ \text{mm}$ under $1.30\ \text{N}$. Find the spring constant and the unstretched length.

The extra force $0.50\ \text{N}$ produces the extra extension $25\ \text{mm} = 0.025\ \text{m}$, so $k = 0.50 / 0.025 = 20\ \text{N m}^{-1}$. At $0.80\ \text{N}$ the extension is $x = F/k = 0.80 / 20 = 0.040\ \text{m}$, so the unstretched length is $90 - 40 = 50\ \text{mm}$. On a graph of length against force the intercept on the length axis is $L_{0}$, the gradient is $1/k$, and the work done in stretching the spring is the triangle between the line and the level $L = L_{0}$, not the whole area down to the force axis: the rectangle below $L_{0}$ is the unstretched length multiplied by the force, which means nothing.

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 $k_{\text{total}} = k_{1} + k_{2}$.

Worked example. Springs of spring constant $6.0\ \text{N cm}^{-1}$ and $4.0\ \text{N cm}^{-1}$ are joined end to end and a load of $80\ \text{N}$ hangs from them. Find the total extension.

Each spring carries the full $80\ \text{N}$, so the extensions are $80/6.0 = 13.3\ \text{cm}$ and $80/4.0 = 20\ \text{cm}$: a total of $33\ \text{cm}$ (the same as $k = 2.4\ \text{N cm}^{-1}$ from $1/k = 1/6.0 + 1/4.0$). Three identical springs give the largest combined constant when all three are in parallel ($3k$) and the smallest when all are in series ($k/3$). When a load is shared by two identical springs in parallel, each carries half the force and so stores a quarter of the energy of one spring holding the whole load; the pair together stores half as much as the single spring did. A spring constant in $\text{N cm}^{-1}$ is $100$ times smaller than the same constant in $\text{N m}^{-1}$; give the unit the question asks for.

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).

For the one-mark definitions: stress is the force per unit cross-sectional area (the force acting normally to the area); strain is the extension per unit original length. The unit of stress, $\text{Pa} = \text{N m}^{-2}$, is $\text{kg m}^{-1}\ \text{s}^{-2}$ in base units. The area of a round wire comes from its diameter $d$: $A = \pi d^{2}/4 = \pi r^{2}$, and using $d$ in place of $r$ makes the area four times too big. Convert $\text{mm}^{2}$ to $\text{m}^{2}$ with $10^{-6}$.

Worked example. A tensile force of $18\ \text{N}$ acts on a wire of cross-sectional area $3.2\ \text{mm}^{2}$. Find the stress.

$\sigma = F/A = 18 / (3.2 \times 10^{-6}) = 5.6 \times 10^{6}\ \text{Pa}$. The same force acts along the whole wire, so a bolt whose diameter is $2d$ at one end and $d$ at the other has four times the stress at the narrow end ($\sigma \propto 1/d^{2}$), and a wire of three times the radius carries one ninth of the stress.

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: $k = EA/L_{0}$.

For the one-mark definition: the Young modulus is the ratio of stress to strain (for a material deformed within its limit of proportionality). Because it belongs to the material, a thicker wire of the same metal has the same Young modulus: it stretches less under the same load because its area is larger, not because the metal is stiffer. A "show that $k = EA/L_{0}$" question wants two lines: $k = F/x$ and $E = FL_{0}/(Ax)$, so $E = kL_{0}/A$. Two things follow: a wire's spring constant stays constant while Hooke's law holds, whatever the force, and a wire of the same metal with twice the diameter has four times the spring constant.

Worked example. A copper wire of length $1.7\ \text{m}$ and diameter $0.64\ \text{mm}$ has Young modulus $1.2 \times 10^{11}\ \text{Pa}$. Find its spring constant.

$A = \pi (0.32 \times 10^{-3})^{2} = 3.2 \times 10^{-7}\ \text{m}^{2}$, so $k = EA/L_{0} = 1.2 \times 10^{11} \times 3.2 \times 10^{-7} / 1.7 = 2.3 \times 10^{4}\ \text{N m}^{-1}$.

On a stress–strain graph the gradient of the straight part is $E$, so of two materials drawn on the same axes the steeper line is the stiffer material, and a wire with double the Young modulus is drawn as a line through the origin with twice the gradient. On a force–extension graph the gradient is $EA/L_{0}$ and the Young modulus is the gradient multiplied by $L_{0}/A$; only a stress–strain graph has a gradient equal to $E$ itself.

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.

A "describe an experiment" answer earns its marks for the quantities measured and how ($L_{0}$ with a metre rule; $d$ with a micrometer at several points; the load from the masses or a newton meter; the extension from a marker read against a fixed scale, or a vernier scale), the graph ($F$ against $x$, gradient $EA/L_{0}$, or stress against strain, gradient $E$) and the precautions: safety goggles in case the wire snaps, a small load first to straighten kinks, and readings taken on unloading as well as loading to check the wire stayed elastic.

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.

In words: elastic deformation means the object returns to its original length (or shape) when the load is removed; plastic deformation means it does not, and a permanent extension remains; the elastic limit is the maximum load (or extension) for which the deformation is still elastic; the limit of proportionality is the point beyond which the extension is no longer proportional to the load. On a graph the limit of proportionality is where the line stops being straight. The elastic limit lies a little beyond it and cannot be read from a loading line alone. So in a "which statement must be correct" item, the end of the straight part is the limit of proportionality, but whether a later point is the elastic limit or the breaking point cannot be told from the shape of the loading line.

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.

A rubber band 橡皮筋 also has different loading and unloading curves, but it returns to its original length: its deformation is elastic. The area between the two curves is again energy dissipated as thermal energy (elastic hysteresis 弹性滞后), and the energy recovered on unloading is the area under the lower curve. The two-mark "explain why the work done in stretching the wire is not equal to the energy recovered when the force is removed" answer says that the wire was taken past its elastic limit, so part of the deformation is plastic and a permanent extension remains, and that part of the work done became thermal energy in the wire.

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.

Why the area: work done is force multiplied by the distance moved, but here the force grows as the material stretches, so the work is the sum of $F\,\Delta x$ over many small extensions, which is the area under the line. Within the limit of proportionality the force rises uniformly from $0$ to $F$, so the average force is $\tfrac{1}{2}F$ and the work is $\tfrac{1}{2}Fx$. On a force–extension graph the gradient is the spring constant.

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.

Worked example. A wire of spring constant $2.0 \times 10^{4}\ \text{N m}^{-1}$ is already extended by $2.0\ \text{mm}$. Find the work done to increase its extension to $3.0\ \text{mm}$.

The stored energy rises from $\tfrac{1}{2} k x_{1}^{2}$ to $\tfrac{1}{2} k x_{2}^{2}$: $W = \tfrac{1}{2} \times 2.0 \times 10^{4} \times \left[(3.0 \times 10^{-3})^{2} - (2.0 \times 10^{-3})^{2}\right] = 0.050\ \text{J}$. This is the trapezium under the line between the two extensions, not $\tfrac{1}{2} k (x_{2} - x_{1})^{2}$. Because $E_{\text{P}} \propto x^{2}$ at fixed $k$, doubling the extension of a wire stores four times the energy ($0.65\ \text{J}$ becomes $2.6\ \text{J}$), and a stored energy gives the extension as $x = \sqrt{2E_{\text{P}}/k}$: a spring of $k = 400\ \text{N m}^{-1}$ storing $0.32\ \text{J}$ is compressed by $\sqrt{2 \times 0.32 / 400} = 0.040\ \text{m}$. For two wires joined end to end the tension is the same in both, so the total energy stored is $\tfrac{1}{2}F x_{1} + \tfrac{1}{2}F x_{2}$ with each wire's own extension.

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. To estimate the work done up to the breaking point, count the squares under the whole curve (part squares as halves) and multiply by the energy one square represents, the force step multiplied by the extension step; saying that the area was found by counting squares is the "explain your reasoning" mark.

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.

A sketch of $E_{\text{P}}$ against extension or compression is a curve through the origin that gets steeper ($E_{\text{P}} \propto x^{2}$), not a straight line. Of two wires of the same length and area under the same load, the one with the smaller Young modulus extends more and so stores more energy.

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.

A definition question is marked against fixed wording. Learn these exactly, and give one answer only.

• 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.

Common mistakes

• Using the diameter as the radius in $A = \pi r^{2}$, or leaving an area in $\text{mm}^{2}$. Halve the diameter first; $1\ \text{mm}^{2} = 10^{-6}\ \text{m}^{2}$.
• Reading the work done off a length–force graph as the whole area down to the axis. Only the triangle above $L_{0}$ is work.
• Treating stored energy as proportional to extension. It goes as $x^{2}$: double the extension, four times the energy; and the work done between two extensions is the difference of two $\tfrac{1}{2}kx^{2}$ values.
• Stating Hooke's law without its condition. "Provided the limit of proportionality is not exceeded" is part of the law.
• Swapping the limit of proportionality (the end of the straight line) and the elastic limit (the end of elastic behaviour, a little beyond it).
• Saying a thicker wire has a larger Young modulus. The modulus is the material's; the thicker wire has a larger spring constant.