Pearson Edexcel · International A-Level · Physics

From a collision to a particle accelerator

A car bumper and a particle detector seem very different. Both use momentum and forces. This unit follows those ideas into circular motion, electric fields, magnetic fields and particle interactions.

WPH14 assesses requirements 81–124 of the Issue 3 specification. It can also use knowledge from Units 1–2. Use the seven skill sheets in order. Check units, directions and the assumptions behind each equation. A familiar equation can give a wrong answer when its model does not fit.

Impulse and two-dimensional collisions

Force changes momentum

Quantité de mouvement 动量 is the vector $\mathbf p=m\mathbf v$. Choose a positive direction before using signs. Impulse 冲量 is the change in momentum. For constant force, or the average force over an interval:

A changing force gives an impulse equal to the signed area under its force–time graph. A triangular pulse has area $\frac12\times\text{base}\times\text{height}$. An area below the time axis is negative. The maximum force is not automatically the average force.

Worked example. A $0.20\ \text{kg}$ ball approaches a wall at $+6.0\ \text{m s}^{-1}$. It rebounds at $-4.0\ \text{m s}^{-1}$. Contact lasts $0.050\ \text{s}$. Find average force on the ball.

• Known: mass, signed initial/final velocities and contact time.
• Why: force impulse changes the ball's momentum; rebound reverses the sign.

The force is opposite the chosen positive direction. The wall feels an opposite force. A deforming bumper or helmet increases stopping time. For the same momentum change, this reduces average force. It does not remove the momentum change.

Conserve each momentum component

An isolated system 孤立系统 has no significant external impulse during the event. Its total momentum is conserved. Internal forces cancel in opposite pairs. Momentum conservation does not require kinetic energy conservation.

For a collision in a plane, resolve every momentum into two perpendicular axes:

Use the angle measured from the chosen axis. Include signs; a downward component is negative if upwards is positive. Do not add momentum magnitudes when directions differ.

Worked example. A $0.10\ \text{kg}$ sphere moving at $2.0\ \text{m s}^{-1}$ strikes two stationary spheres, each $0.10\ \text{kg}$. It stops. The two others leave symmetrically at $1.2\ \text{m s}^{-1}$, at angles $+\theta$ et $-\theta$ to the original direction. Find their separation angle and classify the collision.

• Why: transverse momenta cancel by symmetry; longitudinal momentum gives $\theta$.

An elastic collision 弹性碰撞 conserves total kinetic energy as well as momentum. An inelastic collision 非弹性碰撞 does not conserve kinetic energy; some becomes other forms of energy.

This collision is inelastic. Total energy is still conserved. In an ordinary passive collision, kinetic energy cannot increase without an additional energy source.

For a non-relativistic particle 非相对论粒子, use $p=mv$ et $E_k=\frac12mv^2$. Substituting $v=p/m$ gives:

At equal kinetic energy, $p=\sqrt{2mE_k}$: a larger mass has larger momentum. At equal momentum, a larger mass has smaller kinetic energy. These equations are not valid for a particle moving close to light speed.

Core practicals 9 and 10

Pour core practical 9, use a trolley, force sensor and data logger. Measure trolley mass and velocities just before/after contact using light gates or a motion sensor. Record the force–time pulse at a sampling rate fast enough to resolve it. Zero the force sensor; use consistent axes for force and velocity. Integrate the graph area and compare it with $m(v-u)$. Repeat for different contact forces or speeds. A graph of impulse against momentum change should have gradient close to one. Friction, sensor offsets and missed parts of the pulse can cause disagreement. Secure the track and catch the trolley safely.

Pour core practical 10, record small spheres colliding on a level surface using an overhead camera. Include a length scale in the collision plane and a known frame rate. Keep the camera perpendicular to reduce perspective error. Track centres over several frames before and after the short collision. Convert pixel displacements to metres and frame intervals to seconds. Determine both velocity components, then compare total momentum components and kinetic energies. Repeat with different approach directions. Include uncertainty from positions, timing and scale calibration. A small mismatch within uncertainty is not evidence that momentum fails. Keep spheres contained so they cannot fall or become a slipping hazard.

Mouvement circulaire

Angle, angular speed and tangential speed

An angle of one radian 弧度 subtends an arc equal to the radius. Hence $\theta=s/r$, with angle in radians. One revolution is $2\pi$ radians or $360^\circ$.

Angular velocity 角速度 describes the rate of angular displacement. In the plane problems here, use its signed rotational rate or magnitude as appropriate. Points on one rigid turntable have the same angular speed. Points farther from its centre have larger tangential speed 切向速率. Their velocity directions also vary around the circle.

Worked example. A wheel rotates at $120$ revolutions per minute. Find angular speed and acceleration of a point $0.25\ \text{m}$ from its centre.

Keep the unrounded angular speed during the calculation. Use radius, not diameter.

Why acceleration points inward

Uniform circular motion has constant speed but changing velocity. Draw the initial and final tangential velocity vectors. The change is $\Delta\mathbf v=\mathbf v_2-\mathbf v_1$, not their sum. For a small angular change, the velocity triangle and radius triangle are similar:

The change in velocity points towards the centre in this limit. A resultant centripetal force 向心力 is therefore needed:

Centripetal force names the resultant of real forces. It is not an extra force to add to a force diagram. Friction can provide it on a turntable. Tension can provide it for a ball on a string.

Write the real-force equation

For clothing against the inside of a vertical drum, weight always points down. The drum's normal reaction points towards the centre. At constant speed:

The required resultant has the same magnitude, but the reaction is larger at the bottom. If contact is lost, the surface cannot provide a pulling normal reaction. Check this before using a circular-path model.

For a horizontal turntable, the largest available friction must be at least $mr\omega^2$. If maximum friction is $mg/25$:

Mass cancels because both required and available forces scale with mass. If the inward force disappears, motion initially follows the tangent. Gravity can then curve the later path; “tangent” describes the release direction.

Electric fields and potential

Force and potential describe different things

An electric field 电场 is a region where a charged particle experiences a force. Electric field strength 电场强度 is force per unit positive test charge:

Its units are $\text{N C}^{-1}$ ou $\text{V m}^{-1}$. A positive charge feels force along the field; a negative charge feels force opposite it.

Potentiel électrique 电势 is potential energy per unit charge relative to a chosen reference. For a charge $q$ moved through a potential change:

Potential is a scalar; field is a vector. A point with zero potential need not have zero field. Adding signed potentials is different from adding field vectors.

Radial and uniform fields

For point charges, or outside an isolated charged conducting sphere, with $k=1/(4\pi\varepsilon_0)$:

Here $r$ is separation of charge centres; sphere distances are measured from its centre. Like charges repel, unlike charges attract. Potential takes the sign of $Q$ when zero is at infinity.

Between large parallel plates, away from edges, the field is approximately uniform:

The field points from higher to lower potential. Plate separation $d$ is perpendicular to the plates. An equipotential 等势面 has the same potential everywhere. Field lines cross equipotentials at right angles. Moving along one involves no change in electric potential energy.

The radial field is the negative potential gradient: $E_r=-\mathrm dV/\mathrm dr$. For a positive source, the signed area under its $E$–$r$ curve from $r$ to infinity equals $V(r)$. The area from zero to $r$ is not that potential. In a uniform field, closer equally spaced potential levels mean stronger field.

Worked example. A sphere of radius $0.20\ \text{m}$ has potential $9.0\ \text{kV}$. Find charge and field at $0.30\ \text{m}$ from its centre. Use $k=8.99\times10^9\ \text{N m}^2\text{C}^{-2}$.

Use the surface radius to find charge, then the new centre-distance to find field. Do not reuse the surface distance for the second step.

Compare electrical and gravitational forces

A stationary charged oil drop can satisfy $|q|E=mg$. Use $m=\rho\mathcal V$ if density and volume are given. Divide the charge magnitude by $e=1.60\times10^{-19}\ \text{C}$ to test whether it is close to an integer multiple. Charge sign follows the direction of force required, not its magnitude alone.

To lift a spherical grain, compare $qE$ with $\rho(4\pi r^3/3)g$. Radius is half the listed diameter. A calculated maximum diameter is a threshold: choose the largest listed diameter below it. On an inclined panel, the normal component of weight is $mg\cos\theta$. The electric force perpendicular to the panel need overcome that component to detach a grain; gravity also has a downslope component.

Capacitors and RC circuits

Charge storage and energy

A un condensateur 电容器 stores separated charge on two conductors. They carry equal and opposite charges in the ideal two-plate model. The quoted charge $Q$ is the magnitude on either plate, not the sum of both magnitudes. Capacité 电容 is charge stored per unit p.d.:

One farad is one coulomb per volt. For constant capacitance, a $V$–$Q$ graph is straight through the origin with gradient $1/C$. Its area gives the work needed to transfer charge onto the capacitor:

The factor one half appears because p.d. rises from zero while charge is transferred. It is not $QV$ at the final voltage throughout charging. For discharge between non-zero voltages, subtract the two stored energies:

This is different from $\frac12C(V_{\text{initial}}-V_{\text{final}})^2$.

Charging and discharging

In an ideal series resistor–capacitor charging circuit, initially uncharged, the capacitor p.d. is zero. The resistor initially has the full supply p.d., so current is maximum. As charge builds, capacitor p.d. rises. Resistor p.d. and current fall. At full charge, current is zero and capacitor p.d. equals supply p.d.

For discharge through a fixed resistance, use signed current consistently or work with its magnitude:

The voltage relation follows from $Q=CV$ with constant $C$. Current magnitude follows from $|I|=V/R$ with constant $R$. The time constant 时间常数 is $\tau=RC$. After one time constant, $Q$, $V$ and current magnitude are $e^{-1}\approx0.368$ of their initial values. After three time constants they are about $5\%$, not zero. Stored energy falls faster because it depends on $V^2$.

For charging from an ideal constant supply $V_s$:

A charging capacitor voltage rises towards $V_s$; charging current falls towards zero. Do not sketch the same curve for both quantities. Equal $V_C$ et $V_R$ means each is $V_s/2$, reached at $t=RC\ln2$.

Taking natural logs of the discharge equation gives:

Thus a graph of $\ln V$ contre $t$ has gradient $-1/(RC)$ and intercept $\ln V_0$. The same forms hold for $Q$ and current magnitude. Use consistent measurement units inside the logged numerical values.

Worked example. A $100\ \mu\text{F}$ capacitor discharges from $12.0\ \text{V}$ through $20.0\ \text{k}\Omega$. Find p.d. after $3.0\ \text{s}$ and energy lost.

The lost stored energy becomes mainly heat in the discharge resistance. Use unrounded voltage in the final calculation.

Core practical 11 and useful extensions

Use a low-voltage d.c. supply, resistor, capacitor and switch. Charge through the resistor, then disconnect the supply and discharge through the known resistance. Connect a voltage sensor, data logger or oscilloscope across the capacitor. Record time and p.d.; choose a sampling interval short compared with $RC$. Use an instrument input resistance large compared with the discharge resistance, otherwise it changes the discharge path.

Conserver $R$ et $C$ fixed, repeat after restoring the same initial charge, and compare measured curves with exponential predictions. Determine $RC$ from the $1/e$ level or a log-fit gradient. Avoid taking logs of readings near zero where relative uncertainty is large. For a resistance-tolerance test, calculate $R=-\Delta t/[C\ln(V_2/V_1)]$ during one uninterrupted discharge and compare with both limits of the permitted interval.

Use correct polarity for an electrolytic capacitor and remain below its voltage rating. Discharge it safely through a resistor before altering connections; do not short a charged capacitor. Leakage, sensor loading and resistor heating can change the measured curve.

A smoothing capacitor charges near supply peaks and discharges into the load between them. A smaller load resistance gives a smaller time constant, faster discharge and less smoothing. For two identical series capacitors, the same magnitude of charge appears on each. Their equal p.d.s share the supply, so each stores half the charge of one identical capacitor connected alone. This sharing is a useful bridge to the selected paper; it does not replace the main capacitance model.

Magnetic forces and induction

Directions, flux and force

Magnetic flux density 磁感应强度 $B$ describes the field's force effect. For a straight wire of length $l$ in a uniform field:

Here $\theta$ is between conventional current and field. Only the length inside the field counts. For one charged particle:

Use Fleming's left-hand rule 弗莱明左手定则: first finger along field, second along conventional current, thumb gives force. A positive charge moves with conventional current. Reverse the result for an electron's motion. Dot symbols mean out of the page; crosses mean into it.

Magnetic force is perpendicular to velocity, so it changes direction without doing work. It cannot, alone, increase the particle's kinetic energy. If $\mathbf v$ is parallel to $\mathbf B$, magnetic force is zero.

Magnetic flux 磁通量 through a flat area in a uniform field is:

Here $\alpha$ is between field and the area's normal, not the plane itself. The unit is the weber. Flux linkage 磁通链 is $N\phi$ when the same flux links each of $N$ turns.

Worked example. A coil has 40 turns and radius $0.015\ \text{m}$. A uniform $0.020\ \text{T}$ field is perpendicular to its plane. Find flux linkage.

Use radius, not diameter, and multiply by turns only when finding linkage. A balance under magnets measures the opposite reaction to a force on a separately supported wire. Convert a balance reading change $\Delta m$ to force $\Delta mg$, not $\Delta m$ itself.

A changing linkage induces an e.m.f.

Induction électromagnétique 电磁感应 occurs when flux linkage changes. Relative movement of a coil and magnet can cause it. Rotating the coil or changing current in another linked coil can also cause it. A stationary coil in a constant field has no induced e.m.f. merely because flux is present.

Faraday's law 法拉第电磁感应定律 relates e.m.f. magnitude to rate of flux-linkage change. Lenz's law 楞次定律 gives the opposing direction:

Induced current, if the circuit is closed, produces a magnetic effect opposing the change that caused it. It does not always oppose the existing field: when that field decreases, the induced effect helps maintain it. This opposition is consistent with energy conservation.

Worked example. Flux through each of 50 turns decreases from $6.0\times10^{-5}$ à $2.0\times10^{-5}\ \text{Wb}$ in $0.020\ \text{s}$. Find average induced e.m.f. magnitude.

Faster movement, stronger field, more turns or larger linked area can increase the rate. Explain which quantity actually changes. In a magnetic-strip reader, reversing pole orientation reverses signal direction. For a straight boundary swept across an area of width $L$ at speed $v$, $\Delta A/\Delta t=Lv$, giving $|\mathcal E|=NBLv$ under that uniform-field model.

Wireless charging uses alternating current in one coil, producing changing magnetic field and changing linkage in the phone coil. That induces an e.m.f. The receiving circuit can then charge the battery. Equal turn counts do not ensure equal voltages: some flux may fail to link the other coil, especially with separation or without an iron core.

Nuclear structure, accelerators and tracks

The nucleus and scattering evidence

In $^{A}_{Z}X$, proton number 质子数 $Z$ counts protons and nucleon number 核子数 $A$ counts protons plus neutrons. The neutron count is $A-Z$. A neutral atom has $Z$ electrons; an ion need not.

The older distributed-positive-charge model could explain small deflections but not the observed rare large-angle alpha scattering. Most alpha particles pass nearly straight through thin foil, showing that most of the atom is empty space. A small fraction experience very large deflections. Positive alpha particles feel strong electrostatic repulsion near a small, dense, positive nucleus containing most atomic mass. Few large-angle events show that the nucleus occupies a small fraction of the atom's volume. Keep each observation linked to the particular conclusion it supports.

Make, accelerate and steer a beam

Thermionic emission 热电子发射 releases electrons from a heated metal. Heating supplies energy so some electrons can leave the surface. A positive accelerating electrode attracts them through a vacuum. A potential difference changes kinetic energy by the electric work:

For speed calculations use the positive energy gain $|q|\,|\Delta V|$ only when the particle is accelerated through the field in the appropriate direction. Starting from rest, well below light speed:

A linear accelerator 直线加速器 uses alternating p.d. across gaps between drift tubes. Particles accelerate in gaps and are shielded inside the conducting tubes. The polarity reverses while particles are inside, so the next gap accelerates them again. At constant alternating frequency, increasing speed requires longer tubes to maintain the time between successive gaps. Across $n$ equal accelerating gaps, total kinetic-energy gain is $n|q|V_{\text{gap}}$.

A cyclotron 回旋加速器 uses a magnetic field perpendicular to two hollow dees. Electric field accelerates particles across the gap. Inside a dee, magnetic force curves the path without increasing speed. Polarity reverses during each half-turn, allowing another energy gain at the next crossing. With perpendicular entry:

In the non-relativistic model, the period is independent of speed. Increasing speed increases radius. Time in one dee is $T/2$, not $T$. The formula assumes constant field and fixed mass in this model; synchronisation fails as relativistic effects become important.

Worked example. A proton moves perpendicular to $B=0.40\ \text{T}$ with momentum $2.0\times10^{-20}\ \text{kg m s}^{-1}$. Use proton mass $1.67\times10^{-27}\ \text{kg}$.

When starting instead from a de Broglie wavelength, use $p=h/\lambda$, then $E_k=p^2/(2m)$ et $V=E_k/|q|$. Check that the non-relativistic assumption remains suitable.

Read detector evidence carefully

Charged particles ionise matter along their paths, leaving detectable tracks. A neutral particle does not make a direct ionisation track in this model. Its presence can be inferred from missing momentum or from charged products it later produces.

In a known perpendicular magnetic field, curvature direction together with travel direction can identify charge sign. Without field direction, curvature alone cannot give sign. Radius gives momentum through $p=B|q|r$; comparing radii requires knowledge of charge magnitude. A tightening track often shows momentum being lost. Do not infer travel direction from a constant-radius arc alone.

Conserve charge, total energy and vector momentum at an interaction. If visible outgoing momenta do not add to the incoming momentum, a neutral product may carry the missing component. Draw a vector balance rather than guessing from the number of tracks.

High energies probe nucleon structure in two ways. Large momentum gives a short de Broglie wavelength, allowing small structure to be resolved. High collision energy also allows new massive particles to be created. Quarks are not observed as isolated free particles; do not describe a high-energy collision simply as removing a free quark.

Fast muons can reach the ground because their average lifetime measured in the Earth's frame is increased at speeds close to $c$. This does not mean they exceed light speed. Unit 4 requires the significance of this lifetime increase, not use of relativistic equations.

Particles, antiparticles and conservation

Families and antiparticles

In the quark–lepton model 夸克—轻子模型, baryons and mesons contain quarks. Leptons are fundamental particles in the model.

• A baryon 重子 contains three quarks, for example a proton $uud$ or neutron $udd$. An antibaryon contains three antiquarks.
• A meson 介子 contains one quark and one antiquark, for example a pion.
• A lepton 轻子 is fundamental, for example an electron or neutrino. A positron is an antilepton, not a meson.
• A photon is the quantum of electromagnetic radiation; it is neither a baryon nor a lepton.

Up-type quarks have charge $+2e/3$ and down-type quarks $-e/3$. Antiquarks have opposite charges. For example, $uud$ sums to $+e$ et $udd$ to zero. The symmetry of the quark families predicted a top quark before it was observed. Classification models make testable predictions, not just lists of known particles.

An antiparticle 反粒子 has the same rest mass as its particle and opposite electric charge when charged. Other relevant additive quantum numbers also reverse. An electron has lepton number $+1$; its positron has $-1$. Both have baryon number zero. A neutral antiparticle can differ in quantum numbers despite having no electric charge.

Check conserved totals

Assign baryon number $+1$ to a baryon, $-1$ to an antibaryon, and zero to mesons/leptons/photons. Quarks carry $+1/3$ and antiquarks $-1/3$. Assign lepton number $+1$ to a lepton and $-1$ to an antilepton. Compare total charge, baryon number and lepton number on both sides of a proposed reaction. Also check energy and momentum. Passing the listed conservation checks alone does not prove that an interaction will occur; it establishes that those laws do not forbid it.

Worked example. Test $n\rightarrow p+e^-+\bar\nu_e$.

The antineutrino balances the electron's lepton number. Changing it to a neutrino would make the final total $+2$, so that proposed equation would fail this check. Preserve bars and charge signs when interpreting supplied particle symbols.

Rest energy, creation and annihilation

Mass–energy equivalence 质能等价 relates a rest-mass change to energy:

In pair production 粒子对产生, energy creates a particle–antiparticle pair. The energy must at least supply both rest masses; extra energy can become kinetic energy or recoil. Momentum must also be conserved, so the surrounding interaction matters. In annihilation 湮灭, particle and antiparticle can turn into photons. A pair initially at rest producing two photons gives equal photon energies and opposite momenta. Each photon carries one particle's rest energy for an equal-mass pair; the total is twice that value.

One electronvolt is $e$ joules: $1\ \text{eV}=1.60\times10^{-19}\ \text{J}$. MeV means $10^6$ eV and GeV means $10^9$ eV. A quoted mass in $\text{MeV}/c^2$ is a mass unit, not an energy unit. Multiply by $c^2$ to get its corresponding rest energy.

Worked example. Convert a mass of $140\ \text{MeV}/c^2$ to kilograms.

Use the same units before comparing masses. Percentage difference from an accepted value is $100\times|m_{\text{predicted}}-m_{\text{accepted}}|/m_{\text{accepted}}$. State the reference value. In threshold questions, distinguish rest-energy supply from kinetic energy; do not use $p=mv$ for a photon or a particle close to $c$.

Vérifiez-vous

• Can you use signed impulse and two momentum components, then test kinetic energy separately?
• Can you derive centripetal acceleration and draw only actual forces?
• Can you distinguish field vectors, signed potential and energy changes?
• Can you choose the correct capacitor curve and use log slope or tolerance evidence?
• Can you explain induction as a change in linkage, including the opposing direction?
• Can you separate electric acceleration from magnetic steering and halve a cyclotron period correctly?
• Can you preserve particle symbols, conservation totals and energy/mass units?

The skill sheets develop these methods before authentic past-paper tasks. Original practice also covers requirements not sampled in the two recent papers. A short sample of authentic questions does not define the whole syllabus.

From an ice pack to a star

An ice pack, a swinging bridge and a distant star all involve energy. This unit links molecular motion, nuclear changes, oscillations and astronomical observations. The task is often to test a claim, not just calculate a number: state the model, carry units through the calculation, then compare your result with the claim.

WPH15 covers statements 125–171 of Pearson Issue 3. It also uses Units 1, 2 and 4. Work through sheets 5.1–5.9 before their matching authentic sets. Use the supplied constants consistently: $g=9.81\ \mathrm{m\,s^{-2}}$, $k_B=1.38\times10^{-23}\ \mathrm{J\,K^{-1}}$ et $G=6.67\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$ unless the question gives a different local value. Here $k_B$ is the Boltzmann constant; $k_s$ denotes a spring constant.

Heating, phase change and thermistor calibration

Follow the energy, not only the temperature

Énergie interne 内能 is the total energy in the random motion and interactions of molecules: random kinetic energy plus intermolecular potential energy. It excludes the kinetic energy of the whole object moving together. Temperature relates to average random molecular kinetic energy, not total internal energy. Two samples at the same temperature can have different internal energies because their amounts and states differ.

Specific heat capacity 比热容 $c$ is the energy needed per kilogram for a one-kelvin temperature rise without a change of state. Chaleur latente spécifique 比潜热 $L$ is the energy transferred per kilogram during a particular change of state at constant temperature.

A temperature difference of $1\ ^\circ\mathrm C$ equals a difference of $1\ \mathrm K$. Absolute temperatures in gas and radiation laws must be kelvin. During melting or boiling, energy changes molecular arrangements and potential energy while temperature remains constant in the ideal constant-pressure phase change. An ice pack absorbs energy from the food/surroundings as it melts. It does not send “cold energy” into the food.

At constant heating power, steeper temperature–time gradients mean smaller $mc$, provided heat loss is negligible. A flat section does not mean that energy transfer has stopped. For a heater, input energy is $E_{\rm in}=Pt=VIt$; useful thermal energy can be smaller because energy escapes to surroundings.

Worked example. A $0.40\ \mathrm{kg}$ block of ice at $-10\ ^\circ\mathrm C$ melts to water at $0\ ^\circ\mathrm C$. Use $c_{\rm ice}=2100\ \mathrm{J\,kg^{-1}\,K^{-1}}$ et $L_f=3.4\times10^5\ \mathrm{J\,kg^{-1}}$.

Known: mass, initial/final temperatures and material constants. Why: warming the solid and melting it are separate energy transfers.

Worked example. A $1000\ \mathrm W$ heater boils away $0.20\ \mathrm{kg}$ of water in $600\ \mathrm s$. Find efficiency, using $L_v=2.26\times10^6\ \mathrm{J\,kg^{-1}}$.

For repeated kettle loads, calculate total water volume, number of loads and total operating time. Compare energies over the same stated interval. Standby loss over a day and input needed to boil water are different quantities; state what a particular comparison actually establishes.

CP12: calibrate a thermostat

A thermistor 热敏电阻 has resistance that changes with temperature. The usual negative-temperature-coefficient (NTC) thermistor decreases in resistance when heated. In a potential divider 分压器:

With the thermistor as the lower resistor and output measured across it, warming reduces output voltage. Swapping resistor positions reverses that trend. A thermostat switches at a chosen output threshold; the divider alone is not a complete switching device.

For core practical 12, place the thermistor and a reference thermometer close together in a stirred water bath. Keep electrical contacts dry and insulated. At several temperatures across the intended range, wait for thermal equilibrium and record temperature and divider voltage. Keep supply voltage and fixed resistance unchanged. Use a high-input-resistance voltmeter to reduce loading. Plot output against temperature and interpolate the voltage for the chosen switching temperature. Repeat readings or compare warming/cooling to reveal lag. Do not assume a linear calibration without evidence. Use low-voltage electricity and take care with hot water.

CP13: measure latent heat

Supply measured electrical power to a material during its phase change. Measure mass melted or evaporated over a timed interval after conditions stabilise. A balance measures mass change; do not confuse collected liquid volume with mass. Without loss correction, $L=VIt/\Delta m$ assumes all supplied energy causes the phase change.

A useful correction is to compare two powers at the same steady phase-change temperature. If heat-loss power is approximately unchanged, subtracting the two measurements removes it:

State that constant-loss assumption. A control measurement of melting without the heater is another possible correction in a suitable ice experiment. Reduce heat loss with insulation where safe; avoid splashing, hot surfaces and unsafe mains connections.

Ideal gases and molecular kinetic energy

Kelvin and the ideal-gas model

Absolute zero 绝对零度 is $0\ \mathrm K$, environ $-273\ ^\circ\mathrm C$. It is the lower limit of thermodynamic temperature. In the classical ideal-gas model, average translational kinetic energy tends to zero as temperature tends to zero. Real gases condense before that extrapolation is reached; do not claim molecules in every real material lose all possible energy.

Absolute temperature 绝对温度 is measured in kelvin: $T=\theta_{^\circ\mathrm C}+273$ to the precision normally used here. An ideal gas 理想气体 consists of particles with negligible volume, no intermolecular forces except during elastic collisions, and random motion. Pressure comes from momentum changes at walls.

Here $N$ is the number of molecules, not the number of moles. Use pressure in pascals, volume in cubic metres and temperature in kelvin. For a fixed number of molecules:

At fixed volume, pressure is proportional to kelvin temperature. At fixed temperature, $pV$ is constant. Include external pressure if a gauge reports pressure relative to the atmosphere.

Worked example. A rigid $2.0\times10^{-3}\ \mathrm{m^3}$ vessel contains gas at $1.0\times10^5\ \mathrm{Pa}$ et $300\ \mathrm K$. Find $N$ and pressure after warming to $360\ \mathrm K$.

If one molecule has mass $m_0$, total gas mass is $Nm_0$. For a sphere, first find $V=4\pi r^3/3$; radius is half the diameter.

Derive the molecular energy relation

The kinetic-theory pressure relation is $pV=\tfrac13Nm_0\langle c^2\rangle$, where $\langle c^2\rangle$ is mean square speed 速率平方的平均值. It follows from elastic momentum changes at a wall and equal mean squared motion in the three perpendicular directions. The root mean square speed 方均根速率 is $c_{\rm rms}=\sqrt{\langle c^2\rangle}$; it is not the square of mean speed.

For a cube of side $l$, one molecule with velocity component $c_x$ changes wall momentum by $2m_0c_x$ at each elastic collision. Its return time to that same wall is $2l/c_x$. Hence its average force is $m_0c_x^2/l$. Sum over molecules and divide by wall area $l^2$:

Random motion is equally distributed among perpendicular directions, so $\langle c^2\rangle=3\langle c_x^2\rangle$. This gives the pressure relation above. Equate the two expressions for $pV$, cancel $N$, then rearrange:

At one temperature, different gas species have equal mean translational kinetic energies, but lighter molecules have greater RMS speeds. If speed doubles at the same temperature, molecular mass is one quarter. Heating increases average molecular kinetic energy; it does not give every molecule the same speed.

CP14: pressure and volume at fixed temperature

Trap a fixed amount of gas in a syringe or calibrated cylinder connected to a pressure sensor. Check calibration and use absolute pressure. Include connecting-tube dead volume where significant. Change volume gradually, wait after each change for temperature to return to the surroundings, then record several pressure–volume pairs. Do not let gas escape. A plot of $p$ contre $1/V$ should be straight through the origin within uncertainty. A $p$–$V$ curve is not straight. Avoid excessive pressure and secure connections. Rapid compression heats the gas and breaks the intended constant-temperature condition.

Binding energy, fission and fusion

Mass deficit is an energy difference

A nucleus contains $Z$ protons and $A-Z$ neutrons. Its mass is smaller than the total mass of these free nucleons. This mass deficit 质量亏损 corresponds to the binding energy 结合能 needed to separate the nucleus completely into free nucleons:

Here $c_0$ is the speed of light, distinguished from specific heat capacity. The unified atomic mass unit 统一原子质量单位 is $1\ \mathrm u=1.66\times10^{-27}\ \mathrm{kg}$. Convert all masses to the same unit before subtracting. Nuclear masses and neutral-atom masses are not interchangeable without accounting consistently for electrons. Use the masses the question actually supplies.

Worked example. A nucleus has mass deficit $0.030\ \mathrm u$ et $A=4$. Find binding energy per nucleon.

Why fusion and fission can both release energy

The binding-energy-per-nucleon curve rises steeply for light nuclei, peaks near iron/nickel, then decreases gradually for heavy nuclei. Greater binding energy per nucleon generally means nucleons are more tightly bound. It is the vertical value, not just mass number, that matters when comparing points.

Nuclear fusion 核聚变 combines light nuclei. Nuclear fission 核裂变 splits a heavy nucleus into smaller products, often with neutrons. Energy is released when the products have greater total binding energy and smaller total rest mass than the reactants. For a reaction, calculate the total mass difference, not merely a difference of two per-nucleon values. Total energy is conserved.

Fusion needs very high temperature to give nuclei enough kinetic energy to approach despite electrostatic repulsion, and high density/confinement to make enough close encounters and sustain energy release. Temperature and density have different roles. Merely naming “high pressure” does not explain overcoming repulsion.

Radiation, background and decay

Choose radiation by its interaction

Alpha radiation 阿尔法辐射 consists of helium nuclei: strongly ionising, short range and stopped by paper or a few centimetres of air. Beta radiation 贝塔辐射 consists of electrons or positrons: less ionising, more penetrating, typically stopped by a few millimetres of aluminium. Gamma radiation 伽马辐射 is electromagnetic radiation: weakly ionising and highly penetrating; lead or concrete reduces its intensity but does not provide a sharp stopping thickness.

In a cloud chamber, alpha tracks are thick mainly because of strong ionisation, and relatively straight because the particles have large mass compared with beta particles. Beta tracks are thinner and more easily deflected. Do not swap these explanations. Beta radiation can monitor paper thickness because some passes through and some is absorbed; a change in thickness changes the transmitted count rate.

In nuclear equations 核反应方程, conserve nucleon number and charge number. Beta-minus decay converts a neutron to a proton; the mass number stays the same and atomic number increases by one. Include the antineutrino where required, as in Unit 4:

Alpha decay reduces $A$ by four and $Z$ by two. Gamma emission changes neither $A$ nor $Z$. Nuclear equations conserve more than these two numbers: energy and momentum still matter.

Correct background before applying a model

Background radiation 本底辐射 comes from sources such as rocks, cosmic rays and the environment. Measure it for a sufficiently long time with the test source absent. Subtract its count rate from every source-plus-background reading. Keep counting-time units consistent. Corrected detector count rate is proportional to source activity only when geometry and detection efficiency remain unchanged; it is not automatically the activity in becquerels.

Worked example. A detector reads $62\ \mathrm{min^{-1}}$ at distance $r$, including background $14\ \mathrm{min^{-1}}$. Predict total rate at $2r$ for a small isotropic gamma source with negligible absorption.

Do not divide the background by four: it is not all coming from the test source.

Random events, predictable populations

Désintégration radioactive 放射性衰变 is spontaneous: an unstable nucleus decays without an external trigger. It is random: the exact time for an individual nucleus cannot be predicted. A large population has a predictable statistical decay pattern. Activity 放射性活度 $A$ is the number of decays per second, measured in becquerels 贝可勒尔, $1\ \mathrm{Bq}=1\ \mathrm{s^{-1}}$.

Le decay constant 衰变常数 $\lambda$ is the probability per unit time of decay for a nucleus. The half-life 半衰期 $t_{1/2}$ is the time for the undecayed population, or activity, to halve on average.

At one half-life, $N/N_0=1/2=e^{-\lambda t_{1/2}}$. Taking logs gives:

A log-activity–time line has gradient $-\lambda$. Take logarithms of corrected rates, not source-plus-background readings. Read several half-life intervals from a graph and average; measure from the corrected activity level, not from zero total counts.

Worked example. A source has half-life $3.0\ \mathrm{years}$. When will activity fall to $2.0\%$ of its initial value?

Use a fraction, not $2.0$ inside the logarithm. If calculating power from a radioactive source, $P=A E_{\rm decay}$, with energy per decay in joules. This is released nuclear power; useful electrical output may be smaller. Convert the half-life to seconds when activity must be in becquerels.

CP15: absorption of gamma radiation by lead

Keep source, absorber and detector in fixed positions. Measure background, then counts over equal known intervals for several total lead thicknesses. Repeat or extend the counting time because counts fluctuate. Subtract background rate before comparing transmission. If each equal thickness gives the same fractional reduction, the attenuation is exponential; a graph of log corrected rate against thickness is approximately straight. Three half-value thicknesses transmit $1/8$, not $1/3$.

The material-absorption constant is not the time-decay constant of the source. Avoid readings too close to background, where subtraction gives large relative uncertainty. Handle sources with the specified tools under supervision, keep distance, minimise exposure time and return them to shielding. Never touch a source or direct it at people. Keep lead handling clean and wash hands.

Simple harmonic motion, graphs and energy

The restoring condition

Mouvement harmonique simple 简谐运动 (SHM) occurs when resultant acceleration is proportional to displacement from a fixed equilibrium position and directed towards it:

The negative sign is the restoring direction. Constant speed, repetition alone or a force merely pointing towards a centre does not establish SHM. For a vertically hanging spring, measure $x$ from the loaded equilibrium position. Weight is already balanced there; the resultant for displacement is $-k_sx$.

For release from positive maximum displacement at $t=0$:

Here $A_0$ is amplitude, not radioactive activity. The phase depends on the chosen starting time; a sine displacement can describe the same motion with another origin. Differentiate graphically: the displacement–time gradient is velocity, and velocity–time gradient is acceleration. At maximum displacement, speed is zero and acceleration points back towards equilibrium. At equilibrium, speed is greatest and acceleration is zero.

Worked example. A graph gives amplitude $0.030\ \mathrm m$ and period $0.50\ \mathrm s$. Find maximum speed and acceleration.

Alternatively find maximum speed from a tangent to the displacement graph at equilibrium. Convert centimetres to metres. A negative straight $a$–$x$ gradient is $-\omega^2$, so take its negative before the square root.

Periods and energy

For a mass on an ideal spring and for a simple pendulum at small angles:

Pendulum length is pivot to bob centre. The small-angle period does not depend on bob mass. A period ratio for one unchanged spring gives $T_2/T_1=\sqrt{m_2/m_1}$. Use total new mass before finding added mass.

Worked example. A $0.50\ \mathrm{kg}$ mass has period $0.80\ \mathrm s$. An extra mass increases the period to $1.00\ \mathrm s$ on the same spring.

For an undamped spring oscillator, energy transfers between elastic potential and kinetic stores:

Thus the $E_k$–$x$ graph is an inverted parabola: zero at $\pm A_0$, positive maximum at $x=0$. The $E_p$–$x$ graph opens upwards. Their sum is constant in the undamped model. In a damped system energy is transferred out, so amplitude decreases. Total energy including surroundings remains conserved.

CP16: infer an unknown mass from resonance

Keep the same spring/support system. Attach several known masses, drive with small oscillations and vary frequency to find the largest steady amplitude for each mass. Record the resonant frequency, repeating slowly around each peak. With light damping it approximates natural frequency. Plot $1/f^2$ against total known mass; the spring model gives gradient $4\pi^2/k_s$. Use the calibration to infer an unknown mass from its resonant frequency.

Account for a hanger and, where significant, the effective moving mass of the spring. A nonzero intercept may represent this contribution; do not force the line through the origin without justification. Keep amplitudes small and below the elastic limit. Secure masses and stand, and keep feet clear. Measuring free periods can provide a useful independent comparison, but the specified practical uses resonant frequencies.

Forced oscillations, resonance and damping

A free oscillation 自由振动 follows an initial disturbance without continued periodic driving. A forced oscillation 受迫振动 is maintained by a periodic driving force; its steady frequency is the driving frequency. Fréquence naturelle 固有频率 is the frequency of free oscillation for the system under the stated conditions.

Resonance 共振 occurs when driving frequency is at or near natural frequency, giving efficient energy transfer and a large steady amplitude. It does not mean “maximum frequency”. For a lightly damped system, the amplitude peak is near the undamped natural frequency. Greater damping lowers and broadens the peak; do not present every damped peak as exactly at the undamped natural frequency.

Damping 阻尼 transfers energy away from oscillation, for example through resistive work that increases the thermal energy of dampers and surroundings. Stronger damping can reduce dangerous bridge/building motion. Plastic deformation 塑性形变 of a ductile material also absorbs mechanical energy irreversibly; purely elastic deformation returns stored energy and is not the same mechanism.

Worked explanation. People walking supply a periodic driving force to a bridge. If its frequency is near a natural frequency, energy transfers efficiently into the bridge and amplitude increases. Dampers do work against motion, transferring oscillation energy to thermal stores, so the steady amplitude is limited. Link cause and effect in this order rather than listing “resonance, energy, damping”.

For the same shape and amplitude, adding mass to a light pendulum increases its stored mechanical energy while the small-angle ideal period remains unchanged. Under comparable resistive losses it can lose a smaller fraction of energy per cycle. That contextual damping comparison does not prove that every heavier oscillator always damps more slowly.

Gravitational fields and orbits

Field and potential have different meanings

A gravitational field 引力场 is a region where a mass experiences a force. Gravitational field strength 重力场强度 $g=F/m$ is force per unit test mass. It is a vector pointing towards an isolated source mass. Newton's law for point masses, or outside a spherical symmetric source, is:

$r$ is centre-to-centre distance, not height above the surface. For near-Earth local motion, constant $g$ can be suitable. Over large radial distances it is not constant.

Potentiel gravitationnel 引力势 is potential energy per unit mass, with zero at infinity:

Potential is a scalar and negative at finite distance. Moving outward makes it less negative, so potential energy increases while attractive force weakens. A potential graph value and its corresponding radius give $M=-Vr/G$.

Gravity and electric fields are both radial for isolated point sources and their force magnitudes obey inverse-square laws. Gravity acts on mass and is attractive in this model. Electric force acts on charge and can attract or repel. Electric field direction follows the force on positive test charge; a negative charge is forced oppositely.

Worked example. A $1000\ \mathrm{kg}$ satellite moves from radius $r_1=6.4\times10^6\ \mathrm m$ à $r_2=1.28\times10^7\ \mathrm m$ around Earth, $M=6.0\times10^{24}\ \mathrm{kg}$. Find its potential-energy change.

The positive sign is final minus initial. This is not the total launch energy, since kinetic energy may also change. $mg\Delta h$ with constant surface $g$ is unsuitable over this distance.

Derive a circular orbit

The gravitational force supplies the inward resultant:

Orbiting mass cancels. The period grows with radius. To appear stationary above one point on a rotating planet, a satellite needs a circular equatorial orbit, the same rotational direction and matching angular velocity/period. Matching period alone is insufficient.

Worked example. Pour $GM=4.00\times10^{14}\ \mathrm{m^3\,s^{-2}}$ and period $T=8.64\times10^4\ \mathrm s$, find stationary-orbit radius.

Subtract planet radius only if asked for height above the surface. For a star's surface field, use its radius, not its listed diameter; test a multiple-of-Earth-field claim by calculating the ratio.

Stellar radiation, distances and evolution

Temperature, luminosity and received intensity

A black body 黑体 absorbs all incident electromagnetic radiation and is an ideal thermal emitter. Its continuous spectrum has a characteristic shape. Increasing temperature increases total emitted power per unit area and moves the wavelength peak to shorter wavelength. The area under a spectral-intensity curve represents total intensity only with the appropriate spectral-axis definition.

Luminosity 光度 $L_\star$ is total power emitted. Intensity 强度 $I$ received at distance $d$ is power per unit receiving area. For a spherical black-body star of radius $R$ radiating equally in all directions:

The first area is the emitting surface; the second is the expanding receiving sphere. Do not substitute observer distance into the star's surface area. Real emitters may not be perfect black bodies, and intervening material can absorb radiation.

Wien's stated constant describes a spectrum per unit wavelength. A maximum of a spectrum plotted per unit frequency does not transform into that wavelength maximum simply by replacing $\lambda$ with $c_0/f$. If an exam gives the frequency corresponding to its specified wavelength peak, convert that given wavelength using $c_0=f\lambda$. Do not turn that particular task convention into a general rule for every spectral plot.

Worked example. A star's wavelength spectrum peaks at $580\ \mathrm{nm}$ and its radius is $7.0\times10^8\ \mathrm m$. Find temperature, luminosity and intensity at $1.5\times10^{11}\ \mathrm m$.

Keep unrounded temperature because it is raised to the fourth power. Compare the received intensity with the stated reference before accepting a planetary-temperature or power claim.

Two distance methods

Trigonometric parallax 三角视差 measures a nearby star's apparent angular shift against distant background stars as Earth moves around the Sun. Observations six months apart use a baseline of two Earth-orbit radii. The parallax angle $p$ is half the total shift. For small $p$ in radians, distance is $d\approx r_{\rm orbit}/p$. A parsec 秒差距 is the distance giving parallax $1$ arcsecond; $d_{\rm pc}=1/p_{\rm arcsec}$. Do not use the diameter of Earth or Sun as the orbital baseline.

At large distance, $p$ becomes too small for the instrument to measure accurately. The limitation is angular resolution/relative uncertainty, not necessarily that the star is invisible.

A standard candle 标准烛光 is an object of known luminosity. Identify one in a cluster, measure received intensity and use $d=\sqrt{L_\star/(4\pi I)}$. Its apparent brightness is not assumed known beforehand. Cepheid variable stars have a calibrated relationship between period and luminosity. Measure several complete light-curve periods, divide by their number, read the calibration carefully and convert any solar-luminosity units. A straight line on log axes is not automatically a linear relationship between the raw quantities.

Worked example. A standard candle has $L_\star=4.0\times10^{28}\ \mathrm W$ and received intensity $2.0\times10^{-10}\ \mathrm{W\,m^{-2}}$.

Read the Hertzsprung–Russell diagram

A Hertzsprung–Russell diagram 赫罗图 plots luminosity vertically against surface temperature horizontally, usually with temperature decreasing to the right and logarithmic scales. The main sequence 主序星带 runs from hot, bright stars at upper left to cool, faint stars at lower right. Red giants 红巨星 are cool but luminous because of their large surface areas. White dwarfs 白矮星 are hot but faint because of their small surface areas. Mark the Sun near one solar luminosity and about $5800\ \mathrm K$ on the main sequence.

Stars form when gravity contracts a cloud of gas and dust. The core heats until hydrogen fusion can sustain a main-sequence star; outward pressure balances gravity. In a Sun-like star, core hydrogen eventually runs low, fusion there decreases and gravity contracts the core. Core temperature rises; shell hydrogen burning and later helium fusion are associated with expansion to a red giant. Outer layers are lost, leaving a white-dwarf core with no sustained fusion. It cools over a very long time. The Sun does not become a supernova or a neutron star.

A much more massive star can fuse heavier elements through later stages. When its core can no longer gain energy from fusion, collapse and a supernova can leave a neutron star or black hole, depending on the remnant. A massive main-sequence star has more fuel but consumes it much faster because of its hotter core. Its main-sequence lifetime can therefore be shorter. Explain fusion rate, not just fuel amount. An HR track shows changing temperature/luminosity, not a star travelling across space.

Doppler shifts and cosmology

Compare the same spectral line

Le Effet Doppler 多普勒效应 is a change in observed frequency/wavelength caused by relative motion along the line of sight. Successive wavefronts arrive farther apart from a receding source and closer together from an approaching source. Recession gives longer wavelengths and lower frequencies; approach gives shorter wavelengths and higher frequencies. Motion purely perpendicular to the line of sight is not the recession speed in the simple model.

For electromagnetic radiation at low recession speeds:

Redshift 红移 $z$ is positive for recession. For small shifts, its magnitude also approximately equals the fractional decrease of observed frequency relative to emitted frequency. State the sign convention: $(f_{\rm observed}-f_{\rm rest})/f_{\rm rest}$ is negative for recession. Do not equate a positive redshift to a signed frequency increase. The approximation $v/c_0$ is not a general high-speed relativistic formula.

Worked example. A line emitted at $500\ \mathrm{nm}$ is observed at $505\ \mathrm{nm}$.

To use a star's absorption lines, recall Unit 2: photons are absorbed only when their energies match allowed atomic-level differences. Compare the same identified spectral line, not two different elements.

For opposite limbs of a rotating star, one approaches while the other recedes. If their shifts are equal and opposite, the separation of the two observed wavelengths is twice the shift of either limb. Halve that separation before finding the equatorial speed. Then use $T=2\pi R/v$. This assumes the observed line-of-sight limb speed represents the equatorial rotation speed under the stated viewing geometry.

Hubble law and its limits

On large cosmological scales, recession speed approximately follows Loi de Hubble 哈勃定律:

Le Hubble constant 哈勃常数 $H_0$ has units of inverse time. In $\mathrm{km\,s^{-1}\,Mpc^{-1}}$, convert kilometres to metres and megaparsecs to metres before taking its reciprocal in seconds. Nearby objects may have local motions that do not follow the large-scale relationship.

Worked example. Use $H_0=70\ \mathrm{km\,s^{-1}\,Mpc^{-1}}$ et $1\ \mathrm{Mpc}=3.1\times10^{22}\ \mathrm m$ to estimate an expansion timescale.

The reciprocal is the Hubble time 哈勃时间. Interpreting it as an age assumes a model for the past expansion rate; it is not an exact model-independent age. A larger $H_0$ gives a smaller reciprocal timescale. Evidence for expansion supports an earlier denser, hotter universe; it does not describe an explosion into an already empty centre-surrounding space.

Dark matter 暗物质 is inferred from gravitational effects that visible matter alone does not explain, such as galaxy rotation and motion. It need not emit detectable light. The amount of gravitating matter affects how expansion changes, so uncertainty in matter content and the expansion model affects predictions of the universe's fate. Do not infer that one measurement of $H_0$ alone proves perpetual expansion or future collapse. Modern models also consider dark energy; outcome 171 specifically requires understanding the controversy concerning $H_0$, dark matter, age and fate, rather than memorising one unqualified prediction.

Vérifiez-vous

1. Why can internal energy increase without temperature increasing during a phase change?
2. A fixed amount of gas warms in a rigid vessel. Which temperature scale must be used to compare pressures?
3. Does greater binding energy per nucleon mean that less or more energy is needed to separate each nucleon on average?
4. Why must background be subtracted before using a count-rate ratio or logarithm?
5. An $a$–$x$ line has gradient $-25\ \mathrm{s^{-2}}$. Find angular frequency.
6. Why is maximum SHM speed at equilibrium although acceleration is zero there?
7. Why can a damped bridge be driven at the same frequency but oscillate with a smaller amplitude?
8. Does lifting a satellite to a larger radius increase or decrease its signed gravitational potential energy?
9. A star is hot but faint. Which HR region is plausible, and what can explain its low luminosity?
10. When using the wavelength separation of opposite rotating limbs, why is a factor of two needed?

Answers: (1) molecular potential energy changes; (2) kelvin; (3) more; (4) background does not follow the source's decay/distance law; (5) $\omega=\sqrt{25\ \mathrm{s^{-2}}}=5.0\ \mathrm{rad\,s^{-1}}$; (6) restoring resultant is zero there but energy is mostly kinetic; (7) greater energy loss limits steady amplitude; (8) increases towards zero; (9) white dwarf, small emitting surface; (10) one limb is blueshifted while the other is redshifted.