Unit 4: Further Mechanics, Fields and Particles

Impulse measures the effect of a resultant force acting for a time interval: $\mathbf{J}=\mathbf{F}_{\mathrm{avg}}\Delta t=\Delta\mathbf{p}=m\mathbf{v}-m\mathbf{u}$ for constant mass. Choose a positive direction before subtracting the momenta.

A $0.20\,\mathrm{kg}$ ball reverses from $+6.0$ to $-4.0\,\mathrm{m\,s^{-1}}$. Its impulse is $0.20(-4.0-6.0)=-2.0\,\mathrm{N\,s}$. If contact lasts $0.050\,\mathrm{s}$, the average force is $-40\,\mathrm{N}$; the minus sign gives its direction.

Investigate whether the impulse delivered to a moving object equals its change in momentum. A low-friction trolley of known mass can pass through light gates while a force sensor and data logger record the horizontal force during the same interaction.

Keep the trolley system and mass fixed, minimise friction, and align the sensor with the motion. Plot impulse against change in momentum: agreement is supported by a straight line through the origin with gradient near one, within uncertainty.

For an isolated system, total momentum is a vector conserved through a collision or explosion. Choose perpendicular $x$ and $y$ axes, resolve every velocity before and after, and apply conservation independently to both components.

Assign signs from the chosen axes. For a velocity $v$ at angle $\theta$ from $+x$, use $v_x=v\cos\theta$ and $v_y=v\sin\theta$, changing the sign for leftward or downward components. Solve the two component equations, then recombine unknown components using $v=\sqrt{v_x^2+v_y^2}$ and $\tan\theta=v_y/v_x$, with the quadrant checked.

An equivalent scaled construction places momentum vectors head-to-tail. The before-collision total and after-collision total must be the same resultant; a closed momentum polygon provides a geometrical check.

Use an overhead video or other ICT record to measure the two-dimensional velocities of small spheres immediately before and after a collision. The camera should be fixed perpendicular to a level table, with a length scale in the plane of motion.

Use a level, low-friction surface; avoid spin and glancing vertical motion; select frames close enough to the collision that external impulses remain small. Repeat collisions with different approach directions and quote uncertainty from position and frame resolution.

An isolated collision conserves total momentum whether it is elastic or inelastic. It is elastic only when total kinetic energy is also conserved within experimental uncertainty; otherwise it is inelastic.

If the total kinetic energy is $0.962\,\mathrm{J}$ before and $0.975\,\mathrm{J}$ after, a small difference may be consistent with an elastic collision once measurement uncertainty is considered. A clear decrease means energy has transferred to deformation, internal energy or sound, so the collision is inelastic.

Begin with the classical definitions $E_k=\tfrac12mv^2$ and $p=mv$. Since $v=p/m$, substitution gives $E_k=\tfrac12m(p/m)^2=p^2/(2m)$. The result is valid for a non-relativistic particle of constant rest mass.

For a particle with $p=3.0\times10^{-20}\,\mathrm{kg\,m\,s^{-1}}$ and $m=5.0\times10^{-27}\,\mathrm{kg}$, $E_k=p^2/(2m)=9.0\times10^{-14}\,\mathrm{J}$. Squaring the momentum also squares its power-of-ten factor.

Angular displacement $\theta$ describes the angle swept from a reference direction. In radians, $\theta=s/r$, where $s$ is arc length and $r$ is radius measured in the same length unit; one complete revolution is $2\pi\,\mathrm{rad}=360^\circ$.

$150^\circ=150\pi/180=5\pi/6\,\mathrm{rad}$. Conversely, $1.20\,\mathrm{rad}=1.20(180/\pi)=68.8^\circ$. For an arc of $0.30\,\mathrm{m}$ on radius $0.20\,\mathrm{m}$, $\theta=1.5\,\mathrm{rad}$.

Angular velocity is the rate of change of angular displacement: $\omega=\Delta\theta/\Delta t$, measured in $\mathrm{rad\,s^{-1}}$. In uniform circular motion every point on a rigid body shares $\omega$, while its tangential speed is $v=\omega r$.

A wheel of radius $0.25\,\mathrm{m}$ rotates with period $0.40\,\mathrm{s}$. Then $\omega=2\pi/0.40=15.7\,\mathrm{rad\,s^{-1}}$ and a point on its rim moves at $v=(15.7)(0.25)=3.9\,\mathrm{m\,s^{-1}}$.

At two nearby positions on a circle, draw equal-length velocity vectors tangent to the path. Place their tails together and form $\Delta\mathbf{v}=\mathbf{v}_2-\mathbf{v}_1$. For a small angular displacement $\Delta\theta$, the velocity triangle is similar to the position triangle, so $\Delta v/v\approx\Delta\theta$; $\Delta\mathbf{v}$ points towards the centre.

Divide by the time interval: $a=\Delta v/\Delta t=v(\Delta\theta/\Delta t)=v\omega$. Since $v=r\omega$, this becomes $a=v^2/r=r\omega^2$. The acceleration direction remains radially inward even though its magnitude is constant in uniform circular motion.

Circular motion requires a resultant force directed towards the centre because the velocity direction is continually changing. 'Centripetal force' names this inward resultant; it is supplied by real forces already acting on the object.

Draw only real forces, choose inward as the radial direction, and add their radial components with signs. At the bottom of a vertical circle, an upward tension or reaction must exceed weight to leave an inward resultant; at the top, weight may contribute towards the centre.

After identifying the real forces, set their resultant component towards the centre equal to $ma$: $\sum F_{\mathrm{in}}=ma=mv^2/r=mr\omega^2$. Choose the form containing the quantities actually known.

A $0.50\,\mathrm{kg}$ object moves at $4.0\,\mathrm{m\,s^{-1}}$ in a horizontal circle of radius $2.0\,\mathrm{m}$. The required inward resultant is $mv^2/r=(0.50)(4.0^2)/2.0=4.0\,\mathrm{N}$. The named real force must supply this resultant.

An electric field is a region in which a charged particle experiences an electric force. The field exists because of source charges; a test charge reveals the field but does not create the field being described.

A particle's motion depends on the resultant of electric force and any other forces. Between horizontal plates, for example, a charged particle can accelerate vertically while retaining horizontal velocity, producing a curved path. A stationary charged particle initially accelerates in the force direction.

Electric field strength at a point is $\mathbf{E}=\mathbf{F}/Q$ for a small positive test charge $Q$. It is a vector with unit $\mathrm{N\,C^{-1}}$, equivalent to $\mathrm{V\,m^{-1}}$.

For a particle of charge $q$, the electric force is $\mathbf{F}=q\mathbf{E}$. Use the magnitude $|q|E$ for size; a positive charge is forced along $\mathbf{E}$ and a negative charge in the opposite direction. Combine this force vector with weight or other forces before applying $F=ma$.

A charge of $4.0\times10^{-12}\,\mathrm{C}$ in a field of $8.0\times10^5\,\mathrm{N\,C^{-1}}$ experiences force $F=EQ=3.2\times10^{-6}\,\mathrm{N}$. Whether it rises depends on whether this upward force exceeds its weight.

For two point charges separated by centre-to-centre distance $r$ in free space, the force magnitude is $F=|Q_1Q_2|/(4\pi\varepsilon_0r^2)$. The force lies along the line joining the charges.

The two charges exert equal-magnitude, opposite forces on one another. Doubling one charge doubles $F$; doubling separation reduces $F$ to one quarter. Convert nanocoulombs and centimetres to coulombs and metres before substitution. With several source charges, calculate each force vector and add components.

A point charge $Q$ produces electric field magnitude $E=|Q|/(4\pi\varepsilon_0r^2)$ at distance $r$. The direction is radially outward for positive $Q$ and radially inward for negative $Q$.

At a point between equal positive and negative charges, both field contributions point from the positive charge towards the negative charge, so their magnitudes add. Between two equal positive charges, the midpoint contributions oppose and cancel.

Electric potential is potential energy per unit positive charge; electric field describes how rapidly potential falls with position. Along one dimension, $E=-\mathrm{d}V/\mathrm{d}x$, so the field points from higher potential towards lower potential.

Moving a positive charge along the field lowers its potential energy; moving it against the field requires external work. For a radial field with $V(\infty)=0$, the potential at radius $r$ equals the area under the $E$-against-$r$ curve from $r$ to infinity, with sign set by the source charge.

Between large parallel plates, away from the edges, the electric field is approximately uniform: $E=V/d$, where $V$ is the potential difference and $d$ is the perpendicular plate separation.

For $V=800\,\mathrm{V}$ and $d=0.050\,\mathrm{m}$, $E=1.6\times10^4\,\mathrm{V\,m^{-1}}$. An electron has force magnitude $eE=2.6\times10^{-15}\,\mathrm{N}$ opposite to the field direction.

With zero potential chosen at infinity, a point charge produces $V=Q/(4\pi\varepsilon_0r)$. The sign of $V$ follows the source charge, while $r$ is a positive distance.

Potentials from several charges add algebraically because potential is scalar: calculate each $Q_i/(4\pi\varepsilon_0r_i)$ and sum the signed values. A charge $q$ at potential $V$ has potential energy $U=qV$; a change obeys $\Delta U=q\Delta V$.

For a positive particle approaching a positive nucleus, $V$ and $U$ rise. If it momentarily stops at closest approach, its lost kinetic energy equals the increase in electric potential energy, allowing $r$ to be found from $U=Qq/(4\pi\varepsilon_0r)$.

A positive point charge has straight radial lines directed outward and concentric spherical equipotentials; a negative charge reverses the arrows. A uniform field has parallel, equally spaced field lines and parallel equipotentials perpendicular to them.

Line density is a drawing convention for relative field strength, not a count of physical strands. Curved field lines show the local acceleration direction, but a moving charge with sideways velocity need not follow the line.

Capacitance is $C=Q/V$: the charge stored on either plate per unit potential difference across the capacitor. Its unit is the farad, $1\,\mathrm{F}=1\,\mathrm{C\,V^{-1}}$.

For an ideal fixed capacitor, $C$ is set by its construction; increasing $V$ increases $Q$ in proportion rather than changing $C$. During charging, current transfers charge, so $Q=\int I\,\mathrm{d}t$; the area under an current-time graph gives the delivered charge.

A $220\,\mu\mathrm{F}$ capacitor at $12.0\,\mathrm{V}$ stores $Q=CV=(220\times10^{-6})(12.0)=2.64\times10^{-3}\,\mathrm{C}$. Convert microfarads before calculating.

Charging requires work because the potential difference rises as charge accumulates. For a fixed capacitor, a graph of $V$ against $Q$ is a straight line from the origin to $(Q,V)$, so the stored energy is its triangular area: $W=\tfrac12QV$.

The last two forms follow by substituting $Q=CV$. For $C=47\,\mu\mathrm{F}$ and $V=400\,\mathrm{V}$, $W=\tfrac12(47\times10^{-6})(400)^2=3.76\,\mathrm{J}$.

The time constant is $\tau=RC$. After one time constant, a discharging value is $e^{-1}\approx0.37$ of its initial value; a charging capacitor has reached $1-e^{-1}\approx0.63$ of its final value. A larger $R$ or $C$ stretches the curve horizontally.

During charging, increasing capacitor p.d. leaves decreasing p.d. across the resistor, so current falls. When the capacitor reaches the supply p.d., resistor p.d. and current are zero. During discharge, the capacitor itself drives the current.

Connect a capacitor and resistor in series with a d.c. supply and a switch that selects charging or discharging. Connect an oscilloscope or voltage sensor/data logger in parallel with the capacitor so it records $V_C$ without becoming part of the series current path.

Use a high-input-resistance sensor, record the actual resistor value, and sample much faster than the time constant. Discharge the capacitor safely before rewiring and observe polarity for an electrolytic capacitor.

For discharge through resistance $R$, $Q=Q_0e^{-t/(RC)}$, $V=V_0e^{-t/(RC)}$ and current magnitude $I=I_0e^{-t/(RC)}$. Each quantity falls by the same fractional factor over equal time intervals.

A fitted gradient $m$ gives $RC=-1/m$ and then $C=-1/(mR)$ if $R$ is known. Alternatively, substitute one paired value and time into the exponential equation, retaining consistent units.

The angle $\theta$ is between the magnetic field and the normal to the coil plane. Flux is maximum when the field is perpendicular to the plane and zero when it lies in the plane. For identical turns, linkage multiplies the one-turn flux by $N$.

A 50-turn coil of area $1.2\times10^{-3}\,\mathrm{m^2}$ perpendicular to $B=0.018\,\mathrm{T}$ has $N\Phi=50(0.018)(1.2\times10^{-3})=1.1\times10^{-3}\,\mathrm{Wb}$.

A charge moving through magnetic flux density $B$ experiences force magnitude $F=B|q|v\sin\theta$, where $\theta$ is the angle between velocity and field. The force is zero for parallel motion and maximum at $90^\circ$.

For a positive charge, Fleming's left-hand rule uses first finger for field, second finger for conventional current/positive-charge motion, and thumb for force. Reverse the force direction for a negative charge. The force is perpendicular to both $\mathbf{v}$ and $\mathbf{B}$.

Because the magnetic force is perpendicular to velocity, it does no work and changes direction rather than speed. With $\mathbf{v}\perp\mathbf{B}$ in a uniform field, it can supply centripetal force and produce a circular path. Crossed electric and magnetic forces balance when $qE=Bqv$.

A straight conductor of active length $l$ carrying current $I$ in magnetic flux density $B$ experiences $F=BIl\sin\theta$. Here $\theta$ is the angle between conventional current and the field, and $l$ is only the conductor length inside the field.

Apply Fleming's left-hand rule: first finger field from north to south, second finger conventional current, thumb force. Opposite sides of a current-carrying coil can experience opposite forces, forming a couple and a turning moment.

An e.m.f. is induced only while magnetic flux linkage through the coil changes. Relative motion between magnet and coil changes field strength and geometry at the turns; faster change gives a larger induced e.m.f.

Before the magnet reaches the coil, linkage is nearly constant and e.m.f. is near zero. Approaching and leaving produce opposite polarities because the linkage first changes one way and then the other. A falling magnet can give a larger, narrower later peak because it is moving faster.

Current in a primary coil creates a magnetic field. When that current changes, the field and the flux linkage through a nearby secondary coil change, inducing an e.m.f. in the secondary.

A secondary current flows only if its circuit is complete. A diode may select one polarity so a connected capacitor charges rather than alternately charging and discharging. Switching steady d.c. on or off produces only transient secondary e.m.f.

The combined Faraday-Lenz law is $\mathcal{E}=-\mathrm{d}(N\Phi)/\mathrm{d}t$. The magnitude is the rate of change of flux linkage; the minus sign states that any induced current produces effects opposing the change that caused it.

If linkage changes from $0.012$ to $0.003\,\mathrm{Wb}$ in $0.020\,\mathrm{s}$, the average e.m.f. magnitude is $|0.003-0.012|/0.020=0.45\,\mathrm{V}$. Direction requires a declared positive normal and loop direction, or a clear Lenz-law statement.

A nucleus is described by its nucleon number $A$ and proton number $Z$. The nucleon number, also called mass number, counts all protons and neutrons. The proton number, also called atomic number, counts protons and fixes the element's identity.

For $^{121}_{51}\mathrm{Sb}$, the nucleus contains 51 protons and $121-51=70$ neutrons. An ion of this nuclide still has the same $A$ and $Z$ because gaining or losing electrons does not alter its nucleus.

Rutherford scattering tested the diffuse positive-charge ('plum pudding') model by directing alpha particles at thin metal foil. The observations did not fit diffuse charge, so the atomic model changed to one with a tiny central nucleus.

A large deflection requires a large force acting for a short time. This occurs only when an alpha particle passes very close to a concentrated positive charge. The rarity of these events shows that the nuclear region is tiny compared with the atom.

Heating a metal cathode gives some conduction electrons enough energy to escape its surface: this is thermionic emission. A positively biased anode then attracts the emitted electrons, so the electric field transfers energy and increases their speed.

Across potential difference $V$, an electron gains kinetic energy of magnitude $eV$ when other energy changes are negligible. Electric force is opposite to electric-field direction because the electron is negatively charged.

A magnetic field exerts a force perpendicular to the electron's velocity and the field. It changes the direction of velocity, producing acceleration and a curved path, but it does no work and cannot by itself increase the electron's speed.

In a cyclotron, speed and momentum increase at every gap crossing, so $r=p/(B|Q|)$ increases and the path spirals outward. The magnetic force supplies centripetal acceleration without adding energy.

In a linac, drift-tube lengths increase as a non-relativistic particle speeds up, allowing it to reach each next gap when the electric field again points in the accelerating direction.

For motion perpendicular to a uniform magnetic field, magnetic force supplies centripetal force. Using $B|Q|v=pv/r$ gives $r=p/(B|Q|)$. At non-relativistic speed, $p=mv$, so $r=mv/(B|Q|)$.

An electron with $p=1.50\times10^{-24}\,\mathrm{kg\,m\,s^{-1}}$ in $B=48\,\mu\mathrm{T}$ has $r=p/(Be)=0.195\,\mathrm{m}$. Convert microtesla to tesla before substitution.

At an isolated particle interaction, total electric charge, total energy and total momentum before the vertex equal their totals after it. Momentum is a vector, so balance components rather than adding track momenta as unsigned numbers.

Choose two perpendicular axes through the vertex. Resolve every known momentum, apply $\sum p_x$ and $\sum p_y$ conservation separately, then reconstruct the missing magnitude and direction. Check charge and total energy independently.

To investigate a nucleon's internal structure, the probe must interact on a distance scale smaller than the nucleon. Increasing particle momentum shortens its de Broglie wavelength, so higher-energy beams can resolve finer structure in scattering patterns.

High collision energy can also produce new massive particles through mass-energy conversion. Colliding two equal, opposite beams gives nearly zero total momentum, so less of the available energy must remain as kinetic energy of the products than when a beam hits a stationary target.

Deviations from scattering expected for a structureless target reveal constituent behaviour. The beam energy is therefore selected to make the wavelength small enough and, where relevant, to exceed the energy needed to create detectable products.

Mass-energy equivalence connects a change in system mass with an energy change: $\Delta E=c^2\Delta m$. In particle creation, supplied energy becomes rest energy and usually kinetic energy; in annihilation, particle and antiparticle rest energy becomes radiation and kinetic energy of other products.

Compare the complete initial and final systems. The minimum energy to create particles is their total rest energy, but momentum conservation may require additional kinetic energy. For annihilation of an electron and positron initially at rest, two photons can carry equal and opposite momentum.

Each electron has rest energy $mc^2\approx0.511\,\mathrm{MeV}$. Electron-positron annihilation from rest releases $1.022\,\mathrm{MeV}$ in total, shared by two photons of $0.511\,\mathrm{MeV}$ moving in opposite directions.

Convert energy by multiplying the number of electronvolts by $1.602\times10^{-19}\,\mathrm{J/eV}$. Thus $2.0\,\mathrm{GeV}=2.0\times10^9\times1.602\times10^{-19}=3.2\times10^{-10}\,\mathrm{J}$.

To convert $938\,\mathrm{MeV}/c^2$ to kilograms, first convert $938\,\mathrm{MeV}$ to joules, then divide by $c^2$, giving about $1.67\times10^{-27}\,\mathrm{kg}$. The notation $/c^2$ belongs to a mass unit, not an energy unit.

An unstable particle moving at a speed close to $c$ has a longer measured lifetime in the laboratory frame than its proper lifetime measured in its own rest frame. The effect becomes significant only at relativistic speeds.

Muons formed high in the atmosphere have a proper mean lifetime of about $2.2\,\mu\mathrm{s}$. Even at nearly $c$, a non-relativistic estimate gives only about $660\,\mathrm{m}$ travelled in that time, yet many are detected after travelling several kilometres. Their increased laboratory lifetime explains the observation.

Compare the journey time or observed travel distance with what the proper lifetime would allow. If the speed is a substantial fraction of $c$ and the observed survival is much greater, relativistic lifetime increase cannot be neglected.

Baryons and mesons are hadrons because they are made of quarks. Quarks and leptons are treated as fundamental in this model. A proton or neutron is therefore not fundamental even though it is a subatomic particle.

The pattern of quarks in the model contains paired members. Completing that symmetry led physicists to predict a top quark before it was observed; later experimental detection supported the classification.

Every particle has a corresponding antiparticle with the same rest mass but opposite electric charge and opposite additive quantum numbers such as baryon number and lepton number. The antiparticle is written with a bar or a distinct symbol, for example $e^+$ for the electron's antiparticle.

To deduce antiparticle properties, retain rest mass and reverse charge, baryon number and lepton number. A particle-antiparticle pair can annihilate, provided the products conserve total energy, momentum and all required quantum numbers.

An allowed particle interaction must conserve electric charge $Q$, baryon number $B$ and lepton number $L$. Assign each incoming particle's values, total each column, and compare with the totals for all outgoing particles.

For beta-minus decay $n\rightarrow p+e^-+\bar\nu_e$: charge gives $0=(+1)+(-1)+0$; baryon number gives $1=1+0+0$; lepton number gives $0=0+(+1)+(-1)$. All three ledgers balance.

A particle equation lists every incoming particle to the left of an arrow and every product to the right. Read each supplied symbol first, then use conservation laws to identify a missing symbol or test the completed equation.

The decay $\pi^+\rightarrow\mu^++\nu_\mu$ has charge $+1=(+1)+0$, baryon number $0=0+0$, and lepton number $0=(-1)+(+1)$. The antimuon is an antilepton, while the muon neutrino is a lepton.