A2 Advanced Physics I

The Advanced Physics I exam tests the content on this page as well as the Core Physics I page.

97 Impulse:

• Impulse, $\mathrm{\Delta }p$, is the change in momentum, calculated with $F\mathrm{\Delta }t=\mathrm{\Delta }p$ and therefore measured in $\mathrm{N}\mathrm{s}$ or $\mathrm{k}\mathrm{g}\mathrm{m}{\mathrm{s}}^{-1}$
• It can also be calculated from the area under a force-time graph
• The formula can be rearranged to $F=\frac{\mathrm{\Delta }p}{\mathrm{\Delta }t}$, showing that force is equal to the rate of change of momentum: Newton's second law of motion

98 Investigating the relationship between force exerted and change in momentum (CP9):

• Create a surface for a trolley. Prop one end up with some books to compensate for friction. Once it is being accurately compensated for, the trolley should not accelerate or decelerate after being pushed
• Alternatively, use an air track
• Set up two light gates, about $\frac{1}{4}$^th of the total length away from each end
• Attach a hanging mass to the trolley with a pulley and string, with several individual masses on
• Now release the trolley from the top of the track. It should be accelerated by the mass and pass through the light gates
• Now find $m_{\mathrm{s}\mathrm{y}\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{m}}$. This is the total mass of the system; the sum of the trolley mass, string mass and the hanging mass
• Use the light gates to calculate the change in velocity. Find the change in momentum with $\mathrm{\Delta }{p}_{\mathrm{s}\mathrm{y}\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{m}}=m_{\mathrm{s}\mathrm{y}\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{m}}\mathrm{\Delta }v$
• Now use the time taken for the trolley to travel between the two light gates, $\mathrm{\Delta }t$, to find the rate of change of momentum with $\frac{\mathrm{\Delta }{p}_{\mathrm{s}\mathrm{y}\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{m}}}{\mathrm{\Delta }t}$
• Now calculate the total force acting on the system with $F=Mg$ where $M$ is the mass hanging from the pulley
• Do three repeats
• Now try varying the hanging mass. Do this by moving masses onto the trolley so that the total system mass remains the same
• Plot force acting on the system against the rate of change of momentum of the system. It should be a straight line. This shows that $F=$ $\frac{\mathrm{\Delta }{p}_{\mathrm{s}\mathrm{y}\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{m}}}{\mathrm{\Delta }t}$. This rearranges to $\mathrm{\Delta }p=F\mathrm{\Delta }t$

99 Conservation of linear momentum:

• Momentum is a vector so it can be treated separately for each axis
• Therefore, the vertical momentum before and after the collision should be equal. The same applies for horizontal momentum
• An elastic collision between a moving and stationary object (with equal mass) will cause each to move at right angles to each other. This will not work if the collision is completely head-on, but this is extremely unlikely

100 Analysing collisions between small spheres (CP10):

• Set up a video camera above a table so that it has a view from above. Place metre rulers on the table; one vertical and one horizontal
• Place a stationary sphere (e.g. ball bearing) so that each ruler will read about $50\mathrm{c}\mathrm{m}$
• Start recording and roll a second sphere towards the stationary one
• Use video analysis software to find the distance travelled each frame. If you know the frame period (i.e. 1/framerate), you can calculate the velocity for each sphere immediately before and immediately after the collision
• The total momentum before and after the collision in each dimension and overall should be equal

101 Collisions:

• An elastic collision is one where momentum and kinetic energy is conserved (so no energy is dissipated as other forms of energy)
• In an inelastic collision, momentum is conserved, but kinetic energy isn't

102 Calculating kinetic energy from momentum:

• $E_{\mathrm{k}}=$$\frac{p^2}{2m}$ where $p$ is the momentum and $m$ is the mass of the object
• The formula only works for non-relativistic particles (i.e. particles which have a velocity much less than the speed of light)
• You need to know how to derive this formula:
  - Start with $p=mv$ and $E_{\mathrm{k}}=\frac{1}{2}m{v}^2$
  - To substitute the first into the second, square the first: $p^2=m^2{v}^2$
  - Now rearrange this to $v^2=$ $\frac{p^2}{m^2}$
  - Substitute this into the second equation: $E_{\mathrm{k}}=$ $\frac{1}{2}\frac{m{p}^2}{m^2}$
  - This simplifies to $E_{\mathrm{k}}=$$\frac{p^2}{2m}$

103 Radians:

• Angle in radians $=$ $\frac{\pi }{180}$ $\times$ angle in degrees
• Angles measured in radians usually have the symbol 'rad' - e.g. $1.5\mathrm{r}\mathrm{a}\mathrm{d}$

Angular displacement:

• Angular displacement is the angle through which a rotating object has rotated, usually measured in radians

104 Angular velocity:

• Angular velocity is the angle an object rotates through per second, measured in $\mathrm{r}\mathrm{a}\mathrm{d}{\mathrm{s}}^{-1}$ with the symbol $\omega$
• $\omega =\frac{\theta }{t}$ where $\theta$ the angular displacement (radians), and $t$ is that time period in seconds
• Linear velocity is the actual velocity of a point if a tangent were drawn from it. It has the symbol $v$
• $v=r\omega$ where $r$ is the distance from the centre of the circle to the point which you're calculating the velocity of

Period:

• The period, $T$ is the time taken for a full revolution (in seconds)
• If you know the frequency (revolutions per second), you can use the equation $T=\frac{1}{f}$
• The following can also be derived: $T=$ $\frac{2\pi }{\omega }$

105 Centripetal acceleration:

• If an object is moving in a circle at constant speed, it is still accelerating because the direction component of its velocity is changing. This acceleration is directed towards the centre of the circle and called centripetal acceleration, $a$
• So centripetal acceleration is the rate of change of linear velocity
• $a=$ $\frac{v^2}{r}$ and $a$ = $r{\omega }^2$

Derivation:

• Let $u$ be the initial linear velocity and $v$ be the linear velocity after a movement of $\theta$ radians. Let $x$ be the distance between the reading of $u$ and the reading of $v$ around the circle (shown below):
  
• We can now create a vector triangle containing $u$, $v$ and $\mathrm{\Delta }v$:
  
• This is a sector, so we can use the sector equation for arc length $s=\theta r$ to get $\theta =$ $\frac{x}{r}$
• When $\theta$ is very small, the change in velocity $\mathrm{\Delta }v$ will be approximately equal to the arc length $x$. This can be written as $\theta v=\mathrm{\Delta }v$
• If we combine these two equations, we get $\mathrm{\Delta }v=$ $\frac{x}{r}$ $\times v$
• When $\theta$ is very small, $x=vt$. Therefore, this can be substituted into the above equation to get $\mathrm{\Delta }v=$ $\frac{v^{2}t}{r}$
• If we divide both sides by $t$, we get $a=$ $\frac{v^2}{r}$
• To get $a=r{\omega }^2$, substitute $v=\omega r$ into this formula

106 Centripetal force:

• If an object is accelerating, it must have a force acting on it (Newton's first law)
• Centripetal force acts perpendicular to an object's linear velocity, towards the centre of the circle
• If centripetal force suddenly stopped, the object would fly off at a tangent to its motion when the force stopped

107 Centripetal force:

• You can use the below equations to do centripetal force calculations
• $F=ma=$ $\frac{m{v}^2}{r}$ $=mr{\omega }^2$

108 Electric fields:

• An electric field is a region where a charged particle experiences a force

109 Electric field strength:

• Electric field strength, $E$, is the force per unit of positive charge
• It is a vector in the direction that a positive charge would move
• If a charge $Q$ experiences a force $F$, $E=$ $\frac{F}{Q}$
• From this, we can see that the units for $E$ are $\mathrm{N}{\mathrm{C}}^{-1}$

110 Coulomb's law:

• If two charges are further from each other, the force is lower. This follows an inverse square law:
  $F\propto$ $\frac{1}{r^2}$, where $r$ is the distance between the charges in $\mathrm{m}$
• Force is also proportional to charge ($F\propto Q$). The two equations can be combined to work with two charges:
  $F=$ $\frac{Q_1{Q}_2}{4\pi {\epsilon }_0{r}^2}$
• $F$ is the force of one of the charges. The force on the other is equal but opposite (i.e. $-F$)
• If the force is attractive, $F<0$. If it is repulsive, $F>0$
• ${\epsilon }_0$ is a constant depending on the substance. In air or a vacuum, it's the permittivity of free space: ${\epsilon }_0=8.85\times {10}^{-12}$ $\mathrm{F}{\mathrm{m}}^{-1}$

111 Calculating the electric field strength due to a point charge:

• At a distance $r$ $\mathrm{m}$ from a point charge $Q$, $E=$ $\frac{Q}{4\pi {\epsilon }_0{r}^2}$

112 Electric potential:

• Electric potential at a point in an electric field is the work done per unit charge to move a positively charged particle from infinity to that point
• For example:
  - if you have a negative point charge, call this C
  - pick a random point on its infinitely large electric field, and call this point P
  - the electric potential at point P is the energy which would be needed to attract a positive charge from infinity distance to P

113 Parallel plates:

• With parallel plates with a potential difference between them, there will be an electric field
• This field is constant at all positions between the two plates
• Where $V$ is the potential difference between the plates and $d$ is the distance between them, $E=$ $\frac{V}{d}$

114 Electric potential at a point in a point charge electric field:

• Electric potential (the energy to move a $+1\mathrm{C}$ charge from infinity to a point) is calculated with the equation $V=$ $\frac{Q}{4\pi {\epsilon }_{0}r}$ where $Q$ is the charge of the point charge and $r$ is the distance between the point charge and $+1\mathrm{C}$ charge

115 Electric field lines:

• Electric field lines show the direction that a positive charge would take in an electric field
• The spacing between the lines show the strength of the field - stronger fields have field lines drawn closer together

For parallel plates:

• Parallel plates have a uniform electric field because the charge is uniform on all points between the two plates
• Therefore, the electric field lines are drawn like this (from positive to negative):
  

For point charges:

• Point charges have a non-uniform electric field - the charge on a positive charge would be higher if it is closer to the point charge. On the diagram, the lines are therefore closer together nearer to the point charge:
  

Lines of equipotential:

• Lines of equipotential join up points at the same potential
• They should be drawn at uniform intervals - e.g. with parallel plates with a potential difference of $100\mathrm{V}$, you could draw an equipotential line every $25\mathrm{V}$
  
• With parallel plates, the lines of equipotential are straight and perpendicular to the field lines. They have an equal distance between them
  
  
  
• With point charges, the lines of equipotential are circular. They are closer together closer to the point charge

116 Capacitors:

• A capacitor is a component that can store electrical charge
• They contain two conducting metal plates, separated by an insulating material or gap

Capacitance:

• The amount of charge a capacitor can store is its capacitance, $C$, measured in farads
• $C=$ $\frac{Q}{V}$ where $Q$ is the charge and $V$ is the pd

117 Deriving $W=½QV$:

• The potential difference across a capacitor is directly proportional to the charge stored in it
• Therefore, a graph of potential difference against charge stored would be a straight line through the origin
• Because in a triangle, area $=\frac{1}{2}\times$ base $\times$ height, the energy stored by a capacitor can be calculated with $W=\frac{1}{2}QV$

Derivations from $W=½QV$:

(the colour coding is slightly different here for clarity)
• By combining the equations $Q$ $=CV$ and $W=\frac{1}{2}$$Q$$V$, we get $W=\frac{1}{2}CVV=\frac{1}{2}C{V}^2$
• By combining the equations $V$ $=\frac{Q}{C}$ (rearranged from $Q=CV$) and $W=\frac{1}{2}C$$V$${}^2$, we get $W=$ $\frac{1}{2}\frac{C{Q}^2}{C^2}$ $=$ $\frac{\frac{1}{2}{Q}^2}{C}$
• So with these two methods, we have two further equations:
  $W=\frac{1}{2}C{V}^2$ and $W=$ $\frac{\frac{1}{2}{Q}^2}{C}$

118 Resistor-capacitor charging curves:

• In a circuit with a resistor and capacitor, charging/discharging curves are exponential functions if drawn on a graph of pd/charge/current against time
• Because these graphs are powers of $e$, the first half-life will equal the second half-life and so on. This will also work for any fraction - e.g. quarter-lives. This is due to the constant ratio property of exponential relationships
• The time constant, $RC$ is the time taken for a capacitor's pd/charge/current to fall to $\frac{1}{e}$ (about $37\mathrm{\%}$) of its original value. So after two time constants, it will have fallen to $0.37\times 0.37\approx 14\mathrm{\%}$ of its original pd/charge/current
• As implied by $RC$, it is calculated by multiplying the capacitor's capacitance and the resistance of the resistor (assuming that the rest of the circuit has negligible resistance)
• After five time constants of discharging, the charge in the capacitor has fallen below $1\mathrm{\%}$ of its original charge, which can be considered zero coulombs in most cases

119 Analysing capacitor charging and discharging (CP11):

• If you connect a cathode ray oscilloscope (CRO) in parallel across a capacitor, you can charge it and discharge it through a resistor and view the potential difference on the display
• By setting the timebase (number of milliseconds per division on the horizontal axis) to 0, there'll be a dot in the centre. This will show the potential difference
• Set the gain (number of volts per division on the vertical axis) so that the peak pd will be at the top
• Now use a timer to record the amount of time for it to fall/rise by one division. Do this for several divisions. This can be used to plot a $V$-$t$ graph of charging or discharging
• Instead of a CRO, you could use a data logger. This will connect to an ammeter (in series) and voltage sensor (in parallel) and plug into a computer

120 Capacitor charge equations:

• $Q=Q_0{e}^{-t/RC}$ where
  - $Q$ is the charge at the time $t$
  - $Q_0$ is the initial charge when $t=0$
  - $t$ is the time after the capacitor started charging/discharging that you are investigating
  - $R$ is the resistance of the circuit
  - $C$ is the capacitance of the capacitor
• $\mathrm{\Delta }Q=I\times \mathrm{\Delta }t$ can be substituted into this equation to derive $I=I_0{e}^{-t/RC}$
• The above equation, along with $V=IR$ are used to derive $V=V_0{e}^{-t/RC}$
• By taking natural logs of both sides, $\ln Q=\ln Q_0-\frac{t}{RC}$ can be derived (and the same for $I$ and $V$). When deriving this, remember that the log of $Q_0{e}^{-t/RC}$ is $\ln (Q_0\times e^{\frac{-t}{RC}})$ and not $\ln (Q_0)\times \ln (e^{\frac{-t}{RC}})$

121 Magnetic flux:

• Magnetic flux density, $B$ is a measure of the strength of a magnetic field, measured in $\mathrm{T}$ (Teslas)
• Magnetic flux, $\mathrm{\Phi }$, measures the total magnetic field passing through a given area. It's highest if the field lines are perpendicular to this area (they pass through at a 90° angle). Magnetic flux is calculated with $\mathrm{\Phi }=BA\cos \theta$ [not given in the exam]. $\theta$ is the angle between the field lines and normal of the area. Magnetic flux is measured in Webers ($\mathrm{W}\mathrm{b}$)
• Magnetic flux linkage, $\mathrm{\Phi }N$ is the flux of a coil, $\mathrm{\Phi }$, multiplied by the number of turns, $N$

122 Fleming's left-hand rule for motors:

• ThuMb: Motion
• First finger: Field (N to S)
• Centre finger (middle finger): Current ($+$ to $-$)

Charged particles moving through a magnetic field:

• $F=Bqv\sin \theta$ where
  - $F$ is the force acting on the particle, in $\mathrm{N}$
  - $q$ is the charge of the particle, in $\mathrm{C}$
  - $v$ is the velocity of the charged particle, in $\mathrm{m}{\mathrm{s}}^{-1}$
  - $\theta$ is the angle between the particle's direction and the field lines
• Using Fleming's left-hand rule, you can see that a charged particle will be deflected in a circular path when through a magnetic field

123 Current carrying conductors in a magnetic field:

• $F=BIl\sin \theta$ where
  - $F$ is the force experienced by a wire, in $\mathrm{N}$
  - $B$ is the magnetic flux density of the field, in $\mathrm{T}$
  - $I$ is the current travelling through the wire, in $\mathrm{A}$
  - $l$ is the length of wire, in $\mathrm{m}$
  - $\theta$ is the angle between the current direction and field lines

124 Inducing emf in coils:

• An emf is generated whenever a conductor cuts field lines (or when the flux linkage changes)
• This emf can be increased by
  - increasing magnetic field strength
  - having more turns of the coil
  - increasing the rate of field cutting

125 Mutual inductance:

• When current passes through a coil, a magnetic field is produced. If this is near to another coil, an emf will be induced in the second coil whenever the magnetic field of the first changes
• Therefore, an emf can be induced in a second coil by passing an alternating current through the first
• This property is used in items such as wireless phone chargers (wirelessly transferring power over short distances) and transformers (increasing or decreasing the voltage of AC)
• The induced emf can be increased by
  - connecting the coils with a permanent magnet (e.g. iron rod)
  - decreasing the distance between the two coils
  - increasing the AC power input

126 Lenz's law:

• Lenz's law says that the emf induced is induced in the direction opposite to the change that caused it
• If you consider the direction of the induced emf to be the direction of motion, you can apply Fleming's left-hand rule

127 Faraday's law:

• $\epsilon =$ $\frac{-\mathrm{d}(N\mathrm{\Phi })}{\mathrm{d}t}$, where
  - $\epsilon$ is the induced emf ($\mathrm{V}$)
  - $\mathrm{d}(N\mathrm{\Phi })$ is the change in flux linkage ($\mathrm{W}\mathrm{b}$)
  - $\mathrm{d}t$ is the change in time ($\mathrm{s}$)
• This equation shows that the induced emf is proportional to the rate at which the flux is cut
• The fraction is negative due to Lenz's law

128 Alternating current:

• The period of an AC is the time taken for one complete cycle
• Frequency is the number of complete cycles per second
• The peak value of an AC is the maximum positive/negative value

Root-mean-square value:

• The root-mean-square value, or rms, is a way of measuring the current/voltage of an AC (since they'll vary depending on the current stage in the cycle)
• The rms voltage is the voltage that a DC would need to be at to produce the same power output
• The rms voltage is lower than the peak voltage because an alternating current spends some time at or near to $0\mathrm{V}$ in each cycle
• The rms current is written as $I_{\mathrm{r}\mathrm{m}\mathrm{s}}$ and rms voltage is written as $V_{\mathrm{r}\mathrm{m}\mathrm{s}}$

129 Converting between peak and rms values:

• Peak voltage/current and rms voltage/current are related with the following equations:
  $V_{\mathrm{r}\mathrm{m}\mathrm{s}}=$ $\frac{V_0}{\sqrt{2}}$ and $I_{\mathrm{r}\mathrm{m}\mathrm{s}}=$ $\frac{I_0}{\sqrt{2}}$

130 Proton/nucleon numbers:

• The proton number (or atomic number), $Z$, is the number of protons in the nucleus of an atom/ion. It is equal to the number of electrons if the atom is neutral
• The nucleon number (or mass number), $A$, is the number of protons and neutrons in the nucleus

131 Understanding of atomic structure:

• Dalton: originally suggested that everything is made up of small particles (atoms), each with the shape of a simple sphere
• Thomson: discovered the electron, creating the 'plum pudding' model - negatively charged electrons embedded inside a positively charged sphere
• Rutherford/Geiger/Marsden: fired alpha particles (positive helium nuclei made up of 2 neutrons and 2 protons) through thin gold foil. They noticed that most passed straight through, but some were refracted (due to proton-proton repulsion) and some were reflected (when they were repelled by the nucleus due to being very close). This caused the model to change to the nuclear model, a positive nucleus surrounded by negative electrons

Conclusions from the gold foil experiment:

• Most alpha particles passed straight through, so the atom is mostly empty space
• Some alpha particles were deflected, so the centre of the atom must have a positive charge to repel them
• Very few particles were deflected through angles greater than 90°, so the nucleus must be very small
• Alpha particles would only be deflected greater than 90° if they were scattered by something with a higher mass than themselves, so the nucleus must make up most of the mass of the atom

132 Thermionic emission:

• When a metal is heated, its free electrons gain energy
• With enough energy, they'll overcome the attraction from the positive metal ions and leave the metal. This is called thermionic emission
• Electron guns produce a thin beam of high-speed electrons
  - A high voltage is applied to a heating coil
  - This coil heats a cathode ($-$) (which will have free electrons)
  - Electrons are emitted
  - A short distance from the cathode is a cylindrical anode ($+$) with a small hole
  - This anode attracts the electrons, accelerating them. Those travelling in a straight line will pass through the hole
• Magnetic fields are also sometimes used to change the course or accelerate electrons

133 Linacs:

• A linear particle accelerator (linac) is a linear tube lined with electrodes
• Each electrode has the opposite charge to the adjacent ones
• Charged particles are fired from the start of the tube
• These electrodes are attached to an alternating current, at a frequency so that the particles are always being repelled from the electrode they just left and attracted to the one they are approaching
• This accelerates the particles
• Each electrode is longer than the last so that the particles spend the same amount of time in each

Cyclotrons:

• Cyclotrons accelerate particles in a spiral shape before releasing them
• They contain two thin hollow semicircular electrodes with an alternating (+ and -) potential difference so that the particles are always accelerating
• There is also a magnetic field perpendicular to the electrodes
• Charged particles are fired into an electrode. Due to the magnetic field, it follows a circular path
• The radius slowly increases as the particle accelerates, until it reaches the edge and leaves the cyclotron

Detectors - ionisation:

• Detectors used in particle physics experiments can contain a gas
• This gas is ionised by charged particles
• The gas ions are attracted to wires with a current flowing. The current spike caused by these ions can be detected

Detectors - deflection:

• A strong magnetic field is applied across the detector
• Colliding particles will travel in the same direction so they will not be affected
• Particles produced in collisions will be deflected by the magnetic field, and the deflection can be used to calculate their mass and velocity

134 Finding the radius of a charged particle in a magnetic field:

• $r=$ $\frac{p}{BQ}$. This is derived as follows:
• The centripetal force equation is $F=$ $\frac{m{v}^2}{r}$
• We also have $F=BQv$
• These are both equal to each other, so $\frac{m{v}^2}{r}$ $=BQv$
• Multiplying both sides by $r$: $m{v}^2=BQvr$
• Dividing both sides by $v$: $mv=BQr$
• Dividing both sides by $BQ$: $r=$ $\frac{mv}{BQ}$ $=$ $\frac{p}{BQ}$

135 Interpreting particle tracks:

• Charge, momentum and energy are always maintained in a particle collision
• When a charged particle passes through a substance, it ionises the atoms is passes through. This leaves a trail of ions which can be detected to find the path of electrons
• Cloud chambers contain a supercooled vapour. Ions condense the vapour, leaving a track. More ionising particles will leave short, thick tracks
• Bubble chambers contain hydrogen under pressure to keep it above its boiling point as a liquid. Ions cause gas bubbles to form
• Neutral particles won't show up on a particle track. If they then split into charged particles, those will be visible. This is what causes V shapes - two oppositely charged particles leaving from a decayed neutral particle

136 Why high energies are required in particle physics experiments:

• To get a positively charged particle close to a nucleus, it must have enough energy to overcome the forces of repulsion
• The higher the collision energy, the larger the produced particles will be (due to mass-energy equivalence)
• The more energy that can be given to particles, the shorter their wavelength and the higher the resolution if using them as a probe (e.g. an electron microscope has higher resolution than a light microscope because electrons have a smaller wavelength. The same applies here)

137 Mass-energy equivalence:

• Mass and energy are related, shown in the equation $\mathrm{\Delta }E=c^2\mathrm{\Delta }m$, where
  - $c$ is the speed of light in a vacuum
  - $\mathrm{\Delta }E$ is the energy of the particle in $\mathrm{J}$
  - $\mathrm{\Delta }m$ is the mass in $\mathrm{k}\mathrm{g}$

138 Writing particle masses:

• The equation $\mathrm{\Delta }E=c^2\mathrm{\Delta }m$ can be rearranged to $\mathrm{\Delta }m=$ $\frac{\mathrm{\Delta }E}{c^2}$
• Therefore, the mass of a particle can be expressed in terms of its energy (e.g. the mass of an electron at rest is about $9.11\times {10}^{-31}$ $\mathrm{k}\mathrm{g}$ or 0.512 $\mathrm{M}\mathrm{e}\mathrm{V}/c^2$)
• To convert to $\mathrm{e}\mathrm{V}/c^2$ form from a mass:
  - First, multiply the mass in $\mathrm{k}\mathrm{g}$ by $c^2$, $(3.00\times {10}^8{)}^2$
  - Then divide by the charge of an electron, $1.60\times {10}^{-19}$ $\mathrm{C}$
  - Round to an appropriate number of s.f. and use SI prefixes (e.g. Mega or Tera) to make the number more readable
     - e.g. 512 438 $\mathrm{e}\mathrm{V}/c^2$ → 0.512 $\mathrm{M}\mathrm{e}\mathrm{V}/c^2$

139 Relativistic particles:

• A relativistic speed is a speed close to the speed of light
• At these speeds, particles experience time dilation - they experience time more slowly
• Therefore, they will last longer, and decay more slowly
• Many equations, such as $E_{\mathrm{k}}=\frac{1}{2}m{v}^2$, are not accurate for relativistic particles

140, 141 Baryons (e.g. neutrons and protons):

• There are 6 types of quark; 12 in total including their anti-quarks
  - Up, charm and top quarks have charge $+2e/3$, in ascending mass
  - Down, strange and bottom quarks have charge $-1e/3$, in ascending mass
• Quarks are always part of another particle, they cannot exist on their own
• A baryon is a particle made of three quarks, such as a proton ($uud$) and neutron ($udd$)
• Anti-quarks are written with a line above, e.g. $\overline{u}$ and $\overline{d}$

Antimatter:

• Every fundamental particle has an antimatter version with identical mass and an identical strength, but opposite charge
• Anti-protons are written $p^{-}$ and anti-electrons (positrons) are written $e^{+}$. Otherwise, anti-particles are written with a bar above
• This symmetry is how the top quark was predicted
• Anti-atoms also exist - for example, anti-hydrogen is made of an anti-proton and positron

Mesons:

• Mesons are particles made from a quark and an anti-quark
• Pions are the most common mesons and are written ${\pi }^n$ where $n$ is the charge (either $-$, $+$ or 0)
  - e.g. an up quark and an anti-up quark is called "pi zero" or ${\pi }^0$

Leptons:

• Leptons, e.g. electrons ($e$) and positrons ($e^{-}$), are fundamental particles
• Leptons have $-1e$ charge [and their anti-leptons are $+1e$, indicated with a $+$ or $-$ above them (e.g. ${\mu }^{+}$ is an anti-muon)]

Photons:

• Photons are packets or energy with no mass and no charge

142 Particle interactions:

• On both sides of a particle equation, certain quantities need to be conserved:
  - Baryon number, $B$ (quarks have $B=\frac{1}{3}$, anti-quarks have $B=-\frac{1}{3}$)
  - Lepton number, $L_n$ where $n$ is the generation ($e$, $\mu$ or $\tau$). Leptons have $L=1$, anti-leptons have $L=-1$
  - Charge, $Q$
• Mass and energy don't need to be conserved (due to $\mathrm{\Delta }E=c^2\mathrm{\Delta }m$)