Explain, in terms of a conservation law, why two photons need to be created.

Determine the speed of the incoming electron.

Calculate the energy and the momentum for each photon after the collision.

as the total initial momentum is zero, it must be zero after the collision

2 = ($\gamma$–1)m₀c² = ($\gamma$–1) 0.511

$\gamma$ = 4.91

v = 0.978c

«2 + 2 + 2 × 0.511 = 5.02 MeV» so each photon is 2.51«MeV»

$p=\frac{E}{c}=2.51$ «MeVc^–1»

The gravitational field strength at 20 Mm above the surface of the Earth is about 0.6 N kg^–1. Estimate the time correction per day needed to the time signals, due to the gravitational redshift.

Suggest, whether your answer to (a) underestimates or overestimates the correction required to the time signal.

$\frac{\mathrm{\Delta }f}{f}=\frac{gh}{c^2}$ so $\mathrm{\Delta }f=\frac{0.6\times 20000000}{{\left(3\times {10}^8\right)}^2}=1.3\times {10}^{-10}$

$\frac{\mathrm{\Delta }f}{f}=\frac{\mathrm{\Delta }t}{t}$

1.3 × 10^–10 × 24 × 3600 = 1.15×10^–5 «s» «running fast»

Award [3 max] if for g 0.6 OR 9.8 OR average of 0.6 and 9.8 is used.

ALTERNATIVE 1

g is not constant through ∆h so value determined should be larger

Use ECF from (a)
Accept under or overestimate for SR argument.

ALTERNATIVE 2

the satellite clock is affected by time dilation due to special relativity/its motion

State what is meant by the event horizon of a black hole.

Show that the surface area A of the sphere corresponding to the event horizon is given by

$A=\frac{16\pi G^2{M}^2}{c^4}$.

Suggest why the surface area of the event horizon can never decrease.

The diagram shows a box that is falling freely in the gravitational field of a planet.

A photon of frequency f is emitted from the floor of the box and is received at the ceiling. State and explain the frequency of the photon measured at the ceiling.

the surface at which the escape speed is the speed for light
OR
the surface from which nothing/not even light can escape to the outside
OR
the surface of a sphere whose radius is the Schwarzschild radius

Accept distance as alternative to surface.

[1 mark]

use of $A=4\pi R^2$ and $R=\frac{2GM}{c^2}$

«to get $A=\frac{16\pi G^2{M}^2}{c^4}$»

[1 mark]

since mass and energy can never leave a black hole and $A=\frac{16\pi G^2{M}^2}{c^4}$

OR

some statement that area is increasing with mass

«the area cannot decrease»

[1 mark]

ALTERNATIVE 1 — (student/planet frame):

photon energy/frequency decreases with height
OR
there is a gravitational redshift

detector in ceiling is approaching photons so Doppler blue shift

two effects cancel/frequency unchanged

ALTERNATIVE 2 – (box frame):

by equivalence principle box is an inertial frame

so no force on photons

so no redshift/frequency unchanged

[3 marks]

Calculate the expected shift in frequency between the emitted and the detected gamma rays.

Explain whether the detected frequency would be greater or less than the emitted frequency.

f = «$\frac{E}{h}=$» $\frac{14400\times 1.6\times {10}^{-19}}{6.63\times {10}^{-34}}$ = «3.475 × 10¹⁸ Hz»

Δf = «$\frac{g\times \mathrm{\Delta }h\times f}{c^2}\approx$» 8550 «Hz»

[2 marks]

«as the photon moves away from the Earth, » it has to spend energy to overcome the gravitational field

since E = h$$f, the detected frequency would be lower «than the emitted frequency»

[2 marks]

Define rest mass.

Calculate the total energy of the deuterium particle in $\mathrm{MeV}$.

In relativistic reactions the mass of the products may be less than the mass of the reactants. Suggest what happens to the missing mass.

invariant mass
OR
mass of object when not in motion/in object’s rest frame ✓

«rest energy =» $\left(2.014\times 931.5\right) «\mathrm{MeV}»$ ✓

$«E_{\mathrm{T}}=\mathrm{KE}+\mathrm{rest} \mathrm{energy}=270.0+\left(2.014\times 931.5\right)=» 2146 «\mathrm{MeV}»$ ✓

Final answer accept $\mathit{3}\mathit{.}\mathit{443}\mathit{\times }{\mathit{10}}^{\mathit{-}\mathit{10}}J$ if unit given

Award [2] marks for a bald correct answer.

is converted to energy ✓

as kinetic energy of the products ✓

The definition of rest mass proved to be known by most candidates.

Most were able to score full marks. A few candidates messed up by trying to convert to J, although some were successful with this additional difficulty.

Answers scored at least one or usually both marks.

Calculate the shift in frequency observed by A in terms of Δf.

Calculate the gravitational field strength on the surface of planet X.

                                    The following data is given:

                                    Δf = 170 Hz.

The distance between observer A and B is 10 km. 

Observer A now sends a beam of light initially parallel to the surface of the planet.

Explain why the path of the light is curved.

Δf $\propto$ f

therefore the change is «–»3Δf

[2 marks]

g = «c² $\frac{\mathrm{\Delta }f}{f\mathrm{\Delta }h}$ =» (3 × 10⁸)²$\frac{170}{5.0\times {10}^{14}\times 10000}$

g = 3.1 «ms^–2»

If POT mistake, award [0].

Award [2] for BCA.

[2 marks]

the mass of the planet warps spacetime around itself

the light will follow the shortest path in spacetime «which is curved»

[2 marks]

Calculate the fractional change in frequency of the gamma rays at the detector.

Explain the cause of the frequency shift for the gamma rays in your answer in (a) in the Earth’s gravitational field.

Explain the cause of the frequency shift for the gamma rays in your answer in (a) if the tower and detector were accelerating towards the gamma rays in free space.

$«\frac{∆f}{f}=\frac{g∆h}{c^2}=\frac{9.81\times 22.6}{c^2}»\frac{∆f}{f}=2.46\times {10}^{-15}$ ✓

$GPE$ gained by photons so $E$ increases ✓

$E=hf$, so frequency increases ✓

gamma rays travel at $c$ ✓

detector accelerates towards source so «by Doppler effect» $\lambda$ reduced so frequency increases ✓

Award [1 max] for reference to principle of equivalence without further explanation.

The calculation was easily done by most candidates.

A simple conceptual explanation in terms of the energy changes was not present.

Some candidates did score marks although it was not as simple as anticipated for a classical question.

Calculate the gamma (γ) factor for one of the protons.

Determine, in terms of MeV c^–1, the momentum of the pion.

The diagram shows the paths of the incident protons together with the proton and neutron created in the interaction. On the diagram, draw the path of the pion.

γ «= $\frac{3350}{938}$» = 3.37

[1 mark]

energy of pion = (3350 × 2) – 6200 = 500 «MeV»

500² = p²c² + 140²

p = 480 «MeV c^–1»

[3 marks]

path of pion constructed in direction around 4–5 o’clock by eye

[1 mark]

write down the momentum of the neutrino.

show that the energy of the pion is about 140 MeV.

State the rest mass of the pion with an appropriate unit.

«–»29.8 «MeVc^–1»

[1 mark]

E_π = $\sqrt{p_{\mu }^2{c}^2+m_{\mu }^2{c}^4}$ + p_vc OR E_μ = 109.8 «MeV»

E_π = «$\sqrt{{29.8}^2+{105.7}^2}$ + 29.8 =» 139.6 «MeV»

Final value to at least 3 sig figs required for mark.

[2 marks]

139.6 MeVc^–2

Units required.

Accept 140 MeVc^–2.

[1 mark]

Outline what is meant by the event horizon of a black hole.

Calculate the distance of the event horizon of the black hole from its centre.

                                                                   Mass of Sun = 2 × 10³⁰ kg

Star S-2 is in an elliptical orbit around a black hole. The distance of S-2 from the centre of the black hole varies between a few light-hours and several light-days. A periodic event on S-2 occurs every 5.0 s.

Discuss how the time for the periodic event as measured by an observer on the Earth changes with the orbital position of S-2.

boundary inside which events cannot be communicated to an outside observer

OR

distance/surface at which escape velocity = c

OWTTE

[1 mark]

mass of black hole = 7.2 × 10³⁶ «kg»

«$\frac{2GM}{c^2}$ =» 1 × 10¹⁰ «m»

[2 marks]

wherever S-2 is in orbit, time observed is longer than 5.0 s

when closest to the star S-2 periodic time dilated more than when at greatest distance

Justification using formula or time is more dilated in stronger gravitational fields

[2 marks]

Define total energy.

Determine the momentum of the proton.

Determine the speed of the proton.

Calculate the potential difference V.

total energy is the sum of the rest energy and the kinetic energy ✔

«p² c² = 1050²–938² therefore» p=472«MeVc^–1»✔

$\gamma =\frac{1050}{938}=1.12$  ✔

$v=0.45c$

OR

$V=1.35\times {10}^8$«ms^–1»  ✔

V = 112 «MV» ✔

Generally well answered by most candidates. A common mistake was to define the total energy in the context of classical mechanics.

Most candidates seemed to have the right starting points but mistakes were often made in attempting to convert units. The energy-momentum equation is generally best answered using only ‘MeV’ based units.

An easy calculation, that was generally well answered.

Very few candidates realised that this question required a simple calculation using eV= KE = (γ-1)E.

Show that energy is conserved in this decay.

Calculate the speed of the pion.

momentum of xi baryon is also 289.7«MeVc⁻¹» ✔

total energy of xi baryon and pion is $\sqrt{{289.7}^2+{1321}^2}+\sqrt{{289.7}^2+{135.0}^2}=1672$ «MeV» ✔

which equals the rest energy of the omega ✔

Allow a backwards argument, assuming the energy is equal.

$\gamma «=\frac{\sqrt{{289.7}^2+{135.0}^2}}{135.0}»=2.367$ ✔

$v«=\sqrt{1-\frac{1}{{2.367}^2}}c»=0.903c$ ✔

Award [2] for bald correct answer.

Energy conservation in a decay. This relativistic mechanics question was very challenging. Only a few candidates were able to calculate the total energy of the baryon and pion, despite being able to recognize the correct momentum of the baryon.

Energy conservation in a decay. This relativistic mechanics question was very challenging. Only a few candidates were able to calculate the total energy of the baryon and pion, despite being able to recognize the correct momentum of the baryon. The same for the speed of the pion in b), most of the attempts demonstrated by candidates were not relevant.

Show that the momentum of the electron is 1.41 MeV c^–1.

Explain the origin of each equation.

Calculate, in MeV c^–1, p₁ and p₂.

$pc=\sqrt{E^2-{\left(m{c}^2\right)}^2}=\sqrt{{150}^2-{0.511}^2}$ «= 1.410 MeV» ✔

first equation is due to momentum conservation ✔

second equation is due to total energy conservation ✔

adding 2p₁ = 3.42 MeV c^–1 ⇒ p₁ = 1.17 MeV c^–1 ✔

p₂ = 0.30 MeV c^–1 ✔

State what is meant by the event horizon of a black hole.

The mass of the black hole is 4.0 × 10³⁶kg. Calculate the Schwarzschild radius of the black hole.

The probe is stationary above the event horizon of the black hole in (a). The probe sends a radio pulse every 1.0 seconds (as measured by clocks on the probe). The spacecraft receives the pulses every 2.0 seconds (as measured by clocks on the spacecraft). Determine the distance of the probe from the centre of the black hole.

the distance from the black hole at which the escape speed is the speed of light ✔

R_s = «$\frac{2GM}{c^2}=\frac{2\times 6.67\times {10}^{-11}\times 4.0\times {10}^{36}}{9.0\times {10}^{16}}=$» 5.9 × 10⁹ «m» ✔

$2=\frac{1}{\sqrt{1-\frac{5.9\times {10}^9}{r}}}$ ✔

rearranged to give r

OR

r = 1.33 × 5.9 × 10⁹ «m» ✔

r = 7.9 × 10⁹ «m» ✔

Calculate the potential difference V.

The proton collides with an antiproton moving with the same speed in the opposite direction. As a result both particles are annihilated and two photons of equal energy are produced.

Determine the momentum of one of the photons.

γ = 1.96

E_k = (γ − 1) m₀c² = 900 «Me$$V»

pd ≈ 900 «MV»

Award [2 max] if Energy and Potential difference are not clearly distinguished, eg by the unit.

[3 marks]

energy of proton = γmc² = 1838 «Me$$V»

total energy available = energy of proton + energy of antiproton = 1838 + 1838 = 3676 «Me$$V»

momentum of a one photon = Total energy / 2c = 1838 «Me$$Vc^–1»

[3 marks]

Explain why a change in frequency is expected for the photon detected at the top of the rocket.

Calculate the frequency change.

ALTERNATIVE 1

detector accelerates/moves away from point of photon emission ✔

so Doppler effect / redshift ✔

so f decreases ✔

ALTERNATIVE 2

equivalent to stationary rocket on earth’s surface ✔

photons lose «gravitational» energy as they move upwards ✔

h f OR f decreases ✔

Most candidates answered this question correctly using the equivalence principle and could show that the frequency would decrease.

Arithmetic mistakes were common at the different stages of the calculations even when the process used was correct.

A lambda $\mathrm{\Lambda }$⁰ particle at rest decays into a proton p and a pion ${\pi }^{-}$ according to the reaction

$\mathrm{\Lambda }$⁰ → p + $\pi$^–

where the rest energy of p = 938 MeV and the rest energy of $\pi$^– = 140 MeV.

The speed of the pion after the decay is 0.579c. For this speed $\gamma$ = 1.2265. Calculate the speed of the proton.

pion momentum is $\gamma mv=1.2265\times 140\times 0.579=99.4$ «MeV$$c^–1»

use of momentum conservation to realize that produced particles have equal and opposite momenta

so for proton $\gamma v=\frac{99.4}{938}=0.106c$

solving to get v = 0.105c

Accept pion momentum calculation using E ² = p ²c ² +m ²c ⁴.

Award [2 max] for a non-relativistic answer of v = 0.0864c

[4 marks]

Determine the rest mass of the ${\mathrm{\Lambda }}^0$ particle.

Determine, using your answer to (a), the initial speed of the ${\mathrm{\Lambda }}^0$ particle.

$\mathrm{\Lambda }$ momentum = 900

E_proton = «$\sqrt{p{c}^2+{\left(m{c}^2\right)}^2}=\sqrt{{630}^2+{938}^2}=$» 1130 «MeV»

E_pion = «$\sqrt{{270}^2+{140}^2}=$» 304 «MeV»

so rest mass of $\mathrm{\Lambda }$ = «$\sqrt{{\left(1130+304\right)}^2-{900}^2}=$» 1116 «MeV c^–2»

«E = $\gamma$ mc² so» $\gamma$ = « $\frac{1434}{1116}$ =» 1.28

to give 0.64c

the total energy.

the speed.

Determine the rest mass of X.

neutron energy $=\sqrt{{185}^2+{940}^2}=958$ «MeV» ✔

NOTE: Allow 1.5 × 10^–10 «J»

ALTERNATIVE 1

«use of $E=\gamma E_0$»

«$958=940\gamma$ so» $\gamma =1.019$ ✔

v = 0.193c ✔

ALTERNATIVE 2

«use of $p=\gamma mv$»

$185=940\frac{{\displaystyle \frac{v}{c}}}{\sqrt{1-{\left({\displaystyle \frac{v}{c}}\right)}^2}}$ ✔

v = 0.193c ✔

ALTERNATIVE 3

«use of $p=\gamma mv$»

$v=\frac{pc}{E}$✔

$v=\frac{185}{958}=0.193c$ ✔

NOTE: Allow v = 5.8 × 10⁷«ms^–1» 

momentum of X = 185 «MeV c^–1»✔

energy of X = 1190 – 958 = 232  «MeV»✔

$m_0=«\sqrt{{232}^2-{185}^2}=»140$ «MeV c^–2» ✔

NOTE: Allow mass in kg - gives 2.5 × 10^–28«kg»

Explain why the frequency of the radio waves detected by the observer is lower than $f$.

The probe emits 20 short pulses of these radio waves every minute, according to a clock in the probe. Calculate the time between pulses as measured by the observer.

ALTERNATIVE 1
as the photons move away from the black hole, they lose energy in the gravitational field ✔
since $E=hf$ «the detected frequency is lower than the emitted frequency» ✔

ALTERNATIVE 2
if the observer was accelerating away from the probe, radio waves would undergo Doppler shift towards lower frequency ✔
by the equivalence principle, the gravitational field has the same effect as acceleration ✔

ALTERNATIVE 3
due to gravitational time dilation, time between arrivals of wavefronts is greater for the observer ✔
since $f=\frac{1}{T}$, «the detected frequency is lower than the emitted frequency» ✔

NOTE: The question states that received frequency is lower so take care not to award a mark for simply re-stating this, a valid explanation must be given.

time between pulses = 3 s according to the probe ✔

$∆t=«\frac{3}{\sqrt{1-{\displaystyle \frac{1}{1.5}}}}=» 5.2$ «s» ✔

Outline why the clock near the black hole runs slowly compared to a clock close to the distant observer.

Calculate the number of ticks detected in 10 ks by the distant observer.

this is gravitational time dilation

OR

black hole gives rise to a «strong» gravitational field

clocks in stronger field run more slowly

OR

the clock «signal» is subject to gravitational red-shift

the clock is subject to gravitational red shift

OR

the clock has lost gravitational potential energy in moving close to the black hole

[Max 2 Marks]

ALTERNATIVE 1 (10 ks is in observer frame):

Δt' = $10000\sqrt{1-\frac{6.0\times {10}^5}{7.0\times {10}^8}}$

9995.7 so 9995 «ticks»

Allow 9996

Allow ECF if 10 is used instead of 10000

ALTERNATIVE 2 (10 ks is in rocket frame):

Δt = $\frac{10000}{\sqrt{1-\frac{6.0\times {10}^5}{7.0\times {10}^8}}}$

10004 «ticks»

Allow ECF if 10 is used instead of 10000

State the equivalence principle.

State and explain the path of the light ray according to observer X.

State and explain the path of the light ray according to observer Y.

a freely falling frame in a gravitational field is equivalent to an inertial frame
OR
a frame accelerating in free space is equivalent to a frame at rest in a gravitational field ✔

X is in an inertial frame ✔

so light will follow a straight line path «parallel to the floor of the box» ✔

ALTERNATIVE 1
light must hit right wall of box at same place as determined by X ✔

«but box is accelerating» so path must be curved downward ✔

ALTERNATIVE 2
light is affected by gravity «for the observer at rest to the ground» ✔

so the path is curved downward/toward the ground ✔

Freefall in a gravitational field. Most of the candidates answered well.

Freefall in a gravitational field. Most of the candidates answered well but in b) omitted to mention that X is inertial.

Quite a high number of candidates, in b ii), stated incorrectly, that the light ray will be curved upward.