Full Syllabus Challenge 2 — Paper 4

Cambridge IGCSE Physics 0625 (Extended) — Topics 1–6: Motion, Forces & Energy · Thermal Physics · Waves · Electricity & Magnetism · Nuclear Physics · Space Physics

Duration 1 hour 15 minutes

Total Marks 80

Questions 7

Time Remaining 75:00

Syllabus 0625 / 0972

⚡ Cambridge Challenge Level — Full Syllabus

This paper covers all six topics at real Cambridge IGCSE difficulty. Scoring 50%+ is a solid B, 60%+ is an A, 70%+ is A*. Don't worry if this feels harder -- that's the point!

Instructions

Answer all questions in the spaces provided.
Show all working for calculations. Answers without working may not receive full marks.
You may use a calculator.
Take g = 10 N/kg unless a question states otherwise.
Speed of light c = 3.0 × 10⁸ m/s; speed of sound in air = 340 m/s
Specific heat capacity of water = 4200 J/(kg °C)
1 light-year = 9.5 × 10¹⁵ m; Hubble constant H₀ = 2.2 × 10⁻¹⁸ s⁻¹
Answer all questions first. When finished, click "Submit Exam" at the bottom of Question 7.
Your answers will be automatically graded when you submit the exam. The model answers will be shown for review.

Question Navigation

Question 1 -- Take-off and Landing at Heathrow

Total: 14 marks · Topic 1: Motion, Forces and Energy

An Airbus A320neo prepares to take off from London Heathrow. At take-off its mass is 78 000 kg. Its two engines together produce a thrust of 240 kN, and air resistance plus rolling friction on the runway total 60 kN.

(a) [4 marks]

(i) Calculate the resultant forward force on the aircraft during the take-off run.

(ii) Calculate the acceleration of the aircraft.

(iii) The aircraft must reach a take-off speed of 83 m/s. Assuming the acceleration is uniform, calculate the time needed to reach take-off speed from rest.

Mark Scheme -- Part (a)

(i) Resultant force = 240 − 60 = 180 kN (180 000 N)

(ii) a = F/m stated or used

(ii) a = 180 000 / 78 000 = 2.3 m/s² (accept 2.31)

(iii) t = v/a = 83 / 2.31 = 36 s (accept 35–36 s)

1 mark -- resultant 240 minus 60 equals 180 kN 1 mark -- acceleration equals force divided mass used 1 mark -- 180000 divided 78000 equals 2.3 m/s2 1 mark -- time 83 divided 2.31 equals 36 s

⚠ If you missed marks here: If you got a = 3.1 m/s² you used the full 240 kN thrust — you must subtract the 60 kN of resistive forces first, because a = F/m needs the RESULTANT force. Also convert 180 kN to 180 000 N before dividing by 78 000 kg (leaving it in kN gives a silly 0.0023). For (iii), t = v/a = 83 ÷ 2.31, not 83 × 2.31.

(b) [3 marks]

The runway available is 2.0 km long.

Using v² = u² + 2as (or otherwise), show that the take-off run needs a distance of approximately 1.5 km, and state whether the runway is long enough.

Mark Scheme -- Part (b)

v² = u² + 2as rearranged: s = v² / 2a (u = 0)

s = 83² / (2 × 2.31) = 6889 / 4.62 ≈ 1490 m ≈ 1.5 km

1.5 km is less than the 2.0 km runway, so the runway is long enough (with about 500 m to spare)

1 mark -- s equals v squared divided 2a rearranged 1 mark -- 6889 divided 4.62 approximately 1490 m 1.5 km 1 mark -- less than 2.0 km runway long enough

⚠ If you missed marks here: Two classic slips: forgetting to SQUARE 83 (gives s = 18 m, obviously too small) or dropping the 2 in 2as (gives s ≈ 3.0 km and the wrong conclusion that the runway is too short). And in a "show that" question the final comparison mark is free — always finish with "1.5 km < 2.0 km, so the runway is long enough".

On landing, the aircraft (now of mass 75 000 kg) touches down at 70 m/s. Reverse thrust and wheel brakes slow it to 5.0 m/s in 25 s.

(i) Calculate the change in momentum of the aircraft during this braking.

(ii) Calculate the average resultant force on the aircraft, and explain why slowing down over a longer time reduces the force experienced by the passengers.

Mark Scheme -- Part (c)

(i) Δp = mΔv = 75 000 × (70 − 5.0) = 75 000 × 65 = 4.9 × 10⁶ kg m/s (4 875 000)

(ii) F = Δp/t = 4 875 000 / 25 = 195 000 N (accept 1.95 × 10⁵ N)

(ii) For the same change in momentum, F = Δp/t: a longer stopping time gives a smaller force, so passengers are pushed forward less violently

1 mark -- momentum change 75000 times 65 equals 4875000 kg m/s 1 mark -- force 4875000 divided 25 equals 195000 N 1 mark -- same momentum change longer time smaller force passengers

⚠ If you missed marks here: If you got 5.25 × 10⁶ you used Δv = 70 instead of 70 − 5.0 = 65 m/s — the plane slows TO 5.0 m/s, not to rest (and check you used the landing mass 75 000 kg, not 78 000). For the explanation, "it's gentler" scores nothing: you must say the change in momentum is the SAME, so from F = Δp/t a longer time means a smaller force.

(d) [4 marks]

(i) Calculate the kinetic energy of the 75 000 kg aircraft at its touchdown speed of 70 m/s.

(ii) Most of this energy ends up in the wheel brake discs. State the energy transfer that takes place, and explain why the discs are made from a material with a high specific heat capacity.

Mark Scheme -- Part (d)

(i) KE = ½mv² stated or used

(i) KE = ½ × 75 000 × 70² = 1.8 × 10⁸ J (accept 1.84 × 10⁸ J)

(ii) Kinetic energy is transferred to thermal energy (heat) in the brake discs by friction

(ii) Using ΔT = E/(mc), a high specific heat capacity means the temperature rise for a given energy input is smaller, so the discs do not overheat / warp / catch fire

1 mark -- kinetic energy half m v squared used 1 mark -- half 75000 times 4900 equals 1.8 times 10 to 8 J 1 mark -- kinetic transferred thermal heat brakes friction 1 mark -- high specific heat capacity smaller temperature rise discs not overheat

⚠ If you missed marks here: Forgetting to square 70 gives 2.6 × 10⁶ J and dropping the ½ gives 3.7 × 10⁸ J — the correct KE is ½ × 75 000 × 70² = 1.8 × 10⁸ J. For (ii), "the brakes absorb heat" is not enough: name the transfer (kinetic → thermal by friction) AND link high c to a SMALLER temperature rise for the same energy, so the discs don't overheat.

Question 2 -- The Kulfi Factory

Total: 10 marks · Topic 2: Thermal Physics

A small factory in Old Delhi makes kulfi (Indian frozen dessert). Each mould contains 0.25 kg of kulfi mixture. For the mixture: specific heat capacity = 3900 J/(kg °C); specific latent heat of fusion = 330 000 J/kg; freezing point = 0 °C.

(a) [4 marks]

A mould of mixture enters the freezer at 25 °C.

Calculate the total thermal energy that must be removed to cool the 0.25 kg of mixture from 25 °C to 0 °C and then freeze it completely at 0 °C. Show each stage of your working.

Mark Scheme -- Part (a)

Cooling stage: E₁ = mcΔT = 0.25 × 3900 × 25

E₁ = 24 375 J (accept 24 000–24 400 J)

Freezing stage: E₂ = mL = 0.25 × 330 000 = 82 500 J

Total = 24 375 + 82 500 = 107 000 J (accept 1.07 × 10⁵ J)

1 mark -- cooling E equals mc delta T 0.25 times 3900 times 25 1 mark -- cooling energy 24375 J 1 mark -- freezing E equals mL 0.25 times 330000 equals 82500 J 1 mark -- total 107000 J or 1.07 times 10 to 5

⚠ If you missed marks here: This is a TWO-stage problem — if you only got 24 375 J (cooling) or 82 500 J (freezing) you missed the other stage; the answer must add both. Watch for using water's c = 4200 instead of the mixture's 3900 (gives 26 250 J), and never multiply L by ΔT — latent heat is just E = mL because temperature doesn't change while freezing.

(b) [3 marks]

A temperature probe in one mould shows that the temperature falls steadily, then stays constant at 0 °C for several minutes even though the freezer keeps removing energy, before falling again.

Explain, in terms of particles and energy, why the temperature remains constant while the mixture is freezing.

Mark Scheme -- Part (b)

During freezing the particles are forming bonds and settling into a fixed, ordered arrangement (solid structure)

The energy removed is the latent heat released as these bonds form -- it comes from the particles' potential energy, not their kinetic energy

Temperature measures the average kinetic energy of the particles; since the kinetic energy is unchanged during the change of state, the temperature stays constant

1 mark -- particles forming bonds fixed ordered arrangement solid 1 mark -- energy removed latent heat potential energy not kinetic 1 mark -- temperature average kinetic energy unchanged stays constant

⚠ If you missed marks here: The typical incomplete answer says "the energy is used for freezing instead" — that's only worth 1 mark at best. The examiner wants the PARTICLE story: bonds form as particles settle into a fixed arrangement, the energy removed is POTENTIAL energy (latent heat), and temperature only measures average KINETIC energy, which is unchanged — that's why the reading stays at 0 °C.

The factory delivers kulfi across the city in insulated boxes. Each box has thick expanded-polystyrene walls, a shiny aluminium foil lining, and a tightly fitting lid.

For each of these three features, explain how it reduces thermal energy transfer into the box, naming the method of transfer it reduces.

Mark Scheme -- Part (c)

Expanded polystyrene contains many pockets of trapped air; air is a very poor conductor, so this reduces conduction through the walls

The shiny foil lining is a good reflector / poor absorber of infrared, so it reduces energy entering by radiation

The tight lid stops warm outside air circulating into the box and cold air escaping, so it reduces transfer by convection

1 mark -- polystyrene trapped air poor conductor reduces conduction 1 mark -- shiny foil reflector poor absorber infrared reduces radiation 1 mark -- tight lid stops warm air circulating reduces convection

⚠ If you missed marks here: Each of the three marks needs the feature + HOW it works + the NAMED transfer method: "the foil keeps heat out" scores zero, but "shiny foil reflects infrared, reducing RADIATION" scores. Match them up: polystyrene's trapped air → conduction, shiny foil → radiation, tight lid → convection — and don't say "stops heat" when you mean "reduces".

Question 3 -- Sound, Refraction and Radar

Total: 12 marks · Topic 3: Waves

A music technology student in Kolkata analyses the sound of a tanpura (a stringed instrument) using a microphone connected to an oscilloscope. Take the speed of sound in air as 340 m/s.

(a) [4 marks]

The oscilloscope screen shows exactly 4 complete waves in a time of 10 ms (0.010 s).

(i) Calculate the period and the frequency of the note.

(ii) Calculate the wavelength of this sound in air.

(iii) The player now plucks the same string harder, producing a louder note of the same pitch. Describe how the oscilloscope trace changes.

Mark Scheme -- Part (a)

(i) Period T = 0.010 / 4 = 0.0025 s (2.5 ms)

(i) f = 1/T = 1/0.0025 = 400 Hz

(ii) λ = v/f = 340 / 400 = 0.85 m

(iii) The amplitude (height) of the trace increases; the number of waves on screen / spacing stays the same because the frequency is unchanged

1 mark -- period 0.010 divided 4 equals 0.0025 s 1 mark -- frequency 1 divided period equals 400 Hz 1 mark -- wavelength 340 divided 400 equals 0.85 m 1 mark -- amplitude height increases spacing frequency unchanged

⚠ If you missed marks here: If you got f = 100 Hz you took the whole 10 ms as one period — there are 4 waves on screen, so T = 0.010 ÷ 4 = 0.0025 s and f = 400 Hz. For the wavelength, it's λ = v ÷ f = 340/400 = 0.85 m (multiplying gives a nonsense 136 000 m). In (iii) a louder note of the SAME pitch means only the amplitude grows — if you said the waves get closer together you changed the frequency, which is wrong.

(b) [4 marks]

In an optics lesson, the student shines a laser into a rectangular perspex block of refractive index 1.49 at an angle of incidence of 35°.

(i) Calculate the angle of refraction inside the perspex.

(ii) Calculate the critical angle of the perspex.

(iii) Explain, in terms of wave speed, why the light bends towards the normal as it enters the perspex.

Mark Scheme -- Part (b)

(i) n = sin i / sin r, so sin r = sin 35° / 1.49 = 0.5736 / 1.49 = 0.385

(i) r = 23° (accept 22.6°)

(ii) sin c = 1/1.49 = 0.671, so c = 42° (accept 42.2°)

(iii) Light travels more slowly in perspex than in air; the side of the wavefront entering first slows down first, swinging the wave direction towards the normal

1 mark -- sin r equals sin 35 divided 1.49 equals 0.385 1 mark -- angle refraction 23 degrees 1 mark -- critical angle sin c equals 1 divided 1.49 equals 42 degrees 1 mark -- light slows perspex wavefront enters first slows bends towards normal

⚠ If you missed marks here: If your angle came out near 59° you MULTIPLIED by 1.49 instead of dividing — n = sin i / sin r, so sin r = sin 35° ÷ 1.49 giving r = 23° (smaller than 35°, bending TOWARDS the normal). For the critical angle remember sin c = 1/n, then take sin⁻¹. In (iii) "perspex is denser" scores nothing — you must say light travels SLOWER in perspex and the edge of the wavefront that enters first slows first.

The regional weather office uses a Doppler weather radar that emits microwaves of wavelength 5.6 cm to track monsoon storms.

(i) Calculate the frequency of these microwaves. (c = 3.0 × 10⁸ m/s)

(ii) State where microwaves lie in the electromagnetic spectrum relative to infrared and radio waves.

(iii) Suggest why microwaves are suitable for storm-tracking radar.

Mark Scheme -- Part (c)

(i) f = c/λ = 3.0 × 10⁸ / 0.056

(i) f = 5.4 × 10⁹ Hz (accept 5.36 × 10⁹ Hz)

(ii) Microwaves lie between radio waves (longer wavelength) and infrared (shorter wavelength) in the spectrum

(iii) They travel at the speed of light and pass through the atmosphere/cloud with little absorption, but are partly reflected by rain droplets, so the echoes reveal the position (and motion) of storms

1 mark -- frequency c divided wavelength 3.0 times 10 to 8 divided 0.056 1 mark -- 5.4 times 10 to 9 Hz 1 mark -- microwaves between radio longer infrared shorter 1 mark -- pass through atmosphere cloud reflected rain droplets echoes reveal storms

⚠ If you missed marks here: If you got 5.4 × 10⁷ Hz you divided by 5.6 instead of 0.056 — convert 5.6 cm to metres first, then f = 3.0 × 10⁸ ÷ 0.056 = 5.4 × 10⁹ Hz. For (ii) the order matters: microwaves sit BETWEEN radio (longer λ) and infrared (shorter λ). In (iii) "microwaves are good at detecting rain" is circular — say they pass through the atmosphere but are REFLECTED by rain droplets, so the echo locates the storm.

Question 4 -- Static and Current at the Fuel Depot

Total: 14 marks · Topic 4: Electricity

At a fuel depot near Rotterdam, petrol is pumped from storage tanks into road tankers. Static electricity is a serious fire hazard, and the electric pump systems must be carefully engineered.

(a) [3 marks]

As petrol flows quickly along an insulated plastic pipe, both the fuel and the pipe can become electrically charged.

(i) Explain how the fuel and pipe become charged.

(ii) Explain why this is dangerous, and how connecting a "bonding wire" between the tanker and earth before pumping removes the danger.

Mark Scheme -- Part (a)

(i) Friction between the moving fuel and the pipe transfers electrons from one to the other, leaving one positively and one negatively charged (only electrons move)

(ii) Charge builds up because the plastic is an insulator; the potential difference can grow until a spark jumps, which could ignite the petrol vapour

(ii) The bonding wire is a conductor that lets the charge flow safely to earth as fast as it is generated, so no charge (and no spark) can build up

1 mark -- friction fuel pipe transfers electrons charged 1 mark -- charge builds insulator spark could ignite petrol vapour 1 mark -- bonding wire conductor charge flows safely earth no spark

⚠ If you missed marks here: "Friction creates charge" loses the first mark — friction only TRANSFERS electrons (never protons) from one material to the other, leaving one positive and one negative. For the danger mark you need the full chain: insulator → charge builds up → spark → ignites petrol VAPOUR. And the bonding wire mark needs "conductor lets charge flow to EARTH so it can't build up" — not just "it's safer".

(b) [4 marks]

Part of the depot's control panel is a 6.0 V circuit: a resistor R₁ = 3.0 Ω is connected in series with a parallel combination of R₂ = 4.0 Ω and R₃ = 12 Ω.

Calculate:
(i) the combined resistance of R₂ and R₃ in parallel
(ii) the total resistance of the circuit
(iii) the current supplied by the 6.0 V source
(iv) the current in R₂.

Mark Scheme -- Part (b)

(i) 1/Rp = 1/4.0 + 1/12 = 4/12, so Rp = 3.0 Ω

(ii) Total R = 3.0 + 3.0 = 6.0 Ω

(iii) I = V/R = 6.0 / 6.0 = 1.0 A

(iv) p.d. across parallel pair = 1.0 × 3.0 = 3.0 V, so current in R₂ = 3.0/4.0 = 0.75 A

1 mark -- parallel 1 over 4 plus 1 over 12 equals 3.0 ohm 1 mark -- total resistance 3 plus 3 equals 6.0 ohm 1 mark -- supply current 6 divided 6 equals 1.0 A 1 mark -- pd parallel 3.0 V current R2 0.75 A

⚠ If you missed marks here: Two big traps: adding the parallel pair like series (4.0 + 12 = 16 Ω), or computing 1/Rp = 0.333 correctly but forgetting to FLIP it back — Rp = 3.0 Ω, not 0.33 Ω. In (iv), if you got 1.5 A you put the full 6.0 V across R₂; R₁ drops 3.0 V first, so only 3.0 V sits across the parallel pair, giving 3.0/4.0 = 0.75 A.

The fuel pump motor runs from a 24 V supply and draws a current of 30 A through a long cable of total resistance 0.050 Ω.

(i) Calculate the power wasted as heat in the cable.

(ii) Calculate the potential difference "lost" across the cable.

(iii) Explain why the depot uses a cable with thick copper conductors rather than thin ones.

Mark Scheme -- Part (c)

(i) P = I²R = 30² × 0.050 = 900 × 0.050

(i) P = 45 W

(ii) V = IR = 30 × 0.050 = 1.5 V

(iii) A thicker conductor has a lower resistance, so less power is wasted as heat (P = I²R), less voltage is lost, and the cable does not overheat at high current

1 mark -- power I squared R 900 times 0.05 1 mark -- wasted power 45 W 1 mark -- lost pd 30 times 0.05 equals 1.5 V 1 mark -- thicker lower resistance less power wasted not overheat

⚠ If you missed marks here: If you got 1.5 W you forgot to SQUARE the current — P = I²R = 30² × 0.050 = 45 W; and 720 W means you did P = VI with the full 24 V, which is the whole motor's power, not the cable's waste. For (iii) you must link thick → lower resistance → less I²R heating; "thick wires carry more current" on its own isn't the physics they want.

(d) [3 marks]

The pump runs for 4.0 minutes to fill one tanker, drawing 30 A from the 24 V supply.

(i) Calculate the electric charge that flows through the pump motor in this time.

(ii) Calculate the electrical energy transferred to the pump motor.

Mark Scheme -- Part (d)

(i) Q = It stated or used; t = 4.0 × 60 = 240 s

(i) Q = 30 × 240 = 7200 C

(ii) E = QV = 7200 × 24 = 172 800 J (accept 1.7 × 10⁵ J; or E = VIt = 24 × 30 × 240)

1 mark -- Q equals It time 240 seconds 1 mark -- charge 30 times 240 equals 7200 C 1 mark -- energy 7200 times 24 equals 172800 J

⚠ If you missed marks here: The number-one slip is leaving time in minutes: Q = 30 × 4.0 = 120 C is wrong by a factor of 60 — convert to t = 240 s first, giving Q = 7200 C. Then E = QV = 7200 × 24 = 172 800 J; if your energy is 60× too small, the minutes error carried through.

Question 5 -- The Bicycle Dynamo and the Relay

Total: 10 marks · Topic 4: Electromagnetic Effects

A commuter in Amsterdam has a bicycle with a hub dynamo that powers the lights. Inside the dynamo, a magnet rotates close to a fixed coil of wire wound on an iron core.

(a) [3 marks]

(i) Explain how the dynamo produces an e.m.f. when the wheel turns.

(ii) Explain why the lamp becomes brighter when the cyclist pedals faster.

Mark Scheme -- Part (a)

(i) As the magnet rotates, the magnetic field through the coil is continually changing (field lines are cut by the coil)

(i) A changing magnetic field through the coil induces an e.m.f. in it (electromagnetic induction), which drives a current through the lamp

(ii) Pedalling faster makes the magnet rotate faster, so the field changes at a greater rate and a larger e.m.f. is induced -- the bigger current makes the lamp brighter

1 mark -- rotating magnet field through coil changing field lines cut 1 mark -- changing field induces emf electromagnetic induction drives current 1 mark -- faster rotation greater rate change larger emf brighter

⚠ If you missed marks here: Saying "the magnet spins near the coil so electricity is made" skips the physics: the mark is for the magnetic field through the coil CHANGING (field lines being cut), and a changing field INDUCES an e.m.f. For (ii) the key phrase is "greater RATE of change of field → larger induced e.m.f." — "more electricity because it spins faster" doesn't earn it.

(b) [3 marks]

At a steady cycling speed, the dynamo output is an alternating e.m.f. with peak value 3.0 V and frequency 20 Hz.

(i) Sketch (on paper) a graph of this e.m.f. against time, labelling the peak value and the period.

(ii) On the same axes, sketch the output when the cyclist doubles her speed. Describe two ways the new graph differs from the first.

Mark Scheme -- Part (b)

(i) A smooth alternating (sine-shaped) wave, peak 3.0 V above and below zero, period T = 1/20 = 0.05 s labelled

(ii) At double speed the peak e.m.f. doubles to about 6.0 V (faster rate of field change)

(ii) The frequency also doubles to 40 Hz, so the period halves to 0.025 s (twice as many cycles in the same time)

1 mark -- sine alternating wave peak 3.0 V period 0.05 s 1 mark -- peak emf doubles 6.0 V 1 mark -- frequency doubles 40 Hz period halves 0.025

⚠ If you missed marks here: The most common slip is changing only ONE thing on the doubled-speed sketch. Faster rotation changes BOTH: peak e.m.f. doubles to 6.0 V AND frequency doubles to 40 Hz (period halves to 0.025 s). Also check your first sketch: the wave must go both above AND below zero (it's a.c.) with T = 1/20 = 0.05 s labelled, not 20 s.

At home, the commuter's doorbell system uses a relay: a low-voltage switch circuit at the front door controls a mains-powered chime.

(i) Describe, step by step, how pressing the doorbell switch causes the relay to turn on the mains circuit. Refer to the coil, the iron core, the armature and the contacts.

(ii) State one advantage of using a relay in this situation.

Mark Scheme -- Part (c)

(i) Pressing the switch sends a small current through the relay coil

(i) The current magnetises the iron core (it becomes an electromagnet)

(i) The magnetised core attracts the iron armature, which pivots and pushes the contacts together, completing the mains chime circuit

(ii) The low-voltage door circuit is electrically isolated from the mains circuit, so the switch at the door is safe to touch / a small current can switch a much larger one

1 mark -- pressing switch small current through relay coil 1 mark -- current magnetises iron core electromagnet 1 mark -- core attracts armature pivots pushes contacts together completes mains circuit 1 mark -- low voltage isolated mains safe small current switches larger

⚠ If you missed marks here: This is a sequencing question — each missing link in the chain costs a mark: current in COIL → iron CORE becomes an electromagnet → core attracts the ARMATURE → armature closes the CONTACTS, completing the mains circuit. Answers like "the relay switches on the chime" jump straight to the end and score little. For (ii), say the door circuit is electrically ISOLATED from the mains (safe), not just "it works better".

Question 6 -- Dating Rocks and Radiation in the Sky

Total: 10 marks · Topic 5: Nuclear Physics

A geologist from the University of Nairobi studies volcanic rocks in the East African Rift Valley. When lava solidifies, it contains potassium-40 (half-life 1.3 × 10⁹ years) but no argon-40. As the rock ages, each decaying potassium-40 atom is replaced by a trapped argon-40 atom.

(a) [3 marks]

Analysis of a rock sample shows that for every 1 atom of potassium-40 remaining, there are 3 atoms of argon-40.

Determine the age of the rock. Explain each step of your reasoning.

Mark Scheme -- Part (a)

A ratio of 1 potassium : 3 argon means only 1/4 of the original potassium-40 remains

1 → 1/2 → 1/4 means two half-lives have passed

Age = 2 × 1.3 × 10⁹ = 2.6 × 10⁹ years (2.6 billion years)

1 mark -- ratio 1 to 3 means quarter original potassium remains 1 mark -- quarter means two half lives passed 1 mark -- age 2 times 1.3 equals 2.6 billion years

⚠ If you missed marks here: The trap is reading "1 potassium : 3 argon" as 1/3 remaining. The 3 argon atoms were ALL once potassium, so the original amount was 1 + 3 = 4 and the fraction left is 1/4 — that's exactly two half-lives (1 → ½ → ¼), giving 2 × 1.3 × 10⁹ = 2.6 × 10⁹ years. If you wrote 3.9 × 10⁹ you counted three half-lives.

(b) [3 marks]

The geologist's field detector is calibrated using americium-241 (²⁴¹₉₅Am), which decays by emitting an alpha particle to form an isotope of neptunium (Np).

Write the balanced nuclear equation for this decay, giving the nucleon and proton numbers of the neptunium nuclide and of the alpha particle, and show that the equation balances.

Mark Scheme -- Part (b)

²⁴¹₉₅Am → ²³⁷₉₃Np + ⁴₂He : neptunium is 237/93

The alpha particle is a helium nucleus, ⁴₂He (2 protons + 2 neutrons)

Balance check: nucleon numbers 241 = 237 + 4; proton numbers 95 = 93 + 2

1 mark -- neptunium 237 93 1 mark -- alpha helium nucleus 4 2 protons neutrons 1 mark -- balance 241 equals 237 plus 4 and 95 equals 93 plus 2

⚠ If you missed marks here: An alpha particle is ⁴₂He, so the nucleon number drops by 4 (241 → 237) and the proton number by 2 (95 → 93) — if you wrote neptunium as 239/93 you subtracted 2 from BOTH numbers. The final mark is only earned by explicitly showing both sums balance: 241 = 237 + 4 and 95 = 93 + 2.

Flying home, the geologist reads that airline crew receive a larger annual radiation dose than most ground-based workers.

Explain why the radiation dose is higher at cruising altitude, and state why airlines monitor the flying hours of their crew.

Mark Scheme -- Part (c)

At high altitude there is much less atmosphere above the aircraft to absorb cosmic rays (radiation from space), so the background radiation level is higher than at ground level

Crew fly for many hours each year, so their total (cumulative) dose is significant; monitoring hours limits their dose and reduces the risk of cell damage / cancer

1 mark -- less atmosphere absorb cosmic rays higher background altitude 1 mark -- many hours cumulative dose limited reduce risk cell damage cancer

⚠ If you missed marks here: "You're closer to the Sun/space" is not the reason — the mark is for having LESS ATMOSPHERE above you to absorb cosmic rays, so more of that radiation reaches the aircraft. The second mark needs the idea of a CUMULATIVE dose: many flying hours add up, so airlines cap hours to limit total dose and the risk of cell damage or cancer.

(d) [2 marks]

Nuclear power stations use fission, whereas stars release energy by fusion.

State two differences between nuclear fission and nuclear fusion.

Mark Scheme -- Part (d)

In fission a large, heavy nucleus (e.g. uranium-235) splits into two smaller nuclei; in fusion two light nuclei (e.g. hydrogen) join to form a heavier nucleus

Any second difference, e.g. fusion needs extremely high temperature and pressure and occurs naturally in stars, while fission can be triggered by neutron absorption at ordinary temperatures / fission produces long-lived radioactive waste, fusion does not

1 mark -- fission large heavy nucleus splits fusion light nuclei join 1 mark -- fusion extreme temperature pressure stars fission neutron triggered waste

⚠ If you missed marks here: "Fission splits, fusion joins" alone is too vague for full credit — specify the nuclei: fission is a LARGE, heavy nucleus (like uranium-235) splitting into two smaller ones; fusion is two LIGHT nuclei (like hydrogen) joining into a heavier one. Your second difference must be genuinely different, e.g. fusion needs extreme temperature and pressure (stars) while fission is triggered by neutron absorption, or fission leaves long-lived radioactive waste.

Question 7 -- Lunar Orbits, Star Fates and the Expanding Universe

Total: 10 marks · Topic 6: Space Physics

ISRO's Chandrayaan-3 mission placed a spacecraft in a low orbit around the Moon before landing near the lunar south pole in 2023. The radius of the Moon is 1740 km.

(a) [3 marks]

Before descent, the spacecraft circled the Moon at a height of 100 km above the surface, completing one orbit every 118 minutes.

(i) Show that the orbital radius is 1.84 × 10⁶ m.

(ii) Calculate the orbital speed of the spacecraft using v = 2πr / T.

Mark Scheme -- Part (a)

(i) r = 1740 + 100 = 1840 km = 1.84 × 10⁶ m (radius measured from the Moon's centre)

(ii) T = 118 × 60 = 7080 s; v = 2π × 1.84 × 10⁶ / 7080

(ii) v = 1600 m/s (accept 1630–1640 m/s)

1 mark -- radius 1740 plus 100 equals 1840 km 1.84 times 10 to 6 m centre 1 mark -- T 7080 s v equals 2 pi r divided T substituted 1 mark -- orbital speed 1600 or 1630 m/s

⚠ If you missed marks here: The orbital radius is measured from the Moon's CENTRE, so add the Moon's radius: 1740 + 100 = 1840 km — using just 100 km loses the "show that" mark. Then convert the period: T = 118 × 60 = 7080 s; if your speed came out near 98 000 m/s you left T in minutes. Correct: v = 2π × 1.84 × 10⁶ ÷ 7080 ≈ 1600 m/s.

(b) [3 marks]

The table shows data for three stars studied by the mission's outreach team.

Star	Mass (compared with the Sun)	Colour
Star X	0.4	Red
Star Y	1.1	Yellow
Star Z	25	Blue-white

(i) State which star will eventually explode as a supernova, and explain how its final remnant could be a black hole.

(ii) State the final stage in the life of Star Y after it has passed through its red giant phase.

Mark Scheme -- Part (b)

(i) Star Z: it is far more massive than the Sun, so after its red supergiant stage its core collapses and the outer layers are blown off in a supernova

(i) If the collapsed core left behind is massive enough (more than about 3 solar masses), gravity overwhelms everything and it becomes a black hole (otherwise a neutron star)

(ii) Star Y (Sun-like) sheds its outer layers as a planetary nebula and the core remains as a white dwarf (which slowly cools)

1 mark -- Star Z most massive red supergiant core collapses supernova 1 mark -- core massive enough 3 solar masses gravity black hole otherwise neutron star 1 mark -- Star Y planetary nebula core white dwarf cools

⚠ If you missed marks here: Naming Star Z (25 solar masses) is only the start — the explanation marks need the sequence: massive star → red supergiant → core COLLAPSES → supernova, and then the black hole only forms if the leftover core is massive enough (roughly > 3 solar masses); otherwise it's a neutron star. For Star Y (Sun-like, 1.1 solar masses) the final stage is a WHITE DWARF — if you wrote neutron star or black hole, you mixed up the mass tracks.

A galaxy lies at a distance of 9.0 × 10²⁴ m from Earth.

Use the Hubble equation v = H₀d, with H₀ = 2.2 × 10⁻¹⁸ s⁻¹, to calculate the speed at which this galaxy is receding from us.

Mark Scheme -- Part (c)

v = H₀d = 2.2 × 10⁻¹⁸ × 9.0 × 10²⁴

v = 2.0 × 10⁷ m/s (accept 1.98 × 10⁷ m/s)

1 mark -- v equals H0 times d 2.2 times 10 to minus 18 times 9.0 times 10 to 24 1 mark -- recession speed 2.0 times 10 to 7 m/s

⚠ If you missed marks here: This is a MULTIPLY, not a divide: v = H₀ × d = 2.2 × 10⁻¹⁸ × 9.0 × 10²⁴ — dividing d by H₀ gives an absurd 4 × 10⁴². Handle the powers of ten carefully: −18 + 24 = 6, and 2.2 × 9.0 = 19.8, so v = 19.8 × 10⁶ = 2.0 × 10⁷ m/s (a sensible speed, well below c).

(d) [2 marks]

Mercury completes an orbit of the Sun in 88 days, while Neptune takes 165 years.

Explain two reasons why planets further from the Sun take much longer to complete one orbit.

Mark Scheme -- Part (d)

A more distant planet has a much larger orbit circumference (2πr), so it must travel a far greater distance in each orbit

The Sun's gravitational pull is weaker at greater distance, so the orbital speed of a distant planet is lower -- greater distance covered at lower speed means a much longer period

1 mark -- larger orbit circumference greater distance travel 1 mark -- gravitational pull weaker lower orbital speed longer period

⚠ If you missed marks here: Most students give only ONE reason ("it has further to go") — the question asks for TWO. You need both: the orbit circumference (2πr) is much larger, AND the Sun's gravity is weaker at that distance so the planet moves at a LOWER orbital speed. Longer path at slower speed = a much longer period, like Neptune's 165 years vs Mercury's 88 days.