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IGCSE Physics Paper 4 (Theory / Extended)

Topic 6: Space Physics — Mock Exam 2
1 hour 15 minutes
80
7
75:00
0625 / 0972

Instructions to Candidates

Question 1 — The Earth and Moon
Total: 12 marks
(a) [2]
The Earth rotates on its axis once every 24 hours. Explain how this rotation causes day and night at a location such as Greenwich, London.

Mark Scheme — 1(a)

The Earth rotates on its axis (once every 24 hours) so different parts face towards or away from the Sun [1]
When Greenwich faces towards the Sun it is daytime; when it rotates away from the Sun it is night-time [1]
⚠ If you missed marks here: Most answers get the spinning Earth down and stop there, which is one mark of two. Tie it to the named place: Greenwich is in daylight while it faces the Sun and in darkness once the rotation has carried it to the far side. "The Sun travels across the sky" or "the Sun moves round the Earth" describes an appearance and is rejected as an explanation.
(b) [3]
The Earth's axis is tilted at 23.5° to the perpendicular of its orbital plane. Explain how this tilt causes the United Kingdom to experience longer days and warmer temperatures in summer.

Mark Scheme — 1(b)

The Earth's axis is tilted at 23.5° to the perpendicular of its orbital plane [1]
In summer the Northern Hemisphere (where the UK is located) is tilted towards the Sun, so the Sun is higher in the sky and daylight hours are longer [1]
The Sun's rays strike the surface more directly / at a steeper angle, concentrating energy over a smaller area, so temperatures are higher [1]
⚠ If you missed marks here: "Britain is nearer the Sun in summer" scores nothing — the Earth is actually furthest from the Sun in early July, and the tilt keeps a fixed direction in space all year. Three separate points are wanted: the 23.5° tilt, the Northern Hemisphere leaning towards the Sun so the Sun sits higher and daylight lasts longer, and the rays striking more steeply so the same energy is spread over a smaller area.
(c) [4]
Describe the appearance of the Moon during its cycle of phases. Include in your answer: new Moon, waxing crescent, first quarter, full Moon, waning gibbous, and third quarter. State the approximate time for one complete cycle.

Mark Scheme — 1(c)

New Moon: the Moon is not visible (between Earth and Sun, unlit side faces Earth); waxing crescent: a thin sliver of light appears on the right side, growing larger [1]
First quarter: the right half of the Moon is illuminated; the Moon continues to wax (grow) towards full Moon, where the entire face is illuminated [1]
After full Moon, the illuminated area decreases (waning); waning gibbous: more than half is still lit but decreasing; third quarter: the left half is illuminated [1]
One complete cycle of phases takes approximately 28 days (one lunar month) [1]
⚠ If you missed marks here: The mark most often left unwritten is the last one — about 28 days for a complete cycle — because answers run out of time after listing the phases. Keep waxing and waning the right way round: waxing means the lit area is growing towards full, waning means it is shrinking after full. Calling the gibbous phases "half" merges two named phases and loses the mark for them.
(d) [3]
A student at the Royal Observatory Greenwich notices that the Moon rises approximately 50 minutes later each day. Suggest why this happens, relating your answer to the Moon's orbital period around the Earth.

Mark Scheme — 1(d)

The Moon orbits the Earth in approximately 28 days, moving in the same direction as the Earth's rotation [1]
Each day, the Moon moves a small distance further along its orbit, so the Earth must rotate slightly more than 360° for the same point to face the Moon again [1]
This extra rotation takes approximately 50 minutes (360° ÷ 28 days ≈ 13° per day, which corresponds to about 50 minutes of extra rotation) [1]
⚠ If you missed marks here: A slowing Earth is not the cause and gains nothing. In the 24 hours it takes the Earth to spin once, the Moon has moved about 13° further along its orbit (360° ÷ 28 days) in the same direction as the spin, so the Earth must turn roughly 373° before the Moon is overhead again. That extra 13° takes about 50 minutes, since 13 ÷ 360 × 24 h ≈ 52 min.
Question 2 — The Solar System
Total: 12 marks
(a) [3]
The table below shows data for four planets in the Solar System.
Planet Mass (relative to Earth) Gravitational field strength (N/kg)
Mercury 0.055 3.7
Earth 1.0 9.8
Jupiter 318 24.8
Neptune 17.1 11.2
Compare the gravitational field strengths of these planets and suggest what factor most affects gravitational field strength.

Mark Scheme — 2(a)

Jupiter has the greatest gravitational field strength (24.8 N/kg) and the greatest mass; Mercury has the smallest gravitational field strength (3.7 N/kg) and the smallest mass [1]
In general, planets with greater mass have greater gravitational field strength [1]
However, the size (radius) of the planet also matters — Neptune has more mass than Earth but its larger radius means g is only slightly larger (11.2 vs 9.8 N/kg); gravitational field strength depends on both mass and radius [1]
⚠ If you missed marks here: Stopping at "more mass means more gravity" caps you at two marks — the data contradicts a pure mass rule and mark 3 is for spotting it. Neptune has 17 times Earth's mass yet only 11.2 N/kg against 9.8 N/kg, because its radius is far larger and field strength falls with distance from the centre. Quote figures from the table; a comparison with no numbers in it rarely scores the first mark either.
(b) [3]
Distinguish between the inner (terrestrial) planets and the outer (gas giant) planets. Give two differences in your answer.

Mark Scheme — 2(b)

Inner planets (Mercury, Venus, Earth, Mars) are made of rock and metal (terrestrial/rocky); outer planets (Jupiter, Saturn, Uranus, Neptune) are made mainly of gas and liquid (gas/ice giants) [1]
Difference 1: Inner planets are smaller in size/diameter compared to outer planets which are much larger [1]
Difference 2: Inner planets are closer to the Sun and have shorter orbital periods; outer planets are further from the Sun with longer orbital periods [1]
⚠ If you missed marks here: Two differences means two different properties — "outer planets are large" and "inner planets are small" are one difference written twice and are credited once. Choose from composition (rock and metal against gas and ice), size, and distance with the orbital period that follows from it. Naming the planets in each group is expected in mark 1.
(c) [3]
Describe the accretion model of the formation of the Solar System. Include the role of gravity in your answer.

Mark Scheme — 2(c)

A large cloud of gas and dust (nebula) began to collapse under gravity, forming a spinning disc of material [1]
Dust particles collided and stuck together (accreted) to form larger and larger clumps called planetesimals [1]
As planetesimals grew larger, their gravity attracted more material, eventually forming protoplanets and then the planets we see today; the central mass became the Sun [1]
⚠ If you missed marks here: "The planets broke off the Sun" is a discarded idea and earns nothing. The question asks specifically for the role of gravity, so it must appear twice: gravity collapsing the nebula into a spinning disc, then the growing planetesimals attracting more material because their own gravity has increased. Mark 2 needs the word accretion or a clear description of particles colliding and sticking.
(d) [3]
Calculate the time taken for light to travel from the Sun to Jupiter. (Sun–Jupiter distance = 7.8 × 1011 m, speed of light = 3.0 × 108 m/s). Give your answer in minutes.

Mark Scheme — 2(d)

Correct formula: t = d / v [1]
Substitution: t = 7.8 × 1011 ÷ 3.0 × 108 = 2600 s [1]
Conversion to minutes: 2600 ÷ 60 = 43.3 minutes (accept 43 minutes) [1]
⚠ If you missed marks here: The powers are the danger: 1011 ÷ 108 = 103, giving 2600 s — an answer of 2.6 × 1019 s means the indices were added. Then divide by 60 for 43.3 minutes; multiplying gives 156 000 minutes, and leaving 2600 s as the final answer loses mark 3 because the question specifies minutes.
Question 3 — The Sun and Nuclear Fusion
Total: 12 marks
(a) [2]
State what is meant by 'nuclear fusion' and identify where in the Sun this process occurs.

Mark Scheme — 3(a)

Nuclear fusion is the joining (fusing) of two light nuclei to form a heavier nucleus, releasing energy [1]
This occurs in the core (centre) of the Sun [1]
⚠ If you missed marks here: "Nuclei break apart" is fission and scores zero on a fusion question. Mark 1 wants two light nuclei joining to form a heavier nucleus with energy released — and it is nuclei, not atoms or molecules. Mark 2 needs the core specifically; "in the Sun" or "in the middle of the star" without naming the core is usually not accepted.
(b) [4]
Describe the conditions necessary for nuclear fusion to occur in the Sun and explain why these conditions are required.

Mark Scheme — 3(b)

Extremely high temperature is required (millions of degrees Celsius / approximately 15 million °C in the Sun's core) [1]
High temperature gives the hydrogen nuclei (protons) sufficient kinetic energy to overcome the electrostatic repulsion between the positively charged nuclei [1]
Extremely high pressure / density is also required [1]
High pressure forces the nuclei close enough together so that the strong nuclear force can act, allowing them to fuse [1]
⚠ If you missed marks here: Listing "very hot and very high pressure" answers half the question and takes two marks out of four, because each condition must be given its reason. High temperature supplies the nuclei with enough kinetic energy to overcome the electrostatic repulsion between two positive charges; high pressure and density force them close enough for the strong nuclear force to take hold. Naming 15 million °C helps mark 1.
(c) [3]
The Sun emits infrared, visible light, and ultraviolet radiation. For each type, state one way it affects the Earth.

Mark Scheme — 3(c)

Infrared: warms the Earth's surface / heats the atmosphere / causes thermal effects (e.g. warming oceans, weather patterns) [1]
Visible light: provides light for photosynthesis / enables us to see / illuminates the Earth during daytime [1]
Ultraviolet: can cause sunburn / skin cancer / is partially absorbed by the ozone layer / causes tanning [1]
⚠ If you missed marks here: Each of the three needs its own distinct effect, so writing that visible light "heats the Earth" reuses the infrared point and gains nothing for visible. Safe pairings are infrared with warming of surface, atmosphere or oceans; visible with photosynthesis or seeing; ultraviolet with sunburn, skin cancer or absorption by the ozone layer. An effect with no radiation named beside it cannot be credited.
(d) [3]
Scientists at the Culham Centre for Fusion Energy in Oxfordshire are attempting to replicate nuclear fusion on Earth. Suggest why achieving controlled fusion on Earth is extremely difficult.

Mark Scheme — 3(d)

The plasma must be heated to temperatures of over 100 million °C, which is extremely difficult to achieve and maintain [1]
No known material can contain plasma at such temperatures, so magnetic confinement (e.g. tokamak) must be used to keep the plasma from touching the walls [1]
The energy input currently required to sustain the reaction exceeds the energy output, making it not yet commercially viable / maintaining stable confinement is technically challenging [1]
⚠ If you missed marks here: "We do not have the technology yet" is a restatement of the question and scores nothing. Give the physics: the plasma has to exceed 100 million °C, no solid material can touch it at that temperature so it is held away from the walls by magnetic fields in a tokamak, and the energy put in still exceeds the energy released. Do not drift into radioactive waste or meltdowns — those belong to fission reactors.
Question 4 — Stars
Total: 12 marks
(a) [3]
Describe the formation of a star from a nebula up to the point where it joins the main sequence.

Mark Scheme — 4(a)

A nebula is a large cloud of gas (mainly hydrogen) and dust; gravity causes the cloud to collapse / contract [1]
As the cloud contracts, gravitational potential energy is converted to thermal (kinetic) energy, and the temperature and pressure at the centre increase; this forms a protostar [1]
When the core temperature is high enough (about 15 million °C), nuclear fusion of hydrogen begins; the star reaches a stable state (main sequence) where the outward radiation pressure balances the inward gravitational force [1]
⚠ If you missed marks here: The link that disappears from most answers is the energy conversion in mark 2: as the cloud contracts, gravitational potential energy becomes thermal energy, which is why the centre heats up. "It gets hot" without saying why is weak. Mark 3 needs the main sequence defined as a balance — outward radiation pressure from fusion against inward gravitational collapse — and not simply "it becomes a star".
(b) [3]
Describe what happens to a Sun-like star after it leaves the main sequence. Use the terms: red giant, planetary nebula, and white dwarf.

Mark Scheme — 4(b)

When the hydrogen fuel in the core runs out, the star expands and cools, becoming a red giant [1]
The outer layers of the red giant are ejected into space, forming a planetary nebula [1]
The remaining core contracts under gravity to form a hot, dense white dwarf, which gradually cools over time [1]
⚠ If you missed marks here: A star the mass of the Sun does not explode, so any mention of a supernova here contradicts the physics and costs marks. The trigger for mark 1 is hydrogen in the core running out, after which the star expands and cools into a red giant. Mark 2 is the ejection of the outer layers as a planetary nebula (nothing to do with planets), and mark 3 is the dense remaining core cooling as a white dwarf.
(c) [4]
A star much more massive than the Sun follows a different path after the main sequence. Describe this path, using the terms: red supergiant, supernova, and state the two possible remnants.

Mark Scheme — 4(c)

When a massive star's core fuel is exhausted, it expands to become a red supergiant (much larger than a red giant) [1]
The red supergiant undergoes a catastrophic collapse and then a massive explosion called a supernova, which scatters heavier elements into space [1]
If the remaining core is moderately massive, it collapses to form a neutron star (an extremely dense object made almost entirely of neutrons) [1]
If the remaining core is very massive, it collapses further to form a black hole (an object whose gravitational field is so strong that not even light can escape) [1]
⚠ If you missed marks here: A black hole is not "a hole in space" — describe it as an object whose gravitational field is so strong that not even light can escape, or mark 4 is at risk. The two remnants must be tied to how much core mass survives the explosion, moderately massive giving a neutron star and very massive giving a black hole. Mark 2 needs the word supernova together with heavier elements being scattered into space, not the explosion alone.
(d) [2]
The Crab Nebula, observed through the Lovell Telescope at Jodrell Bank, is the remnant of a supernova recorded by Chinese and Arab astronomers in 1054 AD. The Crab Nebula is approximately 6500 light-years away. Calculate this distance in metres. (1 light-year = 9.5 × 1015 m)

Mark Scheme — 4(d)

Correct method: d = 6500 × 9.5 × 1015 [1]
Correct answer: d = 6.175 × 1019 m ≈ 6.2 × 1019 m [1]
⚠ If you missed marks here: 6500 × 9.5 = 61 750, and 61 750 × 1015 is 6.175 × 1019 m, since converting 61 750 into 6.175 moves the point four places and so adds 4 to the index. Answers of 6.175 × 1018 come from adding only 3. Treating 9.5 × 1015 as kilometres makes the distance a thousand times too large.
Question 5 — Galaxies and the Milky Way
Total: 10 marks
(a) [2]
Define the term 'galaxy' and name the galaxy in which our Solar System is located.

Mark Scheme — 5(a)

A galaxy is a collection of billions of stars held together by gravity [1]
Our Solar System is located in the Milky Way galaxy [1]
⚠ If you missed marks here: Answering "the Universe" or "the Solar System" to the naming part is a scale error and gains nothing — the Solar System sits inside the Milky Way, which is one of billions of galaxies. The definition mark needs both the contents (billions of stars) and what holds them there (gravity); "a large group of stars" alone often misses the gravity point.
(b) [2]
State the approximate diameter of the Milky Way galaxy in light-years.

Mark Scheme — 5(b)

The Milky Way galaxy has a diameter of approximately 100,000 light-years [1]
The Sun is located about 26,000 light-years from the centre of the Milky Way (accept any additional relevant detail about the scale) [1]
⚠ If you missed marks here: 100 000 light-years is a distance; writing "100 000 years" converts it into a time and loses the mark. The second mark is about our place in it — roughly 26 000 light-years out from the centre, not at the centre. Confusing the galaxy's diameter with the distance to the nearest star (4.2 light-years) puts the scale out by a factor of about 24 000.
(c) [3]
The star Proxima Centauri is approximately 4.2 light-years from Earth. Calculate how long it would take light from Proxima Centauri to reach Earth. Then calculate this distance in metres. (1 light-year = 9.5 × 1015 m)

Mark Scheme — 5(c)

Light from Proxima Centauri takes 4.2 years to reach Earth (this is the definition of light-years — the distance is 4.2 light-years, so light takes 4.2 years) [1]
Correct method: d = 4.2 × 9.5 × 1015 [1]
Correct answer: d = 3.99 × 1016 m ≈ 4.0 × 1016 m [1]
⚠ If you missed marks here: Mark 1 requires no calculation at all: 4.2 light-years means the light has been travelling for 4.2 years, straight from the definition. Answers that convert to 1.3 × 108 s and stop have done extra work and still not stated the time in years. For the distance, d = 4.2 × 9.5 × 1015 = 3.99 × 1016 m; dropping the 9.5 gives 4.2 × 1015, more than nine times too small.
(d) [3]
Explain why it is not possible for humans to travel to even the nearest stars with current technology. Use specific numbers in your answer.

Mark Scheme — 5(d)

The nearest star (Proxima Centauri) is 4.2 light-years away, which is approximately 4.0 × 1016 m — an enormous distance [1]
Current spacecraft travel at speeds far less than the speed of light; for example, the fastest spacecraft travel at roughly 70,000 km/h, which would take over 50,000 years to reach Proxima Centauri [1]
No current technology can sustain human life for such a journey (food, fuel, radiation shielding), and no propulsion system can accelerate a crewed spacecraft to a significant fraction of the speed of light [1]
⚠ If you missed marks here: The question says use specific numbers, so "it is much too far away" scores nothing however true it is. Anchor the answer: 4.2 light-years is about 4.0 × 1016 m, and at the roughly 70 000 km/h of the fastest spacecraft that is over 50 000 years of travel. Mark 3 is the human side — no way to carry enough food or fuel, or to shield a crew for that long.
Question 6 — The Expanding Universe
Total: 12 marks
(a) [3]
A galaxy is observed to have a recession velocity of 4.4 × 106 m/s. Using the Hubble constant H₀ = 2.2 × 10−18 s−1 and the equation v = H₀d, calculate the distance to this galaxy in metres and convert to light-years. (1 light-year = 9.5 × 1015 m)

Mark Scheme — 6(a)

Rearranging: d = v / H₀ = 4.4 × 106 ÷ 2.2 × 10−18 [1]
d = 2.0 × 1024 m [1]
In light-years: 2.0 × 1024 ÷ 9.5 × 1015 = 2.1 × 108 light-years ≈ 210 million light-years [1]
⚠ If you missed marks here: Since v = H₀d, the distance is v ÷ H₀; multiplying instead gives 9.7 × 10−12 m, which should look absurd for a galaxy. Dividing subtracts the indices, 6 − (−18) = 24, so the answer is 2.0 × 1024 m — 10−12 means the minus sign was not carried through. Converting to light-years divides by 9.5 × 1015, giving 2.1 × 108; multiplying there gives 1.9 × 1040.
(b) [3]
Explain what is meant by 'redshift' and describe how it is evidence for an expanding Universe.

Mark Scheme — 6(b)

Redshift is the increase in the observed wavelength (or decrease in frequency) of light from distant galaxies, shifting the spectrum towards the red end [1]
This is caused by the galaxies moving away from us (recession); the Doppler effect stretches the wavelength of the light [1]
The further away a galaxy is, the greater its redshift (and recession velocity), which shows that the Universe is expanding — all distant galaxies are moving away from each other [1]
⚠ If you missed marks here: Redshift is not a galaxy glowing red or becoming redder over time — it is the shift of known spectral lines to longer wavelength, and mark 1 wants wavelength increasing or frequency decreasing. Mark 3 is not "galaxies are moving away" on its own; it is the pattern that more distant galaxies show more redshift, and therefore recede faster, which is what points to expanding space.
(c) [3]
Describe the cosmic microwave background radiation (CMBR) and explain why it is considered evidence for the Big Bang theory.

Mark Scheme — 6(c)

CMBR is low-energy microwave radiation that is detected uniformly from all directions in space [1]
According to the Big Bang theory, the early Universe was extremely hot and dense; as the Universe expanded and cooled, the radiation from this early hot period was stretched (redshifted) to microwave wavelengths [1]
The existence and uniformity of CMBR matches the predictions of the Big Bang theory and cannot be easily explained by other models, providing strong evidence that the Universe began from a hot, dense state [1]
⚠ If you missed marks here: Calling the CMBR "background noise" or radiation left over from the Sun and stars scores nothing. Mark 1 is that it arrives as low-energy microwaves uniformly from every direction; mark 2 is its origin as radiation from a hot dense early Universe, stretched to microwave wavelengths as space expanded. Mark 3 is the reasoning step — the observation matches what the Big Bang model predicted, which no rival model explains as neatly.
(d) [3]
Using H₀ = 2.2 × 10−18 s−1, estimate the age of the Universe. Show your working and give your answer in billions of years. (1 year ≈ 3.15 × 107 s)

Mark Scheme — 6(d)

Correct formula: age ≈ 1 / H₀ [1]
Substitution: age = 1 / (2.2 × 10−18) = 4.55 × 1017 s [1]
Conversion: 4.55 × 1017 ÷ 3.15 × 107 = 1.44 × 1010 years ≈ 14.4 billion years [1]
⚠ If you missed marks here: The command is billions of years, so an answer left as 4.55 × 1017 s, or even as 1.44 × 1010 years without the final statement, risks mark 3. Divide by 3.15 × 107; multiplying gives 1.43 × 1025. Note also that 14.4 billion is an estimate because 1/H₀ assumes the expansion rate has never changed, which is why the accepted figure is nearer 13.8 billion.
Question 7 — Orbits and Energy
Total: 10 marks
(a) [3]
Describe the shape of planetary orbits around the Sun. Explain how a planet's speed varies at different points in its orbit, using the principle of conservation of energy.

Mark Scheme — 7(a)

Planetary orbits are elliptical (not circular) with the Sun at one focus of the ellipse [1]
When the planet is closer to the Sun (at perihelion), it moves faster; when it is further from the Sun (at aphelion), it moves slower [1]
By conservation of energy: as the planet moves closer, gravitational potential energy decreases and is converted to kinetic energy (so speed increases); as it moves further away, kinetic energy is converted back to gravitational potential energy (so speed decreases) [1]
⚠ If you missed marks here: The shape mark needs the Sun at one focus of the ellipse, not at the centre — describing the orbit as a circle loses it outright. For mark 3, "the Sun pulls harder when it is near so it goes faster" is not the conservation-of-energy statement being asked for; write it as a transfer, gravitational potential energy converting into kinetic energy on the way in and back again on the way out.
(b) [4]
Mars orbits the Sun at a mean distance of 2.3 × 1011 m. Mars takes 5.94 × 107 s to complete one orbit. Calculate the orbital speed of Mars using v = 2πr / T.

Mark Scheme — 7(b)

Correct formula stated: v = 2πr / T [1]
Correct substitution: v = 2 × π × 2.3 × 1011 / 5.94 × 107 [1]
Numerator calculated: 2π × 2.3 × 1011 = 1.445 × 1012 m (accept 1.44 – 1.45 × 1012) [1]
Final answer: v = 1.445 × 1012 ÷ 5.94 × 107 = 24,300 m/s ≈ 2.4 × 104 m/s [1]
⚠ If you missed marks here: Turning the formula upside down as T / 2πr produces 4.1 × 10−5, a number that cannot be a speed. Keep the indices straight in the last step: 1.445 × 1012 ÷ 5.94 × 107 subtracts to 105, and 0.243 × 105 is 2.43 × 104 m/s — reporting 2.43 × 105 m/s means the leading 0.243 was not rewritten in standard form.
(c) [3]
The gravitational force between the Sun and a planet provides the centripetal force needed for orbital motion. Explain what would happen to a planet's orbital speed if it were moved to an orbit with a larger radius, and why.

Mark Scheme — 7(c)

At a larger orbital radius, the gravitational force (and hence centripetal force) acting on the planet is weaker, because gravitational force decreases with increasing distance [1]
With a weaker centripetal force, the planet needs a lower orbital speed to maintain a stable circular/elliptical orbit [1]
Therefore, the planet's orbital speed decreases as the orbital radius increases; planets further from the Sun orbit more slowly (e.g. Neptune orbits more slowly than Earth) [1]
⚠ If you missed marks here: "It has further to travel so it is slower" describes the orbital period, not the speed, and misses the physics the question asks for. Mark 1 is that gravitational force weakens with distance, so the available centripetal force is smaller; mark 2 is that a smaller centripetal force can only hold a slower orbit. Saying the planet would fly off into space is wrong — a stable orbit still exists at the larger radius, simply at a lower speed, as Neptune shows against Earth.

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IGCSE Physics 0625/0972 — Paper 4 Theory (Extended) — Topic 6: Space Physics — Mock Exam 2
For examination preparation only. Not an official Cambridge Assessment paper.