IGCSE Physics Paper 4 (Theory / Extended)

Topic 6: Space Physics — Mock Exam 2

Duration1 hour 15 minutes

Total Marks80

Questions7

Time Remaining75:00

Syllabus0625 / 0972

Instructions to Candidates

Answer all questions in the spaces provided.
Show all working for calculation questions — marks are awarded for method.
Use the timer to pace yourself: aim for roughly 1 minute per mark.
When you have finished, click Submit Exam to enter marking mode.
In marking mode, compare your answers with the model answers and tick the marks you earned.
Data: speed of light c = 3.0 × 10⁸ m/s | 1 light-year = 9.5 × 10¹⁵ m | H₀ = 2.2 × 10⁻¹⁸ s⁻¹

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]

Mark 1 -- Earth rotates on axis so different parts face towards or away from the Sun Mark 2 -- Daytime when facing the Sun, night-time when rotated away from the Sun

(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]

Mark 3 -- Earth's axis tilted at 23.5 degrees to the perpendicular of its orbital plane Mark 4 -- In summer the Northern Hemisphere is tilted towards the Sun, giving longer daylight hours Mark 5 -- Sun's rays strike more directly, concentrating energy over smaller area, so warmer temperatures

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]

Mark 6 -- New Moon not visible; waxing crescent is thin sliver of light growing on right side Mark 7 -- First quarter has right half illuminated; full Moon has entire face illuminated Mark 8 -- Waning gibbous more than half lit but decreasing; third quarter has left half illuminated Mark 9 -- Complete cycle takes approximately 28 days or one lunar month

(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]

Mark 10 -- Moon orbits Earth in approximately 28 days in same direction as Earth's rotation Mark 11 -- Moon moves further along orbit each day so Earth must rotate more than 360 degrees to face Moon again Mark 12 -- Extra rotation takes approximately 50 minutes, about 13 degrees per day of extra rotation needed

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]

Mark 13 -- Jupiter has greatest g (24.8 N/kg) and greatest mass; Mercury has smallest g (3.7 N/kg) and smallest mass Mark 14 -- Greater mass generally gives greater gravitational field strength Mark 15 -- Size or radius also affects g; gravitational field strength depends on both mass and radius of planet

(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]

Mark 16 -- Inner planets are rocky or made of rock and metal; outer planets are gas or ice giants Mark 17 -- Inner planets are smaller; outer planets are much larger in size Mark 18 -- Inner planets closer to Sun with shorter orbital periods; outer planets further with longer periods

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]

Mark 19 -- Cloud of gas and dust (nebula) collapsed under gravity forming a spinning disc Mark 20 -- Dust particles collided and stuck together (accreted) forming planetesimals Mark 21 -- Planetesimals grew, gravity attracted more material, forming protoplanets then planets; central mass became the Sun

(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]

Mark 22 -- Correct formula used: t = d / v or time = distance divided by speed Mark 23 -- Correct substitution giving 2600 seconds Mark 24 -- Correct conversion to 43.3 minutes or approximately 43 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]

Mark 25 -- Nuclear fusion is joining or fusing of two light nuclei to form a heavier nucleus, releasing energy Mark 26 -- Fusion occurs in the core or centre of the Sun

(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]

Mark 27 -- Extremely high temperature required, millions of degrees in Sun's core Mark 28 -- High temperature gives nuclei sufficient kinetic energy to overcome electrostatic repulsion Mark 29 -- Extremely high pressure or density is also required Mark 30 -- High pressure forces nuclei close enough for the strong nuclear force to act allowing fusion

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]

Mark 31 -- Infrared warms the Earth's surface or heats the atmosphere Mark 32 -- Visible light provides light for photosynthesis or enables vision or illuminates the Earth Mark 33 -- Ultraviolet causes sunburn or skin cancer or is absorbed by ozone layer

(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]

Mark 34 -- Plasma must be heated to over 100 million degrees, extremely difficult to achieve and maintain Mark 35 -- No material can contain plasma at such temperatures; magnetic confinement such as tokamak needed Mark 36 -- Energy input exceeds energy output or maintaining stable confinement is extremely challenging

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]

Mark 37 -- Nebula is large cloud of gas and dust; gravity causes it to collapse or contract Mark 38 -- Gravitational potential energy converts to thermal energy; temperature and pressure increase forming protostar Mark 39 -- Nuclear fusion begins when core is hot enough; radiation pressure balances gravity giving stable main sequence 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]

Mark 40 -- Hydrogen fuel runs out, star expands and cools to become a red giant Mark 41 -- Outer layers ejected into space forming a planetary nebula Mark 42 -- Core contracts under gravity to form a hot dense white dwarf which gradually cools

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]

Mark 43 -- Massive star expands to become a red supergiant when core fuel is exhausted Mark 44 -- Red supergiant undergoes collapse and massive explosion called a supernova Mark 45 -- Remnant 1: neutron star if remaining core is moderately massive Mark 46 -- Remnant 2: black hole if remaining core is very massive, gravity so strong light cannot escape

(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]

Mark 47 -- Correct method: distance = 6500 multiplied by 9.5 times 10 to the 15 Mark 48 -- Correct answer approximately 6.2 times 10 to the 19 metres

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]

Mark 49 -- Galaxy defined as a collection of billions of stars held together by gravity Mark 50 -- Our Solar System is in the Milky Way galaxy

(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]

Mark 51 -- Milky Way diameter approximately 100,000 light-years Mark 52 -- Sun located approximately 26,000 light-years from the centre of the Milky Way or other relevant scale detail

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]

Mark 53 -- Light takes 4.2 years to reach Earth from Proxima Centauri Mark 54 -- Correct method: distance = 4.2 multiplied by 9.5 times 10 to the 15 Mark 55 -- Correct answer approximately 4.0 times 10 to the 16 metres

(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]

Mark 56 -- Nearest star is 4.2 light-years or 4.0 times 10 to the 16 metres away, enormous distance Mark 57 -- Current spacecraft far slower than light; would take tens of thousands of years to reach nearest star Mark 58 -- Cannot sustain human life for such duration or no propulsion to reach significant fraction of light speed

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]

Mark 59 -- Correct rearrangement: d = v divided by H0 Mark 60 -- Correct answer: d = 2.0 times 10 to the 24 metres Mark 61 -- Correct conversion: approximately 2.1 times 10 to the 8 light-years or 210 million light-years

(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]

Mark 62 -- Redshift is increase in observed wavelength of light from distant galaxies, shifted towards red end Mark 63 -- Caused by galaxies moving away; Doppler effect stretches the wavelength Mark 64 -- Further galaxies have greater redshift, showing Universe is expanding with all galaxies moving apart

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]

Mark 65 -- CMBR is low-energy microwave radiation detected uniformly from all directions in space Mark 66 -- Early Universe was hot and dense; radiation stretched to microwave wavelengths as Universe expanded and cooled Mark 67 -- CMBR matches Big Bang predictions; provides strong evidence Universe began from hot dense state

(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]

Mark 68 -- Correct formula: age approximately equals 1 divided by H0 Mark 69 -- Correct calculation: 4.55 times 10 to the 17 seconds Mark 70 -- Correct conversion: approximately 14.4 billion years

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]

Mark 71 -- Orbits are elliptical with the Sun at one focus of the ellipse Mark 72 -- Planet moves faster when closer to Sun (perihelion) and slower when further (aphelion) Mark 73 -- Conservation of energy: closer means less GPE and more KE so faster; further means more GPE and less KE so slower

(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]

Mark 74 -- Correct formula stated: v = 2 pi r divided by T Mark 75 -- Correct substitution of values into formula Mark 76 -- Numerator calculated correctly: approximately 1.45 times 10 to the 12 metres Mark 77 -- Final answer approximately 24300 m/s or 2.4 times 10 to the 4 m/s

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]

Mark 78 -- Gravitational force is weaker at larger radius because gravity decreases with distance Mark 79 -- Weaker centripetal force means lower orbital speed needed for stable orbit Mark 80 -- Orbital speed decreases as orbital radius increases; outer planets orbit more slowly