← Study Hub

IGCSE Physics Paper 4 (Theory / Extended)

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

Instructions

Total Score
0 / 80
Percentage
0%
A* (56+)
A (48-55)
B (40-47)
C (32-39)
D (24-31)
E (16-23)
U (<16)
Question 1: The Earth
12 marks
(a) [2]
Explain why we experience day and night on Earth.

Mark Scheme — 1(a)

The Earth rotates (spins) on its axis once every 24 hours [1]
The side of the Earth facing the Sun experiences daytime; the side facing away from the Sun experiences night-time [1]
⚠ If you missed marks here: "The Earth goes round the Sun" earns nothing here — the orbit takes a year, and it is the Earth spinning on its own axis once every 24 hours that produces day and night. The second mark needs both halves stated together: the side facing the Sun is in daylight while the opposite side is in darkness. "The Sun rises and sets" describes what you see instead of explaining it.
(b) [3]
Explain why the Earth experiences different seasons during its orbit around the Sun.

Mark Scheme — 1(b)

The Earth's axis is tilted at approximately 23.5° to the plane of its orbit (the ecliptic) [1]
When a hemisphere is tilted towards the Sun, it receives more direct (concentrated) sunlight and longer days, producing summer [1]
When a hemisphere is tilted away from the Sun, it receives less direct sunlight and has shorter days, producing winter [1]
⚠ If you missed marks here: The answer examiners see most often is "the Earth is closer to the Sun in summer", and it scores zero — if distance were the cause, both hemispheres would have summer at once. Mark 1 is the tilt of about 23.5° to the orbital plane, and marks 2 and 3 must contrast the two hemispheres: tilted towards the Sun gives more concentrated sunlight and longer days, tilted away gives the reverse.
(c) [3]
Describe the appearance of the Moon at each of the following phases: new Moon, first quarter, full Moon, and third quarter. In each case, describe how much of the Moon's visible surface appears illuminated as seen from Earth.

Mark Scheme — 1(c)

New Moon: the Moon is between the Earth and Sun; the illuminated side faces away from Earth, so the Moon appears dark / not visible [1]
First quarter and third quarter: half of the Moon's visible surface appears illuminated (right half for first quarter, left half for third quarter as seen from the Northern Hemisphere) [1]
Full Moon: the Earth is between the Sun and Moon; the entire visible surface of the Moon is illuminated [1]
⚠ If you missed marks here: "The Earth's shadow covers part of the Moon" is the classic wrong cause — that describes a lunar eclipse, not the phases. Phases depend on how much of the Moon's lit half is turned towards us. Also say "half of the visible face is lit" at first and third quarter; writing "a quarter of the Moon is lit" loses that mark, and full Moon needs the Earth positioned between Sun and Moon.
(d) [4]
The Earth orbits the Sun at a mean distance of 1.5 × 1011 m. The orbital period is 3.15 × 107 s. Calculate the orbital speed of the Earth. Use v = 2πr/T. Show all working.

Mark Scheme — 1(d)

Correct formula stated: v = 2πr / T [1]
Correct substitution: v = (2 × π × 1.5 × 1011) / (3.15 × 107) [1]
Correct evaluation of numerator: 2πr = 9.42 × 1011 m [1]
Correct final answer: v = 29 900 m/s (or 3.0 × 104 m/s) with correct unit [1]
⚠ If you missed marks here: Dropping the 2π and working out r / T gives 4760 m/s instead of 29 900 m/s — a factor of 6.28 out, and the commonest loss on this part. Check the powers as well: 1011 ÷ 107 = 104, so the answer sits in the tens of thousands. The final mark needs m/s written next to the number.
Question 2: The Solar System
12 marks
(a) [2]
List the eight planets of the Solar System in order of increasing distance from the Sun.

Mark Scheme — 2(a)

Inner planets in correct order: Mercury, Venus, Earth, Mars [1]
Outer planets in correct order: Jupiter, Saturn, Uranus, Neptune [1]
⚠ If you missed marks here: Including Pluto costs you the second mark — there are eight planets, and Pluto has been a dwarf planet since 2006. Each mark is all-or-nothing for its group of four, so swapping Uranus and Neptune, or Venus and Mercury, loses that whole mark rather than half of it. Order runs outwards from the Sun, not inwards.
(b) [3]
Compare the properties of the inner planets (Mercury to Mars) with the outer planets (Jupiter to Neptune).

Mark Scheme — 2(b)

Inner planets are smaller in size and composed of rock (solid surfaces) with higher density [1]
Outer planets are much larger (gas giants) composed mainly of hydrogen and helium with lower density [1]
Inner planets have few or no moons; outer planets have many moons and ring systems [1]
⚠ If you missed marks here: A one-sided statement such as "outer planets are big" scores nothing, because the question says compare — every mark needs both halves in the same sentence. Pair them up: small and rocky with high density against large and gaseous with low density, few or no moons against many moons plus rings. "Outer planets are colder" is true but is not one of the three marking points.
(c) [4]
Describe the accretion model of how the Solar System formed from a cloud of gas and dust.

Mark Scheme — 2(c)

A large cloud of gas and dust (nebula) began to collapse under the force of gravity [1]
The cloud began to spin and flattened into a rotating disc (protoplanetary disc) [1]
Particles within the disc collided and stuck together, forming progressively larger clumps (accretion) [1]
The centre of the disc became hot and dense enough for nuclear fusion to begin, forming the Sun; the remaining material in the disc accreted to form the planets [1]
⚠ If you missed marks here: Reaching for the Big Bang here earns nothing — that is the origin of the Universe, not of one solar system, and the two are about 9 billion years apart. Four separate stages are wanted: nebula collapsing under gravity, spinning and flattening into a disc, particles colliding and sticking (accretion), then fusion starting at the hot dense centre to form the Sun. Answers that say "gravity pulled it together and made planets" collapse three marks into one.
(d) [3]
Light travels at 3.0 × 108 m/s. The mean distance from the Sun to Mars is 2.3 × 1011 m. Calculate the time for light to travel from the Sun to Mars. Express your answer in minutes.

Mark Scheme — 2(d)

Correct formula and substitution: t = d / v = (2.3 × 1011) / (3.0 × 108) [1]
Correct calculation: t = 767 s (accept 766.7 s) [1]
Correct conversion to minutes: 767 / 60 = 12.8 minutes (accept 12.7 – 12.8 min) [1]
⚠ If you missed marks here: Multiplying by 60 rather than dividing turns 767 s into 46 000 minutes; the conversion mark needs 767 ÷ 60 = 12.8 min. Watch the formula direction too — v / d instead of d / v gives 1.3 × 10−3, a number far too small to be a travel time. Leaving the answer as 767 with no unit named loses the final mark even though the arithmetic is right.
Question 3: The Sun
12 marks
(a) [2]
State the two main elements that make up the Sun and identify which type of star the Sun is classified as.

Mark Scheme — 3(a)

The Sun is composed mainly of hydrogen and helium [1]
The Sun is classified as a medium-sized star (accept: main sequence star / yellow dwarf) [1]
⚠ If you missed marks here: "The Sun is burning gas" is wrong on both counts — nothing is burning (there is no oxygen and no combustion), and the marking point is that it is made mainly of hydrogen and helium. For the second mark "a star" is too vague; you need medium-sized, main sequence or yellow dwarf. Calling it a giant or a supergiant contradicts the physics of a star still fusing hydrogen steadily.
(b) [4]
Describe the process of nuclear fusion that occurs in the core of the Sun. Include the conditions required and the products formed.

Mark Scheme — 3(b)

Hydrogen nuclei (protons) fuse together in the core of the Sun [1]
Extremely high temperature (~15 million °C) and very high pressure are required to overcome the electrostatic repulsion between nuclei [1]
Hydrogen nuclei fuse to form helium nuclei [1]
Energy is released because the total mass of the products is less than the total mass of the reactants (mass is converted to energy) [1]
⚠ If you missed marks here: If you wrote that nuclei are "split apart", you have described fission, the opposite process, and that sentence scores nothing. Fusion needs hydrogen nuclei joining to make helium, an extremely high temperature (about 15 million °C) and pressure to push past the electrostatic repulsion between two positive charges, and the energy explained as products having less total mass than reactants.
(c) [3]
The Sun emits electromagnetic radiation across a range of wavelengths. Name three types of radiation emitted by the Sun and state one effect of each on Earth.

Mark Scheme — 3(c)

Infrared radiation — causes heating / warms the Earth's surface [1]
Visible light — enables photosynthesis in plants and allows vision [1]
Ultraviolet (UV) radiation — can cause sunburn, skin cancer, or vitamin D production [1]
⚠ If you missed marks here: The usual loss is naming three radiations but attaching effects to only two — each mark needs the type and its effect together in the same statement. Keep the effects distinct: infrared warms the surface, visible light drives photosynthesis and vision, ultraviolet causes sunburn or skin cancer. "UV causes global warming" mixes up two ideas and gains nothing; warming is infrared.
(d) [3]
Tim Peake observed the Sun from the International Space Station. Explain why the Sun appears yellow-white from space but can appear orange or red at sunrise when viewed from the Earth's surface.

Mark Scheme — 3(d)

The Sun emits light across all visible wavelengths; in space there is no atmosphere to scatter light, so it appears yellow-white [1]
At sunrise/sunset, light from the Sun passes through a much greater thickness of the Earth's atmosphere [1]
Shorter wavelengths (blue/violet) are scattered more by molecules in the atmosphere, so mainly the longer wavelengths (orange/red) reach the observer [1]
⚠ If you missed marks here: Saying "the Sun turns red in the evening" gets no credit — the light leaving the Sun is unchanged, and it is the Earth's atmosphere that alters what reaches your eye. Explain the path: near sunrise the light travels through far more atmosphere, and shorter blue wavelengths are scattered away most strongly, leaving the longer orange and red wavelengths to reach you. "The atmosphere reflects red light" is not the scattering point and scores zero.
Question 4: Stars and Stellar Life Cycle
12 marks
(a) [2]
Define the term 'light-year' and state its value in metres.

Mark Scheme — 4(a)

A light-year is the distance that light travels in one year (in a vacuum) [1]
1 light-year = 9.5 × 1015 m [1]
⚠ If you missed marks here: A light-year is a distance, not a time — write "the time light takes to travel in a year" and the first mark goes. State it as the distance light travels in one year in a vacuum. For the value, 9.5 × 1015 m is required; quoting 9.5 × 1015 km is out by a factor of 1000 and 9.5 × 1012 m is the kilometre figure in disguise.
(b) [4]
Describe the complete life cycle of a star with a similar mass to the Sun, starting from a nebula.

Mark Scheme — 4(b)

A nebula (cloud of gas and dust) collapses under gravity; gravitational potential energy is converted to thermal energy, forming a protostar [1]
The protostar becomes hot and dense enough for hydrogen nuclear fusion to begin, and it becomes a stable main sequence star [1]
When hydrogen fuel in the core is depleted, the star expands and cools to become a red giant [1]
The outer layers are ejected as a planetary nebula, leaving a hot, dense white dwarf which gradually cools and fades [1]
⚠ If you missed marks here: Sending a Sun-like star to a supernova and a neutron star describes the wrong life cycle entirely and scores nothing on any of the four marks. The route is nebula → protostar → main sequence → red giant → planetary nebula → white dwarf, and the order carries marks. Mark 1 also wants the energy change — gravitational potential energy converting to thermal energy as the cloud contracts — not merely "gas comes together".
(c) [4]
Describe how the life cycle of a star much more massive than the Sun differs after the main sequence stage.

Mark Scheme — 4(c)

After the main sequence, the massive star expands into a red supergiant (much larger than a red giant) [1]
Heavier elements are fused in the core in successive stages (e.g. helium, carbon, up to iron) [1]
The star explodes violently as a supernova, scattering heavy elements into space [1]
The remaining core collapses to form either a neutron star or, if sufficiently massive, a black hole [1]
⚠ If you missed marks here: Writing "red giant" where the mark scheme wants red supergiant loses mark 1, since the question is specifically about what differs from the Sun's path. The remnant marks depend on the mass of the core left behind: a moderately massive core becomes a neutron star, only a very massive one becomes a black hole, so "all massive stars end as black holes" is wrong. Mark 3 also wants the supernova named and the scattering of heavy elements mentioned.
(d) [2]
Explain why astronomers use light-years rather than kilometres to measure distances between stars.

Mark Scheme — 4(d)

Distances between stars are extremely large; using kilometres would result in inconveniently large numbers that are difficult to comprehend [1]
Light-years provide more manageable figures and also convey how long light takes to travel between objects, giving a sense of scale [1]
⚠ If you missed marks here: "Kilometres are too small" on its own is not enough for either mark — you have to say what goes wrong, that distances between stars run to numbers like 4 × 1013 km, which are unwieldy and hard to picture. The second mark is the extra advantage: a light-year also tells you the travel time of the light, so 4.2 light-years immediately means the light left 4.2 years ago. Answers about light travelling through a vacuum miss the question.
Question 5: Galaxies
10 marks
(a) [2]
State what is meant by a 'galaxy'.

Mark Scheme — 5(a)

A galaxy is a large collection of billions of stars (and gas and dust) [1]
The stars are held together by mutual gravitational attraction [1]
⚠ If you missed marks here: Describing a galaxy as "a group of planets" or "a solar system" is a scale error and gets nothing — a galaxy holds billions of stars, along with gas and dust. The second mark is the mechanism holding it together: mutual gravitational attraction. "It is a part of the Universe" restates the question rather than defining anything.
(b) [3]
Describe the Milky Way galaxy, including its approximate diameter and the position of our Solar System within it.

Mark Scheme — 5(b)

The Milky Way is a spiral galaxy [1]
It has a diameter of approximately 100 000 light-years [1]
Our Solar System is located approximately two-thirds of the way out from the centre, in one of the spiral arms [1]
⚠ If you missed marks here: Placing the Solar System at the centre of the Milky Way costs mark 3 — we sit roughly two-thirds of the way out, inside one of the spiral arms. The shape mark needs the word spiral, and the size mark needs 100 000 light-years; writing 100 000 km or 100 000 years is either the wrong scale or the wrong quantity altogether.
(c) [3]
The nearest large galaxy to the Milky Way is the Andromeda Galaxy, approximately 2.5 million light-years away. Calculate this distance in metres. (1 light-year = 9.5 × 1015 m)

Mark Scheme — 5(c)

Correct substitution: d = 2.5 × 106 × 9.5 × 1015 [1]
Correct multiplication of coefficients: 2.5 × 9.5 = 23.75 [1]
Correct final answer: d = 2.375 × 1022 m (accept 2.4 × 1022 m) [1]
⚠ If you missed marks here: The trap is the power of ten at the end: 2.5 × 9.5 = 23.75, and 106 × 1015 = 1021, so 23.75 × 1021 must become 2.375 × 1022 — moving the point one place left raises the power by one. Answers of 2.375 × 1021 come from forgetting that step. Feeding in 2.5 rather than 2.5 × 106 gives 2.375 × 1016, a million times too close.
(d) [2]
Suggest why astronomers at Jodrell Bank Observatory use radio telescopes rather than optical telescopes to observe distant galaxies.

Mark Scheme — 5(d)

Radio waves can pass through dust clouds and the Earth's atmosphere, which may block or scatter visible light from distant objects [1]
Radio telescopes can detect signals at any time of day or night and are not significantly affected by weather conditions or light pollution [1]
⚠ If you missed marks here: "Radio waves travel faster" is false — every part of the electromagnetic spectrum travels at 3.0 × 108 m/s in a vacuum, so that sentence scores nothing. Mark 1 is that radio waves pass through interstellar dust and the atmosphere where visible light is blocked or scattered; mark 2 is practical, that a radio dish works in daylight, through cloud and despite light pollution.
Question 6: The Expanding Universe
12 marks
(a) [2]
Define the term 'redshift' as observed in the light from distant galaxies.

Mark Scheme — 6(a)

Redshift is an increase in the observed wavelength (decrease in frequency) of light received from a distant galaxy [1]
The absorption/spectral lines in the galaxy's spectrum are shifted towards the red (longer wavelength) end of the electromagnetic spectrum [1]
⚠ If you missed marks here: Redshift is not the galaxy turning red, and it is not the light slowing down — both statements are rejected. Mark 1 is that the observed wavelength increases (frequency decreases), and mark 2 is the observable evidence: known spectral or absorption lines appear shifted towards the red end of the spectrum. Without naming the lines the second mark is hard to secure.
(b) [3]
Explain how observations of redshift from distant galaxies provide evidence that the Universe is expanding.

Mark Scheme — 6(b)

Light from distant galaxies is observed to be redshifted, which shows that the galaxies are moving away from us (recessional velocity) [1]
More distant galaxies show a greater redshift, indicating they are receding at higher speeds (Hubble's law: v = H₀d) [1]
This pattern is observed in all directions, providing evidence that the Universe itself is expanding — space between galaxies is stretching [1]
⚠ If you missed marks here: Concluding that "everything moves away from Earth, so we are at the centre" misreads the evidence and loses mark 3 — the pattern appears in every direction from every galaxy, which is why the interpretation is that space itself is stretching. Mark 2 needs the relationship, not simply motion: the more distant the galaxy, the greater its redshift and recession speed (v = H₀d).
(c) [3]
Describe the Big Bang theory and state one piece of observational evidence that supports it (other than redshift).

Mark Scheme — 6(c)

The Big Bang theory states that the Universe began from a single, extremely hot and dense point (singularity) approximately 13–14 billion years ago and has been expanding and cooling ever since [1]
Cosmic microwave background radiation (CMBR) is detected coming uniformly from all directions in space [1]
CMBR is the remnant thermal radiation from the very early, extremely hot Universe, which has cooled to microwave wavelengths as the Universe expanded [1]
⚠ If you missed marks here: The question rules out redshift, so an answer built on redshift scores nothing at all here. Cosmic microwave background radiation is not radiation from the Sun or from distant stars — the marks want it detected uniformly from every direction, and explained as the leftover radiation from a hot dense early Universe, stretched to microwave wavelengths as space expanded. A date of about 13–14 billion years and the hot dense start are needed for mark 1.
(d) [4]
The Hubble constant H₀ = 2.2 × 10−18 s−1. Use the relationship age ≈ 1/H₀ to estimate the age of the Universe in seconds and convert your answer to years. (1 year = 3.15 × 107 s)

Mark Scheme — 6(d)

Correct formula: age ≈ 1 / H₀ [1]
Correct substitution and calculation: age = 1 / (2.2 × 10−18) = 4.55 × 1017 s [1]
Correct conversion: age in years = (4.55 × 1017) / (3.15 × 107) [1]
Correct final answer: age ≈ 1.44 × 1010 years (approximately 14.4 billion years) [1]
⚠ If you missed marks here: Dividing by a negative power is where this falls apart: 1 ÷ (2.2 × 10−18) = 4.55 × 1017 s, not 4.55 × 10−17 s — the sign of the index flips when it moves out of the denominator. Then divide by 3.15 × 107 to reach 1.44 × 1010 years; multiplying instead gives 1.43 × 1025 years, older than the Universe by a factor of a thousand million million.
Question 7: Orbits and Gravity
10 marks
(a) [3]
Explain why planets orbit the Sun in elliptical paths and describe how a planet's speed changes as it moves along its elliptical orbit. Use the principle of conservation of energy in your answer.

Mark Scheme — 7(a)

The gravitational attraction of the Sun provides the centripetal force that keeps planets in their (elliptical) orbits [1]
A planet moves faster when it is closer to the Sun (at perihelion) and slower when it is farther from the Sun (at aphelion) [1]
By conservation of energy: when the planet is closer to the Sun its gravitational potential energy is lower, so its kinetic energy (and speed) must be higher; when farther away, gravitational PE is higher and KE (speed) is lower [1]
⚠ If you missed marks here: Naming a "centrifugal force" gets no credit — the force acting is the Sun's gravitational attraction, and it acts as the centripetal force towards the Sun. Mark 3 is not "gravity is stronger so it speeds up"; it must be an energy transfer, with gravitational potential energy falling and kinetic energy rising as the planet approaches perihelion, and the reverse out at aphelion.
(b) [4]
The International Space Station (ISS) orbits the Earth at a height of 408 km above the surface. The radius of the Earth is 6370 km. The ISS completes one orbit in 92 minutes. Calculate the orbital speed of the ISS in m/s.

Mark Scheme — 7(b)

Correct orbital radius: r = 6370 + 408 = 6778 km = 6.778 × 106 m [1]
Correct period in seconds: T = 92 × 60 = 5520 s [1]
Correct formula and substitution: v = 2πr / T = (2 × π × 6.778 × 106) / 5520 [1]
Correct final answer: v ≈ 7710 m/s (accept 7700 m/s) with correct unit [1]
⚠ If you missed marks here: Using 408 km as the orbital radius gives about 464 m/s — the radius is measured from the centre of the Earth, so add 6370 km first to get 6778 km. Two unit traps follow: leaving T as 92 instead of 5520 s inflates the answer to roughly 463 000 m/s, and leaving r in kilometres gives 7.71 instead of 7710 m/s. The final mark also requires the unit written.
(c) [3]
Explain why the ISS does not fall to Earth despite being subject to the Earth's gravitational pull.

Mark Scheme — 7(c)

The ISS is in a state of continuous free fall — it is constantly falling towards the Earth due to gravity [1]
However, it has a large forward (tangential) velocity, so its curved path matches the curvature of the Earth's surface [1]
Gravity provides the centripetal force needed to maintain the circular orbit; the ISS moves forward fast enough that it continuously ‘falls around’ the Earth rather than falling towards it [1]
⚠ If you missed marks here: "There is no gravity in space" is the answer to avoid — at 400 km the Earth's gravitational field strength is still about 8.7 N/kg, close to its value on the ground. The marks are that the ISS is in continuous free fall, that its large tangential speed makes its curved path match the curvature of the Earth, and that gravity supplies the centripetal force so it falls around the planet rather than into it.
IGCSE Physics 0625/0972 — Paper 4 Theory (Extended) — Topic 6: Space Physics — Mock Exam 1
For examination preparation only. Not an official Cambridge Assessment paper.