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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]
(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]
(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]
(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]
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]
(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]
(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]
(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]
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]
(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]
(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]
(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]
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]
(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]
(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]
(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]
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]
(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]
(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]
(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]
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]
(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]
(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]
(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]
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]
(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]
(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]
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.