Physics 0625/43 October/November 2025: question paper and mark scheme
Cambridge IGCSE Physics (0625), Paper 4 Extended Theory, October/November 2025. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) A hard ball is dropped vertically downwards.
Fig. 1.1 shows the speed–time graph for the ball from when it is dropped until it hits the ground.
Air resistance is ignored.
8
6
speed
4 m/s
2
0
0 0.2 0.4 0.6 0.8 1.0 time / s
Fig. 1.1
(i) Describe the motion of the ball in Fig. 1.1.
…… [1]
(ii) Show that the distance travelled by the ball is 1.8 m.
[2]
(iii) The mass of the ball is 0.54 kg. Calculate the gravitational potential energy stored in the ball immediately before it is dropped.
gravitational potential energy = …… [2]
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(iv) Determine the kinetic energy of the ball when it has fallen half way to the ground.
kinetic energy = …… [1]
(b) The same ball is dropped from an aeroplane. Air resistance acts on the ball.
Explain how the vertical motion of the ball changes, from when it is dropped to just before it
hits the ground.
Use the idea of forces in your answer.
…… [3]
(c) A soft ball is dropped by a student so that it bounces on the ground.
Fig. 1.2 shows the height–time graph for the ball. The height is measured from the ground.
The ball bounces four times.
A
height
C m
B time / s
Fig. 1.2
Explain why the change in height of the ball between B and C is less than the change in height between A and B.
…… [2]
[Total: 11]
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Mark scheme
1(a)(i) constant acceleration B1
1(a)(ii) area under the graph OR ½ bh OR average speed × time B1
0.5 × 5.9 × 0.6 OR 1.77 (to ⩾ 3 sf) B1
1(a)(iii) 9.5 J A2
(ΔEp =) mg(Δ)h OR 0.54 × 9.8 × 1.8 C1
1(a)(iv) candidate’s 1(a)(iii) value × 0.5 (2sf) A1
1(b) any three from: B3
• (initially there is acceleration due to) an unbalanced force OR resultant force OR downward force OR weight OR
gravitational force acting on the ball
• as speed increases the air resistance increases
• (as air resistance increases) resultant force downwards decreases
OR acceleration decreases
• (eventually) ball falls with constant speed when air resistance = weight OR ball falls with terminal velocity when air
resistance = weight
1(c) (some of the) energy is: B1
transferred to the thermal energy (store) of the ground (when ball hits ground)
OR converted / transferred to thermal energy (store) of the air
OR used to deform ball / transferred to the elastic energy (store) of the ball (as it contacts the ground)
gravitational potential energy (in the store at C) will be less (than at A) B1
OR
(ball has) less kinetic energy (in its store just after B)
A cricket ball is travelling at a speed of 25 m / s horizontally through the air. The mass of the ball is
0.14 kg.
A player catches the ball and the ball comes to rest.
(a) Calculate the change in momentum of the ball during the catch.
change in momentum = …… [2]
(b) The player’s hands are in contact with the moving ball for 9.7 ms.
Calculate the horizontal force exerted by her hands to bring the ball to rest.
force = …… [3]
[Total: 5]
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Mark scheme
2(a) (–) 3.5 kgm / s A2
(Δp =) m(Δ)u OR m{v – u} OR 0.14 × {25 (– 0)} C1
2(b) 360 N A3
0.0097 (s) OR 9.7 × 10–3 (s) OR 3.6 × 10n C1
(F =) (Δ)p / t OR 3.5 / 0.0097 C1
(a) Fig. 3.1 shows a Sankey diagram for the power transfer in a coal‑fired power station.
useful power input power from fuel output
= 3400 MW
power wasted power power used in power used in in station due to power transmission thermal station = 25 MW energy = 40 MW transfers
= 2200 MW
Fig. 3.1
(i) Show that the useful power output from the power station is approximately 1100 MW.
[2]
(ii) Calculate the efficiency of the power station.
efficiency = …… [2]
(b) Describe how useful energy may be obtained from geothermal resources.
…… [2]
[Total: 6]
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Mark scheme
3(a)(i) (wasted power =) 2200 + 40 + 25 OR 2265 B1
OR (useful power output =) power input – wasted power
1140 (MW) (⩾ 3 sf) B1
3(a)(ii) 0.32 OR 32% A2
(efficiency =) {useful power output / total power input} (× 100%) C1
OR 1100 / 3400
3(b) water is pumped into the ground and heated by hot rocks B1
OR
heat from hot rocks underground is used to turn water into steam
steam used to heat buildings B1
OR
steam used to turn generator / turbine / produce electricity
(a) (i) A rectangular glass tank contains 3.0 m3 of liquid. The mass of the liquid is 2800 kg.
Calculate the density of the liquid.
density = …… [1]
(ii) A metal block is placed in the tank and it sinks to the bottom. The top surface of the block is 1.1 m below the surface of the liquid.
Fig. 4.1 shows the block on the bottom of the tank.
liquid
1.1 m
metal block
0.30 m
0.45 m
Fig. 4.1
Calculate the force exerted by the liquid on the top of the block.
force on top of the block = …… [4]
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(b) Fig. 4.2 shows part of a steel railway track. There are small gaps between sections of the rail.
small gap
section of rail
Fig. 4.2
State why gaps are needed between the sections of rail. Explain your answer in terms of particles.
statement ……
explanation ……
[2]
(c) Fig. 4.3 shows a saucepan of boiling water on the heating ring of an electric cooker.
boiling water
plastic handle metal pan
heating ring
Fig. 4.3
The handle is made from plastic.
Explain why this material is suitable.
…… [2]
[Total: 9]
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Mark scheme
4(a)(i) 930 kg / m3 A1
4(a)(ii) 1400 N A4
(∆)p = ρg(∆)h OR 930 × 9.8 × 1.1 OR 10 000 (N / m2) C1
area = length × width OR 0.45 × 0.30 OR 0.14 (m2) C1
(F=) p × A OR 10 000 × 0.14 C1
Alternative method:
1400 N A4
volume (of liquid above block) = length × width × height OR 0.45 × 0.30 × 1.1 OR 0.15 (m3) C1
mass (of liquid above block) = ρ × v OR 930 × 0.15 OR 140 (kg) C1
(F =) mg OR 140 × 9.8 C1
4(b) thermal expansion B1
OR to prevent buckling / bending when heated
any one from: B1
• particles vibrate faster
• particles have greater speed / kinetic energy / momentum
• vibrations (of particles) take up more space
• (average) space / separation between particles increases
4(c) (plastic is a good) insulator / poor conductor B1
(handle) will not get too hot to touch B1
OR reduces heat flow to hand
(a) (i) Electromagnetic waves have many uses.
On Fig. 5.1 draw one line from each use to the region of the electromagnetic (e‑m) spectrum it uses.
Use Region of e‑m spectrum
security marking gamma rays
Bluetooth ultraviolet
optical fibres infrared
detection of cancer radio waves
Fig. 5.1
[2]
(ii) State one advantage of using microwaves compared with radio waves to transmit mobile
(cell) phone signals.
…… [1]
(b) Describe the difference between transverse waves and longitudinal waves.
…… [2]
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(c) Fig. 5.2 shows a simplified diagram of seismic P‑waves travelling through the Earth. The paths of the waves are curved.
Seismic waves are produced by rock movements in earthquakes.
earthquake occurs here
P-waves
(mostly) solid mantle
X
liquid outer core
inner core
Fig. 5.2 (not to scale)
(i) State the type of wave that P‑waves can be modelled as.
…… [1]
(ii) At point X, the P‑wave travels from the solid mantle to the liquid core. There is a sudden change in direction of the path of the wave.
Explain the change in direction of the P‑wave at point X.
…… [2]
(d) The Earth’s surface absorbs incoming radiation from the Sun and also emits thermal radiation from its surface.
Over the past 50 years, scientists estimate that the average temperature of the Earth’s surface has increased by approximately 0.75 °C.
Explain what is causing this average temperature rise.
…… [2]
[Total: 10]
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Mark scheme
5(a)(i) all four correct = 2 marks; any two correct = 1 mark B2
security marking gamma rays
Bluetooth ultraviolet
optical fibres infrared
detection of cancer radio waves
5(a)(ii) any one from: B1
• only require a short aerial / antenna
• can pass through the atmosphere / ionosphere to satellites
• can transmit large amounts of data at high speeds / high rate of data transmission
5(b) any two from: B2
transverse:
• oscillations / vibrations are at right angles to direction of propagation
• medium not required / can travel through a vacuum (for e-m wave)
• have crests / peaks and troughs
longitudinal:
• oscillations / vibrations are parallel to direction of propagation
• medium is required
• have rarefactions and compressions
5(c)(i) longitudinal B1
5(c)(ii) refraction B1
any one from: B1
• change in speed of the wave
• change in density of the medium
5(d) increase in carbon dioxide / greenhouse gases (in the atmosphere) B1
OR
increase in methane / water vapour (in the atmosphere)
any one from: B1
prevents / reduces (re-emitted) radiation escaping (into space)
OR
radiation re-emitted / reflected back to surface (by atmosphere / greenhouse gases)
OR
(amount of) radiation absorbed is greater than (the amount of) radiation leaving the atmosphere
(a) Fig. 6.1 shows a circuit which includes an LDR (light‑dependent resistor).
400 Ω
+
+5.0 V ı
V
Fig. 6.1
The LDR is in a brightly lit room. The voltmeter reads 1.8 V.
(i) Calculate the current in the 400 Ω resistor.
current = …… [3]
(ii) The light level in the room changes from bright to dark.
State and explain the effect on the voltmeter reading.
statement ……
explanation ……
[3]
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(b) Fig. 6.2 shows two light‑emitting diodes (LEDs) connected in parallel in a circuit. R is a red
LED and G is a green LED.
The voltmeter across the LEDs shows a potential difference (p.d.) of 2.0 V.
A
+
+5.0 V
ı
R G V 2.0 V
Fig. 6.2
(i) Fig. 6.3 shows a current‑p.d. graph for the LEDs.
R G
80
60
current
40 mA
20
0
0 1.0 2.0 3.0 4.0 p.d./ V
Fig. 6.3
Use Fig. 6.3 to determine the current in the ammeter.
current = …… [2]
(ii) The connections to the power supply are reversed. State the reading on the ammeter.
current = …… [1]
[Total: 9]
DFD
12
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Mark scheme
6(a)(i) 0.008(0) A OR 8(.0) mA OR 8(.0) × 10–3 A A3
(p.d. across 400 Ω resistor =) 5 – 1.8 OR 3.2 (V) C1
(I =) V / R OR 3.2 / 400 C1
6(a)(ii) increases B1
resistance of LDR increases B1
ratio of potential differences (across each component) equals ratio of the two resistances OR R1 / R2 = V1 / V2 B1
OR
LDR has greater share of the total / supply voltage
OR
potential difference of the source is divided between the resistor (R) and LDR
6(b)(i) 56 mA OR 5.6 × 10–2 A A2
44 (mA) OR 12 (mA) C1
6(b)(ii) 0 (A) B1
Fig. 7.1 shows a simple a.c. generator.
handle coil
N S
slip rings coil
brush front
Fig. 7.1
(a) (i) Explain the function of the slip rings and brushes.
…… [1]
(ii) Describe how an alternating current is generated in the lamp.
…… [3]
(iii) State two possible changes that cause a larger maximum current in the lamp.
1 ……
2 ……
[2]
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(b) Fig. 7.2 is a graph to show how the electromotive force (e.m.f.) varies with time for the coil.
e.m.f.
0
0 A B C D E time
Fig. 7.2
(i) Determine how many revolutions the coil has made, from time = 0 to E.
number of revolutions = …… [1]
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(ii) Fig. 7.3 shows the end view of the position of the plane of the coil at time = 0. The coil is rotating anti‑clockwise.
coil
N S
axis of rotation slip rings
Fig. 7.3
On Fig. 7.4, draw four lines to match the coil position to the times A, B, C and D. The coil is viewed from the front.
Each coil position may be used once, more than once or not at all.
time label coil position from graph
A N S
B N S
C N S
D N S
Fig. 7.4
[3]
[Total: 10]
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Mark scheme
7(a)(i) provide a continuous connection (between the coil and the lamp) B1
OR
prevent the contacts reversing every half turn / 180°
OR
prevent the wires (to the lamp) twisting / tangling
7(a)(ii) any three from: B3
• coil is rotated / moved in magnetic field
• coil cuts (magnetic) field OR changing (magnetic) field through coil
• e.m.f. / voltage induced
• coil forms part of a complete circuit (through the lamp) so there is a current
• as one side of the coil starts to move in the opposite direction, current is induced in the opposite direction
OR direction of current in coil is reversed every half turn / 180°
7(a)(iii) any two from: B2
• increase strength of magnetic field
• increase speed of rotation (of the coil)
• increase number of turns (of coil)
7(b)(i) 2.5 (revolutions) B1
7(b)(ii) any 1 correct = 1 mark B3
any 2 or 3 correct = 2 marks
all 4 correct = 3 marks
Time Label
coil position
f rom graph
A
N S
B N S
C
N S
D N S
(a) Alpha (α) particles are directed at a very thin sheet of gold foil.
The following observations provide evidence for the nuclear model of the atom.
State what conclusions may be made about the model of the atom from these observations.
(i) Observation: most of the particles pass straight through the foil.
conclusion ……
…… [1]
(ii) Observation: a few particles bounce back or are deflected through large angles.
conclusion ……
…… [1]
(b) Fig. 8.1 shows three beams of ionising radiation, alpha, beta and gamma, passing between
two parallel plates. The plates are charged and there is a uniform electric field between them.
charged plate
P
Q
R
charged plate
Fig. 8.1
(i) State and explain which path P, Q or R shows a beam of alpha particles.
path ……
explanation ……
[2]
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(ii) State and explain the direction of the electric field between the plates.
direction ……
explanation ……
[1]
(c) Curium‑242 is a radioactive isotope. It emits alpha particles. It has a half‑life of 160 days.
After 480 days, the mass of curium‑242 in a sample is 2.4 × 10–3 g.
Calculate the initial mass of curium‑242 in the sample.
initial mass of curium‑242 = …… [2]
[Total: 7]
* 0000800000018 * DFD
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Mark scheme
8(a)(i) most of the atom is empty space B1
OR
nucleus is (very) much smaller than the atom
8(a)(ii) nucleus must have a positive charge B1
OR
(the positive) charge is concentrated at the centre of the atom
OR
most of the mass (of the atom) is concentrated in the nucleus
8(b)(i) P B1
alpha particles have greater mass so are deflected less (than the beta particles) B1
8(b)(ii) upwards B1
AND
alpha particles are positively charged so attracted towards negatively charged plate OR direction of field is direction of force
on positive charge
8(c) 1.9 × 10–2 g OR 1.9 × 10–5 kg A2
3 half-lives OR 2.4 × 10–3 × 23 OR 1 / 8 C1
(a) Cosmic microwave background radiation (CMBR) can be observed in every direction in space around us.
(i) State when CMBR was produced.
…… [1]
(ii) State and explain how the wavelength of this radiation has changed since it was formed.
…… [2]
(b) Table 9.1 shows data about planets in our solar system.
Table 9.1
average radius of average surface time for one orbit / planet orbit around the Sun / temperature /
Earth years
106 km °C
Venus 108 0.62 460
Earth 150 1.00 15
Mars 228 1.90 – 65
Jupiter 778 12.00 –110
Saturn 1430 29.00 –140
(i) The orbit of Venus is approximately circular.
Calculate the average orbital speed v of Venus. Give your answer in km / h.
orbital speed = …… km / h [3]
(ii) State how the average surface temperature of a planet changes as its distance from the
Sun increases.
…… [1]
[Total: 7]
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Mark scheme
9(a)(i) (shortly after) Big Bang OR when Universe was formed B1
9(a)(ii) Increased B1
Universe has expanded B1
OR
Universe accelerated outwards
9(b)(i) 120 000 (km / h) OR 1.2 × 105 (km / h) A3
(v =) 2πr / T OR {2 × π × 108 (× 106 ) } / {0.62 × 365 × 24} C1
(T =) 0.62 × 365 × 24 OR (T =) 5431 OR (T =) 226.3 × 24 OR 1.2 × 10n C1
9(b)(ii) decreases B1
(a) Sketch a diagram to show the orbital path of a comet around the Sun.
Mark the position of the Sun on your diagram using an S.
[2]
(b) Draw a cross on your diagram to show where the comet is travelling with the slowest speed.
Label the cross L. [1]
(c) Explain why the speed of a comet changes as it travels around its orbit.
Use the conservation of energy in your explanation.
…… [3]
[Total: 6]
* 0000800000020 * DFD
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Mark scheme
10(a)
S OR S
elliptical shape drawn B1
S at a focus of the ellipse by eye B1
10(b)
S L
L furthest away from their S / Sun B1
10(c) total energy = gravitational (potential) energy + kinetic energy B1
(energy stored as) GPE decreases and (energy stored as) KE increases closer to the Sun B1
when the comet has more kinetic energy it has a greater speed B1
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On this page, free: the mark scheme for IGCSE Physics 0625/43, October/November 2025 (0625_w25_ms_43), and the question paper (0625_w25_qp_43), to read or download, with each question's marking points written out underneath.
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Physics 0625/43 for October/November 2025 is out of 80 marks, with 1 hour 15 minutes to answer it, across 10 questions.
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