IGCSE Physics 0625 Paper 3 Practice Paper (Core Theory) — 80 marks, 1 hour 15 minutes, answers and marking scheme

2026-10-30

IGCSE Physics practice – Core

An original practice paper written to the Cambridge IGCSE Physics 0625 Paper 3 format: Core level, 6 structured questions, 33 parts, 80 marks and 1 hour 15 minutes. All questions are original and are not past paper questions or an official mark scheme.

Total 80 marks · 1 hour 15 minutes · 33 parts

Mark distribution 14 + 13 + 14 + 14 + 13 + 12 = 80

Level Core (same content as Paper 1 and Paper 3)

Paper at a glance

Paper	Level	Time	Marks
Paper 3	Core	1 hour 15 minutes	80 marks

Questions

Total: 80 marks · 1 hour 15 minutes · calculator allowed

Question 1 · Forces and motion [14]

(a) A car travels 150 m at constant speed in 12 s. Calculate its average speed. Show your working. [3]

(b) Convert 12.5 m/s into km/h. Show your working. [3]

(c) A student runs 100 m in 20 s, then walks a further 100 m in 60 s. Calculate the average speed for the whole journey. [3]

(d) A distance-time graph shows a car moving from 0 m to 20 m between 0 s and 4 s, then staying at 20 m from 4 s to 10 s. Calculate the speed in the first 4 s and describe the motion between 4 s and 10 s. [3]

(e) State one change in the motion of a car when the resultant force on it acts in the opposite direction to its motion. [2]

Question 2 · Density and pressure [13]

(a) A piece of metal has a mass of 240 g and a volume of 30 cm3. Calculate the density of the metal. [2]

(b) Convert 8.0 g/cm3 into kg/m3. Show your working. [3]

(c) The metal block is a cuboid measuring 5 cm x 3 cm x 2 cm. It rests on a table on its smallest face (3 cm x 2 cm). Take g = 10 N/kg. Calculate the pressure the block exerts on the table in N/cm2 and in Pa. [3]

(d) The same block is now placed on its largest face (5 cm x 3 cm). State how the pressure changes and explain your answer. [3]

(e) Calculate the pressure in water at a depth of 1.5 m. Density of water = 1000 kg/m3, g = 10 N/kg. [2]

Question 3 · Thermal physics and specific heat capacity [14]

(a) Define specific heat capacity. [2]

(b) A heater raises the temperature of 2 kg of water from 20 °C to 60 °C. The specific heat capacity of water is 4200 J/(kg °C). Calculate the energy gained by the water. [3]

(c) The heater has a power of 700 W. Calculate the time needed to heat the 2 kg of water from 20 °C to 60 °C, giving your answer in seconds and in minutes. [3]

(d) In practice the water takes longer than the calculated time to heat up. Explain why. [2]

(e) Use the motion of particles to explain why a solid expands when it is heated. [2]

(f) At the same room temperature a metal block feels colder than a wooden block. Explain why. [2]

Question 4 · Waves and light [14]

(a) The speed of sound in air is 340 m/s. A sound has a frequency of 500 Hz. Calculate its wavelength. Show your working. [3]

(b) Calculate the time taken for sound to travel 680 m in air. Speed of sound = 340 m/s. [2]

(c) A student claps and hears the echo from a distant wall 0.8 s later. The speed of sound is 340 m/s. Calculate the distance between the student and the wall. [3]

(d) Light passes from air into glass at an angle. State what happens to the speed of the light and to the direction of the ray. [2]

(e) State what is meant by frequency and give its unit. [2]

(f) State what determines the pitch and the loudness of a sound. [2]

Question 5 · Electric circuits and electrical energy [13]

(a) Two resistors of 6 ohm and 2 ohm are connected in series to a 12 V battery. Calculate the total resistance of the circuit. [2]

(b) Calculate the current in the series circuit described in (a). [2]

(c) Calculate the voltage across the 6 ohm resistor and across the 2 ohm resistor, and check that the two voltages add up to the battery voltage. [3]

(d) A second circuit connects two identical 6 ohm resistors in parallel to a 12 V battery. Calculate the total resistance of the parallel combination. [2]

(e) Explain why the total resistance of a parallel combination is smaller than either single resistor. [2]

(f) A 2 kW heater is used for 3 hours. Electricity costs $1.2 per kWh. Calculate the energy used in kWh and the cost. [2]

Question 6 · Radioactivity [12]

(a) An atomic nucleus emits an alpha particle. State how its mass number and its proton (atomic) number change. [2]

(b) A radioactive isotope has a half-life of 8 days and an initial activity of 640 Bq. Calculate the activity after 24 days. [3]

(c) Another sample has a half-life of 6 days. Its activity falls from 800 Bq to 100 Bq. Calculate the time taken. [3]

(d) Explain why alpha particles are more ionising than gamma rays but less penetrating. [2]

(e) State one safety precaution when handling a radioactive source and explain why it is needed. [2]

Answers and marking scheme

Part	Answer	Marking points	Marks
1(a)	Average speed = distance / time = 150 / 12 = 12.5 m/s.	1 mark: correct formula (distance / time). 1 mark: substitution 150 / 12. 1 mark: answer 12.5 m/s with unit.	3
1(b)	12.5 x 3.6 = 45 km/h.	1 mark: method (multiply by 3.6). 1 mark: value 45. 1 mark: unit km/h.	3
1(c)	Total distance 200 m, total time 80 s, average speed = 200 / 80 = 2.5 m/s.	1 mark: total distance 200 m. 1 mark: total time 80 s. 1 mark: answer 2.5 m/s.	3
1(d)	0-4 s: 20 / 4 = 5.0 m/s. 4-10 s: distance does not change, so the car is stationary at a speed of 0.	1 mark: 20 / 4. 1 mark: 5.0 m/s. 1 mark: stationary between 4 s and 10 s (speed 0).	3
1(e)	The car slows down (negative acceleration / deceleration).	1 mark: speed decreases. 1 mark: opposite to the resultant force / deceleration.	2
2(a)	Density = 240 / 30 = 8.0 g/cm3.	1 mark: 240 / 30. 1 mark: 8.0 g/cm3 with unit.	2
2(b)	1 g/cm3 = 1000 kg/m3, so 8.0 x 1000 = 8000 kg/m3.	1 mark: method, multiply by 1000. 1 mark: value 8000. 1 mark: unit kg/m3.	3
2(c)	Volume 5 x 3 x 2 = 30 cm3, consistent with (a); weight = 0.24 kg x 10 N/kg = 2.4 N; contact area 3 x 2 = 6 cm2; pressure = 2.4 / 6 = 0.40 N/cm2 = 4000 Pa.	1 mark: weight 2.4 N. 1 mark: 0.40 N/cm2. 1 mark: 4000 Pa.	3
2(d)	The pressure decreases. The weight is unchanged but the area increases from 6 cm2 to 15 cm2 (2.5 times larger); pressure = force / area, so pressure becomes 2.4 / 15 = 0.16 N/cm2 (1600 Pa).	1 mark: pressure decreases. 1 mark: area rises from 6 cm2 to 15 cm2. 1 mark: 0.16 N/cm2 or 1600 Pa.	3
2(e)	Pressure = 1000 x 10 x 1.5 = 15 000 Pa.	1 mark: correct substitution into p = rho x g x h. 1 mark: 15 000 Pa.	2
3(a)	The energy needed to raise the temperature of 1 kg of a substance by 1 °C.	1 mark: 1 kg of the substance. 1 mark: energy to raise it by 1 °C.	2
3(b)	Temperature change 60 – 20 = 40 °C; energy = 2 x 4200 x 40 = 336 000 J.	1 mark: dT = 40 °C. 1 mark: substitution into E = mc dT. 1 mark: 336 000 J.	3
3(c)	Time = energy / power = 336 000 / 700 = 480 s = 8.0 minutes.	1 mark: method (E / P). 1 mark: 480 s. 1 mark: 8.0 minutes.	3
3(d)	Some energy is lost to the surrounding air and the container, so not all the energy heats the water.	1 mark: energy is lost to the surroundings. 1 mark: not all the energy heats the water.	2
3(e)	The particles vibrate with a larger amplitude, so the average spacing between particles increases and the solid expands.	1 mark: particles vibrate with larger amplitude. 1 mark: average spacing between particles increases.	2
3(f)	Metal is a good conductor, so heat leaves your finger faster than with wood, which makes it feel colder.	1 mark: metal is a good conductor. 1 mark: heat is conducted away from the skin faster.	2
4(a)	Wavelength = speed / frequency = 340 / 500 = 0.68 m.	1 mark: correct formula (v / f). 1 mark: substitution 340 / 500. 1 mark: 0.68 m.	3
4(b)	Time = 680 / 340 = 2.0 s.	1 mark: substitution 680 / 340. 1 mark: 2.0 s.	2
4(c)	The sound travels there and back, so the one-way distance = 340 x 0.8 / 2 = 136 m.	1 mark: recognises the sound travels there and back. 1 mark: 340 x 0.8 = 272. 1 mark: 136 m.	3
4(d)	The light slows down and the ray bends towards the normal (refraction).	1 mark: the light slows down. 1 mark: the ray bends towards the normal.	2
4(e)	Frequency is the number of waves (oscillations) per second; the unit is the hertz (Hz).	1 mark: number of waves per second. 1 mark: unit Hz.	2
4(f)	Pitch is determined by frequency and loudness by amplitude.	1 mark: pitch – frequency. 1 mark: loudness – amplitude.	2
5(a)	Resistance in series = 6 + 2 = 8 ohm.	1 mark: adding the two resistances. 1 mark: 8 ohm.	2
5(b)	Current = voltage / total resistance = 12 / 8 = 1.5 A.	1 mark: substitution 12 / 8. 1 mark: 1.5 A.	2
5(c)	6 ohm: V = IR = 1.5 x 6 = 9 V; 2 ohm: 1.5 x 2 = 3 V; 9 + 3 = 12 V, equal to the battery voltage.	1 mark: 9 V. 1 mark: 3 V. 1 mark: check 9 + 3 = 12 V.	3
5(d)	1/R = 1/6 + 1/6 = 2/6 = 1/3, so R = 3 ohm.	1 mark: substitution 1/R = 1/6 + 1/6. 1 mark: 3 ohm.	2
5(e)	A parallel connection provides an extra path for the current, so the total current is larger; since resistance = voltage / current, the equivalent resistance is smaller.	1 mark: an extra path for the current. 1 mark: larger total current gives a smaller resistance.	2
5(f)	Energy = 2 x 3 = 6 kWh; cost = 6 x 1.2 = $7.20.	1 mark: 6 kWh. 1 mark: $7.20.	2
6(a)	The mass number decreases by 4 and the proton (atomic) number decreases by 2.	1 mark: mass number -4. 1 mark: proton number -2.	2
6(b)	24 / 8 = 3 half-lives; activity 640 -> 320 -> 160 -> 80 Bq.	1 mark: 3 half-lives. 1 mark: repeated halving shown. 1 mark: 80 Bq.	3
6(c)	800 -> 400 -> 200 -> 100 is 3 half-lives; 3 x 6 = 18 days.	1 mark: recognises 3 half-lives. 1 mark: repeated halving shown. 1 mark: 18 days.	3
6(d)	Alpha particles are charged and relatively massive, so they ionise atoms they collide with easily; each collision uses up energy, so they are absorbed within a few centimetres and penetrate poorly.	1 mark: charged and massive, so strongly ionising. 1 mark: loses energy quickly, so easily absorbed.	2
6(e)	Use long tongs and keep the source as far from the body as possible; a greater distance reduces the radiation dose received.	1 mark: a specific precaution (tongs, distance, lead store). 1 mark: reason (reduces exposure/dose).	2

Timing plan

Section	Suggested time	Marks
Questions 1-2	25 min	27 marks
Questions 3-4	26 min	28 marks
Questions 5-6	24 min	25 marks
Checking	remaining time	–

Common pitfalls

Calculations written without working lose every method mark.
Explanations that list key words without a cause-and-effect sentence.
Missing units: N, m s-1, J, V, A, W and ohm.
Core answers may be shorter, but you still need one point per mark available.

How to use this paper Sit the paper closed-book in 75 minutes, then mark point by point with the answer booklet and redo every lost mark.

Syllabus reference Format follows Cambridge IGCSE Physics 0625 (2026-2028) Paper 3 Core: 1 hour 15 minutes, 80 marks.

Frequently asked questions

Who is this paper for?
Students preparing for IGCSE Physics 0625 Core papers (Paper 1 and Paper 3), especially those who need practice in writing structured answers.

Does the answer booklet include a mark scheme?
Yes. Every part lists its marking points so you can mark yourself or a tutor can mark point by point.

Can I print it?
Yes, both the question paper and the answer booklet are A4 PDFs ready to print.

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Originality statement: every question, data set and context in this paper was written by TutorZone. It is not a Cambridge IGCSE Physics 0625 past paper or an official mark scheme.

This paper is not an IGCSE Physics 0625 past paper; only the structure and mark distribution follow the official Paper 3 (Core) format.