Summer 2023 IGCSE Physics Paper 2P

Click "Show Solution" beneath each question to reveal the fully worked answer. Equations are rendered with MathJax — an internet connection is required.

Question 1Moments & Equilibrium

Part (a) – Mass of the pencil

Key concept: Weight = mass × gravitational field strength
\( W = m \times g \), where \( g = 10 \text{ N/kg} \)

Rearrange for mass:
\[ m = \frac{W}{g} = \frac{0.16}{10} = 0.016 \text{ kg} = 16 \text{ g} \]

Answer: B – 16 g

Part (b)(i) – Moment of the weight

Formula: \( \text{moment} = F \times d_\perp \)

The weight acts 3.7 cm from the pivot:
\[ \text{moment} = 0.16 \times 3.7 = 0.592 \text{ N cm} \approx 0.59 \text{ N cm} \]

Moment of weight = 0.59 N cm

Accept 0.00592 N m (unit conversion)

Part (b)(ii) – Moment of force F

Principle of moments: For equilibrium, the total clockwise moment = total anticlockwise moment.

Since the pencil is horizontal and balanced, the moment of F must equal the moment of the weight.

Moment of F = 0.59 N cm (same as part i, by the principle of moments)

Part (b)(iii) – Show that F = 0.080 N

Step 1: The force F acts at the left end, which is 7.4 cm from the pivot.

Step 2: Set the moment of F equal to the moment of the weight:
\[ F \times 7.4 = 0.592 \]

Step 3: Solve for F:
\[ F = \frac{0.592}{7.4} = 0.080 \text{ N} \checkmark \]

F = 0.080 N

Exam tip: In a "show that" question, always write out full working — state the formula, substitute values, and show each step clearly.

Question 2States of Matter & Particle Model

Part (a) – Arrangement and motion at 5°C (solid)

Key concept: At 5°C the chocolate is a solid — below its melting point of 32°C.

Arrangement: Particles are in fixed positions, held in a regular pattern by strong forces.
Motion: Particles can only vibrate about their fixed positions — they cannot move freely.

Fixed positions; vibrating about those positions

Part (b)(i) – Motion at 45°C (liquid)

Key concept: 45°C is above the melting point (32°C) but below the boiling point (55°C) — so the chocolate is a liquid.

Particles move in a random manner.
They are no longer in fixed positions — they can flow and move past each other.
They have more kinetic energy than in the solid state.

Random motion; particles no longer in fixed positions

Part (b)(ii) – Measuring temperature

Answer: D – Thermometer

A balance measures mass, a ruler measures length, a stopwatch measures time. Only a thermometer measures temperature.

Part (b)(iii) – Temperature–time graph (graphical question)

Mark scheme answer:

The correct sketch graph has three distinct sections:

Temperature rises steadily from 5°C up to 32°C (solid heating up).
Temperature stays constant at 32°C for a period — this is the melting point where energy goes into breaking bonds rather than raising temperature.
Temperature rises again from 32°C to 45°C (liquid heating up).

Exam tip: The flat section must be drawn at exactly 32°C. A flat section at 45°C or 55°C would not score marks for this question since the chocolate doesn't boil (max temp is 45°C).

Common mistake: Do not draw a flat section at 5°C — the question says the chocolate is being heated from 5°C, not that it starts at its melting point.

Question 3Oscilloscope, Kinetic Energy & Circular Motion

Part (a)(i) – How the grid became positively charged

Key concept: Objects become positively charged by losing electrons, not by gaining positive particles.

The grid is connected to the positive terminal of a high voltage supply. This causes:

Electrons are transferred away from the grid (to the positive terminal).
The grid is left with more protons than electrons — it becomes positively charged.

Electrons are transferred away from the grid; the grid has lost electrons → becomes positive

Common mistake: Do NOT say "positive particles moved into the grid" — protons do not move in solids. Only electrons are mobile.

Part (a)(ii) – Formula for kinetic energy

\[ KE = \frac{1}{2} m v^2 \]

Part (a)(iii) – Speed of the electron

Given: \( KE = 1.3 \times 10^{-15} \text{ J} \), \( m = 9.1 \times 10^{-31} \text{ kg} \)

Step 1: Write the formula and substitute:
\[ 1.3 \times 10^{-15} = \frac{1}{2} \times 9.1 \times 10^{-31} \times v^2 \]

Step 2: Rearrange for \( v^2 \):
\[ v^2 = \frac{2 \times 1.3 \times 10^{-15}}{9.1 \times 10^{-31}} = \frac{2.6 \times 10^{-15}}{9.1 \times 10^{-31}} = 2.857 \times 10^{15} \]

Step 3: Take the square root:
\[ v = \sqrt{2.857 \times 10^{15}} = 5.35 \times 10^{7} \text{ m/s} \]

Speed = \( 5.3 \times 10^7 \text{ m/s} \)

Exam tip: This is about 18% of the speed of light — electrons are very fast! Always check your powers of ten carefully.

Part (b) – Why electrons spread apart

Key concept: Like charges repel.

All electrons carry a negative charge.
Like charges experience a repulsive force.
So electrons in the beam push each other apart, causing the beam to spread.

Electrons are all negatively charged (like charges) → like charges repel → electrons move apart

Part (c)(i) – Radius of the circle

Using the scale where each grid square = 1 cm, the circle spans 5 grid squares in diameter.

Radius = 2.5 cm (accept 2.4–2.6 cm)

Part (c)(ii) – Orbital speed of electrons

Formula: \( v = \frac{2\pi r}{T} \)

Step 1: Convert units:
\( r = 2.5 \text{ cm} = 2.5 \times 10^{-2} \text{ m} \)
\( T = 24 \text{ ms} = 24 \times 10^{-3} \text{ s} \)

Step 2: Substitute:
\[ v = \frac{2 \times \pi \times 2.5 \times 10^{-2}}{24 \times 10^{-3}} \]

Step 3: Evaluate:
\[ v = \frac{0.15708}{0.024} = 6.54 \text{ m/s} \]

Orbital speed = 6.5 m/s

Common mistake: Forgetting to convert cm → m and ms → s. Always convert to SI units (metres and seconds) before substituting.

Question 4Conservation of Momentum

Part (a) – Show that momentum of bird X ≈ 5 kg m/s

Formula: \( p = m \times v \)

Substitute:
\[ p_X = 0.41 \times 13 = 5.33 \text{ kg m/s} \approx 5 \text{ kg m/s} \checkmark \]

Momentum of bird X = 5.33 kg m/s ≈ 5 kg m/s

"Show that" tip: Show the full multiplication 0.41 × 13 = 5.33 and then state this is approximately 5.

Part (b) – Total momentum just before collision

Both birds travel in the same direction (to the right), so momenta add:
\[ p_{\text{total}} = p_X + p_Y = 5.33 + 0.15 = 5.48 \text{ kg m/s} \]

Total momentum = 5.48 kg m/s

Part (c) – Total momentum just after the catch

Principle of conservation of momentum: In a closed system, total momentum is conserved — it stays the same before and after a collision.

Total momentum after = 5.48 kg m/s (same as before)

Part (d) – Velocity of birds after the catch

Step 1: Find total mass:
\[ m_{\text{total}} = 0.41 + 0.17 = 0.58 \text{ kg} \]

Step 2: Use conservation of momentum:
\[ p = m \times v \implies v = \frac{p}{m} = \frac{5.48}{0.58} \]

Step 3: Evaluate:
\[ v = 9.45 \text{ m/s} \]

Velocity after = 9.4 m/s

Common mistake: Using only bird X's mass (0.41 kg) in the denominator. After the catch, both birds move together so you must use the combined mass (0.58 kg).

Question 5Nuclear Fission Reactor

Part (a) – How a chain reaction occurs

Key concept: A chain reaction is self-sustaining because each fission produces neutrons that can cause further fissions.

A uranium nucleus absorbs a neutron.
The nucleus becomes unstable and splits (fission), releasing energy and further neutrons.
These released neutrons can be absorbed by other uranium nuclei, causing further fissions — the process repeats and multiplies.

Neutron absorbed → nucleus splits → releases more neutrons → those neutrons cause further fissions → chain reaction

Part (b) – Role of control rods

Key concept: Control rods regulate the rate of fission to prevent a runaway reaction.

Control rods absorb neutrons.
This prevents those neutrons from causing further fissions.
By raising or lowering the rods, operators can slow down or speed up the rate of fission, controlling the power output.

Control rods absorb neutrons → fewer neutrons available → rate of fission is controlled/slowed

Part (c) – Fission vs Fusion

These are two very different nuclear processes:

Fission: A large nucleus splits into two smaller nuclei (e.g. uranium-235 splits when it absorbs a neutron).
Fusion: Two small nuclei combine to form a larger nucleus (e.g. two hydrogen nuclei fuse to form helium).

Fission = splitting of a nucleus; Fusion = combining of nuclei

Part (d)(i) – Main energy source in stars

Answer: C – Fusion

Stars fuse hydrogen nuclei into helium, releasing enormous amounts of energy. This is nuclear fusion, not fission.

Part (d)(ii) – Why fusion doesn't happen at low temperatures and pressures

Key concept: Nuclei are positively charged and repel each other.

Protons (in nuclei) all carry positive charge. At low temperatures, nuclei move slowly and the electrostatic repulsion between the positively charged nuclei is strong enough to prevent them from getting close enough to fuse.

At very high temperatures (like inside a star), nuclei move fast enough to overcome this repulsion and get close enough for the strong nuclear force to bind them together.

Electrostatic repulsion between nuclei (protons) prevents fusion at low temperatures — nuclei need very high speed/energy to overcome this repulsion

Question 6National Grid & Transformers

Part (a)(i) – Why a step-up transformer is used

Key concept: Power = current × voltage. To transmit the same power at higher voltage, the current must be lower.

A step-up transformer increases the voltage, which means the current is reduced.
Heating in cables depends on current: \( P_{\text{loss}} = I^2 R \). A lower current means much less energy is wasted as heat in the transmission cables.

Stepping up voltage → reduces current → less heating in the wires → less energy wasted

Part (a)(ii) – Transformer formula

\[ \frac{V_p}{V_s} = \frac{N_p}{N_s} \] Input (primary) voltage ÷ output (secondary) voltage = primary turns ÷ secondary turns

Part (a)(iii) – Number of turns on secondary coil

Given: \( V_p = 15 \text{ kV} \), \( V_s = 340 \text{ kV} \), \( N_p = 1400 \)

Step 1: Write the formula:
\[ \frac{N_p}{N_s} = \frac{V_p}{V_s} \]

Step 2: Rearrange:
\[ N_s = N_p \times \frac{V_s}{V_p} = 1400 \times \frac{340}{15} \]

Step 3: Evaluate:
\[ N_s = 1400 \times 22.67 = 31{,}733 \approx 32{,}000 \text{ turns} \]

Number of turns on secondary coil = 32,000 (accept 31,700–35,000)

Part (b) – Why a transformer won't work with constant d.c.

Key concept: Transformers work by electromagnetic induction, which requires a changing magnetic field.

A constant (d.c.) current creates a constant magnetic field in the primary coil.
Electromagnetic induction requires a changing magnetic field to induce a voltage.
Since the field is not changing, no voltage is induced in the secondary coil — the transformer does not work.

Constant d.c. → constant magnetic field → no changing flux → no induction → no output voltage

Exam tip: Alternating current (a.c.) continuously changes direction, so the magnetic field it creates constantly changes — this is why a.c. is essential for transformers.

Question 7Refraction, Redshift & Big Bang Theory

Part (a)(i) – Equipment to measure angles

Protractor

Part (a)(ii) – Measuring angle of incidence and refraction (graphical question)

Mark scheme answer:

For this question, you need to measure the angles from the diagram using a protractor:

Angle of incidence (i): Measured between the incoming ray and the normal at the point where light enters the prism. The mark scheme accepts 42° to 46°.
Angle of refraction (r): Measured between the refracted ray and the normal inside the prism. The mark scheme accepts 26° to 30°.

Angle of incidence: ~44° | Angle of refraction: ~28°

Part (a)(iii) – Refractive index formula

\[ n = \frac{\sin i}{\sin r} \]

Part (a)(iv) – Calculate refractive index

Using i = 44°, r = 28°:
\[ n = \frac{\sin 44°}{\sin 28°} = \frac{0.6947}{0.4695} = 1.48 \]

Refractive index = ~1.5 (accept 1.34–1.64, depending on measured angles)

Exam tip: Your answer depends on your measured angles. As long as you measure carefully and substitute correctly, you'll score full marks even if your answer differs slightly from 1.5.

Part (b)(i) – Difference in wavelength for galaxy A

\[ \Delta\lambda = 645 - 630 = 15 \text{ nm} \]

Difference in wavelength = 15 nm

Part (b)(ii) – Speed of galaxy A

Formula (redshift): \( \dfrac{\Delta\lambda}{\lambda_0} = \dfrac{v}{c} \)

Step 1: Substitute values:
\[ \frac{15}{630} = \frac{v}{3.0 \times 10^8} \]

Step 2: Rearrange:
\[ v = 3.0 \times 10^8 \times \frac{15}{630} \]

Step 3: Evaluate:
\[ v = 3.0 \times 10^8 \times 0.02381 = 7.14 \times 10^6 \text{ m/s} \]

Speed of galaxy A = \( 7.1 \times 10^6 \text{ m/s} \) (approximately 7,100,000 m/s)

Common mistake: Using 645 nm instead of the reference wavelength 630 nm in the denominator. Always use the original (reference) wavelength \(\lambda_0\) in the bottom of the fraction.

Part (b)(iii) – How these observations support the Big Bang theory

Key concept: The Big Bang theory predicts that the universe is expanding — galaxies should be moving away from us, and more distant galaxies should be moving faster (Hubble's Law).

Galaxy B has twice the redshift of galaxy A → it is moving at twice the speed of galaxy A. This shows that redshift is proportional to speed.
Galaxy B is also twice as far away as galaxy A → more distant galaxies move faster. This supports the idea that galaxies are moving apart from each other (expanding universe).
Both observations are consistent with the Big Bang theory, which says everything started from a single point and has been expanding ever since.

Twice the redshift → twice the speed; twice the distance → twice the speed; galaxies are moving apart — consistent with an expanding universe from the Big Bang

Question 8Specific Heat Capacity of Concrete

Part (a) – Method to find specific heat capacity of concrete

Formula: \( \Delta Q = m \times c \times \Delta T \implies c = \frac{\Delta Q}{m \times \Delta T} \)

Here is a complete, mark-scheme-approved method:

Find the mass of the concrete block using a balance.
Measure the initial temperature of the concrete using the thermometer.
Connect the electrical heater and record the voltmeter reading (V) and ammeter reading (I).
Switch on the heater and record the time (t) using a stopwatch.
Calculate energy supplied to the concrete:
\[ E = V \times I \times t \]
Record the final temperature and calculate temperature change: \(\Delta T = T_{\text{final}} - T_{\text{initial}}\).
(Continue taking temperature readings after switching off the heater and use the maximum temperature reached.)
Rearrange the formula to find specific heat capacity:
\[ c = \frac{E}{m \times \Delta T} = \frac{V \times I \times t}{m \times \Delta T} \]
Repeat the experiment and average the results to improve reliability.

Exam tip: For 5 marks you need at least 5 distinct points. Include: mass measurement, time measurement, temperature change, energy calculation, and rearranging the formula. Mention the insulating material — it reduces heat loss to the surroundings.

Common mistake: Forgetting to say you record temperature after switching off the heater to find the maximum — heat continues to flow from heater to concrete even after switching off.

Part (b) – Advantage of high specific heat capacity for heating water

Definition: Specific heat capacity (SHC) is the energy needed to raise 1 kg of a substance by 1°C. A high SHC means a material can store a large amount of energy per unit mass.

Concrete with a high SHC can absorb and store a large amount of thermal energy when it is heated by the sun.
It can then release this energy slowly to heat the water over a longer period of time.
This means the water temperature is maintained for longer — more efficient heating system.

High SHC → concrete stores lots of energy → releases energy slowly to the water → water stays warm for longer / larger temperature rise in water