IGCSE Physics Paper 4 (Theory/Extended) - Mock 2

Topic 1: Motion, Forces and Energy | Core + Supplement

Duration 75 minutes

Total Marks 80

Questions 7

Time Remaining 75:00

Instructions

Answer ALL questions in the spaces provided.
Show all working for calculation questions. Marks are awarded for method even if the final answer is incorrect.
Take g = 9.8 m/s² unless stated otherwise.
Your answers will be automatically graded when you submit the exam. The model answers will be shown for review.

Question Navigation Score: 0 / 80

Question 1: Measurements and Motion

Total: 12 marks

(a) 2 marks

Define acceleration. Write the equation for acceleration, stating the meaning of each symbol.

Model Answer - Q1(a)

Acceleration is the rate of change of velocity [1]

a = (v - u) / t, where a = acceleration (m/s²), v = final velocity (m/s), u = initial velocity (m/s), t = time (s) [1]

⚠ If you missed marks here: "Acceleration is how fast something speeds up" misses the mark — it is the rate of change of velocity. The second mark is not for writing a = (v − u) / t on its own; the question asks what each symbol means, so u, v, t and their units must all be stated.

Mark 1: Definition - rate of change of velocity

Mark 2: Correct equation with symbols defined

(b) 4 marks

Explain the difference between a scalar quantity and a vector quantity. Give two examples of each.

Model Answer - Q1(b)

A scalar quantity has magnitude only [1]

A vector quantity has both magnitude and direction [1]

Two scalar examples: e.g. speed, mass, distance, time, energy, temperature [1]

Two vector examples: e.g. velocity, force, acceleration, displacement, momentum, weight [1]

⚠ If you missed marks here: The pair most often misclassified is mass and weight: mass is a scalar, weight is a vector because it is a force. Each example mark needs two correct entries, so one of each is half an answer, and the word direction must appear in the vector definition — "a vector is bigger" earns nothing.

Mark 1: Scalar - magnitude only

Mark 2: Vector - magnitude AND direction

Mark 3: Two correct scalar examples

Mark 4: Two correct vector examples

A cyclist starts from rest and accelerates uniformly to 12 m/s in 8 seconds. She then travels at a constant speed of 12 m/s for 20 seconds, before decelerating uniformly to rest in 4 seconds.

Calculate the acceleration of the cyclist during the first 8 seconds.

Model Answer - Q1(c)(i)

a = (v - u) / t = (12 - 0) / 8 [1]

a = 1.5 m/s² [1]

⚠ If you missed marks here: She starts from rest, so u = 0 and the change is the full 12 m/s in 8 s, giving 1.5 m/s². Dividing 12 by the whole 32 s of the journey gives 0.375 m/s². The unit is m/s², not m/s.

Mark 1: Correct substitution into equation

Mark 2: Correct answer 1.5 m/s²

On the axes below, sketch a speed–time graph for the whole 32 seconds of the cyclist’s journey. Mark the speed and the time at each change of motion.

Model Answer - Q1(c)(ii)

Your graph should look like this:

Straight line from (0, 0) to (8, 12), horizontal line from (8, 12) to (28, 12) [1]

Straight line from (28, 12) to (32, 0) with correct labels on axes [1]

⚠ If you missed marks here: The three phases must meet the correct times on the axis: rising to 12 m/s at 8 s, flat until 28 s, then falling to zero at 32 s. Drawing the flat section for 20 s but starting it at 0 s, or ending the graph at 24 s, loses the timing mark even when the shape looks right. Both axes need labels with units.

Mark 1: Correct shape - rising line, flat, falling line with correct values

Mark 2: Correct time values at each phase change and axes labelled

Calculate the total distance travelled by the cyclist during the 32 seconds.

Model Answer - Q1(c)(iii)

Area under graph = (1/2 x 8 x 12) + (20 x 12) + (1/2 x 4 x 12) = 48 + 240 + 24 [1]

Total distance = 312 m [1]

⚠ If you missed marks here: Distance is the area under the graph, so 12 × 32 = 384 m is wrong — that treats the whole journey as constant speed. The two sloping sections are triangles and need the ½: 48 + 240 + 24 = 312 m.

Mark 1: Correct method (area under speed-time graph / three sections)

Mark 2: Correct answer 312 m

Question 2: Density and Floating

Total: 10 marks

(a) 2 marks

Define density. Write the equation for density, including the SI units.

Model Answer - Q2(a)

Density is the mass per unit volume [1]

rho = m / V, where rho is density (kg/m³), m is mass (kg), V is volume (m³) [1]

⚠ If you missed marks here: "Density is how heavy something is" scores zero — it is mass per unit volume. The equation mark asks for units as well, so give ρ = m / V with kg for m, m³ for V and kg/m³ for ρ.

Mark 1: Correct definition - mass per unit volume

Mark 2: Correct equation with SI units (kg/m³)

(b) 4 marks

A solid metal cylinder has a radius of 3.0 cm, a height of 8.0 cm, and a mass of 540 g.

(b)(i) Calculate the volume of the cylinder in cm³. 2 marks

[Volume of a cylinder = pi x r² x h]

Model Answer - Q2(b)(i)

V = pi x r² x h = pi x (3.0)² x 8.0 [1]

V = 226 cm³ (or 72pi = 226.2 cm³) [1]

⚠ If you missed marks here: The 3.0 cm is the radius and it must be squared: π × 3.0² × 8.0 = 226 cm³. Forgetting the square gives 75.4 cm³, and treating 3.0 cm as the diameter (r = 1.5) gives 56.5 cm³.

Mark 1: Correct substitution into V = pi r² h

Mark 2: Correct answer approximately 226 cm³

(b)(ii) Calculate the density of the metal in g/cm³. 2 marks

Model Answer - Q2(b)(ii)

rho = m / V = 540 / 226 [1]

rho = 2.39 g/cm³ (accept 2.4 g/cm³) [1]

⚠ If you missed marks here: Density is mass over volume: 540 / 226 = 2.39 g/cm³. Inverting gives 0.42, and no metal has a density below that of water, so the size of the answer is a check in itself. Leave the units as the question asked — there is no need to convert to kg/m³ here.

Mark 1: Correct substitution into rho = m/V

Mark 2: Correct answer 2.4 g/cm³ (accept 2.39)

Oil with a density of 800 kg/m³ is poured into a beaker of water (density 1000 kg/m³). Describe what happens and explain why.

Model Answer - Q2(c)

The oil floats on top of the water / forms a layer above the water [1]

Because the oil is less dense than the water (800 < 1000 kg/m³) / oil is lighter than the same volume of water [1]

⚠ If you missed marks here: Saying "the oil floats" earns the first mark and nothing more — the explanation mark is the density comparison, 800 kg/m³ against 1000 kg/m³. "Oil is lighter than water" without the numbers or the word density is how this mark is usually lost.

Mark 1: Oil floats on top of water

Mark 2: Oil is less dense than water

(d) 2 marks

Steel has a density of 7800 kg/m³, which is much greater than the density of water (1000 kg/m³). Explain why a ship made of steel is able to float.

Model Answer - Q2(d)

The ship is hollow / has a large volume of air inside / the hull encloses a large volume [1]

This makes the average/overall density of the ship (steel + air) less than the density of water, so it floats / the weight of water displaced equals the weight of the ship [1]

⚠ If you missed marks here: "Because of upthrust" on its own earns nothing here. The marks are for the hull enclosing a large volume of air, and for the average density of the whole ship therefore being less than 1000 kg/m³ — the density of the steel itself never changes.

Mark 1: Ship is hollow / contains air

Mark 2: Average density of ship is less than water / displaces enough water

Question 3: Forces and Newton's Laws

Total: 12 marks

(a) 2 marks

State Newton's Second Law of Motion. Write the equation relating force, mass and acceleration.

Model Answer - Q3(a)

The resultant force acting on an object is equal to the product of its mass and acceleration / the acceleration of an object is proportional to the resultant force and inversely proportional to its mass [1]

F = ma, where F = resultant force (N), m = mass (kg), a = acceleration (m/s²) [1]

⚠ If you missed marks here: The word missing from most answers is resultant: it is the resultant force that equals mass × acceleration. Naming the symbols and their units is the second mark, so a bare F = ma is worth one at most.

Mark 1: Correct statement of Newton's Second Law (must mention resultant force)

Mark 2: F = ma with correct units/symbols

(b) 4 marks

A rocket of mass 5000 kg has an engine that produces an upward thrust of 65 000 N. The weight of the rocket is 49 000 N.

(b) (i) 2 marks

Calculate the resultant force acting on the rocket at lift-off.

Model Answer - Q3(b)(i)

Resultant force = Thrust - Weight = 65 000 - 49 000 [1]

Resultant force = 16 000 N upwards [1]

⚠ If you missed marks here: The two forces act in opposite directions, so subtract: 65 000 − 49 000 = 16 000 N. Adding them gives 114 000 N. Give the direction too — "16 000 N upwards" is what the mark asks for.

Mark 1: Correct subtraction (thrust - weight)

Mark 2: Correct answer 16 000 N (upwards)

(b) (ii) 2 marks

Calculate the initial acceleration of the rocket.

Model Answer - Q3(b)(ii)

a = F / m = 16 000 / 5000 [1]

a = 3.2 m/s² [1]

⚠ If you missed marks here: Use the resultant force, not the thrust: 16 000 / 5000 = 3.2 m/s², whereas 65 000 / 5000 gives 13 m/s². Dividing mass by force, 5000 / 16 000 = 0.31, is the other frequent slip.

Mark 1: Correct use of F = ma rearranged

Mark 2: Correct answer 3.2 m/s²

A ball is dropped from the top of a tall building. Describe and explain the motion of the ball as it falls, in terms of the forces acting, its velocity, and its acceleration. Include the concept of terminal velocity in your answer.

Model Answer - Q3(c)

Initially, the only force is weight (gravity) acting downwards, so the ball accelerates at approximately 9.8 m/s² / freely [1]

As speed increases, air resistance (drag) increases. The resultant downward force decreases, so acceleration decreases (but the ball is still speeding up) [1]

Eventually air resistance equals the weight. The resultant force is zero and the ball travels at constant velocity [1]

This constant maximum velocity is called the terminal velocity. Acceleration is now zero. [1]

⚠ If you missed marks here: The stage most often left out is the middle one: while air resistance is growing the ball is still speeding up, only less quickly each second. "The forces balance so the ball stops" is wrong — at terminal velocity it keeps moving at constant velocity with zero acceleration.

Mark 1: Initially only weight acts / accelerates freely downwards

Mark 2: Air resistance increases with speed / resultant force decreases / acceleration decreases

Mark 3: Air resistance equals weight / resultant force = 0 / constant velocity

Mark 4: This is terminal velocity / acceleration = 0

(d) 2 marks

A car travels around a circular bend on a flat road. Explain what provides the centripetal force and what would happen if the car's speed increases significantly.

Model Answer - Q3(d)

The centripetal force is provided by the friction between the tyres and the road [1]

If speed increases too much, the friction is insufficient to provide the centripetal force needed and the car will skid outwards / leave the circular path / move in a straight line tangentially [1]

⚠ If you missed marks here: Friction between the tyres and the road provides the centripetal force; "centrifugal force" is not a force acting on the car and scores zero. For the second mark say what happens physically: the friction available is too small, so the car carries straight on tangentially and slides outwards off the bend.

Mark 1: Friction between tyres and road provides centripetal force

Mark 2: Car skids / friction not sufficient if speed too great

Question 4: Springs and Moments

Total: 10 marks

(a) 2 marks

Three identical springs each have a spring constant of 20 N/m. A load of 6 N is hung from one of the springs. Calculate the extension of the spring.

Model Answer - Q4(a)

F = k x e, so e = F / k = 6 / 20 [1]

e = 0.3 m (or 30 cm) [1]

⚠ If you missed marks here: Rearrange before substituting: e = F / k = 6 / 20 = 0.3 m. Dividing the other way gives 3.3 m, a spring stretched over three metres. The unit is needed — 0.3 with nothing after it, or 0.3 cm, loses the answer mark.

Mark 1: Correct rearrangement of F = ke

Mark 2: Correct answer 0.3 m or 30 cm

(b) 3 marks

Using the spring from part (a), the following data is recorded: at 0 N the extension is 0 m; at 3 N the extension is 0.15 m; at 6 N the extension is 0.30 m. The spring reaches its limit of proportionality at a load of 6 N.

(i) Plot the first three data points on the axes below.
(ii) Sketch how the graph would continue up to a load of 10 N.

Model Answer - Q4(b)

Your graph should look like this:

Three points plotted correctly: (0, 0), (0.15, 3), (0.30, 6) [1]

Straight line drawn through all three points (showing proportionality up to 6 N) [1]

Beyond 6 N, the line curves away from the straight line (extension increases more for each unit of force / non-linear) showing the spring has exceeded its limit of proportionality [1]

⚠ If you missed marks here: Force belongs on the vertical axis and extension on the horizontal, matching the axes printed on the paper; swapping them puts every point in the wrong place. Draw one ruled straight line through the three points rather than joining them dot to dot, and beyond 6 N the line must bend towards the extension axis — more extension per newton, not less.

Mark 1: Three points correctly plotted

Mark 2: Straight line through points up to 6 N

Mark 3: Curve beyond 6 N showing non-proportional behaviour

A door handle is positioned 0.8 m from the hinge. A person pushes the handle with a force of 15 N perpendicular to the door. Calculate the moment of the force about the hinge.

Model Answer - Q4(c)

Moment = Force x perpendicular distance from pivot = 15 x 0.8 [1]

Moment = 12 N m [1]

⚠ If you missed marks here: Moment = 15 × 0.8 = 12 N m. The unit is the newton metre, N m — N/m is the unit of a spring constant and scores zero. Convert any distance given in centimetres before multiplying.

Mark 1: Correct formula and substitution

Mark 2: Correct answer 12 N m

(d) 3 marks

A uniform plank of length 5.0 m and weight 200 N is supported at both ends (points A and B). A painter weighing 700 N stands on the plank at a point 2.0 m from end A. Calculate the support force at each end.

Model Answer - Q4(d)

Take moments about A: (R_B x 5.0) = (700 x 2.0) + (200 x 2.5) [1]

R_B x 5.0 = 1400 + 500 = 1900, so R_B = 380 N [1]

R_A = total weight - R_B = (700 + 200) - 380 = R_A = 520 N [1]

⚠ If you missed marks here: The plank has its own weight of 200 N acting at its centre, 2.5 m from A; leaving it out gives 280 N instead of 380 N. Take moments about one end so that the unknown force there disappears, then find the other support from the total: R_A = 900 − 380 = 520 N. The two supports must add to 900 N, so use that as a check.

Mark 1: Correct moments equation about one end

Mark 2: Correct calculation of R_B = 380 N

Mark 3: Correct calculation of R_A = 520 N

Question 5: Momentum and Safety

Total: 12 marks

(a) 2 marks

State the principle of conservation of momentum.

Model Answer - Q5(a)

The total momentum of a system of objects remains constant / is conserved [1]

...provided no external resultant force acts on the system / in a closed system [1]

⚠ If you missed marks here: "Momentum before equals momentum after" is one mark, not two. The second mark is the condition, and it is the part most often omitted: provided no external resultant force acts on the system.

Mark 1: Total momentum is conserved / stays the same

Mark 2: Condition: no external resultant force / closed system

(b) 4 marks

A truck of mass 2000 kg is moving at 15 m/s. It collides with a car of mass 1000 kg moving at 5 m/s in the same direction. After the collision, the truck and car stick together and move as one object. Calculate their common velocity after the collision.

Model Answer - Q5(b)

Total momentum before = m1 x v1 + m2 x v2 [1]

= (2000 x 15) + (1000 x 5) = 30 000 + 5 000 = 35 000 kg m/s [1]

Total momentum after = (m1 + m2) x v = (2000 + 1000) x v = 3000v [1]

35 000 = 3000v, so v = 11.7 m/s (accept 11.67 m/s or 35/3 m/s) [1]

⚠ If you missed marks here: Both vehicles are already moving in the same direction, so the momentum before is 30 000 + 5 000 = 35 000 kg m/s — subtracting because they collide gives 25 000 and then 8.3 m/s. Afterwards the mass is the sum, 3000 kg, giving 11.7 m/s; dividing by 2000 kg gives 17.5 m/s.

Mark 1: Correct formula for momentum before

Mark 2: Correct total momentum before = 35 000 kg m/s

Mark 3: Correct expression for momentum after

Mark 4: Correct answer v = 11.7 m/s

A tennis ball of mass 0.060 kg is travelling at 20 m/s towards a racket. The racket hits the ball and it leaves the racket at 30 m/s in the opposite direction. The contact time between the ball and the racket is 0.005 s.

Calculate the change in momentum of the tennis ball. Be careful with direction.

Model Answer - Q5(c)(i)

Change in momentum = m(v - u) = 0.060 x (30 - (-20)) = 0.060 x 50 [1]
(Taking direction towards racket as negative, ball approaches at -20 m/s and leaves at +30 m/s)

Change in momentum = 3.0 kg m/s [1]

⚠ If you missed marks here: The ball reverses, so the two velocities have opposite signs and the change is 30 − (−20) = 50 m/s. Using 30 − 20 = 10 gives 0.6 kg m/s instead of 3.0 kg m/s, and that sign error is the single most common mistake on rebound questions.

Mark 1: Correct method accounting for direction change (v - u = 50)

Mark 2: Correct answer 3.0 kg m/s

Calculate the average force exerted by the racket on the ball.

Model Answer - Q5(c)(ii)

F = change in momentum / time = 3.0 / 0.005 [1]

F = 600 N [1]

⚠ If you missed marks here: Divide the change in momentum by the contact time: 3.0 / 0.005 = 600 N. Multiplying gives 0.015 N, and carrying an incorrect 0.6 kg m/s forward from (i) gives 120 N. Watch the decimal — 0.005 s is five thousandths of a second, not five hundredths.

Mark 1: Correct formula F = impulse / time

Mark 2: Correct answer 600 N

(d) 2 marks

Explain why seat belts are designed to stretch slightly during a crash, rather than being completely rigid.

Model Answer - Q5(d)

The stretching increases the time over which the person decelerates / increases the time for the momentum to change [1]

Since F = change in momentum / time, a longer time means a smaller force acts on the person, reducing injury [1]

⚠ If you missed marks here: "So it does not hurt as much" describes the result rather than the physics and scores nothing. Both marks come from the time: stretching increases the time over which the momentum falls to zero, and since force = change in momentum / time, a longer time gives a smaller force.

Mark 1: Increases the time taken to stop / decelerate

Mark 2: Reduces the force on the person (linked to F = impulse/t)

Question 6: Energy and Work

Total: 12 marks

(a) 3 marks

A high jumper of mass 70 kg clears a bar set at a height of 2.1 m above the ground. Assuming that all of the jumper's kinetic energy at take-off is converted to gravitational potential energy at the top of the jump, calculate the minimum kinetic energy needed at take-off.

[Use g = 9.8 m/s²]

Model Answer - Q6(a)

KE at take-off = GPE at the top of the jump [1]

GPE = mgh = 70 x 9.8 x 2.1 [1]

KE = 1440.6 J (accept 1441 J or 1440 J) [1]

⚠ If you missed marks here: The energy required is the GPE gained at the bar: 70 × 9.8 × 2.1 = 1440.6 J. Leaving g out gives 147 J. The first mark is for stating that the kinetic energy at take-off equals that GPE, so write that line down before calculating.

Mark 1: Correct energy conservation principle stated

Mark 2: Correct substitution into GPE = mgh

Mark 3: Correct answer (approximately 1441 J)

(b) 3 marks

Using your answer to part (a), calculate the minimum take-off speed of the high jumper.

Model Answer - Q6(b)

KE = 1/2 mv², so v² = 2 x KE / m [1]

v² = 2 x 1440.6 / 70 = 41.16 [1]

v = 6.4 m/s (accept 6.42 m/s) [1]

⚠ If you missed marks here: Rearranging ½mv² keeps the 2: v² = 2 × 1440.6 / 70 = 41.2, so v = 6.4 m/s. Dropping the 2 gives 4.5 m/s, and forgetting the square root leaves 41.2 m/s, which is faster than a car.

Mark 1: Correct rearrangement of KE = 1/2 mv²

Mark 2: Correct substitution

Mark 3: Correct answer v = 6.4 m/s

A crane lifts a load of mass 2000 kg vertically through a height of 15 m in 30 s.

Calculate the work done by the crane in lifting the load.

Model Answer - Q6(c)(i)

W = F x d = mgh = 2000 x 9.8 x 15 [1]

W = 294 000 J (294 kJ) [1]

⚠ If you missed marks here: The force being overcome is the weight, so W = mgh = 2000 × 9.8 × 15 = 294 000 J. Using the mass as though it were a force, 2000 × 15 = 30 000 J, is the standard slip — kilograms are never newtons.

Mark 1: Correct formula and substitution

Mark 2: Correct answer 294 000 J or 294 kJ

Calculate the useful output power of the crane.

Model Answer - Q6(c)(ii)

P = W / t = 294 000 / 30 [1]

P = 9800 W (9.8 kW) [1]

⚠ If you missed marks here: Power is work divided by time: 294 000 / 30 = 9800 W. Multiplying by 30 gives 8 820 000 W. The unit is the watt; writing 9800 J loses the mark even though the number is right.

Mark 1: Correct formula P = W / t

Mark 2: Correct answer 9800 W or 9.8 kW

(d) 2 marks

The crane motor uses 450 kJ of electrical energy to lift the load. Calculate the efficiency of the crane.

Model Answer - Q6(d)

Efficiency = (useful energy output / total energy input) x 100 = (294 / 450) x 100 [1]

Efficiency = 65.3% (accept 65%) [1]

⚠ If you missed marks here: Put both energies in the same unit — 450 kJ is 450 000 J, so the ratio is 294 000 / 450 000 = 65.3%. Mixing joules with kilojoules gives 65 300% or 0.065%, and inverting the fraction gives 153%, which no machine can achieve.

Mark 1: Correct formula with correct values substituted

Mark 2: Correct answer 65.3% (accept 65%)

Question 7: Pressure and Energy Resources

Total: 12 marks

(a) 4 marks

Using the relationship between pressure, force and area, explain why:
(i) a tractor has wide tyres
(ii) a nail has a very sharp point

Model Answer - Q7(a)

Tractor: Wide tyres have a large surface area in contact with the ground [1]

Since pressure = force / area, a large area means the pressure on the ground is small, so the tractor does not sink into soft ground [1]

Nail: The sharp point has a very small surface area [1]

Since pressure = force / area, a small area means the pressure is very large, so the nail can penetrate the material easily [1]

⚠ If you missed marks here: Each of the four marks needs the pressure equation applied, so "it spreads the weight out" scores nothing on its own. Say large area → small pressure → the tractor does not sink into soft ground, and small area → large pressure → the nail pushes into the wood.

Mark 1: Tractor - wide tyres = large area

Mark 2: Tractor - large area = low pressure (prevents sinking)

Mark 3: Nail - sharp point = small area

Mark 4: Nail - small area = high pressure (penetrates easily)

(b) 3 marks

A submarine dives to a depth of 200 m below the surface of the sea. The density of seawater is 1030 kg/m³. Calculate the pressure due to the water at this depth.

[Use g = 9.8 m/s²]

Model Answer - Q7(b)

p = rho x g x h [1]

p = 1030 x 9.8 x 200 [1]

p = 2 018 800 Pa (approximately 2.02 x 10⁶ Pa or 2020 kPa) [1]

⚠ If you missed marks here: All three factors are needed: 1030 × 9.8 × 200 = 2 018 800 Pa. Leaving g out gives 206 000 Pa, and using 1000 kg/m³ for fresh water instead of the seawater value given gives 1 960 000 Pa. The depth is already in metres, so no conversion is required.

Mark 1: Correct formula p = rho g h

Mark 2: Correct substitution of values

Mark 3: Correct answer approximately 2 019 000 Pa (accept 2.02 MPa)

Describe how electricity is generated in a hydroelectric dam. Include the energy transformations that take place.

Model Answer - Q7(c)

Water is stored at height behind a dam. When released, the water flows downward and its gravitational potential energy is converted to kinetic energy [1]

The moving water turns a turbine (kinetic energy of water to kinetic energy of turbine) [1]

The turbine is connected to a generator which converts kinetic energy to electrical energy [1]

⚠ If you missed marks here: "The water turns a turbine and makes electricity" is one mark at most. Trace the whole chain: gravitational potential to kinetic as the water falls, kinetic of the water to kinetic of the turbine, then the generator converting kinetic to electrical energy. No chemical or thermal store belongs in this answer.

Mark 1: GPE of water converts to KE as it falls

Mark 2: Water turns a turbine

Mark 3: Turbine drives generator, producing electrical energy

(d) 2 marks

State one advantage and one disadvantage of each of the following energy resources:
(i) Wind energy
(ii) Nuclear energy

Model Answer - Q7(d)

(i) Wind energy:
Advantage: Renewable / no fuel costs / no greenhouse gas emissions during operation / free energy source [any 1]
Disadvantage: Unreliable / depends on wind speed / intermittent / visual pollution / noise / large land area needed [any 1] [1 mark total for one correct advantage AND one correct disadvantage]

(ii) Nuclear energy:
Advantage: Very large energy output from small amount of fuel / reliable / does not produce greenhouse gases during operation / high power output [any 1]
Disadvantage: Produces radioactive waste / risk of nuclear accident / high cost to build and decommission / non-renewable (uranium) [any 1] [1 mark total for one correct advantage AND one correct disadvantage]

⚠ If you missed marks here: Each mark needs an advantage AND a disadvantage for the same resource, so two advantages of wind earns nothing for the other half. Do not claim nuclear power emits greenhouse gases while generating — it does not; its disadvantages are the radioactive waste, the accident risk and the cost of decommissioning.

Mark 1: Wind - one valid advantage AND one valid disadvantage

Mark 2: Nuclear - one valid advantage AND one valid disadvantage

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