IGCSE Physics Paper 4 (Theory / Extended)

Topic 1: Motion, Forces and Energy -- Cambridge Challenge Level (Set 2)

Duration 1 hour 15 minutes

Total Marks 80

Questions 7

Time Remaining 75:00

Syllabus 0625 / 0972

⚡ Cambridge Challenge Level

These questions match real Cambridge IGCSE difficulty. Scoring 50%+ is a solid B, 60%+ is an A, 70%+ is A*. Don't worry if this feels harder -- that's the point!

Instructions

Answer all questions in the spaces provided.
Show all working for calculations. Answers without working may not receive full marks.
You may use a calculator.
Take g = 10 N/kg where needed.
Answer all questions first. When finished, click "Submit Exam" at the bottom of Question 7.
Your answers will be automatically graded when you submit the exam. The model answers will be shown for review.

Question Navigation

Question 1 -- The Time Trial

Total: 12 marks

A cyclist rides a time-trial stage in the Yorkshire Dales. The graph shows her speed during the first 14 s, from the start line to a junction where she brakes. The combined mass of the cyclist and her bicycle is 80 kg.

(a) [2]

Calculate her acceleration during the first 4.0 s.

Model Answer -- 1(a)

a = Δv / Δt = (12 − 0) / 4.0 [1]

a = 3.0 m/s² [1]

⚠ If you missed marks here: Acceleration is the GRADIENT of a speed-time graph: rise 12 m/s over run 4.0 s gives 3.0 m/s². Dividing distance by time here (a common mix-up) gives an average speed, not an acceleration.

Mark 1 -- gradient method change in v over t (1 mark)

Mark 2 -- acceleration 3.0 m per s squared (1 mark)

(b) [2]

Calculate her deceleration between 10 s and 14 s.

Model Answer -- 1(b)

a = (12 − 6) / 4.0 [1]

deceleration = 1.5 m/s² [1]

⚠ If you missed marks here: Speed falls from 12 to 6 m/s in 4 s: 6 ÷ 4 = 1.5 m/s². Note she is still moving at 14 s — the graph ends at 6 m/s, not zero, so use 6 as the final speed, not 0.

Mark 1 -- speed drop 6 over 4 seconds used (1 mark)

Mark 2 -- deceleration 1.5 m per s squared (1 mark)

Calculate the total distance she travels in the 14 s shown.

Model Answer -- 1(c)

Distance = area under the graph [1]

½ × 4 × 12 = 24 m; 6 × 12 = 72 m; ½ × (12 + 6) × 4 = 36 m [1]

total = 24 + 72 + 36 = 132 m [1]

⚠ If you missed marks here: Distance is the AREA under a speed-time graph: triangle (24 m) + rectangle (72 m) + trapezium (36 m) = 132 m. The trapezium is where marks die — the braking section still covers ground, and its area is ½(12+6)×4 = 36 m, not a triangle’s 12 m.

Mark 1 -- area under graph method stated (1 mark)

Mark 2 -- three section areas 24 72 36 computed (1 mark)

Mark 3 -- total distance 132 m (1 mark)

(d) [2]

Between 4 s and 10 s her speed is constant even though she is pedalling hard. Explain this in terms of the forces acting on her.

Model Answer -- 1(d)

At constant speed the forces are balanced / the resultant force is zero [1]

Her forward (driving) force exactly equals the total resistive force (air resistance + friction) [1]

⚠ If you missed marks here: Constant speed does NOT mean no forces — it means the forces cancel. Her pedalling force is exactly matched by air resistance and friction, giving zero resultant. "No forces act" would be Newton’s first law misquoted.

Mark 1 -- resultant force zero at constant speed (1 mark)

Mark 2 -- driving force equals total resistive force (1 mark)

(e) [3]

Calculate the resultant force on the cyclist and bicycle while she is braking.

Model Answer -- 1(e)

F = m a [1]

F = 80 × 1.5 [1]

F = 120 N (opposing the motion) [1]

⚠ If you missed marks here: Use YOUR deceleration from (b): F = ma = 80 × 1.5 = 120 N, directed backwards (against the motion). If you used 12 m/s as an acceleration you mixed up a speed with a rate of change of speed.

Mark 1 -- F equals m a stated (1 mark)

Mark 2 -- substitution 80 times 1.5 (1 mark)

Mark 3 -- resultant force 120 N opposing motion (1 mark)

Question 2 -- The Canal Lock Beam

Total: 11 marks

On the Grand Union Canal, a boater opens a lock gate by pushing on a long white balance beam. She pushes with a force of 250 N at right angles to the beam, at a distance of 2.4 m from the hinge (pivot). The water resists the gate with a force acting at 0.75 m from the hinge.

(a) [2]

Define the moment of a force.

Model Answer -- 2(a)

Moment = force × perpendicular distance [1]

...from the pivot (to the line of action of the force) [1]

⚠ If you missed marks here: All three words earn their keep: force × PERPENDICULAR distance FROM THE PIVOT. Leaving out "perpendicular" or measuring to the wrong point costs the second mark.

Mark 1 -- force times perpendicular distance (1 mark)

Mark 2 -- distance measured from the pivot (1 mark)

(b) [2]

Calculate the moment of her push about the hinge.

Model Answer -- 2(b)

M = F d = 250 × 2.4 [1]

M = 600 N m [1]

⚠ If you missed marks here: 250 × 2.4 = 600 N m. The unit is newton-METRES (N m) — writing "600 N" or "600 J" loses the mark; a moment is not a force and not an energy.

Mark 1 -- moment 250 times 2.4 substituted (1 mark)

Mark 2 -- moment 600 newton metres with unit (1 mark)

The gate is just on the point of opening, so the beam is in equilibrium. State the principle of moments, and use it to calculate the resistive force of the water on the gate.

Model Answer -- 2(c)

Principle of moments: for a body in equilibrium, the sum of clockwise moments about a pivot equals the sum of anticlockwise moments [1]

F × 0.75 = 600 [1]

F = 600 / 0.75 [1]

F = 800 N [1]

⚠ If you missed marks here: Balance the moments, not the forces: her 600 N m must equal the water’s F × 0.75, so F = 800 N. Notice she overcomes an 800 N resistance with a 250 N push — that factor of 3.2 (= 2.4/0.75) is exactly why lock beams are long.

Mark 1 -- principle of moments stated for equilibrium (1 mark)

Mark 2 -- water moment F times 0.75 set equal to 600 (1 mark)

Mark 3 -- rearrangement 600 over 0.75 shown (1 mark)

Mark 4 -- resistive force 800 N calculated (1 mark)

(d) [3]

The beam itself is heavy, yet its weight has no turning effect that the boater must overcome when the beam is horizontal. Explain what is meant by centre of gravity, and why a beam pivoted at its centre of gravity has no resultant moment from its own weight.

Model Answer -- 2(d)

The centre of gravity is the point at which all the weight (of the object) may be considered to act [1]

If the pivot is at the centre of gravity, the weight acts THROUGH the pivot [1]

So its perpendicular distance from the pivot is zero, giving zero moment [1]

⚠ If you missed marks here: Weight acting THROUGH the pivot has zero perpendicular distance, so zero moment — distance, not size of force, is what kills the turning effect. (Real lock beams are actually balanced so the gate’s weight is offset; the physics is the same idea used deliberately.)

Mark 1 -- centre of gravity defined as point where weight acts (1 mark)

Mark 2 -- weight acts through the pivot (1 mark)

Mark 3 -- zero distance so zero moment (1 mark)

Question 3 -- Testing the Climbing Rope Spring

Total: 12 marks

A gear-testing laboratory in Llanberis, Snowdonia, tests a large steel spring used in a climbing-wall safety device. Loads are added and the extension measured:

load / N	0	50	100	150	200	250	300
extension / cm	0	2.0	4.0	6.0	8.0	11.0	15.0

(a) [2]

State Hooke’s law.

Model Answer -- 3(a)

The extension (of a spring) is directly proportional to the load / applied force [1]

...provided the limit of proportionality is not exceeded [1]

⚠ If you missed marks here: The condition is half the law: proportionality holds only UP TO the limit of proportionality. Quoting "extension is proportional to force" alone is one mark, not two.

Mark 1 -- extension proportional to load stated (1 mark)

Mark 2 -- condition limit of proportionality included (1 mark)

(b) [3]

Use the data to calculate the spring constant k in N/m.

Model Answer -- 3(b)

Use a point in the proportional region, e.g. 200 N and 8.0 cm [1]

k = F / x = 200 / 0.080 [1]

k = 2500 N/m [1]

⚠ If you missed marks here: Convert to metres BEFORE dividing: 8.0 cm = 0.080 m, so k = 200/0.080 = 2500 N/m. Using centimetres gives 25 — a plausible-looking wrong answer. And take the point from the straight-line region: the 250 N and 300 N rows are past the limit.

Mark 1 -- point chosen inside proportional region (1 mark)

Mark 2 -- k equals F over x with 0.080 m (1 mark)

Mark 3 -- spring constant 2500 N per m (1 mark)

Show that the spring stops obeying Hooke’s law during the test, and state the load at which the limit of proportionality is reached.

Model Answer -- 3(c)

Up to 200 N every 50 N adds 2.0 cm; beyond that, 50 N adds 3.0 cm then 4.0 cm — extension is no longer proportional to load [1]

Limit of proportionality is at (about) 200 N (8.0 cm) [1]

⚠ If you missed marks here: Test proportionality with the STEPS: equal 50 N steps give equal 2.0 cm steps until 200 N, then bigger and bigger steps. The limit is where the pattern breaks — 200 N — not the largest load in the table.

Mark 1 -- unequal extension steps beyond 200 N shown (1 mark)

Mark 2 -- limit identified at 200 N (1 mark)

(d) [2]

Calculate the extension produced by a load of 120 N.

Model Answer -- 3(d)

x = F / k = 120 / 2500 [1]

x = 0.048 m = 4.8 cm [1]

⚠ If you missed marks here: 120 N is inside the proportional region, so Hooke’s law applies: x = 120/2500 = 0.048 m = 4.8 cm. Sense check against the table: 120 N sits between 100 N (4.0 cm) and 150 N (6.0 cm) — 4.8 cm fits perfectly.

Mark 1 -- extension from F over k substituted (1 mark)

Mark 2 -- extension 4.8 cm or 0.048 m (1 mark)

(e) [3]

In the final device, TWO of these springs are fitted side by side (in parallel), sharing the load equally. Calculate the extension when the pair supports a 200 N load, and explain your reasoning.

Model Answer -- 3(e)

Each spring carries half the load: 200 / 2 = 100 N [1]

Each spring extends by x = 100 / 2500 = 0.040 m [1]

Both stretch together, so the extension of the pair is 4.0 cm — half the single-spring value [1]

⚠ If you missed marks here: Parallel springs SHARE the force: 100 N each, so each (and therefore the pair) extends 4.0 cm — half of the 8.0 cm one spring alone would give. Doubling the springs does not double the stretch; it halves it. (In series the opposite happens.)

Mark 1 -- load shared 100 N per spring (1 mark)

Mark 2 -- each spring extension 0.040 m computed (1 mark)

Mark 3 -- pair extension 4.0 cm which is half of single (1 mark)

Question 4 -- The Tackle at Twickenham

Total: 11 marks

In a rugby match at Twickenham, a 95 kg forward running at 6.0 m/s is tackled head-on by an 85 kg defender running at 4.0 m/s in the opposite direction. They hold on to each other and move together after the tackle. Take the forward’s original direction as positive.

(a) [2]

Calculate the momentum of the forward before the tackle.

Model Answer -- 4(a)

p = m v = 95 × 6.0 [1]

p = 570 kg m/s [1]

⚠ If you missed marks here: p = mv = 95 × 6.0 = 570 kg m/s. Momentum’s unit is kg m/s (or N s) — and it has a DIRECTION; here it is positive by the given convention.

Mark 1 -- momentum m v substituted (1 mark)

Mark 2 -- momentum 570 kg m per s (1 mark)

(b) [4]

Calculate the velocity of the pair immediately after the tackle. State its direction.

Model Answer -- 4(b)

Defender’s momentum = 85 × (−4.0) = −340 kg m/s (opposite direction, so negative) [1]

Total momentum before = 570 − 340 = 230 kg m/s (conserved in the collision) [1]

v = 230 / (95 + 85) = 230 / 180 [1]

v = 1.3 m/s (1.28 m/s) in the forward’s original direction [1]

⚠ If you missed marks here: The sign carries the physics: the defender’s momentum is −340, so the total is 230 (not 910). Momentum is conserved, and the pair (180 kg) moves at 230/180 = 1.3 m/s in the FORWARD’s direction — the sign of your answer tells you who won the collision.

Mark 1 -- defender momentum negative 340 (1 mark)

Mark 2 -- total momentum 230 by conservation (1 mark)

Mark 3 -- divided by combined mass 180 kg (1 mark)

Mark 4 -- velocity 1.3 m per s in forward direction (1 mark)

The tackle lasts 0.40 s. Calculate the average resultant force exerted on the DEFENDER during the tackle.

Model Answer -- 4(c)

Defender’s change in momentum: Δp = 85 × (1.28 − (−4.0)) = 85 × 5.28 = 450 kg m/s [1]

F = Δp / t = 450 / 0.40 [1]

F = 1100 N (1120 N) [1]

⚠ If you missed marks here: The defender goes from −4.0 to +1.28 m/s: a velocity CHANGE of 5.28 m/s (subtracting a negative adds). Δp = 450 kg m/s, so F = 450/0.40 ≈ 1100 N. Using 4.0 − 1.28 = 2.72 misses that his direction reversed.

Mark 1 -- momentum change 450 with sign handled (1 mark)

Mark 2 -- force equals change over time 0.40 s (1 mark)

Mark 3 -- average force about 1100 N (1 mark)

(d) [2]

Players are taught to relax and fall with a tackle, extending the time of the impact. Explain why this reduces the risk of injury.

Model Answer -- 4(d)

The change in momentum is fixed by the collision, so extending the impact TIME reduces the average force (F = Δp/t) [1]

A smaller force on the body means less risk of injury [1]

⚠ If you missed marks here: Δp is decided by the speeds; the only lever left is TIME. Doubling the impact time halves the force for the same momentum change — the same physics as crumple zones, airbags and bending your knees on landing.

Mark 1 -- longer time smaller force for same momentum change (1 mark)

Mark 2 -- smaller force reduces injury risk (1 mark)

Question 5 -- The Loch Diver

Total: 12 marks

A scientific diver surveys the bed of a Scottish freshwater loch. The density of the loch water is 1000 kg/m³, atmospheric pressure at the surface is 100 kPa, and g = 10 N/kg. The diver’s mask has a flat window of area 0.0060 m².

(a) [3]

Before the dive, a water sample is tested in the lab. Describe how to measure the density of the water sample using standard laboratory apparatus.

Model Answer -- 5(a)

Measure the mass of an empty measuring cylinder on a balance, add the water and measure again; subtract to get the water’s mass [1]

Read the water’s volume from the measuring cylinder scale (at eye level) [1]

density = mass / volume [1]

⚠ If you missed marks here: Three steps, three marks: mass by DIFFERENCE (so the cylinder’s own mass is removed), volume from the cylinder scale, then ρ = m/V. Forgetting to subtract the empty cylinder’s mass is the classic practical error.

Mark 1 -- mass found by difference using balance (1 mark)

Mark 2 -- volume read from measuring cylinder (1 mark)

Mark 3 -- density equals mass over volume (1 mark)

(b) [3]

Calculate the pressure due to the water alone at a depth of 25 m.

Model Answer -- 5(b)

Δp = ρ g h [1]

Δp = 1000 × 10 × 25 [1]

Δp = 250 000 Pa = 250 kPa [1]

⚠ If you missed marks here: ρgh = 1000 × 10 × 25 = 250 000 Pa = 250 kPa. This is the pressure due to the WATER only — read the question stem carefully; the atmosphere is added in the next part, not here.

Mark 1 -- pressure formula rho g h stated (1 mark)

Mark 2 -- substitution 1000 times 10 times 25 (1 mark)

Mark 3 -- water pressure 250 kPa calculated (1 mark)

Calculate the TOTAL pressure on the diver at 25 m.

Model Answer -- 5(c)

Total = water pressure + atmospheric pressure = 250 + 100 [1]

total = 350 kPa [1]

⚠ If you missed marks here: The atmosphere presses on the loch surface and that pressure is transmitted all the way down: total = 250 + 100 = 350 kPa. At depth the total is 3.5 atmospheres — which is why dive computers track depth so carefully.

Mark 1 -- atmospheric 100 kPa added to water pressure (1 mark)

Mark 2 -- total pressure 350 kPa (1 mark)

(d) [2]

Calculate the force exerted on the mask window by the water pressure alone at 25 m.

Model Answer -- 5(d)

F = p A = 250 000 × 0.0060 [1]

F = 1500 N [1]

⚠ If you missed marks here: F = pA needs the pressure in PASCALS: 250 000 × 0.0060 = 1500 N — the weight of a small piano on a palm-sized window. Using 250 (kPa) gives 1.5 N and misses why masks are built strong.

Mark 1 -- force equals pressure times area in pascals (1 mark)

Mark 2 -- force 1500 N calculated (1 mark)

(e) [2]

Explain, in terms of the water above the diver, why pressure increases with depth.

Model Answer -- 5(e)

The deeper the diver goes, the greater the weight of water in the column above [1]

That weight acts over the area below it, so the force per unit area (pressure) is greater [1]

⚠ If you missed marks here: Pressure at depth is the weight of everything stacked above, shared over each square metre. More depth = taller column = more weight = more pressure; the ρgh formula is just this argument written as algebra.

Mark 1 -- greater weight of water column above at depth (1 mark)

Mark 2 -- weight per area gives greater pressure (1 mark)

Question 6 -- Electric Mountain

Total: 12 marks

Dinorwig pumped-storage power station in North Wales ("Electric Mountain") stores energy by pumping water to a mountain-top reservoir at night, then releasing it through turbines at moments of peak demand. When generating, water falls through an average height of 500 m at a rate of 4.0 × 10⁵ kg every second. The electrical output is 1.7 GW.

(a) [3]

Calculate the gravitational potential energy lost by the water every second, and hence the maximum possible power.

Model Answer -- 6(a)

ΔE = m g Δh = 4.0 × 10⁵ × 10 × 500 (each second) [1]

ΔE = 2.0 × 10⁹ J per second [1]

So the maximum power available = 2.0 × 10⁹ W = 2.0 GW [1]

⚠ If you missed marks here: mgh with the mass PER SECOND gives energy per second — which IS power: 4.0 × 10⁵ × 10 × 500 = 2.0 × 10⁹ W = 2.0 GW. No separate time step is needed; that is the trick of working per second.

Mark 1 -- m g h with mass per second substituted (1 mark)

Mark 2 -- energy 2.0 GJ per second calculated (1 mark)

Mark 3 -- maximum power 2.0 GW identified (1 mark)

(b) [3]

Calculate the efficiency of the station when generating.

Model Answer -- 6(b)

efficiency = useful power output / total power input [1]

= 1.7 GW / 2.0 GW [1]

= 0.85 = 85% [1]

⚠ If you missed marks here: Useful OUT over total IN: 1.7/2.0 = 85%. Inverting the fraction gives 118% — an instant sign that output and input were swapped, since no machine exceeds 100%.

Mark 1 -- efficiency ratio defined correctly (1 mark)

Mark 2 -- 1.7 over 2.0 substituted (1 mark)

Mark 3 -- efficiency 85 percent (1 mark)

The upper reservoir can deliver 7.0 × 10⁹ kg of water through this height. Calculate the total gravitational potential energy available, in joules and in kWh. (1 kWh = 3.6 × 10⁶ J.)

Model Answer -- 6(c)

E = m g h = 7.0 × 10⁹ × 10 × 500 [1]

E = 3.5 × 10¹³ J [1]

3.5 × 10¹³ / 3.6 × 10⁶ = 9.7 × 10⁶ kWh [1]

⚠ If you missed marks here: The full reservoir: 7.0 × 10⁹ × 10 × 500 = 3.5 × 10¹³ J. Dividing by 3.6 × 10⁶ J/kWh gives 9.7 million kWh — roughly a million homes’ evening. Keep the powers of ten on paper; this is where they slip.

Mark 1 -- total m g h substituted for full reservoir (1 mark)

Mark 2 -- energy 3.5 times 10 to 13 J (1 mark)

Mark 3 -- converted to about 9.7 million kWh (1 mark)

(d) [3]

Pumping the water back up costs MORE energy than the station later generates, yet Dinorwig is highly profitable and useful to the National Grid. Explain why.

Model Answer -- 6(d)

It pumps at night using cheap surplus electricity (when demand is low), and sells at peak times when electricity is expensive/scarce [1]

It can reach full output in seconds, far faster than starting a conventional power station, so it covers sudden surges in demand [1]

So although it is a net CONSUMER of energy, it stores energy usefully — matching supply to demand (and it is the price/timing difference that makes it profitable) [1]

⚠ If you missed marks here: The station sells TIMING, not energy: buy cheap at 3 a.m., sell dear at the half-time kettle surge, and respond in seconds. Saying "it creates energy" fails instantly — the answer must acknowledge the net loss and explain why storage is still worth it.

Mark 1 -- pumps with cheap off-peak surplus electricity (1 mark)

Mark 2 -- responds in seconds to peak demand (1 mark)

Mark 3 -- net consumer but storage matches supply to demand (1 mark)

Question 7 -- The Skydive

Total: 10 marks

A skydiver of mass 70 kg (including kit) jumps from a plane above Salisbury Plain. She falls for some time before opening her parachute. Take g = 10 N/kg.

(a) [4]

Describe and explain how her velocity changes from the moment she jumps until she reaches terminal velocity, in terms of the forces acting.

Model Answer -- 7(a)

At first the only significant force is her weight, so she accelerates downwards (at about 10 m/s²) [1]

As speed increases, air resistance (drag) on her increases [1]

The resultant force (weight − drag) therefore decreases, so her acceleration decreases [1]

Eventually drag equals weight: resultant force zero, acceleration zero — she falls at constant (terminal) velocity [1]

⚠ If you missed marks here: The chain is: weight constant, drag grows with speed, resultant shrinks, acceleration shrinks, until drag = weight and the velocity stops changing. She does NOT stop — terminal velocity is constant speed, and it is the acceleration that has become zero.

Mark 1 -- initial acceleration from weight about 10 m per s squared (1 mark)

Mark 2 -- air resistance increases with speed (1 mark)

Mark 3 -- resultant and acceleration decrease (1 mark)

Mark 4 -- drag equals weight giving constant terminal velocity (1 mark)

(b) [2]

Calculate her weight, and state the size of the air resistance acting on her at terminal velocity.

Model Answer -- 7(b)

W = m g = 70 × 10 = 700 N [1]

At terminal velocity the forces balance, so air resistance = 700 N [1]

⚠ If you missed marks here: W = 70 × 10 = 700 N, and at terminal velocity the drag must EQUAL it — 700 N exactly, no calculation needed beyond the balance argument. Any other value contradicts "terminal".

Mark 1 -- weight 700 N calculated (1 mark)

Mark 2 -- air resistance equal 700 N at terminal velocity (1 mark)

She opens her parachute. Describe and explain her motion from that moment until she reaches a new terminal velocity.

Model Answer -- 7(c)

The parachute’s large area gives a much larger air resistance, now greater than her weight [1]

The resultant force is UPWARDS, so she decelerates (while still moving downwards) [1]

As she slows, drag falls, until drag again equals weight: a new, much LOWER terminal velocity [1]

⚠ If you missed marks here: She never moves upwards — the upward RESULTANT only slows her descent (the "jerked upward" look in videos is the camera operator still falling fast). Deceleration continues until drag shrinks back to 700 N at a much lower speed.

Mark 1 -- larger area gives drag greater than weight (1 mark)

Mark 2 -- upward resultant decelerates her while descending (1 mark)

Mark 3 -- new lower terminal velocity when forces rebalance (1 mark)

(d) [1]

Explain why the second terminal velocity allows her to land safely.

Model Answer -- 7(d)

The low landing speed means little kinetic energy / small momentum change on impact, so the forces on her body at touchdown are small [1]

⚠ If you missed marks here: Safe landing = low speed = small kinetic energy and momentum to get rid of at the ground — the ground supplies far less force over the landing.

Mark 1 -- low speed means small impact forces on landing (1 mark)

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