IGCSE Physics Paper 4 (Theory/Extended) - Cambridge Challenge
Topic 1: Motion, Forces and Energy | Core + Supplement
75 minutes
80
7
75:00
⚡ 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 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: London Underground Braking Analysis
Total: 12 marks
A London Underground train of mass 180 000 kg approaches King's Cross station. The velocity-time graph below shows the train's motion during the final 100 seconds of its journey.
(a)1 mark
Describe the motion of the train during section A (0 to 20 s).
Model Answer - Q1(a)
The train moves at a constant/uniform velocity of 20 m/s (no acceleration) [1]
⚠ If you missed marks here: "Moving steadily" or "constant speed" without a value is risky — the safe answer names constant (uniform) velocity and quotes 20 m/s. Watch out: "constant acceleration" is wrong, because in section A the acceleration is zero.
Mark 1 -- constant velocity 20 m/s or uniform speed
(b)3 marks
Calculate the deceleration of the train during section B (20 to 60 s). Show your working.
Model Answer - Q1(b)
a = (v - u) / t stated or shown [1]
a = (8 - 20) / (60 - 20) = -12 / 40 [1]
a = -0.3 m/s2 (deceleration = 0.3 m/s2) [1]
⚠ If you missed marks here: The classic slip is dividing by 60 s (the clock time) instead of 40 s (how long section B actually lasts, 20 s to 60 s) — that gives 0.2 m/s2 instead of 0.3 m/s2. Also check you used the change in velocity (8 − 20 = −12 m/s), not just 8 or 20 on its own.
Mark 1 -- correct formula a = change in velocity divided by time
Mark 2 -- correct substitution (8 - 20) / 40 or equivalent
Mark 3 -- correct answer 0.3 m/s squared deceleration
(c)3 marks
Calculate the total distance travelled by the train during the 100 seconds shown on the graph.
Model Answer - Q1(c)
Distance = area under velocity-time graph (three sections) [1]
Section A: 20 x 20 = 400 m; Section B: 1/2 x (20 + 8) x 40 = 560 m; Section C: 1/2 x 8 x 40 = 160 m [1]
Total distance = 400 + 560 + 160 = 1120 m [1]
⚠ If you missed marks here: Most lost marks come from using distance = speed × time on the whole graph (20 × 100 = 2000 m — wrong) instead of the area under the graph, or forgetting the ½ in the sloped sections. Section B is a trapezium: ½ × (20 + 8) × 40 = 560 m, not 20 × 40 = 800 m.
Mark 1 -- method using area under graph
Mark 2 -- at least two section areas calculated correctly
Mark 3 -- correct total distance 1120 m
(d)3 marks
Calculate the braking force required during section C to bring the train to rest. Use mass = 180 000 kg.
Model Answer - Q1(d)
First find deceleration in section C: a = (0 - 8) / (100 - 60) = -8 / 40 = -0.2 m/s2 [1]
F = ma = 180 000 x 0.2 [1]
F = 36 000 N (or 36 kN) [1]
⚠ If you missed marks here: The big trap is reusing section B's deceleration (0.3 m/s2) instead of working out section C's own value (8 m/s lost over 40 s = 0.2 m/s2) — that wrong route gives 54 000 N instead of 36 000 N. F = ma needs the acceleration for the section the question actually asks about.
Mark 1 -- correct deceleration 0.2 m/s squared for section C
Mark 2 -- correct use of F = ma with mass 180000
Mark 3 -- correct answer 36000 N or 36 kN
(e)2 marks
The driver applies a gentler braking force in section C than in section B. Suggest why the driver uses two stages of braking instead of one sudden stop from 20 m/s.
Model Answer - Q1(e)
A sudden stop would require a very large deceleration / braking force [1]
This would cause discomfort or injury to standing passengers / large forces on passengers (or: the wheels could lock/skid, making braking less effective) [1]
⚠ If you missed marks here: Just writing "it is safer" or "smoother" scores nothing — you need the physics chain: sudden stop → large deceleration → large force (F = ma) → injury/discomfort to passengers (or wheels skidding). Each mark needs one link of that chain stated explicitly.
Mark 1 -- sudden stop means large deceleration or large force
Mark 2 -- passenger safety or comfort or risk of skidding
Question 2: Cargo Ship Loading
Total: 12 marks
A cargo ship at Southampton port is being loaded with two types of cargo. The ship's hold is a cuboid measuring 40 m long, 10 m wide and 8 m deep. The density of seawater is 1025 kg/m3.
(a)3 marks
Cargo A consists of 800 identical steel crates. Each crate has dimensions 1.0 m x 1.0 m x 0.5 m and a mass of 320 kg.
(i) Calculate the density of one crate. State the unit.
(ii) Calculate the total mass of Cargo A.
Model Answer - Q2(a)
(i) Volume = 1.0 x 1.0 x 0.5 = 0.5 m3; density = mass / volume = 320 / 0.5 [1]
density = 640 kg/m3 [1]
(ii) Total mass of Cargo A = 800 x 320 = 256 000 kg [1]
⚠ If you missed marks here: Watch the crate volume: 1.0 × 1.0 × 0.5 = 0.5 m3, so density = 320 / 0.5 = 640 kg/m3 — if you got 320, you forgot the 0.5 m dimension. And the question says "state the unit": a correct 640 with no kg/m3 still loses that mark.
Mark 1 -- correct volume 0.5 cubic metres and density formula used
Mark 2 -- correct density 640 kg per cubic metre with unit
Mark 3 -- total mass 256000 kg
(b)2 marks
A single crate from Cargo A accidentally falls into the sea. Explain, using the density values, whether the crate will float or sink in the seawater.
Model Answer - Q2(b)
The density of the crate (640 kg/m3) is less than the density of seawater (1025 kg/m3) [1]
Therefore the crate will float, because an object floats when its density is less than that of the surrounding liquid [1]
⚠ If you missed marks here: The intuition trap is "steel sinks" — but the crate is mostly air inside, and the question tells you to use the density values. Quote BOTH numbers (640 kg/m3 vs 1025 kg/m3) for the first mark, then state the rule: less dense than the liquid → it floats.
Mark 1 -- comparison of crate density 640 with seawater density 1025
Mark 2 -- conclusion crate floats because its density is less than seawater
(c)3 marks
The hold is filled completely with a mixture of Cargo A (256 000 kg) and Cargo B. The average density of all cargo in the hold is 750 kg/m3.
Calculate the mass of Cargo B.
Model Answer - Q2(c)
Volume of hold = 40 x 10 x 8 = 3200 m3 [1]
Total mass = density x volume = 750 x 3200 = 2 400 000 kg [1]
Mass of Cargo B = 2 400 000 - 256 000 = 2 144 000 kg (2144 tonnes) [1]
⚠ If you missed marks here: The average density 750 kg/m3 belongs to the WHOLE hold (3200 m3), not to Cargo B alone. The two classic slips: stopping at total mass 2 400 000 kg without subtracting Cargo A's 256 000 kg, or miscalculating the hold volume (40 × 10 × 8 = 3200 m3).
Mark 1 -- hold volume calculated as 3200 cubic metres
Mark 2 -- total mass from density times volume equals 2400000 kg
Mark 3 -- mass of Cargo B equals 2144000 kg
(d)4 marks
Describe an experiment to determine the density of a small, irregularly shaped piece of Cargo B material in a school laboratory. Include:
the apparatus needed
how you would measure the volume
how you would calculate the density
one source of error and how it could be reduced
Model Answer - Q2(d)
Apparatus: balance (to measure mass) and measuring cylinder with water (or eureka/displacement can) [1]
Measure volume by displacement: record initial water level in measuring cylinder, lower the object in, record new water level; volume = final level - initial level [1]
Calculate density = mass / volume using the mass from the balance and volume from displacement [1]
Source of error: parallax error when reading the meniscus on the measuring cylinder; reduced by reading at eye level perpendicular to the scale (OR: air bubbles on the object surface increase apparent volume; reduced by gently tapping the cylinder) [1]
⚠ If you missed marks here: The shape is IRREGULAR, so measuring it with a ruler scores nothing — you must use displacement, and you must mention BOTH water levels (before and after) to earn that mark. The other commonly dropped mark is the last bullet: you need a specific error (parallax, air bubbles) AND how to reduce it, not just "human error".
Mark 1 -- apparatus includes balance and measuring cylinder or displacement can
Mark 2 -- displacement method described with initial and final water levels
Mark 3 -- density equals mass divided by volume stated
Mark 4 -- source of error identified with method to reduce it
Question 3: Car Crash Safety Analysis
Total: 11 marks
A car of mass 1200 kg travelling at 15 m/s collides head-on with a stationary truck of mass 4800 kg. After the collision, the vehicles lock together and move as one combined object.
(a)2 marks
State the principle of conservation of momentum and one condition for it to apply.
Model Answer - Q3(a)
The total momentum of a system of objects is constant / the total momentum before a collision equals the total momentum after [1]
Provided no external (resultant) force acts on the system [1]
⚠ If you missed marks here: Most students state "momentum is conserved" but forget the condition — "provided no external (resultant) force acts" is a whole mark on its own, and the question explicitly asks for it. "Momentum before = momentum after" alone only earns the first mark.
Mark 1 -- total momentum before equals total momentum after or is conserved
Mark 2 -- condition no external resultant force acts
(b)3 marks
Calculate the velocity of the combined vehicles immediately after the collision.
Model Answer - Q3(b)
Total momentum before = (1200 x 15) + (4800 x 0) = 18 000 kg m/s [1]
Total momentum after = (1200 + 4800) x v = 6000v [1]
18 000 = 6000v, so v = 3.0 m/s [1]
⚠ If you missed marks here: The classic slip is dividing 18 000 by the truck's 4800 kg instead of the COMBINED mass 6000 kg (that gives 3.75 m/s instead of 3.0 m/s) — the vehicles lock together, so they move as one 6000 kg object. Also remember the truck starts stationary, so momentum before is just 1200 × 15.
Mark 1 -- momentum before calculated as 18000 kg m/s
Mark 2 -- combined mass 6000 kg used with conservation equation
Mark 3 -- correct answer v = 3.0 m/s
(c)3 marks
The car driver is wearing a seatbelt. The collision brings the driver (mass 70 kg) from 15 m/s to 3.0 m/s in 0.12 s.
(i) Calculate the force exerted on the driver by the seatbelt during the collision.
(ii) Explain how an airbag reduces the risk of injury compared to a seatbelt alone.
Model Answer - Q3(c)
(i) Change in momentum = 70 x (3.0 - 15) = 70 x (-12) = -840 kg m/s; F = change in momentum / time = 840 / 0.12 [1]
F = 7000 N [1]
(ii) The airbag increases the time over which the driver decelerates / the driver's momentum changes over a longer time, so the force on the driver is reduced (F = change in momentum / time, greater time means smaller force) [1]
⚠ If you missed marks here: In (i) you must use the CHANGE in velocity, 15 − 3.0 = 12 m/s — using 3.0 alone gives 1750 N and using 15 gives 8750 N, both wrong; the answer is 70 × 12 / 0.12 = 7000 N. In (ii), "the airbag cushions you" scores zero: the mark is for saying it INCREASES THE TIME of the momentum change, so the force is smaller.
Mark 1 -- change in momentum calculated as 840 kg m/s and force formula used
Mark 2 -- correct force 7000 N
Mark 3 -- airbag increases stopping time so reduces force on driver
(d)3 marks
Calculate the total kinetic energy before the collision and after the collision. Use your answers to explain whether this collision is elastic or inelastic.
Model Answer - Q3(d)
KE before = 1/2 x 1200 x 152 = 135 000 J (truck has zero KE) [1]
KE after = 1/2 x 6000 x 3.02 = 27 000 J [1]
KE is not conserved (135 000 J vs 27 000 J), so this is an inelastic collision; the lost KE is converted to heat, sound and deformation of the vehicles [1]
⚠ If you missed marks here: Check you SQUARED the speeds (½mv2, not ½mv) and used 6000 kg — not 1200 kg — for the KE after. Also beware the reasoning trap: "momentum is conserved so it's elastic" is wrong; momentum is conserved in ALL collisions, but this one is inelastic because KE falls from 135 000 J to 27 000 J.
Mark 1 -- KE before calculated as 135000 J
Mark 2 -- KE after calculated as 27000 J
Mark 3 -- inelastic collision because kinetic energy is not conserved
Question 4: Hydraulic Car Jack
Total: 12 marks
A mechanic uses a hydraulic car jack to lift a car. The jack uses oil (an incompressible liquid) to transmit pressure. The small piston has a cross-sectional area of 5.0 cm2 and the large piston has a cross-sectional area of 200 cm2.
(a)2 marks
The mechanic pushes the small piston with a force of 150 N. Calculate the pressure exerted on the oil by the small piston. Give your answer in Pa.
Model Answer - Q4(a)
Convert area: 5.0 cm2 = 5.0 x 10-4 m2; pressure = force / area = 150 / (5.0 x 10-4) [1]
pressure = 300 000 Pa (or 3.0 x 105 Pa) [1]
⚠ If you missed marks here: The unit conversion is the killer: 5.0 cm2 = 5.0 × 10−4 m2 — you divide by 10 000 (100 × 100), not by 100. If your answer was 3000 Pa your area was 100× too big, and if it was 30 you never converted at all; the question asks for Pa, which needs metres.
Mark 1 -- area converted to square metres and pressure formula used
Mark 2 -- correct pressure 300000 Pa or 3.0 times 10 to the 5 Pa
(b)2 marks
Calculate the upward force exerted by the large piston on the car.
Model Answer - Q4(b)
Pressure is transmitted equally through the liquid: F = pressure x area = 300 000 x (200 x 10-4) [1]
F = 300 000 x 0.02 = 6000 N [1]
⚠ If you missed marks here: It is the PRESSURE (300 000 Pa) that is the same at both pistons, not the force — if you wrote 150 N you assumed equal forces. The large piston's area is 40× bigger (200/5), so the force is 40× bigger: F = p × A = 300 000 × 0.02 m2 = 6000 N.
Mark 1 -- same pressure used with large piston area converted correctly
Mark 2 -- correct force 6000 N
(c)2 marks
Explain why it is important that the oil used in the jack is incompressible.
Model Answer - Q4(c)
If the liquid were compressible, applying a force to the small piston would compress the liquid instead of transmitting pressure [1]
The large piston would not move / the force would not be fully transmitted to the large piston, so the jack would not lift the car effectively [1]
⚠ If you missed marks here: Vague answers like "the oil would leak" or "the jack would break" score nothing. The two marks are for the transmission idea: a compressible liquid would SQUASH instead of passing the pressure on, AND the consequence — the large piston would barely move, so the car would not lift.
Mark 1 -- compressible liquid would absorb force or compress instead of transmitting
Mark 2 -- large piston would not move or pressure not fully transmitted
(d)3 marks
A sealed container of gas is stored in the mechanic's workshop. The gas is at a pressure of 2.5 x 105 Pa and has a volume of 0.040 m3 at a temperature of 20 °C.
The container is accidentally left in direct sunlight and the temperature remains constant but a valve opens, allowing gas to escape into a larger connected chamber. The total volume becomes 0.10 m3.
Calculate the new pressure of the gas. State any law you use.
Model Answer - Q4(d)
At constant temperature, Boyle's law applies: p1V1 = p2V2 [1]
(2.5 x 105) x 0.040 = p2 x 0.10 [1]
p2 = 10 000 / 0.10 = 1.0 x 105 Pa (100 000 Pa) [1]
⚠ If you missed marks here: Naming the law is a mark on its own — write "Boyle's law: p1V1 = p2V2 at constant temperature" before substituting. The classic algebra slip gives 6.25 × 105 Pa (multiplying by 0.10/0.040 instead of 0.040/0.10) — a sanity check catches it: the gas EXPANDED, so the pressure must DROP.
Mark 1 -- Boyle's law stated p1V1 = p2V2 at constant temperature
Mark 2 -- correct substitution of values
Mark 3 -- correct answer 1.0 times 10 to the 5 Pa or 100000 Pa
(e)3 marks
Explain, in terms of molecules, why the pressure decreases when the gas expands into the larger volume at constant temperature.
Model Answer - Q4(e)
Gas molecules move randomly and collide with the walls of the container, creating pressure [1]
When volume increases, the same number of molecules are spread over a larger volume / there are fewer molecules per unit volume [1]
The molecules hit the walls less frequently (temperature is constant so speed is unchanged), so the force per unit area (pressure) decreases [1]
⚠ If you missed marks here: The classic wrong answer says the molecules "slow down" or "lose energy" — they don't, because the temperature (and so their speed) is constant. The marks are for the collision story: molecules hitting the walls cause pressure, the same number of molecules now fill a bigger volume, so they hit the walls LESS OFTEN.
Mark 1 -- molecules collide with walls creating pressure
Mark 2 -- larger volume means fewer molecules per unit volume
Mark 3 -- less frequent collisions with walls so pressure decreases
Question 5: Bungee Jumper Energy Transformations
Total: 11 marks
A bungee jumper of mass 65 kg jumps from a platform 50 m above the ground. The unstretched length of the bungee cord is 20 m. The jumper falls freely for the first 20 m before the cord begins to stretch. She comes momentarily to rest 8.0 m above the ground.
(a)2 marks
Calculate the speed of the jumper at point B, just as the cord becomes taut, after falling freely for 20 m. Use g = 9.8 m/s2.
Model Answer - Q5(a)
Using v2 = u2 + 2as, v2 = 0 + 2 x 9.8 x 20 = 392 [1]
v = 19.8 m/s (accept 20 m/s) [1]
⚠ If you missed marks here: If your answer was 392, you forgot the square root: v2 = 2 × 9.8 × 20 = 392, so v = √392 = 19.8 m/s. And v = u + at is a dead end here because no time is given — use v2 = u2 + 2as or the energy method (½mv2 = mgh).
Mark 1 -- correct use of v squared equals u squared plus 2as or energy method
Mark 2 -- correct speed approximately 19.8 m/s or 20 m/s
(b)3 marks
The jumper falls a total of 42 m from A to C (from 50 m to 8.0 m above the ground).
(i) Calculate the loss in gravitational potential energy between A and C.
(ii) At point C the jumper is momentarily at rest. State where the lost gravitational potential energy has been transferred to.
Model Answer - Q5(b)
(i) Loss in GPE = mgh = 65 x 9.8 x 42 [1]
Loss in GPE = 26 754 J (accept 26 700 J or 26 800 J) [1]
(ii) The GPE has been transferred to elastic (strain) potential energy stored in the stretched bungee cord (and a small amount to thermal energy due to air resistance) [1]
⚠ If you missed marks here: The height trap in (i): use 42 m (50 m down to 8.0 m), not 50 m or 20 m — check whether your answer matches 65 × 9.8 × 42 ≈ 26 750 J. In (ii), "kinetic energy" is wrong because she is momentarily at REST at C; the energy is stored as elastic (strain) potential energy in the stretched cord.
Mark 1 -- correct formula GPE = mgh with values substituted
Mark 2 -- correct answer approximately 26750 J
Mark 3 -- energy stored as elastic potential energy in the bungee cord
(c)3 marks
After reaching point C, the jumper bounces back up but only reaches a height of 38 m above the ground (not back to 50 m).
(i) Explain why she does not return to her original height.
(ii) Calculate the energy dissipated during one complete down-and-up bounce (from A at 50 m to 38 m).
Model Answer - Q5(c)
(i) Energy is dissipated as thermal energy (heat) due to air resistance on the jumper and internal friction/hysteresis in the bungee cord material [1]
(ii) Height difference = 50 - 38 = 12 m; energy dissipated = mgh = 65 x 9.8 x 12 [1]
Energy dissipated = 7644 J (accept 7650 J) [1]
⚠ If you missed marks here: In (ii) the height to use is the DIFFERENCE, 50 − 38 = 12 m — putting 38 m or 50 m into mgh gives a wrong answer several times too big. In (i), "energy is lost" alone doesn't score: say where it went — thermal energy from air resistance and from friction/heating inside the stretching cord.
Mark 1 -- energy lost to thermal energy due to air resistance or cord friction
Mark 2 -- height difference 12 m used with mgh formula
Mark 3 -- correct energy dissipated approximately 7644 J
(d)3 marks
Describe how the velocity, kinetic energy and gravitational potential energy of the jumper change as she falls from A to B (the free-fall section, ignoring air resistance).
Model Answer - Q5(d)
Velocity increases at a constant rate (uniform acceleration of 9.8 m/s2) because the only force acting is gravity [1]
Kinetic energy increases (from zero to maximum at B) as the jumper speeds up [1]
Gravitational potential energy decreases as height decreases; the decrease in GPE equals the increase in KE (energy is conserved, no air resistance) [1]
⚠ If you missed marks here: Saying velocity just "increases" usually drops the first mark — it increases AT A CONSTANT RATE (uniform acceleration 9.8 m/s2) because gravity is the only force. And don't stop at "KE up, GPE down": link them — the loss in GPE equals the gain in KE because energy is conserved in free fall.
Mark 1 -- velocity increases uniformly due to gravitational acceleration
Mark 2 -- kinetic energy increases from zero
Mark 3 -- GPE decreases and is converted to KE energy conserved
Question 6: Wind Turbine Power and Efficiency
Total: 11 marks
A wind turbine near Hampi, Karnataka is used to generate electricity. The turbine blades sweep a circular area of radius 25 m. When the wind speed is 12 m/s, the turbine generates 480 kW of electrical power.
(a)2 marks
State the main energy transformation that takes place in the wind turbine.
Model Answer - Q6(a)
Kinetic energy of the moving air/wind [1]
is transformed into electrical energy (via the generator) [1]
⚠ If you missed marks here: "Wind energy to electricity" is too vague for full marks — the examiner wants the named store: KINETIC energy of the moving air → ELECTRICAL energy. The word "kinetic" is what earns the first mark.
Mark 1 -- kinetic energy of the wind identified as input
Mark 2 -- electrical energy identified as useful output
(b)4 marks
The kinetic energy of the wind passing through the swept area each second can be calculated using:
Power in wind = 1/2 x (air density) x (swept area) x (wind speed)3
The density of air is 1.2 kg/m3. The swept area = πr2.
(i) Calculate the total power available in the wind passing through the turbine blades.
(ii) Calculate the efficiency of the turbine.
Model Answer - Q6(b)
(i) Swept area = π x 252 = 1963.5 m2 (accept 1960 m2) [1]
Power in wind = 1/2 x 1.2 x 1963.5 x 123 = 1/2 x 1.2 x 1963.5 x 1728 = 2 034 000 W (accept 2.03 MW or 2030 kW) [1]
(ii) Efficiency = useful power output / total power input = 480 000 / 2 034 000 [1]
Efficiency = 0.236 or 23.6% (accept 23-24%) [1]
⚠ If you missed marks here: The top error is squaring instead of CUBING the wind speed — 123 = 1728, and using 122 = 144 gives only about 170 kW instead of 2030 kW. In (ii), keep both powers in the same unit (480 000 W / 2 034 000 W ≈ 24%); an efficiency over 100% means you divided the wrong way round.
Mark 1 -- swept area calculated approximately 1960 square metres
Mark 2 -- power in wind approximately 2030 kW or 2034000 W
Mark 3 -- efficiency formula used correctly
Mark 4 -- efficiency approximately 23 to 24 percent
(c)2 marks
Suggest two reasons why the efficiency of the wind turbine is much less than 100%.
Model Answer - Q6(c)
Not all kinetic energy of the wind can be captured; some wind passes through without transferring energy / the wind cannot be brought completely to rest (it must keep flowing past) [1]
Energy is lost as heat due to friction in the bearings/gears/generator, or as sound energy from the blades [1]
⚠ If you missed marks here: "Energy is lost" with no mechanism scores zero — you need two DIFFERENT specific reasons. The one most students miss is that the wind cannot be stopped completely (air must keep flowing past the blades, carrying kinetic energy away); the easier one is friction in bearings/gears/generator or sound from the blades.
Mark 1 -- not all wind kinetic energy captured or wind passes through
Mark 2 -- friction losses in bearings or generator or sound energy
(d)3 marks
The turbine lifts water from a well to a storage tank 15 m above the ground as part of a rural electrification project. The turbine uses 20 kW of its electrical output for this purpose.
Calculate the maximum mass of water that can be lifted to the tank in one hour. State any assumption you make.
Model Answer - Q6(d)
Energy available = power x time = 20 000 x 3600 = 72 000 000 J [1]
E = mgh, so m = E / (g x h) = 72 000 000 / (9.8 x 15) [1]
m = 489 800 kg (accept approximately 490 000 kg or 490 tonnes). Assumption: no energy losses in the pump / 100% efficient pump (or all electrical energy is converted to GPE) [1]
⚠ If you missed marks here: The usual mark-loser is time: one hour = 3600 s, so energy = 20 000 × 3600 = 7.2 × 107 J (using t = 60 makes everything 60× too small). Then m = E/(g × h) = 7.2 × 107 / (9.8 × 15) — and don't forget the question also demands the assumption (100% efficient pump / no losses), which is part of the last mark.
Mark 1 -- energy calculated as 72000000 J from power times time
Mark 2 -- correct rearrangement m = E divided by g times h
Mark 3 -- correct mass approximately 490000 kg with assumption stated
Question 7: Construction Crane - Moments and Equilibrium
Total: 11 marks
A tower crane is used on a construction site in central London. The horizontal jib (beam) is 60 m long and pivots at a point 20 m from one end. A counterweight of mass 12 000 kg is placed at the short end. A load is to be lifted at the far end of the longer arm.
(a)2 marks
(i) Define the moment of a force.
(ii) State the equation for calculating a moment.
Model Answer - Q7(a)
(i) The moment of a force is the turning effect of a force about a point/pivot [1]
(ii) Moment = force x perpendicular distance from the pivot (M = F x d) [1]
⚠ If you missed marks here: The most commonly dropped word is "perpendicular" — moment = force × PERPENDICULAR distance from the pivot; "force × distance" alone can lose the mark. In (i), define it as the TURNING EFFECT of a force, not "a force that turns things".
Mark 1 -- turning effect of a force about a point or pivot
Mark 2 -- moment equals force times perpendicular distance
(b)3 marks
The counterweight has a mass of 12 000 kg and is positioned 20 m from the pivot. Calculate the maximum load (in kg) that can be lifted at the end of the 40 m arm without the crane tipping over. Ignore the weight of the jib.
Model Answer - Q7(b)
For equilibrium: clockwise moment = anticlockwise moment [1]
Counterweight moment = 12 000 x 9.8 x 20 = 2 352 000 N m; Load moment = m x 9.8 x 40 [1]
2 352 000 = m x 9.8 x 40, so m = 2 352 000 / 392 = 6000 kg [1]
⚠ If you missed marks here: The classic slip is swapping the distances — the counterweight sits on the 20 m arm and the load on the 40 m arm, so m = (12 000 × 20)/40 = 6000 kg; swapping gives 24 000 kg. Also state "clockwise moments = anticlockwise moments" explicitly — that principle is the first mark, even though g cancels in the arithmetic.
Mark 1 -- principle of moments stated clockwise equals anticlockwise
Mark 2 -- counterweight moment calculated correctly
Mark 3 -- maximum load 6000 kg
(c)2 marks
An astronaut has a mass of 75 kg on Earth.
(i) Calculate her weight on Earth (g = 9.8 m/s2).
(ii) On the Moon, g = 1.6 m/s2. State her mass on the Moon and explain why it differs from her weight on the Moon.
Model Answer - Q7(c)
(i) W = mg = 75 x 9.8 = 735 N [1]
(ii) Mass on the Moon = 75 kg (mass is the same everywhere); weight on the Moon = 75 x 1.6 = 120 N, which is different because weight depends on gravitational field strength (which is weaker on the Moon), while mass is the amount of matter and does not change with location [1]
⚠ If you missed marks here: The trap is saying her mass changes on the Moon — mass (75 kg) is the amount of matter and NEVER changes with location; only weight changes because W = mg and g is smaller there. For the second mark you need both the unchanged 75 kg AND the reason (weight depends on g, mass does not).
Mark 1 -- weight on Earth 735 N
Mark 2 -- mass stays 75 kg because mass is amount of matter weight depends on gravitational field strength
(d)4 marks
The crane lifts a 4000 kg load vertically through a height of 25 m in 50 s.
(i) Calculate the work done in lifting the load.
(ii) Calculate the power output of the crane motor.
(iii) The crane motor has an input power of 25 kW. Calculate the efficiency of the motor.
Model Answer - Q7(d)
(i) Work done = force x distance = mgh = 4000 x 9.8 x 25 = 980 000 J (980 kJ) [1]
(ii) Power = work done / time = 980 000 / 50 = 19 600 W (19.6 kW) [1]
(iii) Efficiency = useful power output / total power input = 19 600 / 25 000 [1]
Efficiency = 0.784 or 78.4% [1]
⚠ If you missed marks here: In (i), forgetting g gives 100 000 J — work done lifting = mgh = 4000 × 9.8 × 25 = 980 000 J (the lifting force is the WEIGHT, not the mass). In (iii), efficiency = output/input = 19 600 / 25 000 = 78.4%; if you got over 100% you inverted it, and remember 25 kW = 25 000 W before dividing.
Mark 1 -- work done 980000 J using mgh or force times distance
Mark 2 -- power output 19600 W using work divided by time
Mark 3 -- efficiency formula with correct values substituted
Mark 4 -- efficiency 0.784 or 78.4 percent
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C : 32-39
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