Topic 1 · Motion, forces and energy

Marks: 4113
Questions: 105
Multiple choice: 64
Suggested time: 50 minutes

PDFPrint the paperThe paper with writing space, then the mark scheme as a second booklet. Free.PPTXTeach it from the boardOne slide per question with its figure, each followed by its mark scheme. Editable.

Core and Supp.the whole paper · 41 marksCore onlySupp. questions drop out · 13 marks

How hard the questions are

Pitched at the board's own level, not above it. Multiple choice is mostly one step, with a couple of two-step items and one that rewards the candidate who does not take the obvious route. Structured questions open on a 1-mark recall and close on an explanation or a suggestion worth two marks — the marks real candidates most often drop.

11RecallOne idea, one step. The mark is for knowing it.
53RoutineThe standard application — the named equation, the usual graph read.
31DemandingSeveral steps, and you have to choose them. Nothing says which comes first.
10DiscriminatingThe part that separates the top grade: an unfamiliar context, a derivation, or an argument that has to hold together to earn anything.

Multiple choiceRecallCore1.1[1]

Which pair contains only vector quantities?

Adistance and displacement
Benergy and power
Cmass and weight
Dvelocity and weight

Ready to self-mark?Reveal the detailed answerAO11 mark

Answer overview

Dvelocity and weight

Multiple choiceRoutineCore1.2[1]

A cyclist travels 100 m in a straight line in 20 s, then stops at traffic lights for 10 s. What is the average speed over the whole 30 s?

A0 m/s
B3.3 m/s
C5.0 m/s
D10 m/s

Ready to self-mark?Reveal the detailed answerAO21 mark

Answer overview

B3.3 m/s

Multiple choiceRoutineCore1.4[1]

A rectangular metal block measures 2.0 cm × 3.0 cm × 5.0 cm and has a mass of 240 g. What is its density?

Fig. 3.1 An oblique drawing of a solid rectangular metal block. The front face is marked 5.0 cm along its lower edge and 3.0 cm up its left-hand edge, and the edge receding into the page is marked 2.0 cm. The block carries the label mass = 240 g.

A0.13 g/cm³
B8.0 g/cm³
C24 g/cm³
D7200 g/cm³

Ready to self-mark?Reveal the detailed answerAO21 mark

Answer overview

B8.0 g/cm³

Multiple choiceRoutineCore1.5.2[1]

A uniform beam is pivoted at its centre. A weight of 4.0 N hangs 0.80 m from the pivot on the left-hand side. A weight of 8.0 N is hung on the right-hand side so that the beam balances. How far from the pivot is the 8.0 N weight?

Fig. 4.1 A uniform beam rests on a pivot standing on the ground, the pivot placed at the centre of the beam. A 4.0 N weight hangs from the beam at a point 0.80 m to the left of the pivot, and an 8.0 N weight hangs to the right of the pivot at a distance marked d. Both distances are measured along the beam from the pivot. The drawing is not to scale.

A0.20 m
B0.40 m
C0.80 m
D1.6 m

Ready to self-mark?Reveal the detailed answerAO21 mark

Answer overview

B0.40 m

Multiple choiceDemandingSupp.1.7.1[1]

A ball of mass 0.50 kg is dropped from a height of 1.8 m and rebounds to a height of 1.2 m. How much energy is transferred away from the ball during the bounce?

Fig. 5.1 A ball of mass 0.50 kg is drawn twice above level ground. On the left it is at the point of release, with a dimension line marking 1.8 m from the ground up to the ball. On the right it is at the highest point reached after the bounce, with 1.2 m marked in the same way. Dashed vertical lines show the path down and the path back up.

A0.60 J
B2.9 J
C5.9 J
D8.8 J

Ready to self-mark?Reveal the detailed answerAO21 mark

Answer overview

B2.9 J

Multiple choiceRoutineSupp.1.8[1]

A diver is 15 m below the surface of a lake. The density of the water is 1000 kg/m³. What is the increase in pressure on the diver caused by the water above her?

A1.5 × 10³ Pa
B1.5 × 10⁴ Pa
C1.5 × 10⁵ Pa
D1.5 × 10⁶ Pa

Ready to self-mark?Reveal the detailed answerAO21 mark

Answer overview

C1.5 × 10⁵ Pa

StructuredDemandingSupp.1.2 · 1.5.1[9]

A car of mass 1200 kg starts from rest on a straight, level road. It reaches a speed of 24 m/s in 8.0 s. The forward force produced by the engine during this time is 4500 N.

Fig. 7.1 A car on a straight, level road, labelled mass = 1200 kg, with a faint arrow above it showing the direction of travel to the right. A long horizontal arrow points forwards from the front of the car and is labelled forward force 4500 N. A shorter horizontal arrow points backwards from the rear of the car and is labelled total resistive force, with no value given.

(a)
State State the equation linking acceleration, change in velocity and time taken.
[1]
Ready to self-mark?Reveal the detailed answerAO11 mark
Mark-by-mark answer
1. a = Δv / Δt, in words or symbols, with symbols defined or standard
(b)
Calculate Calculate the acceleration of the car.
[2]
Ready to self-mark?Reveal the detailed answerAO22 marks
Mark-by-mark answer
1. a = 24 / 8.0
2. a = 3.0 m/s² (unit required)
(c)
Calculate Calculate the resultant force acting on the car.
[2]
Ready to self-mark?Reveal the detailed answerAO22 marks
Mark-by-mark answer
1. F = ma = 1200 × 3.0
2. F = 3600 N (unit required)
(d)
Determine Determine the total resistive force acting on the car during this time.
[2]
Ready to self-mark?Reveal the detailed answerAO22 marks
Mark-by-mark answer
1. recognises resistive force = forward force − resultant force
2. 4500 − 3600 = 900 N
Examiner insight
The part that separates grades. Candidates who have treated 4500 N as the resultant in (c) cannot recover here, which is exactly what the structure is testing.
(e)
Explain The car later travels along the same road at a constant speed of 30 m/s. Explain, in terms of the forces acting, why the speed is constant.
[2]
Ready to self-mark?Reveal the detailed answerAO12 marks
Mark-by-mark answer
1. the resultant force on the car is zero
2. because the forward force from the engine is equal in size to the total resistive force (and opposite in direction)
Examiner insight
"There is no force" earns nothing. The marks are for a resultant of zero produced by two forces that balance.

StructuredDiscriminatingSupp.1.6 · 1.7.1[9]

Trolley A has a mass of 0.80 kg and moves along a straight, level track at 2.5 m/s. It collides with trolley B, of mass 1.2 kg, which is stationary. After the collision the two trolleys move off joined together. Friction is negligible.

Fig. 8.1 Two panels, one above the other. In the upper panel, labelled before the collision, trolley A of mass 0.80 kg stands on a level track with an arrow showing it moving to the right at 2.5 m/s towards trolley B of mass 1.2 kg, which is at rest. In the lower panel, labelled after the collision, the two trolleys are drawn joined together and moving to the right with a speed marked v.

(a)
State State the principle of conservation of momentum.
[2]
Ready to self-mark?Reveal the detailed answerAO12 marks
Mark-by-mark answer
1. the total momentum of a system (of interacting bodies) before an event equals the total momentum after it
2. provided no external resultant force acts
(b)
Calculate Calculate the speed of the joined trolleys immediately after the collision.
[3]
Ready to self-mark?Reveal the detailed answerAO23 marks
Mark-by-mark answer
1. momentum before = 0.80 × 2.5 = 2.0 kg m/s
2. total mass after = 0.80 + 1.2 = 2.0 kg
3. v = 2.0 / 2.0 = 1.0 m/s
(c)
Show (that) Show that kinetic energy is not conserved in this collision, and determine how much kinetic energy is transferred to other stores.
[3]
Ready to self-mark?Reveal the detailed answerAO23 marks
Mark-by-mark answer
1. E_k before = ½ × 0.80 × 2.5² = 2.5 J
2. E_k after = ½ × 2.0 × 1.0² = 1.0 J
3. 1.5 J transferred away, so kinetic energy is not conserved
Examiner insight
Momentum is conserved and kinetic energy is not: the pairing is the whole point of the question, and candidates who assume both are conserved lose all three marks.
(d)
Suggest Suggest what happens to the kinetic energy that is not accounted for after the collision.
[1]
Ready to self-mark?Reveal the detailed answerAO11 mark
Mark-by-mark answer
1. transferred to the internal (thermal) energy store of the trolleys and surroundings, and/or emitted as sound

StructuredRoutineSupp.1.7.3 · 1.7.4[8]

A wind turbine generates electricity for a school. Over a period of 60 s, the moving air delivers 900 kJ of energy to the turbine blades. The electrical output of the turbine during this time is 9.0 kW.

(a)
State State two renewable energy resources other than wind.
[2]
Ready to self-mark?Reveal the detailed answerAO12 marks
Mark-by-mark answer
1. any one of: solar / hydroelectric / tidal / wave / geothermal / biofuel
2. a second, different one from the same list
Examiner insight
Named fuels such as "gas" or "nuclear" score zero — both are non-renewable.
(b)
Calculate Calculate the input power delivered to the blades by the moving air.
[2]
Ready to self-mark?Reveal the detailed answerAO22 marks
Mark-by-mark answer
1. P = E / t = 900 000 / 60
2. P = 15 000 W = 15 kW
(c)
Calculate Calculate the efficiency of the turbine.
[2]
Ready to self-mark?Reveal the detailed answerAO22 marks
Mark-by-mark answer
1. efficiency = useful output power / total input power = 9.0 / 15
2. = 0.60 or 60%
(d)
Explain Explain why the efficiency of the turbine can never reach 100%.
[2]
Ready to self-mark?Reveal the detailed answerAO12 marks
Mark-by-mark answer
1. some energy is always transferred to non-useful stores — friction in the bearings and gearbox heats the surroundings, and sound is emitted
2. and the air must still be moving after it passes the blades, so it keeps some kinetic energy
Examiner insight
The second marking point is the one that separates a good answer. "Energy is lost" alone is not creditworthy: energy is not destroyed, it is transferred to a store that is not useful.

Practical skillsDemandingCore1.1 · 1.4[9]

A student determines the density of a small irregularly shaped stone. She has an electronic balance, a measuring cylinder, water and a length of thread. The stone fits easily inside the measuring cylinder.

Fig. 10.1 Two measuring cylinders drawn side by side, each with graduations up the wall and a curved meniscus. The first holds water alone, its surface at the 50.0 cm³ mark. The second holds the same water with the stone lowered in on a thread that runs up and out of the cylinder; the stone lies fully submerged near the base and the surface now stands at the 68.0 cm³ mark.

The student's readings
Quantity	Reading
mass of stone m	47.6 g
volume of water alone V₁	50.0 cm³
volume of water and stone V₂	68.0 cm³

(a)
Describe Describe how the student should use the measuring cylinder and thread to obtain the readings V₁ and V₂ accurately.
[3]
Ready to self-mark?Reveal the detailed answerAO33 marks
Mark-by-mark answer
1. part-fill the cylinder with water and record the level V₁, reading the bottom of the meniscus with the eye level with the surface
2. lower the stone in gently on the thread so that no water splashes out and the stone is fully submerged
3. record the new level V₂ once the water is still
Examiner insight
"Eye level with the meniscus" is the mark examiners report as most often missed, and it is worth as much as the method itself.
(b)
Calculate Calculate the density of the stone. Give your answer to an appropriate number of significant figures.
[3]
Ready to self-mark?Reveal the detailed answerAO23 marks
Mark-by-mark answer
1. V = V₂ − V₁ = 68.0 − 50.0 = 18.0 cm³
2. ρ = m / V = 47.6 / 18.0
3. ρ = 2.64 g/cm³ (2.6 g/cm³ also accepted; unit required)
(c)
Suggest The stone is porous and slowly absorbs water. Suggest how this affects the value obtained for the density, and suggest one improvement to the method.
[3]
Ready to self-mark?Reveal the detailed answerAO33 marks
Mark-by-mark answer
1. water enters the stone, so the rise in level V₂ − V₁ is smaller than the true volume of the stone
2. the calculated density is therefore too high
3. improvement: read V₂ immediately on submerging, or seal the stone first, or use a displacement can
Examiner insight
Two marks are for the direction of the error, not just its existence. Answering "it makes it inaccurate" scores one at most.

The cheapest marks on any paper

What the command words are asking for

Every board publishes these and marks to them. A candidate who explains where the question said state has spent four minutes earning one mark; one who states where it said explain has earned none.

Calculate: Work out from given facts, figures or information.
Define: Give the precise meaning.
Describe: State the points of a topic; give characteristics and main features.
Determine: Establish an answer using the information available.
Estimate: Suggest an approximate value.
Explain: Set out purposes or reasons; make relationships evident; give why and/or how.
Give: Produce an answer from a given source or recall.
Identify: Name, select or recognise.
Show (that): Provide structured evidence that leads to a given result.
State: Express in clear terms.
Suggest: Apply knowledge and understanding to situations where there is a range of valid responses in order to make proposals.

What is being tested

IGCSE assessment objectives, and how this paper divides between them

Every candidate sits two theory papers and one practical-skills paper. Core takes Papers 1 and 3 and is capped at grade C; Extended takes Papers 2 and 4 and reaches A*. Both then take either Paper 5 (practical test) or Paper 6 (alternative to practical), which carries 20% either way. Theory papers give you no formula sheet.

AO111 marks · 27%1 mark · 8%

Knowledge with understanding

Recall, describe, explain and use physics ideas, terminology, instruments and conventions.

Across the whole qualification: 50%

AO224 marks · 59%6 marks · 46%

Handling information and problem-solving

Locate and interpret information, translate between forms, calculate, reason, and apply physics to unfamiliar situations.

Across the whole qualification: 30%

AO36 marks · 15%6 marks · 46%

Experimental skills and investigations

Plan, use apparatus, record and present observations, analyse, evaluate, and suggest improvements.

Across the whole qualification: 20%