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Cambridge IGCSE Physics 0625 · Core and Extended · Topic 1

Motion, forces and energy

399 free, high-quality past-paper-style questions on Motion, forces and energy for Cambridge IGCSE Physics, each with a full mark scheme. Covers: Speed, velocity and motion graphs; Forces, friction and springs; Mass, weight and density; and more. For when you have finished the past papers, or want to practise one topic at a time.Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.1,336 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked

1IGPC-035-01-01Speed, velocity and motion graphs Paper CEasy1 mark

A cyclist travels 4.8 km along a straight road in 20 minutes.
What is the average speed of the cyclist?

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Marking pointMarkNotes
Convert to SI units before dividing. Distance = 4.8 km = 4800 m and time = 20 min = 1200 s. Average speed = distance ÷ time = 4800 ÷ 1200 = 4.0 m/s. The value 0.24 comes from 4.8 ÷ 20 with no unit conversion. The value 240 comes from converting km to m but leaving the time in minutes. The value 14 is the speed in km/h (4.8 km ÷ 1/3 h), not in m/s.✓ 1Answer B

Answer: B

Common error: Convert km to m and minutes to s before calculating. Mixing units is the most common way to get a speed that is wrong by a factor of 60 or 1000.

2IGPC-035-01-02Speed, velocity and motion graphs Paper CEasy1 mark

The distance–time graph for a delivery van is a horizontal straight line for the whole of a 5-minute interval.
What does this show about the van during this interval?

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Marking pointMarkNotes
On a distance–time graph the gradient equals the speed. A horizontal line has zero gradient. The distance is not changing, so the van is at rest. A constant speed would need a straight line sloping upwards. Option A confuses this graph with a speed–time graph, where a horizontal line does mean constant speed. Options C and D describe curved lines on a distance–time graph.✓ 1Answer B

Answer: B

Exam tip: Always check the label on the vertical axis first. The same shape of line means different things on distance–time and speed–time graphs.

3IGPC-035-01-03Speed, velocity and motion graphs Paper CEasy1 mark

The speed–time graph for a train is a straight line. The line slopes downwards from 20 m/s to 0 m/s.
Which statement describes the motion of the train?

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Marking pointMarkNotes
The gradient of a speed–time graph is the acceleration. A straight line has a constant gradient. Because the line slopes downwards, the speed is falling at a steady rate, so the deceleration is constant. The speed stays positive throughout, so the train is still moving forwards, not backwards. An increasing rate of deceleration would need a curve that gets steeper.✓ 1Answer C

Answer: C

Common error: A line sloping down on a speed–time graph means slowing down, not moving backwards. Direction changes only show up when velocity becomes negative.

4IGPC-035-01-09Speed, velocity and motion graphs Paper CEasy1 mark

A stone is dropped from rest from a high bridge. Air resistance is negligible.
What is the speed of the stone after 2.0 s? (g = 9.8 m/s2)

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Marking pointMarkNotes
In free fall the acceleration is a constant 9.8 m/s². This means the speed rises by 9.8 m/s every second. After 2.0 s the speed is 9.8 × 2.0 = 19.6 m/s, which is about 20 m/s. The value 9.8 m/s is the speed after only 1 s. The value 4.9 comes from dividing g by the time instead of multiplying. The value 39 comes from multiplying by the time twice.✓ 1Answer C

Answer: C

Exam tip: Rearrange a = Δv ÷ Δt to give Δv = a × Δt. Free-fall questions then become simple multiplications.

5IGPC-035-01-11Speed, velocity and motion graphs Paper CEasy4 marks

A student releases a toy car at the top of a track. She uses a stopwatch to time how long the car takes to travel the full 2.40 m length of the track. She repeats the measurement three times and records 3.1 s, 3.3 s and 3.2 s.

(a)

State the equation that links average speed, distance travelled and time taken.

(1)
(b)

Calculate the mean of the three time readings.

(1)
(c)

Calculate the average speed of the toy car.

(2)
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Marking pointMarkNotes
Part (a)
average speed = distance travelled ÷ time taken✓ 1
Part (b)
(3.1 + 3.3 + 3.2) ÷ 3 = 3.2 s✓ 1
Part (c)
2.40 ÷ 3.2✓ 1
= 0.75 m/s✓ 1

Exam tip: Repeating a timing and taking the mean reduces the effect of random errors, such as reaction time when starting and stopping the stopwatch.

6IGPC-035-01-12Speed, velocity and motion graphs Paper CEasy5 marks

A bus leaves a stop and moves along a straight road.
• From 0 s to 10 s it accelerates uniformly from rest to 15 m/s.
• From 10 s to 50 s it travels at a constant 15 m/s.
• From 50 s to 56 s it decelerates uniformly to rest at the next stop.

10505615time / sspeed / m/s
Graph sketched from the values in the question; not to scale
(a)

Describe the motion of the bus between 10 s and 50 s.

(1)
(b)

Calculate the acceleration of the bus during the first 10 s.

(2)
(c)

Calculate the distance travelled by the bus between 10 s and 50 s.

(2)
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Marking pointMarkNotes
Part (a)
It moves at a constant speed of 15 m/s (zero acceleration)✓ 1
Part (b)
a = (15 − 0) ÷ 10✓ 1
= 1.5 m/s2✓ 1
Part (c)
distance = speed × time = 15 × 40✓ 1
= 600 m✓ 1

Common error: In part (c) the time is 40 s (50 s − 10 s), not 50 s. Always subtract to find the duration of the section you need.

7IGPC-035-03-01Forces, friction and springs Paper CEasy1 mark

A small boat on a lake is pulled east by a rope with a force of 12 N. The wind pushes the boat west with a force of 5 N. What is the resultant force on the boat?

boatrope 12 Nwind 5 N
Diagram not to scale
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The forces act along the same line in opposite directions, so they are subtracted: 12 N − 5 N = 7 N. The resultant acts in the direction of the larger force, which is east. Adding gives 17 N, which treats opposite forces as if they act the same way. 7 N west gives the wrong direction. 2.4 N comes from dividing the forces, which has no physical meaning here.✓ 1Answer B

Answer: B

Exam tip: For forces along one line, choose one direction as positive. The sign of your answer then gives the direction of the resultant.

8IGPC-035-03-02Forces, friction and springs Paper CEasy1 mark

A wooden block is slid across a rough bench and slows down. Which statement about the friction force on the block is correct?

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Marking pointMarkNotes
Friction between surfaces opposes the motion of one surface over the other. Work done against friction transfers energy to the thermal stores of the block and bench, so both get warmer. Friction does not act in the direction of motion. Heating does occur. The vertical upward force on the block is the normal contact force, not friction.✓ 1Answer B

Answer: B

Common error: Do not confuse friction with the normal contact force. Friction acts along the surface, while the normal force acts at right angles to it.

9IGPC-035-03-03Forces, friction and springs Paper CEasy1 mark

A spring has an unstretched length of 8.0 cm. When a load is hung from it, its length becomes 11.5 cm. What is the extension of the spring?

8.0 cmload11.5 cmno loadwith load
Diagram not to scale
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Extension = stretched length − original length = 11.5 cm − 8.0 cm = 3.5 cm. The value 11.5 cm is the new length, not the extension. 8.0 cm is the original length. 19.5 cm comes from adding the two lengths instead of subtracting them.✓ 1Answer A

Answer: A

Common error: Extension is the increase in length, not the total length of the stretched spring.

10IGPC-035-03-10Forces, friction and springs Paper CEasy1 mark

Which of these cannot be caused by a force acting on an object?

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A force can change the size or shape of an object, and it can change its speed or direction of motion. Mass is the quantity of matter in the object, and pushing or pulling on the object does not change it. Options A, B and C are all genuine effects of forces.✓ 1Answer D

Answer: D

Exam tip: Remember the effects of a force: a change in size, shape, speed or direction.

11IGPC-035-03-11Forces, friction and springs Paper CEasy6 marks

A student hangs a spring from a clamp and adds loads to it. She records the length of the spring for each load.
Load / N: 0, 1.0, 2.0, 3.0, 4.0
Length / cm: 5.0, 6.2, 7.4, 8.6, 9.8

(a)

State how the student finds the extension of the spring for each load.

(1)
(b)

Calculate the extension of the spring when the load is 3.0 N.

(1)
(c)

Describe how the results show that the extension is directly proportional to the load.

(2)
(d)

Predict the length of the spring when the load is 2.5 N.

(2)
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Marking pointMarkNotes
Part (a)
She subtracts the original (unloaded) length of 5.0 cm from the measured length.✓ 1
Part (b)
8.6 − 5.0 = 3.6 cm✓ 1
Part (c)
Each extra 1.0 N increases the length, and so the extension, by the same amount (1.2 cm). The extension is zero at zero load, and doubling the load doubles the extension (for example, 2.4 cm at 2.0 N and 4.8 cm at 4.0 N).✓ 2
Part (d)
Extension = 2.5 × 1.2 = 3.0 cm . Length = 5.0 + 3.0 = 8.0 cm✓ 2

Common error: When asked for a length, remember to add the extension back on to the original length.

12IGPC-035-03-12Forces, friction and springs Paper CEasy5 marks

A cyclist rides along a straight, level road. At one moment the forward driving force is 150 N, the air resistance is 90 N and the friction force from the road and bearings is 20 N. Both resistive forces act backwards.

cyclistdriving force 150 Nair resistance 90 Nfriction 20 N
Diagram not to scale
(a)

Calculate the resultant force on the cyclist and state its direction.

(2)
(b)

Describe the motion of the cyclist at this moment.

(1)
(c)

The driving force stays the same and the cyclist's speed increases. State what happens to the air resistance and explain how the cyclist eventually reaches a constant speed.

(2)
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Marking pointMarkNotes
Part (a)
Resultant = 150 − 90 − 20 = 40 N , acting forwards (in the direction of motion)✓ 2
Part (b)
The cyclist is speeding up (accelerating) in the forward direction.✓ 1
Part (c)
The air resistance increases as the speed increases. It increases until the total backward force equals the driving force, so the resultant force becomes zero and the speed stays constant.✓ 2

Exam tip: A resultant force mark usually needs both the size and the direction of the force.

13IGPC-035-02-01Mass, weight and density Paper CEasy1 mark

A bag of rice has a mass of 2.5 kg. The gravitational field strength on Earth is 9.8 N/kg.
What is the weight of the bag of rice?

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Weight = mass × gravitational field strength. W = 2.5 × 9.8 = 24.5 N. Weight is a force, so it is measured in newtons, not kilograms. 2.5 N confuses mass with weight, and 0.26 N comes from dividing the mass by g.✓ 1Answer C

Answer: C

Common error: Weight is a force and is measured in newtons (N). Mass is measured in kilograms (kg). Do not mix up the two units.

14IGPC-035-02-02Mass, weight and density Paper CEasy1 mark

An astronaut travels from the Earth to the Moon. The gravitational field strength on the Moon is smaller than on the Earth.
What happens to the astronaut's mass and to the astronaut's weight?

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Mass is the quantity of matter in the astronaut. This does not change when the astronaut changes location. Weight is the gravitational force on the mass (W = mg). The Moon's g is smaller, so the weight decreases.✓ 1Answer C

Answer: C

Exam tip: Mass depends only on the object itself. Weight also depends on where the object is, because it depends on g.

15IGPC-035-02-03Mass, weight and density Paper CEasy1 mark

A rectangular metal block measures 4.0 cm × 2.5 cm × 2.0 cm. Its mass is 54 g.
What is the density of the metal?

4.0 cm2.0 cm2.5 cm
Diagram not to scale
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The volume is length × width × height: 4.0 × 2.5 × 2.0 = 20 cm3. Density = mass ÷ volume = 54 ÷ 20 = 2.7 g/cm3.
The other options come from common mistakes:
• 0.37 g/cm3 is volume ÷ mass.
• 6.4 g/cm3 comes from adding the three lengths instead of multiplying them.
• 1080 g/cm3 is mass × volume.
✓ 1Answer B

Answer: B

Common error: The volume of a rectangular block is found by multiplying the three dimensions, not adding them.

16IGPC-035-02-13Mass, weight and density Paper CEasy4 marks

The densities of some materials are:
• cork: 0.24 g/cm3
• ice: 0.92 g/cm3
• water: 1.00 g/cm3
• aluminium: 2.70 g/cm3

(a)

State which of cork and aluminium sinks in water.

(1)
(b)

Explain why ice floats on water.

(1)
(c)

A wooden block has a mass of 360 g and a volume of 400 cm3. Calculate its density and state whether it floats in water.

(2)
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Marking pointMarkNotes
Part (a)
Aluminium✓ 1
Part (b)
The density of ice (0.92 g/cm3) is less than the density of water (1.00 g/cm3)✓ 1
Part (c)
ρ = 360 ÷ 400 = 0.90 g/cm3✓ 1
It floats, because 0.90 g/cm3 is less than 1.00 g/cm3✓ 1

Common error: Saying ice floats because it is "light" is not enough. You must compare its density with the density of water.

17IGPC-035-06-01Energy, work and power Paper CEasy1 mark

A child pulls back the rubber band of a catapult and holds it still, ready to fire. In which energy store is the extra energy mainly held?

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A stretched (deformed) object stores energy in its elastic (strain) store. The band is held still, so there is no kinetic energy. The chemical store belongs to the child's muscles, not the band. Thermal energy is not the store that is filled by stretching.✓ 1Answer B

Answer: B

Common error: Do not name the source of the energy (the child's chemical store) when the question asks where the energy is now stored.

18IGPC-035-06-02Energy, work and power Paper CEasy1 mark

A worker pushes a crate across a floor with a constant horizontal force of 40 N. The crate moves 3.0 m in the direction of the force.
How much work does the worker do on the crate?

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Work done = force × distance moved in the direction of the force = 40 N × 3.0 m = 120 J. 0.075 J comes from distance ÷ force, 13 J from force ÷ distance, and 43 J from adding the two values.✓ 1Answer D

Answer: D

Exam tip: Work done W = F × d, where d is the distance moved in the direction of the force. 1 N m = 1 J.

19IGPC-035-06-03Energy, work and power Paper CEasy1 mark

An electric motor does 6000 J of work in 20 s.
What is the power of the motor?

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Marking pointMarkNotes
Power = work done ÷ time taken = 6000 J ÷ 20 s = 300 W. 0.0033 W is time ÷ work (inverted), 6020 W adds the numbers, and 120 000 W multiplies work by time.✓ 1Answer B

Answer: B

Exam tip: Power is the rate of doing work: P = W ÷ t. A watt is one joule per second.

20IGPC-035-06-10Energy, work and power Paper CEasy1 mark

Two students, A and B, have the same weight. They both run up the same flight of stairs. Student A takes 8 s and student B takes 12 s.
Which statement is correct?

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Marking pointMarkNotes
Both students lift the same weight through the same height, so they do the same work. Power = work done ÷ time. A does this work in less time, so A develops the greater power. The time taken does not change the amount of work done.✓ 1Answer B

Answer: B

Exam tip: Work depends on force and distance only. Power also depends on the time taken: same work in less time means more power.

21IGPC-035-06-12Energy, work and power Paper CEasy6 marks

A shopper pushes a loaded trolley along a level supermarket aisle. The trolley moves at a constant speed.

(a)

State the equation that links work done, force and distance moved.

(1)
(b)

The shopper pushes with a constant force of 35 N for a distance of 18 m. Calculate the work done by the shopper.

(2)
(c)

The trolley moves at constant speed, so its kinetic energy does not increase. State where the energy transferred by the shopper goes.

(1)
(d)

The shopper takes 12 s to push the trolley the 18 m. Calculate the power of the shopper's push.

(2)
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Marking pointMarkNotes
Part (a)
work done = force × distance moved in the direction of the force (W = Fd)✓ 1
Part (b)
W = 35 × 18✓ 1
= 630 J✓ 1
Part (c)
To the thermal store of the trolley's wheels/floor/surroundings, because of friction (and air resistance)✓ 1
Part (d)
P = W ÷ t = 630 ÷ 12✓ 1
= 53 W (52.5 W)✓ 1

Common error: Constant speed does not mean no work is done. The work is done against friction, so the energy goes to the thermal store of the surroundings.

22IGPC-035-00-01Physical quantities and measurement Paper CEasy1 mark

A student pours water into a measuring cylinder. The water surface curves downwards in the middle, forming a meniscus.
Which method gives the most accurate reading of the volume?

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For water, the reading is taken at the bottom of the meniscus. The eye must be level with the surface to avoid parallax error. Looking from above or below shifts the reading against the scale, and the top edge of the meniscus is not the correct reference.✓ 1Answer A

Answer: A

Exam tip: Remember two things for a measuring cylinder: eye level with the surface (no parallax) and the bottom of the meniscus.

23IGPC-035-00-02Physical quantities and measurement Paper CEasy1 mark

A student uses a stopwatch to time 20 complete oscillations of a pendulum. The total time is 36.0 s.
What is the period of the pendulum?

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Period = total time ÷ number of oscillations = 36.0 ÷ 20 = 1.80 s. 0.56 s comes from dividing 20 by 36.0. 3.60 s comes from dividing by 10 instead of 20. 720 s comes from multiplying instead of dividing.✓ 1Answer B

Answer: B

Common error: Do not divide the number of oscillations by the time. The period is a time, so it is time ÷ number of oscillations.

24IGPC-035-00-03Physical quantities and measurement Paper CEasy1 mark

A packet contains 250 identical sheets of paper. The stack is 2.5 cm thick.
What is the thickness of one sheet?

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Marking pointMarkNotes
Thickness of one sheet = 2.5 cm ÷ 250 = 0.010 cm. Since 1 cm = 10 mm, this is 0.10 mm. 0.010 mm is the number in cm written with the wrong unit. 1.0 mm and 10 mm come from dividing by 25 or 2.5 instead of 250, or converting units wrongly.✓ 1Answer B

Answer: B

Common error: Do the division in the original unit first, then convert. A common slip is keeping the number in centimetres but writing millimetres.

25IGPC-035-00-12Physical quantities and measurement Paper CEasy4 marks

A student uses a measuring cylinder to find the volume of an irregularly shaped stone.
The cylinder contains water to the 35 cm3 mark. The stone is lowered gently into the water and the level rises to 48 cm3.

35 cm³before48 cm³after
Diagram not to scale
(a)

Describe how the student should read the water level to get an accurate reading.

(2)
(b)

Calculate the volume of the stone.

(1)
(c)

Suggest why the stone must be completely under the water when the second reading is taken.

(1)
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Marking pointMarkNotes
Part (a)
Place the eye level with the water surface, to avoid parallax✓ 1
Read the scale at the bottom of the meniscus✓ 1
Part (b)
48 − 35 = 13 cm3✓ 1
Part (c)
Otherwise only part of the stone's volume displaces water, so the measured volume is too small✓ 1

Common error: The volume of the stone is the rise in water level, not the final reading.

26IGPC-035-04-01Turning effect and centre of gravity Paper CEasy1 mark

A force of 20 N acts on a lever. The perpendicular distance from the pivot to the line of action of the force is 0.30 m.
What is the moment of the force about the pivot?

20 N0.30 m
Diagram not to scale
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Marking pointMarkNotes
Moment = force × perpendicular distance from the pivot = 20 N × 0.30 m = 6.0 N m. Option A divides distance by force, C divides force by distance, and D uses the distance in centimetres (30) instead of metres.✓ 1Answer B

Answer: B

Exam tip: Convert distances to metres before calculating a moment in N m.

27IGPC-035-04-02Turning effect and centre of gravity Paper CEasy1 mark

Which unit is used for the moment of a force?

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Marking pointMarkNotes
A moment is force multiplied by distance, so its unit is the newton metre, N m. N/m comes from wrongly dividing force by distance, N/m2 is the unit of pressure, and kg m/s is the unit of momentum.✓ 1Answer B

Answer: B

Common error: Moment (N m) and pressure (N/m²) both involve newtons and metres. Moment multiplies by distance; pressure divides by area.

28IGPC-035-04-03Turning effect and centre of gravity Paper CEasy1 mark

A door handle is fixed on the edge of the door furthest from the hinges.
Why does this make the door easier to open?

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Marking pointMarkNotes
Moment = force × perpendicular distance. A larger distance from the hinge (the pivot) means the same force produces a larger turning effect. The weight of the door and the friction in the hinges do not change. The moment needed to open the door is also unchanged; what changes is the force needed to provide that moment.✓ 1Answer C

Answer: C

Exam tip: When explaining turning effects, name the pivot and say that the perpendicular distance from it has changed.

29IGPC-035-07-01Pressure in solids and liquids Paper CEasy1 mark

Which unit is equivalent to the pascal (Pa), the unit of pressure?

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Marking pointMarkNotes
Pressure is force per unit area, p = F/A. The unit is therefore newtons divided by square metres, N/m2, which is called the pascal. N m is the unit of moment or work, N/m has no area squared, and kg/m3 is the unit of density.✓ 1Answer C

Answer: C

Common error: Mixing up the units of pressure and density is common. Pressure involves force (N) and area (m²). Density involves mass (kg) and volume (m³).

30IGPC-035-07-02Pressure in solids and liquids Paper CEasy1 mark

A wooden crate of weight 60 N rests on a floor. The base of the crate is a rectangle measuring 0.20 m by 0.30 m.
What is the pressure the crate exerts on the floor?

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Marking pointMarkNotes
The base area is A = 0.20 × 0.30 = 0.060 m2.
The pressure is p = F/A = 60 / 0.060 = 1000 Pa.
3.6 Pa comes from multiplying force by area. 120 Pa comes from adding the two side lengths instead of multiplying them. 300 Pa comes from dividing by only one side length.
✓ 1Answer D

Answer: D

Exam tip: Always work out the area first, length × width in m². Then divide the force by that area.

31IGPC-035-07-03Pressure in solids and liquids Paper CEasy1 mark

A walker wears wide, flat snowshoes and sinks less into soft snow than when wearing ordinary boots.
Why does this happen?

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Marking pointMarkNotes
The walker's weight, which is the force on the snow, is unchanged. The snowshoes spread this force over a much larger area. Because p = F/A, a larger area gives a smaller pressure, so the walker sinks less. The weight and the force do not decrease. The pressure decreases, not increases.✓ 1Answer B

Answer: B

Common error: Saying the force or weight is reduced does not explain the effect. The force stays the same; only the area changes, and that is what changes the pressure.

32IGPC-035-07-11Pressure in solids and liquids Paper CEasy5 marks

A person of weight 600 N stands on a wooden floor wearing shoes with narrow heels.

(a)

State the equation that links pressure, force and area.

(1)
(b)

At one moment the person's whole weight rests on one heel of area 1.5 × 10−4 m2. Calculate the pressure on the floor.

(2)
(c)

Explain why wearing flat trainers is less likely to dent the wooden floor.

(2)
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Marking pointMarkNotes
Part (a)
pressure = force ÷ area, or p = F/A✓ 1
Part (b)
p = 600 / 1.5 × 10−4✓ 1
p = 4.0 × 106 Pa✓ 1
Part (c)
The trainers have a much larger area in contact with the floor, while the force (weight) stays the same✓ 1
so the pressure on the floor is smaller and less likely to dent it✓ 1

Exam tip: In an explanation, state both that the force is unchanged and that the area changes. Then link this to the pressure.

33IGPC-035-02-11Mass, weight and density Paper CEasy5 marks

A traveller packs a suitcase of mass 18 kg. The gravitational field strength on Earth is 9.8 N/kg.

(a)

State what is meant by the mass of an object.

(1)
(b)

Calculate the weight of the suitcase on Earth.

(2)
(c)

The suitcase is taken to a planet where the gravitational field strength is 3.7 N/kg. State the mass of the suitcase on this planet and calculate its weight there.

(2)
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Marking pointMarkNotes
Part (a)
The quantity (amount) of matter in the object (accept: a measure of the object's resistance to a change in its motion)✓ 1
Part (b)
W = mg = 18 × 9.8✓ 1
= 176 N (accept 180 N)✓ 1
Part (c)
Mass = 18 kg, unchanged✓ 1
Weight = 18 × 3.7 = 67 N (accept 66.6 N)✓ 1

Exam tip: When an object moves to a different planet, its mass stays the same. Only its weight changes, because weight depends on the gravitational field strength.

34IGPC-035-07-04Pressure in solids and liquids Paper CEasy1 mark

A tall plastic bottle full of water has three small holes at different heights on one side. Water spurts out of all three holes.
Which statement about the water as it leaves the holes is correct?

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Marking pointMarkNotes
Pressure in a liquid increases with depth below the surface (p = ρgh). The bottom hole is deepest below the water surface, so the pressure there is greatest and water is pushed out of it fastest. The density is the same throughout, but the depth differs at each hole, so the pressures differ. Height above the ground does not affect the pressure in the water. Pressure does not peak halfway down; it keeps increasing all the way to the bottom.✓ 1Answer C

Answer: C

Exam tip: For liquid pressure, measure depth from the liquid surface downwards, not from the bottom of the container upwards.

35IGPC-035-04-12Turning effect and centre of gravity Paper CEasy5 marks

A student wants to find the centre of gravity of a flat, irregularly shaped piece of plywood. She has a clamp stand, a pin, a plumb line and a pencil. There are three small holes near the edge of the plywood.

(a)

State what is meant by the centre of gravity of an object.

(1)
(b)

Describe how the student can find the centre of gravity of the plywood.

(3)
(c)

Suggest one way the student can check her result.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
The point through which the weight of the object appears to act / can be considered to act✓ 1
Part (b)
Hang the plywood from the pin through one hole so that it swings freely, and hang the plumb line from the same pin✓ 1
When both are still, mark the line of the plumb line on the plywood with the pencil✓ 1
Repeat using a second hole; the centre of gravity is where the lines cross✓ 1
Part (c)
Any one of: repeat using the third hole and check that the third line passes through the same point / balance the plywood horizontally on a pencil tip or fingertip placed at the marked point✓ 1

Common error: The plywood must be free to swing on the pin. If it is clamped tightly, it cannot settle with its centre of gravity directly below the pin.

36IGPC-035-07-12Pressure in solids and liquids Paper CEasy5 marks

A concrete dam holds back a deep reservoir of fresh water. The wall of the dam is much thicker at the bottom than at the top.

(a)

State two factors that the pressure in a liquid depends on.

(2)
(b)

Explain why the dam wall is thicker at the bottom.

(2)
(c)

State the direction in which the water pressure acts at a point in the water.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Any two from: depth below the surface of the liquid; density of the liquid; gravitational field strength g—
first correct factor✓ 1
second correct factor✓ 1
Part (b)
The pressure of the water increases with depth✓ 1
so the force on the wall is greatest at the bottom, and a thicker wall is needed there to withstand it✓ 1
Part (c)
in all directions, or equally in every direction✓ 1

Common error: The total amount of water in the reservoir does not set the pressure on the dam. It is the depth of the water that matters.

37IGPC-035-04-11Turning effect and centre of gravity Paper CEasy4 marks

A mechanic uses a wheel brace to undo a tight wheel nut. He pushes down on the end of the brace with a force of 150 N. The force is at right angles to the brace, 0.40 m from the centre of the nut.

150 N0.40 m
Diagram not to scale
(a)

Define the moment of a force about a pivot.

(1)
(b)

Calculate the moment of the 150 N force about the centre of the nut.

(2)
(c)

The nut still does not turn. Suggest how the mechanic could produce a larger moment without increasing his force.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
(Moment =) force × perpendicular distance from the pivot to the line of action of the force✓ 1
Part (b)
Moment = 150 × 0.40✓ 1
= 60 N m✓ 1
Part (c)
Increase the perpendicular distance from the nut to the force, e.g. use a longer brace or fit an extension bar and push further from the nut✓ 1

Exam tip: A definition of moment that leaves out the word perpendicular is incomplete.

38IGPC-035-02-12Mass, weight and density Paper CEasy5 marks

A student wants to find the density of a rectangular block of aluminium.

5.0 cm3.0 cm4.0 cm
Diagram not to scale
(a)

State the equation that links density, mass and volume.

(1)
(b)

Describe how the student can use a ruler to find the volume of the block.

(2)
(c)

The block measures 5.0 cm × 4.0 cm × 3.0 cm and has a mass of 162 g. Calculate the density of aluminium in g/cm3.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
density = mass ÷ volume (ρ = m/V)✓ 1
Part (b)
Measure the length, width and height of the block with the ruler✓ 1
Multiply the three measurements together: V = l × w × h✓ 1
Part (c)
V = 5.0 × 4.0 × 3.0 = 60 cm3✓ 1
ρ = 162 ÷ 60 = 2.7 g/cm3✓ 1

Exam tip: Show the volume as a separate step. This can earn a mark even if the final answer is wrong.

39IGPC-035-06-11Energy, work and power Paper CEasy5 marks

A diesel-powered crane on a building site lifts a load of steel beams from the ground to the top of a building.

(a)

State the principle of conservation of energy.

(1)
(b)

Describe the main useful energy transfer that takes place as the crane lifts the load. Name the stores involved.

(2)
(c)

The crane does 48 000 J of work lifting the load in 16 s. Calculate the power developed by the crane.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Energy cannot be created or destroyed; it can only be transferred from one store to another✓ 1
Accept: the total energy of a closed (isolated) system remains constant.—
Part (b)
Energy is transferred from the chemical store of the diesel fuel✓ 1
to the gravitational potential store of the load✓ 1
Accept 'chemical energy' and 'gravitational potential energy'. Also accept answers that mention an intermediate stage, e.g. mechanical working by the engine or the kinetic store of the moving parts, provided both the chemical and gravitational potential stores are named.—
Part (c)
P = W ÷ t = 48 000 ÷ 16✓ 1
= 3000 W✓ 1

Exam tip: When describing a transfer, name the store the energy starts in and the store it ends in, using the word 'store' each time.

40IGPC3-1-01c-01Physical quantities and measurement Paper CEasy1 mark

A student times 20 complete swings of a pendulum and records 31.0 s. What is the time period of the pendulum?

Show mark scheme
Marking pointMarkNotes
Correct: A. The period is the time for one swing, so divide the total time by the number of swings: 31.0 ÷ 20 = 1.55 s. B is wrong: this is 20 ÷ 31.0, the number of swings per second (the frequency), not the time per swing. C is wrong: this subtracts 20 from 31.0, which has no physical meaning. D is wrong: this multiplies the time by the number of swings instead of dividing.✓ 1Answer A

Answer: A

Exam tip: Timing many swings and dividing reduces the effect of reaction-time error on the period.

41IGPC3-1-01c-02Speed, velocity and motion graphs Paper CEasy1 mark

The distance–time graph for a cyclist consists of a straight sloping line, then a horizontal line, then a steeper straight sloping line. Which description of the motion is correct?

Show mark scheme
Marking pointMarkNotes
Correct: C. A straight sloping line is constant speed, a horizontal line means distance is not changing so the cyclist is stationary, and a steeper line means a greater speed. A is wrong: a horizontal line on a distance–time graph is not constant speed; the distance is not increasing. B is wrong: straight lines on a distance–time graph show constant speed, not acceleration; acceleration would give a curve. D is wrong: this confuses the shape of a distance–time graph with that of a speed–time graph.✓ 1Answer C

Answer: C

Common error: On a distance–time graph the gradient is the speed; a flat line is rest, not steady motion.

42IGPC3-1-01c-03Mass, weight and density Paper CEasy1 mark

A metal block measures 6.0 cm × 5.0 cm × 4.0 cm and has a mass of 540 g. What is the density of the metal?

6.0 cm4.0 cm5.0 cm
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Correct: D. Volume = 6.0 × 5.0 × 4.0 = 120 cm3; density = mass ÷ volume = 540 ÷ 120 = 4.5 g/cm3. A is wrong: this is volume ÷ mass, the equation inverted. B is wrong: this divides by 6.0 + 5.0 + 4.0 = 15 cm3, adding the lengths instead of multiplying. C is wrong: this multiplies mass by volume instead of dividing.✓ 1Answer D

Answer: D

Exam tip: Work out the volume first, then divide the mass by it. Check that the units are consistent (g and cm³).

43IGPC3-1-01c-10Mass, weight and density Paper CEasy8 marks

A student wants to find the density of a small irregular stone. The stone has a mass of 156 g. She has a measuring cylinder, water and a balance.

40 cm³before100 cm³after
Diagram not to scale
(a)

Describe how the student finds the volume of the stone.

(3)
(b)

The water level rises from 40 cm3 to 100 cm3. Calculate the density of the stone.

(3)
(c)

The density of water is 1.0 g/cm3. State and explain whether the stone floats or sinks in water.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
pour water into the measuring cylinder and record the volume✓ 1
lower the stone in carefully so it is fully covered, and record the new reading✓ 1
volume of stone = new reading minus original reading✓ 1
Part (b)
volume = 100 − 40 = 60 cm3✓ 1
density = mass ÷ volume✓ 1
156 ÷ 60 = 2.6 g/cm3✓ 1
Part (c)
it sinks✓ 1
its density is greater than that of water✓ 1

Exam tip: Read the measuring cylinder at the bottom of the meniscus with your eye level with the water surface.

44IGPC3-1-01c-14Mass, weight and density Paper CEasy5 marks

A rover has a mass of 180 kg on Earth, where the gravitational field strength is 9.8 N/kg. The rover is taken to the Moon, where the gravitational field strength is 1.6 N/kg.

(a)

Calculate the weight of the rover on Earth.

(2)
(b)

State the mass of the rover on the Moon.

(1)
(c)

Calculate the weight of the rover on the Moon.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
weight = mass × gravitational field strength✓ 1
180 × 9.8 = 1764 N✓ 1
Part (b)
180 kg✓ 1
Part (c)
180 × 1.6✓ 1
288 N✓ 1

Common error: Mass is the amount of matter and does not change with location. Weight depends on the gravitational field strength.

45IGPC3-1-06r5-01Mass, weight and density Paper CEasy1 mark

An aluminium block for a building facade has a mass of 540 g and a volume of 200 cm3. What is its density?

Show mark scheme
Marking pointMarkNotes
Density = mass ÷ volume = 540 ÷ 200 = 2.7 g/cm3. 0.37 comes from dividing volume by mass. 340 comes from subtracting 200 from 540. 740 comes from adding them.✓ 1Answer B

Answer: B

Common error: Write density as mass ÷ volume and check the units. Never combine mass and volume by adding or subtracting.

46IGPC3-1-06r5-02Mass, weight and density Paper CEasy1 mark

A research submersible has a mass of 80 kg on Earth, where g = 9.8 N/kg. It is transported to Mars, where g = 3.7 N/kg. Which row gives its mass and weight on Mars?

Show mark scheme
Marking pointMarkNotes
Mass is the amount of matter and does not change with location, so it stays at 80 kg. Weight = mass × g = 80 × 3.7 = 296 N. A weight of 784 N is the Earth value. A mass of 30 kg wrongly makes mass depend on g. 22 N comes from dividing 80 by 3.7.✓ 1Answer C

Answer: C

Common error: Mass stays constant everywhere. Only weight changes with the strength of gravity.

47IGPC3-1-06r5-03Pressure in solids and liquids Paper CEasy1 mark

A crate with a weight of 600 N rests on a pallet. The crate's base has an area of 0.15 m2. What pressure does the crate exert on the pallet?

Show mark scheme
Marking pointMarkNotes
Pressure = force ÷ area = 600 ÷ 0.15 = 4000 Pa. 90 Pa comes from multiplying the two values. 2.5 × 10−4 Pa comes from dividing area by force. 40 Pa comes from using 15 instead of 0.15.✓ 1Answer D

Answer: D

Exam tip: Write p = F ÷ A and substitute carefully. Check that the area is in m² before you divide.

48IGPC3-1-06r5-09Pressure in solids and liquids Paper CEasy4 marks

A concrete block for a wall has dimensions 0.50 m × 0.40 m × 0.20 m and a mass of 96 kg.

0.50 m0.20 m0.40 m
Diagram not to scale
(a)

Calculate the volume of the block.

(1)
(b)

Calculate the density of the block.

(2)
(c)

Calculate the weight of the block. Use g = 9.8 N/kg.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
0.50 × 0.40 × 0.20 = 0.040 m3✓ 1
Part (b)
density = mass ÷ volume = 96 ÷ 0.040✓ 1
= 2400 kg/m3 (ECF from (a))✓ 1
Part (c)
96 × 9.8 = 940.8 N (accept 941 N)✓ 1

Exam tip: Weight is a force measured in newtons. Mass is measured in kilograms. Do not give kg as the unit of weight.

49IGPC3-1-06r5-16Speed, velocity and motion graphs Paper CEasy6 marks

A cleaning robot moves up a building facade. It travels 24 m upwards in 30 s at a constant speed. It then stays still for 10 s before returning to its starting point at a constant speed of 1.2 m/s.

(a)

Calculate the speed of the robot during the first 30 s.

(2)
(b)

State what the distance-time graph shows during the 10 s when the robot stays still.

(1)
(c)

State one difference between speed and velocity.

(1)
(d)

Calculate the time taken for the robot to return to its starting point.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
speed = distance ÷ time = 24 ÷ 30✓ 1
= 0.80 m/s✓ 1
Part (b)
a horizontal line (gradient zero)✓ 1
Part (c)
velocity has a direction (it is a vector) but speed does not✓ 1
Part (d)
time = distance ÷ speed = 24 ÷ 1.2✓ 1
= 20 s✓ 1

Exam tip: On a distance-time graph, the gradient gives speed and a flat line means the object is stationary.

50IGPC3-1-06r7-01Mass, weight and density Paper CEasy1 mark

A marine biologist finds the density of a coral fragment of mass 36 g. She places it in a measuring cylinder; the water level rises from 40 cm³ to 52 cm³. What is the density of the coral?

40 cm³before52 cm³after
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
D is correct: volume = 52 − 40 = 12 cm³, so density = 36 / 12 = 3.0 g/cm³. A is wrong because it inverts the equation (12 / 36). B is wrong because it divides by the final reading (36 / 52) rather than the volume of the fragment. C is wrong because it divides by the initial reading (36 / 40).✓ 1Answer D

Answer: D

Common error: The volume of an irregular object is the difference between the two readings, not the final reading.

51IGPC3-1-06r7-02Speed, velocity and motion graphs Paper CEasy1 mark

A delivery van travels 30 km in 45 minutes. What is its average speed?

Show mark scheme
Marking pointMarkNotes
C is correct: 45 min = 0.75 h, so speed = 30 / 0.75 = 40 km/h. A is wrong because it divides by 45 without converting minutes to hours. B is wrong because it inverts the equation (time / distance). D is wrong because it multiplies distance by time.✓ 1Answer C

Answer: C

Exam tip: Convert the time to hours before dividing if the speed is wanted in km/h.

52IGPC3-1-06r7-03Forces, friction and springs Paper CEasy1 mark

A spring in a test rig for a model bridge extends by 4.0 cm when a load of 2.0 N is hung from it. The spring obeys Hooke's law for loads up to 10 N. What is the extension when the load is 6.0 N?

Show mark scheme
Marking pointMarkNotes
C is correct: extension is proportional to load, so tripling the load triples the extension: 4.0 × 3 = 12 cm. A is wrong because it treats extension as inversely proportional to load. B is wrong because it only doubles the extension. D is wrong because it multiplies the extension by the load value (4.0 × 6.0).✓ 1Answer C

Answer: C

Exam tip: Within the limit of proportionality, extension per unit load is constant.

53IGPC3-1-06r7-07Speed, velocity and motion graphs Paper CEasy1 mark

The speed–time graph of a research submersible is a straight line sloping upwards from the origin, followed by a horizontal line. Which statement describes the motion?

Show mark scheme
Marking pointMarkNotes
A is correct: a sloping straight line means constant acceleration and a horizontal line means constant speed. B is wrong because a horizontal speed–time line is not zero speed unless it lies on the axis. C is wrong because the first section is not horizontal. D is wrong because a straight line means constant, not increasing, acceleration.✓ 1Answer A

Answer: A

Common error: A horizontal line on a speed–time graph means constant speed, not rest.

54IGPC3-1-06r7-09Speed, velocity and motion graphs Paper CEasy5 marks

A marine biologist tracks a tagged shark. Its distance from the start is logged: 0 m at 0 s, 12 m at 10 s, 24 m at 20 s and 36 m at 30 s.

(a)

Describe the motion of the shark.

(1)
(b)

Calculate the speed of the shark.

(2)
(c)

Calculate the time the shark takes to travel 90 m at this speed.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
It moves at constant speed (equal distances in equal times)✓ 1
Part (b)
speed = distance / time = 36 / 30✓ 1
1.2 m/s✓ 1
Part (c)
time = distance / speed = 90 / 1.2 (ECF from (b))✓ 1
75 s✓ 1

Exam tip: Check units before calculating: here the answers are in m/s and s.

55IGPC3-1-07r8-01Mass, weight and density Paper CEasy1 mark

A crop-spraying drone has a mass of 15 kg. The gravitational field strength is 9.8 N/kg. What is the weight of the drone?

Show mark scheme
Marking pointMarkNotes
Weight = mass × g = 15 × 9.8 = 147 N, so C. A (1.5 N) comes from dividing mass by g. B (15 N) treats the mass in kg as a weight in newtons. D (1470 N) is a power-of-ten slip.✓ 1Answer C

Answer: C

Common error: Mass is measured in kg and weight in N; the two are linked by W = mg.

56IGPC3-1-07r8-02Mass, weight and density Paper CEasy1 mark

A sample of grain has a mass of 0.84 kg and occupies a volume of 0.0012 m3. What is the density of the grain?

Show mark scheme
Marking pointMarkNotes
Density = mass ÷ volume = 0.84 ÷ 0.0012 = 700 kg/m3, so C. A comes from inverting the formula (volume ÷ mass). B and D are decimal-place slips when dividing by 0.0012.✓ 1Answer C

Answer: C

Exam tip: Write the formula density = mass ÷ volume first, then check that your answer is a sensible size for the material.

57IGPC3-1-07r8-09Mass, weight and density Paper CEasy6 marks

A light aircraft has a fuel tank holding 0.25 m3 of aviation fuel of density 800 kg/m3. Use g = 9.8 N/kg.

(a)

Calculate the mass of fuel in the tank.

(2)
(b)

Calculate the weight of this fuel.

(2)
(c)

A different fuel has density 720 kg/m3. State whether 0.25 m3 of it has a greater or smaller mass than the fuel above and give a reason.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
density = mass ÷ volume, so mass = 800 × 0.25✓ 1
200 kg✓ 1
Part (b)
W = mg = 200 × 9.8 (ECF from (a))✓ 1
1960 N✓ 1
Part (c)
smaller mass✓ 1
same volume but lower density, so less mass per unit volume✓ 1

Exam tip: Write the rearranged formula before substituting, and keep the units of each quantity in kg, m and m³.

58IGPC3-1-07r8-10Speed, velocity and motion graphs Paper CEasy6 marks

A tractor travels 4500 m along a farm track in 30 minutes at a steady speed.

(a)

Calculate the speed of the tractor in m/s.

(2)
(b)

The tractor continues at this speed for a further 12 minutes. Calculate the distance it travels in these 12 minutes.

(2)
(c)

The tractor then goes round a bend at constant speed. Explain why its velocity changes.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
speed = distance ÷ time = 4500 ÷ (30 × 60)✓ 1
2.5 m/s✓ 1
Part (b)
distance = speed × time = 2.5 × (12 × 60)✓ 1
1800 m✓ 1
Part (c)
velocity is a vector (has speed and direction)✓ 1
the direction changes even though the speed stays the same✓ 1

Common error: Convert minutes to seconds before using speed in m/s.

59IGPC3-1-07r8-11Pressure in solids and liquids Paper CEasy5 marks

A guitar amplifier of mass 25 kg stands on a flat floor. Use g = 9.8 N/kg.

(a)

Calculate the weight of the amplifier.

(1)
(b)

The base of the amplifier measures 0.50 m by 0.20 m. Calculate the pressure it exerts on the floor.

(2)
(c)

The amplifier is stood on its end, which measures 0.20 m by 0.25 m. Calculate the new pressure on the floor.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
W = mg = 25 × 9.8 = 245 N✓ 1
Part (b)
pressure = force ÷ area = 245 ÷ (0.50 × 0.20)✓ 1
2450 Pa✓ 1
Part (c)
area = 0.20 × 0.25 = 0.050 m2 and p = 245 ÷ 0.050 (ECF from (a))✓ 1
4900 Pa✓ 1

Exam tip: The force on the floor is the weight, and the area is the area in contact with the floor, not the largest face.

60IGPC3-1-05r15-01Physical quantities and measurement Paper CEasy1 mark

A pharmacist stacks 25 identical tablets and measures the height of the stack with a ruler as 17.5 mm. What is the thickness of one tablet?

Show mark scheme
Marking pointMarkNotes
Thickness of one tablet = 17.5 ÷ 25 = 0.70 mm, so A is correct. B comes from dividing 25 by 17.5. C is a decimal-place slip. D multiplies the height by the number of tablets.✓ 1Answer A

Answer: A

Exam tip: Measuring a stack of identical items and dividing reduces the effect of the reading error on one item.

61IGPC3-1-05r15-02Speed, velocity and motion graphs Paper CEasy1 mark

In a video game the distance travelled by a character is plotted against time. The graph is a straight line from the origin to 20 m at 4.0 s, followed by a horizontal line until 10 s. What is the average speed of the character over the whole 10 s?

41020time / sdistance / m
Graph sketched from the values in the question; not to scale
Show mark scheme
Marking pointMarkNotes
Average speed = total distance ÷ total time = 20 ÷ 10 = 2.0 m/s, so B is correct. A is the speed during the first 4.0 s only. C is the speed during the horizontal section. D divides by only 6.0 s, ignoring the time spent moving.✓ 1Answer B

Answer: B

Common error: Average speed uses the whole time interval, including any time spent stationary.

62IGPC3-1-05r15-07Mass, weight and density Paper CEasy1 mark

A box of medicine has mass 0.40 kg on Earth. It is taken to the Moon, where the gravitational field strength is 1.6 N/kg. On Earth g = 9.8 N/kg. What are the mass and weight of the box on the Moon?

Show mark scheme
Marking pointMarkNotes
Mass does not change with location, so it is 0.40 kg. Weight = mg = 0.40 × 1.6 = 0.64 N, so D is correct. A uses the Earth weight. B wrongly lets the mass fall with g. C changes the mass and keeps the Earth weight.✓ 1Answer D

Answer: D

Common error: Mass is constant everywhere; only weight depends on gravitational field strength.

63IGPC3-1-05r15-09Physical quantities and measurement Paper CEasy7 marks

A pharmacy technician checks the thickness of aluminium foil used to seal tablet packs.

(a)

Name the instrument best suited to measuring the thickness of a stack of ten sheets of foil to 0.01 mm.

(1)
(b)

With the jaws fully closed the instrument reads +0.04 mm. State the name for this type of error.

(1)
(c)

With the ten sheets between the jaws the instrument reads 0.84 mm. Determine the correct thickness of the ten sheets.

(2)
(d)

Suggest why the technician measures ten sheets together instead of one sheet.

(1)
(e)

Calculate the thickness of one sheet of foil, using the corrected value from part (c).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
micrometer (screw gauge)✓ 1
Part (b)
zero error✓ 1
Part (c)
subtract the zero error: 0.84 − 0.04✓ 1
0.80 mm✓ 1
Part (d)
one sheet is too thin to measure accurately / the error in the reading is a smaller fraction of the larger total thickness✓ 1
Part (e)
0.80 ÷ 10✓ 1
0.080 mm✓ 1

Common error: Zero errors must be subtracted from readings; a positive zero error makes every reading too large.

64IGPC3-1-05r15-13Forces, friction and springs Paper CEasy7 marks

A pharmacy pill dispenser uses a spring. Extensions for different loads are recorded: load/N 0, 2.0, 4.0, 6.0, 8.0, 10.0; extension/cm 0, 2.0, 4.0, 6.0, 8.0, 12.0.

(a)

State the relationship between extension and load for loads up to 8.0 N.

(1)
(b)

State the largest load in the table for which the spring obeys Hooke's law.

(1)
(c)

The spring's natural length is 12.0 cm. Calculate its length when the load is 3.5 N.

(2)
(d)

A tray of mass 0.60 kg is pulled along a bench with a force of 4.0 N. The friction force is 2.5 N. Calculate the acceleration of the tray.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
extension is directly proportional to load✓ 1
Part (b)
8.0 N✓ 1
Part (c)
extension is 3.5 cm (1 cm per newton)✓ 1
length = 12.0 + 3.5 = 15.5 cm✓ 1
Part (d)
resultant force = 4.0 − 2.5 = 1.5 N✓ 1
a = F/m = 1.5 ÷ 0.60✓ 1
2.5 m/s²✓ 1

Common error: Use the resultant force, not the pulling force alone, when calculating acceleration with friction present.

65IGPC3-1-06r18-01Mass, weight and density Paper CEasy1 mark

A cube-shaped component for a space probe has sides of 0.10 m and a mass of 2.7 kg. What is the density of the material?

Show mark scheme
Marking pointMarkNotes
Volume = 0.10 × 0.10 × 0.10 = 1.0 × 10−3 m3, so density = 2.7 / 1.0 × 10−3 = 2700 kg/m3. A (27) divides by the side length 0.10 m instead of the volume. B (270) divides by the area 0.10 × 0.10 m2. D (3.7 × 10−4) inverts the formula (volume ÷ mass).✓ 1Answer C

Answer: C

Common error: Volume of a cube is side cubed. Check that the units of your answer, kg/m³, match the calculation you did.

66IGPC3-1-06r18-02Mass, weight and density Paper CEasy1 mark

A probe has a mass of 450 kg. On the Moon the gravitational field strength is 1.6 N/kg. What is the weight of the probe on the Moon?

Show mark scheme
Marking pointMarkNotes
Weight = mass × gravitational field strength = 450 × 1.6 = 720 N. A (281 N) divides mass by g. B (450 N) confuses weight with mass (mass does not change). D (4410 N) uses Earth's g = 9.8 N/kg instead of the Moon's value.✓ 1Answer C

Answer: C

Exam tip: Mass is constant everywhere; weight depends on the local gravitational field strength. Use W = mg with the g given for the place in the question.

67IGPC3-1-06r18-05Pressure in solids and liquids Paper CEasy1 mark

A steel pad that supports part of a bridge deck carries a load of 2.0 × 105 N. The area of the pad in contact with the ground is 4000 cm2 (0.40 m2). What is the pressure on the ground?

Show mark scheme
Marking pointMarkNotes
Pressure = force ÷ area = 2.0 × 105 / 0.40 = 5.0 × 105 Pa. A (50 Pa) divides by 4000, the area in cm2, without converting to m2. B (8.0 × 104 Pa) multiplies force by area. D (5.0 × 106 Pa) converts cm2 to m2 by dividing by 100 instead of 10 000.✓ 1Answer C

Answer: C

Common error: 1 m² = 10 000 cm², not 100 cm². Convert area to m² before using pressure = force ÷ area in pascals.

68IGPC3-1-06r18-09Speed, velocity and motion graphs Paper CEasy5 marks

A rover on Mars drives in a straight line across level ground. It travels 18 m in 120 s at a constant speed.

(a)

Calculate the speed of the rover.

(2)
(b)

Calculate the time the rover needs to travel 45 m at this speed.

(2)
(c)

State the difference between speed and velocity.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 speed = distance ÷ time = 18 / 120✓ 1
A1 0.15 m/s✓ 1
Part (b)
M1 time = distance ÷ speed = 45 / 0.15✓ 1
A1 300 s (5.0 minutes)✓ 1
Part (c)
B1 velocity has a direction (is a vector) whereas speed has magnitude only (is a scalar)✓ 1

Exam tip: Write the formula, substitute the numbers with units, then give the answer with its unit. Rearranging speed = distance ÷ time gives time = distance ÷ speed.

69IGPC3-1-06r18-11Energy, work and power Paper CEasy5 marks

The drum of a coffee roaster is turned by an electric motor. The motor receives energy at a rate of 150 W and delivers 105 W of useful mechanical power to the drum.

(a)

Calculate the energy transferred to the motor in 15 minutes.

(2)
(b)

Calculate the efficiency of the motor.

(2)
(c)

Name the form of energy in which most of the wasted energy leaves the motor.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 energy = power × time with time = 15 × 60 = 900 s✓ 1
A1 135 000 J (1.35 × 105 J)✓ 1
Part (b)
M1 efficiency = useful output ÷ total input × 100%✓ 1
A1 105 / 150 × 100 = 70%✓ 1
Part (c)
B1 thermal (heat) energy✓ 1

Common error: Convert minutes to seconds before using energy = power × time. Efficiency compares useful output with total input and cannot be greater than 100%.

70IGPC3-1-07r20-01Pressure in solids and liquids Paper CEasy1 mark

A steel counterweight block in a theatre flying system rests on the stage floor. Its weight is 480 N and the area in contact with the floor is 0.20 m2. What pressure does the block exert on the floor?

Show mark scheme
Marking pointMarkNotes
Pressure = force ÷ area = 480 ÷ 0.20 = 2400 Pa, so C is correct. A multiplies force by area. B gives the force and ignores the area. D divides area by force, the wrong way round.✓ 1Answer C

Answer: C

Common error: Pressure is force divided by area; if your answer gets bigger when the contact area gets bigger, the division is upside down.

71IGPC3-1-07r20-03Speed, velocity and motion graphs Paper CEasy1 mark

A ferry leaves a harbour and accelerates uniformly from rest to 8.0 m/s in 5.0 s. It then travels at a constant 8.0 m/s for a further 7.0 s. What total distance does the ferry travel in the 12 s?

5128time / sspeed / m/s
Graph sketched from the values in the question; not to scale
Show mark scheme
Marking pointMarkNotes
Distance = area under the speed–time graph = ½ × 5.0 × 8.0 + 8.0 × 7.0 = 20 + 56 = 76 m, so C is correct. A is ½ × 12 × 8.0, treating the whole journey as uniform acceleration. B counts only the constant-speed stage. D is 8.0 × 12, treating the whole journey as constant speed.✓ 1Answer C

Answer: C

Exam tip: Split a speed–time graph into a triangle and a rectangle and add the areas.

72IGPC3-1-07r20-09Physical quantities and measurement Paper CEasy5 marks

A stage lamp hangs from a long chain and swings gently from side to side. A student wants to find the time for one complete swing (the period).

(a)

Explain why the student times 20 complete swings rather than a single swing.

(2)
(b)

The time for 20 complete swings is 31.4 s. Calculate the period of the lamp.

(2)
(c)

State one further step the student should take to make the value of the period more reliable.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 the reaction-time error in starting and stopping the stopwatch is about the same for either measurement✓ 1
B1 so it is a much smaller fraction of the total time for 20 swings / the percentage uncertainty is smaller✓ 1
Part (b)
M1 period = 31.4 ÷ 20✓ 1
A1 1.57 s✓ 1
Part (c)
B1 repeat the timing and calculate the average (of the periods)✓ 1

Exam tip: Timing many oscillations reduces the effect of reaction time on the measured period.

73IGPC3-1-07r20-10Mass, weight and density Paper CEasy6 marks

A volcanologist wants to find the density of a small piece of volcanic rock of irregular shape.

(a)

Describe how to find the volume of the rock using a measuring cylinder and water.

(3)
(b)

The rock has a mass of 84 g and a volume of 35 cm3. Calculate its density.

(2)
(c)

The density of water is 1.0 g/cm3. State whether the rock floats or sinks in water and give a reason.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 partly fill the cylinder with water and record the volume V1✓ 1
B1 lower the rock fully into the water and record the new volume V2✓ 1
B1 volume of rock = V2 − V1✓ 1
Part (b)
M1 density = mass ÷ volume = 84 ÷ 35✓ 1
A1 2.4 g/cm3✓ 1
Part (c)
B1 sinks, because its density (2.4 g/cm3) is greater than that of water✓ 1

Common error: Read the water level at the bottom of the meniscus, at eye level, before and after the rock is added.

74IGPC3-1-08r20-01Mass, weight and density Paper CEasy1 mark

A marine biologist deploys a tagging buoy of mass 5.0 kg. The gravitational field strength is 9.8 N/kg. What is the weight of the buoy?

Show mark scheme
Marking pointMarkNotes
Weight = mass × gravitational field strength = 5.0 × 9.8 = 49 N, so C is correct. A divides mass by g. B gives the mass as if it were the weight. D is a power-of-ten slip.✓ 1Answer C

Answer: C

Common error: Mass is in kg and weight is a force in N; convert with W = mg.

75IGPC3-1-08r20-03Turning effect and centre of gravity Paper CEasy1 mark

A fitter tightens a bolt on a steel truss using a spanner. A force of 60 N is applied perpendicular to the spanner at a distance of 0.25 m from the bolt. What is the moment of the force about the bolt?

60 N0.25 m
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Moment = force × perpendicular distance = 60 × 0.25 = 15 N m, so B is correct. A divides distance by force. C gives the force alone. D divides force by distance.✓ 1Answer B

Answer: B

Common error: A moment is force multiplied by distance, so a longer spanner needs a smaller force for the same moment.

76IGPC3-1-08r20-05Speed, velocity and motion graphs Paper CEasy1 mark

A delivery lorry travels 90 km in 1 hour 15 minutes. What is its average speed?

Show mark scheme
Marking pointMarkNotes
Time = 1.25 h; speed = 90 ÷ 1.25 = 72 km/h, so B is correct. A treats 1 h 15 min as 1.5 h. C ignores the extra 15 minutes. D multiplies distance by time.✓ 1Answer B

Answer: B

Common error: Convert minutes to a decimal fraction of an hour (15 min = 0.25 h), not 0.15 h.

77IGPC3-1-08r22-01Physical quantities and measurement Paper CEasy1 mark

An archaeologist times a swinging plumb line. Twenty complete swings take 31.4 s. What is the time for one swing (the period)?

Show mark scheme
Marking pointMarkNotes
Period = total time ÷ number of swings = 31.4 ÷ 20 = 1.57 s. 0.637 s is the number of swings per second (the frequency), found by dividing the wrong way round. 11.4 s subtracts 20 from 31.4. 628 s multiplies the two numbers.✓ 1Answer B

Answer: B

Common error: Time many swings and divide by the number of swings; dividing the other way round gives frequency, not period.

78IGPC3-1-08r22-02Mass, weight and density Paper CEasy1 mark

A bronze artefact has a mass of 420 g. Its volume, found by displacement of water, is 50 cm3. What is the density of the bronze?

Show mark scheme
Marking pointMarkNotes
Density = mass ÷ volume = 420 ÷ 50 = 8.4 g/cm3. 0.12 g/cm3 comes from dividing volume by mass. 370 comes from subtracting the numbers and 470 from adding them, which has no physical meaning.✓ 1Answer C

Answer: C

Exam tip: Write the formula density = mass ÷ volume before substituting so the units come out consistently.

79IGPC3-1-08r22-09Speed, velocity and motion graphs Paper CEasy6 marks

A cyclist rides 4.5 km in 15 minutes. Later, in a separate ride, she accelerates uniformly from rest to 6.0 m/s in 8.0 s.

(a)

Calculate her average speed for the first ride in m/s.

(2)
(b)

Calculate her acceleration in the second ride.

(2)
(c)

Calculate the distance she travels in the 8.0 s of the second ride.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
average speed = distance ÷ time = 4500 ÷ 900✓ 1
5.0 m/s✓ 1
Part (b)
acceleration = change in speed ÷ time = 6.0 ÷ 8.0✓ 1
0.75 m/s2✓ 1
Part (c)
average speed = 3.0 m/s, or area under speed–time graph, distance = ½ × 6.0 × 8.0✓ 1
24 m✓ 1

Common error: Convert km to m and minutes to seconds before calculating speed in m/s.

80IGPC3-1-08r22-10Mass, weight and density Paper CEasy6 marks

An archaeologist finds a small pottery shard and wants to know its density.

(a)

Describe how the volume of a small irregular object can be found using a measuring cylinder and water.

(2)
(b)

The shard has a mass of 87 g and a volume of 30 cm3. Calculate its density in g/cm3.

(2)
(c)

Convert the density to kg/m3.

(1)
(d)

The density of water is 1000 kg/m3. State whether the shard floats or sinks in water and give a reason.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
read the water level in the measuring cylinder before and after fully lowering the object in✓ 1
volume = difference between the two readings✓ 1
Part (b)
density = mass ÷ volume = 87 ÷ 30✓ 1
2.9 g/cm3✓ 1
Part (c)
2.9 × 1000 = 2900 kg/m3✓ 1
ECF from (b)—
Part (d)
sinks, because its density is greater than that of water✓ 1

Exam tip: 1 g/cm³ equals 1000 kg/m³; multiply by 1000 when converting.

81IGPC3-1-08r22-16Mass, weight and density Paper CEasy7 marks

A technician measures the density of a liquid coolant used in a data centre. An empty bottle has a mass of 0.18 kg. When 250 cm3 of coolant is poured in, the bottle and coolant have a mass of 0.53 kg. Use g = 9.8 N/kg.

(a)

Calculate the mass of the coolant.

(1)
(b)

Calculate the density of the coolant in kg/m3.

(2)
(c)

Calculate the weight of the coolant in the bottle.

(2)
(d)

The bottle of coolant is taken to the Moon, where g is smaller. State what happens to the mass and to the weight of the coolant.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
0.53 − 0.18 = 0.35 kg✓ 1
Part (b)
250 cm3 = 2.5 × 10−4 m3, density = 0.35 ÷ (2.5 × 10−4)✓ 1
1400 kg/m3✓ 1
Part (c)
weight = mass × g = 0.35 × 9.8✓ 1
3.43 N✓ 1
Part (d)
mass stays the same✓ 1
weight is smaller✓ 1

Common error: Mass is measured in kg and does not change with place; weight is a force in N and depends on g.

82IGPC3-1-07r25-01Physical quantities and measurement Paper CEasy1 mark

A beekeeper stacks 20 identical sheets of wax foundation and measures the height of the stack as 15.0 mm. What is the thickness of one sheet?

Show mark scheme
Marking pointMarkNotes
Thickness of one sheet = 15.0 mm ÷ 20 = 0.75 mm, because the total is shared equally between 20 sheets. 7.5 mm comes from a misplaced decimal point. 300 mm comes from multiplying by 20 instead of dividing. 1.3 mm comes from dividing 20 by 15.0, which inverts the calculation.✓ 1Answer B

Answer: B

Exam tip: Measuring a stack and dividing is a good way to reduce the effect of the instrument's resolution on a very small thickness.

83IGPC3-1-07r25-07Speed, velocity and motion graphs Paper CEasy1 mark

A hospital lift starts from rest and speeds up uniformly to 3.0 m/s in 2.0 s. It then moves at a constant 3.0 m/s for 5.0 s. What is the total distance it travels in these 7.0 s?

273time / sspeed / m/s
Graph as described; not to scale
Show mark scheme
Marking pointMarkNotes
Distance during speeding up = ½ × 3.0 × 2.0 = 3.0 m. Distance at constant speed = 3.0 × 5.0 = 15 m. Total = 18 m. The value 15 m ignores the first stage. The value 21 m treats the whole 7.0 s as being at 3.0 m/s. The value 10.5 m uses an average speed of 1.5 m/s for the whole 7.0 s.✓ 1Answer A

Answer: A

Common error: Different stages need different calculations: use the area of a triangle while speed is changing uniformly and a rectangle while it is constant.

84IGPC3-1-07r25-08Forces, friction and springs Paper CEasy1 mark

A spring in a beehive scale has a length of 20 cm when unloaded. A load of 4.0 N makes its length 26 cm. The spring obeys Hooke's law and the limit of proportionality is not exceeded. What is the length of the spring when the load is 10 N?

20 cm4.0 N26 cm10 N? cm
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
A 4.0 N load gives an extension of 6 cm, so a 10 N load gives 6 × 10 ÷ 4.0 = 15 cm. The length is 20 + 15 = 35 cm. The value 15 cm is the extension, not the length. The value 65 cm treats the length as proportional to the load (26 × 10 ÷ 4.0). The value 50 cm treats the unloaded length of 20 cm as proportional to the load.✓ 1Answer C

Answer: C

Common error: Hooke's law relates extension, not total length, to the load.

85IGPC3-1-07r25-15Forces, friction and springs Paper CEasy7 marks

A beekeeper's hanging scale contains a spring. The table shows the extension of the spring for different loads.
Load / N: 2.0, 4.0, 6.0, 8.0, 10.0, 12.0
Extension / cm: 3.0, 6.0, 9.0, 12.0, 15.0, 21.0
Use g = 9.8 N/kg.

(a)

Calculate the extension per newton for loads up to 10.0 N.

(2)
(b)

State which load is beyond the limit of proportionality of the spring and give a reason.

(2)
(c)

A frame of mass 0.45 kg is hung from the spring. Calculate the extension of the spring. Give your answer to 2 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
extension ÷ load, for example 3.0 ÷ 2.0✓ 1
1.5 cm/N✓ 1
Part (b)
12.0 N✓ 1
the extension is not in proportion: it should be 18 cm but is 21 cm✓ 1
Part (c)
W = mg = 0.45 × 9.8 = 4.41 N✓ 1
extension = 4.41 × 1.5✓ 1
6.6 cm✓ 1

Common error: Hooke's law only applies up to the limit of proportionality. Convert a mass to a weight before using a force–extension relationship.

86IGPC3-1-10r27-01Physical quantities and measurement Paper CEasy1 mark

In a coffee roastery, a technician times 20 complete swings of a pendulum clock and records 36 s. What is the period of one swing?

Show mark scheme
Marking pointMarkNotes
Period = total time / number of swings = 36 / 20 = 1.8 s. 0.56 s comes from dividing the number of swings by the time (the frequency, in the wrong form). 18 s divides by 2 instead of 20. 720 s multiplies the time by the number of swings.✓ 1Answer C

Answer: C

Exam tip: Time many swings and divide: this reduces the percentage error in the period.

87IGPC3-1-10r27-02Mass, weight and density Paper CEasy1 mark

A steel block for a bridge model has a mass of 15.6 kg and a volume of 2000 cm3. What is the density of the steel in g/cm3?

Show mark scheme
Marking pointMarkNotes
Density = mass / volume = 15 600 g / 2000 cm3 = 7.8 g/cm3. 0.13 is volume divided by mass (inverted). 7800 is the same density in kg/m3, so the unit does not match. 0.0078 comes from using the mass in kg with the volume in cm3 and then dividing again by 1000.✓ 1Answer A

Answer: A

Common error: Check units before dividing: kg with m³, or g with cm³.

88IGPC3-1-10r27-04Forces, friction and springs Paper CEasy1 mark

A spring in a roasting-machine scale extends by 4.0 cm when the load is 6.0 N. The spring obeys Hooke's law for all loads up to 20 N. What is its extension when the load is 15 N?

Show mark scheme
Marking pointMarkNotes
Extension is directly proportional to load: 4.0 × (15 / 6.0) = 10 cm. 1.6 cm treats extension as inversely proportional to the load. 9.0 cm subtracts the loads instead of scaling. 19 cm adds the load in newtons to the extension in centimetres.✓ 1Answer B

Answer: B

Common error: Extension is proportional to load; it is not found by adding or subtracting loads.

89IGPC3-1-10r27-09Speed, velocity and motion graphs Paper CEasy5 marks

A walker takes 3 min 20 s to cross a straight bridge of length 240 m at a constant speed.

(a)

Calculate the walker's speed.

(2)
(b)

A car crosses the same bridge at a constant 15 m/s. Calculate the time the car takes. Ignore the length of the car.

(2)
(c)

State one difference between speed and velocity.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
speed = distance / time = 240 / 200✓ 1
1.2 m/s✓ 1
Part (b)
time = distance / speed = 240 / 15✓ 1
16 s✓ 1
Part (c)
velocity has direction (is a vector) but speed does not (is a scalar)✓ 1

Common error: Convert minutes to seconds before using speed = distance / time.

90IGPC3-1-10r27-10Mass, weight and density Paper CEasy5 marks

A coffee roaster measures the density of coffee beans before and after roasting. 12.0 kg of green beans fill a container of volume 0.016 m3. After roasting, the beans have a mass of 10.2 kg and occupy a volume of 0.024 m3.

(a)

Calculate the density of the green beans.

(2)
(b)

Calculate the density of the roasted beans.

(2)
(c)

Explain why the density decreases during roasting.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
density = mass / volume = 12.0 / 0.016✓ 1
750 kg/m3✓ 1
Part (b)
10.2 / 0.024✓ 1
425 kg/m3✓ 1
Part (c)
the mass decreases (water lost) while the volume increases, so mass / volume is smaller✓ 1

Exam tip: Always write the formula and the substituted numbers before the answer.

91IGPC3-1-10r28-01Physical quantities and measurement Paper CEasy1 mark

An archaeologist measures the thickness of a pottery fragment with a micrometer screw gauge. With the jaws closed the gauge reads 0.04 mm. With the fragment between the jaws it reads 3.46 mm. What is the true thickness of the fragment?

Show mark scheme
Marking pointMarkNotes
The gauge reads 0.04 mm when it should read zero, so every reading is 0.04 mm too high. The true thickness is 3.46 − 0.04 = 3.42 mm. 3.46 mm ignores the zero error. 3.50 mm adds the zero error instead of subtracting it. 3.04 mm subtracts 0.4 mm, a slip in place value.✓ 1Answer C

Answer: C

Common error: A positive zero error means readings are too large, so subtract it from every reading.

92IGPC3-1-10r28-02Speed, velocity and motion graphs Paper CEasy1 mark

A cyclist's speed increases uniformly from 0 to 12 m/s in 8.0 s. What distance does the cyclist travel in this time?

Show mark scheme
Marking pointMarkNotes
With uniform acceleration the average speed is (0 + 12) / 2 = 6.0 m/s, so the distance is 6.0 × 8.0 = 48 m, which is also the area under the speed–time graph. 96 m uses the final speed for the whole time. 12 m is just the final speed. 1.5 m is the acceleration in m/s2, not a distance.✓ 1Answer B

Answer: B

Common error: Do not use the final speed for the whole journey when the speed is changing: use the average speed or the area under the graph.

93IGPC3-1-10r28-04Physical quantities and measurement Paper CEasy1 mark

An irregular stone from a dig has a mass of 156 g. When it is lowered into a measuring cylinder, the water level rises by 60 cm3. What is the density of the stone?

Show mark scheme
Marking pointMarkNotes
Density = mass / volume = 156 / 60 = 2.6 g/cm3. 0.38 results from dividing volume by mass. 96 comes from subtracting the volume from the mass, which has no physical meaning. About 9400 comes from multiplying mass by volume.✓ 1Answer A

Answer: A

Common error: Density is mass divided by volume: check that your answer makes sense (stone is denser than water, so greater than 1 g/cm³).

94IGPC3-1-10r28-08Speed, velocity and motion graphs Paper CEasy1 mark

A cyclist rides once around a circular track at a constant speed of 10 m/s. Which statement is correct?

Show mark scheme
Marking pointMarkNotes
Velocity is a vector, so a change of direction is a change of velocity even at constant speed. The velocity is therefore not constant, and the cyclist is accelerating. Saying velocity is constant confuses speed with velocity. The velocity is not zero at every instant: only the overall displacement over the lap is zero. Saying there is no acceleration ignores that acceleration is the rate of change of velocity, which includes direction.✓ 1Answer A

Answer: A

Common error: Speed is a scalar, velocity is a vector: changing direction changes the velocity.

95IGPC3-1-10r28-10Speed, velocity and motion graphs Paper CEasy7 marks

A cyclist starts from rest and accelerates uniformly for 10 s until reaching a speed of 8.0 m/s. The cyclist then rides at this constant speed for 30 s and finally slows uniformly to rest in 5.0 s.

1040458time / sspeed / m/s
Graph as described; not to scale
(a)

Calculate the acceleration of the cyclist during the first 10 s.

(2)
(b)

Calculate the total distance travelled by the cyclist.

(3)
(c)

Calculate the average speed for the whole journey. Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 acceleration = change in speed ÷ time = 8.0 ÷ 10✓ 1
A1 0.80 m/s2✓ 1
Part (b)
M1 distance = area under the speed–time graph, each stage found✓ 1
A1 stages 40 m, 240 m and 20 m✓ 1
A1 total 300 m✓ 1
Part (c)
M1 total time = 10 + 30 + 5.0 = 45 s and speed = 300 ÷ 45✓ 1
A1 6.67 m/s (ECF from (b))✓ 1

Common error: Average speed is total distance divided by total time, not the average of the speeds in each stage.

96IGPC3-1-07r30-03Speed, velocity and motion graphs Paper CEasy1 mark

A porter pushes a hospital trolley along a corridor. On a distance–time graph, the motion is a straight line sloping upwards from the origin for 10 s, followed by a horizontal straight line lasting 5 s. Which statement describes the motion?

Show mark scheme
Marking pointMarkNotes
On a distance–time graph the gradient is the speed. A straight sloping line means a constant, non-zero speed, and a horizontal line means the distance is not changing, so the trolley is stationary. Constant acceleration for 10 s confuses the graph with a speed–time graph. A horizontal line cannot mean a different non-zero speed, because the distance would still be increasing. A straight line is not speeding up; that would need a curve.✓ 1Answer C

Answer: C

Common error: A horizontal line on a distance–time graph means the object is stationary; on a speed–time graph it means constant speed.

97IGPC3-1-07r30-08Physical quantities and measurement Paper CEasy1 mark

A technician in a hospital physics laboratory times 20 complete swings of a pendulum and records 32.4 s. What is the time for one swing?

Show mark scheme
Marking pointMarkNotes
Time for one swing = total time ÷ number of swings = 32.4 / 20 = 1.62 s. 648 s multiplies instead of dividing. 0.617 s divides the other way round (20 / 32.4). 16.2 s divides by 2 instead of 20.✓ 1Answer D

Answer: D

Exam tip: Timing many swings and dividing reduces the effect of reaction-time error on the measurement.

98IGPC3-1-07r30-11Mass, weight and density Paper CEasy7 marks

A beekeeper measures a jar of honey. The empty jar has a mass of 120 g. The jar with the honey has a mass of 540 g. The honey has a volume of 300 cm3.

(a)

Calculate the mass of the honey.

(1)
(b)

Calculate the density of the honey in g/cm3.

(2)
(c)

Calculate the volume occupied by 1000 g of this honey. Give your answer to 3 significant figures.

(2)
(d)

Calculate the weight of the 420 g of honey. Use g = 9.8 N/kg and give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
540 − 120 = 420 g✓ 1
Part (b)
density = mass / volume✓ 1
= 420 / 300 = 1.4 g/cm3✓ 1
Part (c)
volume = mass / density = 1000 / 1.4✓ 1
= 714 cm3✓ 1
Part (d)
mass = 0.42 kg and W = mg = 0.42 × 9.8✓ 1
= 4.116 = 4.12 N✓ 1

Common error: Weight is a force measured in newtons; mass in grams must be changed to kilograms before using W = mg.

99IGPC3-1-07r30-16Physical quantities and measurement Paper CEasy5 marks

A nurse checks an intravenous drip. She times 10 drops falling and finds that they take 24 s. Each drop has a volume of 0.050 cm3.

(a)

Calculate the average time between drops.

(1)
(b)

Suggest why she times 10 drops rather than a single drop.

(1)
(c)

Calculate the volume of liquid delivered in one hour.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
24 / 10 = 2.4 s✓ 1
Part (b)
the error from reaction time is a smaller fraction of the longer time, so the average is more accurate✓ 1
Part (c)
time in seconds = 3600 s✓ 1
number of drops = 3600 / 2.4 = 1500✓ 1
volume = 1500 × 0.050 = 75 cm3✓ 1

Exam tip: Convert the time to the same unit (seconds) as the drop interval before dividing.

100IGPC3-1-08r31-01Physical quantities and measurement Paper CEasy1 mark

A stack of 20 identical steel plates for a bridge test has a total thickness of 15.0 mm, measured with a micrometer. What is the thickness of one plate?

Show mark scheme
Marking pointMarkNotes
One plate is 15.0 mm ÷ 20 = 0.75 mm. 1.3 mm comes from dividing 20 by 15.0, the wrong way round. 7.5 mm is a decimal-place error. 300 mm comes from multiplying instead of dividing.✓ 1Answer B

Answer: B

Exam tip: Measuring many identical items together and dividing reduces the effect of the measuring error on each one.

101IGPC3-1-08r31-06Physical quantities and measurement Paper CEasy1 mark

Which list contains only vector quantities?

Show mark scheme
Marking pointMarkNotes
Velocity, force and momentum all have direction as well as size. The list with speed includes a scalar, although force and weight are vectors. The list with mass includes a scalar, although acceleration and velocity are vectors. The list with distance and speed contains two scalars, so it fails even though displacement is a vector.✓ 1Answer C

Answer: C

Common error: Speed and velocity are not the same: speed is a scalar and velocity is a vector.

102IGPC3-1-08r31-09Physical quantities and measurement Paper CEasy5 marks

A student finds the density of a steel sample used for a bridge cable. The sample has mass 158 g. It is lowered into a measuring cylinder containing 40 cm3 of water, and the water level rises to 60 cm3.

(a)

State the volume of the steel sample.

(1)
(b)

Calculate the density of the sample in g/cm3.

(2)
(c)

Convert your answer to (b) into kg/m3.

(1)
(d)

Suggest how the student should read the measuring cylinder to avoid a parallax error.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 20 cm3✓ 1
Part (b)
M1 density = mass / volume = 158 / 20✓ 1
A1 7.9 g/cm3 (ECF from (a))✓ 1
Part (c)
A1 7900 kg/m3 (7.9 × 1000)✓ 1
Part (d)
B1 with the eye level with the water surface (bottom of the meniscus)✓ 1

Exam tip: 1 g/cm³ is equal to 1000 kg/m³.

103IGPC3-1-08r31-10Momentum and impulse Paper CEasy8 marks

A rover from a space probe drives across a planet's surface. Its distance from the start is recorded.
Time / s: 0, 300, 400, 700
Distance / m: 0, 18, 18, 49

(a)

Calculate the speed of the rover from 0 to 300 s.

(2)
(b)

State the speed from 300 s to 400 s and give a reason.

(2)
(c)

Calculate the speed of the rover from 400 s to 700 s. Give your answer to 2 significant figures.

(2)
(d)

Calculate the average speed of the rover for the whole 700 s.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 speed = distance / time = 18 / 300✓ 1
A1 0.060 m/s✓ 1
Part (b)
B1 0 m/s✓ 1
B1 the distance does not change, so the rover is stationary✓ 1
Part (c)
M1 speed = (49 − 18) / 300✓ 1
A1 0.10 m/s✓ 1
Part (d)
M1 average speed = total distance / total time = 49 / 700✓ 1
A1 0.070 m/s✓ 1

Common error: Average speed uses the total distance divided by the total time, including any time when the object is stationary.

104IGPC3-1-08r31-11Forces, friction and springs Paper CEasy6 marks

A spring in a coffee-scoop balance is loaded and the extension is measured.
Load / N: 0, 2.0, 4.0, 6.0, 8.0, 10.0, 12.0
Extension / cm: 0, 1.0, 2.0, 3.0, 4.0, 5.0, 6.5

(a)

State whether the spring has gone beyond its limit of proportionality by a load of 12 N. Give a reason using the data.

(2)
(b)

Determine the extension for a load of 7.0 N, assuming the spring is within its limit of proportionality.

(2)
(c)

Determine the load that gives an extension of 4.5 cm.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 yes✓ 1
B1 from 10 N to 12 N the extension increases by 1.5 cm, but it increased by 1.0 cm for each earlier 2.0 N✓ 1
Part (b)
M1 extension is 1.0 cm per 2.0 N, so 7.0 / 2.0 × 1.0✓ 1
A1 3.5 cm✓ 1
Part (c)
M1 load = 4.5 / 1.0 × 2.0✓ 1
A1 9.0 N✓ 1

Exam tip: Within the limit of proportionality, extension is directly proportional to load, so use the ratio from a point in the table.

105IGPC3-1-09r32-01Speed, velocity and motion graphs Paper CEasy1 mark

A space probe coasts at a constant speed of 15 km/s for 2.0 hours. What distance does it travel in this time?

Show mark scheme
Marking pointMarkNotes
Convert the time to seconds: 2.0 h = 7200 s. Distance = speed × time = 15 × 7200 = 1.08 × 105 km. 1800 km uses 120 (minutes) instead of seconds. 30 km multiplies by 2 and ignores the conversion from hours to seconds. 1.08 × 108 km applies an unnecessary extra factor of 1000.✓ 1Answer B

Answer: B

Common error: Always convert the time into the same unit as the one in the speed (seconds here) before multiplying.

106IGPC3-1-09r32-02Mass, weight and density Paper CEasy1 mark

A sample of green coffee beans has mass 0.150 kg and occupies a volume of 2.0 × 10−4 m3. After roasting, the sample has mass 0.120 kg and volume 3.0 × 10−4 m3. What is the density of the roasted beans?

Show mark scheme
Marking pointMarkNotes
Density = mass ÷ volume using the roasted values: 0.120 ÷ 3.0 × 10−4 = 400 kg/m3. 600 kg/m3 combines the roasted mass with the original volume. 500 kg/m3 combines the original mass with the roasted volume. 2.5 × 10−3 comes from dividing volume by mass.✓ 1Answer C

Answer: C

Common error: When a sample changes, take both mass and volume from the same state of the sample.

107IGPC3-1-09r32-05Momentum and impulse Paper CEasy1 mark

A bridge-inspection drone of mass 0.40 kg flies horizontally at 6.0 m/s. What is its kinetic energy?

Show mark scheme
Marking pointMarkNotes
Kinetic energy = ½mv2 = ½ × 0.40 × 6.02 = 7.2 J. 14.4 J leaves out the ½. 1.2 J uses ½mv instead of squaring the speed. 2.4 J is the momentum mv (in kg m/s), not an energy.✓ 1Answer A

Answer: A

Common error: Square the speed only, not the mass, and remember the factor ½ in kinetic energy.

108IGPC3-1-09r32-09Physical quantities and measurement Paper CEasy6 marks

A student investigates a model bridge deck that swings from side to side like a pendulum. Using a stopwatch, the student finds that 20 complete oscillations take 31.4 s.

(a)

Calculate the time for one oscillation (the period).

(2)
(b)

Explain why the student timed 20 oscillations rather than timing one.

(2)
(c)

Use T = 1.57 s to calculate the frequency of the swing, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 period = total time ÷ number of oscillations = 31.4 ÷ 20✓ 1
A1 1.57 s✓ 1
Part (b)
the error from starting and stopping the stopwatch (reaction time) is about the same size whatever number of swings is timed✓ 1
so it is a much smaller fraction of the longer total time, giving a more accurate value of the period✓ 1
Part (c)
M1 f = 1 ÷ T = 1 ÷ 1.57✓ 1
A1 0.637 Hz✓ 1

Exam tip: Timing many repeats and dividing reduces the effect of human reaction time on a single measurement.

109IGPC3-1-09r32-10Mass, weight and density Paper CEasy5 marks

A space probe has a mass of 460 kg. The gravitational field strength is 9.8 N/kg on Earth and 3.7 N/kg on the surface of Mars.

(a)

Calculate the weight of the probe on Earth.

(2)
(b)

State the mass of the probe on Mars.

(1)
(c)

Calculate the weight of the probe on the surface of Mars.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 W = mg = 460 × 9.8✓ 1
A1 4508 N✓ 1
Part (b)
460 kg, because mass does not depend on the gravitational field strength✓ 1
Part (c)
M1 W = mg = 460 × 3.7✓ 1
A1 1702 N✓ 1

Common error: Mass (in kg) stays the same everywhere; only the weight (in N) changes with the gravitational field strength.

110IGPC3-1-08r38-01Physical quantities and measurement Paper CEasy1 mark

An archaeologist finds a stack of 25 identical Roman coins with a total height of 40 mm. What is the thickness of one coin?

Show mark scheme
Marking pointMarkNotes
Thickness of one coin = total height ÷ number of coins = 40 mm ÷ 25 = 1.6 mm. Dividing the number of coins by the height gives 0.63, subtracting the numbers gives 15, and multiplying them gives 1000.✓ 1Answer C

Answer: C

Exam tip: Measuring a stack of identical objects and dividing reduces the percentage uncertainty in one object's thickness.

111IGPC3-1-08r38-02Speed, velocity and motion graphs Paper CEasy1 mark

A cyclist rides 4500 m in 15 minutes at a steady speed. What is the speed in m/s?

Show mark scheme
Marking pointMarkNotes
Convert the time to 900 s, then speed = distance ÷ time = 4500 ÷ 900 = 5.0 m/s. Using 15 without converting gives 300, dividing by 60 only gives 75, and dividing time by distance gives 0.20.✓ 1Answer A

Answer: A

Common error: Convert minutes to seconds before using a speed in m/s.

112IGPC3-1-08r38-13Pressure in solids and liquids Paper CEasy5 marks

A stone slab of mass 60 kg is displayed on a museum floor. Use g = 9.8 N/kg. The slab is a cuboid and rests on a face measuring 0.40 m by 0.30 m.

(a)

Calculate the weight of the slab.

(2)
(b)

Calculate the pressure that the slab exerts on the floor.

(2)
(c)

The slab is turned onto a smaller face. State what happens to the pressure on the floor.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 weight = mass × g = 60 × 9.8✓ 1
A1 588 N✓ 1
Part (b)
M1 area = 0.40 × 0.30 = 0.12 m2 and p = F ÷ A✓ 1
A1 4900 Pa (ECF from (a))✓ 1
Part (c)
B1 the pressure increases because the same force acts over a smaller area✓ 1

Exam tip: Use the weight in newtons, not the mass in kilograms, when calculating pressure from force and area.

113IGPC3-1-08r41-03Mass, weight and density Paper CEasy1 mark

A pharmacist measures out 40 cm3 of a syrup. The syrup has a mass of 52 g. What is the density of the syrup?

Show mark scheme
Marking pointMarkNotes
Density = mass ÷ volume = 52 ÷ 40 = 1.3 g/cm³. 0.77 g/cm³ comes from dividing volume by mass. 12 g/cm³ comes from subtracting volume from mass. 2080 g/cm³ comes from multiplying mass by volume.✓ 1Answer A

Answer: A

Exam tip: Write the formula density = mass ÷ volume before substituting, and keep the units g and cm³ together so the answer comes out in g/cm³.

114IGPC3-1-08r41-05Forces, friction and springs Paper CEasy1 mark

A spring extends by 4.0 cm when a load of 6.0 N is hung from it. The spring obeys Hooke’s law over this range. What is the extension when the load is 15 N, assuming the limit of proportionality is not exceeded?

Show mark scheme
Marking pointMarkNotes
Extension is directly proportional to load, so the extension is 4.0 × (15 ÷ 6.0) = 10 cm. 1.6 cm inverts the ratio. 13 cm adds the extra 9 N load as if it were 9 cm. 60 cm multiplies 4.0 by 15.✓ 1Answer B

Answer: B

Exam tip: For a spring within its limit of proportionality, extension ÷ load stays the same, so scale the extension by the same factor as the load.

115IGPC3-1-08r41-07Energy, work and power Paper CEasy1 mark

A hand pump on a water supply is pushed with a force of 85 N through 0.40 m on each stroke. A worker makes 30 strokes. How much work does the worker do?

Show mark scheme
Marking pointMarkNotes
Work = force × distance. One stroke does 85 × 0.40 = 34 J, so 30 strokes do 34 × 30 = 1020 J. 34 J is the work in one stroke only. 2550 J multiplies the force by the number of strokes and ignores distance. 6375 J divides by the distance instead of multiplying.✓ 1Answer A

Answer: A

Exam tip: Work done is force × distance moved in the direction of the force; add up the work for every stroke.

116IGPC3-1-08r41-09Mass, weight and density Paper CEasy4 marks

A pharmacist wants the density of a coated tablet. She puts 25 identical tablets into a measuring cylinder containing 40.0 cm3 of water. The water level rises to 52.5 cm3. The total mass of the 25 tablets is 31.5 g.

(a)

Calculate the volume of one tablet.

(2)
(b)

Calculate the density of the tablet material. Give your answer in g/cm³ to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Volume of 25 tablets = 52.5 − 40.0 = 12.5 cm³✓ 1
Volume of one tablet = 12.5 ÷ 25 = 0.50 cm³✓ 1
Part (b)
density = mass ÷ volume = 31.5 ÷ 12.5✓ 1
2.52 g/cm³✓ 1

Exam tip: The rise in water level, not the final reading, gives the volume of the objects placed in the cylinder.

117IGPC3-1-09r44-01Mass, weight and density Paper CEasy1 mark

A beekeeper fills a jar with 0.50 litre of honey. The density of honey is 1400 kg/m3. 1 litre = 1.0 × 10−3 m3.
What is the mass of honey in the jar?

Show mark scheme
Marking pointMarkNotes
Volume = 0.50 × 10−3 m3, so mass = density × volume = 1400 × 5.0 × 10−4 = 0.70 kg. 700 kg comes from using 0.50 as the volume in m3 without converting from litres. 2.8 × 106 kg comes from dividing density by volume. 3.6 × 10−7 kg comes from dividing volume by density.✓ 1Answer B

Answer: B

Common error: Convert litres to m³ before using density in kg/m³. Check that the size of your answer is sensible for a jar of honey.

118IGPC3-1-09r44-02Speed, velocity and motion graphs Paper CEasy1 mark

A Formula 1 car accelerates uniformly from rest to 50 m/s in 5.0 s. It then travels at a constant 50 m/s for a further 10 s.
What is the total distance travelled in the 15 s?

51550time / sspeed / m/s
Graph as described; not to scale
Show mark scheme
Marking pointMarkNotes
Distance while accelerating = area of a triangle on the speed–time graph = ½ × 50 × 5.0 = 125 m. Distance at constant speed = 50 × 10 = 500 m. Total = 625 m. 750 m assumes the car travels at 50 m/s for all 15 s. 500 m ignores the acceleration stage. 375 m uses the average of 0 and 50 m/s for the whole 15 s, which is only valid while the acceleration is uniform.✓ 1Answer D

Answer: D

Exam tip: On a speed–time graph, split the motion into stages and add the areas under each stage.

119IGPC3-1-09r44-06Forces, friction and springs Paper CEasy1 mark

A spring extends by 4.0 cm when a 2.0 N load hangs from it. A beekeeper hangs a hive frame of weight 5.0 N from the same spring. The limit of proportionality is not exceeded.
What is the extension of the spring?

Show mark scheme
Marking pointMarkNotes
Extension is directly proportional to load, so extension = 4.0 × (5.0 ÷ 2.0) = 10 cm. 20 cm multiplies 4.0 by 5.0 and ignores the 2.0 N. 1.6 cm inverts the ratio of the loads. 7.0 cm adds the 3.0 N increase in load to the extension in cm, which mixes up the units.✓ 1Answer D

Answer: D

Exam tip: Within the limit of proportionality, doubling the load doubles the extension. Set up a ratio before calculating.

120IGPC3-1-09r44-09Physical quantities and measurement Paper CEasy5 marks

A beekeeper wants to find the thickness of one sheet of beeswax foundation. She stacks 20 identical sheets and measures the height of the stack with a millimetre ruler. The stack is 14.0 mm high.

(a)

Calculate the thickness of one sheet.

(2)
(b)

Explain why measuring the stack gives a more reliable value than measuring a single sheet with the same ruler.

(2)
(c)

Name an instrument that could measure the thickness of a single sheet more precisely than a ruler.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 thickness = total height ÷ number of sheets = 14.0 ÷ 20✓ 1
A1 0.70 mm✓ 1
Part (b)
B1 one sheet is about 0.7 mm, which is smaller than the 1 mm smallest division of the ruler✓ 1
B1 so the percentage uncertainty would be very large for one sheet, while the stack gives a larger reading and a much smaller percentage uncertainty✓ 1
Part (c)
B1 micrometer screw gauge (accept vernier calipers)✓ 1

Common error: Divide the total height by the number of sheets, not by the number of gaps between them.

121IGPC3-1-09r44-10Mass, weight and density Paper CEasy5 marks

A hospital technician finds the density of a steel surgical clamp. The mass of the clamp is 156 g. The clamp is lowered into a measuring cylinder containing 50 cm3 of water. The water level rises to 70 cm3.

(a)

Calculate the volume of the clamp.

(1)
(b)

Calculate the density of the steel in g/cm3.

(2)
(c)

Convert your answer to (b) into kg/m3. Use 1 g/cm3 = 1000 kg/m3.

(1)
(d)

State why the clamp must be completely under the water.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 70 − 50 = 20 cm3✓ 1
Part (b)
M1 density = mass ÷ volume = 156 ÷ 20 (ECF from (a))✓ 1
A1 7.8 g/cm3✓ 1
Part (c)
A1 7800 kg/m3 (7.8 × 1000, ECF from (b))✓ 1
Part (d)
B1 so that the volume of water displaced equals the whole volume of the clamp✓ 1

Exam tip: Read the measuring cylinder at eye level at the bottom of the meniscus, and subtract the starting reading from the final reading.

122IGPC3-1-09r45-01Speed, velocity and motion graphs Paper CEasy1 mark

A Formula 1 car leaves the pit lane. It accelerates uniformly from rest to 24 m/s in 8.0 s and then travels at a constant 24 m/s for a further 20 s. What is the total distance travelled in the 28 s?

82824time / sspeed / m/s
Graph as described; not to scale
Show mark scheme
Marking pointMarkNotes
The distance is the area under the speed–time graph. Acceleration stage: ½ × 8.0 × 24 = 96 m. Constant-speed stage: 24 × 20 = 480 m. Total = 96 + 480 = 576 m. 480 m leaves out the distance covered while accelerating. 672 m treats the car as if it moved at 24 m/s for the whole 28 s. 336 m uses ½ × 28 × 24, as if the whole journey were one triangle.✓ 1Answer C

Answer: C

Exam tip: On a speed–time graph, split the area into triangles and rectangles and add them.

123IGPC3-1-09r45-02Mass, weight and density Paper CEasy1 mark

A steel surgical tray in a hospital has a mass of 640 g and a volume of 80 cm3. What is the density of the steel?

Show mark scheme
Marking pointMarkNotes
Density = mass ÷ volume = 640 ÷ 80 = 8.0 g/cm3. 0.125 g/cm3 comes from dividing volume by mass. 720 g/cm3 adds the two numbers. 51 200 g/cm3 multiplies them.✓ 1Answer B

Answer: B

Common error: Dividing the wrong way round gives a density far too small for a metal; check your answer is sensible.

124IGPC3-1-09r45-03Forces, friction and springs Paper CEasy1 mark

A beekeeper uses a spring balance to weigh a feeder. A load of 2.0 N extends the spring by 3.0 cm. The spring obeys Hooke's law and its limit of proportionality is not exceeded. What is the extension when the load is 5.0 N?

Show mark scheme
Marking pointMarkNotes
Extension is directly proportional to load, so extension = 3.0 × (5.0 ÷ 2.0) = 7.5 cm. 1.2 cm comes from inverting the ratio (3.0 × 2.0 ÷ 5.0). 5.0 cm adds the extra 2.0 cm to 3.0 cm, treating the increase in load as the increase in extension. 15 cm multiplies 3.0 by 5.0.✓ 1Answer A

Answer: A

Exam tip: Doubling the load doubles the extension only while the spring is within its limit of proportionality.

125IGPC3-1-09r45-09Pressure in solids and liquids Paper CEasy7 marks

A beekeeper lifts a hive box containing frames of honey. The box has a mass of 12.0 kg. The gravitational field strength is 9.8 N/kg. The base of the box is a rectangle measuring 0.30 m by 0.40 m.

(a)

Calculate the weight of the box.

(2)
(b)

The box rests on the ground on its rectangular base. Calculate the pressure it exerts on the ground. Use the unrounded weight from (a).

(3)
(c)

The box is turned so that it stands on a smaller face of area 0.060 m2. State how the pressure on the ground changes and explain why.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 weight = mass × gravitational field strength✓ 1
A1 W = 12.0 × 9.8 = 117.6 N (accept 118 N)✓ 1
Part (b)
B1 area = 0.30 × 0.40 = 0.12 m2✓ 1
M1 pressure = force ÷ area✓ 1
A1 117.6 ÷ 0.12 = 980 Pa✓ 1
Part (c)
B1 pressure increases✓ 1
B1 the same force (weight) acts on a smaller area✓ 1

Common error: Convert all lengths to metres before finding an area in m², and use the weight in newtons, not the mass in kg, as the force.

126IGPC3-1-09r45-15Forces, friction and springs Paper CEasy6 marks

A beekeeper tests a spring used in a hive scale by adding loads and measuring the extension.
Load / N: 0, 2.0, 4.0, 6.0, 8.0, 10.0
Extension / cm: 0, 1.5, 3.0, 4.5, 6.0, 9.0

(a)

State whether the spring obeys Hooke's law for loads up to 8.0 N. Give evidence from the data.

(2)
(b)

Calculate the extension that a load of 5.0 N would produce, assuming the spring still obeys Hooke's law. Give your answer to 3 significant figures.

(2)
(c)

Explain what the result for the 10.0 N load shows.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 yes✓ 1
B1 the extension is directly proportional to the load, for example it doubles when the load doubles, or extension ÷ load is constant at 0.75 cm/N✓ 1
Part (b)
M1 extension = 0.75 × 5.0 (or 1.5 × 5.0 ÷ 2.0)✓ 1
A1 3.75 cm✓ 1
Part (c)
B1 the extension of 9.0 cm is larger than the 7.5 cm expected from proportionality✓ 1
B1 the limit of proportionality has been exceeded✓ 1

Common error: Hooke's law only applies up to the limit of proportionality; beyond it, extension grows faster than load.

127IGPC-035-01-04Speed, velocity and motion graphs Paper CMedium1 mark

A car starts from rest and accelerates uniformly. After 8.0 s its speed is 12 m/s.
How far does the car travel in this time?

Show mark scheme
Marking pointMarkNotes
Distance is the area under the speed–time graph. Here the area is a triangle: ½ × 8.0 s × 12 m/s = 48 m. The value 96 m comes from using speed × time without the ½, which would only be correct at a constant 12 m/s. The value 1.5 is the gradient, which gives the acceleration in m/s², not a distance. The value 20 comes from adding the numbers, which has no physical meaning.✓ 1Answer C

Answer: C

Exam tip: When a body accelerates uniformly from rest, the distance equals the average speed (half the final speed) multiplied by the time. This is the same as the triangle area.

128IGPC-035-01-05Speed, velocity and motion graphs Paper CMedium1 mark

A train slows down uniformly from 25 m/s to 10 m/s in 30 s.
What is the acceleration of the train?

Show mark scheme
Marking pointMarkNotes
Use a = Δv ÷ Δt = (final speed − initial speed) ÷ time. This gives (10 − 25) ÷ 30 = −15 ÷ 30 = −0.50 m/s². The negative sign shows a deceleration. +0.50 has the right size but ignores the fact that the train is slowing. −0.83 uses only the initial speed (25 ÷ 30). −1.2 adds the two speeds instead of subtracting them (35 ÷ 30).✓ 1Answer C

Answer: C

Common error: Always work out final minus initial. A deceleration is a negative acceleration, so the sign carries information.

129IGPC-035-01-07Speed, velocity and motion graphs Paper CMedium1 mark

A distance–time graph for a walker is made of three straight sections:
• from 0 s to 10 s, the distance rises from 0 m to 50 m
• from 10 s to 30 s, the distance stays at 50 m
• from 30 s to 40 s, the distance rises from 50 m to 110 m
Which statement is correct?

10304050110time / sdistance / m
Graph sketched from the values in the question; not to scale
Show mark scheme
Marking pointMarkNotes
In the final section the walker covers (110 − 50) m = 60 m in (40 − 30) s = 10 s, so the speed is 6.0 m/s. In the first section the speed is 50 ÷ 10 = 5.0 m/s, not 50 m/s. Between 10 s and 30 s the distance does not change, so the walker is at rest. The average speed is total distance ÷ total time = 110 ÷ 40 = 2.75 m/s. The value 3.7 m/s comes from averaging the three section speeds (5, 0 and 6), which is not a valid method.✓ 1Answer C

Answer: C

Common error: Average speed is never the mean of the section speeds. Always divide the total distance by the total time.

130IGPC-035-01-15Speed, velocity and motion graphs Paper CMedium7 marks

A runner's position along a straight path is recorded every 20 s.
time / s: 0, 20, 40, 60, 80
distance / m: 0, 100, 200, 260, 320

(a)

State how a distance–time graph shows that a runner is moving at constant speed.

(1)
(b)

Calculate the speed of the runner between 0 s and 40 s.

(2)
(c)

Describe how the motion of the runner between 40 s and 80 s differs from the motion between 0 s and 40 s.

(2)
(d)

Calculate the average speed of the runner for the whole 80 s.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The line is straight and sloping (constant gradient)✓ 1
Part (b)
speed = 200 ÷ 40✓ 1
= 5.0 m/s✓ 1
Part (c)
the runner moves more slowly (the gradient is smaller)✓ 1
the speed is still constant, at 120 ÷ 40 = 3.0 m/s✓ 1
Part (d)
average speed = 320 ÷ 80✓ 1
= 4.0 m/s✓ 1

Common error: In part (d), averaging 5.0 m/s and 3.0 m/s happens to give the right answer here only because the two sections last equally long. Always use total distance ÷ total time.

131IGPC-035-03-06Forces, friction and springs Paper CMedium1 mark

A delivery van travels along a straight, level road at a constant speed of 25 m/s. Which statement about the forces on the van is correct?

Show mark scheme
Marking pointMarkNotes
The van moves in a straight line at constant speed, so its velocity is not changing. When velocity is unchanged, the resultant force is zero. The forward driving force is therefore balanced by the resistive forces. A forward resultant would make the van accelerate. Resistive forces are still present, but they are balanced.✓ 1Answer C

Answer: C

Common error: Many students think a moving object must have a forward resultant force. A resultant force is needed only to change speed or direction.

132IGPC-035-03-14Forces, friction and springs Paper CMedium5 marks

A worker pushes a heavy wooden crate across a concrete floor. The crate moves to the right.

(a)

State the direction of the friction force on the crate.

(1)
(b)

The worker pushes horizontally with a force of 45 N and the crate moves at a constant speed. State the size of the friction force and explain your answer.

(2)
(c)

Suggest one way of reducing the friction between the crate and the floor.

(1)
(d)

After a while, the base of the crate feels slightly warm. Explain why.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
To the left, opposite to the direction of motion.✓ 1
Part (b)
45 N . The speed is constant, so the resultant force is zero and friction balances the push.✓ 2
Part (c)
Any one of: put rollers or wheels under the crate; lubricate the floor; use a smoother surface.✓ 1
Part (d)
Work is done against friction, which transfers energy to the thermal store of the crate and floor.✓ 1

Common error: Constant speed means the forces are balanced. It does not mean that the push is bigger than friction.

133IGPC-035-02-04Mass, weight and density Paper CMedium1 mark

A student pours water into a measuring cylinder. The reading is 35 cm3. She lowers a stone of mass 30 g into the water until it is fully covered. The reading rises to 47 cm3.
What is the density of the stone?

35 cm³before47 cm³after
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
The volume of the stone equals the volume of water it displaces: 47 − 35 = 12 cm3. Density = 30 ÷ 12 = 2.5 g/cm3.
The other options come from common mistakes:
• 0.64 g/cm3 uses the final reading as the volume.
• 0.86 g/cm3 uses the initial reading as the volume.
• 0.40 g/cm3 is volume ÷ mass.
✓ 1Answer D

Answer: D

Exam tip: In a displacement method, the volume of the object is the change in the reading, not either reading on its own.

134IGPC-035-02-05Mass, weight and density Paper CMedium1 mark

The density of water is 1.0 g/cm3. The masses and volumes of four solid objects are shown.
Which object floats in water?

Show mark scheme
Marking pointMarkNotes
An object floats if its density is less than the density of the liquid. The densities of the objects are:
• 1.2 g/cm3 (120 g, 100 cm3)
• 1.2 g/cm3 (18 g, 15 cm3)
• 0.80 g/cm3 (80 g, 100 cm3)
• 1.5 g/cm3 (30 g, 20 cm3)
Only the object of density 0.80 g/cm3 is less dense than water, so only it floats. The 18 g object is the lightest, but it still sinks because its density is greater than that of water.
✓ 1Answer C

Answer: C

Common error: Floating depends on density, not on mass alone. A light object can sink, and a heavy object can float.

135IGPC-035-02-07Mass, weight and density Paper CMedium1 mark

A bag of sugar is measured on the Earth and then on the Moon, using two instruments:
• a spring balance (newton meter)
• a beam balance, which compares the bag with standard masses.
How do the readings on the Moon compare with the readings on the Earth?

Show mark scheme
Marking pointMarkNotes
A spring balance measures weight, and weight is smaller on the Moon because g is smaller, so its reading decreases.
A beam balance compares the weight of the bag with the weight of standard masses. Both are in the same gravitational field, so both weights fall by the same factor. The balance still balances with the same standard masses, so the reading is unchanged.
✓ 1Answer B

Answer: B

Exam tip: A balance that compares two weights in the same field compares their masses. Its reading does not depend on g.

136IGPC-035-02-09Mass, weight and density Paper CMedium1 mark

An empty beaker has a mass of 85 g. A student pours 150 cm3 of a liquid into the beaker. The total mass is now 205 g.
What is the density of the liquid?

Show mark scheme
Marking pointMarkNotes
The mass of the liquid is 205 − 85 = 120 g. Density = 120 ÷ 150 = 0.80 g/cm3.
The other options come from common mistakes:
• 1.37 g/cm3 includes the mass of the beaker.
• 0.57 g/cm3 uses the mass of the beaker alone.
• 1.25 g/cm3 is volume ÷ mass.
✓ 1Answer B

Answer: B

Common error: Remember to subtract the mass of the empty container to find the mass of the liquid alone.

137IGPC-035-02-14Mass, weight and density Paper CMedium6 marks

A student wants to find the density of a small brass key ring, which has an irregular shape. She has a measuring cylinder, water, a length of thin thread and an electronic balance.

(a)

Describe how she can use the measuring cylinder to find the volume of the key ring.

(3)
(b)

The readings are 42.0 cm3 and 48.5 cm3. The mass of the key ring is 55.9 g. Calculate the density of the brass.

(2)
(c)

Suggest one precaution that would make her measurement of volume more accurate.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Partly fill the cylinder with water and record the volume reading✓ 1
Lower the key ring on the thread until it is completely under the water✓ 1
Record the new reading; the volume of the key ring is the new reading minus the first reading✓ 1
Part (b)
V = 48.5 − 42.0 = 6.5 cm3✓ 1
ρ = 55.9 ÷ 6.5 = 8.6 g/cm3✓ 1
Part (c)
Any one of the following :✓ 1
• read the scale at eye level, at the bottom of the meniscus—
• make sure no air bubbles cling to the key ring—
• use a narrower measuring cylinder with finer scale divisions—

Exam tip: In a method description, say that the object must be fully submerged. Saying it is just "put in the water" is often not enough for the mark.

138IGPC-035-06-06Energy, work and power Paper CMedium1 mark

Which energy resource is renewable and does not get its energy from radiation from the Sun?

Show mark scheme
Marking pointMarkNotes
Tides are caused mainly by the gravitational pull of the Moon, so tidal energy does not come from the Sun's radiation, and the tides will continue, so it is renewable. Wind is caused by the Sun heating the atmosphere unevenly. Hydroelectric energy relies on rain from water evaporated by the Sun. Coal formed from plants that used sunlight, and it is non-renewable.✓ 1Answer C

Answer: C

Exam tip: The resources that do not come from the Sun's radiation are geothermal, nuclear and tidal. Learn these three.

139IGPC-035-06-07Energy, work and power Paper CMedium1 mark

A pendulum is set swinging in air. After several minutes it has stopped moving.
Which statement about the energy the pendulum had at the start is correct?

Show mark scheme
Marking pointMarkNotes
Energy cannot be created or destroyed. Air resistance and friction at the pivot do work against the motion, and this transfers the energy to the thermal store of the surroundings. The bob ends at its lowest point, so it has not gained gravitational potential energy. No chemical change happens in the string.✓ 1Answer C

Answer: C

Common error: Never say energy is 'lost' or 'used up' in the sense of disappearing. Say it is transferred to the thermal store of the surroundings, which is a wasted transfer.

140IGPC-035-00-04Physical quantities and measurement Paper CMedium1 mark

A measuring cylinder contains water to the 30.0 cm3 mark. Five identical marbles are lowered into the water and sink completely. The water level rises to 42.5 cm3.
What is the volume of one marble?

30.0 cm³before42.5 cm³after
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Volume of the five marbles = 42.5 − 30.0 = 12.5 cm3. Volume of one marble = 12.5 ÷ 5 = 2.5 cm3. 12.5 cm3 is the volume of all five marbles. 8.5 cm3 divides the final reading by 5 without subtracting the starting volume. 6.0 cm3 divides the starting volume by 5.✓ 1Answer A

Answer: A

Exam tip: Displacement method: find the rise in level first (final − initial), then divide by the number of objects.

141IGPC-035-00-08Physical quantities and measurement Paper CMedium1 mark

A student wants to find the period of a pendulum. She times 20 oscillations instead of a single oscillation.
Why does this give a more accurate value for the period?

Show mark scheme
Marking pointMarkNotes
Starting and stopping the stopwatch adds a reaction-time error of a similar size whatever is being timed. When the total time is much longer, this error is a much smaller fraction of the measurement. Dividing by 20 then gives a more accurate period. Errors are reduced, not removed. The amplitude and the period do not change because of how many swings are timed.✓ 1Answer D

Answer: D

Exam tip: For 'why time many oscillations', say that the reaction-time error becomes a smaller fraction (percentage) of the total time.

142IGPC-035-00-09Physical quantities and measurement Paper CMedium1 mark

A student times 10 oscillations of a pendulum three times. The results are 15.2 s, 15.6 s and 15.4 s.
What is the best value for the period of the pendulum?

Show mark scheme
Marking pointMarkNotes
Mean time for 10 oscillations = (15.2 + 15.6 + 15.4) ÷ 3 = 15.4 s. Period = 15.4 ÷ 10 = 1.54 s. 15.4 s is the time for 10 oscillations, not one. 4.62 s divides the total of all three readings by 10. 0.154 s divides by 100.✓ 1Answer B

Answer: B

Exam tip: Average the repeated readings first, then divide by the number of oscillations in each reading.

143IGPC-035-04-04Turning effect and centre of gravity Paper CMedium1 mark

A child of weight 300 N sits 2.0 m from the pivot of a see-saw. A second child of weight 400 N sits on the other side so that the see-saw balances. The weight of the see-saw acts at the pivot.
How far from the pivot is the second child?

300 N400 N2.0 md
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Principle of moments: 300 × 2.0 = 400 × d, so d = 600 / 400 = 1.5 m. The heavier child must sit closer to the pivot. Option D comes from inverting the ratio (400 × 2.0 / 300), option C assumes equal distances, and option A subtracts values instead of using moments.✓ 1Answer B

Answer: B

Common error: The heavier person always sits closer to the pivot. If your answer puts them further away, you have inverted the ratio.

144IGPC-035-04-05Turning effect and centre of gravity Paper CMedium1 mark

A desk lamp falls over easily when it is knocked.
Which change to its design would make it most stable?

Show mark scheme
Marking pointMarkNotes
An object topples when the line of action of its weight passes outside its base. A low centre of gravity and a wide base mean the lamp must be tilted through a larger angle before this happens, so it is more stable. Each of the other options includes at least one change that reduces stability.✓ 1Answer B

Answer: B

Exam tip: Stability depends on two factors: the height of the centre of gravity and the width of the base. Mention both.

145IGPC-035-04-06Turning effect and centre of gravity Paper CMedium1 mark

Which statement describes an object that is in equilibrium?

Show mark scheme
Marking pointMarkNotes
Equilibrium requires both conditions: no resultant force, so there is no change in motion, and no resultant moment, so there is no turning. A resultant moment alone would make the object rotate, and a resultant force alone would make it accelerate. Option D confuses forces with moments.✓ 1Answer C

Answer: C

Common error: Equal forces on each side of a pivot do not guarantee balance. The moments must be equal.

146IGPC-035-04-09Turning effect and centre of gravity Paper CMedium1 mark

A student finds the centre of gravity of an irregular piece of card. She hangs it freely from a pin through a hole and draws a vertical line on it using a plumb line. She repeats this using a second hole.
Where is the centre of gravity?

Show mark scheme
Marking pointMarkNotes
A freely suspended object comes to rest with its centre of gravity directly below the point of suspension, so the centre of gravity lies somewhere on each vertical line. The only point on both lines is where they cross. The other options are not on both lines in general.✓ 1Answer B

Answer: B

Exam tip: A third hole and line gives a useful check: all three lines should meet at one point.

147IGPC-035-04-14Turning effect and centre of gravity Paper CMedium5 marks

A gardener lifts the handles of a loaded wheelbarrow so that it is just off the ground and held steady. The wheel axle acts as the pivot. The combined weight of the barrow and its load is 600 N. This weight acts 0.40 m horizontally from the axle. The gardener applies a vertical upward force at the handles, 1.2 m horizontally from the axle.

600 NF0.40 m1.2 m
Wheelbarrow modelled as a lever (axle = pivot); not to scale
(a)

Calculate the moment of the 600 N weight about the axle.

(2)
(b)

Calculate the upward force the gardener must apply at the handles.

(2)
(c)

Suggest why the gardener should place heavy loads near the front of the barrow, close to the wheel.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Moment = 600 × 0.40✓ 1
= 240 N m✓ 1
Part (b)
F × 1.2 = 240✓ 1
F = 200 N✓ 1
Part (c)
This reduces the distance of the weight from the pivot, so its moment is smaller and a smaller lifting force is needed✓ 1

Exam tip: Write out the principle of moments as an equation with numbers substituted before solving. This can earn a mark even if the final answer is wrong.

148IGPC-035-07-05Pressure in solids and liquids Paper CMedium1 mark

A rectangular brick is placed on a table in three different ways: on its largest face, on its middle-sized face and on its smallest face.
Which statement is correct?

Show mark scheme
Marking pointMarkNotes
The force on the table equals the weight of the brick, so it does not depend on which face is down. Pressure is p = F/A. With the same force, the smallest contact area gives the greatest pressure. So the force is constant, and the pressure is greatest on the smallest face.✓ 1Answer D

Answer: D

Common error: Do not confuse force with pressure. Turning an object onto a different face changes the pressure, but its weight stays the same.

149IGPC-035-07-08Pressure in solids and liquids Paper CMedium1 mark

Three containers of different shapes are filled with the same liquid to the same depth. Container X is narrow, container Y is wide, and container Z gets wider towards the top.
How do the pressures due to the liquid at the bottom of the containers compare?

XYZsame depth
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
The pressure in a liquid depends only on the depth and the density of the liquid. It does not depend on the shape of the container or the volume of liquid. All three containers have the same liquid at the same depth, so the pressure at the bottom is the same in each.✓ 1Answer D

Answer: D

Common error: A greater volume of liquid does not mean a greater pressure at the base. Only depth and density matter.

150IGPC-035-07-13Pressure in solids and liquids Paper CMedium6 marks

A farm tractor has a mass of 5000 kg. It has four tyres, and each tyre has an area of 0.35 m2 in contact with the ground. Take g = 9.8 N/kg.

(a)

Calculate the weight of the tractor.

(2)
(b)

Calculate the pressure the tractor exerts on the ground.

(3)
(c)

Suggest why tractors used on muddy fields are fitted with very wide tyres.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
W = mg = 5000 × 9.8✓ 1
W = 49 000 N✓ 1
Part (b)
total area = 4 × 0.35 = 1.4 m2✓ 1
p = F/A = 49 000 / 1.4✓ 1
p = 35 000 Pa✓ 1
Part (c)
A larger contact area gives a lower pressure, so the tractor does not sink into the mud✓ 1

Common error: Using the area of only one tyre gives a pressure four times too large. Add up the areas of all the surfaces in contact.

151IGPC-035-07-15Pressure in solids and liquids Paper CMedium6 marks

A solid concrete block measures 0.50 m × 0.40 m × 0.20 m. The density of concrete is 2400 kg/m3. Take g = 9.8 N/kg.

0.50 m0.20 m0.40 m
Diagram not to scale
(a)

Calculate the volume of the block.

(1)
(b)

Calculate the weight of the block.

(2)
(c)

The block can be placed on any of its faces. Calculate the greatest pressure it can exert on the ground.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
V = 0.50 × 0.40 × 0.20 = 0.040 m3✓ 1
Part (b)
m = ρV = 2400 × 0.040 = 96 kg✓ 1
W = 96 × 9.8 = 940 N (941 N)✓ 1
Part (c)
The greatest pressure is on the smallest face, area 0.40 × 0.20 = 0.080 m2✓ 1
p = 940.8 / 0.080✓ 1
p = 1.2 × 104 Pa (11 760 Pa)✓ 1

Exam tip: Keep unrounded values in your calculator between steps. Round only the final answer, to 2 or 3 significant figures.

152IGPC-035-06-14Energy, work and power Paper CMedium5 marks

A coastal region plans to replace an old coal-fired power station with a group of wind turbines.

(a)

State what is meant by a renewable energy resource.

(1)
(b)

State one advantage and one disadvantage of generating electricity using wind turbines rather than coal.

(2)
(c)

The wind turbines produce an average total power output of 2.0 MW. Calculate the energy they produce in 24 hours.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A resource that is replaced (by natural processes) at least as fast as it is used, so it will not run out✓ 1
Part (b)
Advantage: any one of: renewable / produces no carbon dioxide or polluting gases when operating / no fuel needed, so no fuel costs✓ 1
Disadvantage: any one of: output depends on the wind, so it is unreliable or not always available / each turbine has a small output, so many turbines and a large area are needed / visual or noise impact✓ 1
Part (c)
t = 24 × 3600 = 86 400 s and E = P × t = 2.0 × 106 × 86 400✓ 1
= 1.7 × 1011 J (accept 1.73 × 1011 J)✓ 1

Common error: Convert MW to W (× 10⁶) and hours to seconds (× 3600) before using E = Pt. Otherwise the answer is not in joules.

153IGPC-035-00-05Physical quantities and measurement Paper CMedium1 mark

The end of a wooden metre rule is worn, so a student measures a pencil starting from the 2.0 cm mark. The other end of the pencil lines up with the 9.4 cm mark.
What is the length of the pencil?

Show mark scheme
Marking pointMarkNotes
Length = final reading − starting reading = 9.4 cm − 2.0 cm = 7.4 cm, so 7.4 cm is correct. 9.4 cm assumes the pencil started at the zero mark. 11.4 cm adds the two readings instead of subtracting. 74 cm is 74 mm with the wrong unit, a mix-up between millimetres and centimetres.✓ 1Answer A

Answer: A

Exam tip: Starting from a mark other than zero is good practice with a damaged rule, as long as you subtract the starting reading from the final reading.

154IGPC3-1-01c-04Forces, friction and springs Paper CMedium1 mark

A spring extends by 4.0 cm when a load of 6.0 N hangs from it. The spring obeys Hooke's law for all loads up to 20 N. What is the extension when the load is 15 N?

Show mark scheme
Marking pointMarkNotes
Correct: B. Extension is proportional to load: 4.0 × (15 ÷ 6.0) = 10 cm. A is wrong: this scales the wrong way round, 4.0 × 6.0 ÷ 15, treating extension as inversely proportional to load. C is wrong: this adds the extra 9 N as 9 cm to the original 4.0 cm, ignoring proportionality. D is wrong: this uses 6.0 × 15 ÷ 4.0, a ratio that does not represent extension.✓ 1Answer B

Answer: B

Exam tip: If extension is proportional to load, a load 2.5 times bigger gives an extension 2.5 times bigger.

155IGPC3-1-01c-05Momentum and impulse Paper CMedium1 mark

A 2.0 kg ball is dropped from rest from a height of 5.0 m. Air resistance is negligible. What is the kinetic energy of the ball just before it hits the ground? (g = 9.8 N/kg)

Show mark scheme
Marking pointMarkNotes
Correct: C. Gravitational potential energy lost = mgh = 2.0 × 9.8 × 5.0 = 98 J, all transferred to kinetic energy. A is wrong: this uses ½mgh, wrongly applying the ½ from the kinetic energy equation to the potential energy. B is wrong: this multiplies mass by height and leaves out g. D is wrong: this doubles the answer, equal to mv2 instead of ½mv2.✓ 1Answer C

Answer: C

Exam tip: Energy conservation avoids needing the speed: the kinetic energy gained equals the potential energy lost.

156IGPC3-1-01c-08Turning effect and centre of gravity Paper CMedium1 mark

A tall, narrow cupboard on a ramp is slowly tilted. At which point does it begin to topple?

Show mark scheme
Marking pointMarkNotes
Correct: D. The weight then produces a turning effect that rotates the cupboard about its lower edge instead of restoring it. A is wrong: toppling depends on the shape of the object and the position of the centre of gravity, not on a fixed angle for every object. B is wrong: friction would affect sliding, not toppling, and the weight does not change as the ramp tilts. C is wrong: weight acts at all times; only the line of action relative to the base matters.✓ 1Answer D

Answer: D

Common error: A low centre of gravity and a wide base make an object harder to topple.

157IGPC3-1-01c-12Energy, work and power Paper CMedium7 marks

A crane lifts a concrete beam of mass 250 kg vertically through 12 m in 20 s at a steady speed. The motor of the crane has an input power of 2100 W. (g = 9.8 N/kg)

(a)

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

(2)
(b)

Calculate the useful power output of the crane.

(2)
(c)

State what happens to the energy that is not usefully transferred.

(1)
(d)

Calculate the efficiency of the crane.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
weight = 250 × 9.8 = 2450 N✓ 1
work = force × distance = 2450 × 12 = 29400 J✓ 1
Part (b)
power = work ÷ time✓ 1
29400 ÷ 20 = 1470 W✓ 1
Part (c)
it is transferred to the surroundings as thermal energy (and sound)✓ 1
Part (d)
efficiency = useful power output ÷ input power (×100%)✓ 1
1470 ÷ 2100 × 100 = 70%✓ 1

Exam tip: Efficiency compares useful output with total input. It can never be more than 100%.

158IGPC3-1-06r5-04Speed, velocity and motion graphs Paper CMedium1 mark

A warehouse trolley starts from rest and accelerates uniformly to 12 m/s in 4 s. It travels at 12 m/s for 6 s and then slows uniformly to rest in 2 s. What is the total distance travelled?

4101212time / sspeed / m/s
Graph sketched from the values in the question; not to scale
Show mark scheme
Marking pointMarkNotes
The distance is the area under the speed-time graph: ½ × 4 × 12 = 24 m, 6 × 12 = 72 m and ½ × 2 × 12 = 12 m, giving 108 m. 84 m omits the acceleration stage. 96 m omits the slowing stage. 144 m treats the whole 12 s as steady motion at 12 m/s.✓ 1Answer C

Answer: C

Common error: When a motion has several stages, find each area separately. Triangles need the factor ½.

159IGPC3-1-06r5-07Speed, velocity and motion graphs Paper CMedium1 mark

The distance-time graph for a cargo drone is a straight line sloping upwards for 20 s, then horizontal for 10 s, then a straight line sloping downwards that is steeper than the first line. Which description matches this motion?

Show mark scheme
Marking pointMarkNotes
On a distance-time graph, gradient is speed. A straight sloping line means constant speed, a horizontal line means stationary, and a downward line means heading back, with a steeper gradient meaning greater speed. A straight line does not show acceleration. A horizontal line is not constant speed.✓ 1Answer B

Answer: B

Common error: A horizontal line on a distance-time graph means the object is at rest, not moving at constant speed.

160IGPC3-1-06r5-10Speed, velocity and motion graphs Paper CMedium8 marks

A delivery van accelerates uniformly from rest to 15 m/s in 10 s. It then travels at a constant speed for 30 s and finally decelerates uniformly to rest in 10 s.

10405015time / sspeed / m/s
Graph sketched from the values in the question; not to scale
(a)

Describe the motion of the van between 10 s and 40 s.

(1)
(b)

Calculate the acceleration of the van during the first 10 s.

(2)
(c)

Calculate the total distance travelled by the van.

(3)
(d)

Calculate the average speed of the van for the whole journey.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
it moves at constant speed (15 m/s)✓ 1
Part (b)
acceleration = change in speed ÷ time = 15 ÷ 10✓ 1
= 1.5 m/s2✓ 1
Part (c)
distance while accelerating = ½ × 10 × 15 = 75 m✓ 1
distance at constant speed = 30 × 15 = 450 m✓ 1
distance while decelerating = 75 m, so total = 600 m✓ 1
Part (d)
total time = 50 s✓ 1
600 ÷ 50 = 12 m/s (ECF from (c))✓ 1

Common error: Average speed is total distance ÷ total time. It is not the average of the highest and lowest speeds.

161IGPC3-1-06r5-13Mass, weight and density Paper CMedium5 marks

A student measures the density of a piece of coral rock. Its mass is 91 g. The reading on a measuring cylinder containing water rises from 60 cm3 to 95 cm3 when the coral is fully immersed.

60 cm³before95 cm³after
Diagram not to scale
(a)

Calculate the volume of the coral.

(1)
(b)

Calculate the density of the coral rock.

(2)
(c)

The density of sea water is 1.03 g/cm3. State whether the coral rock floats or sinks in sea water and give a reason.

(1)
(d)

Suggest a way to find the volume of a piece of coral that is too large for the measuring cylinder.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
95 − 60 = 35 cm3✓ 1
Part (b)
density = mass ÷ volume = 91 ÷ 35✓ 1
= 2.6 g/cm3✓ 1
Part (c)
it sinks because its density is greater than that of sea water✓ 1
Part (d)
immerse it in a full overflow (displacement) can and collect and measure the volume of water displaced✓ 1

Exam tip: The volume of an irregular solid is the change in water level, not the final reading.

162IGPC3-1-06r7-04Pressure in solids and liquids Paper CMedium1 mark

An architect's building has a mass of 2.0×105 kg and rests on a flat foundation of area 100 m². Use g = 9.8 N/kg. What is the average pressure on the ground?

Show mark scheme
Marking pointMarkNotes
C is correct: weight = 2.0×105 × 9.8 = 1.96×106 N, and p = F / A = 1.96×106 / 100 = 19 600 Pa = 19.6 kPa. A is wrong because it divides area by force. B is wrong because it uses mass instead of weight (2.0×105 / 100 = 2000 Pa). D is wrong because it multiplies force by area.✓ 1Answer C

Answer: C

Common error: Pressure needs the weight in newtons, not the mass in kilograms.

163IGPC3-1-06r7-10Mass, weight and density Paper CMedium5 marks

A marine biologist collects a 250 cm³ sample of seawater. Its mass is 257.5 g. Use g = 9.8 N/kg.

(a)

Calculate the density of the seawater. Give your answer in g/cm³.

(2)
(b)

Calculate the weight of the sample. Give your answer in newtons to 3 significant figures.

(2)
(c)

State what happens to the mass and the weight of the sample if it is taken to the Moon.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
density = mass / volume = 257.5 / 250✓ 1
1.03 g/cm³✓ 1
Part (b)
mass = 0.2575 kg✓ 1
W = mg = 0.2575 × 9.8 = 2.52 N✓ 1
Part (c)
The mass stays the same and the weight is smaller (the gravitational field strength is less)✓ 1

Common error: Convert grams to kilograms before using W = mg.

164IGPC3-1-06r7-11Forces, friction and springs Paper CMedium5 marks

An architect tests a spring for a vibration damper. The unloaded length is 20.0 cm. A load of 2.0 N gives a length of 22.4 cm, 4.0 N gives 24.8 cm and 6.0 N gives 27.2 cm.

(a)

State the extension when the load is 6.0 N.

(1)
(b)

Show that the spring obeys Hooke's law for these loads.

(2)
(c)

Predict the length of the spring when the load is 7.0 N, assuming it is still within the limit of proportionality.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
27.2 − 20.0 = 7.2 cm✓ 1
Part (b)
The extension for 2.0 N, 4.0 N and 6.0 N is 2.4 cm, 4.8 cm and 7.2 cm✓ 1
Doubling or tripling the load doubles or triples the extension, so extension is proportional to load✓ 1
Part (c)
Extension per newton = 1.2 cm/N, so extension = 1.2 × 7.0 = 8.4 cm✓ 1
Length = 20.0 + 8.4 = 28.4 cm✓ 1

Common error: Hooke's law relates the extension, not the total length, to the load.

165IGPC3-1-06r7-12Forces, friction and springs Paper CMedium6 marks

A warehouse worker pushes a crate of mass 45 kg along a level floor. At first the crate moves at constant speed when the push is 180 N.

cratepush 180 Nfriction F
Diagram not to scale
(a)

State the size of the resultant force on the crate and the size of the friction force.

(2)
(b)

The worker increases the push to 225 N. Explain what happens to the motion of the crate.

(2)
(c)

Calculate the acceleration of the crate with the 225 N push. Assume the friction force is unchanged.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The resultant force is zero✓ 1
Friction force = 180 N✓ 1
Part (b)
The resultant force is no longer zero (it is forwards)✓ 1
The crate accelerates✓ 1
Part (c)
resultant force = 225 − 180 = 45 N✓ 1
a = F / m = 45 / 45 = 1.0 m/s²✓ 1

Exam tip: Find the resultant force first, then use F = ma.

166IGPC3-1-07r8-03Speed, velocity and motion graphs Paper CMedium1 mark

The speed of an aircraft on a runway increases uniformly from 0 to 60 m/s in 20 s. It then stays at 60 m/s for a further 10 s. What is the total distance travelled in the 30 s?

203060time / sspeed / m/s
Graph sketched from the values in the question; not to scale
Show mark scheme
Marking pointMarkNotes
Distance = area under the speed-time graph = (½ × 20 × 60) + (10 × 60) = 600 + 600 = 1200 m, so C. A counts only the accelerating stage. B treats the whole 30 s as one triangle (½ × 30 × 60). D uses the maximum speed for the whole time (60 × 30).✓ 1Answer C

Answer: C

Common error: Do not use the final speed for the whole journey; split the motion into stages and find each area separately.

167IGPC3-1-07r8-07Forces, friction and springs Paper CMedium1 mark

A spring in a seed dispenser obeys Hooke's law within its limit of proportionality. It extends by 3.0 cm when a load of 2.0 N is hung from it. What is the extension when the load is 5.0 N?

Show mark scheme
Marking pointMarkNotes
Extension is directly proportional to load: 3.0 × (5.0 ÷ 2.0) = 7.5 cm, so D. A inverts the ratio (3.0 × 2.0 ÷ 5.0). B equates the extension in cm with the load in N. C adds the 7.5 cm extension to the 3.0 cm already present.✓ 1Answer D

Answer: D

Exam tip: Use the ratio load ÷ extension (the same constant for each load below the limit) rather than adding values.

168IGPC3-1-07r8-08Forces, friction and springs Paper CMedium1 mark

A farm worker pushes a 20 kg crate across a level floor with a constant horizontal force of 50 N. The crate moves at a constant speed. The gravitational field strength is 9.8 N/kg. What is the friction force on the crate?

cratepush 50 Nfriction F
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Constant speed means the resultant force is zero, so friction equals the 50 N push, B. A assumes no net force means no friction at all. C subtracts the push from the weight (196 − 50). D uses the weight of the crate, which acts vertically and is balanced by the normal force.✓ 1Answer B

Answer: B

Common error: Constant speed does not mean no forces; it means the forces balance.

169IGPC3-1-07r8-12Forces, friction and springs Paper CMedium5 marks

A spring is used in a balance for weighing sacks of seed. Its natural length is 20.0 cm. When a load of 4.0 N is hung from it, its length is 28.0 cm. The spring obeys Hooke's law for all loads in this question. Use g = 9.8 N/kg.

20.0 cm4.0 N28.0 cmno loadwith load
Diagram not to scale
(a)

Calculate the extension of the spring when the load is 4.0 N.

(1)
(b)

State the relationship between the extension and the load for this spring.

(1)
(c)

A bag of seed of mass 1.5 kg is hung from the spring. Calculate the length of the spring.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
28.0 − 20.0 = 8.0 cm✓ 1
Part (b)
extension is directly proportional to load✓ 1
Part (c)
weight = 1.5 × 9.8 = 14.7 N✓ 1
extension = 14.7 × (8.0 ÷ 4.0) = 29.4 cm✓ 1
length = 20.0 + 29.4 = 49.4 cm✓ 1

Common error: Extension is the increase in length, not the total length of the spring.

170IGPC3-1-07r8-15Speed, velocity and motion graphs Paper CMedium7 marks

A small aircraft taxis along a runway. Its speed increases uniformly from 0 to 12 m/s in 8.0 s. It then travels at a constant 12 m/s for 20 s, and finally slows uniformly to rest in 6.0 s.

8283412time / sspeed / m/s
Graph sketched from the values in the question; not to scale
(a)

Calculate the acceleration during the first 8.0 s.

(2)
(b)

Calculate the total distance travelled.

(3)
(c)

Calculate the average speed over the whole journey, to 2 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
acceleration = change in speed ÷ time = 12 ÷ 8.0✓ 1
1.5 m/s2✓ 1
Part (b)
first stage: ½ × 8.0 × 12 = 48 m and last stage: ½ × 6.0 × 12 = 36 m✓ 1
constant stage: 12 × 20 = 240 m✓ 1
total = 324 m✓ 1
Part (c)
average speed = total distance ÷ total time = 324 ÷ 34 (ECF from (b))✓ 1
9.5 m/s✓ 1

Exam tip: Average speed is total distance divided by total time, not the mean of the highest and lowest speeds.

171IGPC3-1-05r15-03Mass, weight and density Paper CMedium1 mark

A syringe is filled with 7.5 cm3 of liquid medicine of density 0.80 g/cm3. What is the mass of the liquid?

Show mark scheme
Marking pointMarkNotes
mass = density × volume = 0.80 × 7.5 = 6.0 g, so C is correct. A is volume ÷ density. B is density ÷ volume. D adds the two numbers.✓ 1Answer C

Answer: C

Exam tip: Check the rearrangement by units: g/cm³ × cm³ gives g.

172IGPC3-1-05r15-04Pressure in solids and liquids Paper CMedium1 mark

Water in a supply pipe exerts a pressure of 3.0×105 Pa on a valve cover of area 0.040 m2. What is the force on the cover due to the water?

Show mark scheme
Marking pointMarkNotes
force = pressure × area = 3.0×105 × 0.040 = 1.2×104 N, so B is correct. A is pressure ÷ area. C is a power-of-ten slip. D is area ÷ pressure.✓ 1Answer B

Answer: B

Common error: Rearranging p = F/A incorrectly is common; F = p × A.

173IGPC3-1-05r15-08Forces, friction and springs Paper CMedium1 mark

A spring in a game-controller trigger extends by 2.0 cm when the force on it is 5.0 N. The limit of proportionality is not exceeded. What is the extension when the force is 8.0 N?

Show mark scheme
Marking pointMarkNotes
Extension is proportional to force, so extension = 2.0 × (8.0 ÷ 5.0) = 3.2 cm and C is correct. A adds the 3.0 N difference to 2.0 cm. B divides 8.0 by 2.0 (wrong). D inverts the ratio (2.0 × 5.0 ÷ 8.0).✓ 1Answer C

Answer: C

Exam tip: Use a ratio for proportional relationships: new extension = old extension × new force ÷ old force.

174IGPC3-1-05r15-14Mass, weight and density Paper CMedium7 marks

A bottle has a mass of 90 g when empty. When filled with 125 cm³ of syrup the total mass is 250 g. Use g = 9.8 N/kg.

(a)

Calculate the mass of the syrup.

(1)
(b)

Calculate the density of the syrup in g/cm³.

(2)
(c)

A sealed capsule of density 1.05 g/cm³ is placed in the syrup. State whether it floats or sinks and explain.

(2)
(d)

Calculate the weight of the full bottle on Earth, using a mass of 0.250 kg.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
250 − 90 = 160 g✓ 1
Part (b)
density = mass/volume = 160 ÷ 125✓ 1
1.28 g/cm³✓ 1
Part (c)
floats✓ 1
capsule's density is less than the density of the syrup✓ 1
Part (d)
W = mg = 0.250 × 9.8✓ 1
2.45 N✓ 1

Exam tip: Convert grams to kilograms before using W = mg.

175IGPC3-1-06r18-03Forces, friction and springs Paper CMedium1 mark

A spring balance used to weigh sacks of coffee beans contains a spring that obeys Hooke's law. A load of 4.0 N stretches the spring by 5.0 cm. The limit of proportionality is not exceeded. What is the extension when the load is 10 N?

Show mark scheme
Marking pointMarkNotes
Extension is directly proportional to load, so extension = 5.0 × (10 / 4.0) = 12.5 cm. A (2.0 cm) inverts the ratio. B (11 cm) adds the extra 6.0 N load directly to 5.0 cm, ignoring proportionality. D (50 cm) multiplies 10 by 5.0 without using the 4.0 N reference load.✓ 1Answer C

Answer: C

Common error: If extension is proportional to load, multiply the known extension by the ratio of the loads, with the new load on top.

176IGPC3-1-06r18-04Energy, work and power Paper CMedium1 mark

A crane on a bridge lifts a 1000 kg deck section vertically through 6.0 m in 15 s at a steady speed. What is the useful power output of the crane? (g = 9.8 N/kg)

Show mark scheme
Marking pointMarkNotes
Work done = weight × height = 1000 × 9.8 × 6.0 = 58 800 J. Power = work ÷ time = 58 800 / 15 = 3920 W. A (400 W) leaves out g, using 1000 × 6.0 / 15. B (24 500 W) multiplies by the time and divides by the height instead of dividing by time. C (58 800 W) is the work done in joules; it has not been divided by 15 s.✓ 1Answer D

Answer: D

Exam tip: Find the work (energy transferred) first, then divide by time. Do not forget to convert mass to weight with g.

177IGPC3-1-06r18-08Speed, velocity and motion graphs Paper CMedium1 mark

The distance–time graph for a coffee-delivery trolley on a bridge walkway consists of three straight sections: from 0 s to 20 s the distance rises from 0 m to 30 m; from 20 s to 35 s the distance stays at 30 m; from 35 s to 45 s the distance rises from 30 m to 60 m. What is the average speed of the trolley over the whole 45 s?

2035453060time / sdistance / m
Graph sketched from the values in the question; not to scale
Show mark scheme
Marking pointMarkNotes
Average speed = total distance ÷ total time = 60 / 45 = 1.33 m/s, including the 15 s when the trolley is stationary. B (1.50) is the speed during the first 20 s only. C (2.25) is the mean of the two moving speeds, 1.5 and 3.0 m/s. D (3.00) is the speed during the last 10 s only.✓ 1Answer A

Answer: A

Exam tip: For average speed on a distance–time graph, use total distance ÷ total time, including any time spent stationary.

178IGPC3-1-06r18-12Forces, friction and springs Paper CMedium5 marks

An inspection platform of mass 90 kg is pulled along a level bridge deck by a cable. The cable pulls with a horizontal force of 250 N and the friction force on the platform is 180 N.

platformcable pull 250 Nfriction 180 N
Diagram not to scale
(a)

Calculate the resultant force on the platform.

(1)
(b)

Calculate the acceleration of the platform.

(2)
(c)

The pull of the cable is now reduced to 180 N while the platform is moving. Describe the motion of the platform.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 250 − 180 = 70 N✓ 1
Part (b)
M1 a = F ÷ m = 70 / 90✓ 1
A1 0.78 m/s2 (ECF from (a))✓ 1
Part (c)
B1 resultant force is zero (forces balanced)✓ 1
B1 platform continues to move at a constant speed in a straight line✓ 1

Common error: Use the resultant force, not the pull of the cable alone, in F = ma. Zero resultant force means no acceleration, not that the object stops.

179IGPC3-1-06r18-16Mass, weight and density Paper CMedium6 marks

A sample of green coffee beans has a mass of 240 g and a volume of 300 cm3. After roasting, the beans have lost 15% of their mass and their volume has increased to 340 cm3.

(a)

Calculate the density of the green beans.

(2)
(b)

Calculate the density of the roasted beans.

(2)
(c)

Explain, in terms of mass and volume, why the roasted beans are less dense than the green beans.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 density = mass ÷ volume = 240 / 300✓ 1
A1 0.80 g/cm3✓ 1
Part (b)
M1 roasted mass = 0.85 × 240 = 204 g, then 204 / 340✓ 1
A1 0.60 g/cm3✓ 1
Part (c)
B1 mass decreases (water and other material is lost)✓ 1
B1 volume increases (beans swell), so mass per unit volume is smaller✓ 1

Exam tip: When a question gives a percentage loss, find the new mass first, then use density = mass ÷ volume. Keep units consistent (g and cm³ together).

180IGPC3-1-07r20-04Mass, weight and density Paper CMedium1 mark

A volcanologist finds that a piece of lava rock has a mass of 150 g. When it is fully submerged in a measuring cylinder, the water level rises by 60 cm3. What is the density of the rock?

Show mark scheme
Marking pointMarkNotes
Density = mass ÷ volume = 150 ÷ 60 = 2.5 g/cm3, so B is correct. A divides volume by mass. C subtracts volume from mass. D adds them.✓ 1Answer B

Answer: B

Common error: Density is mass divided by volume; check the units g/cm³ are what you would get from g ÷ cm³.

181IGPC3-1-07r20-05Forces, friction and springs Paper CMedium1 mark

A spring has an unloaded length of 12 cm. A load of 4.0 N makes its length 20 cm. The spring is not stretched beyond its limit of proportionality. What is the length of the spring when the load is 6.0 N?

12 cm4.0 N20 cmno loadwith load
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
The extension for 4.0 N is 20 − 12 = 8.0 cm, so 2.0 cm per newton. At 6.0 N the extension is 12 cm and the length is 12 + 12 = 24 cm, so A is correct. B adds 6.0 to 20. C scales the total length (20 ÷ 4.0 × 6.0). D is the extension alone, not the length.✓ 1Answer A

Answer: A

Common error: Hooke's law links load and extension, not load and total length; always subtract the unloaded length first.

182IGPC3-1-07r20-07Turning effect and centre of gravity Paper CMedium1 mark

A light beam is pivoted at its middle and is in equilibrium. A 24 N load hangs 0.25 m to the left of the pivot. A downward force of 40 N is applied to the right of the pivot. How far from the pivot is the 40 N force applied?

24 N40 N0.25 md
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Clockwise moment = anticlockwise moment: 40 × d = 24 × 0.25 = 6.0 N m, so d = 0.15 m and A is correct. B assumes equal distances. C inverts the ratio (40 × 0.25 ÷ 24). D gives the moment in N m rather than a distance.✓ 1Answer A

Answer: A

Common error: Equal moments do not mean equal distances; the larger force acts at the smaller distance.

183IGPC3-1-07r20-12Speed, velocity and motion graphs Paper CMedium7 marks

A scenery platform on a theatre stage starts from rest and accelerates uniformly for 4.0 s until its speed is 0.80 m/s. It then moves at a constant speed for 10 s, and finally decelerates uniformly to rest in 2.0 s.

414160.8time / sspeed / m/s
Graph sketched from the values in the question; not to scale
(a)

Calculate the acceleration of the platform during the first 4.0 s.

(2)
(b)

Calculate the total distance travelled by the platform.

(3)
(c)

Calculate the average speed of the platform over the whole 16 s.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 acceleration = change in speed ÷ time = 0.80 ÷ 4.0✓ 1
A1 0.20 m/s2✓ 1
Part (b)
M1 distance = area under the speed–time graph (sum of three stages)✓ 1
A1 ½ × 4.0 × 0.80 = 1.6 m, 0.80 × 10 = 8.0 m and ½ × 2.0 × 0.80 = 0.80 m✓ 1
A1 total distance = 10.4 m✓ 1
Part (c)
M1 average speed = total distance ÷ total time = 10.4 ÷ 16 (ECF from (b))✓ 1
A1 0.65 m/s✓ 1

Common error: Average speed is total distance divided by total time; do not average the maximum speed with zero.

184IGPC3-1-07r20-14Forces, friction and springs Paper CMedium6 marks

A stage crew pushes a crate of mass 50 kg across a level floor. The pushing force is 240 N and the friction force on the crate is 90 N.

cratepush 240 Nfriction 90 N
Diagram not to scale
(a)

Calculate the resultant horizontal force on the crate.

(2)
(b)

Calculate the acceleration of the crate.

(2)
(c)

The crew stops pushing while the crate is moving. Describe what happens to the speed of the crate and explain why.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 resultant = pushing force − friction = 240 − 90✓ 1
A1 150 N✓ 1
Part (b)
M1 a = F ÷ m = 150 ÷ 50 (ECF from (a))✓ 1
A1 3.0 m/s2✓ 1
Part (c)
B1 the crate slows down (and stops)✓ 1
B1 friction is now the only horizontal force, so the resultant force acts against the motion✓ 1

Common error: Use the resultant force, not the pushing force alone, in F = ma.

185IGPC3-1-07r20-15Turning effect and centre of gravity Paper CMedium6 marks

A uniform lighting rail has length 4.0 m and weight 120 N. It rests on a pivot that is 0.80 m from its left end. A lamp of weight W hangs from the left end, and the rail is in equilibrium and horizontal.

W120 N0.80 m4.0 m
Diagram not to scale
(a)

State the position of the centre of gravity of the uniform rail.

(1)
(b)

State the principle of moments.

(2)
(c)

Calculate W.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 at its midpoint, 2.0 m from either end✓ 1
Part (b)
B1 for a body in equilibrium the sum of the clockwise moments about any point✓ 1
B1 equals the sum of the anticlockwise moments about the same point✓ 1
Part (c)
M1 the weight of the rail acts 1.2 m from the pivot, giving a moment of 120 × 1.2 = 144 N m✓ 1
M1 W × 0.80 = 144✓ 1
A1 W = 180 N✓ 1

Exam tip: Take moments about the pivot, so the unknown pivot force has no moment and drops out.

186IGPC3-1-08r20-02Pressure in solids and liquids Paper CMedium1 mark

An architect specifies a rectangular concrete block measuring 0.50 m × 0.40 m × 0.20 m. The block has a mass of 96 kg. What is the density of the concrete?

0.50 m0.20 m0.40 m
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Volume = 0.50 × 0.40 × 0.20 = 0.040 m3; density = 96 ÷ 0.040 = 2400 kg/m3, so C is correct. A multiplies mass by volume. B divides by the area of one face (0.20 m2). D divides volume by mass.✓ 1Answer C

Answer: C

Exam tip: Find the volume from all three dimensions before using density = mass ÷ volume.

187IGPC3-1-08r20-06Forces, friction and springs Paper CMedium1 mark

A spring balance used for weighing fishing nets obeys Hooke's law. A load of 15 N produces an extension of 6.0 cm. What load produces an extension of 10 cm?

Show mark scheme
Marking pointMarkNotes
Extension is proportional to load: 15 × 10 ÷ 6.0 = 25 N, so C is correct. A inverts the ratio (15 × 6.0 ÷ 10). B adds 6.0 to 10. D multiplies the load by the extension of the first test (15 × 6.0).✓ 1Answer C

Answer: C

Exam tip: Use the ratio load ÷ extension, which is constant within the limit of proportionality.

188IGPC3-1-08r20-10Mass, weight and density Paper CMedium6 marks

A marine biologist collects 800 cm3 of sea water. The density of the sea water is 1.03 g/cm3. g = 9.8 N/kg.

(a)

Calculate the mass of the sample in grams.

(2)
(b)

Calculate the weight of the sample in newtons. Use the mass 824 g exactly.

(2)
(c)

Explain the difference between mass and weight.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 mass = density × volume = 1.03 × 800✓ 1
A1 824 g✓ 1
Part (b)
M1 convert to 0.824 kg and use W = mg = 0.824 × 9.8 (ECF from (a))✓ 1
A1 8.08 N (8.0752 N)✓ 1
Part (c)
B1 mass is the amount of matter in an object (measured in kg) and is the same everywhere✓ 1
B1 weight is the gravitational force on the object (measured in N) and depends on the gravitational field strength✓ 1

Exam tip: Convert grams to kilograms before using W = mg to get the weight in newtons.

189IGPC3-1-08r20-11Pressure in solids and liquids Paper CMedium6 marks

A stone statue weighs 7200 N and stands on a square base of side 0.60 m on level ground.

(a)

Calculate the area of the base.

(1)
(b)

Calculate the pressure that the statue exerts on the ground.

(2)
(c)

The soil can safely withstand a pressure of only 12 000 Pa. Calculate the minimum area of base needed.

(2)
(d)

Explain why wider foundations reduce the risk of a building sinking into soft ground.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 0.60 × 0.60 = 0.36 m2✓ 1
Part (b)
M1 pressure = force ÷ area = 7200 ÷ 0.36 (ECF from (a))✓ 1
A1 20 000 Pa✓ 1
Part (c)
M1 area = force ÷ pressure = 7200 ÷ 12 000✓ 1
A1 0.60 m2✓ 1
Part (d)
B1 the same weight acts over a larger area, so the pressure on the ground is smaller✓ 1

Exam tip: Rearrange p = F ÷ A to A = F ÷ p when you are given a limit on the pressure.

190IGPC3-1-08r20-14Speed, velocity and motion graphs Paper CMedium6 marks

A delivery truck makes a journey of three stages. In the first 1.5 h it travels 90 km. It then stops for 0.5 h. Finally it travels a further 60 km in 1.0 h.

(a)

Calculate the speed of the truck in the first stage.

(2)
(b)

State what a distance–time graph of the journey would show during the stop.

(1)
(c)

Calculate the average speed for the whole journey.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 speed = distance ÷ time = 90 ÷ 1.5✓ 1
A1 60 km/h✓ 1
Part (b)
B1 a horizontal line (the gradient is zero)✓ 1
Part (c)
B1 total distance = 90 + 60 = 150 km✓ 1
M1 total time = 1.5 + 0.5 + 1.0 = 3.0 h (including the stop)✓ 1
A1 average speed = 150 ÷ 3.0 = 50 km/h✓ 1

Common error: Include the time spent stopped when finding the average speed for the whole journey.

191IGPC3-1-08r20-15Forces, friction and springs Paper CMedium6 marks

An architect tests a model of a hanging support using a spring of unloaded length 0.200 m. The table shows the length of the spring for each load.
Load / N: 0, 2.0, 4.0, 6.0, 8.0
Length / m: 0.200, 0.230, 0.260, 0.290, 0.320

(a)

Calculate the extension of the spring for a load of 6.0 N.

(1)
(b)

Use the data to show that the extension is proportional to the load.

(2)
(c)

Calculate the length of the spring for a load of 5.0 N.

(2)
(d)

State what is meant by the limit of proportionality.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 0.290 − 0.200 = 0.090 m✓ 1
Part (b)
B1 extensions are 0.030, 0.060, 0.090 and 0.120 m for 2.0, 4.0, 6.0 and 8.0 N✓ 1
B1 extension ÷ load is constant (0.015 m/N), or doubling the load doubles the extension✓ 1
Part (c)
M1 extension = 5.0 × 0.015 = 0.075 m✓ 1
A1 length = 0.200 + 0.075 = 0.275 m✓ 1
Part (d)
B1 the load beyond which the extension is no longer proportional to the load✓ 1

Exam tip: Always calculate extension (current length minus unloaded length) before comparing it with the load.

192IGPC3-1-08r22-03Speed, velocity and motion graphs Paper CMedium1 mark

A cyclist's speed increases uniformly from 0 to 8.0 m/s in 10 s. She then rides at a constant 8.0 m/s for 20 s, and finally slows uniformly to rest in 5.0 s. What is the total distance travelled?

1030358time / sspeed / m/s
Graph as described; not to scale
Show mark scheme
Marking pointMarkNotes
Distance is the area under the speed–time graph: ½ × 8.0 × 10 = 40 m, plus 8.0 × 20 = 160 m, plus ½ × 8.0 × 5.0 = 20 m, giving 220 m. 160 m counts only the constant-speed stage. 280 m uses 8.0 m/s for the whole 35 s. 140 m is ½ × 8.0 × 35, treating the whole journey as one triangle.✓ 1Answer D

Answer: D

Common error: Do not use the maximum speed for the whole time; split the journey into stages and find each area.

193IGPC3-1-08r22-04Pressure in solids and liquids Paper CMedium1 mark

A server cabinet in a data centre has a weight of 4500 N. It rests on four identical feet, each with a contact area of 25 cm2. What is the pressure the cabinet exerts on the floor?

Show mark scheme
Marking pointMarkNotes
Total area = 4 × 25 cm2 = 100 cm2 = 0.0100 m2. Pressure = 4500 ÷ 0.0100 = 450000 Pa = 4.5 × 105 Pa. 1.8 × 106 Pa uses the area of only one foot. 45 Pa uses 100 cm2 without converting to m2. 11 Pa multiplies force by area instead of dividing.✓ 1Answer A

Answer: A

Exam tip: Convert areas from cm² to m² (divide by 10 000) before calculating pressure in pascals.

194IGPC3-1-08r22-05Forces, friction and springs Paper CMedium1 mark

A spring in a bicycle saddle obeys Hooke's law. A load of 30 N compresses it by 4.5 cm. What is the compression when the load is 80 N?

Show mark scheme
Marking pointMarkNotes
For a spring within its limit of proportionality, compression is proportional to load. Compression = 4.5 × (80 ÷ 30) = 12 cm. 1.7 cm comes from inverting the ratio. 54.5 cm adds the 50 N difference in load to 4.5 cm. 530 cm divides 80 by 0.15 cm per newton instead of multiplying.✓ 1Answer B

Answer: B

Common error: Extension is proportional to load, so scale by the ratio of loads rather than adding or subtracting values.

195IGPC3-1-08r22-11Forces, friction and springs Paper CMedium5 marks

A student tests a spring from a bicycle brake lever. The table shows the load and the extension of the spring.
Load / N: 0, 2.0, 4.0, 6.0, 8.0
Extension / cm: 0, 2.0, 4.0, 6.0, 9.0

(a)

State the extension of the spring when the load is 6.0 N.

(1)
(b)

The spring goes beyond its limit of proportionality at some load. State whether this happens at 8.0 N and explain how the data show this.

(2)
(c)

Calculate the spring constant for the proportional part of the data, in N/m.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
6.0 cm✓ 1
Part (b)
yes, beyond the limit at 8.0 N✓ 1
extension is no longer proportional to load (the first four loads give 1.0 cm per newton, but 8.0 N gives 9.0 cm)✓ 1
Part (c)
k = force ÷ extension, e.g. 6.0 ÷ 0.060✓ 1
100 N/m✓ 1

Exam tip: Use an extension in metres when the spring constant is required in N/m.

196IGPC3-1-08r22-12Pressure in solids and liquids Paper CMedium5 marks

A server cabinet in a data centre has a mass of 640 kg. It rests on four identical feet, each of area 8.0 cm2. Use g = 9.8 N/kg and give the final pressure to 3 significant figures.

(a)

Calculate the weight of the cabinet.

(1)
(b)

Show that the total area in contact with the floor is 3.2 × 10−3 m2.

(1)
(c)

Calculate the pressure on the floor. Use the unrounded weight from (a).

(2)
(d)

The technician places a flat steel plate under the feet. Explain why this reduces the pressure on the floor.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
weight = mass × g = 640 × 9.8 = 6272 N✓ 1
Part (b)
4 × 8.0 = 32 cm2 = 32 × 10−4 m2 = 3.2 × 10−3 m2✓ 1
Part (c)
pressure = force ÷ area = 6272 ÷ (3.2 × 10−3)✓ 1
1.96 × 106 Pa (= 1960000 Pa)✓ 1
Part (d)
the same force acts over a larger area, so pressure (= force ÷ area) is smaller✓ 1

Common error: Convert the area to m² before dividing; pascals need newtons per square metre.

197IGPC3-1-07r25-02Mass, weight and density Paper CMedium1 mark

A beekeeper pours honey into an empty jar of mass 0.150 kg. The jar then has a total mass of 0.640 kg. The honey fills a volume of 350 cm3. What is the density of the honey?

Show mark scheme
Marking pointMarkNotes
Mass of honey = 0.640 − 0.150 = 0.490 kg. Volume = 350 × 10−6 m3. Density = 0.490 ÷ (350 × 10−6) = 1400 kg/m3. The value 1830 kg/m3 wrongly uses the mass of the jar plus honey. The value 1.4 kg/m3 is the number in g/cm3 given the wrong unit. The value 1.4 × 106 kg/m3 comes from converting cm3 to m3 using 10−9 instead of 10−6.✓ 1Answer D

Answer: D

Common error: 1 m³ is 10⁶ cm³, not 10⁹ cm³. Subtract the mass of the container before finding density.

198IGPC3-1-07r25-03Pressure in solids and liquids Paper CMedium1 mark

A hospital trolley has a weight of 600 N. It rests on four wheels. Each wheel is in contact with the floor over an area of 5.0 cm2, and the weight is shared equally. What is the pressure exerted by the trolley on the floor?

Show mark scheme
Marking pointMarkNotes
Total contact area = 4 × 5.0 cm2 = 20 cm2 = 2.0 × 10−3 m2. Pressure = force ÷ area = 600 ÷ (2.0 × 10−3) = 3.0 × 105 Pa. The value 1.2 × 106 Pa uses the area of one wheel only. The value 30 Pa divides by 20 without converting cm2 to m2. The value 1.2 Pa multiplies force by area instead of dividing.✓ 1Answer A

Answer: A

Common error: Pressure in pascals needs area in m². Count every point of contact when finding the total area.

199IGPC3-1-07r25-11Turning effect and centre of gravity Paper CMedium7 marks

A school in a hospital wants to find the centre of gravity of an irregular piece of thin card.

loadweight 1.2 Npivot5.0 cm30 cm100 cm
Metre rule. Diagram not to scale
(a)

Describe how the centre of gravity of the card can be found. You may use a pin, a plumb line (a string with a weight on it) and a pencil.

(4)
(b)

A uniform metre rule of weight 1.2 N is balanced horizontally on a pivot at the 30 cm mark. A load is hung from the 5.0 cm mark. Calculate the weight of the load.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Make a small hole near the edge and hang the card freely from a pin✓ 1
Hang the plumb line from the same pin and, when it is still, mark its line on the card✓ 1
Repeat for at least two other holes✓ 1
The centre of gravity is where the lines cross✓ 1
Part (b)
The weight of the rule acts at the 50 cm mark, 20 cm from the pivot✓ 1
Clockwise moment = anticlockwise moment: 1.2 × 20 = W × 25✓ 1
W = 0.96 N✓ 1

Exam tip: Take moments about the pivot, and measure each distance from the pivot to the line of action of its force.

200IGPC3-1-07r25-12Mass, weight and density Paper CMedium7 marks

A hospital pharmacy stores a liquid medicine. A volume of 250 cm3 of the medicine has a mass of 0.280 kg. Use g = 9.8 N/kg.

(a)

Calculate the density of the medicine in g/cm3.

(2)
(b)

Describe how the pharmacist could find the density of a small irregular metal valve that sinks in water.

(3)
(c)

Calculate the weight of 0.280 kg of the medicine. Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
density = mass ÷ volume = 280 ÷ 250✓ 1
1.12 g/cm3✓ 1
Part (b)
Measure the mass of the valve using a balance✓ 1
Find its volume by the rise in water level in a measuring cylinder when the valve is fully submerged✓ 1
Density = mass ÷ volume✓ 1
Part (c)
W = mg = 0.280 × 9.8✓ 1
2.74 N✓ 1

Common error: Convert the mass to kg before using W = mg, and do not mix g/cm³ with kg/m³.

201IGPC3-1-10r27-03Speed, velocity and motion graphs Paper CMedium1 mark

A small rover on a distant moon speeds up uniformly from rest to 12 m/s in 8.0 s. It then moves at a constant 12 m/s for a further 10 s. What is the total distance travelled in the 18 s?

81812time / sspeed / m/s
Graph as described; not to scale
Show mark scheme
Marking pointMarkNotes
Distance during speeding up = average speed × time = 6 × 8 = 48 m (area of the triangle under a speed-time graph); distance at constant speed = 12 × 10 = 120 m; total = 168 m. 120 m ignores the speeding-up stage. 216 m uses 12 m/s for all 18 s. 240 m uses 12 m/s for 20 s by mistake.✓ 1Answer D

Answer: D

Exam tip: Split the motion into stages and find the area under the speed-time graph for each stage.

202IGPC3-1-10r27-11Speed, velocity and motion graphs Paper CMedium7 marks

A space probe approaches the surface of a moon. During a 10 s descent burn its speed falls steadily from 20 m/s to 0. The probe has mass 300 kg.

1020time / sspeed / m/s
Graph as described; not to scale
(a)

Calculate the magnitude of the probe's acceleration.

(2)
(b)

Calculate the distance the probe travels during the 10 s.

(2)
(c)

State the direction of the resultant force on the probe relative to its direction of motion.

(1)
(d)

Calculate the magnitude of the resultant force on the probe.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
acceleration = change in speed / time = 20 / 10✓ 1
2.0 m/s2✓ 1
Part (b)
average speed 10 m/s (or area under speed-time graph = ½ × 20 × 10)✓ 1
distance = 100 m✓ 1
Part (c)
opposite to the direction of motion✓ 1
Part (d)
F = ma = 300 × 2.0 (ECF from (a))✓ 1
600 N✓ 1

Exam tip: Use the magnitude of the acceleration in F = ma and give the direction separately.

203IGPC3-1-10r27-12Forces, friction and springs Paper CMedium5 marks

Students hang loads from a spring in a model of a suspension bridge. The spring's length is 20.0, 24.0, 28.0, 32.0, 36.0 and 44.0 cm for loads of 0, 2.0, 4.0, 6.0, 8.0 and 10.0 N respectively.

(a)

State the largest load in the readings for which the spring obeys Hooke's law.

(1)
(b)

Calculate the spring constant, in N/m, using the readings up to the limit of proportionality.

(2)
(c)

Calculate the length of the spring when the load is 5.0 N.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
8.0 N✓ 1
Part (b)
extension at 8.0 N = 36.0 − 20.0 = 16.0 cm = 0.160 m, k = F / x = 8.0 / 0.160✓ 1
50 N/m✓ 1
Part (c)
extension = 5.0 / 50 = 0.10 m (10 cm) (ECF from (b))✓ 1
length = 30 cm (0.30 m)✓ 1

Common error: Hooke's law uses the extension, not the total length of the spring.

204IGPC3-1-10r28-03Pressure in solids and liquids Paper CMedium1 mark

A server cabinet in a data centre weighs 4200 N. It stands on four identical feet, each in contact with the floor over an area of 25 cm2. What is the pressure that the cabinet exerts on the floor?

Show mark scheme
Marking pointMarkNotes
The total contact area is 4 × 25 = 100 cm2 = 0.010 m2, so p = 4200 / 0.010 = 4.2 × 105 Pa. 42 Pa results from using the area in cm2 instead of m2. 1.05 × 103 is the force on one foot (4200 / 4 = 1050 N) with the area left out. 1.7 × 106 Pa uses the area of only one foot, 4200 / 0.0025.✓ 1Answer D

Answer: D

Exam tip: Convert cm² to m² by dividing by 10 000, and use the total contact area.

205IGPC3-1-10r28-05Forces, friction and springs Paper CMedium1 mark

A spring obeys Hooke's law. A force of 18 N produces an extension of 7.5 cm. What extension does a force of 30 N produce, assuming the limit of proportionality is not reached?

Show mark scheme
Marking pointMarkNotes
Extension is proportional to force, so the extension is 7.5 × (30 / 18) = 12.5 cm. 19.5 cm adds the extra 12 N as if it were 12 cm to the original extension. 4.5 cm inverts the ratio, 7.5 × (18 / 30). 25 cm doubles the answer for no reason; the force is not doubled.✓ 1Answer C

Answer: C

Exam tip: With Hooke's law, set up the ratio of forces: the extension scales by the same factor.

206IGPC3-1-10r28-07Energy, work and power Paper CMedium1 mark

A motor lifts a server rack of mass 25 kg through a height of 3.0 m in 5.0 s at constant speed. What is the useful power output of the motor? (g = 9.8 N/kg)

Show mark scheme
Marking pointMarkNotes
Work done = mgh = 25 × 9.8 × 3.0 = 735 J, and power = 735 / 5.0 = 147 W. 15 W omits g, using 25 × 3.0 / 5.0. 735 W is the work done in joules, not divided by the time. 3700 W multiplies the work by the time instead of dividing.✓ 1Answer B

Answer: B

Exam tip: Power is work divided by time: find the work in joules first.

207IGPC3-1-10r28-09Mass, weight and density Paper CMedium7 marks

An archaeologist records a rectangular stone block of dimensions 0.40 m × 0.25 m × 0.15 m. The mass of the block is 38 kg. (g = 9.8 N/kg)

(a)

Calculate the volume of the block.

(2)
(b)

Calculate the density of the stone. Give your answer to 3 significant figures.

(2)
(c)

The block rests on the ground on its largest face. Calculate the pressure it exerts on the ground. Give your answer to 3 significant figures and use unrounded values.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 volume = 0.40 × 0.25 × 0.15✓ 1
A1 0.015 m3✓ 1
Part (b)
M1 density = mass ÷ volume = 38 ÷ 0.015✓ 1
A1 2530 kg/m3✓ 1
Part (c)
M1 weight = 38 × 9.8 = 372.4 N and area = 0.40 × 0.25 = 0.10 m2✓ 1
M1 pressure = force ÷ area = 372.4 ÷ 0.10✓ 1
A1 3720 Pa✓ 1

Exam tip: The largest face gives the smallest pressure: choose the area that matches the face in contact with the ground.

208IGPC3-1-10r28-13Forces, friction and springs Paper CMedium7 marks

A cyclist and bicycle have a total mass of 82 kg. The cyclist pedals so that the driving force on the bicycle is 150 N. The resistive forces on the bicycle are 90 N.

82 kgdriving 150 Nresistive 90 N
Diagram not to scale
(a)

Calculate the resultant force on the bicycle.

(1)
(b)

Calculate the acceleration of the cyclist. Give your answer to 2 significant figures.

(2)
(c)

After a long time the cyclist reaches a constant speed with the driving force still 150 N. State the size of the resistive force and explain your answer.

(2)
(d)

Using an acceleration of 0.73 m/s2, calculate the speed of the cyclist 6.0 s after starting from rest.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 150 − 90 = 60 N✓ 1
Part (b)
M1 a = F ÷ m = 60 ÷ 82✓ 1
A1 0.73 m/s2✓ 1
Part (c)
150 N✓ 1
the forces are balanced, so the resultant force is zero and there is no acceleration✓ 1
Part (d)
M1 v = a × t = 0.73 × 6.0✓ 1
A1 4.38 m/s✓ 1

Common error: Constant speed does not mean no forces act: it means the forces are balanced.

209IGPC3-1-10r28-15Energy, work and power Paper CMedium6 marks

A pump in a data centre raises 600 kg of cooling water every minute through a height of 18 m to a roof tank. The pump is 70% efficient. (g = 9.8 N/kg)

(a)

Calculate the gravitational potential energy gained by 600 kg of water raised through 18 m. Give your answer to 3 significant figures.

(2)
(b)

Calculate the useful power output of the pump. Use the unrounded energy and give your answer to 3 significant figures.

(2)
(c)

Calculate the electrical power input to the pump. Use the unrounded useful power.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 E = mgh = 600 × 9.8 × 18✓ 1
A1 1.06 × 105 J✓ 1
Part (b)
M1 P = E ÷ t = 105 840 ÷ 60✓ 1
A1 1760 W✓ 1
Part (c)
M1 input power = useful power ÷ efficiency = 1764 ÷ 0.70✓ 1
A1 2520 W✓ 1

Common error: Efficiency is useful output divided by total input, so the input power is larger than the output power.

210IGPC3-1-07r30-04Mass, weight and density Paper CMedium1 mark

A jar holds 250 cm3 of honey. The density of the honey is 1.4 g/cm3. What is the weight of the honey? (g = 9.8 N/kg)

Show mark scheme
Marking pointMarkNotes
Mass = density × volume = 1.4 × 250 = 350 g = 0.35 kg, so weight = 0.35 × 9.8 = 3.43 N. 1.75 N divides the volume by the density (250 / 1.4 = 179 g) instead of multiplying. 0.35 N treats the mass in kg as if it were the weight. 3430 N multiplies the mass in grams by 9.8 without converting to kg.✓ 1Answer D

Answer: D

Common error: Convert grams to kilograms before using W = mg.

211IGPC3-1-07r30-06Forces, friction and springs Paper CMedium1 mark

A beekeeper’s spring balance contains a spring that obeys Hooke’s law within its limit. A load of 2.0 N extends the spring by 4.0 cm. What is the extension for a load of 5.0 N, assuming the limit is not exceeded?

Show mark scheme
Marking pointMarkNotes
Extension is directly proportional to load, so the extension is 4.0 × (5.0 / 2.0) = 10 cm. 7.0 cm adds the extra 3.0 N to the extension as if the numbers were in the same units. 1.6 cm inverts the ratio (4.0 × 2.0 / 5.0). 20 cm multiplies 4.0 by 5.0.✓ 1Answer B

Answer: B

Exam tip: If extension is proportional to load, scale the extension by the same ratio as the load.

212IGPC3-1-07r30-07Energy, work and power Paper CMedium1 mark

A pit crew raises a Formula 1 car of mass 750 kg through a height of 0.20 m using a jack. What is the gain in gravitational potential energy of the car? (g = 9.8 N/kg)

Show mark scheme
Marking pointMarkNotes
Gain in gravitational potential energy = mgh = 750 × 9.8 × 0.20 = 1470 J. 150 J leaves out g (750 × 0.20). 147 000 J uses the height as 20 (centimetres) instead of 0.20 m. 7350 J is the weight in newtons, which is a force, not an energy.✓ 1Answer C

Answer: C

Common error: Write heights in metres before using mgh; the unit of energy is the joule, not the newton.

213IGPC3-1-07r30-09Speed, velocity and motion graphs Paper CMedium7 marks

A Formula 1 car and driver, total mass 750 kg, accelerate in a straight line from rest. The speed is recorded every second.
Time / s: 0, 1, 2, 3, 4
Speed / m/s: 0, 12, 24, 36, 48
After 4.0 s the car continues at a constant speed of 48 m/s for a further 5.0 s.

(a)

State how the data show that the acceleration is constant.

(1)
(b)

Calculate the acceleration of the car.

(2)
(c)

Calculate the resultant force on the car while it is accelerating.

(2)
(d)

Calculate the distance travelled by the car during the 5.0 s at constant speed.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
the speed increases by the same amount (12 m/s) in every second✓ 1
Part (b)
a = Δv / t✓ 1
= 48 / 4.0 = 12 m/s2✓ 1
Part (c)
F = ma = 750 × 12✓ 1
= 9000 N (ECF from (b))✓ 1
Part (d)
distance = speed × time✓ 1
= 48 × 5.0 = 240 m✓ 1

Exam tip: Acceleration is the change in speed per second, so its unit is m/s²; show the substitution before the answer.

214IGPC3-1-07r30-12Forces, friction and springs Paper CMedium6 marks

A beekeeper hangs different loads from a spring. The natural length of the spring is 12.0 cm.
Load / N: 0, 2.0, 4.0, 6.0, 8.0, 10.0, 12.0
Length / cm: 12.0, 14.4, 16.8, 19.2, 21.6, 24.0, 28.0

(a)

Calculate the extension of the spring when the load is 6.0 N.

(1)
(b)

State the relationship between extension and load for loads up to 10.0 N.

(1)
(c)

Predict the length of the spring when the load is 7.0 N. Assume the spring is still obeying the same relationship.

(2)
(d)

For a load of 12.0 N the spring is longer than the same relationship would predict. Explain why.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
19.2 − 12.0 = 7.2 cm✓ 1
Part (b)
extension is directly proportional to load✓ 1
Part (c)
extension per newton = 7.2 / 6.0 = 1.2 cm/N (or equivalent)✓ 1
length = 12.0 + 7.0 × 1.2 = 20.4 cm✓ 1
Part (d)
the limit of proportionality has been exceeded✓ 1
so the extension is no longer proportional to the load / the spring stretches more per newton✓ 1

Exam tip: Extension = stretched length − natural length; use extension, not total length, when checking for proportionality.

215IGPC3-1-07r30-15Energy, work and power Paper CMedium7 marks

In a pit garage, a motor-driven jack raises a Formula 1 car of mass 720 kg through a height of 0.25 m in 2.0 s. The motor takes in electrical energy at a rate of 1400 W. Use g = 9.8 N/kg.

(a)

Calculate the weight of the car.

(1)
(b)

Calculate the work done by the jack on the car.

(2)
(c)

Calculate the useful power output of the jack.

(2)
(d)

Calculate the efficiency of the jack and motor.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
720 × 9.8 = 7056 N✓ 1
Part (b)
W = Fd✓ 1
= 7056 × 0.25 = 1764 J✓ 1
Part (c)
P = W / t✓ 1
= 1764 / 2.0 = 882 W✓ 1
Part (d)
efficiency = useful power output / power input × 100%✓ 1
= 882 / 1400 × 100 = 63% (ECF from (c))✓ 1

Common error: Efficiency compares useful output with total input in the same units, and cannot be more than 100%.

216IGPC3-1-08r31-02Speed, velocity and motion graphs Paper CMedium1 mark

A space probe lander slows down at a uniform rate from 12 m/s to rest in 8.0 s. What distance does it travel while slowing down?

Show mark scheme
Marking pointMarkNotes
The distance is the average speed multiplied by time: ½ × 12 × 8.0 = 48 m, which is the area under the speed–time graph. 1.5 m is the deceleration (12 ÷ 8.0), not a distance. 20 m comes from adding the speed and the time, which have different units. 96 m uses the initial speed for the whole time, which ignores the slowing down.✓ 1Answer D

Answer: D

Common error: With uniform acceleration, use the average of the initial and final speeds, not the initial speed alone.

217IGPC3-1-08r31-03Mass, weight and density Paper CMedium1 mark

A sample of green coffee beans has mass 200 g and volume 250 cm3. After roasting, the beans have mass 170 g and the sample occupies 340 cm3. What is the density of the roasted sample?

Show mark scheme
Marking pointMarkNotes
Density = mass ÷ volume using the roasted values: 170 ÷ 340 = 0.50 g/cm3. 0.80 g/cm3 is the density before roasting. 0.68 g/cm3 combines the roasted mass with the unroasted volume. 2.0 g/cm3 is volume ÷ mass, which is upside down.✓ 1Answer A

Answer: A

Common error: Use the mass and volume from the same state of the material when calculating density.

218IGPC3-1-08r31-07Turning effect and centre of gravity Paper CMedium1 mark

A uniform plank 6.0 m long is balanced on a pivot at its centre in a bridge-model test. A load of 240 N rests 1.5 m from the pivot on one side. A load of 360 N is placed on the other side so that the plank balances. How far from the pivot is the 360 N load?

240 N360 Npivot1.5 m? m
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
For balance, clockwise moment = anticlockwise moment: 360 × d = 240 × 1.5, so d = (240 × 1.5) ÷ 360 = 1.0 m. The larger load must sit closer to the pivot. 2.25 m comes from swapping the loads in the equation. 1.5 m assumes equal distances balance unequal loads. 0.67 m is the ratio of the loads, not a moment calculation.✓ 1Answer A

Answer: A

Exam tip: Write the moment on each side as force × distance before rearranging.

219IGPC3-1-08r31-14Energy, work and power Paper CMedium7 marks

In a coffee roastery, a worker lifts a 20 kg sack of green beans through a height of 3.0 m at a steady speed in 4.0 s. (g = 9.8 N/kg.)

(a)

Calculate the weight of the sack.

(2)
(b)

Calculate the work done in lifting the sack.

(2)
(c)

State the gain in gravitational potential energy of the sack.

(1)
(d)

Calculate the useful power developed by the worker.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 W = mg = 20 × 9.8✓ 1
A1 196 N✓ 1
Part (b)
M1 work = force × distance = 196 × 3.0✓ 1
A1 588 J (ECF from (a))✓ 1
Part (c)
B1 588 J (equal to the work done; ECF from (b))✓ 1
Part (d)
M1 power = work / time = 588 / 4.0✓ 1
A1 147 W (ECF from (b))✓ 1

Exam tip: At constant speed the lifting force equals the weight, so the work done equals the gain in gravitational energy.

220IGPC3-1-09r32-03Turning effect and centre of gravity Paper CMedium1 mark

A uniform horizontal plank is pivoted at its centre. A load of 300 N sits 2.0 m to the left of the pivot. A load of 400 N is placed on the right so that the plank balances. How far from the pivot is the 400 N load?

300 N400 Npivot2.0 m? m
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
At balance, clockwise moment = anticlockwise moment: 400 × d = 300 × 2.0, so d = 1.5 m. 2.0 m assumes equal distances balance unequal loads. 2.7 m comes from inverting the ratio (400 × 2.0 ÷ 300). 0.75 m results from dividing the moment by 400 and then needlessly halving it.✓ 1Answer D

Answer: D

Exam tip: Write moment = force × perpendicular distance for each side, then equate them; a heavier load must sit closer to the pivot.

221IGPC3-1-09r32-07Speed, velocity and motion graphs Paper CMedium1 mark

A freight train crossing a bridge brakes uniformly from 20 m/s to rest in 4.0 s. What distance does it travel while braking?

Show mark scheme
Marking pointMarkNotes
With uniform deceleration the average speed is (20 + 0) ÷ 2 = 10 m/s, so distance = 10 × 4.0 = 40 m. 80 m uses the initial speed for the whole time. 160 m uses ½at2 with a = 20 m/s2 (the speed mistaken for the acceleration). 5.0 is the size of the deceleration in m/s2, not a distance.✓ 1Answer C

Answer: C

Common error: Distance is not initial speed × time unless the speed is constant; use the average speed or the area under the speed–time graph.

222IGPC3-1-09r32-11Mass, weight and density Paper CMedium5 marks

A technician needs the density of a steel bolt from a bridge. The bolt has mass 78.0 g. It is lowered into a measuring cylinder containing 50.0 cm3 of water and the water level rises to 60.0 cm3.

(a)

Determine the volume of the bolt.

(2)
(b)

Calculate the density of the steel in g/cm3.

(2)
(c)

Convert your answer to kg/m3.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 volume = rise in water level = 60.0 − 50.0✓ 1
A1 10.0 cm3✓ 1
Part (b)
M1 density = mass ÷ volume = 78.0 ÷ 10.0✓ 1
A1 7.80 g/cm3 (ECF from (a))✓ 1
Part (c)
A1 7800 kg/m3 (since 1 g/cm3 = 1000 kg/m3)✓ 1

Exam tip: The volume of an irregular solid is the difference between the two readings, not the final reading.

223IGPC3-1-09r32-13Speed, velocity and motion graphs Paper CMedium7 marks

A maintenance train on a bridge starts from rest and accelerates uniformly to 8.0 m/s in 10 s. It then travels at constant speed for 40 s before decelerating uniformly to rest in 20 s.

1050708time / sspeed / m/s
Graph as described; not to scale
(a)

Calculate the acceleration during the first 10 s.

(2)
(b)

Calculate the total distance travelled.

(3)
(c)

Calculate the average speed for the whole journey, to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 a = change in speed ÷ time = 8.0 ÷ 10✓ 1
A1 0.80 m/s2✓ 1
Part (b)
M1 distance while accelerating = ½ × 8.0 × 10 = 40 m✓ 1
M1 distance at constant speed = 8.0 × 40 = 320 m and while decelerating = ½ × 8.0 × 20 = 80 m✓ 1
A1 total 440 m✓ 1
Part (c)
M1 average speed = total distance ÷ total time = 440 ÷ 70✓ 1
A1 6.29 m/s (ECF from (b))✓ 1

Exam tip: Distance is the area under a speed–time graph; split trapeziums into triangles and rectangles and add them.

224IGPC3-1-08r38-03Mass, weight and density Paper CMedium1 mark

An irregular stone from a dig has a mass of 270 g. It is lowered into a measuring cylinder and the water level rises from 50 cm3 to 140 cm3. What is the density of the stone?

Show mark scheme
Marking pointMarkNotes
The volume of the stone is the rise in water level, 140 − 50 = 90 cm3, so density = 270 ÷ 90 = 3.0 g/cm3. Using the final reading of 140 gives 1.9, inverting the formula gives 0.33, and using the initial reading of 50 gives 5.4.✓ 1Answer D

Answer: D

Common error: The volume of an irregular object is the change in the water level, not the final reading.

225IGPC3-1-08r38-04Forces, friction and springs Paper CMedium1 mark

A spring in a server-rack latch extends by 4.0 cm when the load is 6.0 N. The spring obeys Hooke's law. What is its extension when the load is 15 N?

Show mark scheme
Marking pointMarkNotes
Extension is directly proportional to load, so the extension is 4.0 × (15 ÷ 6.0) = 10 cm. Using the inverse ratio gives 1.6, adding the extra load as extra centimetres gives 13, and multiplying the numbers gives 60.✓ 1Answer B

Answer: B

Exam tip: If load increases by a factor of 2.5, the extension increases by the same factor, provided the limit of proportionality is not passed.

226IGPC3-1-08r38-07Energy, work and power Paper CMedium1 mark

A cyclist and bicycle have a total mass of 75 kg. They climb a hill with a vertical height of 20 m in 50 s. Use g = 9.8 N/kg. What is the average useful power output?

Show mark scheme
Marking pointMarkNotes
Work done against gravity = mgh = 75 × 9.8 × 20 = 14 700 J, and power = work ÷ time = 14 700 ÷ 50 = 294 W, Omitting g gives 30 W, using just the weight 75 × 9.8 = 735 ignores both the height and the time, and forgetting to divide by the time gives 14 700.✓ 1Answer B

Answer: B

Common error: Power is work done divided by time; a value in joules has not yet been divided by time.

227IGPC3-1-08r38-10Mass, weight and density Paper CMedium5 marks

An archaeologist wants to identify the metal of a small statuette of mass 0.45 kg. Use g = 9.8 N/kg. The statuette is lowered into a measuring cylinder; the water level rises from 200 cm3 to 250 cm3.

(a)

Calculate the weight of the statuette. Give your answer to 2 significant figures.

(2)
(b)

Calculate the density of the metal in g/cm3. The mass of the statuette is 450 g.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 weight = mass × g = 0.45 × 9.8✓ 1
A1 4.4 N✓ 1
Part (b)
B1 volume = 250 − 200 = 50 cm3✓ 1
M1 density = mass ÷ volume✓ 1
A1 9.0 g/cm3✓ 1

Common error: Mass is measured in kg or g, weight in N; do not use a mass value as if it were a weight.

228IGPC3-1-08r38-11Forces, friction and springs Paper CMedium6 marks

A student tests a spring from a cable-tension gauge in a data centre. The unstretched length of the spring is 10.0 cm.

Load / N: 0, 2.0, 4.0, 6.0, 8.0
Length / cm: 10.0, 12.5, 15.0, 17.5, 20.0

(a)

Show that the spring obeys Hooke's law for these loads.

(2)
(b)

Calculate the spring constant in N/m.

(2)
(c)

Calculate the load that produces an extension of 15 cm. Assume the spring is still within its limit of proportionality.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 extensions calculated: 0, 2.5, 5.0, 7.5, 10.0 cm✓ 1
A1 extension is proportional to load (the ratio load ÷ extension is constant at 0.80 N/cm)✓ 1
Part (b)
M1 k = F ÷ x using F = 8.0 N and x = 0.100 m✓ 1
A1 80 N/m✓ 1
Part (c)
M1 F = kx = 80 × 0.15✓ 1
A1 12 N✓ 1

Common error: Convert centimetres to metres before using a spring constant in N/m.

229IGPC3-1-08r38-16Turning effect and centre of gravity Paper CMedium5 marks

An archaeologist balances a uniform wooden beam of length 3.0 m on a pivot at its midpoint. A load of 150 N hangs 1.0 m to the left of the pivot. A second load of 100 N is hung on the right to keep the beam balanced.

150 N100 Npivot1.0 m? m
Diagram not to scale
(a)

State the position of the centre of gravity of the uniform beam.

(1)
(b)

State the principle of moments.

(1)
(c)

Calculate the distance of the 100 N load from the pivot.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 at its midpoint (centre)✓ 1
Part (b)
B1 for a body in equilibrium the sum of the clockwise moments about a point equals the sum of the anticlockwise moments about the same point✓ 1
Part (c)
M1 anticlockwise moment = 150 × 1.0 = 150 N m✓ 1
M1 100 × d = 150✓ 1
A1 1.5 m✓ 1

Exam tip: Take moments about the pivot, so the weight of a beam pivoted at its centre has no moment.

230IGPC3-1-08r41-02Speed, velocity and motion graphs Paper CMedium1 mark

A character in a computer game accelerates uniformly from rest to 6.0 m/s in 3.0 s, moves at a constant 6.0 m/s for 4.0 s, then decelerates uniformly to rest in 2.0 s. What is the total distance travelled by the character?

3796time / sspeed / m/s
Graph as described; not to scale
Show mark scheme
Marking pointMarkNotes
Distance is the area under the speed–time graph: ½ × 6.0 × 3.0 = 9 m while accelerating, 6.0 × 4.0 = 24 m at constant speed and ½ × 6.0 × 2.0 = 6 m while decelerating, giving 39 m. 33 m leaves out the deceleration stage. 54 m multiplies the top speed by the whole time of 9 s. 48 m forgets the factor ½ for the accelerating and decelerating stages.✓ 1Answer C

Answer: C

Common error: Speeds that change uniformly give triangles on the graph, whose area needs the factor ½; only the constant-speed stage is a plain rectangle.

231IGPC3-1-08r41-10Speed, velocity and motion graphs Paper CMedium5 marks

A water-delivery truck is travelling at 15 m/s in a straight line. The driver brakes and the truck slows down uniformly, coming to rest in 6.0 s.

(a)

Calculate the deceleration of the truck.

(2)
(b)

Calculate the distance the truck travels while braking.

(2)
(c)

When the truck is full of water the braking distance is greater, even though the braking force is the same. Suggest why.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
deceleration = change in speed ÷ time = 15 ÷ 6.0✓ 1
2.5 m/s²✓ 1
Part (b)
distance = average speed × time = 7.5 × 6.0, or area under speed–time graph ½ × 15 × 6.0✓ 1
45 m✓ 1
Part (c)
larger mass so the same braking force gives a smaller deceleration, so the truck takes longer and travels further to stop✓ 1

Exam tip: When a question gives start speed, end speed and time, use acceleration = change in speed ÷ time and keep the sign of the change in mind.

232IGPC3-1-08r41-15Turning effect and centre of gravity Paper CMedium6 marks

A uniform metre rule is used as a balance beam in a pharmacy. It is pivoted at its centre, the 50 cm mark. The rule weighs 1.0 N. A 2.0 N object hangs from the 20 cm mark.

2.0 Nrule 1.0 N3.0 Npivot (50 cm)30 cm? cm
Metre rule. Diagram not to scale
(a)

Calculate the moment of the 2.0 N weight about the pivot.

(2)
(b)

A 3.0 N weight is hung on the other side of the pivot so that the rule balances. Calculate its distance from the pivot.

(2)
(c)

Calculate the upward force of the pivot on the rule.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
distance from pivot = 0.30 m✓ 1
moment = 2.0 × 0.30 = 0.60 N m✓ 1
Part (b)
clockwise moment = anticlockwise moment: 3.0 × d = 0.60✓ 1
d = 0.20 m✓ 1
Part (c)
upward force = sum of the downward forces✓ 1
2.0 + 3.0 + 1.0 = 6.0 N✓ 1

Exam tip: For a balanced beam, total clockwise moment equals total anticlockwise moment, and the upward force equals the total downward force.

233IGPC3-1-09r44-03Turning effect and centre of gravity Paper CMedium1 mark

A light rod of negligible weight is used as a balance arm in a hospital. It is pivoted at point P. A bag of fluid weighing 60 N hangs 0.50 m to the left of P. A weight hangs 1.0 m to the right of P and the rod balances horizontally.
What is the weight on the right?

60 N?P0.50 m1.0 m
Light rod, negligible weight. Diagram not to scale
Show mark scheme
Marking pointMarkNotes
For balance, clockwise moment = anticlockwise moment: W × 1.0 = 60 × 0.50, so W = 30 N. 60 N assumes equal weights balance whatever the distances. 120 N comes from inverting the distances (60 × 1.0 ÷ 0.50). 90 N comes from using the total distance 1.5 m instead of the distance from the pivot.✓ 1Answer A

Answer: A

Common error: Each moment uses the perpendicular distance from the pivot to that force, not the distance between the two forces.

234IGPC3-1-09r44-07Energy, work and power Paper CMedium1 mark

An electric lift in a hospital raises a bed and patient of total mass 150 kg through a height of 8.0 m in 12 s. g = 9.8 N/kg.
What is the useful power output of the lift?

Show mark scheme
Marking pointMarkNotes
Work done = mgh = 150 × 9.8 × 8.0 = 11 760 J. Power = work ÷ time = 11 760 ÷ 12 = 980 W. 100 W comes from leaving out g. 11 760 W is the work done, with no division by time. 141 120 W multiplies by the time instead of dividing.✓ 1Answer A

Answer: A

Common error: Power is energy transferred per second. Divide the work by the time taken, and include g when converting mass to weight.

235IGPC3-1-09r44-11Speed, velocity and motion graphs Paper CMedium7 marks

A Formula 1 car completes one lap of a circuit of length 5040 m in 84.0 s.

(a)

Calculate the average speed of the car for the lap.

(2)
(b)

At one point on a straight the car is travelling at 85 m/s. Explain why this is greater than the average speed for the lap.

(1)
(c)

On the straight the car accelerates uniformly from 20 m/s to 80 m/s in 3.0 s. Calculate the acceleration.

(2)
(d)

Calculate the distance travelled by the car in these 3.0 s.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 average speed = distance ÷ time = 5040 ÷ 84.0✓ 1
A1 60.0 m/s✓ 1
Part (b)
B1 the car travels slower than this in the corners or when braking, so the average over the whole lap is lower✓ 1
Part (c)
M1 acceleration = change in speed ÷ time = (80 − 20) ÷ 3.0✓ 1
A1 20 m/s2✓ 1
Part (d)
M1 distance = average speed × time = ((20 + 80) ÷ 2) × 3.0 (accept area under a speed–time graph)✓ 1
A1 150 m✓ 1

Exam tip: For uniform acceleration the average speed is halfway between the initial and final speeds, so distance = average speed × time.

236IGPC3-1-09r44-12Forces, friction and springs Paper CMedium7 marks

A beekeeper pushes a hive of mass 30 kg across level, rough ground. Use g = 9.8 N/kg.

30 kgpush 180 Nfriction 120 N
Diagram not to scale
(a)

Calculate the weight of the hive.

(2)
(b)

She pushes horizontally with a force of 180 N and the friction force is 120 N. Calculate the resultant horizontal force on the hive.

(1)
(c)

Calculate the acceleration of the hive.

(2)
(d)

The hive is now moving. She reduces her push to 120 N, with the friction force still 120 N. Describe the motion of the hive and explain your answer.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 W = mg = 30 × 9.8✓ 1
A1 294 N✓ 1
Part (b)
A1 180 − 120 = 60 N✓ 1
Part (c)
M1 a = F ÷ m = 60 ÷ 30 (ECF from (b))✓ 1
A1 2.0 m/s2✓ 1
Part (d)
B1 the hive continues to move at constant speed in a straight line✓ 1
B1 because the forces are balanced, so the resultant force is zero✓ 1

Common error: Use the resultant force in F = ma, not the push alone. A zero resultant force means no acceleration, not no motion.

237IGPC3-1-09r44-13Turning effect and centre of gravity Paper CMedium6 marks

A hospital drip stand has a tall pole fixed to a wide, heavy circular base.

(a)

State what is meant by centre of gravity.

(1)
(b)

Explain why the heavy, wide base makes the stand difficult to knock over.

(3)
(c)

Describe how to find the centre of gravity of an irregular flat piece of card.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 the point at which the weight of the object appears to act (accept: the point through which the whole weight acts)✓ 1
Part (b)
B1 the heavy base lowers the centre of gravity of the stand✓ 1
B1 the wide base means the stand must be tilted through a larger angle before the vertical line through the centre of gravity falls outside the base✓ 1
B1 the stand only topples when that line is outside the base, so it takes more tilt (or force) to topple it✓ 1
Part (c)
B1 hang the card from a pin through a hole near its edge, hang a plumb line from the same pin and mark the vertical line on the card✓ 1
B1 repeat with at least one other hole; the centre of gravity is where the lines cross✓ 1

Exam tip: Link stability to the vertical line through the centre of gravity and whether it stays inside the base.

238IGPC3-1-09r44-16Pressure in solids and liquids Paper CMedium7 marks

A hospital bed and patient have a total mass of 160 kg. The bed rests on four wheels. Each wheel is in contact with the floor over an area of 4.0 cm2. Use g = 9.8 N/kg.

(a)

Calculate the total weight of the bed and patient.

(2)
(b)

Calculate the total area in contact with the floor, in m2. 1 cm2 = 1.0 × 10−4 m2.

(2)
(c)

Calculate the pressure exerted on the floor by the bed.

(2)
(d)

Explain why wheels with a larger contact area would reduce the pressure on the floor.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 W = mg = 160 × 9.8✓ 1
A1 1568 N✓ 1
Part (b)
M1 4 × 4.0 = 16 cm2✓ 1
A1 1.6 × 10−3 m2✓ 1
Part (c)
M1 p = F ÷ A = 1568 ÷ (1.6 × 10−3) (ECF from (a) and (b))✓ 1
A1 980 000 Pa (9.8 × 105 Pa)✓ 1
Part (d)
B1 the same force (weight) acts over a larger area, so pressure = force ÷ area is smaller✓ 1

Common error: Convert cm² to m² using 1 cm² = 10⁻⁴ m², not 10⁻² m².

239IGPC3-1-09r45-06Energy, work and power Paper CMedium1 mark

A beekeeper lifts a box of honey frames of mass 20 kg through a height of 1.2 m at a steady speed in 4.0 s. The gravitational field strength is 9.8 N/kg. What is the useful power developed?

Show mark scheme
Marking pointMarkNotes
Work done = weight × height = 20 × 9.8 × 1.2 = 235.2 J. Power = work ÷ time = 235.2 ÷ 4.0 = 58.8 W, 235 W is the work done, not divided by time. 6.0 W uses mass instead of weight (20 × 1.2 ÷ 4.0). 941 W multiplies the work by the time instead of dividing.✓ 1Answer A

Answer: A

Exam tip: Power is energy transferred per second: always divide the work done by the time taken.

240IGPC3-1-09r45-07Turning effect and centre of gravity Paper CMedium1 mark

A uniform plank of length 3.0 m is balanced on a pivot at its centre. A beekeeper hangs a 30 N hive part 0.80 m to the left of the pivot. At what distance to the right of the pivot must a 20 N weight be hung so that the plank stays horizontal?

30 N20 Npivot0.80 m? m
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
The weight of the plank acts at the pivot, so it has no moment. For balance, clockwise moment = anticlockwise moment: 20 × d = 30 × 0.80, so d = 24 ÷ 20 = 1.2 m. 0.53 m comes from putting the 30 N force on the wrong side of the equation (20 × 0.80 ÷ 30). 0.80 m wrongly assumes equal distances balance unequal forces. 1.5 m is simply the end of the plank.✓ 1Answer D

Answer: D

Exam tip: For a balanced object, total clockwise moment equals total anticlockwise moment about the pivot.

241IGPC3-1-09r45-10Speed, velocity and motion graphs Paper CMedium8 marks

A Formula 1 car starts from rest. Its speed is recorded every 2.0 s.
Time / s: 0, 2.0, 4.0, 6.0, 8.0
Speed / m/s: 0, 10, 20, 30, 40
After 8.0 s the car continues at a constant speed of 40 m/s for a further 12 s.

(a)

Describe the motion of the car between 0 and 8.0 s.

(2)
(b)

Calculate the acceleration of the car between 0 and 8.0 s.

(2)
(c)

Calculate the distance travelled by the car in the first 8.0 s.

(2)
(d)

Calculate the average speed of the car over the whole 20 s.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 the speed increases✓ 1
B1 by equal amounts in equal times, so constant (uniform) acceleration✓ 1
Part (b)
M1 acceleration = change in speed ÷ time = 40 ÷ 8.0✓ 1
A1 5.0 m/s2✓ 1
Part (c)
M1 distance = average speed × time = 20 × 8.0 (accept area under the graph ½ × 8.0 × 40)✓ 1
A1 160 m✓ 1
Part (d)
M1 total distance = 160 + (40 × 12) = 640 m (ECF from (c)), and average speed = total distance ÷ total time✓ 1
A1 640 ÷ 20 = 32 m/s✓ 1

Exam tip: Average speed is total distance divided by total time; it is not the mean of the highest and lowest speeds when the motion has more than one stage.

242IGPC3-1-09r45-12Mass, weight and density Paper CMedium7 marks

A beekeeper pours honey into a jar. The empty jar has a mass of 0.35 kg. When filled with honey the jar and honey have a total mass of 1.07 kg. The honey fills a volume of 500 cm3.

(a)

Calculate the mass of the honey.

(1)
(b)

Calculate the density of the honey in g/cm3.

(2)
(c)

A drop of honey is placed in a beaker of water of density 1.00 g/cm3. Predict what happens to the drop and explain your answer.

(2)
(d)

Convert the density of the honey to kg/m3. Use 1 m3 = 1 000 000 cm3.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 1.07 − 0.35 = 0.72 kg (or 720 g)✓ 1
Part (b)
M1 density = mass ÷ volume = 720 ÷ 500✓ 1
A1 1.44 g/cm3✓ 1
Part (c)
B1 the drop sinks✓ 1
B1 honey is denser than water (1.44 g/cm3 compared with 1.00 g/cm3)✓ 1
Part (d)
M1 0.72 kg ÷ 0.000500 m3 (or 1.44 × 1000)✓ 1
A1 1440 kg/m3✓ 1

Common error: 1 g/cm³ equals 1000 kg/m³, not 100 kg/m³; the volume conversion involves a factor of 10⁶.

243IGPC-035-01-10Speed, velocity and motion graphs Paper CHard1 mark

A scooter moves at a constant speed of 6.0 m/s for 10 s. It then decelerates uniformly and comes to rest 4.0 s later.
What is the average speed of the scooter for the whole 14 s?

10146time / sspeed / m/s
Graph sketched from the values in the question; not to scale
Show mark scheme
Marking pointMarkNotes
First find the total distance from the area under the speed–time graph. Constant-speed section: 6.0 × 10 = 60 m. Braking section: ½ × 4.0 × 6.0 = 12 m. Total distance = 72 m, so average speed = 72 ÷ 14 = 5.1 m/s. The value 3.0 is the average of the start and end speeds of the braking section only. The value 4.5 averages the two section speeds (6.0 and 3.0) without weighting them by time. The value 6.0 ignores the braking section.✓ 1Answer C

Answer: C

Common error: Do not average speeds directly when the sections last for different times. Find the total distance first, then divide by the total time.

244IGPC-035-02-10Mass, weight and density Paper CHard1 mark

A solid steel block has a density of 8000 kg/m3. Its weight on Earth is 39.2 N. The gravitational field strength is 9.8 N/kg.
What is the volume of the block?

Show mark scheme
Marking pointMarkNotes
First find the mass: m = W ÷ g = 39.2 ÷ 9.8 = 4.0 kg. Then volume = mass ÷ density = 4.0 ÷ 8000 = 5.0 × 10−4 m3.
The other options come from common mistakes:
• 4.9 × 10−3 m3 uses the weight in place of the mass.
• 2.0 × 103 m3 is density ÷ mass.
• 3.2 × 104 m3 is mass × density.
✓ 1Answer A

Answer: A

Common error: The density equation uses mass in kg, not weight in N. Convert weight to mass first using m = W/g.

245IGPC-035-04-08Turning effect and centre of gravity Paper CHard1 mark

A uniform beam AB is 3.0 m long and has a weight of 120 N. It rests on a pivot 0.50 m from end A.
What downward force applied at end A holds the beam horizontal?

F120 N0.50 m3.0 mAB
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
The weight of a uniform beam acts at its centre, 1.5 m from A. This point is 1.5 − 0.50 = 1.0 m from the pivot.
F × 0.50 = 120 × 1.0, so F = 240 N.
Option A inverts the distance ratio, option B assumes equal forces balance, and option D places the weight at end B.
✓ 1Answer C

Answer: C

Exam tip: For a uniform beam, the weight acts at the midpoint. Find the distance from this point to the pivot, not to the end of the beam.

246IGPC-035-04-10Turning effect and centre of gravity Paper CHard1 mark

A force of 40 N is applied to the end of a spanner, 0.25 m from the centre of a nut. The force is not at right angles to the spanner. The perpendicular distance from the centre of the nut to the line of action of the force is 0.15 m.
What is the moment of the force about the nut?

nut40 N0.25 m0.15 m
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Moment = force × perpendicular distance from the pivot to the line of action of the force = 40 × 0.15 = 6.0 N m. Option C uses the length of the spanner (0.25 m), which is not perpendicular to the force. Option B uses the difference of the distances, and option D divides the force by the distance.✓ 1Answer A

Answer: A

Common error: Use the perpendicular distance to the line of action of the force, not simply the distance to where the force is applied.

247IGPC-035-01-16Speed, velocity and motion graphs Paper CHard8 marks

A car is travelling at 20 m/s when the driver sees a hazard ahead. The car continues at 20 m/s for the driver's reaction time of 0.70 s. The brakes are then applied, and the car decelerates uniformly at 5.0 m/s2 until it stops.

(a)

Calculate the distance travelled during the reaction time.

(1)
(b)

Calculate the time taken for the car to stop once the brakes are applied.

(2)
(c)

Calculate the distance travelled while the car is braking.

(2)
(d)

Calculate the average speed of the car from the moment the driver sees the hazard until the car stops.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
distance = 20 × 0.70 = 14 m✓ 1
Part (b)
t = change in speed ÷ deceleration = 20 ÷ 5.0✓ 1
= 4.0 s✓ 1
Part (c)
Any valid method, e.g. average speed × time = ½(20 + 0) × 4.0, or area under the speed–time graph = ½ × 4.0 × 20, or v2 = u2 + 2as giving s = 202 ÷ (2 × 5.0)✓ 1
= 40 m✓ 1
Part (d)
total distance = 14 + 40 = 54 m✓ 1
total time = 0.70 + 4.0 = 4.7 s✓ 1
average speed = 54 ÷ 4.7 = 11.5 m/s (accept 11 m/s)✓ 1

Exam tip: Picture the speed–time graph: a horizontal line for the reaction time, then a straight line down to zero. The distances are the areas of a rectangle and a triangle. Average speed is total distance ÷ total time, not the mean of the start and end speeds.

248IGPC-035-00-16Physical quantities and measurement Paper CHard7 marks

A student has a bag of identical small steel ball bearings and a 100 cm3 measuring cylinder. The cylinder is marked in divisions of 1 cm3.

50 cm³before58 cm³after
Diagram not to scale
(a)

Explain why adding a single ball bearing to water in this cylinder would not give an accurate value for its volume.

(2)
(b)

The student adds 50 ball bearings to water in the cylinder. The water level rises from 50.0 cm3 to 58.0 cm3.
Calculate the volume of one ball bearing.

(2)
(c)

The student then places 10 ball bearings touching in a straight line beside a ruler. The row is 6.8 cm long.
Calculate the diameter of one ball bearing.

(1)
(d)

The volume V of a sphere of diameter d is given by \(V = \frac{\pi d^3}{6}\).
Use the diameter you calculated from the row of ball bearings to calculate the volume of one ball bearing. Hence show that your result agrees with the volume you found using the measuring cylinder.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The volume of one ball bearing is smaller than one scale division✓ 1
so the rise in water level is too small to read, and the reading uncertainty is larger than the volume being measured✓ 1
Part (b)
Volume of 50 ball bearings = 58.0 − 50.0 = 8.0 cm3✓ 1
Volume of one = 8.0 ÷ 50 = 0.16 cm3✓ 1
Part (c)
6.8 ÷ 10 = 0.68 cm✓ 1
Part (d)
V = π × 0.683 ÷ 6 = 0.1646 cm3 (≈ 0.165 cm3)✓ 1
0.1646 cm3 is 0.16 cm3 to 2 significant figures, which is the same as the value from the measuring cylinder✓ 1

Exam tip: Measuring many identical objects and dividing works for both lengths and volumes. It is the same idea as timing many oscillations.

249IGPC3-1-01c-15Turning effect and centre of gravity Paper CHard7 marks

A uniform plank of length 4.0 m and weight 120 N is balanced on a pivot placed 1.0 m from its left end. A child of weight 300 N sits at the left end of the plank. An adult of weight 450 N sits on the right-hand side of the pivot.

300 N120 N450 N1.0 md4.0 m
Diagram not to scale
(a)

State the principle of moments.

(2)
(b)

Calculate the moment of the weight of the plank about the pivot.

(2)
(c)

Calculate the distance d from the pivot at which the adult must sit so that the plank is balanced.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
for a body in equilibrium, the sum of the clockwise moments about a pivot✓ 1
equals the sum of the anticlockwise moments about the same pivot✓ 1
Part (b)
weight acts at the centre of the plank, 1.0 m from the pivot✓ 1
moment = 120 × 1.0 = 120 N m✓ 1
Part (c)
anticlockwise moment = 300 × 1.0 = 300 N m✓ 1
300 = 120 + 450 × d✓ 1
d = 0.40 m✓ 1

Exam tip: Draw the pivot and mark each force's distance from it. The weight of a uniform plank acts at its centre, which here is not at the pivot.

250IGPC3-1-06r7-08Pressure in solids and liquids Paper CHard1 mark

A solid concrete cube of side L rests on a level floor. It is replaced by a solid cube of the same concrete with side 2L. How does the pressure on the floor change?

Show mark scheme
Marking pointMarkNotes
B is correct: weight rises by a factor of 8 and base area by a factor of 4, so pressure rises by 8 / 4 = 2 (equivalently p = ρgh depends only on height). A is wrong because the heights differ. C is wrong because it uses only the area ratio. D is wrong because it uses only the weight ratio.✓ 1Answer B

Answer: B

Common error: When all lengths double, volume (and weight) increases eightfold but area only fourfold.

251IGPC3-1-06r7-16Speed, velocity and motion graphs Paper CHard8 marks

A logistics drone moves along a straight path. Its speed increases uniformly from 0 to 9.0 m/s in 6.0 s, stays at 9.0 m/s for the next 4.0 s, then decreases uniformly to 0 over a further 4.0 s.

610149time / sspeed / m/s
Graph sketched from the values in the question; not to scale
(a)

Calculate the acceleration of the drone during the first 6.0 s.

(2)
(b)

Calculate the total distance travelled in the 14 s.

(4)
(c)

Calculate the average speed of the drone over the whole 14 s. Give your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
a = change in speed / time = 9.0 / 6.0✓ 1
1.5 m/s²✓ 1
Part (b)
Distance in first 6.0 s = ½ × 6.0 × 9.0 = 27 m✓ 1
Distance at constant speed = 9.0 × 4.0 = 36 m✓ 1
Distance while slowing = ½ × 4.0 × 9.0 = 18 m✓ 1
Total distance = 27 + 36 + 18 = 81 m✓ 1
Part (c)
average speed = total distance / total time = 81 / 14 (ECF from (b))✓ 1
5.79 m/s✓ 1

Exam tip: Average speed uses total distance over total time, not the average of the greatest and least speeds.

252IGPC3-1-08r20-16Turning effect and centre of gravity Paper CHard8 marks

A uniform plank of length 5.0 m and weight 400 N rests horizontally on two supports. Support A is 0.50 m from the left end and support B is 0.50 m from the right end. A worker of weight 700 N stands on the plank 1.5 m from the left end.

700 Nweight 400 NAB??0.50 m1.5 m0.50 m
Diagram not to scale
(a)

State the two conditions for the plank to be in equilibrium.

(2)
(b)

Calculate the upward force at support B by taking moments about A.

(3)
(c)

Calculate the upward force at support A.

(2)
(d)

Explain what happens to the force at B as the worker walks towards B.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 the resultant force on the plank is zero (upward forces equal downward forces)✓ 1
B1 the resultant moment about any point is zero (clockwise moments equal anticlockwise moments)✓ 1
Part (b)
M1 the distances from A are 2.0 m (plank's weight) and 1.0 m (worker); moments 400 × 2.0 = 800 N m and 700 × 1.0 = 700 N m✓ 1
M1 B × 4.0 = 800 + 700✓ 1
A1 B = 375 N✓ 1
Part (c)
M1 A + B = 400 + 700 = 1100 N (ECF from (b))✓ 1
A1 A = 725 N✓ 1
Part (d)
B1 it increases because the worker's moment about A becomes larger (and the force at A decreases)✓ 1

Exam tip: Taking moments about one support removes its unknown force from the equation, leaving one unknown.

253IGPC3-1-08r22-13Speed, velocity and motion graphs Paper CHard10 marks

A survey drone at an archaeological site accelerates uniformly from rest at 0.80 m/s2 until it reaches 8.0 m/s. It then flies at constant speed, and finally decelerates uniformly at 1.6 m/s2 to rest. The total distance flown is 260 m.

(a)

Calculate the time taken to reach 8.0 m/s.

(2)
(b)

Calculate the distance flown while accelerating.

(2)
(c)

Calculate the time taken and the distance flown while decelerating.

(3)
(d)

Calculate the total time of the flight.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
time = change in speed ÷ acceleration = 8.0 ÷ 0.80✓ 1
10 s✓ 1
Part (b)
distance = average speed × time = 4.0 × 10, or area under graph ½ × 8.0 × 10✓ 1
40 m✓ 1
Part (c)
time = 8.0 ÷ 1.6 = 5.0 s✓ 1
distance = ½ × 8.0 × 5.0✓ 1
20 m✓ 1
Part (d)
distance at constant speed = 260 − 40 − 20 = 200 m✓ 1
time at constant speed = 200 ÷ 8.0 = 25 s✓ 1
total time = 10 + 25 + 5.0 = 40 s✓ 1
ECF from (a), (b) and (c)—

Exam tip: For a multi-stage journey, find each stage separately and subtract known stages from the total.

254IGPC3-1-10r27-07Energy, work and power Paper CHard1 mark

A motor raises a 60 kg sack of coffee beans through a height of 4.0 m in 12 s. The motor is 40% efficient. The gravitational field strength is 9.8 N/kg. What is the input power to the motor?

Show mark scheme
Marking pointMarkNotes
Useful work = mgh = 60 × 9.8 × 4.0 = 2352 J, so useful power = 2352 / 12 = 196 W. Input power = useful power / efficiency = 196 / 0.40 = 490 W. 196 W is the useful output only. 78 W multiplies by the efficiency instead of dividing. 5880 W divides the energy by efficiency but never divides by the time.✓ 1Answer B

Answer: B

Common error: Input power is larger than output power: divide the useful power by the efficiency.

255IGPC3-1-10r27-15Turning effect and centre of gravity Paper CHard8 marks

A uniform horizontal bridge section of length 8.0 m and mass 200 kg rests on two supports, one at each end. A lorry of weight 5880 N stands with its centre of gravity 2.0 m from the left-hand end. The weight of the bridge section acts at its centre. The gravitational field strength is 9.8 N/kg.

lorry 5880 Nweight of sectionleftright??2.0 m8.0 m
Diagram not to scale
(a)

State the principle of moments.

(2)
(b)

Calculate the upward force at the right-hand support by taking moments about the left-hand end.

(3)
(c)

Calculate the upward force at the left-hand support.

(2)
(d)

The lorry moves so that its centre of gravity is at the middle of the section. Calculate the force at each support.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
for a body in equilibrium, the sum of the clockwise moments about any point✓ 1
equals the sum of the anticlockwise moments about that point✓ 1
Part (b)
weight of section = 200 × 9.8 = 1960 N✓ 1
clockwise moments: 1960 × 4.0 + 5880 × 2.0 = R × 8.0✓ 1
R = 19 600 / 8.0 = 2450 N✓ 1
Part (c)
total upward force = total downward force = 1960 + 5880 = 7840 N✓ 1
7840 − 2450 = 5390 N (ECF from (b))✓ 1
Part (d)
7840 / 2 = 3920 N at each support✓ 1

Exam tip: Take moments about a support: the unknown force there has no moment, which removes one unknown.

256IGPC3-1-08r31-16Speed, velocity and motion graphs Paper CHard9 marks

The speed of a probe's lander during the final part of its descent is recorded. Between the times shown the speed changes at a uniform rate.
Time / s: 0, 10, 18, 28
Speed / m/s: 20, 20, 4.0, 4.0

101828420time / sspeed / m/s
Graph as described; not to scale
(a)

Calculate the deceleration of the lander between 10 s and 18 s.

(2)
(b)

Calculate the distance travelled between 10 s and 18 s.

(2)
(c)

Calculate the total distance travelled between 0 and 28 s.

(3)
(d)

Calculate the average speed of the lander over the 28 s.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 deceleration = change in speed / time = (20 − 4.0) / 8.0✓ 1
A1 2.0 m/s2✓ 1
Part (b)
M1 distance = average speed × time = ½ × (20 + 4.0) × 8.0 (or area under the graph)✓ 1
A1 96 m✓ 1
Part (c)
M1 distance from 0 to 10 s = 20 × 10 = 200 m✓ 1
M1 distance from 18 s to 28 s = 4.0 × 10 = 40 m✓ 1
A1 336 m (ECF from (b))✓ 1
Part (d)
M1 average speed = total distance / total time = 336 / 28✓ 1
A1 12 m/s (ECF from (c))✓ 1

Common error: Split a speed–time graph into sections, and find each distance from the area under that section.

257IGPC-035-00-06Physical quantities and measurement Paper EEasy1 mark

Which list contains only vector quantities?

Show mark scheme
Marking pointMarkNotes
Force, acceleration and momentum all have both magnitude and direction, so they are vectors. Distance, time and temperature are all scalars. In option C, speed and energy are scalars. In option D, mass is a scalar, even though weight is a vector.✓ 1Answer B

Answer: B

Common error: Mass and weight are often confused. Mass (kg) is a scalar, but weight is a force, so it is a vector.

258IGPC-035-05-01Momentum and impulse Paper EEasy1 mark

A car of mass 1200 kg travels along a straight road at 15 m/s.
What is the momentum of the car?

Show mark scheme
Marking pointMarkNotes
Momentum p = mv = 1200 × 15 = 18 000 kg m/s. 9000 kg m/s wrongly includes a factor of ½. 135 000 kg m/s is the kinetic energy ½mv² in joules, not momentum. 80 kg m/s comes from dividing mass by speed.✓ 1Answer A

Answer: A

Common error: Momentum is mv. It has no ½ and no squared term. Those belong to kinetic energy.

259IGPC-035-05-02Momentum and impulse Paper EEasy1 mark

Which unit can be used for impulse?

Show mark scheme
Marking pointMarkNotes
Impulse = force × time, so its unit is the newton second (N s). This is equivalent to kg m/s, the unit of momentum. N/s would come from dividing force by time. J is the unit of energy. kg m/s2 is equivalent to the newton, the unit of force.✓ 1Answer B

Answer: B

Exam tip: N s and kg m/s are equivalent units. Either can be used for impulse or change in momentum.

260IGPC-035-05-03Momentum and impulse Paper EEasy1 mark

Which expression is equal to the resultant force acting on an object?

Show mark scheme
Marking pointMarkNotes
Resultant force equals the rate of change of momentum: F = Δp/Δt. Change in momentum × time has no physical meaning here. Momentum ÷ mass gives velocity. Momentum × velocity has the units of energy, not force.✓ 1Answer B

Answer: B

Exam tip: F = Δp/Δt is the general form of F = ma. Rearranged, it gives impulse: FΔt = Δp.

261IGPC-035-05-04Momentum and impulse Paper EEasy1 mark

Two ice skaters collide and move off separately. No external resultant force acts on them.
Which quantity must be the same before and after the collision?

Show mark scheme
Marking pointMarkNotes
When no external resultant force acts, the total momentum of the system is conserved. The momentum and velocity of each individual skater can change, because they exert forces on each other. Kinetic energy is usually not conserved, because some is transferred to thermal energy and sound.✓ 1Answer D

Answer: D

Common error: Conservation of momentum applies to the total for the whole system, not to each object separately.

262IGPC-035-05-11Momentum and impulse Paper EEasy4 marks

A skater of mass 55 kg glides across an ice rink in a straight line at a constant speed of 6.0 m/s.

(a)

State the equation that defines momentum.

(1)
(b)

Calculate the momentum of the skater.

(2)
(c)

Momentum is a vector quantity. State what this means.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
momentum = mass × velocity (p = mv)✓ 1
Part (b)
p = 55 × 6.0✓ 1
= 330 kg m/s✓ 1
Part (c)
It has both a magnitude (size) and a direction✓ 1

Exam tip: Always include a unit with a calculated answer. Either kg m/s or N s is acceptable for momentum.

263IGPC-035-05-12Momentum and impulse Paper EEasy5 marks

A tennis player serves a ball. The ball is momentarily at rest at the top of its throw when the racket hits it. The racket exerts an average force of 120 N on the ball for 0.0050 s. The mass of the ball is 0.060 kg.

(a)

State the equation that links impulse, force and time.

(1)
(b)

Calculate the impulse on the ball.

(2)
(c)

Calculate the speed of the ball as it leaves the racket.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
impulse = force × time (FΔt)✓ 1
Part (b)
impulse = 120 × 0.0050✓ 1
= 0.60 N s✓ 1
Part (c)
impulse = change in momentum, so 0.60 = 0.060 × v✓ 1
v = 10 m/s✓ 1

Exam tip: Impulse equals change in momentum. If the object starts from rest, the impulse equals its final momentum.

264IGPC-035-01-06Speed, velocity and motion graphs Paper EMedium1 mark

An athlete runs exactly one lap of a 400 m track in 80 s. She finishes at the point where she started.
What are her average speed and the magnitude of her average velocity for the lap?

Show mark scheme
Marking pointMarkNotes
Average speed uses the total distance travelled: 400 m ÷ 80 s = 5.0 m/s. Average velocity uses the displacement, which is the change in position in a given direction. She ends where she started, so her displacement is zero and her average velocity is 0 m/s. Option A treats speed and velocity as identical. Options C and D wrongly suggest that returning to the start cancels the distance run.✓ 1Answer B

Answer: B

Exam tip: Speed is a scalar and is based on distance. Velocity is a vector and is based on displacement. On a closed path the displacement is zero.

265IGPC-035-01-14Speed, velocity and motion graphs Paper EMedium6 marks

A ball is thrown vertically upwards with a speed of 14.7 m/s. Air resistance is negligible. The acceleration of free fall is 9.8 m/s2.

(a)

Calculate the time taken for the ball to reach its highest point.

(2)
(b)

Use the area under the speed–time graph to calculate the maximum height reached by the ball.

(2)
(c)

Upwards is taken as the positive direction. Explain why the velocity of the ball is negative as it falls back down, while its speed is positive.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
speed falls by 9.8 m/s each second, so t = 14.7 ÷ 9.8✓ 1
= 1.5 s✓ 1
Part (b)
height = ½ × 1.5 × 14.7✓ 1
= 11 m (11.0 m)✓ 1
Part (c)
velocity includes direction, and the ball is now moving downwards, opposite to the positive direction✓ 1
speed is only the magnitude, with no direction, so it cannot be negative✓ 1

Exam tip: At the highest point the velocity is zero, but the acceleration is still 9.8 m/s² downwards.

266IGPC-035-03-04Forces, friction and springs Paper EMedium1 mark

A load of 6.0 N stretches a spring by 4.0 cm. The spring obeys Hooke's law. What is the spring constant of the spring?

6.0 N4.0 cmno loadwith load
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
k = F/x. Converting the extension to metres, x = 4.0 cm = 0.040 m, so k = 6.0/0.040 = 150 N/m. The value 1.5 N/m uses the extension in cm without converting to metres. The value 0.67 N/m is x/F, which is the inverted ratio. The value 24 N/m comes from multiplying F by x.✓ 1Answer D

Answer: D

Exam tip: Always check the unit of extension before calculating k. If the answer must be in N/m, the extension must be in metres.

267IGPC-035-03-05Forces, friction and springs Paper EMedium1 mark

A car of mass 1200 kg has a forward driving force of 3000 N. The total resistive force on the car is 600 N. What is the acceleration of the car?

cardriving force 3000 Nresistive force 600 N
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Resultant force = 3000 N − 600 N = 2400 N. Using a = F/m gives 2400/1200 = 2.0 m/s². The value 2.5 m/s² ignores the resistive force. The value 3.0 m/s² adds the resistive force instead of subtracting it. The value 0.50 m/s² is m/F, which is the inverted ratio.✓ 1Answer B

Answer: B

Exam tip: In F = ma, F is always the resultant force, not just the driving force.

268IGPC-035-03-07Forces, friction and springs Paper EMedium1 mark

A student adds loads to a steel spring and measures its extension. Which statement describes what happens once the load goes beyond the limit of proportionality?

Show mark scheme
Marking pointMarkNotes
Up to the limit of proportionality, extension is directly proportional to load, and the load–extension graph is a straight line through the origin. Beyond this limit the graph curves, so the extension is no longer proportional to the load. The spring still stretches, usually by more per newton, so it does not stop stretching and its stiffness does not become zero. There is no fixed doubling rule.✓ 1Answer A

Answer: A

Exam tip: On a load–extension graph, the limit of proportionality is the point where the straight line first begins to curve.

269IGPC-035-03-09Forces, friction and springs Paper EMedium1 mark

A ball on a string is whirled in a horizontal circle at constant speed. What is the direction of the resultant force on the ball?

Show mark scheme
Marking pointMarkNotes
The ball's speed is constant, but its direction keeps changing, so its velocity is changing. This requires a resultant force. The force acts at right angles to the motion, towards the centre of the circle, and it is provided by the tension in the string. An outward force is a common misconception. A force along the direction of motion would change the speed, not the direction.✓ 1Answer B

Answer: B

Common error: Constant speed does not mean constant velocity. A change of direction needs a resultant force towards the centre.

270IGPC-035-03-13Forces, friction and springs Paper EMedium7 marks

A spring has an unstretched length of 12.0 cm. A mass of 0.40 kg is hung from it and the length of the spring becomes 15.5 cm. The spring obeys Hooke's law throughout. Use g = 9.8 N/kg.

12.0 cm0.40 kg15.5 cmno loadwith load
Diagram not to scale
(a)

Calculate the weight of the 0.40 kg mass.

(1)
(b)

Calculate the extension of the spring.

(1)
(c)

Calculate the spring constant of the spring in N/m.

(2)
(d)

An extra mass is added and the length of the spring becomes 18.3 cm. Calculate the size of the extra mass.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
W = mg = 0.40 × 9.8 = 3.92 N (accept 3.9 N)✓ 1
Part (b)
15.5 − 12.0 = 3.5 cm (= 0.035 m)✓ 1
Part (c)
k = F/x = 3.92/0.035 = 112 N/m (accept 110 N/m)✓ 2
Part (d)
New extension = 18.3 − 12.0 = 6.3 cm = 0.063 m . Total force = kx = 112 × 0.063 = 7.06 N, so total mass = 7.06/9.8 = 0.72 kg . Extra mass = 0.72 − 0.40 = 0.32 kg✓ 3

Exam tip: Convert cm to m before using F = kx when k is in N/m.

271IGPC-035-02-06Mass, weight and density Paper EMedium1 mark

On a distant planet, an object of mass 3.0 kg has a weight of 11.1 N. The planet has no atmosphere.
The object is dropped near the planet's surface. What is its acceleration?

Show mark scheme
Marking pointMarkNotes
The gravitational field strength is g = W ÷ m = 11.1 ÷ 3.0 = 3.7 N/kg. Gravitational field strength is equivalent to the acceleration of free fall, so the acceleration is 3.7 m/s2.
The other options come from common mistakes:
• 9.8 m/s2 wrongly assumes Earth's value.
• 33 m/s2 comes from multiplying instead of dividing.
• 0.27 m/s2 is m ÷ W.
✓ 1Answer B

Answer: B

Exam tip: 1 N/kg is the same as 1 m/s². With no air resistance, the value of g is the acceleration of a falling object.

272IGPC-035-02-15Mass, weight and density Paper EMedium6 marks

A space probe lands on a moon that has no atmosphere. A test mass of 2.5 kg on the probe has a weight of 9.25 N on this moon.

(a)

Define gravitational field strength.

(1)
(b)

Calculate the gravitational field strength on the surface of the moon.

(2)
(c)

Explain why a beam balance would show the same mass for the test mass on this moon as on Earth.

(2)
(d)

A small rock falls from rest near the moon's surface. State its acceleration.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
The (gravitational) force per unit mass✓ 1
Part (b)
g = W ÷ m = 9.25 ÷ 2.5✓ 1
= 3.7 N/kg✓ 1
Part (c)
The test mass and the standard masses are in the same gravitational field, so their weights change by the same factor✓ 1
The balance therefore compares masses, and mass does not change with location✓ 1
Part (d)
3.7 m/s2✓ 1

Exam tip: The gravitational field strength in N/kg is equal to the free-fall acceleration in m/s², provided there is no air resistance.

273IGPC-035-06-05Energy, work and power Paper EMedium1 mark

A delivery worker lifts a box of mass 25 kg from the floor onto a shelf 1.6 m higher. The gravitational field strength g is 9.8 N/kg.
What is the increase in the gravitational potential energy of the box?

Show mark scheme
Marking pointMarkNotes
ΔEp = mgΔh = 25 × 9.8 × 1.6 = 392 J. 40 J uses m × h and leaves out g, 153 J divides the weight by the height, and 245 J is just the weight (in N) with no height included.✓ 1Answer D

Answer: D

Common error: Mass in kg must be multiplied by g to give weight in N before multiplying by the height gained.

274IGPC-035-06-09Energy, work and power Paper EMedium1 mark

A lamp has an input power of 60 W. Its efficiency is 15%.
How much energy does the lamp waste in 1.0 minute?

Show mark scheme
Marking pointMarkNotes
The total energy input in 60 s is 60 W × 60 s = 3600 J. The useful output is 15% of this, which is 540 J. So the wasted energy = 3600 − 540 = 3060 J (or 85% of 3600 J). 51 J is the wasted energy in only 1 s, 540 J is the useful energy, and 3600 J is the total input.✓ 1Answer C

Answer: C

Common error: Check whether the question asks for useful or wasted energy, and convert minutes to seconds before using E = Pt.

275IGPC-035-00-07Physical quantities and measurement Paper EMedium1 mark

Two forces act on a crate. One force is 5.0 N and the other is 12 N. The two forces are at right angles to each other.
What is the magnitude of the resultant force?

12 N5.0 NR
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
For forces at right angles, use Pythagoras: resultant = √(5.0² + 12²) = √169 = 13 N. 17 N would be correct only if the forces acted in the same direction. 7.0 N would be correct only if they acted in opposite directions. 8.5 N is simply the average of the two forces.✓ 1Answer C

Answer: C

Common error: Vectors at right angles cannot be added like ordinary numbers. Use Pythagoras or a scale drawing.

276IGPC-035-00-14Physical quantities and measurement Paper EMedium7 marks

Physical quantities can be classified as scalars or vectors.

9.0 N (east)12 N (north)Rθ
Diagram not to scale
(a)

State the difference between a scalar quantity and a vector quantity.

(1)
(b)

Classify each of these quantities as a scalar or a vector:
distance, momentum, temperature, gravitational field strength.

(2)
(c)

A small boat is pulled by two ropes. One rope pulls due east with a force of 9.0 N and the other pulls due north with a force of 12 N.
Calculate the magnitude of the resultant force on the boat.

(2)
(d)

Calculate the angle between the resultant force and the direction due east.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A scalar has magnitude only, but a vector has both magnitude and direction✓ 1
Part (b)
Momentum and gravitational field strength are vectors✓ 1
Distance and temperature are scalars✓ 1
Part (c)
Resultant = √(9.0² + 12²)✓ 1
= 15 N✓ 1
Part (d)
tan θ = 12 ÷ 9.0✓ 1
θ = 53° (north of east)✓ 1

Exam tip: A scale drawing is also acceptable for a right-angled resultant. State your scale and measure the angle carefully.

277IGPC-035-04-07Turning effect and centre of gravity Paper EMedium1 mark

A uniform metre rule is pivoted at its 50 cm mark. A 2.0 N weight hangs at the 10 cm mark and a 1.0 N weight hangs at the 20 cm mark.
What downward force at the 90 cm mark makes the rule balance?

2.0 N1.0 NF10 cm20 cm50 cm90 cm
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Distances are measured from the pivot, not from the end of the rule.
Anticlockwise moments: 2.0 × 0.40 + 1.0 × 0.30 = 1.10 N m.
Clockwise moment: F × 0.40.
F = 1.10 / 0.40 = 2.75 N ≈ 2.8 N.
Option A uses the scale readings as distances, option B ignores the 1.0 N weight, and option D simply adds the forces.
✓ 1Answer C

Answer: C

Common error: Scale readings on a metre rule are not distances from the pivot. Subtract the pivot position first.

278IGPC-035-07-06Pressure in solids and liquids Paper EMedium1 mark

A diver is 15 m below the surface of a freshwater lake. The density of the water is 1000 kg/m3 and g = 9.8 N/kg.
What is the pressure due to the water at this depth?

water surfacediver15 m
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Δp = ρgΔh = 1000 × 9.8 × 15 = 147 000 Pa = 1.47 × 105 Pa.
147 Pa comes from using a density of 1 (g/cm3) instead of 1000 kg/m3. 1.5 × 104 Pa leaves out g. 2.48 × 105 Pa adds atmospheric pressure, but the question asks only for the pressure due to the water.
✓ 1Answer C

Answer: C

Exam tip: Read carefully whether the question asks for the pressure due to the liquid alone or the total pressure including the atmosphere.

279IGPC-035-07-07Pressure in solids and liquids Paper EMedium1 mark

The pressure due to water at a depth of 4.0 m is measured. Oil has a density of 800 kg/m3 and water has a density of 1000 kg/m3.
At what depth in the oil is the pressure due to the oil the same as this?

water4.0 moilh
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
For equal pressures, ρoilghoil = ρwaterghwater.
hoil = 4.0 × 1000 / 800 = 5.0 m.
The oil is less dense, so a greater depth is needed for the same pressure. 3.2 m multiplies by 0.8 instead of dividing. 4.0 m ignores density. 1.25 m is just the ratio of the densities with no depth.
✓ 1Answer D

Answer: D

Exam tip: Check that your answer makes sense: a less dense liquid must be deeper to produce the same pressure.

280IGPC-035-05-05Momentum and impulse Paper EMedium1 mark

A ball of mass 0.15 kg moves horizontally at 8.0 m/s towards a wall. It rebounds along the same line at 6.0 m/s.
What is the magnitude of the change in momentum of the ball?

Show mark scheme
Marking pointMarkNotes
Take the direction towards the wall as positive.
Initial momentum = 0.15 × 8.0 = +1.2 kg m/s.
Final momentum = 0.15 × (−6.0) = −0.90 kg m/s.
Change = −0.90 − 1.2 = −2.1 kg m/s, so the magnitude is 2.1 kg m/s.
0.30 kg m/s ignores the reversal of direction. 1.2 kg m/s is only the initial momentum. 0 assumes the momentum cannot change.
✓ 1Answer C

Answer: C

Common error: When an object rebounds, the speeds add when finding the change in velocity, because the direction reverses.

281IGPC-035-05-06Momentum and impulse Paper EMedium1 mark

A trolley of mass 2.0 kg moves at 3.0 m/s. It collides with a stationary trolley of mass 1.0 kg. The two trolleys stick together.
What is their speed immediately after the collision?

2.0 kg1.0 kg3.0 m/sat rest
Before the collision
Show mark scheme
Marking pointMarkNotes
Total momentum before = 2.0 × 3.0 = 6.0 kg m/s.
After the collision the combined mass is 3.0 kg, so v = 6.0 ÷ 3.0 = 2.0 m/s.
3.0 m/s assumes the speed is unchanged. 1.5 m/s halves the speed instead of using the combined mass. 6.0 m/s is the value of the momentum, not the speed.
✓ 1Answer A

Answer: A

Exam tip: When objects stick together, divide the total momentum by the combined mass.

282IGPC-035-05-07Momentum and impulse Paper EMedium1 mark

A footballer kicks a stationary ball of mass 0.45 kg. The ball leaves the boot at 20 m/s. The boot is in contact with the ball for 0.015 s.
What is the average force on the ball?

Show mark scheme
Marking pointMarkNotes
Change in momentum = 0.45 × 20 = 9.0 kg m/s.
F = Δp ÷ Δt = 9.0 ÷ 0.015 = 600 N.
9.0 N is the change in momentum, not divided by the time. 0.135 N multiplies by the time instead of dividing. 300 N uses the average speed of 10 m/s instead of the change in speed.
✓ 1Answer D

Answer: D

Common error: Force = change in momentum ÷ time. A short contact time gives a large force.

283IGPC-035-05-08Momentum and impulse Paper EMedium1 mark

A rifle of mass 4.0 kg is at rest. It fires a bullet of mass 0.010 kg at 400 m/s.
What is the recoil speed of the rifle?

Show mark scheme
Marking pointMarkNotes
Total momentum before firing is zero, so the rifle's momentum after firing is equal and opposite to the bullet's.
Bullet momentum = 0.010 × 400 = 4.0 kg m/s.
Rifle speed = 4.0 ÷ 4.0 = 1.0 m/s.
400 m/s assumes both move at the same speed. 4.0 m/s is the momentum value, not the speed. 1600 m/s multiplies by the rifle's mass instead of dividing.
✓ 1Answer C

Answer: C

Exam tip: In an explosion from rest, the total momentum stays zero. The two parts move in opposite directions with equal-sized momenta.

284IGPC-035-05-13Momentum and impulse Paper EMedium7 marks

In a safety test, a car of mass 900 kg travelling at 12 m/s hits a barrier and comes to rest. The car is fitted with a crumple zone at the front and an airbag for the driver.

(a)

Calculate the magnitude of the change in momentum of the car.

(2)
(b)

The crumple zone makes the car take 0.40 s to stop. Calculate the average force on the car.

(2)
(c)

Explain how the airbag reduces the force on the driver during the collision.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Δp = 900 × 12✓ 1
= 10 800 kg m/s✓ 1
Part (b)
F = Δp ÷ Δt = 10 800 ÷ 0.40✓ 1
= 27 000 N✓ 1
Part (c)
The driver's change in momentum is the same with or without the airbag✓ 1
The airbag increases the time taken for the driver to stop✓ 1
Force = change in momentum ÷ time, so the force is smaller✓ 1

Exam tip: In safety-feature explanations, state that the change in momentum is fixed, that the time increases, and that therefore the force decreases.

285IGPC-035-05-14Momentum and impulse Paper EMedium8 marks

In a railway yard, truck A of mass 8000 kg rolls at 2.5 m/s along a straight, level track. It collides with a stationary truck B of mass 12 000 kg. The trucks couple together and move off together.

A 8000 kgB 12 000 kg2.5 m/sat rest
Before the collision
(a)

State the principle of conservation of momentum.

(1)
(b)

Calculate the speed of the coupled trucks just after the collision.

(3)
(c)

Calculate the kinetic energy lost in the collision.

(3)
(d)

Suggest what happens to the kinetic energy that is lost.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
When no external resultant force acts on a system, its total momentum stays constant (total momentum before = total momentum after)✓ 1
Part (b)
momentum before = 8000 × 2.5 = 20 000 kg m/s✓ 1
20 000 = (8000 + 12 000) × v✓ 1
v = 1.0 m/s✓ 1
Part (c)
KE before = ½ × 8000 × 2.52 = 25 000 J✓ 1
KE after = ½ × 20 000 × 1.02 = 10 000 J✓ 1
KE lost = 25 000 − 10 000 = 15 000 J (1.5 × 104 J)✓ 1
Part (d)
Any one of: transferred to thermal (internal) energy of the trucks and surroundings / transferred as sound / used in deforming the trucks or couplings✓ 1

Common error: Momentum is conserved in the collision, but kinetic energy is not. Do not use kinetic energy to find the speed after a collision.

286IGPC-035-05-15Momentum and impulse Paper EMedium5 marks

An astronaut is at rest in space, 5.0 m from a space station. The astronaut is holding a tool of mass 3.0 kg. The mass of the astronaut and spacesuit, not including the tool, is 90 kg. To return to the station, the astronaut throws the tool directly away from the station at 6.0 m/s, measured relative to the station.

(a)

Explain why the astronaut moves towards the space station.

(2)
(b)

Calculate the speed of the astronaut after the throw.

(2)
(c)

The astronaut moves at this constant speed. Calculate the time the astronaut takes to reach the station.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
There is no external force, so momentum is conserved and the total momentum is zero both before and after the throw✓ 1
The tool gains momentum away from the station, so the astronaut gains an equal momentum in the opposite direction, towards the station✓ 1
Part (b)
90 × v = 3.0 × 6.0✓ 1
v = 0.20 m/s✓ 1
Part (c)
t = 5.0 ÷ 0.20 = 25 s✓ 1

Exam tip: In a push-apart or throw that starts from rest, state that the total momentum is zero both before and after. Then equate the sizes of the two momenta.

287IGPC-035-01-13Speed, velocity and motion graphs Paper EMedium7 marks

A lift (elevator) moves upwards from the ground floor of a building.
• It accelerates uniformly from rest to 2.0 m/s in 1.5 s.
• It then travels at a constant 2.0 m/s for 6.0 s.
• Finally it decelerates uniformly to rest in 2.5 s.

1.57.5102time / sspeed / m/s
Graph sketched from the values in the question; not to scale
(a)

Calculate the total distance travelled by the lift.

(3)
(b)

Calculate the deceleration of the lift in the final stage.

(2)
(c)

Calculate the average speed of the lift for the whole journey.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
uses area under speed–time graph; e.g. accelerating section ½ × 1.5 × 2.0 = 1.5 m✓ 1
constant-speed section 2.0 × 6.0 = 12 m and decelerating section ½ × 2.5 × 2.0 = 2.5 m✓ 1
total = 16 m✓ 1
Part (b)
deceleration = change in speed ÷ time = 2.0 ÷ 2.5✓ 1
= 0.80 m/s2 (accept acceleration = −0.80 m/s2)✓ 1
Part (c)
total time = 10.0 s; average speed = total distance ÷ total time = 16 ÷ 10.0 (allow their distance from the first part)✓ 1
= 1.6 m/s✓ 1

Exam tip: Split the area under a speed–time graph into triangles and rectangles. Calculate each one separately, then add them.

288IGPC-035-04-15Turning effect and centre of gravity Paper EMedium7 marks

A uniform metre rule of weight 1.2 N is balanced on a pivot at its 50 cm mark. A 0.50 N weight hangs at the 15 cm mark and a 0.30 N weight hangs at the 30 cm mark. An unknown weight W hangs at the 80 cm mark and the rule is horizontal and balanced.

0.50 N0.30 N1.2 NW15 cm30 cm50 cm80 cm
Diagram not to scale
(a)

State the principle of moments.

(1)
(b)

Calculate W.

(3)
(c)

Explain why the weight of the rule does not need to be included when taking moments about the pivot to calculate W.

(1)
(d)

Calculate the upward force exerted by the pivot on the rule.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
For an object in equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments about that point✓ 1
Part (b)
Anticlockwise moments about the pivot = 0.50 × 0.35 + 0.30 × 0.20 = 0.235 N m✓ 1
W × 0.30 = 0.235✓ 1
W = 0.78 N (accept 0.783 N)✓ 1
Part (c)
The rule is uniform, so its weight acts at the 50 cm mark. This is the pivot, so the weight is at zero distance from the pivot and has no moment✓ 1
Part (d)
There is no resultant force, so the upward force equals the total downward force: 0.50 + 0.30 + 0.78 + 1.2✓ 1
= 2.8 N✓ 1

Common error: The pivot force must also support the weight of the rule itself. Leaving out the 1.2 N gives 1.6 N instead of 2.8 N.

289IGPC-035-06-13Energy, work and power Paper EMedium6 marks

A diver of mass 55 kg steps off a platform 10 m above the surface of a swimming pool. She starts from rest. The gravitational field strength g is 9.8 N/kg.

pooldiver10 m
Diagram not to scale
(a)

Calculate the decrease in the gravitational potential energy of the diver as she falls from the platform to the water surface.

(2)
(b)

Air resistance is negligible. Calculate the speed of the diver as she reaches the water.

(3)
(c)

In practice, the diver's speed as she reaches the water is slightly less than the value you calculated in the previous part. Suggest why.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
ΔEp = mgΔh = 55 × 9.8 × 10✓ 1
= 5390 J (accept 5400 J)✓ 1
Part (b)
Gain in Ek = loss in Ep, so ½mv2 = 5390 J (or ½v2 = gh)✓ 1
v2 = 2 × 5390 ÷ 55 = 196 m2/s2 (or v2 = 2 × 9.8 × 10 = 196)✓ 1
v = 14 m/s✓ 1
Part (c)
Air resistance acts on her and does work against her motion, so some energy is transferred as thermal energy to the air and her surroundings rather than to her kinetic store✓ 1

Exam tip: Write down the energy equation (loss of Ep = gain of Ek) before substituting numbers. This earns credit even if a later step goes wrong.

290IGPC3-1-01c-07Pressure in solids and liquids Paper EMedium1 mark

A scuba diver is 20 m below the surface of a lake. The density of the water is 1000 kg/m3 and g = 9.8 N/kg. What is the pressure on the diver due to the water alone?

water surfacediver20 m
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Correct: B. p = ρgh = 1000 × 9.8 × 20 = 196 000 Pa = 196 kPa. A is wrong: this leaves out g, giving ρh = 20 000 Pa. C is wrong: this adds atmospheric pressure of about 100 kPa, which the question excludes. D is wrong: this halves the answer, treating the pressure as the average between the surface and the diver's depth.✓ 1Answer B

Answer: B

Exam tip: Pressure from a liquid depends on density, g and depth, not on the shape or volume of the lake.

291IGPC3-1-01c-11Momentum and impulse Paper EMedium7 marks

A goalkeeper catches a football of mass 0.43 kg moving horizontally at 18 m/s. The ball is brought to rest in 0.045 s.

(a)

Calculate the momentum of the ball just before it is caught.

(2)
(b)

Calculate the average force exerted by the goalkeeper's gloves on the ball.

(3)
(c)

Explain why wearing gloves that increase the time taken to stop the ball reduces the force on the goalkeeper's hands.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
momentum = mass × velocity✓ 1
0.43 × 18 = 7.74 kg m/s✓ 1
Part (b)
force = change in momentum ÷ time✓ 1
7.74 ÷ 0.045✓ 1
172 N✓ 1
Part (c)
the change in momentum is the same✓ 1
a longer time means a smaller rate of change of momentum, so a smaller force✓ 1

Exam tip: Force equals the rate of change of momentum, so for a fixed change in momentum a longer stopping time means a smaller force.

292IGPC3-1-01c-13Forces, friction and springs Paper EMedium7 marks

A student investigates a spring of original length 10.0 cm by adding loads and measuring the length. Results: 0 N, 10.0 cm; 1.0 N, 12.5 cm; 2.0 N, 15.0 cm; 3.0 N, 17.5 cm; 4.0 N, 20.0 cm; 5.0 N, 28.0 cm.

(a)

Calculate the extension of the spring when the load is 3.0 N.

(2)
(b)

Between which two loads does the spring pass its limit of proportionality? Give a reason for your answer.

(2)
(c)

Calculate the spring constant of the spring. Give the unit.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
extension = 17.5 − 10.0✓ 1
= 7.5 cm✓ 1
Part (b)
between 4.0 N and 5.0 N✓ 1
the extension up to 4.0 N is proportional to load (2.5 cm per newton), but at 5.0 N it is 18.0 cm instead of the expected 12.5 cm✓ 1
Part (c)
use a load and extension in the proportional region, e.g. 2.0 N and 5.0 cm = 0.050 m✓ 1
k = force ÷ extension = 2.0 ÷ 0.050✓ 1
40 N/m✓ 1

Exam tip: Convert extension to metres before calculating the spring constant in N/m.

293IGPC3-1-06r5-05Momentum and impulse Paper EMedium1 mark

A sled of mass 24 kg slides at 5.0 m/s on a smooth surface. It picks up a parcel of mass 16 kg that was at rest, and they move off together. What is their speed?

sled 24 kgparcel 16 kg5.0 m/sat rest
Before the collision
Show mark scheme
Marking pointMarkNotes
Momentum is conserved: 24 × 5.0 = (24 + 16) × v, so v = 120 ÷ 40 = 3.0 m/s. 5.0 m/s ignores the added mass. 2.0 m/s uses the parcel's mass in place of the sled's. 2.5 m/s halves the speed without using the masses.✓ 1Answer B

Answer: B

Exam tip: Write total momentum before and after the interaction, then solve for the unknown speed.

294IGPC3-1-06r5-06Pressure in solids and liquids Paper EMedium1 mark

A sea lion dives to a depth of 30 m in sea water. Take the density of sea water as 1000 kg/m3 and g = 9.8 N/kg. What is the pressure due to the water at this depth?

water surfacesea lion30 m
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
p = ρgh = 1000 × 9.8 × 30 = 294 000 Pa = 294 kPa. 0.294 kPa leaves out the density. 30 kPa leaves out g. 394 kPa adds atmospheric pressure, which the question did not ask for.✓ 1Answer A

Answer: A

Common error: The pressure due to a liquid is ρgh. Atmospheric pressure is added only when the total pressure is requested.

295IGPC3-1-06r5-11Momentum and impulse Paper EMedium8 marks

A marine biologist of mass 60 kg stands on a stationary research raft of mass 240 kg floating on calm water and jumps horizontally off it at 2.0 m/s. Ignore the water's resistance.

(a)

State the principle of conservation of momentum.

(2)
(b)

Calculate the momentum of the biologist just after leaving the raft.

(1)
(c)

Calculate the speed at which the raft moves backwards.

(2)
(d)

Calculate the total kinetic energy gained by the biologist and the raft.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
the total momentum of a system before an interaction equals the total momentum after✓ 1
provided no external force acts on the system✓ 1
Part (b)
60 × 2.0 = 120 kg m/s✓ 1
Part (c)
momentum of raft = 120 kg m/s in the opposite direction✓ 1
speed = 120 ÷ 240 = 0.50 m/s✓ 1
Part (d)
KE of biologist = ½ × 60 × 2.02 = 120 J✓ 1
KE of raft = ½ × 240 × 0.502 = 30 J✓ 1
total = 150 J✓ 1

Exam tip: In recoil questions the two momenta are equal in size but opposite in direction, while the kinetic energies are not equal.

296IGPC3-1-06r7-05Momentum and impulse Paper EMedium1 mark

A 25 kg trolley in a warehouse moves at 3.0 m/s. A constant force of 150 N acts against its motion until it stops. For how long does the force act?

Show mark scheme
Marking pointMarkNotes
B is correct: change of momentum = 25 × 3.0 = 75 kg m/s, and t = 75 / 150 = 0.50 s. A is wrong because it divides speed by force (3.0 / 150). C is wrong because it uses kinetic energy (112.5 J) in place of momentum. D is wrong because it inverts the equation (150 / 75).✓ 1Answer B

Answer: B

Exam tip: Force equals change of momentum divided by time, so time equals change of momentum divided by force.

297IGPC3-1-06r7-06Pressure in solids and liquids Paper EMedium1 mark

A sealed sample bottle is lowered 25 m below the surface of the sea. Take the density of water to be 1000 kg/m³ and g = 9.8 N/kg. What is the increase in pressure on the bottle compared with the surface?

water surfacebottle25 m
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
C is correct: Δp = ρgh = 1000 × 9.8 × 25 = 245 000 Pa = 245 kPa. A is wrong because it omits the density (9.8 × 25 = 245 Pa). B is wrong because it omits g (1000 × 25 = 25 000 Pa). D is wrong because it adds atmospheric pressure, but the question asks only for the increase.✓ 1Answer C

Answer: C

Common error: Read whether the question asks for the pressure due to the water or the total pressure including the atmosphere.

298IGPC3-1-06r7-14Momentum and impulse Paper EMedium6 marks

A 60 kg box slides along a conveyor at 4.0 m/s and is brought to rest by a rubber buffer in 0.20 s.

(a)

Calculate the change in momentum of the box.

(2)
(b)

Calculate the average force exerted by the buffer on the box.

(2)
(c)

Explain why a softer buffer that stops the box more slowly exerts a smaller force.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Δp = mv = 60 × 4.0✓ 1
240 kg m/s✓ 1
Part (b)
F = Δp / t = 240 / 0.20 (ECF from (a))✓ 1
1200 N✓ 1
Part (c)
The change in momentum is the same✓ 1
F = Δp / t, so a longer time gives a smaller force✓ 1

Exam tip: Use impulse: force multiplied by time equals change in momentum.

299IGPC3-1-06r7-15Pressure in solids and liquids Paper EMedium6 marks

A submersible has a flat window of area 0.020 m² at a depth of 120 m in seawater of density 1030 kg/m³. Use g = 9.8 N/kg.

(a)

Calculate the pressure due to the water at the window. Give your answer to 2 significant figures.

(2)
(b)

Use a pressure of 1.2×106 Pa to calculate the force of the water on the window. Give your answer in newtons.

(2)
(c)

A second window of the same area is at a depth of 240 m. State and explain how the force on it compares with the force on the first window.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
p = ρgh = 1030 × 9.8 × 120✓ 1
1.2×106 Pa✓ 1
Part (b)
F = pA✓ 1
F = 1.2×106 × 0.020 = 24000 N✓ 1
Part (c)
The force is twice as large✓ 1
The pressure is proportional to depth so doubling the depth doubles the pressure (and the force)✓ 1

Common error: Do not use the rounded pressure in a later step unless the question says to; here the question tells you which value to use.

300IGPC3-1-07r8-04Pressure in solids and liquids Paper EMedium1 mark

A diver descends in a freshwater lake from a depth of 5.0 m to a depth of 10 m. Ignoring the atmosphere, what happens to the pressure due to the water on the diver?

water surfacediver5.0 mdiver10 m
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Pressure due to a liquid is p = ρgh, so it is directly proportional to depth when ρ and g are constant. Doubling the depth doubles the pressure, so C. A treats pressure as inversely related to depth. B assumes constant density means constant pressure. D squares the depth ratio.✓ 1Answer C

Answer: C

Common error: Liquid pressure depends on depth, density and g, not on the shape of the lake or the volume of water around the diver.

301IGPC3-1-07r8-05Momentum and impulse Paper EMedium1 mark

In a drum test, a drumstick of mass 0.050 kg strikes a drumhead at 6.0 m/s and is brought to rest in 0.020 s. What is the average force exerted on the stick by the drumhead?

Show mark scheme
Marking pointMarkNotes
Force = change in momentum ÷ time = (0.050 × 6.0) ÷ 0.020 = 15 N, so C. A multiplies the momentum by the time instead of dividing. B is the change in momentum (0.30 kg m/s) with the division by time forgotten. D uses v ÷ t and leaves out the mass.✓ 1Answer C

Answer: C

Exam tip: Find the change in momentum first, then divide by the time taken; check that the units of your answer are newtons.

302IGPC3-1-07r8-13Momentum and impulse Paper EMedium5 marks

An aircraft flies at 80 m/s. A bird of mass 1.5 kg, effectively at rest, hits the windscreen and comes to rest relative to the aircraft in 0.0040 s.

(a)

Calculate the change in momentum of the bird relative to the aircraft.

(2)
(b)

Calculate the average force on the bird.

(2)
(c)

Suggest how the force on the bird would change if the collision lasted longer, with the same change in momentum.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
change in momentum = mass × change in velocity = 1.5 × 80✓ 1
120 kg m/s✓ 1
Part (b)
force = change in momentum ÷ time = 120 ÷ 0.0040 (ECF from (a))✓ 1
30000 N✓ 1
Part (c)
force would be smaller (force is inversely proportional to time for the same momentum change)✓ 1

Exam tip: Impulse, force × time, equals the change in momentum, so a longer stopping time means a smaller average force.

303IGPC3-1-05r15-05Momentum and impulse Paper EMedium1 mark

In an air-hockey game a puck of mass 0.30 kg moving at 1.5 m/s hits a stationary puck of mass 0.60 kg. The two pucks stick together. What is their speed immediately after the collision?

0.30 kg0.60 kg1.5 m/sat rest
Before the collision
Show mark scheme
Marking pointMarkNotes
Momentum is conserved: 0.30 × 1.5 = 0.45 kg m/s = (0.30 + 0.60) × v, so v = 0.50 m/s and D is correct. A assumes the speed is unchanged. B divides by only the mass of the second puck (0.60). C is the initial momentum, not divided by the total mass.✓ 1Answer D

Answer: D

Exam tip: After pucks stick together, divide the total momentum by the combined mass.

304IGPC3-1-05r15-10Speed, velocity and motion graphs Paper EMedium8 marks

In a racing game a car starts from rest. Its speed increases uniformly to 24 m/s in 6.0 s. It then travels at constant speed for 10 s, and finally slows down uniformly to rest in 8.0 s.

6162424time / sspeed / m/s
Graph sketched from the values in the question; not to scale
(a)

Calculate the acceleration of the car during the first 6.0 s.

(2)
(b)

Calculate the distance travelled at constant speed.

(2)
(c)

Describe the shape of the speed–time graph during the final 8.0 s.

(1)
(d)

Calculate the total distance travelled by the car, using the area under the speed–time graph.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
a = Δv/t = 24 ÷ 6.0✓ 1
4.0 m/s²✓ 1
Part (b)
distance = speed × time = 24 × 10✓ 1
240 m✓ 1
Part (c)
straight line sloping downwards from 24 m/s to zero✓ 1
Part (d)
area of first stage ½ × 6.0 × 24 = 72 m and last stage ½ × 8.0 × 24 = 96 m✓ 1
total = 72 + 240 + 96 , ECF from (b)✓ 1
408 m✓ 1

Exam tip: For distance from a speed–time graph, split the graph into triangles and rectangles and add the areas.

305IGPC3-1-06r18-10Speed, velocity and motion graphs Paper EMedium7 marks

A conveyor belt carries green coffee beans into a roasting plant. Its speed–time graph has three straight sections. The speed increases uniformly from 0 to 0.60 m/s in 5.0 s, stays constant at 0.60 m/s for 20 s, then decreases uniformly to 0 in 10 s.

525350.6time / sspeed / m/s
Graph sketched from the values in the question; not to scale
(a)

Calculate the acceleration of the belt during the first 5.0 s.

(2)
(b)

Calculate the magnitude of the deceleration during the final 10 s.

(2)
(c)

Calculate the total distance moved by the belt during the 35 s.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 acceleration = change in speed ÷ time = 0.60 / 5.0✓ 1
A1 0.12 m/s2✓ 1
Part (b)
M1 0.60 / 10✓ 1
A1 0.060 m/s2✓ 1
Part (c)
M1 distance = area under the speed–time graph✓ 1
M1 area of the two triangles ½ × 5.0 × 0.60 = 1.5 m and ½ × 10 × 0.60 = 3.0 m, and rectangle 0.60 × 20 = 12 m✓ 1
A1 16.5 m✓ 1

Common error: Distance on a speed–time graph is the area under the line, not the gradient. Split a trapezium-shaped graph into triangles and rectangles.

306IGPC3-1-06r18-14Pressure in solids and liquids Paper EMedium6 marks

A cold-brew coffee plant stores water in a tank. The water has density 1000 kg/m3 and is 1.5 m deep. A circular valve in the base of the tank has an area of 0.020 m2. Ignore atmospheric pressure. (g = 9.8 N/kg)

1.5 mwatervalve, area 0.020 m²
Diagram not to scale
(a)

Calculate the pressure due to the water at the base of the tank.

(2)
(b)

Calculate the force of the water on the valve.

(2)
(c)

The tank is emptied and refilled with water until the pressure at the base is 24 500 Pa. Calculate the new depth of water.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 p = ρgh = 1000 × 9.8 × 1.5✓ 1
A1 14 700 Pa✓ 1
Part (b)
M1 force = pressure × area = 14 700 × 0.020✓ 1
A1 294 N (ECF from (a))✓ 1
Part (c)
M1 h = p ÷ (ρg) = 24 500 / (1000 × 9.8)✓ 1
A1 2.5 m✓ 1

Exam tip: In liquid pressure calculations, depth is the vertical height of the liquid above the point, not the volume or the slant length. Rearrange p = ρgh to find h.

307IGPC3-1-07r20-02Momentum and impulse Paper EMedium1 mark

Two scenery trolleys on a flat stage floor are pushed together. Trolley A has mass 3.0 kg and moves at 4.0 m/s towards trolley B, which has mass 1.0 kg and is at rest. The trolleys couple together on collision. What is their speed immediately afterwards?

A 3.0 kgB 1.0 kg4.0 m/sat rest
Before the collision
Show mark scheme
Marking pointMarkNotes
Momentum before = 3.0 × 4.0 = 12 kg m/s. After the collision the mass is 4.0 kg, so v = 12 ÷ 4.0 = 3.0 m/s, so B is correct. A is the mean of 4.0 and 0. C assumes A keeps its speed. D gives the momentum, not the speed.✓ 1Answer B

Answer: B

Common error: After a coupling collision the total mass is the sum of both masses; divide the total momentum by that, not by one mass.

308IGPC3-1-07r20-08Momentum and impulse Paper EMedium1 mark

A volcanic bomb of mass 2.0 kg lands on soft ash at 20 m/s and is brought to rest in 0.080 s. What is the average resultant force on the bomb while it stops?

Show mark scheme
Marking pointMarkNotes
Force = change in momentum ÷ time = (2.0 × 20) ÷ 0.080 = 500 N, so C is correct. A multiplies momentum by time. B is the momentum change (impulse) in N s, not a force. D divides the kinetic energy (400 J) by the time.✓ 1Answer C

Answer: C

Exam tip: Force = change in momentum ÷ time; the change in momentum must be divided by the stopping time.

309IGPC3-1-07r20-11Momentum and impulse Paper EMedium7 marks

A barge A of mass 6000 kg moves at 1.5 m/s along a straight canal towards a stationary barge B of mass 9000 kg. The barges lock together on collision and move off as one. Ignore resistive forces from the water.

A 6000 kgB 9000 kg1.5 m/sat rest
Before the collision
(a)

State the principle of conservation of momentum.

(2)
(b)

Calculate the speed of the barges immediately after they lock together.

(3)
(c)

The barges take 4.5 s to lock together. Use your unrounded or rounded answer to (b) (both give the same result) to calculate the average force on barge B during this time.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 the total momentum of a system of objects before a collision equals the total momentum after✓ 1
B1 provided no external (resultant) force acts on the system✓ 1
Part (b)
B1 momentum before = 6000 × 1.5 = 9000 kg m/s✓ 1
M1 9000 = (6000 + 9000) × v✓ 1
A1 v = 0.60 m/s✓ 1
Part (c)
M1 force = change in momentum of B ÷ time = (9000 × 0.60) ÷ 4.5 (ECF from (b))✓ 1
A1 1200 N✓ 1

Exam tip: Apply the force equation to one object only: use that object's own mass and its own change in velocity.

310IGPC3-1-08r20-08Speed, velocity and motion graphs Paper EMedium1 mark

A delivery robot moves 30 m east and then 40 m north in a total time of 50 s. What is the magnitude of its average velocity?

start30 m east40 m northN
Diagram not to scale (not a scale drawing of the path)
Show mark scheme
Marking pointMarkNotes
The displacement is √(302 + 402) = 50 m, so average velocity = 50 ÷ 50 = 1.0 m/s, so C is correct. A uses only the eastward distance. B uses only the northward distance. D is the average speed (70 ÷ 50), which uses total distance.✓ 1Answer C

Answer: C

Common error: Average velocity uses displacement (straight-line distance and direction), not the total distance travelled.

311IGPC3-1-08r22-06Momentum and impulse Paper EMedium1 mark

A ball of mass 0.40 kg hits a wall at 6.0 m/s and rebounds along the same line at 4.0 m/s. What is the magnitude of the change in momentum of the ball?

Show mark scheme
Marking pointMarkNotes
Momentum is a vector, so the direction reverses: change = 0.40 × (6.0 − (−4.0)) = 0.40 × 10 = 4.0 kg m/s. 0.80 kg m/s subtracts the speeds (6.0 − 4.0) and so ignores the reversal. 1.6 kg m/s is only the momentum after the rebound. 2.4 kg m/s is only the momentum before.✓ 1Answer C

Answer: C

Common error: When an object rebounds, add the magnitudes of the speeds because the velocity changes sign.

312IGPC3-1-08r22-07Momentum and impulse Paper EMedium1 mark

On an archaeological site a 3.0 kg survey trolley starts from rest on smooth rails. A constant horizontal force of 12 N acts on it for 2.5 s. Friction is negligible. What is the speed of the trolley at the end of this time?

Show mark scheme
Marking pointMarkNotes
Impulse = force × time = 12 × 2.5 = 30 N s, and this equals the change in momentum, so v = 30 ÷ 3.0 = 10 m/s. 30 m/s forgets to divide by the mass. 4.0 m/s uses force ÷ mass, the acceleration, and ignores the time. 0.10 m/s divides mass by impulse.✓ 1Answer D

Answer: D

Exam tip: Impulse equals force × time and also equals change in momentum, mass × change in velocity.

313IGPC3-1-08r22-14Momentum and impulse Paper EMedium7 marks

In a data centre a maintenance robot of mass 40 kg moves at 1.5 m/s towards a stationary trolley of mass 20 kg. The two collide and couple together, moving off in the same direction. Ignore friction.

40 kg1.5 m/s20 kgat rest
Diagram not to scale
(a)

Calculate the momentum of the robot before the collision.

(1)
(b)

Calculate the speed of the robot and trolley immediately after the collision.

(2)
(c)

Calculate the kinetic energy lost in the collision.

(3)
(d)

State what happens to the lost kinetic energy.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
40 × 1.5 = 60 kg m/s✓ 1
Part (b)
momentum is conserved, so 60 = (40 + 20) × v✓ 1
v = 1.0 m/s✓ 1
Part (c)
kinetic energy before = ½ × 40 × 1.52 = 45 J✓ 1
kinetic energy after = ½ × 60 × 1.02 = 30 J✓ 1
loss = 15 J✓ 1
ECF from (b)—
Part (d)
transferred to thermal energy and sound✓ 1

Common error: Momentum is conserved in a collision but kinetic energy usually is not.

314IGPC3-1-07r25-04Pressure in solids and liquids Paper EMedium1 mark

A saline drip bag is hung above a patient's arm. The saline has density 1020 kg/m3. The pressure due to the saline at the vein must be 4.0 kPa. The gravitational field strength is 9.8 N/kg. What height of the saline surface above the vein gives this pressure?

salineneedlepatient's armh = ?
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Pressure due to a liquid column p = ρgh, so h = p ÷ (ρg) = 4000 ÷ (1020 × 9.8) = 0.40 m. The value 3.9 m leaves out g. The value 38 m multiplies by g instead of dividing by it. The value 4.0 × 10−4 m treats 4.0 kPa as 4.0 Pa instead of 4000 Pa.✓ 1Answer C

Answer: C

Exam tip: Rearrange p = ρgh before substituting, and convert kPa to Pa first.

315IGPC3-1-07r25-05Momentum and impulse Paper EMedium1 mark

A Formula 1 car of mass 800 kg slows from 70 m/s to 30 m/s in a time of 4.0 s. What is the average resultant force acting on the car?

Show mark scheme
Marking pointMarkNotes
Force = change in momentum ÷ time = 800 × (70 − 30) ÷ 4.0 = 8000 N. The value 6000 N uses only the final velocity. The value 14 000 N uses only the initial velocity. The value 10 N is the deceleration in m/s2 and has not been multiplied by the mass.✓ 1Answer D

Answer: D

Common error: Use the change in velocity, not either velocity on its own, when finding a change in momentum.

316IGPC3-1-07r25-16Pressure in solids and liquids Paper EMedium7 marks

A honey storage tank is filled to a depth of 1.5 m with honey of density 1400 kg/m3. Use g = 9.8 N/kg.

honey1.5 mtap cover, area 4.0 cm²
Diagram not to scale
(a)

Calculate the pressure due to the honey at the base of the tank. Give your answer to 3 significant figures.

(2)
(b)

A circular tap cover of area 4.0 cm2 is fitted at the base of the tank. Calculate the force of the honey on the cover. Give your answer to 2 significant figures.

(3)
(c)

Explain why the pressure halfway up the tank is less than the pressure at the base.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
p = ρgh = 1400 × 9.8 × 1.5✓ 1
2.06 × 104 Pa (accept 20 600 Pa)✓ 1
Part (b)
area = 4.0 × 10−4 m2✓ 1
F = pA = 20 580 × 4.0 × 10−4 (ECF from (a); using 20 600 also gives the same answer to 2 s.f.)✓ 1
8.2 N✓ 1
Part (c)
There is less depth (and so less liquid above that point)✓ 1
pressure = ρgh, so pressure is proportional to the depth✓ 1

Exam tip: State that pressure in a liquid depends on depth, density and g. The width of the container does not matter.

317IGPC3-1-10r27-05Momentum and impulse Paper EMedium1 mark

A space probe of mass 600 kg moves at 2.0 m/s and docks with a stationary module of mass 200 kg. The two stay joined. What is their common velocity?

600 kg2.0 m/s200 kgat rest
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Momentum is conserved: 600 × 2.0 = (600 + 200) × v, so v = 1200 / 800 = 1.5 m/s. 1.0 m/s is the mean of 2.0 and 0, which ignores the masses. 2.0 m/s assumes the probe's velocity is unchanged. 6.0 m/s divides by the module's mass only.✓ 1Answer A

Answer: A

Exam tip: After joining, the total mass moves together, so divide by the sum of the masses.

318IGPC3-1-10r27-06Pressure in solids and liquids Paper EMedium1 mark

A diver inspects a bridge pier in a river at a depth of 5.0 m. The density of the water is 1000 kg/m3 and g = 9.8 N/kg. What is the pressure due to the water at this depth?

surfacediver5.0 m
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Pressure = ρgh = 1000 × 9.8 × 5.0 = 49 000 Pa = 4.9 × 104 Pa. 5.0 × 103 Pa leaves out g (ρ × h). 2.0 × 103 Pa divides instead of multiplying (ρg / h). 4.9 × 103 Pa is a factor of ten too small, from a slip with the density.✓ 1Answer C

Answer: C

Exam tip: For liquid pressure use p = ρgh with density in kg/m³ and depth in m.

319IGPC3-1-10r27-08Momentum and impulse Paper EMedium1 mark

A thruster on a space probe of mass 800 kg gives a constant force of 0.40 N for 50 s. No other forces act. What is the change in speed of the probe?

Show mark scheme
Marking pointMarkNotes
Impulse = F t = 0.40 × 50 = 20 N s = change in momentum, so change in speed = 20 / 800 = 0.025 m/s. 5.0 × 10−4 m/s ignores the time (0.40 / 800). 20 m/s is the impulse itself, forgetting to divide by the mass. 1.6 × 104 m/s multiplies by the mass instead of dividing.✓ 1Answer D

Answer: D

Exam tip: Change in momentum = force × time, then divide by the mass to get the change in velocity.

320IGPC3-1-10r27-13Momentum and impulse Paper EMedium7 marks

A space probe of mass 900 kg moves at 5.0 m/s and catches up with a drifting satellite of mass 600 kg moving in the same direction at 2.0 m/s. The two couple together.

900 kg5.0 m/s600 kg2.0 m/s
Diagram not to scale
(a)

State the principle of conservation of momentum.

(1)
(b)

Calculate the total momentum before the coupling.

(2)
(c)

Calculate the common velocity after the coupling.

(2)
(d)

The coupling takes 0.50 s. Calculate the average force on the satellite during the coupling. Use the value of velocity from (c).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
in a closed system (no external force) the total momentum before an interaction equals the total momentum after✓ 1
Part (b)
900 × 5.0 + 600 × 2.0✓ 1
5700 kg m/s✓ 1
Part (c)
v = 5700 / (900 + 600)✓ 1
3.8 m/s✓ 1
Part (d)
F = change in momentum / time = 600 × (3.8 − 2.0) / 0.50✓ 1
2160 N✓ 1

Exam tip: For the force on one object, use only that object's change in momentum.

321IGPC3-1-10r28-12Forces, friction and springs Paper EMedium7 marks

A museum hangs fossil specimens from a spring balance. A student tests the spring by adding loads and measuring its extension.

Load / N02.04.06.08.010.0
Extension / cm01.63.24.86.49.5

(a)

State the largest load in the table for which the spring obeys Hooke's law.

(1)
(b)

Calculate the spring constant of the spring. Give your answer in N/m.

(3)
(c)

A fossil hung on the spring produces an extension of 5.6 cm. Calculate the mass of the fossil. Use k = 125 N/m and g = 9.8 N/kg and give your answer to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 8.0 N✓ 1
Part (b)
M1 convert 6.4 cm to 0.064 m✓ 1
M1 k = F ÷ x = 8.0 ÷ 0.064✓ 1
A1 125 N/m✓ 1
Part (c)
M1 weight = kx = 125 × 0.056 = 7.0 N✓ 1
M1 mass = weight ÷ g = 7.0 ÷ 9.8✓ 1
A1 0.714 kg✓ 1

Common error: Convert extension from cm to m before using k in N/m.

322IGPC3-1-10r28-14Pressure in solids and liquids Paper EMedium6 marks

A data centre cools its servers by immersing them in a tank of oil of density 850 kg/m3. The oil is 0.80 m deep. (g = 9.8 N/kg) Ignore atmospheric pressure in this question.

oil0.80 mhatch, area 0.30 m²
Diagram not to scale
(a)

Calculate the pressure due to the oil at the bottom of the tank. Give your answer to 3 significant figures.

(2)
(b)

A rectangular hatch of area 0.30 m2 lies flat on the bottom of the tank. Calculate the force on the hatch due to the oil. Use the unrounded pressure and give your answer to 3 significant figures.

(2)
(c)

Explain why the pressure in a liquid increases with depth.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 p = ρgh = 850 × 9.8 × 0.80✓ 1
A1 6660 Pa✓ 1
Part (b)
M1 F = pA = 6664 × 0.30✓ 1
A1 2.00 × 103 N✓ 1
Part (c)
the liquid at greater depth supports the weight of more liquid above it✓ 1
so the force per unit area is larger✓ 1

Exam tip: Pressure in a liquid depends on depth and density, not on the shape of the container.

323IGPC3-1-07r30-02Momentum and impulse Paper EMedium1 mark

A Formula 1 car and driver have a total mass of 800 kg. During a straight-line test the speed increases from 20 m/s to 50 m/s in 3.0 s. What is the average resultant force on the car?

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Marking pointMarkNotes
Force = change in momentum ÷ time = 800 × (50 − 20) / 3.0 = 8000 N. 13 300 N uses only the final momentum (800 × 50) divided by the time. 24 000 N finds the change in momentum but forgets to divide by the time. 72 000 N multiplies by the time instead of dividing.✓ 1Answer B

Answer: B

Exam tip: Resultant force equals the rate of change of momentum, so use the change in speed (final − initial), not the final speed alone.

324IGPC3-1-07r30-10Momentum and impulse Paper EMedium8 marks

A porter pushes a trolley carrying a patient at 1.5 m/s into a stationary, empty hospital bed. The trolley and patient have a total mass of 80 kg. The bed has a mass of 70 kg. The two lock together and move off as one.

80 kg1.5 m/s70 kgat rest
Diagram not to scale
(a)

Calculate the momentum of the trolley and patient before the collision.

(2)
(b)

Calculate the speed of the trolley and bed immediately after the collision.

(2)
(c)

Calculate the kinetic energy lost in the collision.

(3)
(d)

State what happens to the kinetic energy that is lost.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
p = mv✓ 1
= 80 × 1.5 = 120 kg m/s✓ 1
Part (b)
momentum is conserved, so total momentum after = 120 kg m/s✓ 1
v = 120 / (80 + 70) = 0.80 m/s (ECF from (a))✓ 1
Part (c)
KE before = ½ × 80 × 1.52 = 90 J✓ 1
KE after = ½ × 150 × 0.802 = 48 J✓ 1
energy lost = 90 − 48 = 42 J (ECF from (b))✓ 1
Part (d)
transferred to thermal energy / sound / deformation of the beds✓ 1

Common error: Momentum is conserved in a collision when no external force acts, but kinetic energy is not conserved unless the collision is elastic.

325IGPC3-1-08r31-04Momentum and impulse Paper EMedium1 mark

A space probe of mass 800 kg moves at 0.60 m/s and docks with a stationary module of mass 1200 kg. The two then move together. What is their speed?

800 kg0.60 m/s1200 kgat rest
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Momentum is conserved: 800 × 0.60 = (800 + 1200) × v, so v = 480 ÷ 2000 = 0.24 m/s. 0.60 m/s ignores the extra mass. 0.40 m/s divides the initial momentum by the module mass only. 0.36 m/s multiplies the module mass by the speed instead of the probe mass.✓ 1Answer C

Answer: C

Exam tip: After a collision where objects join, the total mass is the sum of both masses.

326IGPC3-1-08r31-05Pressure in solids and liquids Paper EMedium1 mark

A bridge pier is built in a river. At a point 5.0 m below the surface, what is the pressure due to the water alone? (Density of water = 1000 kg/m3, g = 9.8 N/kg.)

Show mark scheme
Marking pointMarkNotes
Pressure = ρgh = 1000 × 9.8 × 5.0 = 49 000 Pa = 49 kPa. 5.0 kPa leaves out g. 149 kPa adds atmospheric pressure, but only the water was asked for. 0.51 kPa comes from dividing by g instead of multiplying.✓ 1Answer B

Answer: B

Common error: Check whether the question asks for pressure due to the liquid alone or the total pressure including the atmosphere.

327IGPC3-1-08r31-12Momentum and impulse Paper EMedium5 marks

A space probe of mass 600 kg is moving at 2.5 m/s. A thruster then gives a constant force of 40 N for 15 s in the direction of motion. Assume the mass of the probe stays constant.

(a)

Calculate the impulse given to the probe.

(2)
(b)

Calculate the change in velocity of the probe.

(2)
(c)

Calculate the final speed of the probe.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 impulse = force × time = 40 × 15✓ 1
A1 600 N s✓ 1
Part (b)
M1 change in velocity = impulse / mass = 600 / 600✓ 1
A1 1.0 m/s (ECF from (a))✓ 1
Part (c)
A1 2.5 + 1.0 = 3.5 m/s (ECF from (b))✓ 1

Exam tip: Impulse equals change in momentum, so divide by the mass to find the change in velocity.

328IGPC3-1-09r32-04Momentum and impulse Paper EMedium1 mark

A thruster on a space probe of mass 400 kg exerts a constant force of 20 N for 50 s. No other forces act and the mass stays constant. What is the change in speed of the probe?

Show mark scheme
Marking pointMarkNotes
Impulse = force × time = 20 × 50 = 1000 N s, and impulse = change in momentum = m Δv, so Δv = 1000 ÷ 400 = 2.5 m/s. 0.40 m/s inverts the calculation (400 ÷ 1000). 1000 m/s forgets to divide by the mass. 0.050 is the acceleration in m/s2 (20 ÷ 400), which is not a change in speed.✓ 1Answer A

Answer: A

Common error: The impulse is not the change in speed: divide the impulse by the mass to get Δv.

329IGPC3-1-09r32-06Pressure in solids and liquids Paper EMedium1 mark

A remotely operated vehicle inspects a bridge pier in a river at a depth of 12 m. The density of the river water is 1000 kg/m3 and g = 9.8 N/kg. What is the pressure due to the water at this depth (not including the atmosphere)?

Show mark scheme
Marking pointMarkNotes
p = ρgh = 1000 × 9.8 × 12 = 1.18 × 105 Pa. 1.22 × 103 divides by g instead of multiplying. 9.8 × 103 omits the depth (ρg only). 1.18 × 102 omits the density (gh only).✓ 1Answer D

Answer: D

Exam tip: Check that all three factors (density, g and depth) appear; the answer for a 12 m depth in water should be of order 10⁵ Pa.

330IGPC3-1-09r32-12Forces, friction and springs Paper EMedium5 marks

A suspension-bridge model uses a steel spring of unstretched length 12.0 cm. A force of 5.0 N stretches the spring by 2.0 cm. The limit of proportionality is not exceeded.

12.0 cm5.0 N14.0 cm8.0 N? cm
Diagram not to scale
(a)

Calculate the spring constant of the spring, in N/m.

(2)
(b)

Calculate the length of the spring when the force is 8.0 N.

(2)
(c)

State what is meant by the limit of proportionality.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 k = F ÷ x = 5.0 ÷ 0.020 (extension in metres)✓ 1
A1 250 N/m✓ 1
Part (b)
M1 extension = 8.0 ÷ 250 = 0.032 m (3.2 cm)✓ 1
A1 length = 12.0 + 3.2 = 15.2 cm✓ 1
Part (c)
the point beyond which the extension is no longer directly proportional to the force applied✓ 1

Common error: Convert the extension to metres before using k = F ÷ x, and add the extension to the original length for the stretched length.

331IGPC3-1-09r32-14Momentum and impulse Paper EMedium7 marks

A space probe of mass 600 kg moves at 3.0 m/s towards a stationary service module of mass 400 kg. They dock and move off together.

600 kg3.0 m/s400 kgat rest
Diagram not to scale
(a)

Calculate the momentum of the probe before docking.

(2)
(b)

Calculate the speed of the combined spacecraft after docking.

(2)
(c)

Calculate the kinetic energy lost in the docking. Use v = 1.8 m/s.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 p = mv = 600 × 3.0✓ 1
A1 1800 kg m/s✓ 1
Part (b)
M1 momentum is conserved so v = 1800 ÷ (600 + 400)✓ 1
A1 1.8 m/s✓ 1
Part (c)
M1 KE before = ½ × 600 × 3.02 = 2700 J✓ 1
M1 KE after = ½ × 1000 × 1.82 = 1620 J✓ 1
A1 energy lost = 1080 J✓ 1

Common error: Momentum is conserved in a docking collision, but kinetic energy is not: the missing energy is transferred to other forms such as thermal energy and sound.

332IGPC3-1-08r38-05Momentum and impulse Paper EMedium1 mark

In a data centre a trolley of mass 40 kg moving at 1.5 m/s collides with a stationary cart of mass 60 kg, and the two stick together. What is their speed immediately after the collision?

40 kg1.5 m/s60 kgat rest
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Momentum is conserved: 40 × 1.5 = 60 kg m/s, which is shared by a total mass of 100 kg, so v = 60 ÷ 100 = 0.60 m/s. A speed of 0.75 would need equal masses, 0.90 comes from multiplying the initial speed by the wrong mass ratio, and 1.5 ignores the extra mass that must be moved.✓ 1Answer A

Answer: A

Common error: After a collision where objects stick together, divide the total momentum by the combined mass.

333IGPC3-1-08r38-09Speed, velocity and motion graphs Paper EMedium6 marks

On a straight road a cyclist accelerates uniformly from rest to 9.0 m/s in 6.0 s, then continues at a constant speed of 9.0 m/s for a further 4.0 s.

6109time / sspeed / m/s
Graph as described; not to scale
(a)

Calculate the acceleration during the first 6.0 s.

(2)
(b)

Calculate the total distance travelled in the 10 s.

(3)
(c)

Calculate the average speed over the 10 s.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 acceleration = change in speed ÷ time✓ 1
A1 1.5 m/s2✓ 1
Part (b)
M1 distance in the first 6.0 s = average speed × time = 4.5 × 6.0 = 27 m (or the area of a triangle)✓ 1
M1 distance at constant speed = 9.0 × 4.0 = 36 m✓ 1
A1 63 m✓ 1
Part (c)
A1 63 ÷ 10 = 6.3 m/s (ECF from (b))✓ 1

Exam tip: Split the motion into stages and find the distance for each stage separately before adding.

334IGPC3-1-08r41-04Pressure in solids and liquids Paper EMedium1 mark

The surface of the water in a storage tower is 25 m above a tap in a house. The density of water is 1000 kg/m3 and g = 9.8 N/kg. What is the pressure due to the water at the tap, ignoring atmospheric pressure?

watertap in house25 m
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Pressure due to a liquid column = ρgh = 1000 × 9.8 × 25 = 245 000 Pa = 2.45 × 10⁵ Pa. 245 Pa leaves out the density. 25 000 Pa leaves out g. 2.45 × 10⁷ Pa uses the height in cm instead of m.✓ 1Answer D

Answer: D

Common error: Convert every quantity to SI units before using p = ρgh; the height must be in metres and the density in kg/m³.

335IGPC3-1-08r41-06Momentum and impulse Paper EMedium1 mark

A trolley of mass 12 kg moves at 1.5 m/s along a pharmacy corridor and collides with a stationary trolley of mass 18 kg. The trolleys stick together. What is their speed just after the collision?

12 kg1.5 m/s18 kgat rest
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Momentum is conserved: 12 × 1.5 = 18 kg m/s = (12 + 18) × v, so v = 0.60 m/s. 0.75 m/s takes the mean of 1.5 and 0. 0.90 m/s multiplies the initial speed by 18/30 instead of 12/30. 1.5 m/s assumes speed is unchanged although mass has increased.✓ 1Answer C

Answer: C

Common error: After trolleys stick together, divide the total momentum by the combined mass, not by the mass of one trolley.

336IGPC3-1-08r41-13Pressure in solids and liquids Paper EMedium5 marks

A water supply tap in a house is 18 m below the surface of the water in a storage tank. The density of water is 1000 kg/m3 and g = 9.8 N/kg. Ignore atmospheric pressure.

watertap18 m
Diagram not to scale
(a)

Calculate the pressure due to the water at the tap. Give your answer to 3 significant figures.

(2)
(b)

The tap contains a circular seal of area 4.0 × 10−4 m2. Calculate the force the water exerts on the seal. Use the unrounded value of the pressure and give your answer to 3 significant figures.

(2)
(c)

A tap on a higher floor, nearer the surface of the water in the tank, has a lower pressure. Explain why.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
p = ρgh = 1000 × 9.8 × 18✓ 1
1.76 × 10⁵ Pa (176 400 Pa)✓ 1
Part (b)
force = pressure × area = 176 400 × 4.0 × 10⁻⁴✓ 1
70.6 N✓ 1
Part (c)
smaller depth below the water surface, so a shorter column of water above the tap✓ 1

Exam tip: Pressure in a liquid depends on the vertical depth below the surface, not on the pipe length or the volume of the tank.

337IGPC3-1-09r44-15Energy, work and power Paper EMedium7 marks

An electric hoist in a hospital lifts a patient of mass 80 kg through a height of 1.5 m. Use g = 9.8 N/kg.

(a)

Calculate the gain in gravitational potential energy of the patient.

(2)
(b)

The hoist motor takes 1960 J of electrical energy from the supply. Calculate the efficiency of the hoist.

(2)
(c)

State what happens to the energy that is not usefully transferred.

(1)
(d)

The lift takes 6.0 s. Calculate the useful power output of the hoist. Use your answer to (a).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 ΔE = mgΔh = 80 × 9.8 × 1.5✓ 1
A1 1176 J✓ 1
Part (b)
M1 efficiency = useful energy output ÷ total energy input = 1176 ÷ 1960 (ECF from (a))✓ 1
A1 0.60 or 60%✓ 1
Part (c)
B1 it is dissipated as thermal energy (heating of the motor and moving parts by friction) and sound to the surroundings✓ 1
Part (d)
M1 P = E ÷ t = 1176 ÷ 6.0✓ 1
A1 196 W✓ 1

Common error: Efficiency compares useful output with total input, so it can never be more than 100%. Do not use the wasted energy in the ratio.

338IGPC3-1-09r45-04Momentum and impulse Paper EMedium1 mark

A 3.0 kg hospital supply trolley moving at 4.0 m/s collides with a stationary 1.0 kg trolley. The two trolleys lock together. What is their speed immediately after the collision?

3.0 kg4.0 m/s1.0 kgat rest
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Momentum is conserved: 3.0 × 4.0 = (3.0 + 1.0) × v, so v = 12 ÷ 4.0 = 3.0 m/s. 4.0 m/s ignores the extra mass that must now be moved. 2.0 m/s averages the two initial speeds (4.0 and 0). 12 m/s is the initial momentum, not divided by the combined mass.✓ 1Answer C

Answer: C

Common error: Use the combined mass after the collision when the objects stick together.

339IGPC3-1-09r45-11Momentum and impulse Paper EMedium7 marks

A porter stops a hospital bed that is moving along a corridor. The bed and patient have a total mass of 90 kg and are moving at 1.5 m/s.

(a)

Calculate the momentum of the bed and patient.

(2)
(b)

The porter brings the bed to rest in 1.2 s. Calculate the average force the porter applies.

(2)
(c)

Calculate the average force needed if the bed is brought to rest from the same speed in 3.0 s.

(2)
(d)

Suggest why a longer stopping time is safer for a patient on the bed.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 momentum = mass × velocity✓ 1
A1 90 × 1.5 = 135 kg m/s✓ 1
Part (b)
M1 force = change in momentum ÷ time✓ 1
A1 135 ÷ 1.2 = 112.5 N (accept 113 N)✓ 1
Part (c)
M1 force = 135 ÷ 3.0✓ 1
A1 45 N✓ 1
Part (d)
B1 the force on the patient (and bed) is smaller, so there is less risk of injury or discomfort✓ 1

Exam tip: The same change in momentum over a longer time needs a smaller average force: F = Δp ÷ t.

340IGPC3-1-09r45-14Pressure in solids and liquids Paper EMedium7 marks

A hospital bed is raised by a hydraulic system. A nurse pushes on a small piston of area 4.0 × 10−3 m2 with a force of 60 N. The liquid transmits the pressure to a large piston of area 0.12 m2 under the bed.

60 Nforce = ?small piston: 4.0 × 10⁻³ m²large piston: 0.12 m²large piston
Diagram not to scale
(a)

Calculate the pressure that the nurse produces in the liquid.

(2)
(b)

State the pressure acting on the large piston and calculate the force that the liquid exerts on it.

(3)
(c)

Explain why the hydraulic system uses a liquid and not a gas.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 pressure = force ÷ area = 60 ÷ 0.0040✓ 1
A1 15 000 Pa✓ 1
Part (b)
B1 15 000 Pa (the same pressure is transmitted through the liquid)✓ 1
M1 force = pressure × area✓ 1
A1 15 000 × 0.12 = 1800 N✓ 1
Part (c)
B1 a liquid is (almost) incompressible, whereas a gas is easily compressed✓ 1
B1 so the pressure is transmitted at once and the large piston moves, rather than the gas just being squashed✓ 1

Exam tip: In a hydraulic system the pressure is the same everywhere in the liquid; a larger area therefore gives a larger force.

341IGPC-035-01-08Speed, velocity and motion graphs Paper EHard1 mark

A skydiver jumps from a stationary helicopter and falls for a long time before opening her parachute.
Which description best matches her speed–time graph from the moment she jumps until just before the parachute opens?

Show mark scheme
Marking pointMarkNotes
At first the only significant force is her weight, so she accelerates at about 9.8 m/s² and the graph starts steep. As her speed increases, air resistance increases. This reduces the resultant force, so her acceleration falls and the gradient decreases. When air resistance equals her weight, the resultant force is zero and she falls at a constant terminal velocity, shown by a horizontal line. Option A would only apply with no air resistance. Option B wrongly assumes the acceleration grows. Option D would need an upward resultant force large enough to stop her, which does not happen before the parachute opens.✓ 1Answer C

Answer: C

Exam tip: Link each change in gradient to a change in resultant force. Larger air resistance means a smaller resultant force, which means a smaller acceleration.

342IGPC-035-03-08Forces, friction and springs Paper EHard1 mark

A spring extends by 4.0 cm when a load of 6.0 N hangs from it alone. Two of these identical springs are joined end to end, one below the other, and a load of 9.0 N is hung from the bottom. The springs obey Hooke's law and their weights can be ignored. What is the total extension?

9.0 Nspring 1spring 2
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Each spring carries the full 9.0 N load. One spring extends 4.0 cm per 6.0 N, so under 9.0 N it extends 9.0 × 4.0/6.0 = 6.0 cm. There are two springs in the chain, so the total extension is 2 × 6.0 = 12.0 cm. The value 3.0 cm assumes the springs share the load side by side. The value 6.0 cm counts only one spring. The value 8.0 cm doubles the original extension without allowing for the larger load.✓ 1Answer D

Answer: D

Common error: When springs hang one below another, each spring carries the whole load. The extensions then add together.

343IGPC-035-03-15Forces, friction and springs Paper EHard7 marks

A skydiver of total mass 75 kg jumps from an aircraft. Use g = 9.8 N/kg.

(a)

Calculate the weight of the skydiver.

(1)
(b)

At one moment during the fall, the air resistance on the skydiver is 480 N. Calculate the acceleration of the skydiver at this moment.

(2)
(c)

Explain why the acceleration of the skydiver decreases as the fall continues.

(2)
(d)

The parachute opens and, at that instant, the air resistance is 1200 N. Calculate the deceleration of the skydiver at that instant.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
W = 75 × 9.8 = 735 N✓ 1
Part (b)
Resultant force = 735 − 480 = 255 N . a = F/m = 255/75 = 3.4 m/s² (downwards)✓ 2
Part (c)
As the speed increases, the air resistance increases. The resultant downward force therefore decreases, so the acceleration (a = F/m) decreases.✓ 2
Part (d)
Resultant = 1200 − 735 = 465 N upwards . a = 465/75 = 6.2 m/s² (deceleration)✓ 2

Common error: When the parachute opens, the skydiver is still moving downwards but is slowing down. The resultant force is upwards; the skydiver does not move upwards.

344IGPC-035-03-16Forces, friction and springs Paper EHard6 marks

A car of mass 900 kg is travelling at 20 m/s on a dry, straight road. The driver brakes and the car comes to rest in 4.0 s with a constant deceleration.

(a)

Calculate the deceleration of the car.

(1)
(b)

Calculate the braking force on the car.

(1)
(c)

On an icy road, the largest friction force the tyres can provide is 1800 N. Calculate the shortest time in which the car can stop from 20 m/s on this road.

(2)
(d)

Later, the car travels around a curved bend at a constant speed. State the direction of the resultant force on the car and suggest what happens if the friction between the tyres and the road is too small.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
a = Δv/t = 20/4.0 = 5.0 m/s²✓ 1
Part (b)
F = ma = 900 × 5.0 = 4500 N✓ 1
Part (c)
Maximum deceleration = 1800/900 = 2.0 m/s² . t = 20/2.0 = 10 s✓ 2
Part (d)
The resultant force acts towards the centre of the curve, at right angles to the motion. If friction is too small, the required inward force cannot be provided, so the car skids outwards and moves in a straighter path, off the curve.✓ 2

Exam tip: The maximum friction force sets the maximum deceleration. Use a = F/m first, then find the time.

345IGPC-035-02-08Mass, weight and density Paper EHard1 mark

Three liquids that do not mix are poured into a tall beaker:
• oil, density 0.92 g/cm3
• water, density 1.00 g/cm3
• glycerol, density 1.26 g/cm3
The liquids settle into layers. A small plastic bead of density 1.10 g/cm3 is then dropped in.
Where does the bead come to rest?

glycerol 1.26 g/cm³water 1.00 g/cm³oil 0.92 g/cm³
Diagram not to scale (bead not shown)
Show mark scheme
Marking pointMarkNotes
The liquids form layers in order of density, with the least dense on top: oil, then water, then glycerol at the bottom.
The bead (1.10 g/cm3) is denser than the oil and denser than the water, so it sinks through both of them. It is less dense than the glycerol, so it cannot sink into it. The bead therefore rests at the water–glycerol boundary.
✓ 1Answer C

Answer: C

Exam tip: An object sinks through every liquid that is less dense than it is. It stops on top of the first liquid that is denser than it is.

346IGPC-035-06-08Energy, work and power Paper EHard1 mark

A roller-coaster car starts from rest at a height of 20 m above the ground. It rolls down the track to a point 5.0 m above the ground. Friction and air resistance are negligible. The gravitational field strength g is 9.8 N/kg.
What is the speed of the car at the lower point?

start (at rest)lower point20 m5.0 m
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
The height lost is 20 − 5.0 = 15 m. The loss of Ep equals the gain of Ek: mgΔh = ½mv2, so v = √(2gΔh) = √(2 × 9.8 × 15) = √294 = 17 m/s. 9.9 m/s uses the final height of 5.0 m, 20 m/s uses the full 20 m, and 12 m/s leaves out the factor of 2.✓ 1Answer C

Answer: C

Exam tip: Use the change in height, not either height on its own. The mass cancels, so the speed does not depend on the mass of the car.

347IGPC-035-06-16Energy, work and power Paper EHard8 marks

A car of mass 900 kg is travelling at 20 m/s on a level road. The driver applies the brakes, and the car stops. Assume that the braking force is constant at 6000 N and that all of the car's kinetic energy is removed by the work done by the brakes.

(a)

Calculate the kinetic energy of the car before braking.

(2)
(b)

Calculate the braking distance of the car.

(2)
(c)

The same car travels at 40 m/s and the same braking force is applied. Explain why the braking distance is not simply doubled, and calculate the new braking distance.

(3)
(d)

State the energy transfer that takes place in the brakes as the car slows down.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Ek = ½mv2 = 0.5 × 900 × 202✓ 1
= 180 000 J✓ 1
Part (b)
work done by brakes = Ek, so F × d = 180 000 J✓ 1
d = 180 000 ÷ 6000 = 30 m✓ 1
Part (c)
Ek is proportional to v2, so doubling the speed makes the kinetic energy 4 times as large (720 000 J)✓ 1
With the same braking force, the brakes must do 4 times the work, so the distance is 4 times as large✓ 1
new braking distance = 720 000 ÷ 6000 = 120 m✓ 1
Part (d)
Energy is transferred from the kinetic store of the car to the thermal store of the brakes (and surroundings)✓ 1

Common error: Kinetic energy depends on speed squared. Doubling the speed does not double the energy or the braking distance; it multiplies both by four.

348IGPC-035-00-10Physical quantities and measurement Paper EHard1 mark

An aircraft flies due north at 80 m/s relative to the air. A wind blows due east at 60 m/s.
What is the resultant velocity of the aircraft?

60 m/s80 m/svNE
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Magnitude = √(80² + 60²) = √10000 = 100 m/s. The angle from north is given by tan θ = 60 ÷ 80, so θ = 37° east of north. 53° is the angle measured from east, not from north. 140 m/s adds the speeds as if they were in the same direction. 20 m/s subtracts them.✓ 1Answer B

Answer: B

Common error: Check which direction your angle is measured from. Here tan θ = opposite/adjacent with north as the adjacent side.

349IGPC-035-07-09Pressure in solids and liquids Paper EHard1 mark

A diver swims down from a depth of 2.0 m to a depth of 12.0 m in seawater of density 1030 kg/m3. Take g = 9.8 N/kg.
By how much does the pressure on the diver increase?

water surfacediver2.0 mdiver12.0 m
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
The change in depth is Δh = 12.0 − 2.0 = 10.0 m.
Δp = ρgΔh = 1030 × 9.8 × 10.0 = 100 940 Pa ≈ 1.01 × 105 Pa.
1.21 × 105 Pa uses the final depth of 12 m instead of the change in depth. 2.02 × 104 Pa uses the starting depth of 2 m. 1.03 × 104 Pa leaves out g.
✓ 1Answer C

Answer: C

Exam tip: In Δp = ρgΔh, the Δh is the change in depth between the two positions, not the depth of either position.

350IGPC-035-07-10Pressure in solids and liquids Paper EHard1 mark

A U-tube contains water of density 1000 kg/m3. One end of the tube is connected to a gas supply and the other end is open to the air. The water level on the open side is 0.25 m higher than on the gas side. Atmospheric pressure is 1.00 × 105 Pa and g = 9.8 N/kg.
What is the pressure of the gas?

0.25 mto gas supplyopen to airwater
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
The gas has pushed the water up the open side, so the gas pressure is greater than atmospheric pressure.
The extra pressure is ρgh = 1000 × 9.8 × 0.25 = 2450 Pa.
Gas pressure = 1.00 × 105 + 2450 = 1.0245 × 105 Pa ≈ 1.02 × 105 Pa.
2.45 × 103 Pa is only the excess pressure. 0.98 × 105 Pa subtracts the excess instead of adding it. 1.25 × 105 Pa wrongly treats 0.25 m as 0.25 × 105 Pa.
✓ 1Answer C

Answer: C

Common error: Decide first whether the gas pressure is above or below atmospheric pressure. The liquid is always pushed away from the side with the higher pressure.

351IGPC-035-05-09Momentum and impulse Paper EHard1 mark

A trolley X of mass 0.50 kg moves at 4.0 m/s towards a stationary trolley Y of mass 1.5 kg. After the collision, X rebounds along the same line at 2.0 m/s.
What is the speed of Y after the collision?

X 0.50 kgY 1.5 kg4.0 m/sat rest
Before the collision
Show mark scheme
Marking pointMarkNotes
Take X's initial direction as positive.
Momentum before = 0.50 × 4.0 = 2.0 kg m/s.
Momentum after = 0.50 × (−2.0) + 1.5v = −1.0 + 1.5v.
Setting these equal: 2.0 = −1.0 + 1.5v, so 1.5v = 3.0 and v = 2.0 m/s.
0.67 m/s treats X's rebound as a forward motion. 1.3 m/s ignores X's momentum after the collision. 1.0 m/s subtracts the momenta but does not divide by Y's mass.
✓ 1Answer D

Answer: D

Common error: Give a rebounding velocity a negative sign before applying conservation of momentum.

352IGPC-035-05-10Momentum and impulse Paper EHard1 mark

An object of mass 2.5 kg moves at 3.0 m/s. A constant force of 50 N then acts on it for 0.20 s, in the direction of its motion.
What is the speed of the object when the force stops acting?

Show mark scheme
Marking pointMarkNotes
Impulse = FΔt = 50 × 0.20 = 10 N s, and this equals the change in momentum.
Change in velocity = 10 ÷ 2.5 = 4.0 m/s.
Final speed = 3.0 + 4.0 = 7.0 m/s.
4.0 m/s forgets the initial speed. 1.0 m/s treats the force as acting against the motion. 10 m/s uses the impulse value as if it were a speed.
✓ 1Answer C

Answer: C

Exam tip: Impulse gives the change in momentum. Add it to the initial momentum to find the final momentum.

353IGPC-035-05-16Momentum and impulse Paper EHard7 marks

A ball of mass 0.25 kg is dropped from rest from a height of 1.8 m onto a hard floor. It rebounds vertically to a height of 0.80 m. The ball is in contact with the floor for 0.020 s. Ignore air resistance. Use g = 9.8 N/kg.

released from rest1.8 mhighest point after rebound0.80 m
Diagram not to scale
(a)

Calculate the speed of the ball just before it hits the floor.

(2)
(b)

Calculate the speed of the ball just after it leaves the floor.

(1)
(c)

Calculate the magnitude of the change in momentum of the ball during its contact with the floor.

(2)
(d)

Calculate the magnitude of the average resultant force on the ball while it is in contact with the floor.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
mgh = ½mv2, so v = √(2 × 9.8 × 1.8)✓ 1
v = 5.9 m/s✓ 1
Part (b)
v = √(2 × 9.8 × 0.80) = 4.0 m/s✓ 1
Part (c)
The direction reverses, so the speeds add: Δp = 0.25 × (5.94 + 3.96)✓ 1
= 2.5 kg m/s (accept 2.47–2.48 kg m/s)✓ 1
Part (d)
Resultant force = change in momentum ÷ time: F = 2.475 ÷ 0.020 (follow-through from the previous part, so 2.5 ÷ 0.020 is also accepted)✓ 1
F = 124 N (accept any value from 120 N to 125 N). This is the net force on the ball, so the weight is not added or subtracted.✓ 1

Common error: The ball reverses direction, so its change in momentum uses the sum of the two speeds, not their difference.

354IGPC-035-00-15Physical quantities and measurement Paper EHard7 marks

A boat crosses a river 45 m wide. The boat's engine gives it a velocity of 3.0 m/s relative to the water, pointing straight across the river (due north). The river flows due east at 1.2 m/s.

3.0 m/s1.2 m/svθ45 mriver flow →north banksouth bank
Diagram not to scale
(a)

Calculate the magnitude of the resultant velocity of the boat.

(2)
(b)

Calculate the angle between the boat's actual path and the direction due north.

(2)
(c)

Calculate the time taken for the boat to cross the river.

(2)
(d)

Calculate how far downstream from its starting point the boat lands on the opposite bank.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
v = √(3.02 + 1.22)✓ 1
= 3.2 m/s✓ 1
Part (b)
tan θ = 1.2 ÷ 3.0 (or sin θ = 1.2 ÷ 3.23)✓ 1
θ = 22° (east of north)✓ 1
Part (c)
Time = 45 ÷ 3.0, since only the northward component carries the boat across (or path length 48.5 m ÷ resultant speed 3.23 m/s)✓ 1
= 15 s✓ 1
Part (d)
Distance = 1.2 × 15 (or 45 × 1.2 ÷ 3.0) = 18 m; allow a correct value using the candidate's time from the previous part✓ 1

Common error: To find the crossing time, divide the river width by the speed across the river (3.0 m/s), not by the resultant speed.

355IGPC-035-04-16Turning effect and centre of gravity Paper EHard8 marks

A uniform plank AB is 4.0 m long and weighs 300 N. It rests horizontally on two supports, one at end A and one at end B. A painter of weight 700 N stands on the plank 1.0 m from end A.

700 N300 N1.0 m4.0 mAB
Diagram not to scale
(a)

State the two conditions that must be satisfied for the plank to be in equilibrium.

(2)
(b)

By taking moments about end A, calculate the upward force on the plank from the support at B.

(3)
(c)

Calculate the upward force on the plank from the support at A.

(1)
(d)

The painter walks slowly towards end B. Describe and explain how the force from the support at A changes.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
There is no resultant force on the plank✓ 1
There is no resultant moment (turning effect) on the plank✓ 1
Part (b)
Total moment of the weights about A = 300 × 2.0 + 700 × 1.0 = 1300 N m✓ 1
FB × 4.0 = 1300✓ 1
FB = 325 N (accept 330 N)✓ 1
Part (c)
FA = 1000 − 325 = 675 N (accept 680 N, or a value consistent with the answer to the previous part)✓ 1
Part (d)
The force at A decreases✓ 1
Explanation, either: the painter's moment about B decreases as his distance from B gets smaller, so a smaller upward force at A is needed to balance moments about B; or: the painter's moment about A increases, so FB increases, and since FA + FB = 1000 N, FA decreases✓ 1

Exam tip: Taking moments about one support removes that support's unknown force from the equation. Then use the zero resultant force condition to find the other support force.

356IGPC-035-07-16Pressure in solids and liquids Paper EHard7 marks

A glass tank with a base area of 0.20 m2 contains a 0.50 m deep layer of water (density 1000 kg/m3). A 0.30 m deep layer of oil (density 800 kg/m3) floats on top of the water. The two liquids do not mix. Take g = 9.8 N/kg.

water0.50 moil0.30 m
Diagram not to scale
(a)

Suggest why the oil forms the top layer.

(1)
(b)

Calculate the pressure due to the oil at the boundary between the oil and the water.

(2)
(c)

Calculate the total pressure due to both liquids at the base of the tank.

(2)
(d)

Calculate the force on the base of the tank due to the liquids.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The oil has a lower density than water✓ 1
Part (b)
p = ρgh = 800 × 9.8 × 0.30✓ 1
p = 2400 Pa (accept 2350 Pa or 2352 Pa)✓ 1
Part (c)
pressure due to the water = 1000 × 9.8 × 0.50 = 4900 Pa✓ 1
total = 2352 + 4900 = 7300 Pa (accept 7250 Pa or 7252 Pa; allow error carried forward from (b))✓ 1
Part (d)
F = pA = 7252 × 0.20✓ 1
F = 1500 N (accept 1450 N or 1460 N; allow error carried forward from (c))✓ 1

Common error: Do not use a single density for the whole 0.80 m depth. Work out the pressure from each layer using its own density and depth, then add them.

357IGPC-035-02-16Mass, weight and density Paper EHard7 marks

A rectangular glass tank has internal dimensions of 50 cm × 40 cm × 30 cm. It is filled completely with oil of density 850 kg/m3. The gravitational field strength is 9.8 N/kg.

50 cm30 cm40 cm
Diagram not to scale
(a)

Calculate the volume of oil in m3.

(2)
(b)

Calculate the mass of the oil.

(2)
(c)

Calculate the weight of the oil in the full tank.

(1)
(d)

Some of the oil is then removed, and water of density 1000 kg/m3 is poured in. The oil and the water do not mix. State and explain which liquid forms the upper layer.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Convert the dimensions to metres: 0.50 m × 0.40 m × 0.30 m✓ 1
V = 0.060 m3✓ 1
Part (b)
m = ρV = 850 × 0.060✓ 1
= 51 kg✓ 1
Part (c)
W = mg = 51 × 9.8 = 500 N (accept 499.8 N)✓ 1
Part (d)
Oil forms the upper layer✓ 1
because its density (850 kg/m3) is less than the density of water (1000 kg/m3)✓ 1

Common error: Convert centimetres to metres before multiplying. If you multiply in cm³ and then divide by 100, the volume is wrong by a factor of 10 000, because 1 m³ = 10⁶ cm³.

358IGPC3-1-01c-06Momentum and impulse Paper EHard1 mark

A 1500 kg van moving at 12 m/s collides with a stationary 1000 kg car. The two vehicles lock together and move off as one. What is their speed immediately after the collision?

van 1500 kgcar 1000 kg12 m/sat rest
Before the collision
Show mark scheme
Marking pointMarkNotes
Correct: A. Momentum is conserved: 1500 × 12 = 18 000 kg m/s = (1500 + 1000) × v, so v = 7.2 m/s. B is wrong: this takes the mean of 12 m/s and 0 m/s, ignoring the different masses. C is wrong: this assumes the speed is unchanged, ignoring the extra mass that must be moved. D is wrong: this divides the van's speed by the car's mass ratio only (12 × 1000 ÷ 2500), not using the van's momentum.✓ 1Answer A

Answer: A

Exam tip: Before the collision only the van has momentum. After, the combined mass moves together.

359IGPC3-1-01c-16Pressure in solids and liquids Paper EHard9 marks

A cylindrical storage tank has a flat base of area 0.50 m2. It contains oil of density 800 kg/m3 to a depth of 1.5 m. (g = 9.8 N/kg)

1.5 moilbase area 0.50 m²
Diagram not to scale
(a)

Calculate the pressure due to the oil at the base of the tank.

(2)
(b)

Calculate the force on the base of the tank due to the oil.

(2)
(c)

Explain why the pressure at the same depth in water would be greater than in the oil.

(2)
(d)

A pressure sensor in the oil reads 5880 Pa. Calculate its depth below the surface.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
pressure = density × g × depth✓ 1
800 × 9.8 × 1.5 = 11760 Pa✓ 1
Part (b)
force = pressure × area✓ 1
11760 × 0.50 = 5880 N✓ 1
Part (c)
pressure is proportional to density✓ 1
water has a greater density than the oil✓ 1
Part (d)
depth = pressure ÷ (density × g)✓ 1
5880 ÷ (800 × 9.8)✓ 1
0.75 m✓ 1

Exam tip: Liquid pressure depends on depth and density only. The width of the tank makes no difference to the pressure at a given depth.

360IGPC3-1-06r5-08Momentum and impulse Paper EHard1 mark

A buoy of mass 0.40 kg hits a harbour wall at 6.0 m/s and rebounds at 4.0 m/s. The contact lasts 0.020 s. What is the average force exerted on the buoy by the wall?

Show mark scheme
Marking pointMarkNotes
Take the rebound direction as positive: change in momentum = 0.40 × (4.0 − (−6.0)) = 4.0 kg m/s. Force = 4.0 ÷ 0.020 = 200 N. 40 N uses the difference in speeds (2.0). 80 N uses only the rebound speed. 120 N uses only the incoming speed.✓ 1Answer D

Answer: D

Common error: On a rebound the velocity reverses, so add the two speeds when finding the change in momentum.

361IGPC3-1-06r5-14Speed, velocity and motion graphs Paper EHard7 marks

A delivery van of mass 1500 kg travels at 18 m/s and brakes to rest in 4.5 s with a constant braking force.

(a)

Calculate the momentum of the van before braking.

(1)
(b)

Calculate the average braking force.

(2)
(c)

Calculate the distance travelled by the van while braking.

(2)
(d)

The van now carries an extra load of 500 kg. It travels at the same speed and the same braking force acts. Calculate the time taken to stop.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
1500 × 18 = 27 000 kg m/s✓ 1
Part (b)
force = change in momentum ÷ time = 27 000 ÷ 4.5✓ 1
= 6000 N✓ 1
Part (c)
average speed = 9 m/s, so distance = 9 × 4.5✓ 1
= 40.5 m✓ 1
Part (d)
new momentum = 2000 × 18 = 36 000 kg m/s✓ 1
time = 36 000 ÷ 6000 = 6.0 s✓ 1

Common error: The braking force, mass and speed together fix the stopping time. A heavier vehicle at the same speed with the same force takes longer to stop.

362IGPC3-1-06r7-13Momentum and impulse Paper EHard8 marks

A freight wagon of mass 3.0×103 kg moves at 2.0 m/s towards a stationary wagon of mass 1.0×103 kg on a level track. They collide and couple together.

3.0×10³ kg1.0×10³ kg2.0 m/sat rest
Before the collision
(a)

Calculate the momentum of the moving wagon before the collision.

(2)
(b)

Determine the speed of the coupled wagons immediately after the collision.

(3)
(c)

Calculate the kinetic energy lost in the collision. Give your answer in joules.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
p = mv✓ 1
p = 3.0×103 × 2.0 = 6000 kg m/s✓ 1
Part (b)
Momentum is conserved so total momentum after = 6000 kg m/s✓ 1
Total mass = 4.0×103 kg✓ 1
v = 6000 / 4000 = 1.5 m/s✓ 1
Part (c)
Before: ½ × 3000 × 2.0² = 6000 J✓ 1
After: ½ × 4000 × 1.5² = 4500 J✓ 1
Energy lost = 6000 − 4500 = 1500 J✓ 1

Common error: Momentum is conserved in a collision like this, but kinetic energy is not.

363IGPC3-1-07r8-06Momentum and impulse Paper EHard1 mark

A tractor of mass 3000 kg moves at 4.0 m/s and collides with a stationary trailer of mass 1000 kg. The two couple together and move off as one. How much kinetic energy is transferred to other forms (such as thermal and sound energy) in the collision?

Show mark scheme
Marking pointMarkNotes
Momentum: 3000 × 4.0 = 4000 × v, so v = 3.0 m/s. KE before = ½ × 3000 × 4.0² = 24000 J; KE after = ½ × 4000 × 3.0² = 18000 J; difference = 6000 J, so B. A is the KE the trailer would have at 4.0 m/s. C is the KE remaining after the collision. D assumes all the initial KE is lost.✓ 1Answer B

Answer: B

Common error: Momentum is conserved in a collision, but kinetic energy is not unless the collision is elastic.

364IGPC3-1-07r8-14Momentum and impulse Paper EHard7 marks

Two carts move along a straight, level farm track. Cart A has mass 400 kg and moves at 3.0 m/s. Cart B has mass 200 kg and moves in the same direction at 1.0 m/s. Cart A catches up with cart B and they collide. After the collision cart A moves at 2.0 m/s in the same direction.

A 400 kgB 200 kg3.0 m/s1.0 m/s
Before the collision
(a)

Calculate the total momentum of the two carts before the collision.

(2)
(b)

Calculate the velocity of cart B after the collision.

(2)
(c)

Determine whether the collision is elastic. Support your answer with a calculation of the kinetic energy lost or gained.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A: 400 × 3.0 = 1200 kg m/s and B: 200 × 1.0 = 200 kg m/s✓ 1
total = 1400 kg m/s✓ 1
Part (b)
1400 = 400 × 2.0 + 200 × v (momentum conserved)✓ 1
v = 3.0 m/s✓ 1
Part (c)
KE before = ½ × 400 × 3.02 + ½ × 200 × 1.02 = 1900 J✓ 1
KE after = ½ × 400 × 2.02 + ½ × 200 × 3.02 = 1700 J✓ 1
200 J of KE is lost so the collision is not elastic✓ 1

Common error: Check kinetic energy separately; conserved momentum does not prove that a collision is elastic.

365IGPC3-1-07r8-16Pressure in solids and liquids Paper EHard6 marks

A farm water tank is filled with water to a depth of 3.2 m. The density of water is 1000 kg/m3 and g = 9.8 N/kg. Atmospheric pressure is 1.0 × 105 Pa.

3.2 mwatercircular valve in the base
Diagram not to scale
(a)

Calculate the pressure due to the water at the base of the tank.

(2)
(b)

State the total pressure at the base of the tank, to 3 significant figures.

(1)
(c)

A circular valve of area 0.0040 m2 is fitted in the base. Calculate the force on the valve due to the water alone, to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
p = ρgh = 1000 × 9.8 × 3.2✓ 1
31360 Pa✓ 1
Part (b)
1.31 × 105 Pa✓ 1
Part (c)
F = pA✓ 1
31360 × 0.0040 (ECF from (a))✓ 1
125 N✓ 1

Common error: Use only the pressure due to the water when asked for the force due to the water; atmospheric pressure acts on both sides in this situation.

366IGPC3-1-05r15-06Pressure in solids and liquids Paper EHard1 mark

The water in a supply tank is 10 m deep. The density of water is 1000 kg/m3 and g = 9.8 N/kg. What is the pressure due to the water at the bottom of the tank?

Show mark scheme
Marking pointMarkNotes
p = ρgh = 1000 × 9.8 × 10 = 9.8×104 Pa, so C is correct. A uses the density as 1.0 (g/cm3) without converting. B omits g. D adds atmospheric pressure, which the question does not ask for.✓ 1Answer C

Answer: C

Common error: Use density in kg/m³ with p = ρgh; 1.0 g/cm³ is 1000 kg/m³.

367IGPC3-1-05r15-11Momentum and impulse Paper EHard9 marks

In an arcade bowling game a ball of mass 4.0 kg moves at 5.0 m/s towards a stationary pin of mass 2.0 kg. After the collision the ball continues in the same direction at 3.0 m/s and the pin moves off in the same direction.

ball 4.0 kgpin 2.0 kg5.0 m/sat rest
Before the collision
(a)

Calculate the momentum of the ball before the collision.

(1)
(b)

Calculate the speed of the pin immediately after the collision.

(3)
(c)

The ball is in contact with the pin for 0.040 s. Calculate the average force on the pin, using the speed from (b).

(2)
(d)

Calculate the kinetic energy lost in the collision.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
p = mv = 4.0 × 5.0 = 20 kg m/s✓ 1
Part (b)
total momentum before = total momentum after✓ 1
20 = 4.0 × 3.0 + 2.0 × v✓ 1
v = 4.0 m/s✓ 1
Part (c)
F = Δp/t with Δp = 2.0 × 4.0 = 8.0 kg m/s✓ 1
8.0 ÷ 0.040 = 200 N✓ 1
Part (d)
KE before = ½ × 4.0 × 5.0² = 50 J✓ 1
KE after = ½ × 4.0 × 3.0² + ½ × 2.0 × 4.0² = 18 + 16 = 34 J✓ 1
loss = 16 J✓ 1

Exam tip: Write the momentum equation with every object before and after the collision, then solve for the unknown speed.

368IGPC3-1-05r15-15Momentum and impulse Paper EHard7 marks

A pinball of mass 0.080 kg moves towards a flipper at 2.0 m/s. The flipper hits it and the ball leaves in the opposite direction at 4.0 m/s.

(a)

State the equation for momentum.

(1)
(b)

Calculate the change in momentum of the ball.

(2)
(c)

The ball is in contact with the flipper for 0.012 s. Calculate the average force on the ball.

(2)
(d)

Explain why a longer contact time would give a smaller force for the same change in momentum.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
momentum = mass × velocity✓ 1
Part (b)
takes direction into account: change in velocity = 4.0 − (−2.0) = 6.0 m/s✓ 1
Δp = 0.080 × 6.0 = 0.48 kg m/s✓ 1
Part (c)
F = Δp/t = 0.48 ÷ 0.012✓ 1
40 N✓ 1
Part (d)
force = rate of change of momentum / F = Δp/t✓ 1
the same Δp divided by a larger t gives a smaller F✓ 1

Common error: Momentum is a vector: when a ball rebounds, add the two speeds to find the change in velocity.

369IGPC3-1-06r18-06Pressure in solids and liquids Paper EHard1 mark

A sealed instrument housing is lowered 20 m below the surface of a lake of density 1030 kg/m3. What is the increase in pressure on the housing due to the water alone? (g = 9.8 N/kg)

water surfacehousing20 m
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Pressure due to a liquid = ρgh = 1030 × 9.8 × 20 = 2.02 × 105 Pa. A (505 Pa) divides by the depth instead of multiplying (ρg ÷ h). B (2.06 × 104 Pa) omits g (ρ × h). D (3.02 × 105 Pa) adds atmospheric pressure (about 1.0 × 105 Pa), but the question asks only for the pressure due to the water.✓ 1Answer C

Answer: C

Common error: The pressure caused by a liquid column is ρgh. Atmospheric pressure acts in addition at the surface, so add it only when the total pressure is asked for.

370IGPC3-1-06r18-07Momentum and impulse Paper EHard1 mark

A 30 kg sample-collecting probe hits an asteroid surface at 2.0 m/s and rebounds directly away from the surface at 0.50 m/s. What is the magnitude of the change in momentum of the probe?

Show mark scheme
Marking pointMarkNotes
Take the direction towards the surface as positive: initial momentum = +60 kg m/s, final momentum = −15 kg m/s. Change = −15 − 60 = −75, so the magnitude is 75 kg m/s. A (15) is only the momentum after rebound. B (45) subtracts the speeds, ignoring that the direction reverses. C (60) is only the momentum before impact.✓ 1Answer D

Answer: D

Common error: Momentum is a vector. When an object rebounds, add the magnitudes of the momentum before and after; do not subtract them.

371IGPC3-1-06r18-15Momentum and impulse Paper EHard8 marks

A space probe of mass 640 kg moves at 4.5 m/s in a straight line and catches up with a module of mass 960 kg moving at 1.5 m/s in the same direction. The two docking mechanisms lock and the combined spacecraft moves on as one body. No external forces act on the spacecraft during the docking.

probe 640 kgmodule 960 kg4.5 m/s1.5 m/s
Before the collision
(a)

Calculate the total momentum of the two bodies before docking.

(2)
(b)

Calculate the speed of the combined spacecraft immediately after docking.

(2)
(c)

The combined spacecraft then fires a thruster that gives a constant forward force of 800 N for 3.0 s. Calculate the impulse of the force.

(2)
(d)

Calculate the speed of the spacecraft at the end of the 3.0 s. Use your answers to (b) and (c).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 momentum = mass × velocity for each, 640 × 4.5 = 2880 kg m/s and 960 × 1.5 = 1440 kg m/s✓ 1
A1 4320 kg m/s✓ 1
Part (b)
M1 momentum conserved: v = 4320 ÷ (640 + 960)✓ 1
A1 2.7 m/s✓ 1
Part (c)
M1 impulse = force × time = 800 × 3.0✓ 1
A1 2400 N s✓ 1
Part (d)
M1 change in momentum = impulse, so change in speed = 2400 ÷ 1600 = 1.5 m/s✓ 1
A1 4.2 m/s (2.7 + 1.5)✓ 1

Common error: In a docking or collision, add the masses after the bodies join. Impulse (force × time) equals the change in momentum, so divide by the total mass to find the change in speed.

372IGPC3-1-07r20-06Pressure in solids and liquids Paper EHard1 mark

Two pressure sensors are fixed to the vertical side of a ship's hull, 15 m and 40 m below the surface of the sea. The density of sea water is 1030 kg/m3 and g = 9.8 N/kg. What is the difference between the pressures due to the water at the two sensors?

water surfacesensor15 msensor40 m
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
The pressure difference depends on the difference in depth: Δp = ρgΔh = 1030 × 9.8 × (40 − 15) = 2.5 × 105 Pa, so B is correct. A uses only 15 m, C uses only 40 m, and D uses the sum of the depths (55 m).✓ 1Answer B

Answer: B

Exam tip: For a pressure difference between two depths, use the vertical separation of the two points in p = ρgh.

373IGPC3-1-07r20-13Pressure in solids and liquids Paper EHard8 marks

An inspection hatch of area 0.040 m2 is fixed in the vertical side of a ship's hull. Its centre is 6.0 m below the surface of the sea. Density of sea water = 1030 kg/m3; g = 9.8 N/kg. Treat the pressure as the value at the centre of the hatch.

hullsea surfacehatch, area 0.040 m²6.0 m
Diagram not to scale
(a)

Calculate the pressure due to the water at the hatch.

(2)
(b)

Calculate the force on the hatch due to the water. Use the unrounded value from (a).

(2)
(c)

The ship is loaded and sinks lower so that the hatch is 12 m below the surface. Calculate the new pressure due to the water at the hatch.

(2)
(d)

Explain why the pressure due to a liquid at a given depth is greater in sea water than in fresh water.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 p = ρgh = 1030 × 9.8 × 6.0✓ 1
A1 6.06 × 104 Pa (60564 Pa)✓ 1
Part (b)
M1 F = pA = 60564 × 0.040 (ECF from (a))✓ 1
A1 2.42 × 103 N (2423 N)✓ 1
Part (c)
M1 p = 1030 × 9.8 × 12 (or doubling the pressure in (a) because depth has doubled)✓ 1
A1 1.21 × 105 Pa (121128 Pa)✓ 1
Part (d)
B1 sea water has a greater density✓ 1
B1 so the weight of the column of liquid above a given area is greater (p = ρgh)✓ 1

Exam tip: In p = ρgh, h is the vertical depth below the surface, and the pressure due to the liquid does not include the atmosphere unless the question says so.

374IGPC3-1-07r20-16Energy, work and power Paper EHard8 marks

A winch on a volcano observatory platform lifts an instrument package of mass 250 kg vertically through 12 m in 30 s at constant speed. g = 9.8 N/kg.

(a)

Calculate the weight of the package.

(1)
(b)

Calculate the work done on the package.

(2)
(c)

Calculate the useful power output of the winch.

(2)
(d)

The winch motor is 75% efficient. Calculate the electrical power supplied to the motor. Give your answer to 3 significant figures.

(2)
(e)

State what happens to the energy that is wasted by the motor.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 250 × 9.8 = 2450 N✓ 1
Part (b)
M1 work = force × distance = 2450 × 12✓ 1
A1 29400 J✓ 1
Part (c)
M1 power = work ÷ time = 29400 ÷ 30 (ECF from (b))✓ 1
A1 980 W✓ 1
Part (d)
M1 input power = useful power ÷ efficiency = 980 ÷ 0.75 (ECF from (c))✓ 1
A1 1310 W (1306.67 W)✓ 1
Part (e)
B1 it is transferred as thermal energy (and sound) to the surroundings✓ 1

Common error: Efficiency compares useful output with total input: divide the useful power by the efficiency to get the input power.

375IGPC3-1-08r20-07Momentum and impulse Paper EHard1 mark

A researcher of mass 80 kg stands on a stationary raft of mass 200 kg floating on calm water. She steps off horizontally and moves at 1.5 m/s relative to the water. Ignore water resistance. What is the speed of the raft immediately afterwards?

Show mark scheme
Marking pointMarkNotes
Total momentum stays zero: 80 × 1.5 = 200 × v, so v = 0.60 m/s opposite to her motion, so B is correct. A divides by the combined mass and gets the direction wrong. C assumes the raft moves at the same speed as the researcher. D inverts the mass ratio (200 ÷ 80 × 1.5).✓ 1Answer B

Answer: B

Common error: Momentum is a vector: the raft's momentum must be equal and opposite to the researcher's, so the lighter object moves faster.

376IGPC3-1-08r20-13Momentum and impulse Paper EHard8 marks

Two carts run on a straight horizontal rail in a warehouse. Cart A has mass 5.0 kg and moves to the right at 2.4 m/s. Cart B has mass 3.0 kg and moves to the left at 1.6 m/s. They collide head-on and stick together. Take right as positive.

A 5.0 kgB 3.0 kg2.4 m/s1.6 m/s
Before the collision
(a)

Calculate the velocity of the carts after the collision.

(3)
(b)

Calculate the kinetic energy lost in the collision.

(4)
(c)

State whether the collision is elastic and give a reason.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 total momentum before = 5.0 × 2.4 − 3.0 × 1.6 = 7.2 kg m/s (cart B has negative momentum)✓ 1
M1 7.2 = (5.0 + 3.0) × v✓ 1
A1 v = 0.90 m/s to the right✓ 1
Part (b)
M1 initial KE of A = ½ × 5.0 × 2.42 = 14.4 J✓ 1
A1 initial KE of B = ½ × 3.0 × 1.62 = 3.84 J✓ 1
A1 final KE = ½ × 8.0 × 0.902 = 3.24 J (ECF from (a))✓ 1
A1 energy lost = 18.24 − 3.24 = 15.0 J✓ 1
Part (c)
B1 not elastic, because kinetic energy is not conserved (it is transferred to other forms such as thermal and sound)✓ 1

Common error: Treat velocity as a vector in momentum calculations: an object moving in the opposite direction has negative momentum.

377IGPC3-1-08r22-08Pressure in solids and liquids Paper EHard1 mark

A diver works on a sunken jetty in a harbour. The total pressure on the diver is 1.98 × 105 Pa. Atmospheric pressure is 1.0 × 105 Pa. The density of the water is 1000 kg/m3 and g = 9.8 N/kg. What is the depth of the diver?

surfacediverdepth = ?
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
The pressure due to the water is 1.98 × 105 − 1.0 × 105 = 9.8 × 104 Pa. Depth = p ÷ (ρg) = 98000 ÷ (1000 × 9.8) = 10 m. 20 m ignores atmospheric pressure. 98 m divides only by the density, forgetting g. 198 m divides the total pressure by the density only.✓ 1Answer A

Answer: A

Common error: Total pressure at depth includes atmospheric pressure; subtract it before using p = ρgh.

378IGPC3-1-08r22-15Momentum and impulse Paper EHard8 marks

A cyclist and bicycle of total mass 80 kg travel at 9.0 m/s and run into a line of soft straw bales. They are brought to rest uniformly in 0.60 s.

(a)

Calculate the change in momentum of the cyclist and bicycle.

(2)
(b)

Calculate the average force exerted by the bales.

(2)
(c)

Calculate the distance travelled while stopping.

(2)
(d)

Show that the work done by the bales equals the initial kinetic energy of the cyclist and bicycle.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
change in momentum = mass × change in velocity = 80 × 9.0✓ 1
720 kg m/s✓ 1
Part (b)
force = change in momentum ÷ time = 720 ÷ 0.60✓ 1
1200 N✓ 1
Part (c)
average speed = 4.5 m/s✓ 1
distance = 4.5 × 0.60 = 2.7 m✓ 1
Part (d)
work done = force × distance = 1200 × 2.7 = 3240 J✓ 1
kinetic energy = ½ × 80 × 9.02 = 3240 J, so they are equal✓ 1

Exam tip: A longer stopping time for the same change in momentum means a smaller average force.

379IGPC3-1-07r25-06Momentum and impulse Paper EHard1 mark

Trolley A of mass 2.0 kg moves at 3.0 m/s along a smooth horizontal track towards a stationary trolley B of mass 1.0 kg. After the collision, A continues in the same direction at 1.0 m/s. What is the speed of B after the collision?

before2.0 kg3.0 m/s1.0 kgat restafter2.0 kg1.0 m/s1.0 kg? m/s
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Momentum is conserved. Before: 2.0 × 3.0 = 6.0 kg m/s. After: 2.0 × 1.0 + 1.0 × v. So 6.0 = 2.0 + v and v = 4.0 m/s. The value 2.0 m/s divides the remaining momentum by 2.0 kg instead of 1.0 kg. The value 6.0 m/s ignores the momentum A still has. The value 5.0 m/s subtracts A's speed from the initial momentum, mixing speed with momentum.✓ 1Answer B

Answer: B

Exam tip: Write the momentum before and after as full products (mass × velocity) for every object, then equate the totals.

380IGPC3-1-07r25-13Pressure in solids and liquids Paper EHard9 marks

A Formula 1 car of mass 800 kg rests on four identical tyres. The area of each tyre in contact with the ground is 0.020 m2. Use g = 9.8 N/kg.

50 Nforce = ?small piston: 4.0 cm²large piston: 160 cm²large piston (car)
Diagram not to scale
(a)

Calculate the weight of the car.

(1)
(b)

Calculate the pressure exerted by the tyres on the ground.

(2)
(c)

In the garage the car is lifted by a hydraulic jack. A force of 50 N is applied to a small piston of area 4.0 cm2. Calculate the pressure in the oil.

(2)
(d)

The large piston has an area of 160 cm2. Assume the pressure is the same throughout the oil. Calculate the force on the large piston.

(2)
(e)

Calculate the minimum force on the small piston needed to lift the full weight of the car.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
800 × 9.8 = 7840 N✓ 1
Part (b)
total area = 4 × 0.020 = 0.080 m2 and p = F ÷ A = 7840 ÷ 0.080✓ 1
9.8 × 104 Pa (ECF from (a))✓ 1
Part (c)
p = F ÷ A = 50 ÷ (4.0 × 10−4)✓ 1
1.25 × 105 Pa✓ 1
Part (d)
F = pA = 1.25 × 105 × 160 × 10−4✓ 1
2000 N✓ 1
Part (e)
pressure needed = 7840 ÷ (160 × 10−4) = 4.9 × 105 Pa, or force ratio 160 ÷ 4.0 = 40✓ 1
F = 7840 ÷ 40 = 196 N (ECF from (a))✓ 1

Common error: Convert cm² to m² (× 10⁻⁴) before finding pressure in pascals. A hydraulic system multiplies force, so the small piston needs a smaller force.

381IGPC3-1-10r27-16Momentum and impulse Paper EHard7 marks

A lander of mass 250 kg touches down on a moon at a vertical speed of 3.0 m/s and comes to rest in 0.15 s as its legs crumple. Assume the force from the legs is constant and ignore the weight of the lander during the landing.

(a)

Calculate the momentum of the lander just before touchdown.

(1)
(b)

Calculate the average force exerted by the legs.

(2)
(c)

Calculate the kinetic energy of the lander just before touchdown.

(2)
(d)

Calculate the distance by which the legs crumple. Assume constant deceleration.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
250 × 3.0 = 750 kg m/s✓ 1
Part (b)
F = change in momentum / time = 750 / 0.15✓ 1
5000 N✓ 1
Part (c)
KE = ½ m v2 = 0.5 × 250 × 3.02✓ 1
1125 J✓ 1
Part (d)
average speed during the stop = 3.0 / 2 = 1.5 m/s✓ 1
distance = 1.5 × 0.15 = 0.225 m✓ 1

Exam tip: With constant deceleration to rest, the average speed is half of the initial speed.

382IGPC3-1-10r28-11Momentum and impulse Paper EHard8 marks

A cyclist of total mass 70 kg (rider and bicycle) travelling at 6.0 m/s collides with a stationary cyclist of total mass 55 kg. The two cyclists move off together in a straight line.

70 kg6.0 m/s55 kgat rest
Diagram not to scale
(a)

State the principle of conservation of momentum.

(2)
(b)

Calculate the speed of the two cyclists immediately after the collision.

(3)
(c)

Calculate the kinetic energy transferred to other forms in the collision. Give your answer to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
in a collision between bodies, the total momentum before equals the total momentum after✓ 1
provided no external resultant force acts on the system✓ 1
Part (b)
M1 momentum before = 70 × 6.0 = 420 kg m/s✓ 1
M1 total mass = 125 kg, speed = 420 ÷ 125✓ 1
A1 3.36 m/s✓ 1
Part (c)
M1 initial KE = ½ × 70 × 6.02 = 1260 J✓ 1
M1 final KE = ½ × 125 × 3.362 = 705.6 J✓ 1
A1 energy transferred = 554 J✓ 1

Exam tip: Momentum is conserved in a collision, but kinetic energy generally is not: check with both calculations.

383IGPC3-1-10r28-16Momentum and impulse Paper EHard9 marks

A cycle helmet is tested by dropping a headform of mass 5.0 kg onto a hard anvil. The headform hits the anvil at 6.0 m/s and is brought to rest. Without a helmet, the headform stops in 8.0 ms.

(a)

Calculate the change in momentum of the headform.

(2)
(b)

Calculate the average force on the headform without a helmet.

(2)
(c)

Explain how the foam lining of a helmet reduces the force on the head in a collision.

(2)
(d)

With a helmet, the headform stops in 20 ms. Calculate the percentage reduction in the average force compared with no helmet.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 Δp = mv = 5.0 × 6.0✓ 1
A1 30 kg m/s✓ 1
Part (b)
M1 F = Δp ÷ t = 30 ÷ 0.0080✓ 1
A1 3750 N✓ 1
Part (c)
the foam crushes, increasing the time taken for the head to stop✓ 1
the change in momentum is the same, so the average force F = Δp ÷ t is smaller✓ 1
Part (d)
M1 force with helmet = 30 ÷ 0.020 = 1500 N✓ 1
M1 reduction = (3750 − 1500) ÷ 3750✓ 1
A1 60%✓ 1

Exam tip: For a collision, use F = Δp ÷ t: a longer stopping time gives a smaller force for the same momentum change.

384IGPC3-1-07r30-05Pressure in solids and liquids Paper EHard1 mark

In a hospital laboratory a pressure sensor is placed at a depth of 0.60 m in water of density 1000 kg/m3. The same sensor is then lowered into oil of density 750 kg/m3. At what depth in the oil does the sensor give the same pressure reading from the liquid?

watersensoroilsensor0.60 m? m
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Pressure from a liquid is p = ρgh, so for equal pressure ρwhw = ρoho and ho = 0.60 × 1000 / 750 = 0.80 m. The less dense oil needs a greater depth. 0.45 m multiplies by the density ratio instead of dividing. 0.60 m assumes depth alone decides the pressure. 1.07 m applies the density ratio twice.✓ 1Answer A

Answer: A

Exam tip: For two liquids at equal pressure, the less dense liquid must be deeper: set ρ₁h₁ = ρ₂h₂.

385IGPC3-1-07r30-13Pressure in solids and liquids Paper EHard7 marks

A hospital hangs a bag of saline solution so that the surface of the liquid is 0.80 m above the needle in a patient’s arm. The saline has a density of 1000 kg/m3. The blood in the vein has a pressure 2000 Pa above atmospheric pressure. Use g = 9.8 N/kg.

salineneedlepatient's arm0.80 m
Diagram not to scale
(a)

Calculate the pressure due to the saline at the needle.

(2)
(b)

Calculate the difference between this pressure and the blood pressure.

(1)
(c)

The nurse lowers the bag until the flow just stops. Calculate the height of the liquid surface above the needle at this point. Give your answer to 3 significant figures.

(2)
(d)

A second bag holds twice the volume of saline in a wider container, with the surface also 0.80 m above the needle. State and explain how the pressure at the needle compares with the first bag.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
p = ρgh✓ 1
= 1000 × 9.8 × 0.80 = 7840 Pa✓ 1
Part (b)
7840 − 2000 = 5840 Pa (ECF from (a))✓ 1
Part (c)
pressure from the saline equals the blood pressure = 2000 Pa✓ 1
h = p / (ρg) = 2000 / (1000 × 9.8) = 0.204 m✓ 1
Part (d)
the pressure is the same✓ 1
pressure depends on depth (height), density and g, not on the volume or width of the container✓ 1

Common error: Pressure in a liquid depends on the vertical height of the liquid, not on how much liquid there is or the shape of its container.

386IGPC3-1-07r30-14Turning effect and centre of gravity Paper EHard7 marks

A uniform wooden plank is 3.0 m long and has a weight of 150 N. It rests horizontally on a pivot placed 1.0 m from its left end. A hive of weight 300 N is placed on the plank to the left of the pivot so that the plank balances.

hive 300 Nweight 150 Npivot1.0 m? m3.0 m
Diagram not to scale
(a)

State the position of the centre of gravity of the plank.

(1)
(b)

Calculate the moment of the weight of the plank about the pivot.

(2)
(c)

Calculate the distance from the pivot to the hive.

(2)
(d)

Calculate the upward force of the pivot on the plank.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
at the midpoint, 1.5 m from either end✓ 1
Part (b)
distance from pivot = 1.5 − 1.0 = 0.50 m✓ 1
moment = 150 × 0.50 = 75 N m✓ 1
Part (c)
sum of clockwise moments = sum of anticlockwise moments, so 300 × d = 75✓ 1
d = 0.25 m (ECF from (b))✓ 1
Part (d)
plank is in equilibrium, so upward force = total downward force✓ 1
150 + 300 = 450 N✓ 1

Exam tip: Take moments about the pivot so that the unknown pivot force has no moment and drops out of the equation.

387IGPC3-1-08r31-08Forces, friction and springs Paper EHard1 mark

A bridge-inspection platform of mass 150 kg is lifted by a single cable and accelerates upwards at 2.0 m/s2. What is the tension in the cable? (g = 9.8 N/kg.)

150 kgtension T = ?weighta = 2.0 m/s²
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
The resultant force upwards is T − mg = ma, so T = m(g + a) = 150 × 11.8 = 1770 N. 300 N is the resultant force only, which leaves out the weight. 1170 N uses g − a, which would be the case for downward acceleration. 1470 N is just the weight, which is the tension only at constant speed.✓ 1Answer D

Answer: D

Common error: When accelerating upwards, the cable must support the weight and also provide the resultant force.

388IGPC3-1-08r31-15Turning effect and centre of gravity Paper EHard6 marks

A uniform plank of length 8.0 m and weight 2400 N forms a footbridge. It rests on a support at each end, A (left) and B (right). A person of weight 800 N stands 2.0 m from end A.

800 Nweight 2400 NAB??2.0 m8.0 m
Diagram not to scale
(a)

State the principle of moments.

(1)
(b)

Calculate the upward force at support A by taking moments about support B.

(3)
(c)

Calculate the upward force at support B.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 for a body in equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments about that point✓ 1
Part (b)
M1 take moments about B (the plank's weight acts at its centre, 4.0 m from B)✓ 1
M1 RA × 8.0 = 2400 × 4.0 + 800 × 6.0✓ 1
A1 1800 N✓ 1
Part (c)
M1 total upward force = total downward force, RA + RB = 2400 + 800✓ 1
A1 1400 N (ECF from (b))✓ 1

Exam tip: Take moments about one support so that its unknown force has no moment and drops out of the equation.

389IGPC3-1-09r32-08Energy, work and power Paper EHard1 mark

A winch lifts a 500 kg bridge section through a height of 3.0 m in 20 s. The winch is 60% efficient. g = 9.8 N/kg. What is the input power to the winch?

Show mark scheme
Marking pointMarkNotes
Useful power = mgh ÷ t = 500 × 9.8 × 3.0 ÷ 20 = 735 W. Efficiency = useful power ÷ input power, so input = 735 ÷ 0.60 = 1225 W. 735 W ignores the efficiency. 441 W multiplies by 0.60 instead of dividing. 14 700 is the energy gained (in J), not a power.✓ 1Answer B

Answer: B

Common error: Input is always larger than useful output: divide the useful power by the efficiency, never multiply.

390IGPC3-1-09r32-16Pressure in solids and liquids Paper EHard7 marks

A sensor is fixed to the bed of a river beneath a bridge, at a depth of 6.5 m. The density of the water is 1000 kg/m3, g = 9.8 N/kg and atmospheric pressure is 1.0 × 105 Pa. The sensor has a flat window of area 2.5 × 10−3 m2.

surfacesensor6.5 m
River; diagram not to scale
(a)

Calculate the pressure due to the water at the sensor.

(2)
(b)

Calculate the total pressure on the window of the sensor.

(2)
(c)

Calculate the force on the window, to 3 significant figures. Use the unrounded total pressure.

(2)
(d)

State how the pressure due to the water changes if the depth is doubled.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 p = ρgh = 1000 × 9.8 × 6.5✓ 1
A1 63 700 Pa✓ 1
Part (b)
M1 add atmospheric pressure: 63 700 + 1.0 × 105✓ 1
A1 1.637 × 105 Pa (accept 1.6 × 105 Pa)✓ 1
Part (c)
M1 F = pA = 1.637 × 105 × 2.5 × 10−3✓ 1
A1 409 N (ECF from (b))✓ 1
Part (d)
it doubles, because the pressure is directly proportional to the depth✓ 1

Common error: The total pressure under water includes atmospheric pressure; the ρgh term alone gives only the pressure due to the water.

391IGPC3-1-08r38-06Pressure in solids and liquids Paper EHard1 mark

The density of fresh water in a lake is 1000 kg/m3 and g = 9.8 N/kg. An archaeologist's diver moves from a depth of 8.0 m to a depth of 20 m. What is the increase in water pressure?

surfacediver (8.0 m)diver (20 m)8.0 m20 m
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
The change in depth is 12 m, so Δp = ρgΔh = 1000 × 9.8 × 12 = 1.2 × 105 Pa. Adding the two depths gives 2.7 × 105, omitting g gives 1.2 × 104, and using only the final depth gives 2.0 × 105.✓ 1Answer D

Answer: D

Common error: A difference in pressure depends on the difference in depth, not on either depth alone.

392IGPC3-1-08r38-14Momentum and impulse Paper EHard6 marks

In a data centre an automated trolley of mass 120 kg travels at 2.0 m/s and hits a stationary trolley of mass 80 kg. The two trolleys couple together and move off together.

120 kg2.0 m/s80 kgat rest
Diagram not to scale
(a)

Calculate the speed of the trolleys immediately after the collision.

(3)
(b)

Calculate the change in momentum of the 80 kg trolley.

(1)
(c)

The collision lasts 0.40 s. Calculate the average force on the 80 kg trolley.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 momentum before = 120 × 2.0 = 240 kg m/s✓ 1
M1 momentum conserved: 240 = (120 + 80)v✓ 1
A1 1.2 m/s✓ 1
Part (b)
A1 80 × 1.2 = 96 kg m/s (ECF from (a))✓ 1
Part (c)
M1 force = change in momentum ÷ time = 96 ÷ 0.40✓ 1
A1 240 N✓ 1

Exam tip: Force equals change in momentum divided by time, so a longer contact time means a smaller force for the same change.

393IGPC3-1-08r38-15Pressure in solids and liquids Paper EHard6 marks

A diver works on an archaeological site on the sea bed at a depth of 18 m. The density of sea water is 1030 kg/m3, g = 9.8 N/kg and atmospheric pressure is 1.0 × 105 Pa. Give answers to 2 significant figures. You may use rounded or unrounded intermediate values.

surfacediver on sea bed18 m
Diagram not to scale
(a)

Calculate the pressure due to the water at the depth of 18 m.

(2)
(b)

Calculate the total pressure on the diver.

(1)
(c)

A recovery hatch has an area of 0.25 m2. Calculate the force exerted on one side of the hatch by the water alone.

(2)
(d)

State why the pressure at the same depth in a fresh-water lake would be smaller.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 p = ρgh = 1030 × 9.8 × 18✓ 1
A1 1.8 × 105 Pa✓ 1
Part (b)
A1 1.8 × 105 + 1.0 × 105 = 2.8 × 105 Pa (ECF from (a))✓ 1
Part (c)
M1 F = pA = 1.8 × 105 × 0.25✓ 1
A1 4.5 × 104 N✓ 1
Part (d)
B1 fresh water has a lower density✓ 1

Common error: The pressure due to the liquid alone is ρgh; add atmospheric pressure only if the total pressure is asked for.

394IGPC3-1-08r41-08Momentum and impulse Paper EHard1 mark

A capsule falls through the air in a pharmacy test rig. Its speed increases from 2.0 m/s to 6.0 m/s. By what factor does its kinetic energy increase?

Show mark scheme
Marking pointMarkNotes
Kinetic energy = ½mv², so it is proportional to v². The speed is multiplied by 3, so the energy is multiplied by 3² = 9. A factor of 3 treats energy as proportional to speed. A factor of 4 uses the difference between the speeds. A factor of 36 squares the final speed without comparing it with the initial value.✓ 1Answer D

Answer: D

Common error: Kinetic energy depends on the square of speed, so tripling the speed multiplies the energy by nine, not three.

395IGPC3-1-08r41-14Momentum and impulse Paper EHard5 marks

In a pinball game a ball of mass 0.080 kg moves at 2.5 m/s towards a bumper. It rebounds along the same line at 1.5 m/s, in the opposite direction. The ball is in contact with the bumper for 0.020 s.

(a)

Calculate the change in momentum of the ball.

(2)
(b)

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

(2)
(c)

State the direction of this force relative to the initial motion of the ball.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
take the initial direction as positive: change = 0.080 × (−1.5) − 0.080 × 2.5✓ 1
magnitude 0.32 kg m/s✓ 1
Part (b)
force = change in momentum ÷ time = 0.32 ÷ 0.020✓ 1
16 N✓ 1
Part (c)
opposite to the initial direction of motion (accept: away from the bumper)✓ 1

Common error: Momentum has direction: when a ball rebounds, add the magnitudes of the two momenta rather than subtracting them.

396IGPC3-1-09r44-05Momentum and impulse Paper EHard1 mark

Two Formula 1 cars collide on a straight track and lock together. Car A has mass 800 kg and travels at 30 m/s. It catches car B, of mass 1200 kg, travelling in the same direction at 10 m/s.
What is the speed of the joined cars immediately after the collision?

A 800 kg30 m/sB 1200 kg10 m/s
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Momentum is conserved: total momentum before = 800 × 30 + 1200 × 10 = 36 000 kg m/s. Total mass = 2000 kg, so speed = 36 000 ÷ 2000 = 18 m/s. 20 m/s is the mean of the two speeds, which ignores the different masses. 6.0 m/s subtracts the momenta as if the cars were moving towards each other. 12 m/s uses only the momentum of car A.✓ 1Answer B

Answer: B

Common error: Add momenta in the same direction and subtract them only when the objects move in opposite directions. The total mass is used after they join.

397IGPC3-1-09r44-08Energy, work and power Paper EHard1 mark

A Formula 1 car brakes with a constant braking force and stops in a distance d from speed v. The same car brakes with the same constant force from a speed of 1.5v.
What is the new stopping distance?

Show mark scheme
Marking pointMarkNotes
Work done by the braking force = kinetic energy lost, so F × d = ½mv2. Distance is proportional to v2, so the new distance is 1.52 d = 2.25d. 1.5d wrongly assumes distance is proportional to speed. 3.4d uses speed cubed. 0.67d inverts the ratio.✓ 1Answer C

Answer: C

Exam tip: Kinetic energy depends on speed squared, so any quantity that comes from it changes by the square of the speed ratio.

398IGPC3-1-09r45-05Pressure in solids and liquids Paper EHard1 mark

A hospital saline bag hangs from a stand and is vented so that the surface of the saline is at atmospheric pressure, 1.0 × 105 Pa. The surface is 1.5 m above the needle in the patient's arm. The density of the saline is 1000 kg/m3 and the gravitational field strength is 9.8 N/kg. By how much does the pressure in the saline at the needle exceed atmospheric pressure?

salineneedlepatient's arm1.5 m
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
The extra pressure is due only to the column of saline: p = ρgh = 1000 × 9.8 × 1.5 = 14 700 Pa = 14.7 kPa. 115 kPa is the total pressure at the needle, because atmospheric pressure has been added on. 1.5 kPa leaves out g (1000 × 1.5). 9.8 kPa leaves out the height (1000 × 9.8).✓ 1Answer D

Answer: D

Common error: Pressure due to a liquid column is ρgh; do not add atmospheric pressure when the question asks for the excess over it.

399IGPC3-1-09r45-13Energy, work and power Paper EHard9 marks

A Formula 1 car has a mass of 800 kg including the driver. On a straight it accelerates from 20 m/s to 80 m/s.

(a)

Calculate the gain in kinetic energy of the car.

(3)
(b)

The engine delivers a useful power of 600 kW. Ignoring all resistive forces, calculate the time taken for this gain in kinetic energy.

(2)
(c)

In practice the car takes 5.0 s. Calculate the percentage of the energy supplied by the engine in the 5.0 s that becomes kinetic energy. Use 2 400 000 J for the gain in kinetic energy.

(3)
(d)

Suggest where the rest of the energy goes.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 kinetic energy = ½mv2✓ 1
M1 change = ½ × 800 × (802 − 202) (or 2 560 000 − 160 000)✓ 1
A1 2 400 000 J (2.4 MJ)✓ 1
Part (b)
M1 time = energy ÷ power = 2 400 000 ÷ 600 000 (ECF from (a))✓ 1
A1 4.0 s✓ 1
Part (c)
B1 energy supplied = 600 000 × 5.0 = 3 000 000 J✓ 1
M1 percentage = useful energy ÷ energy supplied × 100✓ 1
A1 2 400 000 ÷ 3 000 000 × 100 = 80%✓ 1
Part (d)
B1 work done against air resistance and friction, transferred as thermal energy (heating of the air, tyres and track)✓ 1

Exam tip: Power is energy ÷ time, so energy supplied = power × time; efficiency compares useful energy with the total supplied.