Original exam-style questions with full mark schemes. Use the filters to choose level and difficulty, and tick questions to build a printable test.
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Original questions in the style of each exam boardEvery answer re-solved independently before publishingCreated by Evan Kotronis with AI assistanceUpdated 15 October 2026 · new questions added continuouslyFree access until 30 November 2026How every question is checked →
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1IGPC-035-01-01Speed, velocity and motion graphsPaper 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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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.
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Answer 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 graphsPaper 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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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.
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Answer 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 graphsPaper 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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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.
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Answer 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 graphsPaper 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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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.
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Answer 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 graphsPaper 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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Part (a)
average speed = distance travelled ÷ time taken
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Part (b)
(3.1 + 3.3 + 3.2) ÷ 3 = 3.2 s
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Part (c)
2.40 ÷ 3.2
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= 0.75 m/s
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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 graphsPaper 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.
(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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Part (a)
It moves at a constant speed of 15 m/s (zero acceleration)
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Part (b)
a = (15 − 0) ÷ 10
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= 1.5 m/s2
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Part (c)
distance = speed × time = 15 × 40
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= 600 m
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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 springsPaper 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?
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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.
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Answer 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 springsPaper 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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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.
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Answer 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 springsPaper 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?
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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.
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Answer 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 springsPaper 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.
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Answer 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 springsPaper 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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Part (a)
She subtracts the original (unloaded) length of 5.0 cm from the measured length.
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Part (b)
8.6 − 5.0 = 3.6 cm
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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).
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Part (d)
Extension = 2.5 × 1.2 = 3.0 cm . Length = 5.0 + 3.0 = 8.0 cm
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Common error: When asked for a length, remember to add the extension back on to the original length.
12IGPC-035-03-12Forces, friction and springsPaper 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.
(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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Part (a)
Resultant = 150 − 90 − 20 = 40 N , acting forwards (in the direction of motion)
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Part (b)
The cyclist is speeding up (accelerating) in the forward direction.
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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.
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Exam tip: A resultant force mark usually needs both the size and the direction of the force.
13IGPC-035-02-01Mass, weight and densityPaper 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.
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Answer 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 densityPaper 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.
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Answer 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 densityPaper 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?
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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.
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Answer 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 densityPaper 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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Part (a)
Aluminium
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Part (b)
The density of ice (0.92 g/cm3) is less than the density of water (1.00 g/cm3)
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Part (c)
ρ = 360 ÷ 400 = 0.90 g/cm3
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It floats, because 0.90 g/cm3 is less than 1.00 g/cm3
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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 powerPaper 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.
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Answer 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 powerPaper 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.
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Answer 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 powerPaper CEasy1 mark
An electric motor does 6000 J of work in 20 s. What is the power of the motor?
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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.
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Answer 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 powerPaper 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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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.
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Answer 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 powerPaper 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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Part (a)
work done = force × distance moved in the direction of the force (W = Fd)
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Part (b)
W = 35 × 18
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= 630 J
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Part (c)
To the thermal store of the trolley's wheels/floor/surroundings, because of friction (and air resistance)
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Part (d)
P = W ÷ t = 630 ÷ 12
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= 53 W (52.5 W)
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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 measurementPaper 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.
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Answer 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 measurementPaper 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.
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Answer 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 measurementPaper 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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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.
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Answer 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 measurementPaper 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.
(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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Part (a)
Place the eye level with the water surface, to avoid parallax
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Read the scale at the bottom of the meniscus
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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
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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 gravityPaper 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?
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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.
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Answer B
Answer: B
Exam tip: Convert distances to metres before calculating a moment in N m.
27IGPC-035-04-02Turning effect and centre of gravityPaper CEasy1 mark
Which unit is used for the moment of a force?
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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.
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Answer 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 gravityPaper 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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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.
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Answer 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 liquidsPaper CEasy1 mark
Which unit is equivalent to the pascal (Pa), the unit of pressure?
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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.
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Answer 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 liquidsPaper 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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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.
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Answer 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 liquidsPaper 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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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.
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Answer 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 liquidsPaper 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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Part (a)
pressure = force ÷ area, or p = F/A
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Part (b)
p = 600 / 1.5 × 10−4
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p = 4.0 × 106 Pa
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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
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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 densityPaper 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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Part (a)
The quantity (amount) of matter in the object (accept: a measure of the object's resistance to a change in its motion)
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Part (b)
W = mg = 18 × 9.8
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= 176 N (accept 180 N)
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Part (c)
Mass = 18 kg, unchanged
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Weight = 18 × 3.7 = 67 N (accept 66.6 N)
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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 liquidsPaper 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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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.
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Answer 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 gravityPaper 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)
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Part (a)
The point through which the weight of the object appears to act / can be considered to act
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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
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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
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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 liquidsPaper 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)
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Part (a)
Any two from: depth below the surface of the liquid; density of the liquid; gravitational field strength g
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first correct factor
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second correct factor
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Part (b)
The pressure of the water increases with depth
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so the force on the wall is greatest at the bottom, and a thicker wall is needed there to withstand it
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Part (c)
in all directions, or equally in every direction
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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 gravityPaper 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.
(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)
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Part (a)
(Moment =) force × perpendicular distance from the pivot to the line of action of the force
✓ 1
Part (b)
Moment = 150 × 0.40
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= 60 N m
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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
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Exam tip: A definition of moment that leaves out the word perpendicular is incomplete.
38IGPC-035-02-12Mass, weight and densityPaper CEasy5 marks
A student wants to find the density of a rectangular block of aluminium.
(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)
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Part (a)
density = mass ÷ volume (ρ = m/V)
✓ 1
Part (b)
Measure the length, width and height of the block with the ruler
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Multiply the three measurements together: V = l × w × h
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Part (c)
V = 5.0 × 4.0 × 3.0 = 60 cm3
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ρ = 162 ÷ 60 = 2.7 g/cm3
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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 powerPaper 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)
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Part (a)
Energy cannot be created or destroyed; it can only be transferred from one store to another
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Accept: the total energy of a closed (isolated) system remains constant.
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Part (b)
Energy is transferred from the chemical store of the diesel fuel
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to the gravitational potential store of the load
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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.
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Part (c)
P = W ÷ t = 48 000 ÷ 16
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= 3000 W
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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.
40IGPC-035-01-04Speed, velocity and motion graphsPaper 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?
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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.
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Answer 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.
41IGPC-035-01-05Speed, velocity and motion graphsPaper 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?
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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).
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Answer C
Answer: C
Common error: Always work out final minus initial. A deceleration is a negative acceleration, so the sign carries information.
42IGPC-035-01-07Speed, velocity and motion graphsPaper 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?
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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.
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Answer C
Answer: C
Common error: Average speed is never the mean of the section speeds. Always divide the total distance by the total time.
43IGPC-035-01-15Speed, velocity and motion graphsPaper 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)
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Part (a)
The line is straight and sloping (constant gradient)
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Part (b)
speed = 200 ÷ 40
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= 5.0 m/s
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Part (c)
the runner moves more slowly (the gradient is smaller)
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the speed is still constant, at 120 ÷ 40 = 3.0 m/s
✓ 1
Part (d)
average speed = 320 ÷ 80
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= 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.
44IGPC-035-03-06Forces, friction and springsPaper 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?
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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.
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Answer 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.
45IGPC-035-03-14Forces, friction and springsPaper 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)
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Part (a)
To the left, opposite to the direction of motion.
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Part (b)
45 N . The speed is constant, so the resultant force is zero and friction balances the push.
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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.
46IGPC-035-02-04Mass, weight and densityPaper 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?
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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.
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Answer 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.
47IGPC-035-02-05Mass, weight and densityPaper 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?
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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.
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Answer C
Answer: C
Common error: Floating depends on density, not on mass alone. A light object can sink, and a heavy object can float.
48IGPC-035-02-07Mass, weight and densityPaper 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?
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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.
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Answer 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.
49IGPC-035-02-09Mass, weight and densityPaper 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?
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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.
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Answer B
Answer: B
Common error: Remember to subtract the mass of the empty container to find the mass of the liquid alone.
50IGPC-035-02-14Mass, weight and densityPaper 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)
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Part (a)
Partly fill the cylinder with water and record the volume reading
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Lower the key ring on the thread until it is completely under the water
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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
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• make sure no air bubbles cling to the key ring
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• use a narrower measuring cylinder with finer scale divisions
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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.
51IGPC-035-06-06Energy, work and powerPaper CMedium1 mark
Which energy resource is renewable and does not get its energy from radiation from the Sun?
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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.
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Answer C
Answer: C
Exam tip: The resources that do not come from the Sun's radiation are geothermal, nuclear and tidal. Learn these three.
52IGPC-035-06-07Energy, work and powerPaper 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?
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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.
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Answer 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.
53IGPC-035-00-04Physical quantities and measurementPaper 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?
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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.
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Answer A
Answer: A
Exam tip: Displacement method: find the rise in level first (final − initial), then divide by the number of objects.
54IGPC-035-00-08Physical quantities and measurementPaper 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?
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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.
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Answer 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.
55IGPC-035-00-09Physical quantities and measurementPaper 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?
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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.
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Answer B
Answer: B
Exam tip: Average the repeated readings first, then divide by the number of oscillations in each reading.
56IGPC-035-04-04Turning effect and centre of gravityPaper 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?
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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.
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Answer 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.
57IGPC-035-04-05Turning effect and centre of gravityPaper CMedium1 mark
A desk lamp falls over easily when it is knocked. Which change to its design would make it most stable?
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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.
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Answer 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.
58IGPC-035-04-06Turning effect and centre of gravityPaper CMedium1 mark
Which statement describes an object that is in equilibrium?
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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.
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Answer C
Answer: C
Common error: Equal forces on each side of a pivot do not guarantee balance. The moments must be equal.
59IGPC-035-04-09Turning effect and centre of gravityPaper 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?
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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.
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Answer B
Answer: B
Exam tip: A third hole and line gives a useful check: all three lines should meet at one point.
60IGPC-035-04-14Turning effect and centre of gravityPaper 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.
(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)
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Part (a)
Moment = 600 × 0.40
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= 240 N m
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Part (b)
F × 1.2 = 240
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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.
61IGPC-035-07-05Pressure in solids and liquidsPaper 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?
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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.
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Answer 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.
62IGPC-035-07-08Pressure in solids and liquidsPaper 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?
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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.
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Answer D
Answer: D
Common error: A greater volume of liquid does not mean a greater pressure at the base. Only depth and density matter.
63IGPC-035-07-13Pressure in solids and liquidsPaper 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)
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Part (a)
W = mg = 5000 × 9.8
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W = 49 000 N
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Part (b)
total area = 4 × 0.35 = 1.4 m2
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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.
64IGPC-035-07-15Pressure in solids and liquidsPaper 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.
(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)
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Part (a)
V = 0.50 × 0.40 × 0.20 = 0.040 m3
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Part (b)
m = ρV = 2400 × 0.040 = 96 kg
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W = 96 × 9.8 = 940 N (941 N)
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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.
65IGPC-035-06-14Energy, work and powerPaper 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)
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Part (a)
A resource that is replaced (by natural processes) at least as fast as it is used, so it will not run out
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Part (b)
Advantage: any one of: renewable / produces no carbon dioxide or polluting gases when operating / no fuel needed, so no fuel costs
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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.
66IGPC-035-00-05Physical quantities and measurementPaper 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?
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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.
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Answer 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.
67IGPC-035-01-10Speed, velocity and motion graphsPaper 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?
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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.
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Answer 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.
68IGPC-035-02-10Mass, weight and densityPaper 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?
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Notes
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.
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Answer 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.
69IGPC-035-04-08Turning effect and centre of gravityPaper 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?
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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.
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Answer 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.
70IGPC-035-04-10Turning effect and centre of gravityPaper 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?
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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.
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Answer 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.
71IGPC-035-01-16Speed, velocity and motion graphsPaper 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)
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Part (a)
distance = 20 × 0.70 = 14 m
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Part (b)
t = change in speed ÷ deceleration = 20 ÷ 5.0
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= 4.0 s
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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)
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= 40 m
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Part (d)
total distance = 14 + 40 = 54 m
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total time = 0.70 + 4.0 = 4.7 s
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average speed = 54 ÷ 4.7 = 11.5 m/s (accept 11 m/s)
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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.
72IGPC-035-00-16Physical quantities and measurementPaper 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.
(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)
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Part (a)
The volume of one ball bearing is smaller than one scale division
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so the rise in water level is too small to read, and the reading uncertainty is larger than the volume being measured
0.1646 cm3 is 0.16 cm3 to 2 significant figures, which is the same as the value from the measuring cylinder
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Exam tip: Measuring many identical objects and dividing works for both lengths and volumes. It is the same idea as timing many oscillations.
73IGPC-035-00-06Physical quantities and measurementPaper EEasy1 mark
Which list contains only vector quantities?
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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.
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Answer 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.
74IGPC-035-05-01Momentum and impulsePaper EEasy1 mark
A car of mass 1200 kg travels along a straight road at 15 m/s. What is the momentum of the car?
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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.
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Answer A
Answer: A
Common error: Momentum is mv. It has no ½ and no squared term. Those belong to kinetic energy.
75IGPC-035-05-02Momentum and impulsePaper EEasy1 mark
Which unit can be used for impulse?
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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.
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Answer B
Answer: B
Exam tip: N s and kg m/s are equivalent units. Either can be used for impulse or change in momentum.
76IGPC-035-05-03Momentum and impulsePaper EEasy1 mark
Which expression is equal to the resultant force acting on an object?
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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.
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Answer B
Answer: B
Exam tip: F = Δp/Δt is the general form of F = ma. Rearranged, it gives impulse: FΔt = Δp.
77IGPC-035-05-04Momentum and impulsePaper 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?
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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.
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Answer D
Answer: D
Common error: Conservation of momentum applies to the total for the whole system, not to each object separately.
78IGPC-035-05-11Momentum and impulsePaper 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)
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Part (a)
momentum = mass × velocity (p = mv)
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Part (b)
p = 55 × 6.0
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= 330 kg m/s
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Part (c)
It has both a magnitude (size) and a direction
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Exam tip: Always include a unit with a calculated answer. Either kg m/s or N s is acceptable for momentum.
79IGPC-035-05-12Momentum and impulsePaper 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)
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Part (a)
impulse = force × time (FΔt)
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Part (b)
impulse = 120 × 0.0050
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= 0.60 N s
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Part (c)
impulse = change in momentum, so 0.60 = 0.060 × v
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v = 10 m/s
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Exam tip: Impulse equals change in momentum. If the object starts from rest, the impulse equals its final momentum.
80IGPC-035-01-06Speed, velocity and motion graphsPaper 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?
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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.
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Answer 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.
81IGPC-035-01-14Speed, velocity and motion graphsPaper 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)
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Part (a)
speed falls by 9.8 m/s each second, so t = 14.7 ÷ 9.8
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= 1.5 s
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Part (b)
height = ½ × 1.5 × 14.7
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= 11 m (11.0 m)
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Part (c)
velocity includes direction, and the ball is now moving downwards, opposite to the positive direction
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speed is only the magnitude, with no direction, so it cannot be negative
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Exam tip: At the highest point the velocity is zero, but the acceleration is still 9.8 m/s² downwards.
82IGPC-035-03-04Forces, friction and springsPaper 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?
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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.
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Answer 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.
83IGPC-035-03-05Forces, friction and springsPaper 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?
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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.
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Answer B
Answer: B
Exam tip: In F = ma, F is always the resultant force, not just the driving force.
84IGPC-035-03-07Forces, friction and springsPaper 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?
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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.
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Answer 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.
85IGPC-035-03-09Forces, friction and springsPaper 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?
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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.
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Answer B
Answer: B
Common error: Constant speed does not mean constant velocity. A change of direction needs a resultant force towards the centre.
86IGPC-035-03-13Forces, friction and springsPaper 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.
(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)
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Part (a)
W = mg = 0.40 × 9.8 = 3.92 N (accept 3.9 N)
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Part (b)
15.5 − 12.0 = 3.5 cm (= 0.035 m)
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Part (c)
k = F/x = 3.92/0.035 = 112 N/m (accept 110 N/m)
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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
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Exam tip: Convert cm to m before using F = kx when k is in N/m.
87IGPC-035-02-06Mass, weight and densityPaper 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?
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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.
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Answer 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.
88IGPC-035-02-15Mass, weight and densityPaper 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)
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Part (a)
The (gravitational) force per unit mass
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Part (b)
g = W ÷ m = 9.25 ÷ 2.5
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= 3.7 N/kg
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Part (c)
The test mass and the standard masses are in the same gravitational field, so their weights change by the same factor
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The balance therefore compares masses, and mass does not change with location
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Part (d)
3.7 m/s2
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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.
89IGPC-035-06-05Energy, work and powerPaper 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?
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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.
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Answer D
Answer: D
Common error: Mass in kg must be multiplied by g to give weight in N before multiplying by the height gained.
90IGPC-035-06-09Energy, work and powerPaper 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?
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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.
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Answer C
Answer: C
Common error: Check whether the question asks for useful or wasted energy, and convert minutes to seconds before using E = Pt.
91IGPC-035-00-07Physical quantities and measurementPaper 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?
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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.
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Answer C
Answer: C
Common error: Vectors at right angles cannot be added like ordinary numbers. Use Pythagoras or a scale drawing.
92IGPC-035-00-14Physical quantities and measurementPaper EMedium7 marks
Physical quantities can be classified as scalars or vectors.
(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)
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Part (a)
A scalar has magnitude only, but a vector has both magnitude and direction
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Part (b)
Momentum and gravitational field strength are vectors
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Distance and temperature are scalars
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Part (c)
Resultant = √(9.0² + 12²)
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= 15 N
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Part (d)
tan θ = 12 ÷ 9.0
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θ = 53° (north of east)
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Exam tip: A scale drawing is also acceptable for a right-angled resultant. State your scale and measure the angle carefully.
93IGPC-035-04-07Turning effect and centre of gravityPaper 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?
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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.
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Answer C
Answer: C
Common error: Scale readings on a metre rule are not distances from the pivot. Subtract the pivot position first.
94IGPC-035-07-06Pressure in solids and liquidsPaper 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?
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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.
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Answer 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.
95IGPC-035-07-07Pressure in solids and liquidsPaper 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?
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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.
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Answer D
Answer: D
Exam tip: Check that your answer makes sense: a less dense liquid must be deeper to produce the same pressure.
96IGPC-035-05-05Momentum and impulsePaper 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?
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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.
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Answer C
Answer: C
Common error: When an object rebounds, the speeds add when finding the change in velocity, because the direction reverses.
97IGPC-035-05-06Momentum and impulsePaper 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?
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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.
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Answer A
Answer: A
Exam tip: When objects stick together, divide the total momentum by the combined mass.
98IGPC-035-05-07Momentum and impulsePaper 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?
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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.
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Answer D
Answer: D
Common error: Force = change in momentum ÷ time. A short contact time gives a large force.
99IGPC-035-05-08Momentum and impulsePaper 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?
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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.
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Answer 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.
100IGPC-035-05-13Momentum and impulsePaper 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)
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Part (a)
Δp = 900 × 12
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= 10 800 kg m/s
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Part (b)
F = Δp ÷ Δt = 10 800 ÷ 0.40
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= 27 000 N
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Part (c)
The driver's change in momentum is the same with or without the airbag
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The airbag increases the time taken for the driver to stop
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Force = change in momentum ÷ time, so the force is smaller
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Exam tip: In safety-feature explanations, state that the change in momentum is fixed, that the time increases, and that therefore the force decreases.
101IGPC-035-05-14Momentum and impulsePaper 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)
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)
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Part (a)
When no external resultant force acts on a system, its total momentum stays constant (total momentum before = total momentum after)
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Part (b)
momentum before = 8000 × 2.5 = 20 000 kg m/s
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20 000 = (8000 + 12 000) × v
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v = 1.0 m/s
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Part (c)
KE before = ½ × 8000 × 2.52 = 25 000 J
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KE after = ½ × 20 000 × 1.02 = 10 000 J
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KE lost = 25 000 − 10 000 = 15 000 J (1.5 × 104 J)
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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
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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.
102IGPC-035-05-15Momentum and impulsePaper 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)
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Part (a)
There is no external force, so momentum is conserved and the total momentum is zero both before and after the throw
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The tool gains momentum away from the station, so the astronaut gains an equal momentum in the opposite direction, towards the station
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Part (b)
90 × v = 3.0 × 6.0
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v = 0.20 m/s
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Part (c)
t = 5.0 ÷ 0.20 = 25 s
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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.
103IGPC-035-01-13Speed, velocity and motion graphsPaper 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.
(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)
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Part (a)
uses area under speed–time graph; e.g. accelerating section ½ × 1.5 × 2.0 = 1.5 m
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constant-speed section 2.0 × 6.0 = 12 m and decelerating section ½ × 2.5 × 2.0 = 2.5 m
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total = 16 m
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Part (b)
deceleration = change in speed ÷ time = 2.0 ÷ 2.5
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= 0.80 m/s2 (accept acceleration = −0.80 m/s2)
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Part (c)
total time = 10.0 s; average speed = total distance ÷ total time = 16 ÷ 10.0 (allow their distance from the first part)
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= 1.6 m/s
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Exam tip: Split the area under a speed–time graph into triangles and rectangles. Calculate each one separately, then add them.
104IGPC-035-04-15Turning effect and centre of gravityPaper 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.
(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)
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Part (a)
For an object in equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments about that point
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Part (b)
Anticlockwise moments about the pivot = 0.50 × 0.35 + 0.30 × 0.20 = 0.235 N m
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W × 0.30 = 0.235
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W = 0.78 N (accept 0.783 N)
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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
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Part (d)
There is no resultant force, so the upward force equals the total downward force: 0.50 + 0.30 + 0.78 + 1.2
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= 2.8 N
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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.
105IGPC-035-06-13Energy, work and powerPaper 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.
(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)
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Part (a)
ΔEp = mgΔh = 55 × 9.8 × 10
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= 5390 J (accept 5400 J)
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Part (b)
Gain in Ek = loss in Ep, so ½mv2 = 5390 J (or ½v2 = gh)
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.
106IGPC-035-01-08Speed, velocity and motion graphsPaper 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?
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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.
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Answer 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.
107IGPC-035-03-08Forces, friction and springsPaper 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?
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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.
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Answer D
Answer: D
Common error: When springs hang one below another, each spring carries the whole load. The extensions then add together.
108IGPC-035-03-15Forces, friction and springsPaper 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)
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Part (a)
W = 75 × 9.8 = 735 N
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Part (b)
Resultant force = 735 − 480 = 255 N . a = F/m = 255/75 = 3.4 m/s² (downwards)
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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.
109IGPC-035-03-16Forces, friction and springsPaper 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)
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Part (a)
a = Δv/t = 20/4.0 = 5.0 m/s²
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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.
110IGPC-035-02-08Mass, weight and densityPaper 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?
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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.
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Answer 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.
111IGPC-035-06-08Energy, work and powerPaper 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?
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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.
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Answer 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.
112IGPC-035-06-16Energy, work and powerPaper 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)
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Part (a)
Ek = ½mv2 = 0.5 × 900 × 202
✓ 1
= 180 000 J
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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.
113IGPC-035-00-10Physical quantities and measurementPaper 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?
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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.
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Answer B
Answer: B
Common error: Check which direction your angle is measured from. Here tan θ = opposite/adjacent with north as the adjacent side.
114IGPC-035-07-09Pressure in solids and liquidsPaper 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?
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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.
✓ 1
Answer 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.
115IGPC-035-07-10Pressure in solids and liquidsPaper 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?
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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.
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Answer 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.
116IGPC-035-05-09Momentum and impulsePaper 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?
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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.
✓ 1
Answer D
Answer: D
Common error: Give a rebounding velocity a negative sign before applying conservation of momentum.
117IGPC-035-05-10Momentum and impulsePaper 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?
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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.
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Answer C
Answer: C
Exam tip: Impulse gives the change in momentum. Add it to the initial momentum to find the final momentum.
118IGPC-035-05-16Momentum and impulsePaper 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.
(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)
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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.
119IGPC-035-00-15Physical quantities and measurementPaper 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.
(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)
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Part (a)
v = √(3.02 + 1.22)
✓ 1
= 3.2 m/s
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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.
120IGPC-035-04-16Turning effect and centre of gravityPaper 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.
(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)
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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.
121IGPC-035-07-16Pressure in solids and liquidsPaper 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.
(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)
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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.
122IGPC-035-02-16Mass, weight and densityPaper 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.
(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)
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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.
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