4.5 Forces — revision question pack

25 specification points · notes, questions, answers and worked methods

Checked against AQA 8463 section 4.5. Review basis: the qualification registry sourced from the AQA GCSE Physics (8463) specification; registry verification recorded 17 July 2026.

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4.5.1.1 · Scalar and vector quantities

Explanation

A scalar quantity has magnitude only, whereas a vector quantity has both magnitude and an associated direction.
Classify a stated quantity by checking whether its direction is needed to specify it completely; force, weight and velocity are vectors, while mass, energy and speed are scalars.
A vector can be shown by an arrow: its length represents the magnitude and its arrowhead shows the direction.
Do not call a quantity a vector merely because it can be large or negative; a vector must include direction.

Worked example

A trolley has a mass of $6.0\,\text{kg}$ and moves with a velocity of $2.5\,\text{m/s}$ east. State which of these two quantities is a vector.

1.Mass needs only a magnitude, so it is scalar.
2.The velocity includes the direction east, so velocity is the vector.

Answer: Velocity is the vector quantity.

Common mistakes

Don't fall into the trap of calling speed a vector because an object can travel in a stated direction.
Don't fall into the trap of describing a vector only by its magnitude and omitting the direction.

Exam tip

For ‘state whether scalar or vector’, decide whether direction is needed to specify the quantity completely.

Tier 1 · Easy

Explain why a speedometer reading alone cannot give a vehicle's velocity.
[2 marks]
Total for this question: 2
An annotated force arrow is $4.0\,\text{cm}$ long and points west. The diagram scale is $1.0\,\text{cm}$ for $6.0\,\text{N}$ , but the label says ‘ $18\,\text{N}$ east’. Identify both errors and give the correct label.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A scale drawing uses $1.0\,\text{cm}$ to represent $4.0\,\text{N}$ . Describe the arrow that represents a force of $20\,\text{N}$ acting vertically downwards.
[2 marks]
Total for this question: 2
A journey is described by a distance of $2.4\,\text{km}$ , a displacement of $1.8\,\text{km}$ north, a speed of $6.0\,\text{m/s}$ and a velocity of $4.5\,\text{m/s}$ north. Give the two scalar quantities and the two vector quantities, and justify your answer.
[3 marks]
Total for this question: 3
A thermometer records a temperature of $-5\,^\circ\text{C}$ . Taking forwards as positive, a force sensor records $-12\,\text{N}$ , meaning a $12\,\text{N}$ force acting backwards. Classify temperature and force as scalar or vector quantities, and explain what the minus sign means in each case.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A cyclist's speedometer reads $11\,\text{m/s}$ while the cyclist travels south-west. Give the cyclist's speed and velocity, and explain why these are different quantities even though their magnitudes match.
[3 marks]
Total for this question: 3
On a map, a displacement arrow is $7.2\,\text{cm}$ long on a bearing of $135^\circ$ . The scale is $1.0\,\text{cm}$ for $150\,\text{m}$ . Determine the displacement, then explain why the distance travelled cannot be found from this arrow alone.
[4 marks]
Total for this question: 4
Two force arrows start at the same point and have equal lengths. One points north-east and the other points south-west. Compare their magnitudes and directions, then explain why equal arrow lengths do not mean that the vectors are equal.
[3 marks]
Total for this question: 3
Two force sensors measure the same towing force. Sensor A defines east as positive and reads $-36\,\text{N}$ . Sensor B defines west as positive and reads $+36\,\text{N}$ . Determine the force vector reported by each sensor and explain why the two readings describe the same vector.
[4 marks]
Total for this question: 4
Three diagrams use different arrow scales. Diagram A shows a $6.0\,\text{cm}$ arrow east with $1.0\,\text{cm}=4.0\,\text{N}$ . Diagram B shows a $3.0\,\text{cm}$ arrow east with $1.0\,\text{cm}=8.0\,\text{N}$ . Diagram C shows a $5.0\,\text{cm}$ arrow west with $1.0\,\text{cm}=5.0\,\text{N}$ . Determine each force and compare the three vectors.
[5 marks]
Total for this question: 5

4.5.1.2 · Contact and non-contact forces

Explanation

A force is a push or pull caused by an interaction between objects; force is a vector quantity.
Decide whether the objects must touch: friction, air resistance, tension and normal contact force are contact forces, while gravitational, electrostatic and magnetic forces are non-contact forces.
An interaction produces a force on each object; represent each force with a vector arrow on the object that experiences it.
Do not describe air resistance as non-contact just because air is hard to see: collisions with air particles make it a contact force.

Worked example

State whether friction and gravitational force are contact or non-contact forces.

1.Friction requires touching surfaces.
2.Gravitational force acts between separated masses, so it does not require contact.

Answer: Friction is contact; gravitational force is non-contact.

Common mistakes

Don't fall into the trap of classifying air resistance as non-contact because air cannot easily be seen.
Don't fall into the trap of naming friction without stating the two surfaces whose contact produces it.

Exam tip

For each force, ask whether the interacting objects must touch before classifying it.

Tier 1 · Easy

Air resistance acts on a cyclist. State whether air resistance is a contact or non-contact force and explain your choice.
[2 marks]
Total for this question: 2
A charged balloon pulls a strip of paper across a gap. A student's annotation calls the force friction. Identify the error and give the correct force. State whether the force is a contact or a non-contact force, and justify your answer.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A magnet attracts an iron pin across a small gap. Describe the force interaction between the magnet and the pin.
[2 marks]
Total for this question: 2
In trial A, an iron ball rolls towards a magnet while a $5\,\text{mm}$ gap remains. In trial B, a rubber block slows only while it touches a mat. Name the force responsible in each trial. State whether each force is a contact or a non-contact force, using the observations as evidence.
[4 marks]
Total for this question: 4
A ball rests on a tray. The tray is suddenly moved downwards so that it loses contact with the ball. State which force on the ball disappears and which force remains, explaining each answer using contact and non-contact interactions.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A parachutist is falling through the air while attached to an open parachute by cords. Name one non-contact force and two different contact forces acting in this situation, giving the object on which each named force acts.
[3 marks]
Total for this question: 3
A steel washer hangs from a cotton thread above an electromagnet without touching it. For the forces on the washer due to Earth, the thread and the electromagnet, name each force and state the interacting object pair. State whether each force is a contact or a non-contact force.
[6 marks]
Total for this question: 6
A charged plastic rod is held near a small oppositely charged metal sphere suspended by an insulating thread. The sphere moves towards the rod before they touch. Use the observation to identify the force on the sphere, classify it, and explain why neither friction nor tension caused the initial sideways motion.
[4 marks]
Total for this question: 4
A charged oil droplet falls through air between two charged plates without touching either plate. At one instant, Earth pulls the droplet down, the electric field pulls it up and the air resists its downward motion. Name the three forces, give the interacting object pair for each and state whether each force is contact or non-contact.
[6 marks]
Total for this question: 6
Two carts collide head-on and then separate. While their buffers touch, cart P pushes cart Q to the right and cart Q pushes cart P to the left. After contact ends, magnets on the carts repel across a gap. Compare the two interactions by naming the force type, giving whether it is contact or non-contact and giving the force direction on each cart.
[5 marks]
Total for this question: 5

4.5.1.3 · Gravity

Explanation

Weight is the force on an object due to gravity, while mass measures the amount of matter and does not change when gravitational field strength changes.
Calculate weight using $W=mg$ , with $W$ in newtons, $m$ in kilograms and $g$ in newtons per kilogram; use the value of $g$ supplied in the question.
For example, a $3.0\,\text{kg}$ object where $g=9.8\,\text{N/kg}$ has weight $3.0\times9.8=29.4\,\text{N}$ , acting through its centre of mass.
Do not give weight in kilograms or assume it is constant everywhere; weight changes with $g$ and is measured with a calibrated newtonmeter.

Worked example

A $7.5\,\text{kg}$ bag is in a region where $g=9.8\,\text{N/kg}$ . Calculate the bag's weight.

1.Use $W=mg$ : $W=7.5\times9.8=73.5\,\text{N}$ .

Answer: $73.5\,\text{N}$

Common mistakes

Don't fall into the trap of giving weight in kilograms instead of newtons.
Don't fall into the trap of saying mass changes on another planet when it is the gravitational field strength and weight that change.

Exam tip

In a $W=mg$ calculation, use mass in kilograms and the value of $g$ supplied in the question.

Tier 1 · Easy

An astronaut travels from Earth to the Moon, where gravitational field strength is smaller. State what happens to the astronaut's mass and weight.
[2 marks]
Total for this question: 2
The same sealed sample gives newtonmeter readings of $14.4\,\text{N}$ at site A and $9.6\,\text{N}$ at site B. Without calculating, identify the site with the greater gravitational field strength and explain what happens to the sample's mass between the sites.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A sample has mass $2.4\,\text{kg}$ . Its weight is $23.5\,\text{N}$ at location A and $3.8\,\text{N}$ at location B. Calculate the gravitational field strength at each location.
[3 marks]
Total for this question: 3
A student measures mass and weight at one location. The results are $1.2\,\text{kg}$ and $4.08\,\text{N}$ , $2.0\,\text{kg}$ and $6.80\,\text{N}$ , and $2.8\,\text{kg}$ and $10.5\,\text{N}$ . Use $W=mg$ to identify the anomalous pair and determine the gravitational field strength from the consistent results.
[4 marks]
Total for this question: 4
A weight arrow on a diagram of a uniform rectangular block starts at an upper corner and points horizontally. State where the arrow should start and which way it should point, explaining both corrections.
[3 marks]
Total for this question: 3

Tier 3 · Hard

An explorer has mass $68\,\text{kg}$ . A newtonmeter would show $666\,\text{N}$ for the explorer on planet P and $177\,\text{N}$ on moon Q. Determine $g$ at P and Q, then explain what happens to the explorer's mass during the journey.
[4 marks]
Total for this question: 4
An empty capsule weighs $48.6\,\text{N}$ at a test site on another planet. Adding a $750\,\text{g}$ sample makes the reading $55.35\,\text{N}$ . Use $W=mg$ to determine the gravitational field strength at the site and the mass of the empty capsule.
[4 marks]
Total for this question: 4
A lander and its cargo have a combined weight of $18.0\,\text{kN}$ where $g=4.0\,\text{N/kg}$ . The empty lander has mass $3.20\times10^3\,\text{kg}$ . Calculate the cargo mass, then calculate the combined weight where $g=9.8\,\text{N/kg}$ .
[5 marks]
Total for this question: 5
A weight–mass graph is a straight line through the origin. Two points on it are $(0.50\,\text{kg},\ 4.90\,\text{N})$ and $(1.80\,\text{kg},\ 17.64\,\text{N})$ . Determine the gravitational field strength from the gradient, predict the weight of a $3.25\,\text{kg}$ object and explain the significance of the line passing through the origin.
[5 marks]
Total for this question: 5
A newtonmeter has a positive zero error: it reads $1.4\,\text{N}$ when unloaded. On Earth, where $g=9.8\,\text{N/kg}$ , it reads $30.8\,\text{N}$ with a rock attached. Determine the rock's true weight and mass. The same meter and rock are taken to a place where $g=3.7\,\text{N/kg}$ . Determine the reading if the zero error is unchanged.
[5 marks]
Total for this question: 5

4.5.1.4 · Resultant forces

Explanation

The resultant force is the single force that has the same effect as all the forces acting together. For forces along one straight line, choose a positive direction, give opposite forces opposite signs and add them.
For example, $18\,\text{N}$ right and $11\,\text{N}$ left give a resultant of $18-11=7\,\text{N}$ to the right; equal opposing forces give zero resultant. Higher tier: use a free-body diagram to isolate an object or system and show every force as a labelled arrow; several forces may combine to give a non-zero resultant or balance to zero.
A single force can be resolved into two components at right angles whose combined effect is the original force.
In a scale vector diagram, draw force arrows to scale and in the correct directions, place them head-to-tail, then measure the resultant from the start of the first arrow to the end of the last; a closed diagram represents equilibrium.
Do not add magnitudes when forces oppose, and do not omit the direction of a non-zero resultant because force is a vector.

Opposing forces and their resultant.

Worked example

Two horizontal forces act on a crate: $16\,\text{N}$ east and $9\,\text{N}$ west. Calculate the resultant force.

1.The forces oppose, so subtract their magnitudes: $16-9=7\,\text{N}$ .
2.The larger force acts east, so the resultant is east.

Answer: $7\,\text{N}$ east

Common mistakes

Don't fall into the trap of adding the magnitudes of forces that act in opposite directions.
Don't fall into the trap of giving only $7\,\text{N}$ and omitting the direction of the resultant force.

Exam tip

Choose a positive direction, combine signed forces, then state the resultant's magnitude and direction.

Tier 1 · Easy

A stationary box has two equal horizontal forces acting on it in opposite directions. State the resultant force and the effect on the box's motion.
[2 marks]
Total for this question: 2
A diagram shows $31\,\text{N}$ acting right and $12\,\text{N}$ acting left. A student labels the resultant as $43\,\text{N}$ right. Identify the error and give the correct resultant force.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A model boat is pulled forwards by $42\,\text{N}$ . Water resistance is $27\,\text{N}$ and air resistance is $6\,\text{N}$ , both backwards. Determine the resultant force on the boat.
[2 marks]
Total for this question: 2
In test A, an eastward thruster provides $64\,\text{N}$ while an unknown force acts west, producing a resultant of $18\,\text{N}$ east. In test B, the same westward force acts but the eastward thrust is reduced by $10\,\text{N}$ . Determine the unknown force and the resultant in test B.
[4 marks]
Total for this question: 4
A crate has a $20\,\text{N}$ force acting east. A westward force is increased through $12\,\text{N}$ , $20\,\text{N}$ and $27\,\text{N}$ . Determine the resultant force for each value, including its direction, and identify when the forces are balanced.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Forces on a rail cart are $85\,\text{N}$ east, $34\,\text{N}$ west and $19\,\text{N}$ west. Calculate the resultant, then state the additional single force needed to make the forces balanced.
[3 marks]
Total for this question: 3
Higher only: Two cables pull a ring with forces of $6.0\,\text{N}$ east and $8.0\,\text{N}$ north. Using a scale of $1.0\,\text{cm}$ for $2.0\,\text{N}$ , a student draws a $3.0\,\text{cm}$ east arrow followed head-to-tail by an $8.0\,\text{cm}$ north arrow. Identify and correct the error, then determine the resultant's magnitude and direction.
[5 marks]
Total for this question: 5
Higher only: Two ropes pull a ring: a $12\,\text{N}$ force due east and a $9.0\,\text{N}$ force north-east. Using a scale of $1.0\,\text{cm} = 2.0\,\text{N}$ , draw a scale vector diagram to find the magnitude and direction of the resultant force.
[4 marks]
Total for this question: 4
Higher only: A lift moves upwards at constant velocity. Its cable tension is $6200\,\text{N}$ upwards and its weight is $5800\,\text{N}$ downwards. Air resistance also acts. Describe a free-body diagram for the isolated lift and determine the air resistance. The tension then increases to $6600\,\text{N}$ while the other forces are unchanged. Determine the new resultant and describe the change in motion.
[6 marks]
Total for this question: 6
Higher only: Two ropes pull a ring with forces of $10\,\text{N}$ east and $24\,\text{N}$ north. Using only a scale drawing with $1.0\,\text{cm}=4.0\,\text{N}$ , determine the magnitude and direction of the resultant. State the single force that would produce equilibrium.
[5 marks]
Total for this question: 5

4.5.2 · Work done and energy transfer

Explanation

Work is done when a force causes a displacement, transferring energy between stores; work done against friction raises temperature.
Use $W=Fs$ , where $s$ is the distance moved along the force's line of action; $W$ is in joules, $F$ in newtons and $s$ in metres.
For example, a $25\,\text{N}$ force moving an object $3.0\,\text{m}$ along its line of action does $25\times3.0=75\,\text{J}$ of work.
Do not multiply by a distance perpendicular to the force, and remember that $1\,\text{J}=1\,\text{N m}$ .
In an exam, apply the named relationship to the quantities, units and direction given in the question.

Worked example

A horizontal force of $35\,\text{N}$ moves a box $4.0\,\text{m}$ horizontally. Calculate the work done by the force.

1.The movement is along the force's line of action, so $W=Fs=35\times4.0=140\,\text{J}$ .

Answer: $140\,\text{J}$

Common mistakes

Don't fall into the trap of using a distance perpendicular to the force in $W=Fs$ .
Don't fall into the trap of saying friction destroys energy instead of transferring it to thermal energy stores.

Exam tip

Use the distance moved along the force's line of action and give work done in joules.

Tier 1 · Easy

A person pushes horizontally on a rigid wall, but the wall does not move. Calculate the work done on the wall and explain your answer.
[2 marks]
Total for this question: 2
A porter carries a suitcase horizontally at constant height. Explain why the upward supporting force from the porter does no work on the suitcase during the horizontal movement.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A winch lifts a load vertically through $6.5\,\text{m}$ using a constant upward force of $480\,\text{N}$ . Calculate the work done and identify the energy store that increases.
[3 marks]
Total for this question: 3
A trolley travels from the same storeroom to the same loading bay by either route A, which is $18\,\text{m}$ long, or route B, which is $24\,\text{m}$ long. Its motor exerts a constant $75\,\text{N}$ force along the direction of motion on either route. A technician claims that the motor does the same work because both routes have the same displacement. Calculate the work done on each route and evaluate the claim.
[4 marks]
Total for this question: 4
A constant force acts along the motion of a trolley. Work done is $360\,\text{J}$ after $2.0\,\text{m}$ , $900\,\text{J}$ after $5.0\,\text{m}$ and $1440\,\text{J}$ after $8.0\,\text{m}$ . Use the data to determine the force and predict the work done after $6.5\,\text{m}$ .
[4 marks]
Total for this question: 4

Tier 3 · Hard

A powered trolley moves $12\,\text{m}$ along a level floor. Its motor provides a forward force of $180\,\text{N}$ while friction is $140\,\text{N}$ . Calculate the work done by the motor, the energy transferred thermally by friction and the remaining energy transferred to the trolley's kinetic energy store.
[5 marks]
Total for this question: 5
A braking force transfers $18\,\text{kJ}$ from a vehicle's kinetic energy store while stopping it over $24\,\text{m}$ . On a second surface, the same average braking force transfers $7.5\,\text{kJ}$ . Use $W=Fs$ to determine the braking force and the second stopping distance, and state the thermal effect of the work against friction.
[5 marks]
Total for this question: 5
A cart is pushed $5.0\,\text{m}$ with a force of $120\,\text{N}$ and then a further $8.0\,\text{m}$ with a different constant force. The total work done is $1.32\,\text{kJ}$ . Calculate the second force and state what the work done represents.
[5 marks]
Total for this question: 5
A sledge is pulled by a constant $85\,\text{N}$ force acting along its motion against a constant friction force of $55\,\text{N}$ . Its kinetic energy store increases by $900\,\text{J}$ . Calculate the resultant force, determine the distance travelled and calculate the energy transferred thermally by friction.
[5 marks]
Total for this question: 5
A winch exerts a constant $1.5\,\text{kN}$ force along the motion of a cable for $18\,\text{m}$ . A logger records $27\,\text{kJ}$ of work. Verify the logger reading and express this work in newton-metres. In a second test, the same force transfers $4.8\,\text{kJ}$ . Determine the cable movement and explain why joules and newton-metres have the same value for work.
[5 marks]
Total for this question: 5

4.5.3 · Forces and elasticity

Explanation

A stationary object needs more than one force to change shape: one unbalanced force would accelerate the object as a whole, whereas forces acting at different positions can change the separation of its parts. Equal outward pulls stretch an object, forces that turn different parts in opposite directions bend it, and inward pushes from opposite ends compress it.
Elastic deformation is reversed when the forces are removed; inelastic deformation leaves the object permanently changed. Up to the limit of proportionality use $F=ke$ , measuring extension $e$ from the original length; a straight force-extension graph through the origin represents direct proportion.
For the required practical, clamp a spring beside a millimetre ruler and record its unloaded length. Add known masses one at a time, let the spring stop moving, record each new length, calculate extension and convert each mass to force using $W=mg$ ;
repeat readings, plot force against extension and unload the spring to check that it returns to its original length. For example, a spring with $k=160\,\text{N/m}$ extended by $0.050\,\text{m}$ needs $F=160\times0.050=8.0\,\text{N}$ and stores $E_e=\frac12ke^2=0.20\,\text{J}$ .
Do not substitute the spring's total length for extension, and do not apply the linear relationship beyond the limit of proportionality.

Force–extension graph with its linear region.

Worked example

A spring's length changes from $0.18\,\text{m}$ to $0.23\,\text{m}$ . Calculate its extension.

1.Extension is stretched length minus original length: $e=0.23-0.18=0.050\,\text{m}$ .

Answer: $0.050\,\text{m}$

Common mistakes

Don't fall into the trap of using the spring's total length as $e$ instead of calculating its extension.
Don't fall into the trap of extending the straight-line $F=ke$ relationship beyond the limit of proportionality.

Exam tip

Calculate extension as stretched length minus original length before using $F=ke$ .

Tier 1 · Easy

A wire stays longer after the stretching forces are removed. State the type of deformation and explain how the observation identifies it.
[2 marks]
Total for this question: 2
An unloaded spring is $12.0\,\text{cm}$ long. A student plots spring length against force, sees that the graph starts at $12.0\,\text{cm}$ rather than the origin, and concludes that the spring is inelastically deformed. Explain the student's error and state what should be plotted instead of spring length.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Within its linear region, a spring extends by $0.075\,\text{m}$ when a force of $18\,\text{N}$ is applied. Calculate the spring constant and predict the extension produced by $12\,\text{N}$ .
[3 marks]
Total for this question: 3
In a force–extension investigation, forces of $1.0\,\text{N}$ , $2.0\,\text{N}$ and $3.0\,\text{N}$ produce extensions of $0.80\,\text{cm}$ , $1.60\,\text{cm}$ and $2.40\,\text{cm}$ . Determine the spring constant from the gradient and describe one repeat check that would improve the data.
[4 marks]
Total for this question: 4
A spring has a natural length of $10.0\,\text{cm}$ . With forces of $2.0$ , $4.0$ , $6.0$ and $8.0\,\text{N}$ , its loaded lengths are $11.2$ , $12.4$ , $13.6$ and $15.4\,\text{cm}$ . Identify the greatest force for which extension is proportional to force, and justify your answer using the data.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A spring of constant $240\,\text{N/m}$ is stretched by $0.18\,\text{m}$ without exceeding its limit of proportionality. Calculate the applied force and the elastic potential energy stored. State what observation after unloading would show that the spring had instead been inelastically deformed.
[5 marks]
Total for this question: 5
A student clamps a spring beside a ruler, adds several masses at once, records each length while the spring is oscillating, plots mass against length and stops after the final loaded reading. Describe a corrected procedure for required practical 6 that produces valid force–extension data and checks for inelastic deformation.
[5 marks]
Total for this question: 5
A spring stores $1.80\,\text{J}$ when its extension is $0.120\,\text{m}$ . Both this extension and an extension of $0.200\,\text{m}$ are within the proportional region. Calculate the energy stored at $0.200\,\text{m}$ and determine the percentage increase in stored energy.
[5 marks]
Total for this question: 5
A spring buffer stores $1.44\,\text{J}$ when compressed by $0.080\,\text{m}$ within its proportional region. Use the elastic potential energy equation to determine the spring constant and the compressing force. Calculate the energy stored when the compression is reduced to $0.050\,\text{m}$ .
[5 marks]
Total for this question: 5
A spring has a natural length of $15.0\,\text{cm}$ . Forces of $3.0$ , $6.0$ , $9.0$ and $12.0\,\text{N}$ produce loaded lengths of $16.5$ , $18.0$ , $19.5$ and $21.8\,\text{cm}$ . After all forces are removed, the spring is $15.4\,\text{cm}$ long. Determine the spring constant in the proportional region, calculate the energy stored at $9.0\,\text{N}$ , compare the measured extension at $12.0\,\text{N}$ with the proportional prediction and use the unloading evidence to explain what happened.
[6 marks]
Total for this question: 6

4.5.4 · Moments, levers and gears (physics only)

Explanation

A moment is the turning effect of a force about a pivot, and its size is $M=Fd$ where $d$ is the perpendicular distance to the force's line of action. For a balanced object, choose one pivot and set the total clockwise moment equal to the total anticlockwise moment.
For example, $15\,\text{N}$ acting $0.40\,\text{m}$ from a pivot produces a moment of $6.0\,\text{N m}$ .
A lever transmits a rotational effect about its pivot: for the same input force, applying it farther from the pivot gives a larger moment, so a long lever can produce a larger output force nearer the pivot.
Meshing gear teeth transmit a tangential force and rotation; when a smaller driving gear turns a larger gear, the larger gear turns more slowly and in the opposite direction but delivers a larger moment because the force acts at a larger radius.
Do not use a sloping distance measured to the point where the force is applied; the equation requires the perpendicular distance to the line of action.

A force acting on a lever at a perpendicular distance from the pivot.

Worked example

A force of $28\,\text{N}$ acts perpendicular to a handle $0.35\,\text{m}$ from its pivot. Calculate the moment.

1.Use $M=Fd$ : $M=28\times0.35=9.8\,\text{N m}$ .

Answer: $9.8\,\text{N m}$

Common mistakes

Don't fall into the trap of using the full lever length instead of the perpendicular distance from pivot to force line.
Don't fall into the trap of giving the moment in newtons rather than newton metres.

Exam tip

For equilibrium, compare total clockwise and anticlockwise moments about the same pivot.

Tier 1 · Easy

Explain why applying the same perpendicular force at the end of a longer spanner produces a greater turning effect.
[2 marks]
Total for this question: 2
A $30\,\text{N}$ force is applied at the end of an angled spanner. The hand is $0.50\,\text{m}$ from the pivot, but the shortest distance between the pivot and the line along which the force acts is $0.42\,\text{m}$ . A student calculates a moment of $15\,\text{N m}$ . Identify the error and calculate the correct moment.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A seesaw is balanced. A child of weight $360\,\text{N}$ sits $1.5\,\text{m}$ to the left of the pivot. Calculate how far to the right of the pivot a child of weight $450\,\text{N}$ must sit.
[3 marks]
Total for this question: 3
The same tangential force of $45\,\text{N}$ acts where two meshing gears touch. The driving gear has radius $0.030\,\text{m}$ and the other gear has radius $0.090\,\text{m}$ . Calculate the moment on each gear, then compare the gears' directions of rotation and rotational speeds.
[4 marks]
Total for this question: 4
A latch needs a moment of at least $25\,\text{N m}$ . Design A applies a perpendicular force of $60\,\text{N}$ at $0.45\,\text{m}$ from the pivot. Design B applies $80\,\text{N}$ at $0.30\,\text{m}$ . Calculate both moments and select every design that opens the latch.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A uniform $4.0\,\text{m}$ beam of weight $220\,\text{N}$ is supported at its left end and at a point $3.2\,\text{m}$ from the left end. A $300\,\text{N}$ load is placed at the right end. Calculate the upward force from the support at $3.2\,\text{m}$ .
[4 marks]
Total for this question: 4
A uniform $3.0\,\text{m}$ beam is pivoted $1.0\,\text{m}$ from its left end, so its weight acts $0.50\,\text{m}$ to the right of the pivot. In configuration A, a $180\,\text{N}$ load at the left end balances a $60\,\text{N}$ load at the right end. Determine the beam's weight. In configuration B, the $60\,\text{N}$ load is removed; determine where a $75\,\text{N}$ load must be placed to balance the beam.
[5 marks]
Total for this question: 5
Two light levers that turn freely transmit a force. On lever A, a $150\,\text{N}$ input acts $0.24\,\text{m}$ from the pivot and the output acts $0.060\,\text{m}$ from it. That output becomes the input to lever B at $0.15\,\text{m}$ from its pivot; lever B's output acts $0.050\,\text{m}$ from the pivot. Calculate the final output force.
[4 marks]
Total for this question: 4
A light lever that turns freely has a $110\,\text{N}$ input force acting $0.40\,\text{m}$ from its pivot. Its output force acts $0.055\,\text{m}$ from the pivot and then acts normally on a door $0.12\,\text{m}$ from the door hinge. Calculate the lever's output force and the moment on the door. A moment of at least $90\,\text{N m}$ opens the door. Determine whether it opens.
[5 marks]
Total for this question: 5
A light lifting arm that turns freely is pivoted between a counterweight and a load. A $2400\,\text{N}$ counterweight acts $1.8\,\text{m}$ to the left of the pivot. On the right, a cable pulls down with $300\,\text{N}$ at $0.60\,\text{m}$ and a $1200\,\text{N}$ load acts at an unknown distance. Determine the load distance for balance. The load is then moved $0.20\,\text{m}$ farther from the pivot. Determine the resultant moment and its direction.
[5 marks]
Total for this question: 5

4.5.5.1.1 · Pressure in a fluid 1 (physics only)

Explanation

A fluid is a liquid or a gas, and fluid pressure produces a force normal, or at right angles, to a surface.
Calculate pressure using $p=F/A$ , with normal force $F$ in newtons, surface area $A$ in square metres and pressure $p$ in pascals.
For example, a normal force of $90\,\text{N}$ on $0.030\,\text{m}^2$ produces $p=90\div0.030=3000\,\text{Pa}$ .
Do not use an area in $\text{cm}^2$ without converting it to $\text{m}^2$ , and use only the component of force normal to the surface.
In an exam, apply the named relationship to the quantities, units and direction given in the question.

Worked example

A gas pushes normally on a hatch with force $240\,\text{N}$ . The hatch area is $0.080\,\text{m}^2$ . Calculate the pressure.

1.Use $p=F/A$ : $p=240\div0.080=3000\,\text{Pa}$ .

Answer: $3000\,\text{Pa}$

Common mistakes

Don't fall into the trap of using area in $\text{cm}^2$ without converting it to $\text{m}^2$ .
Don't fall into the trap of multiplying force by area instead of dividing force by area.

Exam tip

For pressure, use the force normal to the surface and give the answer in pascals.

Tier 1 · Easy

The same normal force acts on two flat surfaces, but surface $B$ has twice the area of surface $A$ . Compare the pressures on the two surfaces.
[2 marks]
Total for this question: 2
A person stands evenly on two identical snowshoes. A student calculates the pressure by dividing the person's whole weight by the area of one snowshoe. Explain the error and give two equivalent correct calculation routes.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A liquid exerts a normal force of $54\,\text{N}$ on a sensor of area $30\,\text{cm}^2$ . Calculate the pressure on the sensor.
[3 marks]
Total for this question: 3
The same pressure sensor gives these force and pressure pairs: $80\,\text{N}$ with $20\,\text{kPa}$ , $120\,\text{N}$ with $30\,\text{kPa}$ , and $160\,\text{N}$ with $36\,\text{kPa}$ . Use $p=F/A$ to determine the sensor area, identify the anomalous pair and give its corrected pressure.
[4 marks]
Total for this question: 4
A force of $1200\,\text{N}$ acts normally on an area of $0.050\,\text{m}^2$ . A student gives the pressure as $24000\,\text{N/m}$ . Check the number and correct the unit, explaining why the metre must be squared.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A sealed chamber produces the same normal force of $1260\,\text{N}$ on either of two removable panels. Panel X measures $0.30\,\text{m}$ by $0.20\,\text{m}$ and panel Y measures $0.45\,\text{m}$ by $0.35\,\text{m}$ . Calculate the pressure on each panel and determine which panel experiences the lower pressure.
[5 marks]
Total for this question: 5
A platform is supported evenly by two rectangular pads, each $18\,\text{cm}$ by $6.0\,\text{cm}$ . The pressure beneath the pads is $24\,\text{kPa}$ . Calculate the platform's weight. Then determine the pressure if the same weight is supported by one pad only.
[5 marks]
Total for this question: 5
A machine of weight $18.6\,\text{kN}$ is supported evenly by identical pads, each of area $0.025\,\text{m}^2$ . The floor's pressure limit is $120\,\text{kPa}$ . Determine the minimum number of pads required and verify that your choice keeps the pressure below the limit.
[5 marks]
Total for this question: 5
A gas in a sealed container is at a pressure of $18\,\text{kPa}$ . It touches a horizontal face of area $40\,\text{cm}^2$ and a vertical face of area $65\,\text{cm}^2$ . Calculate the force on each face, describe the direction of each force and explain why neither force acts along its surface.
[5 marks]
Total for this question: 5
The liquid in a syringe is at a pressure $250\,\text{kPa}$ greater than its surroundings. The plunger area is $1.8\,\text{cm}^2$ . An operator applies a $36\,\text{N}$ force inwards, normal to the plunger. Calculate the outward pressure force, determine the resultant force and state the additional inward force needed to hold the plunger stationary.
[4 marks]
Total for this question: 4

4.5.5.1.2 · Pressure in a fluid 2 (physics only) (HT only)

Explanation

Higher tier: in a liquid, pressure increases with the height of liquid above the point and with the liquid's density. Use $p=h\rho g$ for pressure due to a liquid column, with $h$ in metres, $\rho$ in $\text{kg/m}^3$ and $g$ in $\text{N/kg}$ ; subtract depths before using it for a pressure difference.
For example, in water with $\rho=1000\,\text{kg/m}^3$ and $g=9.8\,\text{N/kg}$ , a $0.50\,\text{m}$ depth change gives $\Delta p=0.50\times1000\times9.8=4900\,\text{Pa}$ . Upthrust arises because the bottom of a submerged object is at greater pressure than its top.
An object rises when upthrust is greater than its weight, sinks when its weight is greater than the upthrust, and floats at rest when upthrust balances its weight.
Equivalently, an object with a lower average density than the fluid floats, one with a greater average density sinks, and one with the same density can remain suspended.
Do not include horizontal position in $p=h\rho g$ .

Pressure difference across a submerged object.

Worked example

Oil has density $820\,\text{kg/m}^3$ . Calculate the pressure due to a $1.5\,\text{m}$ column of the oil when $g=9.8\,\text{N/kg}$ .

1.Use $p=h\rho g=1.5\times820\times9.8=12054\,\text{Pa}$ , which is $1.21\times10^4\,\text{Pa}$ to three significant figures.

Answer: $1.21\times10^4\,\text{Pa}$

Common mistakes

Don't fall into the trap of measuring depth from the bottom of the liquid instead of down from its surface.
Don't fall into the trap of saying the upward force is larger because the bottom of the object has a larger area when the areas are equal.

Exam tip

In $p=h\rho g$ , use vertical depth below the liquid surface and explain upthrust using the pressure difference.

Tier 1 · Easy

Explain why a fully submerged object experiences an upward force due to the liquid pressure on it.
[2 marks]
Total for this question: 2
Two sensors are at the same depth in a tank of still water, one near the centre and one near a side wall. A student claims the wall sensor reads a higher pressure because it is closer to the wall. Evaluate the claim.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Two pressure sensors are $0.40\,\text{m}$ and $2.10\,\text{m}$ below the surface of a liquid of density $960\,\text{kg/m}^3$ . Calculate the pressure difference when $g=9.8\,\text{N/kg}$ .
[3 marks]
Total for this question: 3
In liquid A, moving a sensor $0.90\,\text{m}$ deeper increases its pressure by $7056\,\text{Pa}$ . In liquid B, moving it $0.75\,\text{m}$ deeper increases its pressure by $8820\,\text{Pa}$ . When $g=9.8\,\text{N/kg}$ , calculate the density of each liquid and identify the denser liquid.
[4 marks]
Total for this question: 4
A cuboid has dimensions $0.30\,\text{m}$ by $0.20\,\text{m}$ by $0.10\,\text{m}$ and is fully submerged in a liquid of density $700\,\text{kg/m}^3$ , where $g=10\,\text{N/kg}$ . Calculate the pressure difference and upthrust when the vertical side is first $0.30\,\text{m}$ and then $0.10\,\text{m}$ .
[4 marks]
Total for this question: 4

Tier 3 · Hard

A fully submerged cuboid is $0.30\,\text{m}$ high and has horizontal top and bottom areas of $0.012\,\text{m}^2$ . It is in water of density $1000\,\text{kg/m}^3$ , where $g=9.8\,\text{N/kg}$ . Calculate the pressure difference between its bottom and top, use this to calculate the upthrust, and predict its initial motion if its weight is $30\,\text{N}$ .
[5 marks]
Total for this question: 5
A floating raft has a horizontal base area of $0.75\,\text{m}^2$ and weight $1470\,\text{N}$ . The raft is a rectangular block with vertical sides. It floats in water of density $1000\,\text{kg/m}^3$ , where $g=9.8\,\text{N/kg}$ . Use upthrust and $p=h\rho g$ to calculate the initial depth of the raft's base. A $441\,\text{N}$ payload is then added. Calculate the new depth and the change in depth.
[5 marks]
Total for this question: 5
An incompressible liquid has constant density. A fully submerged cuboid is moved deeper without changing its orientation. A student claims its upthrust increases because the liquid pressure is greater. Evaluate the claim by comparing the pressures on the cuboid's top and bottom faces.
[4 marks]
Total for this question: 4
A $6.0\,\text{kg}$ cuboid is fully submerged and held stationary by a vertical string pulling upwards. Its height is $0.25\,\text{m}$ and its horizontal face area is $0.020\,\text{m}^2$ . The liquid density is $850\,\text{kg/m}^3$ . Take $g=9.8\,\text{N/kg}$ . Determine the bottom-to-top pressure difference, the upthrust and the string tension. Predict the initial motion if the string is cut.
[6 marks]
Total for this question: 6
A rectangular cuboid has horizontal base area $0.16\,\text{m}^2$ , height $0.25\,\text{m}$ and weight $320\,\text{N}$ . It is placed first in liquid A of density $700\,\text{kg/m}^3$ and then in liquid B of density $1000\,\text{kg/m}^3$ . Use $g=9.8\,\text{N/kg}$ . Determine whether it floats or sinks in each liquid. If it floats, calculate the equilibrium depth of its base below the surface.
[6 marks]
Total for this question: 6

4.5.5.2 · Atmospheric pressure (physics only)

Explanation

Atmospheric pressure is produced by air molecules colliding with surfaces; the atmosphere is a thin layer of air around Earth.
As altitude increases, there are fewer air molecules above a surface and the air is less dense, so atmospheric pressure decreases.
A pressure difference across a surface produces a resultant normal force; calculate it by rearranging $p=F/A$ to $F=pA$ .
Do not say that atmospheric pressure becomes zero on a mountain; it decreases with height because there is less air above, but an atmosphere remains.

Worked example

State what microscopic event produces atmospheric pressure on a window.

1.Use the particle model: moving air molecules repeatedly strike the surface, and their collisions create pressure.

Answer: Air molecules collide with the window surface.

Common mistakes

Don't fall into the trap of saying atmospheric pressure is caused by the weight of one air molecule rather than many molecular collisions.
Don't fall into the trap of predicting atmospheric pressure increases with altitude even though fewer air molecules are above the surface.

Exam tip

For ‘explain atmospheric pressure’, link moving air molecules colliding with a surface to a force per unit area.

Tier 1 · Easy

Explain why atmospheric pressure is lower at a higher altitude.
[2 marks]
Total for this question: 2
A sealed flexible food packet expands as it is carried from sea level to a mountain. Explain the change using the atmospheric pressures inside and outside the packet.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Explain why a barometer records a lower atmospheric pressure at the top of a tall mountain than at sea level.
[3 marks]
Total for this question: 3
A pressure sensor records $96\,\text{kPa}$ at an altitude of $0.4\,\text{km}$ and $84\,\text{kPa}$ at $1.6\,\text{km}$ . At $2.8\,\text{km}$ , predict which reading is most likely: $76\,\text{kPa}$ , $84\,\text{kPa}$ or $92\,\text{kPa}$ . Justify your prediction using air molecules.
[3 marks]
Total for this question: 3
A student drinks through a straw by lowering the air pressure inside the straw. Explain why the liquid rises. Your answer should refer to the pressure on the liquid surface and the pressure inside the straw.
[3 marks]
Total for this question: 3

Tier 3 · Hard

At high altitude, the pressure inside a sealed case is $101\,\text{kPa}$ and the atmospheric pressure outside is $75\,\text{kPa}$ . A flat lid has area $0.012\,\text{m}^2$ . Calculate the resultant force on the lid and state its direction.
[4 marks]
Total for this question: 4
A sealed piston contains gas at an unknown pressure. The piston is locked in position, so the trapped gas stays at the same pressure at both sites. At a site where atmospheric pressure is $100\,\text{kPa}$ , the resultant force on the piston is $320\,\text{N}$ inwards. At another site, atmospheric pressure is $84\,\text{kPa}$ and the resultant force is $160\,\text{N}$ inwards. Use $F=\Delta pA$ to determine the piston area and the gas pressure.
[5 marks]
Total for this question: 5
Plan an investigation using a portable pressure sensor to test how atmospheric pressure changes with altitude in a tall building. State the independent and dependent variables, describe how to improve the reliability of the data and give the expected trend.
[4 marks]
Total for this question: 4
A calibrated pressure sensor A and sensor B are compared at three altitudes. Their readings in $\text{kPa}$ are A: $102$ , $90$ , $81$ and B: $106$ , $94$ , $85$ . At a fourth, higher site only B is available and reads $78\,\text{kPa}$ . Identify B's systematic error, correct the final reading and explain the overall pressure trend using air molecules.
[4 marks]
Total for this question: 4
A suction hook on a horizontal ceiling traps air at $61\,\text{kPa}$ above an area of $18\,\text{cm}^2$ . The surrounding atmospheric pressure below it is $99\,\text{kPa}$ . Calculate the upward force holding the hook to the ceiling, determine the maximum mass it can support when $g=9.8\,\text{N/kg}$ and explain why a small air leak can make the hook fall.
[5 marks]
Total for this question: 5

4.5.6.1.1 · Distance and displacement

Explanation

Distance is the total length of the path travelled and is scalar; displacement is the straight-line change from start to finish and includes direction, so it is vector.
Add every part of a route to find distance, but use only the start and finish positions to find displacement.
For example, travelling $9\,\text{m}$ east and then $4\,\text{m}$ west gives distance $13\,\text{m}$ and displacement $5\,\text{m}$ east.
Do not report displacement without a direction, and do not assume distance and displacement are equal unless the path is straight without reversing.

Distance follows the route; displacement joins start to finish.

Worked example

A walker travels $14\,\text{m}$ north and then $5\,\text{m}$ south. Determine the distance and displacement.

1.Distance adds both path lengths: $14+5=19\,\text{m}$ .
2.Taking north as positive, displacement is $14-5=9\,\text{m}$ north.

Answer: Distance $19\,\text{m}$ ; displacement $9\,\text{m}$ north.

Common mistakes

Don't fall into the trap of adding outward and return distances to calculate displacement.
Don't fall into the trap of giving a displacement magnitude without its direction.

Exam tip

Track total path length for distance, but compare final and initial positions for displacement.

Tier 1 · Easy

An athlete completes one $400\,\text{m}$ lap and finishes at the starting point. State the distance travelled and the displacement.
[2 marks]
Total for this question: 2
A hiker walks $50\,\text{m}$ east and then $70\,\text{m}$ west. A student reports a distance of $-20\,\text{m}$ and a displacement of $20\,\text{m}$ . Identify the error and give the correct distance and displacement, including the displacement direction.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A robot moves $6.0\,\text{m}$ east and then $8.0\,\text{m}$ north. Calculate its distance. Then, using a scale of $1.0\,\text{cm}=2.0\,\text{m}$ , draw a scale vector diagram to find the magnitude and direction of its displacement.
[4 marks]
Total for this question: 4
Route A has a path length of $900\,\text{m}$ and finishes $600\,\text{m}$ north of its start. Route B has a path length of $700\,\text{m}$ and finishes $700\,\text{m}$ east of its start. Give the distance and displacement for each route, then identify the route that could be a straight journey without reversing.
[4 marks]
Total for this question: 4
During a journey, an odometer increases by $1.7\,\text{km}$ while a position sensor shows that the vehicle finishes $1.2\,\text{km}$ west of its starting point. State the distance and displacement, and explain why their magnitudes differ.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A survey drone flies $600\,\text{m}$ east, $250\,\text{m}$ west and then $120\,\text{m}$ north. Calculate the total distance and the net eastward displacement before the northward flight. Then use a scale of $1.0\,\text{cm}=50\,\text{m}$ to draw a scale vector diagram and find the magnitude and direction of the displacement from launch.
[5 marks]
Total for this question: 5
On a map, a route goes $3.0\,\text{cm}$ east, $4.0\,\text{cm}$ north and then $3.0\,\text{cm}$ west. The scale is $1.0\,\text{cm}$ for $250\,\text{m}$ . Calculate the distance travelled and the magnitude and direction of the displacement.
[4 marks]
Total for this question: 4
A tracker records a runner's position every $20\,\text{s}$ . Adding the straight-line gaps between recorded positions gives $1.84\,\text{km}$ , while the final position is $1.20\,\text{km}$ east of the start. The runner follows curved paths between readings. Explain which value gives the displacement and why $1.84\,\text{km}$ underestimates the true distance travelled.
[4 marks]
Total for this question: 4
A walker travels $400\,\text{m}$ on a bearing of $030^\circ$ and then $400\,\text{m}$ on a bearing of $120^\circ$ . Using only a head-to-tail scale drawing with $1.0\,\text{cm}=100\,\text{m}$ , determine the walker's distance and displacement from the starting point.
[5 marks]
Total for this question: 5
East is positive on a straight track. A vehicle starts at position $-120\,\text{m}$ , reaches $+260\,\text{m}$ and finishes at $-40\,\text{m}$ . Determine the direction and length of each stage, the total distance, the displacement and the ratio of distance to displacement magnitude.
[5 marks]
Total for this question: 5

4.5.6.1.2 · Speed

Explanation

Speed is a scalar rate of change of distance; typical values are about $1.5\,\text{m/s}$ for walking, $3\,\text{m/s}$ for running, $6\,\text{m/s}$ for cycling and $330\,\text{m/s}$ for sound in air.
Typical transport-system speeds are about $13$ – $30\,\text{m/s}$ for a car, $55\,\text{m/s}$ for a train and $250\,\text{m/s}$ for a passenger plane.
Use $s=vt$ for constant speed and rearrange to $v=s/t$ ; for non-uniform motion, average speed is total distance divided by total time.
For example, $450\,\text{m}$ travelled in $30\,\text{s}$ gives an average speed of $450\div30=15\,\text{m/s}$ .
Do not average two speeds unless the time spent at each speed is equal; instead use the complete distance and complete time, including stops when appropriate.

Worked example

A toy car travels $24\,\text{m}$ in $6.0\,\text{s}$ at constant speed. Calculate its speed.

1.Use $v=s/t$ : $v=24\div6.0=4.0\,\text{m/s}$ .

Answer: $4.0\,\text{m/s}$

Common mistakes

Don't fall into the trap of averaging two speeds without accounting for the different times spent at each speed.
Don't fall into the trap of using displacement instead of total distance when calculating average speed.

Exam tip

Average speed is total distance divided by total time, not usually the mean of the stated speeds.

Tier 1 · Easy

A runner covers $120\,\text{m}$ in $20\,\text{s}$ , rests for $10\,\text{s}$ , then covers $180\,\text{m}$ in $30\,\text{s}$ . Calculate the average speed for the whole interval.
[3 marks]
Total for this question: 3
A data logger reports that a walking student covered $90\,\text{m}$ in $6.0\,\text{s}$ . Calculate the speed and compare it with a typical walking speed of about $1.5\,\text{m/s}$ to evaluate the reading.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A runner covers $300\,\text{m}$ in $48\,\text{s}$ , rests for $12\,\text{s}$ and then covers another $200\,\text{m}$ in $32\,\text{s}$ . Calculate the average speed for the whole interval.
[3 marks]
Total for this question: 3
A toy car is timed over a $1.20\,\text{m}$ track. Four times are $0.84\,\text{s}$ , $0.82\,\text{s}$ , $1.48\,\text{s}$ and $0.83\,\text{s}$ . Identify the anomalous result, calculate the mean of the valid times and use it to determine the car's speed.
[4 marks]
Total for this question: 4
Two light gates are $12.0\,\text{m}$ apart. A vehicle takes $0.75\,\text{s}$ to travel between them. The speed limit is $54\,\text{km/h}$ . Calculate the vehicle's speed in metres per second and determine whether it exceeds the limit.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A delivery drone travels $1.8\,\text{km}$ to a site in $2\,\text{min}\ 20\,\text{s}$ . It waits for $30\,\text{s}$ , then returns along the same route at a constant $15\,\text{m/s}$ . Calculate its average speed for the complete trip from departure to return.
[5 marks]
Total for this question: 5
A courier travels at $12\,\text{m/s}$ for $45\,\text{s}$ , waits for $30\,\text{s}$ , then covers an unknown distance in $75\,\text{s}$ . The average speed for the complete $150\,\text{s}$ interval is $7.5\,\text{m/s}$ . Determine the unknown distance and the speed during the final stage.
[5 marks]
Total for this question: 5
An object travels $80\,\text{m}$ forwards in the first $10\,\text{s}$ , $20\,\text{m}$ back along the same route in the next $10\,\text{s}$ , then $60\,\text{m}$ forwards in the final $10\,\text{s}$ . A student plots cumulative distance values of $0$ , $80$ , $60$ and $120\,\text{m}$ at $0$ , $10$ , $20$ and $30\,\text{s}$ . Correct the two erroneous distances, calculate the speed in each interval and identify the fastest interval.
[5 marks]
Total for this question: 5
Plan an investigation to determine the average speed of a remote-control car over a non-uniform journey. State how distance and time are measured, how the average speed is calculated and two steps to improve repeatability and reduce measurement error.
[5 marks]
Total for this question: 5
A bus travels $900\,\text{m}$ at $15\,\text{m/s}$ , then $900\,\text{m}$ at $10\,\text{m/s}$ . It waits for $30\,\text{s}$ before travelling a final $1.2\,\text{km}$ in $80\,\text{s}$ . Calculate the average speed for the whole journey and evaluate a student's claim that the answer is the mean of the three moving speeds.
[5 marks]
Total for this question: 5

4.5.6.1.3 · Velocity

Explanation

Velocity is speed in a stated direction, so it is a vector quantity; speed is scalar.
For motion over an interval, use displacement rather than distance when finding average velocity, then give the direction of the displacement.
For example, a displacement of $60\,\text{m}$ west in $15\,\text{s}$ gives an average velocity of $4.0\,\text{m/s}$ west.
Higher tier: an object moving in a circle can have constant speed but changing velocity because its direction of motion changes continuously.
Do not use total path length to calculate average velocity, and do not omit direction from a velocity value.

Worked example

A train moves north at a speed of $18\,\text{m/s}$ . State its velocity.

1.Velocity is speed with direction, so attach the stated direction north to the magnitude $18\,\text{m/s}$ .

Answer: $18\,\text{m/s}$ north

Common mistakes

Don't fall into the trap of giving a speed as the velocity without adding a direction.
Don't fall into the trap of calling an object at constant speed constant velocity while its direction is changing.

Exam tip

A complete velocity answer needs both the speed and the direction of motion.

Tier 1 · Easy

A car turns from travelling east to travelling north without changing its speed. Explain why its velocity changes.
[2 marks]
Total for this question: 2
An arrow points south and is labelled $9.0\,\text{m/s}$ . A student calls the arrow a speed vector. Identify the error and give the correct description, speed and velocity represented.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A cart moves $50\,\text{m}$ east and then $20\,\text{m}$ west in a total time of $10\,\text{s}$ . Calculate its average velocity.
[3 marks]
Total for this question: 3
A trolley is $12\,\text{m}$ east of an origin at $0\,\text{s}$ , $32\,\text{m}$ east at $5.0\,\text{s}$ and $20\,\text{m}$ east at $11.0\,\text{s}$ . Calculate its average velocity during each of the two time intervals.
[4 marks]
Total for this question: 4
East is defined as the positive direction. A trolley travels with constant velocity $-6.0\,\text{m/s}$ for $12\,\text{s}$ . State its direction of motion and calculate its displacement.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A rescue vehicle travels $80\,\text{m}$ east and then $150\,\text{m}$ north in $30\,\text{s}$ . Using a scale of $1.0\,\text{cm}=20\,\text{m}$ , draw a scale vector diagram to find its displacement. Hence calculate the magnitude and direction of its average velocity, and compare this with its average speed.
[5 marks]
Total for this question: 5
A ship travels $5.6\,\text{km}$ on a bearing of $060^\circ$ in $16\,\text{min}$ , then moves $1.4\,\text{km}$ back along the same line in $4.0\,\text{min}$ and waits for $10\,\text{min}$ . Calculate its average velocity and average speed for the whole interval, giving the velocity direction.
[5 marks]
Total for this question: 5
A tracker reports that a vehicle travels $450\,\text{m}$ and has a displacement of $150\,\text{m}$ west during $60\,\text{s}$ . It gives an average speed of $7.5\,\text{m/s}$ and an average velocity of $7.5\,\text{m/s}$ west. Check both reported values, correct any error and explain why the two averages differ.
[4 marks]
Total for this question: 4
Higher only: A car moves around a circular track at constant speed. At the easternmost point it travels north, and at the westernmost point it travels south. Compare its speed and velocity at the two points, and explain why a resultant force is needed even though the speed is constant.
[4 marks]
Total for this question: 4
Vehicles A, B and C each travel $600\,\text{m}$ in $80\,\text{s}$ . A has a displacement of $600\,\text{m}$ east, B returns to its starting point, and C has a displacement of $400\,\text{m}$ north. Calculate each average speed and average velocity, then explain why equal average speeds do not imply equal average velocities.
[6 marks]
Total for this question: 6

4.5.6.1.4 · The distance–time relationship

Explanation

A distance–time graph shows distance travelled on the vertical axis and time on the horizontal axis.
Its gradient is speed: $v=\dfrac{\Delta s}{\Delta t}$ .
A straight sloping section represents constant speed, a steeper section represents a greater speed, and a horizontal section represents an object at rest.
For an average speed, use the total distance travelled divided by the total time.
Higher tier: find an instantaneous speed on a curved distance–time graph by drawing a tangent at the required point and calculating the tangent's gradient.

A distance–time graph with constant-speed, stationary and faster sections.

Worked example

A distance–time graph rises from $30\,\text{m}$ at $8\,\text{s}$ to $102\,\text{m}$ at $20\,\text{s}$ . Calculate the speed on this section.

1.Distance change $=102-30=72\,\text{m}$ .
2.Time change $=20-8=12\,\text{s}$ .
3.Gradient $=\dfrac{72}{12}=6.0\,\text{m/s}$ .

Answer: $6.0\,\text{m/s}$

Common mistakes

Don't fall into the trap of using $102/20$ instead of the changes in coordinates between the two chosen points.
Don't fall into the trap of calling a horizontal section constant speed when its zero gradient means the object is stationary.
Don't fall into the trap of drawing a chord through a curve instead of a tangent at the stated instant (Higher tier).

Exam tip

For a graph calculation, mark a large gradient triangle and show both coordinate differences before dividing.

Tier 1 · Easy

A runner travels $156\,\text{m}$ in $48\,\text{s}$ at constant speed. Calculate the gradient of the distance–time graph.
[2 marks]
Total for this question: 2
A distance–time graph passes through $0\,\text{m}$ at $0\,\text{s}$ , $24\,\text{m}$ at $6\,\text{s}$ and $19\,\text{m}$ at $10\,\text{s}$ . The vertical axis shows total distance travelled. Identify the impossible section and explain your answer.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A distance–time graph is a straight line from $(0\,\text{s},0\,\text{m})$ to $(24\,\text{s},120\,\text{m})$ . It is then horizontal for $8\,\text{s}$ . State what happens during the horizontal section and calculate the speed during the first section.
[3 marks]
Total for this question: 3
A distance–time graph has straight-line sections for two intervals. A walker covers $90\,\text{m}$ in the first $18\,\text{s}$ and another $64\,\text{m}$ in the next $8\,\text{s}$ . A student says the walker was faster in the first interval because more distance was covered. Use the two graph gradients to evaluate the claim.
[4 marks]
Total for this question: 4
A straight section of a distance–time graph starts at $(5.0\,\text{s},28\,\text{m})$ . The object then moves at $6.5\,\text{m s}^{-1}$ until $13.0\,\text{s}$ . Determine the distance coordinate at the end of the section.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A robot moves along a straight track. Its distance from the start is $0\,\text{m}$ at $0\,\text{s}$ , $54\,\text{m}$ at $12\,\text{s}$ , $150\,\text{m}$ at $28\,\text{s}$ and $150\,\text{m}$ at $40\,\text{s}$ . The graph joins these points with straight lines. Calculate the speed in each time interval and the average speed over all $40\,\text{s}$ .
[5 marks]
Total for this question: 5
Higher only: A tangent drawn to a curved distance–time graph at $7.0\,\text{s}$ passes through $(3.0\,\text{s},18\,\text{m})$ and $(11.0\,\text{s},86\,\text{m})$ . Determine the instantaneous speed at $7.0\,\text{s}$ .
[3 marks]
Total for this question: 3
A courier travels $180\,\text{m}$ at a constant $6.0\,\text{m s}^{-1}$ , remains stationary, then travels another $210\,\text{m}$ at a constant $7.0\,\text{m s}^{-1}$ . The average speed for the whole journey is $5.2\,\text{m s}^{-1}$ . Determine how long the courier remains stationary.
[4 marks]
Total for this question: 4
Walker A leaves a checkpoint and travels at a constant $3.2\,\text{m s}^{-1}$ . Walker B leaves the same checkpoint $25\,\text{s}$ later and follows the same route at a constant $5.2\,\text{m s}^{-1}$ . Determine the time after A starts when B catches A, the distance from the checkpoint, and how the two straight lines on a distance–time graph show which walker is faster.
[5 marks]
Total for this question: 5
A cyclist travels $100\,\text{m}$ at $4.0\,\text{m s}^{-1}$ and then another $100\,\text{m}$ at $10\,\text{m s}^{-1}$ . A student calculates the average speed as $(4.0+10)/2=7.0\,\text{m s}^{-1}$ . Calculate the actual average speed and explain why the student's method gives the wrong value.
[4 marks]
Total for this question: 4

4.5.6.1.5 · Acceleration

Explanation

Acceleration is the rate at which velocity changes. Calculate it using $a=\dfrac{v-u}{t}$ , where $u$ is initial velocity, $v$ is final velocity and $t$ is the time for the change; acceleration is measured in $\text{m/s}^2$ .
On a velocity–time graph, acceleration is the gradient: a horizontal section has zero acceleration and a negative gradient means acceleration in the negative direction.
Direction matters because velocity is a vector.
Uniform acceleration also obeys $v^2-u^2=2as$ ; this is supplied on the equation sheet and is examined on both tiers.
Higher tier only: finding the distance travelled from the area under a velocity–time graph.

A velocity–time graph showing acceleration, constant velocity and negative acceleration.

Worked example

A cyclist's velocity changes from $4.0\,\text{m/s}$ to $16.0\,\text{m/s}$ in $6.0\,\text{s}$ . Calculate the average acceleration.

1.Change in velocity $=16.0-4.0=12.0\,\text{m/s}$ .
2. $a=\dfrac{v-u}{t}=\dfrac{12.0}{6.0}$ .

Answer: $2.0\,\text{m/s}^2$

Common mistakes

Don't fall into the trap of dividing the final velocity by time instead of first calculating $v-u$ .
Don't fall into the trap of reading the height of a velocity–time graph as acceleration instead of calculating its gradient.
Don't fall into the trap of using the graph's gradient when the question asks for distance from the area beneath it (Higher tier).

Exam tip

Write the signed velocity change explicitly before substituting into an acceleration calculation.

Tier 1 · Easy

A scooter increases its velocity from $3.0\,\text{m s}^{-1}$ to $15.0\,\text{m s}^{-1}$ in $8.0\,\text{s}$ . Calculate its average acceleration.
[2 marks]
Total for this question: 2
In one interval a cyclist's velocity increases by $6.0\,\text{m s}^{-1}$ in $5.0\,\text{s}$ . In a second interval it increases by $9.0\,\text{m s}^{-1}$ in $3.0\,\text{s}$ . Calculate both accelerations and identify the greater one.
[3 marks]
Total for this question: 3

Tier 2 · Standard

A velocity–time graph rises uniformly from $0$ to $12\,\text{m s}^{-1}$ in $4.0\,\text{s}$ , stays horizontal for $6.0\,\text{s}$ , then falls uniformly to $0$ in $3.0\,\text{s}$ . Determine the acceleration in each section.
[4 marks]
Total for this question: 4
A horizontal section of a velocity–time graph lies at $-4.0\,\text{m s}^{-1}$ . A student says the object is stationary because the graph is horizontal. Explain why the claim is wrong and state the object's acceleration.
[3 marks]
Total for this question: 3
An object has an initial velocity of $+14\,\text{m s}^{-1}$ and a constant acceleration of $-2.5\,\text{m s}^{-2}$ . Calculate the time taken to reach zero velocity and determine its velocity after $8.0\,\text{s}$ .
[4 marks]
Total for this question: 4

Tier 3 · Hard

A train accelerates uniformly from $8.0\,\text{m s}^{-1}$ to $20.0\,\text{m s}^{-1}$ while travelling $168\,\text{m}$ . Calculate its acceleration and the time taken.
[5 marks]
Total for this question: 5
Higher only: A velocity–time graph joins the points $(0\,\text{s},2.0\,\text{m s}^{-1})$ , $(5.0\,\text{s},14\,\text{m s}^{-1})$ , $(14\,\text{s},14\,\text{m s}^{-1})$ and $(18\,\text{s},2.0\,\text{m s}^{-1})$ with straight lines. Calculate the distance travelled during the complete motion.
[5 marks]
Total for this question: 5
During a $4.0\,\text{s}$ interval, a test trolley accelerates uniformly. Its initial velocity is $3.0\,\text{m s}^{-1}$ and its final velocity is $15.0\,\text{m s}^{-1}$ . A sensor report states that the trolley travels $45\,\text{m}$ during this interval. Determine its acceleration, then use the Physics Equations Sheet to determine the distance it travels and test whether the reported distance is consistent.
[5 marks]
Total for this question: 5
A vehicle accelerates uniformly from $6.0\,\text{m s}^{-1}$ to $18\,\text{m s}^{-1}$ in $4.0\,\text{s}$ . It then brakes uniformly to rest over $54\,\text{m}$ . Calculate the acceleration during the first stage, the acceleration during braking and the total time for both stages. Use equations from the Physics Equations Sheet.
[6 marks]
Total for this question: 6
Take forward as positive. An object has initial velocity $+18\,\text{m s}^{-1}$ and accelerates uniformly at $-2.5\,\text{m s}^{-2}$ for $4.0\,\text{s}$ . Determine its final velocity and displacement using equations from the Physics Equations Sheet. Explain how the signs show the object is slowing down while still moving forward.
[5 marks]
Total for this question: 5

4.5.6.2.1 · Newton's First Law

Explanation

Newton's First Law states that if the resultant force on an object is zero, an object at rest remains at rest and a moving object continues at constant velocity.
Constant velocity includes constant speed in a constant direction.
Forces can act while the resultant is zero: a car travelling steadily may have its driving force balanced by resistive forces.
If the resultant force is not zero, the object's velocity changes by speeding up, slowing down or changing direction.
Higher tier: inertia is the tendency of an object to continue in its state of rest or uniform motion.

Worked example

A car travels east at constant velocity. Its engine provides a $760\,\text{N}$ force east. Determine the total resistive force.

1.Constant velocity means the resultant force is zero.
2.The resistive force must balance the engine force in the opposite direction.

Answer: $760\,\text{N}$ west

Common mistakes

Don't fall into the trap of claiming that no forces act because the car moves at constant velocity.
Don't fall into the trap of saying a forward resultant force is needed to maintain a steady speed on a straight road.
Don't fall into the trap of describing inertia as a force rather than a tendency to resist a change in motion (Higher tier).

Exam tip

When the question states constant velocity, begin by writing that the resultant force is zero.

Tier 1 · Easy

A boat moves at constant velocity. Its propeller provides a forward force of $680\,\text{N}$ . Determine the total resistive force.
[2 marks]
Total for this question: 2
A puck is moving north when the resultant force on it becomes zero. A student predicts that the puck will slow down and stop. Use Newton's First Law to explain what is wrong with the prediction.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A van travels in a straight line at a steady $18\,\text{m s}^{-1}$ . The engine force is $920\,\text{N}$ . Explain the motion in terms of the forces and state the resultant force.
[3 marks]
Total for this question: 3
A probe is moving west at $3.0\,\text{m s}^{-1}$ . Two horizontal forces act on it: $48\,\text{N}$ west and $48\,\text{N}$ east. Predict its subsequent motion and justify your answer.
[3 marks]
Total for this question: 3
A crate is initially at rest with a force of $85\,\text{N}$ east and a force of $85\,\text{N}$ west acting on it. The eastward force is removed. Determine the new resultant force and predict the crate's motion.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A lift moves upward. The motor force is $6200\,\text{N}$ upward, its weight is $5800\,\text{N}$ and friction is $400\,\text{N}$ downward. Describe its motion. The motor force then falls to $5000\,\text{N}$ while the lift is still moving upward. Calculate the new resultant force and describe the change in motion.
[5 marks]
Total for this question: 5
Higher only: An empty shopping trolley and a loaded trolley are initially at rest. Explain, using inertia, why the loaded trolley is harder to set moving and harder to stop once moving.
[3 marks]
Total for this question: 3
A probe is moving east. During interval A the resultant force is east, during interval B the resultant force is zero, and during interval C the resultant force is west. The probe is still moving east at the end of interval C. Describe how its velocity changes in each interval and justify each description in terms of the resultant force.
[5 marks]
Total for this question: 5
A $45\,\text{kg}$ crate is at rest on a horizontal floor. Take $g=9.8\,\text{N kg}^{-1}$ . Calculate the crate's weight and the normal contact force from the floor. A person then pulls vertically upward on the crate with a force of $200\,\text{N}$ , but the crate remains at rest. Calculate the new normal contact force and the resultant force on the crate.
[5 marks]
Total for this question: 5
A wheeled cart is towed to the right at constant velocity. The tow force is $240\,\text{N}$ and the rolling resistance is $240\,\text{N}$ to the left. The tow line is released. A student claims that the cart will pass through rest and then accelerate to the left because the resistance acts left. Evaluate the claim using Newton's First Law and the fact that rolling resistance is zero once the cart is stationary.
[4 marks]
Total for this question: 4

4.5.6.2.2 · Newton's Second Law

Explanation

Newton's Second Law links resultant force, mass and acceleration: $F=ma$ . Acceleration is proportional to resultant force, so doubling the resultant force doubles the acceleration for a fixed mass.
Acceleration is inversely proportional to mass, so a larger mass gives a smaller acceleration for the same resultant force.
Combine all forces first, then use force in newtons, mass in kilograms and acceleration in $\text{m/s}^2$ .
In the required practical, change force while keeping total mass constant, or change mass while keeping force constant, measure acceleration and repeat.
Higher tier: inertial mass is $m=F/a$ .

Worked example

A $920\,\text{kg}$ car has a driving force of $3100\,\text{N}$ and resistive forces totalling $1260\,\text{N}$ . Calculate its acceleration.

1.Resultant force $=3100-1260=1840\,\text{N}$ forwards.
2. $a=\dfrac{F}{m}=\dfrac{1840}{920}$ .

Answer: $2.0\,\text{m/s}^2$ forwards

Common mistakes

Don't fall into the trap of substituting the driving force into $F=ma$ without subtracting the resistive forces.
Don't fall into the trap of changing pulling force in the practical by adding mass, so the total moving mass also changes.
Don't fall into the trap of treating inertial mass as a force that opposes motion (Higher tier).

Exam tip

For a multi-force calculation, state the resultant force and its direction before applying $F=ma$ .

Tier 1 · Easy

A resultant force accelerates a $12\,\text{kg}$ object at $1.5\,\text{m s}^{-2}$ . Calculate the resultant force.
[2 marks]
Total for this question: 2
A resultant force of $2.4\,\text{N}$ gives an object an acceleration of $1.5\,\text{m s}^{-2}$ . Determine the object's mass.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A student uses a wheeled cart, a pulley and slotted masses to investigate how force affects acceleration while total mass stays constant. Describe how the student should change the force, measure the acceleration and improve the reliability of the results. State the expected relationship.
[5 marks]
Total for this question: 5
In the force–acceleration required practical, a student tries to investigate acceleration as mass changes while force stays constant. For each run, the student transfers slotted masses from the hanger to the trolley. Explain why this does not keep the force constant and describe a valid way to change mass while keeping the force constant. Include one reliability improvement.
[5 marks]
Total for this question: 5
A $4.0\,\text{kg}$ object has a resultant force of $12\,\text{N}$ and an acceleration of $3.0\,\text{m s}^{-2}$ . The force is then doubled while the object's mass is tripled. Calculate the new acceleration and compare it with the original acceleration.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A $1250\,\text{kg}$ car increases its velocity from $8.0\,\text{m s}^{-1}$ to $26.0\,\text{m s}^{-1}$ in $9.0\,\text{s}$ . The resistive forces total $700\,\text{N}$ . Calculate the acceleration, the resultant force and the engine force.
[5 marks]
Total for this question: 5
Higher only: The same object accelerates at $1.2\,\text{m s}^{-2}$ under a resultant force of $18\,\text{N}$ and at $2.0\,\text{m s}^{-2}$ under $30\,\text{N}$ . Use both observations to determine its inertial mass, then predict the resultant force needed for $2.8\,\text{m s}^{-2}$ .
[5 marks]
Total for this question: 5
In a force–acceleration experiment, the pulling weight is kept at $2.8\,\text{N}$ . The measured pairs of total moving mass and acceleration are $(1.0\,\text{kg},2.4\,\text{m s}^{-2})$ , $(1.5\,\text{kg},1.6\,\text{m s}^{-2})$ and $(2.0\,\text{kg},1.2\,\text{m s}^{-2})$ . Calculate the resultant force for every pair. Explain what the results show and why the resultant force is smaller than the pulling weight.
[6 marks]
Total for this question: 6
A trolley experiences a constant resistive force. An applied force of $5.4\,\text{N}$ produces an acceleration of $1.2\,\text{m s}^{-2}$ , while an applied force of $9.0\,\text{N}$ produces an acceleration of $3.0\,\text{m s}^{-2}$ . Determine the trolley's mass and the resistive force. Then calculate its acceleration when the applied force is $12.6\,\text{N}$ .
[6 marks]
Total for this question: 6
Friction is negligible in a trolley experiment. The applied-force and acceleration pairs are $(2.1\,\text{N},1.4\,\text{m s}^{-2})$ , $(3.6\,\text{N},2.4\,\text{m s}^{-2})$ , $(5.1\,\text{N},3.4\,\text{m s}^{-2})$ and $(6.6\,\text{N},2.8\,\text{m s}^{-2}\text{ reported})$ . Use the first three pairs to determine the trolley's mass. Predict the acceleration for $6.6\,\text{N}$ and evaluate the reported result.
[6 marks]
Total for this question: 6

4.5.6.2.3 · Newton's Third Law

Explanation

Newton's Third Law states that whenever two objects interact, they exert forces on each other that are equal in magnitude and opposite in direction.
The two forces are the same type, occur together and act on different objects, so they do not cancel on a force diagram for one object.
A swimmer pushes water backwards while the water pushes the swimmer forwards.
Identify a pair by naming both objects and reversing which object exerts the force: the force of A on B pairs with the force of B on A.

A Newton's Third Law pair acting on two different interacting objects.

Worked example

A book pulls the Earth upwards gravitationally with a force of $12\,\text{N}$ . State the paired force.

1.Reverse the two interacting objects while keeping the force gravitational.
2.Use equal magnitude and opposite direction.

Answer: The Earth pulls the book downwards gravitationally with a force of $12\,\text{N}$ .

Common mistakes

Don't fall into the trap of pairing the book's weight with the normal contact force even though both act on the book.
Don't fall into the trap of saying one force occurs first and the reaction force follows later.
Don't fall into the trap of drawing both forces of the pair on the same object.

Exam tip

Name both objects in each force statement to show that a Third Law pair acts on different objects.

Tier 1 · Easy

A hammer exerts a downward force on a nail. State the corresponding Newton's Third Law force.
[2 marks]
Total for this question: 2
A parachutist is descending at constant velocity. Weight acts downward and air resistance acts upward. Explain why these two balanced forces are not a Newton's Third Law pair.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A book rests on a table. The Earth pulls the book downward. Identify the Newton's Third Law partner to this force and explain why it is not the table's upward force on the book.
[3 marks]
Total for this question: 3
A magnet pulls an iron block horizontally with a force of $36\,\text{N}$ . State the Newton's Third Law partner force, including the objects, magnitude and direction. Explain why the pair cannot cancel on one object's force diagram.
[4 marks]
Total for this question: 4
During a collision, cart A exerts a force of $42\,\text{N}$ east on cart B. State the force exerted by cart B on cart A. Explain why the two carts can have different accelerations even though these interaction forces are equal in magnitude.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A propeller pushes water backward with a force of $4.2\,\text{kN}$ . The water resistance on the boat is $1.8\,\text{kN}$ backward and the boat's mass is $1200\,\text{kg}$ . State the third-law force exerted by the water because of the propeller interaction, then calculate the boat's acceleration.
[5 marks]
Total for this question: 5
A hanging lamp is stationary. The Earth pulls the lamp downward with $48\,\text{N}$ and a cable pulls the lamp upward with $48\,\text{N}$ . A student calls these a Newton's Third Law pair. Explain what is wrong with the student's claim by naming the partner force to each of the two stated forces, and distinguish interaction pairs from balanced forces.
[5 marks]
Total for this question: 5
A $60\,\text{kg}$ student and a stationary $240\,\text{kg}$ wheeled platform are on a level surface with negligible resistance. The student pushes away, exerting an average force of $180\,\text{N}$ on the platform. Calculate the acceleration magnitude of each object during the push and explain the relationship between the directions of their accelerations.
[5 marks]
Total for this question: 5
Tug A tows barge B to the right with a horizontal rope force of $320\,\text{N}$ . A has mass $500\,\text{kg}$ , an engine force of $920\,\text{N}$ right and water resistance of $200\,\text{N}$ left. B has mass $300\,\text{kg}$ and water resistance of $80\,\text{N}$ left. State the Newton's Third Law rope-force pair, then calculate the resultant force and acceleration of each vessel.
[6 marks]
Total for this question: 6
Two magnets on separate carts repel. The magnetic force on cart A is $0.90\,\text{N}$ east. Cart A has mass $0.40\,\text{kg}$ and accelerates east at $1.5\,\text{m s}^{-2}$ . Cart B has mass $0.60\,\text{kg}$ and accelerates west at $0.80\,\text{m s}^{-2}$ . Determine the magnetic force on B and the resistive force on each cart. Explain why the acceleration magnitudes are different.
[6 marks]
Total for this question: 6

4.5.6.3.1 · Stopping distance

Explanation

A vehicle's stopping distance is the thinking distance plus the braking distance. Thinking distance is travelled during the driver's reaction time, before the brakes act; calculate it using $s=vt$ if speed is constant.
Braking distance is travelled after the brakes are applied until the vehicle stops.
Both usually increase with initial speed.
AQA may ask you to estimate values from typical stopping-distance data, so read the correct speed and keep units consistent.
Reaction-time factors affect thinking distance, while road, tyre and brake conditions affect braking distance.

Worked example

A car travels at $18\,\text{m/s}$ . Its driver's reaction time is $0.65\,\text{s}$ and its braking distance is $24\,\text{m}$ . Calculate the stopping distance.

1.Thinking distance $=vt=18\times0.65=11.7\,\text{m}$ .
2.Stopping distance $=11.7+24=35.7\,\text{m}$ .

Answer: $35.7\,\text{m}$

Common mistakes

Don't fall into the trap of adding a reaction time in seconds directly to a braking distance in metres.
Don't fall into the trap of using braking distance alone when the question asks for total stopping distance.
Don't fall into the trap of assigning tiredness to braking distance rather than to reaction time and thinking distance.

Exam tip

Write ‘stopping = thinking + braking’ before using the data so that neither stage is omitted.

Tier 1 · Easy

A driver's thinking distance is $12\,\text{m}$ and the braking distance is $28\,\text{m}$ . Calculate the stopping distance.
[1 mark]
Total for this question: 1
A student adds a reaction time of $0.60\,\text{s}$ directly to a braking distance of $24\,\text{m}$ to obtain a stopping distance. Explain the error and state the extra measurement needed.
[3 marks]
Total for this question: 3

Tier 2 · Standard

A car travels at $15\,\text{m s}^{-1}$ . The driver's reaction time is $0.64\,\text{s}$ and the braking distance is $19\,\text{m}$ . Calculate the thinking distance and the stopping distance.
[3 marks]
Total for this question: 3
A car's stopping distance is $43\,\text{m}$ and its braking distance is $31\,\text{m}$ . The car travels at $24\,\text{m s}^{-1}$ during the driver's reaction time. Determine the reaction time.
[3 marks]
Total for this question: 3
The speed of a car is $20\,\text{m s}^{-1}$ when its driver sees an obstacle $45\,\text{m}$ ahead. The driver's reaction time is $0.70\,\text{s}$ and the braking distance is $34\,\text{m}$ . Calculate the thinking distance and stopping distance, then determine whether the car fails to stop in time.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A delivery van has a reaction time of $0.75\,\text{s}$ . At $20\,\text{m s}^{-1}$ its braking distance is $32\,\text{m}$ ; at $25\,\text{m s}^{-1}$ its braking distance is $50\,\text{m}$ . Calculate the stopping distance at each speed and the increase in stopping distance.
[5 marks]
Total for this question: 5
A van travels at $72\,\text{km h}^{-1}$ . Convert this speed to metres per second. Its braking distance is $30\,\text{m}$ . Calculate the stopping distance for a reaction time of $0.60\,\text{s}$ , then calculate how much the stopping distance increases if distraction raises the reaction time to $0.92\,\text{s}$ .
[6 marks]
Total for this question: 6
For one car in fixed conditions, braking distance is proportional to speed squared. Its braking distance is $25\,\text{m}$ at $20\,\text{m s}^{-1}$ . The same car travels at $25\,\text{m s}^{-1}$ and the driver's reaction time is $0.68\,\text{s}$ . Calculate the new stopping distance and decide whether the car can stop before an obstacle $55\,\text{m}$ away.
[6 marks]
Total for this question: 6
The table gives typical thinking and braking distances at three speeds. At $13\,\text{m s}^{-1}$ they are $9\,\text{m}$ and $14\,\text{m}$ ; at $22\,\text{m s}^{-1}$ they are $15\,\text{m}$ and $38\,\text{m}$ ; at $31\,\text{m s}^{-1}$ they are $21\,\text{m}$ and $75\,\text{m}$ . Calculate the total stopping distance at each speed, then explain why the braking-distance column grows far faster than the thinking-distance column.
[5 marks]
Total for this question: 5
A car travels at $24\,\text{m s}^{-1}$ and the driver's reaction time is $0.55\,\text{s}$ . The braking distance is $36\,\text{m}$ on a dry road. On a wet road it is $50\%$ greater, while reaction time is unchanged. Calculate both stopping distances and determine whether a clear distance of $65\,\text{m}$ is sufficient on the wet road.
[5 marks]
Total for this question: 5

4.5.6.3.2 · Reaction time

Explanation

Human reaction times vary, with typical values from $0.2\,\text{s}$ to $0.9\,\text{s}$ .
Tiredness, alcohol, some drugs and distractions can increase reaction time, so a moving vehicle covers a greater thinking distance before braking begins.
Reaction time can be measured with a ruler-drop test or computer test.
To investigate a factor, change only that independent variable, control the release method and other conditions, repeat each measurement, identify anomalies and compare mean reaction times.
A conclusion should be based on the pattern and spread of repeated results, not one reading.

Worked example

A car travels at $22\,\text{m/s}$ and the driver reacts in $0.48\,\text{s}$ . Calculate the thinking distance.

1.During the reaction time the car continues at $22\,\text{m/s}$ .
2. $s=vt=22\times0.48$ .

Answer: $10.56\,\text{m}$

Common mistakes

Don't fall into the trap of using one ruler-drop result instead of repeating and calculating a mean.
Don't fall into the trap of changing both the distraction and the ruler release position between conditions.
Don't fall into the trap of claiming an anomalous result proves the investigated factor has an effect.

Exam tip

For an ‘evaluate’ question, discuss repeats, anomalies and the difference between the two mean reaction times.

Tier 1 · Easy

State the typical range of human reaction times and give one factor that can increase a driver's reaction time.
[2 marks]
Total for this question: 2
A student measures reaction time with a falling ruler. State why the ruler should be released without warning and why several readings should be taken.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A driver travels at $22\,\text{m s}^{-1}$ and reacts in $0.48\,\text{s}$ . Calculate the thinking distance.
[2 marks]
Total for this question: 2
A student tests whether background conversation affects reaction time. Every no-conversation trial is completed before every conversation trial, and only one reading is taken in each condition. Identify two weaknesses and describe one improvement for each.
[4 marks]
Total for this question: 4
A ruler-drop lookup table gives: $5\,\text{cm}\to0.10\,\text{s}$ , $10\,\text{cm}\to0.14\,\text{s}$ , $15\,\text{cm}\to0.17\,\text{s}$ and $20\,\text{cm}\to0.20\,\text{s}$ . A student's catch distances are $14$ , $16$ and $15\,\text{cm}$ . Calculate the mean distance, use the table to estimate the reaction time, comment on the result using the typical human range and give one possible reason for it.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A student measures reaction time four times without a distraction and obtains $0.31$ , $0.29$ , $0.30$ and $0.89\,\text{s}$ . With a distraction the results are $0.44$ , $0.47$ , $0.45$ and $0.46\,\text{s}$ . Identify the anomalous result, calculate suitable mean times and evaluate the effect of the distraction.
[5 marks]
Total for this question: 5
A driver's reaction times without conversation are $0.24$ , $0.26$ and $0.25\,\text{s}$ ; with conversation they are $0.39$ , $0.35$ and $0.37\,\text{s}$ . At $90\,\text{km h}^{-1}$ , first convert the speed to metres per second. Calculate the two mean reaction times, the corresponding thinking distances and the increase caused by conversation. Give one limitation of the evidence.
[6 marks]
Total for this question: 6
Four participants are tested without and with a distraction. Their paired reaction times, in seconds, are A: $0.28$ and $0.39$ ; B: $0.35$ and $0.44$ ; C: $0.24$ and $0.38$ ; D: $0.31$ and $0.42$ . Calculate the increase for each participant and the mean increase. Explain one advantage of using paired results and give one limitation of this evidence.
[6 marks]
Total for this question: 6
Plan an investigation using a computer reaction-time test to determine whether repeated practice changes reaction time. Include the independent and dependent variables, controls, a method for collecting enough data and how the results should be analysed.
[6 marks]
Total for this question: 6
For one driver, thinking distances are $4.8\,\text{m}$ at $12\,\text{m s}^{-1}$ , $7.2\,\text{m}$ at $18\,\text{m s}^{-1}$ and $9.6\,\text{m}$ at $24\,\text{m s}^{-1}$ . After taking a medicine that can slow responses, the thinking distance at $18\,\text{m s}^{-1}$ is $10.8\,\text{m}$ . Determine the original reaction time from every pair, the later reaction time and the percentage increase in reaction time.
[5 marks]
Total for this question: 5

4.5.6.3.3 · Factors affecting braking distance 1

Explanation

Braking distance is affected by the vehicle's speed, road conditions and the condition of its tyres and brakes.
Wet or icy roads and worn tyres reduce grip, so the frictional braking force and deceleration are smaller and the vehicle travels farther before stopping.
Poor brakes can also reduce the braking force.
A greater initial speed gives the vehicle more kinetic energy, so more work must be done to stop it and the braking distance increases substantially.
Keep this separate from thinking distance: tiredness, alcohol, drugs and distractions affect reaction time rather than the braking process.

Worked example

The average braking force is $6200\,\text{N}$ on a dry road and $3100\,\text{N}$ on a wet road. A car stops in $18\,\text{m}$ on the dry road from the same initial speed. Estimate its wet-road braking distance.

1.The same kinetic energy must be removed, so braking work $Fs$ is unchanged.
2.The force halves, so the distance doubles: $18\times\dfrac{6200}{3100}$ .

Answer: $36\,\text{m}$

Common mistakes

Don't fall into the trap of saying tiredness directly increases braking distance rather than thinking distance.
Don't fall into the trap of claiming worn tyres increase friction and therefore shorten braking distance.
Don't fall into the trap of assuming braking distance increases in direct proportion to speed in otherwise fixed conditions.

Exam tip

In an ‘explain’ answer, link poor grip to smaller frictional force, smaller deceleration and a longer braking distance.

Tier 1 · Easy

Give one example of poor vehicle condition that increases braking distance and explain why it does so.
[2 marks]
Total for this question: 2
A driver says tiredness makes the car's braking distance longer, while icy conditions make its thinking distance longer. Explain what is wrong with both parts of the statement.
[3 marks]
Total for this question: 3

Tier 2 · Standard

The average braking force on a car is $6200\,\text{N}$ on a dry road and $3100\,\text{N}$ on a wet road. The car has the same initial speed in both tests and stops in $18\,\text{m}$ on the dry road. Estimate the wet-road braking distance.
[3 marks]
Total for this question: 3
At the same initial speed, a car records these braking distances: new tyres on dry road, $21\,\text{m}$ ; worn tyres on dry road, $28\,\text{m}$ ; new tyres on wet road, $35\,\text{m}$ ; worn tyres on wet road, $49\,\text{m}$ . Use the data to give two conclusions about factors affecting braking distance.
[4 marks]
Total for this question: 4
A car has a braking distance of $18\,\text{m}$ at $12\,\text{m s}^{-1}$ . In unchanged conditions its braking distance later measures $50\,\text{m}$ . Assuming braking distance is proportional to speed squared, estimate the later speed.
[3 marks]
Total for this question: 3

Tier 3 · Hard

For one car in fixed conditions, measured braking distances are $6.2\,\text{m}$ at $10\,\text{m s}^{-1}$ , $13.9\,\text{m}$ at $15\,\text{m s}^{-1}$ and $24.8\,\text{m}$ at $20\,\text{m s}^{-1}$ . Show that the data are consistent with braking distance being proportional to speed squared, then estimate the distance at $25\,\text{m s}^{-1}$ .
[4 marks]
Total for this question: 4
The same car, driver, initial speed and dry road are used before and after its brakes are serviced. Before servicing, repeated braking distances are $31$ , $48$ , $33$ and $32\,\text{m}$ . Afterwards they are $22$ , $21$ , $23$ and $22\,\text{m}$ . Identify any anomalous result, calculate suitable mean braking distances and use them to evaluate the effect of servicing the brakes.
[5 marks]
Total for this question: 5
Design an investigation using the same wheeled trolley to compare braking distance on two track surfaces. Describe how to make the trolley start braking at the same speed, state two other control variables, explain how braking distance is measured, and describe how the results should be made reliable.
[6 marks]
Total for this question: 6
A car's braking distance is $20\,\text{m}$ at $16\,\text{m s}^{-1}$ on a dry road. On a wet road at $24\,\text{m s}^{-1}$ , its braking distance is $63\,\text{m}$ . Assuming braking distance is proportional to speed squared in fixed conditions and the wet-road factor is constant across these speeds, calculate the distance expected at $24\,\text{m s}^{-1}$ on the dry road, determine the wet-road factor, and predict the wet-road distance at $20\,\text{m s}^{-1}$ .
[6 marks]
Total for this question: 6
At one fixed speed, a car's kinetic energy store is $168\,\text{kJ}$ . The average braking force is $7000\,\text{N}$ with new tyres on a dry road, $5600\,\text{N}$ with worn tyres on a dry road, and $4200\,\text{N}$ with new tyres on a wet road. Calculate the braking distance in each case, and state whether changing only to worn tyres or changing only to a wet road is the more dangerous single change. Justify your answer in terms of grip, friction, deceleration and braking distance.
[5 marks]
Total for this question: 5

4.5.6.3.4 · Factors affecting braking distance 2

Explanation

When a vehicle brakes, friction does work and transfers energy from the vehicle's kinetic energy store to thermal energy stores of the brakes and surroundings. The work done by the braking force is $W=Fs$ .
A faster or more massive vehicle has more kinetic energy, so stopping it over the same distance requires a greater braking force.
From $F=ma$ , a greater braking force on the same mass produces a greater deceleration.
Very large decelerations can cause brakes to overheat or the driver to lose control.
Energy is transferred during braking; it is not destroyed.

Worked example

A $1100\,\text{kg}$ car travels at $18\,\text{m/s}$ and is stopped by an average braking force of $6600\,\text{N}$ . Calculate its braking distance.

1. $E_k=\dfrac{1}{2}mv^2=0.5\times1100\times18^2=178200\,\text{J}$ .
2.Set braking work equal to the energy transferred: $Fs=178200$ .
3. $s=178200\div6600=27\,\text{m}$ .

Answer: $27\,\text{m}$

Common mistakes

Don't fall into the trap of saying the kinetic energy is destroyed rather than transferred to thermal energy stores.
Don't fall into the trap of using the vehicle's weight instead of the braking force in $W=Fs$ .
Don't fall into the trap of claiming a greater braking force gives a smaller deceleration for the same mass.

Exam tip

For a braking calculation, connect kinetic energy lost to work done by the braking force before rearranging.

Tier 1 · Easy

Describe the main energy transfer when friction in a vehicle's brakes brings the vehicle to rest.
[2 marks]
Total for this question: 2
A student claims that increasing a vehicle's braking force is always safer. Explain why a much greater braking force can create a danger even though it can reduce braking distance.
[3 marks]
Total for this question: 3

Tier 2 · Standard

A $1100\,\text{kg}$ car travels at $18\,\text{m s}^{-1}$ . Its average braking force is $6600\,\text{N}$ . Calculate its initial kinetic energy and the braking distance, assuming all of this energy is removed by the braking force.
[4 marks]
Total for this question: 4
A car loses $240\,\text{kJ}$ from its kinetic energy store while braking over $30\,\text{m}$ . Convert the energy to joules and determine the average braking force.
[4 marks]
Total for this question: 4
Vehicle A has mass $1200\,\text{kg}$ and speed $15\,\text{m s}^{-1}$ . Vehicle B has mass $675\,\text{kg}$ and speed $20\,\text{m s}^{-1}$ . Calculate the kinetic energy of each vehicle. Both are stopped by the same average braking force of $4500\,\text{N}$ ; determine and compare their braking distances.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A $1500\,\text{kg}$ car travelling at $22\,\text{m s}^{-1}$ is stopped by an average braking force of $8250\,\text{N}$ . Calculate the braking distance and the magnitude of the deceleration. Explain one danger of increasing the braking force substantially.
[6 marks]
Total for this question: 6
Higher only: A $1400\,\text{kg}$ car travelling at $90\,\text{km h}^{-1}$ stops in $3.2\,\text{s}$ . Convert the speed to metres per second, then estimate the magnitudes of the average deceleration and average resultant force. State why the force is an estimate.
[6 marks]
Total for this question: 6
Higher only: A $1200\,\text{kg}$ vehicle travels at $20\,\text{m s}^{-1}$ . For controlled braking, the magnitude of its average deceleration must not exceed $6.0\,\text{m s}^{-2}$ . Calculate the greatest permitted average braking force and the corresponding minimum braking distance. Decide whether stopping within $28\,\text{m}$ meets the deceleration limit.
[6 marks]
Total for this question: 6
A $1000\,\text{kg}$ car travelling at $18\,\text{m s}^{-1}$ is brought to rest in two braking stages. In the first stage, the average braking force is $4000\,\text{N}$ over $18\,\text{m}$ . Calculate the initial kinetic energy, the energy transferred in the first stage, and the average braking force needed over a further $15\,\text{m}$ to remove the remaining kinetic energy.
[6 marks]
Total for this question: 6
Use the Physics Equations Sheet. A $1200\,\text{kg}$ car slows from $20\,\text{m s}^{-1}$ to $10\,\text{m s}^{-1}$ . $60\%$ of the decrease in kinetic energy heats brake discs of total mass $24\,\text{kg}$ and specific heat capacity $450\,\text{J kg}^{-1}\,{}^{\circ}\text{C}^{-1}$ . Calculate the kinetic energy decrease, the energy transferred to the discs and their temperature rise. Describe how the remaining energy is transferred.
[6 marks]
Total for this question: 6

4.5.7.1 · Momentum is a property of moving objects (HT only)

Explanation

Higher tier: momentum is a property of a moving object and is calculated using $p=mv$ .
Here $p$ is momentum in $\text{kg m/s}$ , $m$ is mass in kilograms and $v$ is velocity in $\text{m/s}$ .
Momentum is a vector, so it has the same direction as the velocity.
For motion along one line, choose a positive direction and give momentum in the opposite direction a negative sign.
A stationary object has zero momentum, while increasing either mass or speed increases the magnitude of momentum.

Worked example

A $1350\,\text{kg}$ car travels west at $16\,\text{m/s}$ . Calculate its momentum.

1. $p=mv=1350\times16=21600\,\text{kg m/s}$ .
2.Momentum has the same direction as velocity, so it is westward.

Answer: $2.16\times10^4\,\text{kg m/s}$ west

Common mistakes

Don't fall into the trap of using a mass in grams without converting it to kilograms.
Don't fall into the trap of calculating the correct momentum magnitude but omitting its direction.
Don't fall into the trap of treating speed as signed while failing to define a positive direction.

Exam tip

State a positive direction before combining momenta from objects moving in opposite directions.

Tier 1 · Easy

A $0.18\,\text{kg}$ ball moves at $12\,\text{m s}^{-1}$ . Calculate its momentum.
[2 marks]
Total for this question: 2
Two objects have equal masses and move in opposite directions. Object X moves twice as fast as object Y. Compare the magnitudes and directions of their momenta.
[3 marks]
Total for this question: 3

Tier 2 · Standard

A $1350\,\text{kg}$ car travels west at $16\,\text{m s}^{-1}$ . Calculate its momentum, including direction.
[2 marks]
Total for this question: 2
A puck of mass $65\,\text{g}$ has momentum $0.390\,\text{kg m s}^{-1}$ east. Convert its mass to kilograms and determine its velocity.
[4 marks]
Total for this question: 4
A graph of momentum against velocity for one object is a straight line through the origin and the point $(+7.0\,\text{m s}^{-1},+5.6\,\text{kg m s}^{-1})$ . Determine the object's mass and predict its momentum at $-9.0\,\text{m s}^{-1}$ .
[4 marks]
Total for this question: 4

Tier 3 · Hard

Vehicle A has mass $720\,\text{kg}$ and travels east at $18\,\text{m s}^{-1}$ . Vehicle B has mass $1080\,\text{kg}$ and travels west at $11\,\text{m s}^{-1}$ . Taking east as positive, calculate each momentum and determine which has the greater momentum magnitude and by how much.
[5 marks]
Total for this question: 5
Cart A has mass $0.84\,\text{kg}$ and velocity $+3.5\,\text{m s}^{-1}$ . Cart B moves at $-2.1\,\text{m s}^{-1}$ and has momentum equal in magnitude but opposite in direction to Cart A. Calculate both signed momenta and determine the mass of Cart B.
[5 marks]
Total for this question: 5
An object's measured mass lies between $0.395$ and $0.405\,\text{kg}$ , and its measured speed lies between $2.9$ and $3.1\,\text{m s}^{-1}$ . Calculate the smallest and largest possible momentum magnitudes. Use the range to assess a reported momentum of $1.30\,\text{kg m s}^{-1}$ .
[5 marks]
Total for this question: 5
At one instant, cart A has mass $0.75\,\text{kg}$ and velocity $+4.8\,\text{m s}^{-1}$ , while cart B has mass $1.20\,\text{kg}$ and velocity $-1.5\,\text{m s}^{-1}$ . Cart C has mass $0.60\,\text{kg}$ . Calculate the signed momenta of A and B, then determine the velocity C must have for the total signed momentum of the three carts to be zero.
[5 marks]
Total for this question: 5
Vehicle A has mass $900\,\text{kg}$ and speed $20\,\text{m s}^{-1}$ . Vehicle B has mass $1600\,\text{kg}$ and the same kinetic energy as A. Calculate the common kinetic energy, B's speed and both momentum magnitudes. Compare the momentum magnitudes.
[6 marks]
Total for this question: 6

4.5.7.2 · Conservation of momentum (HT only)

Explanation

Higher tier: in a closed system, total momentum before an event equals total momentum after it.
Choose one direction as positive, calculate each momentum using $p=mv$ , and include negative signs for objects moving in the opposite direction.
In a collision where objects stick together, add their masses because they share one final velocity: final momentum is $(m_1+m_2)v$ .
In an explosion or recoil, an initially stationary system has zero total momentum, so the products have equal and opposite total momenta.
Momentum is conserved, but kinetic energy need not be conserved in a collision.

Worked example

A $2.0\,\text{kg}$ trolley moving right at $3.0\,\text{m/s}$ hits a stationary $1.0\,\text{kg}$ trolley. They stick together. Calculate their velocity.

1.Initial momentum $=(2.0\times3.0)+(1.0\times0)=6.0\,\text{kg m/s}$ .
2.Joined mass $=2.0+1.0=3.0\,\text{kg}$ .
3. $6.0=3.0v$ , so $v=2.0\,\text{m/s}$ .

Answer: $2.0\,\text{m/s}$ to the right

Common mistakes

Don't fall into the trap of adding speeds instead of adding the signed momenta of the objects.
Don't fall into the trap of forgetting to add the masses when the colliding objects stick together.
Don't fall into the trap of assuming kinetic energy must be conserved because momentum is conserved.

Exam tip

Write a complete ‘total before = total after’ momentum equation before rearranging for the unknown.

Tier 1 · Easy

A $2.0\,\text{kg}$ trolley moving at $3.0\,\text{m s}^{-1}$ collides with a stationary $1.0\,\text{kg}$ trolley. They stick together. Calculate their common velocity.
[3 marks]
Total for this question: 3
The total momentum of a closed system is $+4.0\,\text{kg m s}^{-1}$ before an event. Afterwards, object A has momentum $-1.5\,\text{kg m s}^{-1}$ . Determine the momentum of object B.
[3 marks]
Total for this question: 3

Tier 2 · Standard

A $0.75\,\text{kg}$ trolley moving right at $6.4\,\text{m s}^{-1}$ catches a $1.25\,\text{kg}$ trolley moving right at $1.6\,\text{m s}^{-1}$ . The trolleys lock together. Determine their final velocity.
[4 marks]
Total for this question: 4
Two carts collide. Before the collision, cart A has mass $0.50\,\text{kg}$ and velocity $+1.6\,\text{m s}^{-1}$ ; cart B has mass $0.30\,\text{kg}$ and velocity $-1.0\,\text{m s}^{-1}$ . Afterwards their measured momenta are $+0.175$ and $+0.315\,\text{kg m s}^{-1}$ . Calculate the total momentum before and after. Evaluate whether the measurements support conservation of momentum.
[6 marks]
Total for this question: 6
A $1.5\,\text{kg}$ trolley moves right at $2.4\,\text{m s}^{-1}$ before a spring separates it into two parts. A $0.60\,\text{kg}$ part moves left at $1.5\,\text{m s}^{-1}$ . Calculate the velocity of the other part.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A launcher of mass $3.8\,\text{kg}$ and a $0.20\,\text{kg}$ projectile are initially at rest. The projectile is fired horizontally at $32\,\text{m s}^{-1}$ . Calculate the launcher's recoil velocity.
[4 marks]
Total for this question: 4
Cart A of mass $0.80\,\text{kg}$ moves at $+5.0\,\text{m s}^{-1}$ and collides with cart B of mass $1.20\,\text{kg}$ moving at $-1.0\,\text{m s}^{-1}$ . After the collision, cart A moves at $-0.50\,\text{m s}^{-1}$ . Calculate cart B's final velocity.
[5 marks]
Total for this question: 5
Cart X has mass $0.60\,\text{kg}$ and approaches a stationary $0.40\,\text{kg}$ cart at $4.0\,\text{m s}^{-1}$ . After impact they remain joined and travel at $2.4\,\text{m s}^{-1}$ . Calculate the system's total momentum and kinetic energy before and after impact. Use the results to compare what is conserved.
[6 marks]
Total for this question: 6
A stationary $1.50\,\text{kg}$ object separates into three fragments while external forces are negligible. A $0.30\,\text{kg}$ fragment moves at $+8.0\,\text{m s}^{-1}$ and a $0.50\,\text{kg}$ fragment moves at $-3.0\,\text{m s}^{-1}$ . Calculate the mass and velocity of the third fragment, taking the positive direction as the direction of the first fragment.
[5 marks]
Total for this question: 5
A $3.0\,\text{kg}$ platform carrying two $1.0\,\text{kg}$ parcels moves at $+4.0\,\text{m s}^{-1}$ while external forces are negligible. One parcel is released with velocity $-1.0\,\text{m s}^{-1}$ relative to the ground. Later, the second parcel is released with velocity $-0.75\,\text{m s}^{-1}$ relative to the ground. Determine the velocity of the platform-plus-parcel system after the first release and the platform's velocity after the second release.
[6 marks]
Total for this question: 6

4.5.7.3 · Changes in momentum (physics only) (HT only)

Explanation

Higher tier: force equals the rate of change of momentum, so $F=\dfrac{\Delta p}{\Delta t}$ .
For constant mass, $\Delta p=m(v-u)$ .
Choose a positive direction and use signed velocities; an object that rebounds has final velocity with the opposite sign, so its velocity change is larger than either speed alone.
Rearranging gives $F\Delta t=\Delta p$ : for the same momentum change, increasing the collision time reduces the average force.
This is how airbags, helmets, crash mats and crumple zones reduce forces on people.

Worked example

A $0.16\,\text{kg}$ ball travels towards a wall at $12\,\text{m/s}$ and rebounds at $8.0\,\text{m/s}$ . Contact lasts $0.050\,\text{s}$ . Calculate the average force.

1.Take towards the wall as positive: $u=+12\,\text{m/s}$ and $v=-8.0\,\text{m/s}$ .
2. $\Delta p=m(v-u)=0.16(-8.0-12)=-3.2\,\text{kg m/s}$ .
3. $F=\dfrac{-3.2}{0.050}=-64\,\text{N}$ .

Answer: $64\,\text{N}$ away from the wall

Common mistakes

Don't fall into the trap of using $12-8=4\,\text{m/s}$ when the ball reverses direction.
Don't fall into the trap of dividing momentum rather than change in momentum by the contact time.
Don't fall into the trap of saying an airbag reduces momentum change instead of increasing the time for that change.

Exam tip

For a rebound, assign opposite signs to the initial and final velocities before calculating $v-u$ .

Tier 1 · Easy

A $0.35\,\text{kg}$ ball moving at $8.0\,\text{m s}^{-1}$ is brought to rest in $0.10\,\text{s}$ . Calculate the magnitude of the average force.
[3 marks]
Total for this question: 3
Two helmet linings bring a cyclist's head to rest with the same change in momentum. Lining B doubles the stopping time compared with lining A. Compare their average forces and explain which lining is safer.
[3 marks]
Total for this question: 3

Tier 2 · Standard

A $0.16\,\text{kg}$ ball travels toward a wall at $12\,\text{m s}^{-1}$ and rebounds at $8.0\,\text{m s}^{-1}$ . Contact lasts $0.050\,\text{s}$ . Calculate the magnitude and direction of the average force on the ball.
[4 marks]
Total for this question: 4
Two pads each produce a momentum change of $3.6\,\text{kg m s}^{-1}$ . Pad P acts for $0.012\,\text{s}$ and pad Q for $0.030\,\text{s}$ . Calculate the average force from each pad and identify the design that reduces the force more.
[4 marks]
Total for this question: 4
An average force of $+48\,\text{N}$ acts on an object for $0.075\,\text{s}$ . Its velocity changes from $-2.0\,\text{m s}^{-1}$ to $+4.0\,\text{m s}^{-1}$ . Determine the object's mass.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A $950\,\text{kg}$ vehicle travelling at $14\,\text{m s}^{-1}$ stops in a collision. A rigid structure would stop it in $0.080\,\text{s}$ , while a crumple zone increases the stopping time to $0.32\,\text{s}$ . Calculate the average force magnitude in each case and explain the safety benefit of the crumple zone.
[6 marks]
Total for this question: 6
A $0.12\,\text{kg}$ ball approaches a bat at $+18\,\text{m s}^{-1}$ and leaves at $-12\,\text{m s}^{-1}$ . Take the initial direction as positive. Calculate the signed change in momentum. If the average force is $-90\,\text{N}$ , determine the contact time. A padded bat produces the same momentum change in $0.10\,\text{s}$ ; calculate its average force and compare the two forces.
[6 marks]
Total for this question: 6
An object of mass $0.50\,\text{kg}$ has an initial speed of $8.0\,\text{m s}^{-1}$ and protective material brings it to rest. The average force magnitude must not exceed $40\,\text{N}$ . A thicker material gives a longer contact time. The measured contact times are $0.040\,\text{s}$ for material A, $0.15\,\text{s}$ for B and $0.20\,\text{s}$ for C. Calculate the momentum change magnitude and minimum contact time, choose the least thick suitable material and calculate its average force magnitude.
[6 marks]
Total for this question: 6
A $0.80\,\text{kg}$ trolley initially moves at $-1.0\,\text{m s}^{-1}$ . An average force of $+6.0\,\text{N}$ acts for $0.30\,\text{s}$ , followed by an average force of $+2.0\,\text{N}$ for $0.50\,\text{s}$ . Calculate the momentum change in each interval, then determine the trolley's final momentum and velocity.
[5 marks]
Total for this question: 5
Every $0.50\,\text{s}$ , $2.4\,\text{kg}$ of water reaches a vertical wall at $6.0\,\text{m s}^{-1}$ and is stopped by the wall. Calculate the horizontal momentum change of this water, the average horizontal force on the water, and the average force exerted by the water on the wall.
[5 marks]
Total for this question: 5

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

4.5.1.1 · Scalar and vector quantities

Tier 1 · Easy

Mark scheme for 4.5.1.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	A speedometer gives only the magnitude of the motion. Velocity is a vector and also requires the direction of motion.	Distinguish the scalar from the vector. Speed is fully specified by magnitude, but velocity is not complete until a direction is included.	2
Total Question 1			2
02.1	The stated magnitude is wrong: $4.0\times6.0=24\,\text{N}$ . The stated direction is wrong: the correct label is $24\,\text{N}$ west.	Use the scale to convert the arrow length to a magnitude: $4.0\,\text{cm}\times6.0\,\text{N/cm}=24\,\text{N}$ . Read the direction from the arrowhead, which points west.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.5.1.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	The arrow is $5.0\,\text{cm}$ long. Its arrowhead points vertically downwards.	The required length is $20\div4.0=5.0\,\text{cm}$ . Draw that length vertically and put the arrowhead at the downward end to show the force direction.	2
Total Question 1			2
02.1	Distance and speed are scalars. Displacement and velocity are vectors. Scalars have magnitude only, whereas vectors have both magnitude and direction.	Distance and speed are fully specified by their numerical sizes and units, so they are scalars. Displacement and velocity also require the stated direction north, so they are vectors.	3
Total Question 2			3
03.1	Temperature is scalar; $-5\,^\circ\text{C}$ is a value below the zero point on the temperature scale, not a direction. Force is a vector because it has magnitude and direction. The backward force could be written as $-12\,\text{N}$ because it acts opposite to the chosen positive direction.	Classify each quantity by whether direction is required. A negative scalar temperature locates the reading below the scale's zero, whereas a negative sign on the force would encode its direction relative to the chosen positive axis.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.5.1.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Speed: $11\,\text{m/s}$ . Velocity: $11\,\text{m/s}$ south-west. Speed is scalar, while velocity includes direction and is vector.	Keep only the magnitude for speed. Attach the stated direction to the same magnitude for velocity. The direction is what makes the velocity a different, vector quantity.	3
Total Question 1			3
02.1	The displacement magnitude is $1080\,\text{m}$ , or $1.08\,\text{km}$ . The displacement is on a bearing of $135^\circ$ . Distance cannot be found because the arrow gives only the straight-line change from start to finish, not the path length.	Apply the map scale: $7.2\times150=1080\,\text{m}=1.08\,\text{km}$ . A vector needs magnitude and direction, so keep the $135^\circ$ bearing. Distance is scalar path length and may exceed the displacement magnitude, so it is not encoded by this arrow.	4
Total Question 2			4
03.1	The forces have equal magnitudes because the arrows have equal lengths. Their directions are opposite: north-east and south-west. The vectors are not equal because equal vectors must have the same magnitude and direction.	Read arrow length as magnitude and the arrowhead orientation as direction. Matching only the lengths is insufficient: the opposite directions make the vectors unequal.	3
Total Question 3			3
04.1	Sensor A reports a force of magnitude $36\,\text{N}$ acting west. Sensor B also reports a force of magnitude $36\,\text{N}$ acting west. The signs differ because the sensors use opposite positive directions. The vectors are equal because they have the same magnitude and direction.	Interpret each sign using that sensor's chosen positive direction. A negative reading for east-positive means west, while a positive reading for west-positive also means west. In both cases the magnitude is the positive size $36\,\text{N}$ .	4
Total Question 4			4
05.1	Diagram A represents $24\,\text{N}$ east. Diagram B represents $24\,\text{N}$ east. Diagram C represents $25\,\text{N}$ west. The vectors in A and B are equal despite their different drawn lengths because both magnitude and direction match. The vector in C is different in both magnitude and direction.	Convert each arrow length using its own scale: $6.0\times4.0=24\,\text{N}$ , $3.0\times8.0=24\,\text{N}$ and $5.0\times5.0=25\,\text{N}$ . Compare the physical magnitudes and directions, not the lengths on diagrams that use different scales.	5
Total Question 5			5

4.5.1.2 · Contact and non-contact forces

Tier 1 · Easy

Mark scheme for 4.5.1.2 Tier 1 · Easy
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01.1	Air resistance is a contact force. It is caused by collisions between air particles and the cyclist.	Ask whether matter must touch the cyclist. Air particles collide with the cyclist's surface, so the interaction requires contact even though the air is not easily visible.	2
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02.1	The interaction is an electrostatic, non-contact force. The balloon and paper are physically separated when the force acts.	The observation shows that the force acts before the objects touch. Friction requires touching surfaces, whereas an electrostatic force can act across the gap.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.5.1.2 Tier 2 · Standard
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01.1	The magnetic force is non-contact. The magnet pulls the pin and the pin pulls the magnet, with the forces directed towards each other.	The objects are separated, so the interaction is non-contact. The interaction acts on both objects: draw one arrow on the pin towards the magnet and one on the magnet towards the pin.	2
Total Question 1			2
02.1	Trial A: magnetic force, which is non-contact because it acts across the gap. Trial B: friction, which is contact because it acts only while the block and mat touch.	Use physical separation as the test. Motion towards the magnet before contact identifies a magnetic non-contact interaction. Slowing only during surface contact identifies friction, a contact force.	4
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03.1	The normal contact force from the tray disappears when contact is lost. The ball's weight, or gravitational force, remains. The normal force requires contact, whereas gravity is a non-contact interaction with Earth.	Track what changes when the objects separate. Removing tray–ball contact removes the normal force, but it does not remove the Earth–ball gravitational interaction.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.5.1.2 Tier 3 · Hard
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01.1	Gravitational force acts on the parachutist or parachute and is non-contact. Air resistance acts on the parachute or parachutist and is contact. Tension acts through the cords on the parachutist or parachute and is contact.	Identify gravity as acting without physical contact. Then identify collisions with air as air resistance and pulling by the cords as tension; both require contact with the object experiencing the force.	3
Total Question 1			3
02.1	The force due to Earth is gravitational force. The interacting pair is Earth–washer; this is non-contact. The force due to the thread is tension. The interacting pair is thread–washer; this is contact. The force due to the electromagnet is magnetic force. The interacting pair is electromagnet–washer; this is non-contact.	Match each effect to its interacting objects. Gravity and magnetism act while the objects are separated, whereas the thread must touch the washer to exert tension.	6
Total Question 2			6
03.1	An electrostatic force pulls the oppositely charged sphere towards the rod. It is a non-contact force because it acts while there is a gap. Friction cannot cause the motion because the rod and sphere are not touching. Initially the thread is vertical, so its tension does not provide the sideways pull towards the rod.	Use the across-gap motion as evidence for a non-contact electrostatic interaction. Eliminate friction because contact is absent, and eliminate the initial thread tension because its line of action is along the vertical thread.	4
Total Question 3			4
04.1	The first force is the gravitational force, or weight. It acts between Earth and the droplet and is non-contact. The second force is the electrostatic force. It acts between the charged plates and the droplet and is non-contact. The third force is air resistance, or drag. It acts between the air and the droplet and is contact.	Match each force to the objects that interact. Earth and the charged plates can exert forces across a separation, whereas air resistance is produced by air particles contacting the moving droplet.	6
Total Question 4			6
05.1	During the collision, the buffer forces are normal contact forces. The force on P is left and the force on Q is right while the buffers touch. After separation, the interaction is a non-contact magnetic force. The magnetic force on P is left and the magnetic force on Q is right across the gap. Each interaction produces a force on both carts, but only the buffer interaction requires physical contact.	Use the touching buffers as evidence for a contact interaction and the gap as evidence for a non-contact magnetic interaction. Put each force arrow on the cart that experiences it and point the paired arrows away from one another in this repulsion scenario.	5
Total Question 5			5

4.5.1.3 · Gravity

Tier 1 · Easy

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01.1	The astronaut's mass is unchanged. The astronaut's weight decreases because $W=mg$ and $g$ is smaller.	Mass is the amount of matter and does not depend on location. Weight is the gravitational force, so reducing gravitational field strength reduces weight for the same mass.	2
Total Question 1			2
02.1	Site A has the greater gravitational field strength because the same mass has greater weight there. The sample's mass is unchanged between the sites.	For the same sample, weight is directly proportional to gravitational field strength. The larger newtonmeter reading therefore identifies the larger $g$ , while changing location does not change mass.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.5.1.3 Tier 2 · Standard
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01.1	At A, $g=9.79\,\text{N/kg}$ . At B, $g=1.58\,\text{N/kg}$ .	Rearrange $W=mg$ to $g=W/m$ . At A, $g=23.5\div2.4=9.79\,\text{N/kg}$ . At B, $g=3.8\div2.4=1.58\,\text{N/kg}$ .	3
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02.1	The $2.8\,\text{kg}$ , $10.5\,\text{N}$ pair is anomalous. $g=3.40\,\text{N/kg}$ from each consistent pair.	Rearrange to $g=W/m$ . The first two values give $4.08\div1.2=3.40\,\text{N/kg}$ and $6.80\div2.0=3.40\,\text{N/kg}$ . The third gives $10.5\div2.8=3.75\,\text{N/kg}$ , so it is inconsistent.	4
Total Question 2			4
03.1	The arrow should start at the centre of mass, which is the geometric centre for this uniform block. It should point vertically downwards, towards the centre of Earth. Weight is the gravitational force and is treated as acting through the object's centre of mass.	Correct the point of application first: a uniform block's centre of mass is at its centre. Then correct the force direction: near Earth's surface, weight acts vertically downwards.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.5.1.3 Tier 3 · Hard
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01.1	At P, $g=9.79\,\text{N/kg}$ . At Q, $g=2.60\,\text{N/kg}$ . The mass remains $68\,\text{kg}$ because changing gravitational field strength changes weight, not mass.	Use $g=W/m$ . For P, $666\div68=9.79\,\text{N/kg}$ ; for Q, $177\div68=2.60\,\text{N/kg}$ . Mass is independent of location, so it stays $68\,\text{kg}$ while the weight changes.	4
Total Question 1			4
02.1	The gravitational field strength is $9.0\,\text{N/kg}$ . The empty capsule has mass $5.4\,\text{kg}$ .	Convert $750\,\text{g}$ to $0.750\,\text{kg}$ . Its added weight is $55.35-48.6=6.75\,\text{N}$ , so $g=6.75\div0.750=9.0\,\text{N/kg}$ . Then $m=48.6\div9.0=5.4\,\text{kg}$ .	4
Total Question 2			4
03.1	The combined mass is $4500\,\text{kg}$ . The cargo mass is $1300\,\text{kg}$ . The combined weight at the second site is $4.41\times10^4\,\text{N}$ , or $44.1\,\text{kN}$ .	Convert $18.0\,\text{kN}$ to $18000\,\text{N}$ . Use $m=W/g=18000\div4.0=4500\,\text{kg}$ , then subtract $3200\,\text{kg}$ to get $1300\,\text{kg}$ . Mass is unchanged, so the new combined weight is $4500\times9.8=44100\,\text{N}$ .	5
Total Question 3			5
04.1	The gradient, and hence $g$ , is $9.8\,\text{N/kg}$ . The predicted weight is $31.85\,\text{N}$ , or $31.9\,\text{N}$ to three significant figures. A line through the origin shows that weight is directly proportional to mass. It also shows that zero mass corresponds to zero weight in this model.	Calculate the gradient as $(17.64-4.90)\div(1.80-0.50)=9.8\,\text{N/kg}$ . Then apply $W=mg=3.25\times9.8=31.85\,\text{N}$ . A straight line through the origin is the graph signature of direct proportionality.	5
Total Question 4			5
05.1	The rock's true weight on Earth is $29.4\,\text{N}$ . The rock's mass is $3.0\,\text{kg}$ . Its true weight at the second place is $11.1\,\text{N}$ . The meter reads $12.5\,\text{N}$ after the unchanged $1.4\,\text{N}$ zero error is added.	Correct the Earth reading first: $30.8-1.4=29.4\,\text{N}$ . Use $m=W/g=29.4\div9.8=3.0\,\text{kg}$ . The mass stays constant, so the new true weight is $3.0\times3.7=11.1\,\text{N}$ and the displayed reading is $11.1+1.4=12.5\,\text{N}$ .	5
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4.5.1.4 · Resultant forces

Tier 1 · Easy

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01.1	The resultant force is $0\,\text{N}$ . The box remains stationary.	Equal forces in opposite directions cancel when combined. With no resultant force and with the box initially at rest, its motion does not change.	2
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02.1	The student added forces that act in opposite directions. The correct resultant is $31-12=19\,\text{N}$ right.	Choose right as positive and combine signed forces: $31+(-12)=19\,\text{N}$ . The positive sign shows that the resultant acts right.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.5.1.4 Tier 2 · Standard
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01.1	$9\,\text{N}$ forwards	The total backward force is $27+6=33\,\text{N}$ . Therefore the resultant is $42-33=9\,\text{N}$ forwards.	2
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02.1	The unknown force is $46\,\text{N}$ west. The eastward thrust in test B is $54\,\text{N}$ . The resultant in test B is $8\,\text{N}$ east.	Test A gives $64-F=18$ , so $F=46\,\text{N}$ west. In test B the eastward force is $64-10=54\,\text{N}$ , giving $54-46=8\,\text{N}$ east.	4
Total Question 2			4
03.1	With $12\,\text{N}$ west, the resultant is $8\,\text{N}$ east. With $20\,\text{N}$ west, the resultant is $0\,\text{N}$ and the forces are balanced. With $27\,\text{N}$ west, the resultant is $7\,\text{N}$ west.	Subtract the smaller opposing force from the larger and keep the larger force's direction. The resultant changes direction after the westward force passes $20\,\text{N}$ ; equality at $20\,\text{N}$ gives balance.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.5.1.4 Tier 3 · Hard
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01.1	The resultant is $32\,\text{N}$ east. An additional force of $32\,\text{N}$ west is needed.	Combine the westward forces: $34+19=53\,\text{N}$ . The resultant is $85-53=32\,\text{N}$ east. A balancing force must be equal and opposite, so it is $32\,\text{N}$ west.	3
Total Question 1			3
02.1	The north arrow is mis-scaled: it should be $4.0\,\text{cm}$ long, not $8.0\,\text{cm}$ . Keep the arrows head-to-tail and draw the resultant from the tail of the east arrow to the head of the corrected north arrow. The resultant measures $5.0\,\text{cm}$ , so its magnitude is $10\,\text{N}$ ; accept $9.6$ – $10.4\,\text{N}$ . Its direction is about $53^\circ$ north of east; accept $51$ – $55^\circ$ .	The scale converts $8.0\,\text{N}$ to $8.0\div2.0=4.0\,\text{cm}$ , so the student's $8.0\,\text{cm}$ second arrow is too long. Complete the corrected head-to-tail scale drawing, measure the resultant as $5.0\,\text{cm}$ and use a protractor to measure about $53^\circ$ north of east. The scale gives $5.0\times2.0=10\,\text{N}$ .	5
Total Question 2			5
03.1	Draw a $6.0\,\text{cm}$ arrow due east and a $4.5\,\text{cm}$ arrow north-east, using a parallelogram or head-to-tail construction. The resultant arrow measures about $9.7\,\text{cm}$ , giving a magnitude of about $19\,\text{N}$ ; accept $18.5$ – $20.0\,\text{N}$ . Its direction is about $19^\circ$ north of east; accept $17$ – $21^\circ$ north of east.	Use the stated scale and preserve the $45^\circ$ north-east direction. Complete a parallelogram or place the second vector head-to-tail, then draw the resultant from the starting point. Measure its length and angle, and convert the measured length using $1.0\,\text{cm}=2.0\,\text{N}$ .	4
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04.1	The free-body diagram has an upward cable-tension arrow and downward arrows for weight and air resistance, all acting on the lift. Constant velocity means the initial resultant is $0\,\text{N}$ . The air resistance is $6200-5800=400\,\text{N}$ downwards. The new resultant is $6600-5800-400=400\,\text{N}$ upwards. The lift accelerates upwards, so its upward speed increases.	Isolate only the lift and include every external force on it. Balance the initial upward and downward forces because its velocity is constant. After the tension changes, combine the signed vertical forces to obtain the upward resultant.	6
Total Question 4			6
05.1	Draw a $2.5\,\text{cm}$ east arrow followed head-to-tail by a $6.0\,\text{cm}$ north arrow. The resultant measures about $6.5\,\text{cm}$ , giving about $26\,\text{N}$ ; accept $25$ – $27\,\text{N}$ . Its direction is about $67^\circ$ north of east; accept $65$ – $69^\circ$ north of east. Equilibrium needs a force of about $26\,\text{N}$ in the opposite direction, about $67^\circ$ south of west, with the same acceptance bands.	Convert the two forces to $2.5\,\text{cm}$ and $6.0\,\text{cm}$ arrows and place them head-to-tail at right angles. Draw the resultant from the start of the first arrow to the end of the second, then measure its length and direction. Reverse that measured vector to give the equilibrant.	5
Total Question 5			5

4.5.2 · Work done and energy transfer

Tier 1 · Easy

Mark scheme for 4.5.2 Tier 1 · Easy
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01.1	The work done on the wall is $0\,\text{J}$ . There is no displacement along the force's line of action.	Use $W=Fs$ . The wall's displacement is zero, so multiplying the applied force by $0\,\text{m}$ gives zero work done on the wall.	2
Total Question 1			2
02.1	The suitcase has no displacement along the upward force's line of action. The horizontal movement is perpendicular to the supporting force, so the work done by that force is zero.	Work uses the distance moved along the force's line of action. The supporting force is vertical while the displacement is horizontal, so the relevant distance is zero.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.5.2 Tier 2 · Standard
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01.1	$3.12\times10^3\,\text{J}$ The gravitational potential energy store of the load increases.	The displacement is along the upward force, so $W=480\times6.5=3120\,\text{J}$ . Lifting transfers energy to the load's gravitational potential energy store.	3
Total Question 1			3
02.1	Route A: $W=75\times18=1350\,\text{J}$ . Route B: $W=75\times24=1800\,\text{J}$ . The claim is incorrect: the motor does $450\,\text{J}$ more work on route B. Work depends on the distance moved along the force's line of action, so the longer route requires more work even though the start and finish points, and hence the displacement, are the same.	Apply $W=Fs$ to each route because the motor force acts along the motion: $W_A=75\times18=1350\,\text{J}$ and $W_B=75\times24=1800\,\text{J}$ . Their difference is $1800-1350=450\,\text{J}$ . Equal displacement does not make the route lengths equal, so it does not make these work values equal.	4
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03.1	Each data pair gives a force of $180\,\text{N}$ . The predicted work after $6.5\,\text{m}$ is $1170\,\text{J}$ .	For a constant force along the motion, $W/s$ should be constant. The pairs give $360/2.0=900/5.0=1440/8.0=180\,\text{N}$ . Therefore $W=Fs=180\times6.5=1170\,\text{J}$ .	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.5.2 Tier 3 · Hard
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01.1	Motor work: $2160\,\text{J}$ . Thermal transfer: $1680\,\text{J}$ . Increase in kinetic energy store: $480\,\text{J}$ .	The motor does $W=180\times12=2160\,\text{J}$ . Work against friction is $140\times12=1680\,\text{J}$ and is transferred thermally. The remainder is $2160-1680=480\,\text{J}$ , transferred to the kinetic energy store.	5
Total Question 1			5
02.1	The average braking force is $750\,\text{N}$ . The second stopping distance is $10\,\text{m}$ . Work against friction raises the temperature of the brakes, tyres, road or surroundings.	Convert $18\,\text{kJ}$ to $18000\,\text{J}$ , then $F=W/s=18000\div24=750\,\text{N}$ . Convert $7.5\,\text{kJ}$ to $7500\,\text{J}$ , so $s=W/F=7500\div750=10\,\text{m}$ . The frictional work transfers energy thermally.	5
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03.1	The work in the first stage is $600\,\text{J}$ . The work in the second stage is $720\,\text{J}$ . The second force is $90\,\text{N}$ . The work done is the energy transferred by the forces.	Calculate $W_1=120\times5.0=600\,\text{J}$ and convert the total to $1320\,\text{J}$ . Therefore $W_2=1320-600=720\,\text{J}$ and $F_2=W_2/s_2=720\div8.0=90\,\text{N}$ .	5
Total Question 3			5
04.1	The resultant force along the motion is $85-55=30\,\text{N}$ . The distance travelled is $900/30=30\,\text{m}$ . The energy transferred thermally is $55\times30=1650\,\text{J}=1.65\,\text{kJ}$ .	The resultant force is $85-55=30\,\text{N}$ . The work done by this resultant force equals the increase in kinetic energy, so $s=900/30=30\,\text{m}$ . Friction transfers $E=Fs=55\times30=1650\,\text{J}=1.65\,\text{kJ}$ thermally.	5
Total Question 4			5
05.1	The force is $1500\,\text{N}$ , so the first work value is $1500\times18=27000\,\text{J}=27\,\text{kJ}$ ; the logger is correct. The first work value is also $27000\,\text{N m}$ . The second cable movement is $4800/1500=3.2\,\text{m}$ . One joule is the work done when one newton moves an object one metre along the force's line of action, so $1\,\text{J}=1\,\text{N m}$ .	Convert $1.5\,\text{kN}$ to $1500\,\text{N}$ and apply $W=Fs$ . This gives $27000\,\text{J}$ in the first test. Since a joule is a newton-metre, the numerical value is unchanged in $\text{N m}$ . For the second test, convert $4.8\,\text{kJ}$ to $4800\,\text{J}$ and use $s=W/F=4800/1500=3.2\,\text{m}$ .	5
Total Question 5			5

4.5.3 · Forces and elasticity

Tier 1 · Easy

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01.1	The wire has undergone inelastic deformation. It does not return to its original length when the forces are removed.	Use what happens after unloading. Elastic deformation is reversed when the forces are removed; a permanent change in length therefore identifies inelastic deformation.	2
Total Question 1			2
02.1	The non-zero intercept is the spring's unloaded length, not evidence of inelastic deformation. The student should plot extension, found by subtracting $12.0\,\text{cm}$ from each loaded length, against force.	Direct proportionality applies to force and extension, not force and total spring length. A length graph therefore starts at the original length even for an elastic spring.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.5.3 Tier 2 · Standard
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01.1	$k=240\,\text{N/m}$ Extension at $12\,\text{N}$ is $0.050\,\text{m}$ .	Rearrange $F=ke$ to $k=F/e=18\div0.075=240\,\text{N/m}$ . Still within the linear region, $e=F/k=12\div240=0.050\,\text{m}$ .	3
Total Question 1			3
02.1	The spring constant is $125\,\text{N/m}$ . Repeat each length reading at the same force and calculate a mean extension after checking for anomalies.	Convert extension to metres. For example, $1.60\,\text{cm}=0.0160\,\text{m}$ , so the force–extension gradient is $k=2.0\div0.0160=125\,\text{N/m}$ . Repeated readings reduce random uncertainty and expose anomalous results.	4
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03.1	The extensions are $1.2$ , $2.4$ , $3.6$ and $5.4\,\text{cm}$ . Extension is proportional to force up to $6.0\,\text{N}$ . At $8.0\,\text{N}$ , proportionality would predict $4.8\,\text{cm}$ rather than the actual $5.4\,\text{cm}$ , so the limit of proportionality has been exceeded above $6.0\,\text{N}$ .	Subtract the natural length from every loaded length. The first three extension-to-force ratios are all $0.60\,\text{cm/N}$ , while the final ratio is $5.4/8.0=0.675\,\text{cm/N}$ . Therefore $6.0\,\text{N}$ is the greatest listed force in the proportional region.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.5.3 Tier 3 · Hard
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01.1	Applied force: $43.2\,\text{N}$ . Elastic potential energy: $3.89\,\text{J}$ . Inelastic deformation would be shown if the spring did not return to its original length.	Use $F=ke=240\times0.18=43.2\,\text{N}$ . Then $E_e=\frac12ke^2=0.5\times240\times0.18^2=3.888\,\text{J}$ , which rounds to $3.89\,\text{J}$ . A permanent extension after removing the force indicates inelastic deformation.	5
Total Question 1			5
02.1	Record the unloaded length beside a fixed millimetre ruler. Add known masses one at a time and let the spring stop moving before reading its new length. Repeat readings, check for anomalies and calculate mean extensions from the unloaded length. Convert each mass to force using $W=mg$ and plot force against extension. Remove the masses and check that the spring returns to its original length.	RP₆ changes load in controlled steps, measures settled lengths against the same ruler and converts the raw values into force and extension. Repeats improve reliability, and the unloading check distinguishes elastic from inelastic deformation.	5
Total Question 2			5
03.1	The energy at $0.200\,\text{m}$ extension is $5.00\,\text{J}$ . The stored energy increases by about $178\%$ .	For the same spring, $E_e$ is proportional to $e^2$ . Thus $E_2=1.80(0.200/0.120)^2=5.00\,\text{J}$ . The percentage increase is $(5.00-1.80)/1.80\times100=177.8\%$ , about $178\%$ .	5
Total Question 3			5
04.1	The spring constant is $450\,\text{N/m}$ . The compressing force at $0.080\,\text{m}$ is $36\,\text{N}$ . The energy stored at $0.050\,\text{m}$ is exactly $0.5625\,\text{J}$ ; accept $0.56\,\text{J}$ to two significant figures or $0.562\,\text{J}$ / $0.563\,\text{J}$ .	Rearrange $E_e=\frac12ke^2$ to $k=2E_e/e^2=2.88/0.080^2=450\,\text{N/m}$ . Then $F=ke=450\times0.080=36\,\text{N}$ . At the smaller compression, $E_e=0.5\times450\times0.050^2=0.5625\,\text{J}$ .	5
Total Question 4			5
05.1	The extensions are $1.5$ , $3.0$ , $4.5$ and $6.8\,\text{cm}$ . The first three points give $k=F/e=3.0/0.015=200\,\text{N/m}$ . At $9.0\,\text{N}$ , $E_e=0.5\times200\times0.045^2=0.2025\,\text{J}$ , or $0.20\,\text{J}$ to two significant figures. If proportionality continued, $12.0\,\text{N}$ would produce a $6.0\,\text{cm}$ extension, but the measured extension is $6.8\,\text{cm}$ . After unloading, the spring retains a $0.4\,\text{cm}$ extension; this permanent set is evidence of inelastic deformation.	Subtract the $15.0\,\text{cm}$ natural length to obtain each extension. In the linear region, $k=F/e=3.0/0.015=200\,\text{N/m}$ . Use $E_e=0.5\times200\times0.045^2=0.2025\,\text{J}$ . The $12.0\,\text{N}$ point exceeds the proportional prediction, and the $0.4\,\text{cm}$ extension remaining after unloading confirms a permanent change.	6
Total Question 5			6

4.5.4 · Moments, levers and gears (physics only)

Tier 1 · Easy

Mark scheme for 4.5.4 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	The force acts at a greater perpendicular distance from the pivot. Since $M=Fd$ , the greater distance produces a greater moment for the same force.	The pivot is the centre of the nut. Increasing the perpendicular distance $d$ while keeping $F$ constant increases the product $Fd$ , so the turning effect is larger.	2
Total Question 1			2
02.1	The student used the pivot-to-hand distance instead of the perpendicular distance to the force's line of action. The correct moment is $30\times0.42=12.6\,\text{N m}$ .	Use $M=Fd$ with $d=0.42\,\text{m}$ , because that is the perpendicular distance from the pivot to the force's line of action. This gives $M=30\times0.42=12.6\,\text{N m}$ .	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.5.4 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$1.2\,\text{m}$	Balance clockwise and anticlockwise moments: $360\times1.5=450d$ . Therefore $d=540\div450=1.2\,\text{m}$ .	3
Total Question 1			3
02.1	Driving-gear moment: $1.35\,\text{N m}$ . Other-gear moment: $4.05\,\text{N m}$ . The gears rotate in opposite directions, and the larger gear rotates more slowly.	Apply $M=Fd$ at each radius: $45\times0.030=1.35\,\text{N m}$ and $45\times0.090=4.05\,\text{N m}$ . Meshing teeth reverse the rotation direction; the larger gear turns through fewer rotations for the same number of teeth passing the contact point.	4
Total Question 2			4
03.1	Design A produces $27\,\text{N m}$ . Design B produces $24\,\text{N m}$ . Only design A reaches the $25\,\text{N m}$ threshold and opens the latch.	Use $M=Fd$ for each design: $M_A=60\times0.45=27\,\text{N m}$ and $M_B=80\times0.30=24\,\text{N m}$ . Compare each result with the required minimum rather than choosing the larger force alone.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.5.4 Tier 3 · Hard
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01.1	$513\,\text{N}$ upward	The beam's weight acts at its centre, $2.0\,\text{m}$ from the left end. Taking moments about the left support gives $R\times3.2=(220\times2.0)+(300\times4.0)=1640\,\text{N m}$ . Hence $R=1640\div3.2=512.5\,\text{N}$ , or $513\,\text{N}$ upward.	4
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02.1	The beam's weight is $120\,\text{N}$ . The $75\,\text{N}$ load must be $1.6\,\text{m}$ to the right of the pivot.	In A, the left load gives $180\times1.0=180\,\text{N m}$ anticlockwise. Balance gives $180=(60\times2.0)+(W\times0.50)$ , so $W=120\,\text{N}$ . In B, $180=(120\times0.50)+75d$ , so $d=(180-60)\div75=1.6\,\text{m}$ to the right.	5
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03.1	Lever A produces an output force of $600\,\text{N}$ . The final output force from lever B is $1800\,\text{N}$ .	For lever A, balance moments: $150\times0.24=F_A\times0.060$ , so $F_A=600\,\text{N}$ . This is lever B's input, so $600\times0.15=F_B\times0.050$ . Hence $F_B=1800\,\text{N}$ .	4
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04.1	The input moment is $44\,\text{N m}$ . The lever's output force is $800\,\text{N}$ . The moment on the door is $96\,\text{N m}$ . The door opens because $96\,\text{N m}$ is $6\,\text{N m}$ above the required moment.	Balance moments on the freely turning light lever: $110\times0.40=F\times0.055$ , so $F=800\,\text{N}$ . The force is normal to the door lever arm, giving $M=800\times0.12=96\,\text{N m}$ . Compare this with $90\,\text{N m}$ .	5
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05.1	The counterweight moment is $4320\,\text{N m}$ anticlockwise. The cable moment is $180\,\text{N m}$ clockwise. The load must act $3.45\,\text{m}$ from the pivot for balance. After the move, the clockwise moment exceeds the anticlockwise moment by $240\,\text{N m}$ . The resultant moment is $240\,\text{N m}$ clockwise.	For balance, $2400\times1.8=(300\times0.60)+(1200d)$ , so $d=(4320-180)/1200=3.45\,\text{m}$ . At $3.65\,\text{m}$ , the clockwise total is $180+(1200\times3.65)=4560\,\text{N m}$ , which is $240\,\text{N m}$ greater than the anticlockwise moment.	5
Total Question 5			5

4.5.5.1.1 · Pressure in a fluid 1 (physics only)

Tier 1 · Easy

Mark scheme for 4.5.5.1.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	The pressure on surface $B$ is half the pressure on surface $A$ . For the same force, $p=F/A$ , so doubling the area halves the pressure.	Hold the normal force constant in $p=F/A$ . Replacing $A$ with $2A$ gives $p=F/(2A)$ , which is half the original pressure.	2
Total Question 1			2
02.1	The student pairs the total force with only half of the total contact area, so the pressure is twice the correct value. Either divide the whole weight by the combined area of both snowshoes, or divide half the weight by the area of one snowshoe.	In $p=F/A$ , the force and area must refer to the same contact. Even loading means each shoe supports half the weight, while both shoes together support the whole weight.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.5.5.1.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$1.8\times10^4\,\text{Pa}$	Convert the area: $30\,\text{cm}^2=30\times10^{-4}=0.0030\,\text{m}^2$ . Then $p=54\div0.0030=18000\,\text{Pa}$ .	3
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02.1	The sensor area is $0.0040\,\text{m}^2$ . The $160\,\text{N}$ , $36\,\text{kPa}$ pair is anomalous. Its corrected pressure is $40\,\text{kPa}$ .	Convert pressure to pascals. The first pair gives $A=80\div20000=0.0040\,\text{m}^2$ and the second confirms $120\div30000=0.0040\,\text{m}^2$ . Hence the third pressure should be $160\div0.0040=40000\,\text{Pa}=40\,\text{kPa}$ .	4
Total Question 2			4
03.1	The numerical value $24000$ is correct. The correct unit is $\text{N/m}^2$ , equivalent to pascals, so the pressure is $24000\,\text{Pa}$ . Area is measured in square metres, so dividing force by area leaves metres squared in the denominator.	Apply $p=F/A$ : $1200/0.050=24000$ . The denominator is the area unit $\text{m}^2$ , not a length unit $\text{m}$ , giving $24000\,\text{N/m}^2=24000\,\text{Pa}$ .	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.5.5.1.1 Tier 3 · Hard
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01.1	Panel X: $2.10\times10^4\,\text{Pa}$ . Panel Y: $8.00\times10^3\,\text{Pa}$ . Panel Y experiences the lower pressure.	Panel X has area $0.30\times0.20=0.060\,\text{m}^2$ , so $p_X=1260\div0.060=21000\,\text{Pa}$ . Panel Y has area $0.45\times0.35=0.1575\,\text{m}^2$ , so $p_Y=1260\div0.1575=8000\,\text{Pa}$ . The same force spread over Y's larger area gives the lower pressure.	5
Total Question 1			5
02.1	The platform's weight is $518\,\text{N}$ . The pressure on one pad is $48\,\text{kPa}$ .	One pad has area $108\,\text{cm}^2=108\times10^{-4}=0.0108\,\text{m}^2$ , so two pads have area $0.0216\,\text{m}^2$ . Convert $24\,\text{kPa}$ to $24000\,\text{Pa}$ and use $F=pA=24000\times0.0216=518.4\,\text{N}$ . On one pad, $p=518.4\div0.0108=48000\,\text{Pa}=48\,\text{kPa}$ .	5
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03.1	At least $6.2$ pads would be required mathematically, so the minimum whole number is $7$ pads. With $7$ pads, the pressure is about $106\,\text{kPa}$ , which is below $120\,\text{kPa}$ . Six pads would produce $124\,\text{kPa}$ , so six is not sufficient.	Convert $18.6\,\text{kN}$ to $18600\,\text{N}$ and $120\,\text{kPa}$ to $120000\,\text{Pa}$ . One pad can support $pA=120000\times0.025=3000\,\text{N}$ at the limit, so $18600/3000=6.2$ and round up to $7$ . Check: $p=18600/(7\times0.025)=106286\,\text{Pa}$ , about $106\,\text{kPa}$ .	5
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04.1	The pressure is $18000\,\text{Pa}$ and the face areas are $0.0040\,\text{m}^2$ and $0.0065\,\text{m}^2$ . The force on the horizontal face is $72\,\text{N}$ , acting vertically and normal to that face. The force on the vertical face is $117\,\text{N}$ , acting horizontally and normal to that face. Fluid pressure produces force at right angles to a surface, not along the surface.	Convert $18\,\text{kPa}$ to $18000\,\text{Pa}$ and the areas to $0.0040\,\text{m}^2$ and $0.0065\,\text{m}^2$ . Apply $F=pA$ to obtain $72\,\text{N}$ and $117\,\text{N}$ . Rotate the force direction with each surface so that it remains normal.	5
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05.1	The outward pressure force is $45\,\text{N}$ . The resultant force is $9\,\text{N}$ outwards. An additional $9\,\text{N}$ inward force is needed for equilibrium.	Convert $250\,\text{kPa}$ to $250000\,\text{Pa}$ and $1.8\,\text{cm}^2$ to $1.8\times10^{-4}\,\text{m}^2$ . Then $F=pA=250000\times1.8\times10^{-4}=45\,\text{N}$ . Combine the opposing forces and add an equal opposite force to remove the $9\,\text{N}$ outward resultant.	4
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4.5.5.1.2 · Pressure in a fluid 2 (physics only) (HT only)

Tier 1 · Easy

Mark scheme for 4.5.5.1.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Liquid pressure is greater at the object's bottom than at its top because the bottom is deeper. The greater upward force on the bottom produces a resultant upward force, or upthrust.	Compare the two horizontal faces. Greater depth gives greater pressure, so the upward pressure force on the bottom exceeds the downward pressure force on the top; their difference is upthrust.	2
Total Question 1			2
02.1	The claim is incorrect: both sensors read the same pressure. In the same liquid, pressure depends on vertical depth, density and gravitational field strength, not horizontal position.	Both sensors have the same liquid height above them and share the same $\rho$ and $g$ . Their horizontal locations therefore do not change the liquid pressure.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.5.5.1.2 Tier 2 · Standard
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01.1	$1.60\times10^4\,\text{Pa}$	The depth difference is $2.10-0.40=1.70\,\text{m}$ . Hence $\Delta p=\Delta h\rho g=1.70\times960\times9.8=15993.6\,\text{Pa}$ , or $1.60\times10^4\,\text{Pa}$ .	3
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02.1	Liquid A has density $800\,\text{kg/m}^3$ . Liquid B has density $1200\,\text{kg/m}^3$ and is denser.	Rearrange $\Delta p=\Delta h\rho g$ to $\rho=\Delta p/(\Delta h g)$ . For A, $\rho=7056\div(0.90\times9.8)=800\,\text{kg/m}^3$ . For B, $\rho=8820\div(0.75\times9.8)=1200\,\text{kg/m}^3$ .	4
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03.1	With the $0.30\,\text{m}$ side vertical, the pressure difference is $2100\,\text{Pa}$ and the upthrust is $42\,\text{N}$ . With the $0.10\,\text{m}$ side vertical, the pressure difference is $700\,\text{Pa}$ and the upthrust is again $42\,\text{N}$ .	First orientation: $\Delta p=0.30\times700\times10=2100\,\text{Pa}$ and horizontal area is $0.20\times0.10=0.020\,\text{m}^2$ , so $U=2100\times0.020=42\,\text{N}$ . Second orientation: $\Delta p=0.10\times700\times10=700\,\text{Pa}$ and area is $0.30\times0.20=0.060\,\text{m}^2$ , so $U=700\times0.060=42\,\text{N}$ .	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.5.5.1.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Pressure difference: $2940\,\text{Pa}$ . Upthrust: $35.3\,\text{N}$ . It initially accelerates upwards because upthrust exceeds weight.	The depth difference is the cuboid height, so $\Delta p=h\rho g=0.30\times1000\times9.8=2940\,\text{Pa}$ . The force difference is $F=\Delta pA=2940\times0.012=35.28\,\text{N}$ . Since $35.28\,\text{N}>30\,\text{N}$ , the resultant is upward and the cuboid initially accelerates upward.	5
Total Question 1			5
02.1	The initial base depth is $0.200\,\text{m}$ . The new base depth is $0.260\,\text{m}$ . The base moves $0.060\,\text{m}$ deeper.	For floating equilibrium, upthrust equals weight. Initially $p=F/A=1470\div0.75=1960\,\text{Pa}$ , so $h=1960\div(1000\times9.8)=0.200\,\text{m}$ . With the payload, $F=1470+441=1911\,\text{N}$ and $p=1911\div0.75=2548\,\text{Pa}$ , giving $h=2548\div9800=0.260\,\text{m}$ . The increase is $0.060\,\text{m}$ .	5
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03.1	The claim is false for the stated constant-density liquid and fully submerged cuboid. Moving deeper increases the pressure on both the top and bottom faces by the same amount. Their vertical separation, the liquid density and $g$ are unchanged, so the bottom-to-top pressure difference is unchanged. The face area is unchanged, so the upthrust produced by that pressure difference is unchanged.	Use $\Delta p=h\rho g$ across the cuboid itself. Its height $h$ , the liquid density and $g$ do not change with its overall depth, so $\Delta p$ and therefore $U=\Delta pA$ stay constant even though both face pressures are larger.	4
Total Question 3			4
04.1	The pressure difference is $2082.5\,\text{Pa}$ , or $2.08\times10^3\,\text{Pa}$ . The upthrust is $41.65\,\text{N}$ ; accept $41.6$ – $41.7\,\text{N}$ . The weight is $58.8\,\text{N}$ . The string tension is $17.15\,\text{N}$ upwards; accept $17.1$ – $17.2\,\text{N}$ . When the string is cut, the initial resultant is $17.15\,\text{N}$ downwards; accept $17.1$ – $17.2\,\text{N}$ . The cuboid initially accelerates downwards.	Use $\Delta p=h\rho g=0.25\times850\times9.8=2082.5\,\text{Pa}$ and $U=\Delta pA=2082.5\times0.020=41.65\,\text{N}$ . The weight is $6.0\times9.8=58.8\,\text{N}$ . Equilibrium requires $T+U=W$ , so $T=58.8-41.65=17.15\,\text{N}$ .	6
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05.1	The maximum upthrust in liquid A is $274.4\,\text{N}$ . The cuboid sinks in liquid A because its $320\,\text{N}$ weight is greater than the maximum upthrust. The maximum upthrust in liquid B is $392\,\text{N}$ , so floating equilibrium is possible. In liquid B, the required pressure difference is $320/0.16=2000\,\text{Pa}$ . The equilibrium depth is $2000/(1000\times9.8)=0.204\,\text{m}$ , which is less than the cuboid's height.	For full submersion, calculate $U=h\rho gA$ . Liquid A gives $0.25\times700\times9.8\times0.16=274.4\,\text{N}$ , which is $45.6\,\text{N}$ or $14.25\%$ below the weight. Liquid B can provide up to $392\,\text{N}$ . For floating equilibrium in B, set $h\rho gA=320$ and solve for $h=0.204\,\text{m}$ .	6
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4.5.5.2 · Atmospheric pressure (physics only)

Tier 1 · Easy

Mark scheme for 4.5.5.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	There are fewer air molecules above a surface at higher altitude. The air is less dense, so fewer molecular collisions produce a smaller pressure.	At greater altitude the column of atmosphere above the surface is smaller and the air is less dense. Fewer air-particle collisions per unit area therefore produce a lower pressure.	2
Total Question 1			2
02.1	Atmospheric pressure outside the packet decreases with altitude. The greater pressure inside produces an outward resultant force, so the flexible packet expands until the pressures are closer to balance.	The sealed packet initially traps air at the lower-altitude pressure. Moving higher reduces the external pressure, so the pressure difference pushes the flexible sides outward.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.5.5.2 Tier 2 · Standard
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01.1	There are fewer air molecules above the mountain-top instrument. The air is less dense at the greater altitude. There are fewer molecular collisions with the instrument, so the pressure is lower.	Link greater altitude to a smaller amount and weight of air above the surface. This means lower air density and fewer collisions per unit area, giving lower pressure.	3
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02.1	The most likely reading is $76\,\text{kPa}$ because it is lower than $84\,\text{kPa}$ . At the greater altitude there are fewer air molecules above the sensor and the air is less dense. Fewer molecular collisions with the sensor per unit area produce a lower pressure.	The pair of readings shows atmospheric pressure falling as altitude rises. The $2.8\,\text{km}$ site is higher still, so choose the only value below $84\,\text{kPa}$ : $76\,\text{kPa}$ . A smaller, less-dense column of air above the sensor causes fewer collisions per unit area and therefore lower pressure.	3
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03.1	The atmospheric pressure on the liquid surface is greater than the pressure inside the straw. This pressure difference produces an upward resultant force on the liquid in the straw. The liquid rises until the pressure forces balance at the new height.	Compare the pressures rather than saying that suction pulls the liquid. The greater external atmospheric pressure pushes liquid into and up the lower-pressure straw.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.5.5.2 Tier 3 · Hard
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01.1	$312\,\text{N}$ outwards	The pressure difference is $101-75=26\,\text{kPa}=26000\,\text{Pa}$ . Using $F=\Delta pA$ gives $F=26000\times0.012=312\,\text{N}$ . The larger pressure is inside, so the resultant force is outwards.	4
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02.1	The piston area is $0.010\,\text{m}^2$ . The gas pressure is $68\,\text{kPa}$ .	Subtract the two force equations: $320-160=(100000-84000)A$ , so $A=160\div16000=0.010\,\text{m}^2$ . At the first site, the pressure difference is $320\div0.010=32000\,\text{Pa}=32\,\text{kPa}$ . Because the force is inward, the outside pressure is larger, so the gas pressure is $100-32=68\,\text{kPa}$ .	5
Total Question 2			5
03.1	The independent variable is altitude or height in the building, measured at several known levels. The dependent variable is the atmospheric pressure recorded by the same pressure sensor. Take readings during a rapid ascent and descent, repeat at each level and average paired readings so changes in weather over time do not masquerade as an altitude effect. Atmospheric pressure is expected to decrease as altitude increases.	Vary only the measurement height and use one calibrated sensor. Collect repeated readings at each level on both an ascent and a prompt descent to control weather-related pressure drift. A plot of mean pressure against altitude should show the predicted decreasing trend.	4
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04.1	Sensor B reads $4\,\text{kPa}$ too high at every shared site. The corrected final reading is $78-4=74\,\text{kPa}$ . At greater altitude there are fewer air molecules above the sensor and the air is less dense. Fewer molecular collisions per unit area produce a lower atmospheric pressure.	Compare the paired readings: each B value exceeds A by the same $4\,\text{kPa}$ , identifying a constant offset. Subtract that offset from B's final reading, then link the decreasing corrected values to fewer air-particle collisions at increasing altitude.	4
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05.1	The pressure difference is $38\,\text{kPa}$ and the holding force is $68.4\,\text{N}$ . The maximum supported mass is $68.4/9.8=7.0\,\text{kg}$ . A small leak makes the trapped pressure rise towards atmospheric pressure, so the pressure difference and holding force fall; the hook can then fall.	Convert $18\,\text{cm}^2$ to $0.0018\,\text{m}^2$ . The holding force is $F=(99000-61000)\times0.0018=68.4\,\text{N}$ , so $m=F/g=68.4/9.8=6.98\,\text{kg}$ , or $7.0\,\text{kg}$ . A leak admits outside air, raising the trapped pressure towards $99\,\text{kPa}$ and reducing both $\Delta p$ and $F=\Delta pA$ .	5
Total Question 5			5

4.5.6.1.1 · Distance and displacement

Tier 1 · Easy

Mark scheme for 4.5.6.1.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Distance travelled $=400\,\text{m}$ Displacement $=0\,\text{m}$	Distance is the complete path length, so it is one $400\,\text{m}$ lap. Displacement compares the finish with the start; these positions coincide, so the displacement is zero.	2
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02.1	The distance is $50+70=120\,\text{m}$ . The displacement is $20\,\text{m}$ west.	Distance adds both positive path lengths. Taking east as positive gives displacement $50-70=-20\,\text{m}$ ; the negative sign means $20\,\text{m}$ west.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.5.6.1.1 Tier 2 · Standard
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01.1	Distance $14.0\,\text{m}$ . The displacement arrow measures about $5.0\,\text{cm}$ , giving $10.0\,\text{m}$ at about $53^\circ$ north of east; accept $9.6$ – $10.4\,\text{m}$ and $51$ – $55^\circ$ .	The path length is $6.0+8.0=14.0\,\text{m}$ . Draw a $3.0\,\text{cm}$ east arrow followed head-to-tail by a $4.0\,\text{cm}$ north arrow. Draw the displacement from the start to the finish, then measure its length and angle; about $5.0\,\text{cm}$ represents $10.0\,\text{m}$ at about $53^\circ$ north of east.	4
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02.1	Route A: distance $900\,\text{m}$ ; displacement $600\,\text{m}$ north. Route B: distance $700\,\text{m}$ ; displacement $700\,\text{m}$ east. Route B could be straight without reversing because its distance equals its displacement magnitude.	Path length is distance, while the stated straight-line change from the start is displacement. Equality of distance and displacement magnitude is possible for straight motion in one direction, which fits route B.	4
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03.1	The distance is $1.7\,\text{km}$ . The displacement is $1.2\,\text{km}$ west. Distance measures the whole path, whereas displacement is the straight-line change in position with direction.	Take the odometer change as path length. Take the position-sensor result as the vector change from start to finish. A route that is not a single unreversed straight line can make distance exceed displacement magnitude.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.5.6.1.1 Tier 3 · Hard
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01.1	Distance $970\,\text{m}$ . The net eastward displacement before the northward flight is $350\,\text{m}$ . The final displacement arrow measures about $7.4\,\text{cm}$ , giving about $370\,\text{m}$ at about $19^\circ$ north of east; accept $355$ – $385\,\text{m}$ and $17$ – $21^\circ$ .	Distance is $600+250+120=970\,\text{m}$ and the opposing east-west stages give $600-250=350\,\text{m}$ east. Draw a $7.0\,\text{cm}$ east arrow followed by a $2.4\,\text{cm}$ north arrow. Measure the start-to-finish arrow as about $7.4\,\text{cm}$ and its angle as about $19^\circ$ north of east; the scale gives about $370\,\text{m}$ .	5
Total Question 1			5
02.1	The distance travelled is $2500\,\text{m}$ , or $2.50\,\text{km}$ . The displacement is $1000\,\text{m}$ north, or $1.00\,\text{km}$ north.	The map-route length is $3.0+4.0+3.0=10.0\,\text{cm}$ , giving $10.0\times250=2500\,\text{m}$ . The east and west sections cancel, leaving a $4.0\,\text{cm}$ north displacement, or $4.0\times250=1000\,\text{m}$ north.	4
Total Question 2			4
03.1	The displacement is $1.20\,\text{km}$ east because it comes from the final position relative to the start. The $1.84\,\text{km}$ sum uses straight lines between samples rather than the curved paths actually followed. Each straight chord is shorter than the curved path between the same recorded positions, so their sum underestimates the true distance travelled. More frequent position readings would improve the distance estimate.	Use only the start and finish positions for displacement. For distance, recognise that joining separated samples with straight chords underestimates the curved route; reducing the time between samples makes the sum follow the path more closely.	4
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04.1	The distance is $800\,\text{m}$ . Draw two $4.0\,\text{cm}$ arrows head-to-tail on bearings $030^\circ$ and $120^\circ$ . The start-to-finish arrow measures about $5.7\,\text{cm}$ , giving a displacement of about $570\,\text{m}$ ; accept $550$ – $590\,\text{m}$ . The displacement bearing is about $075^\circ$ ; accept $073^\circ$ – $077^\circ$ .	Add the two route lengths for distance. For displacement, draw both $4.0\,\text{cm}$ route vectors at the stated bearings, head-to-tail, and measure the straight arrow from the original start to the final point. Convert its measured length with the given scale and read its bearing with a protractor.	5
Total Question 4			5
05.1	The first stage is $380\,\text{m}$ east. The second stage is $300\,\text{m}$ west. The total distance is $680\,\text{m}$ . The displacement is $80\,\text{m}$ east. The ratio of distance to displacement magnitude is $680:80=8.5:1$ .	Subtract successive positions: $260-(-120)=380\,\text{m}$ east and $-40-260=-300\,\text{m}$ , or $300\,\text{m}$ west. Add stage lengths for distance. Compare only final and initial positions for displacement: $-40-(-120)=80\,\text{m}$ east.	5
Total Question 5			5

4.5.6.1.2 · Speed

Tier 1 · Easy

Mark scheme for 4.5.6.1.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Total distance $=300\,\text{m}$ Total time $=60\,\text{s}$ Average speed $=5.0\,\text{m/s}$	Include the rest in the complete time: $20+10+30=60\,\text{s}$ . The total distance is $120+180=300\,\text{m}$ , so average speed is $300/60=5.0\,\text{m/s}$ .	3
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02.1	The calculated speed is $90\div6.0=15\,\text{m/s}$ . This is about ten times the typical walking speed, so the distance or time reading is probably wrong.	Use speed $=$ distance $\div$ time. Comparing $15\,\text{m/s}$ with $1.5\,\text{m/s}$ shows a factor-of-ten mismatch, which is evidence of a measurement or recording error.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.5.6.1.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$5.43\,\text{m/s}$	Total distance is $300+200=500\,\text{m}$ . Total time includes the rest: $48+12+32=92\,\text{s}$ . Average speed is $500\div92=5.43\,\text{m/s}$ .	3
Total Question 1			3
02.1	$1.48\,\text{s}$ is anomalous. The mean valid time is $0.83\,\text{s}$ . The car's speed is $1.4\,\text{m/s}$ ; accept $1.45\,\text{m/s}$ .	Exclude $1.48\,\text{s}$ because it is far from the other repeats. The mean is $(0.84+0.82+0.83)\div3=0.83\,\text{s}$ . Then $v=s/t=1.20\div0.83=1.44578\ldots\,\text{m/s}$ , which is $1.4\,\text{m/s}$ to two significant figures; accept $1.45\,\text{m/s}$ .	4
Total Question 2			4
03.1	The vehicle's speed is $16\,\text{m/s}$ . The limit is $15\,\text{m/s}$ . The vehicle exceeds the limit by $1\,\text{m/s}$ .	Use $v=s/t=12.0/0.75=16\,\text{m/s}$ . Convert the limit using $54\,\text{km}=54000\,\text{m}$ and $1\,\text{h}=3600\,\text{s}$ , so $54000/3600=15\,\text{m/s}$ . The measured speed is therefore higher.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.5.6.1.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	$12.4\,\text{m/s}$	Convert $1.8\,\text{km}$ to $1800\,\text{m}$ and the outward time to $140\,\text{s}$ . The return time is $1800\div15=120\,\text{s}$ . Total distance is $3600\,\text{m}$ and total time is $140+30+120=290\,\text{s}$ , so average speed is $3600\div290=12.4\,\text{m/s}$ .	5
Total Question 1			5
02.1	The unknown distance is $585\,\text{m}$ . The speed during the final stage is $7.8\,\text{m/s}$ .	The total distance is $7.5\times150=1125\,\text{m}$ . The first stage covers $12\times45=540\,\text{m}$ and the wait adds no distance, so the final stage covers $1125-540=585\,\text{m}$ . Its speed is $585\div75=7.8\,\text{m/s}$ .	5
Total Question 2			5
03.1	Cumulative distance cannot decrease, so the correct values at $20\,\text{s}$ and $30\,\text{s}$ are $100\,\text{m}$ and $160\,\text{m}$ . The interval speeds are $8.0\,\text{m/s}$ , $2.0\,\text{m/s}$ and $6.0\,\text{m/s}$ . The first interval is the fastest.	Distance travelled adds path length even during backtracking: $80+20=100\,\text{m}$ and $100+60=160\,\text{m}$ . The gradients over the three $10\,\text{s}$ intervals are $80/10=8.0$ , $20/10=2.0$ and $60/10=6.0\,\text{m/s}$ .	5
Total Question 3			5
04.1	Measure the full marked route with a tape measure or metre ruler. Record the time from crossing the start line to crossing the finish line, preferably using video or light gates to reduce reaction-time effects. Calculate average speed as total distance divided by total time. Repeat the complete journey, identify any anomalous time and calculate a mean from valid repeats. Keep the route, car settings and start procedure the same for each repeat.	The required average uses the complete route length and complete journey time even though speed changes during the run. Use fixed start and finish markers, reduce timing error with an automatic or video method, and repeat under controlled conditions before calculating a mean.	5
Total Question 4			5
05.1	The first two travel times are $60\,\text{s}$ and $90\,\text{s}$ . The total distance is $3000\,\text{m}$ and the total time is $260\,\text{s}$ . The average speed is $11.5\,\text{m/s}$ to three significant figures. The student's method is incorrect because the stages last for different times and the waiting time must be included. The simple mean of the three moving speeds would not represent total distance divided by total time.	Calculate $900/15=60\,\text{s}$ and $900/10=90\,\text{s}$ , then include the $30\,\text{s}$ wait and $80\,\text{s}$ final stage. The total is $260\,\text{s}$ over $900+900+1200=3000\,\text{m}$ , so $v_{\text{average}}=3000/260=11.538\ldots\,\text{m/s}$ .	5
Total Question 5			5

4.5.6.1.3 · Velocity

Tier 1 · Easy

Mark scheme for 4.5.6.1.3 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Velocity includes direction as well as magnitude. The direction changes from east to north, so the velocity changes even though the speed is constant.	Treat velocity as a vector. Its magnitude stays the same because the speed is unchanged, but its direction changes, so the velocity is different after the turn.	2
Total Question 1			2
02.1	Speed is scalar, so ‘speed vector’ is incorrect. The speed is $9.0\,\text{m/s}$ and the velocity is $9.0\,\text{m/s}$ south.	The arrow supplies a direction, so the complete vector is velocity. Removing the direction leaves the scalar speed magnitude.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.5.6.1.3 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$3.0\,\text{m/s}$ east	The displacement is $50-20=30\,\text{m}$ east. Average velocity is displacement divided by time: $30\div10=3.0\,\text{m/s}$ east.	3
Total Question 1			3
02.1	From $0$ to $5.0\,\text{s}$ : $4.0\,\text{m/s}$ east. From $5.0$ to $11.0\,\text{s}$ : $2.0\,\text{m/s}$ west.	Take east as positive. The first displacement is $32-12=20\,\text{m}$ in $5.0\,\text{s}$ , giving $4.0\,\text{m/s}$ east. The second is $20-32=-12\,\text{m}$ in $6.0\,\text{s}$ , giving $-2.0\,\text{m/s}$ , or $2.0\,\text{m/s}$ west.	4
Total Question 2			4
03.1	The trolley moves west. Its displacement is $-72\,\text{m}$ , or $72\,\text{m}$ west.	The negative sign means motion opposite to the defined positive east direction, so the trolley moves west. Use $s=vt=(-6.0)\times12=-72\,\text{m}$ .	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.5.6.1.3 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	The displacement arrow measures about $8.5\,\text{cm}$ , giving about $170\,\text{m}$ at about $62^\circ$ north of east; accept $163$ – $177\,\text{m}$ and $60$ – $64^\circ$ . Average velocity about $5.7\,\text{m/s}$ at about $62^\circ$ north of east; accept $5.4$ – $5.9\,\text{m/s}$ . Average speed $7.7\,\text{m/s}$ , which is greater because it uses distance rather than displacement.	Draw a $4.0\,\text{cm}$ east arrow followed head-to-tail by a $7.5\,\text{cm}$ north arrow. The measured start-to-finish arrow is about $8.5\,\text{cm}$ at about $62^\circ$ north of east, representing about $170\,\text{m}$ . Average velocity is $170\div30=5.7\,\text{m/s}$ in that direction. Distance is $80+150=230\,\text{m}$ , so average speed is $230\div30=7.7\,\text{m/s}$ .	5
Total Question 1			5
02.1	The average velocity is $2.33\,\text{m/s}$ on a bearing of $060^\circ$ . The average speed is $3.89\,\text{m/s}$ .	The total time is $30\,\text{min}=1800\,\text{s}$ . Displacement is $5.6-1.4=4.2\,\text{km}=4200\,\text{m}$ on the original bearing, so average velocity is $4200\div1800=2.333\ldots\,\text{m/s}$ , or $2.33\,\text{m/s}$ , on $060^\circ$ . Distance is $5.6+1.4=7.0\,\text{km}=7000\,\text{m}$ , so average speed is $7000\div1800=3.888\ldots\,\text{m/s}$ , or $3.89\,\text{m/s}$ .	5
Total Question 2			5
03.1	The reported average speed is correct: $7.5\,\text{m/s}$ . The average velocity should be $2.5\,\text{m/s}$ west. Average speed uses distance, whereas average velocity uses displacement and includes direction.	Check speed with total distance: $450/60=7.5\,\text{m/s}$ . Check velocity with displacement: $150/60=2.5\,\text{m/s}$ west. The values differ because the path length is greater than the displacement magnitude.	4
Total Question 3			4
04.1	The speed has the same value at both points. The velocities are different because their directions are north and south. Velocity changes continuously around the circular track because direction changes continuously. A changing velocity means the car accelerates, so a resultant force is required. Stating that this force acts towards the centre is a correct extra detail but is not required.	Separate the scalar magnitude from the vector direction. Constant speed fixes only the magnitude; the opposite directions make the two velocities different. Continuous direction change is acceleration, which requires a resultant force.	4
Total Question 4			4
05.1	All three average speeds are $600/80=7.5\,\text{m/s}$ . A's average velocity is $7.5\,\text{m/s}$ east. B's average velocity is $0\,\text{m/s}$ because its displacement is zero. C's average velocity is $5.0\,\text{m/s}$ north. Average speed uses distance, whereas average velocity uses displacement and direction.	Use the common distance and time for speed. For velocity, divide each stated displacement by $80\,\text{s}$ and retain its direction. The different displacements produce different average velocities even though all three path lengths and times match.	6
Total Question 5			6

4.5.6.1.4 · The distance–time relationship

Tier 1 · Easy

Mark scheme for 4.5.6.1.4 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	$3.25\,\text{m s}^{-1}$	The graph gradient is speed. Use $v=\Delta s/\Delta t=156/48=3.25\,\text{m s}^{-1}$ .	2
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02.1	The section from $6$ to $10\,\text{s}$ is impossible because total distance travelled cannot decrease.	The plotted total falls from $24\,\text{m}$ to $19\,\text{m}$ . Total distance travelled is cumulative, so it can stay constant while an object rests but cannot fall as time increases.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.5.6.1.4 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	The object is stationary for $8\,\text{s}$ . $5.0\,\text{m s}^{-1}$	A horizontal line has zero gradient, so the distance does not change and the object is stationary. For the first section, $v=120/24=5.0\,\text{m s}^{-1}$ .	3
Total Question 1			3
02.1	First interval: $5.0\,\text{m s}^{-1}$ Second interval: $8.0\,\text{m s}^{-1}$ The claim is false; the walker was faster in the second interval.	The first gradient is $90/18=5.0\,\text{m s}^{-1}$ . The second gradient uses the distance and time changes for that interval: $64/8=8.0\,\text{m s}^{-1}$ . Speed depends on distance per unit time, so the smaller distance does not make the second interval slower.	4
Total Question 2			4
03.1	$80\,\text{m}$	The time change is $13.0-5.0=8.0\,\text{s}$ . A gradient of $6.5\,\text{m s}^{-1}$ gives a distance change of $6.5\times8.0=52\,\text{m}$ . The final coordinate is $28+52=80\,\text{m}$ .	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.5.6.1.4 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	$4.5\,\text{m s}^{-1}$ from $0$ to $12\,\text{s}$ $6.0\,\text{m s}^{-1}$ from $12$ to $28\,\text{s}$ $0\,\text{m s}^{-1}$ from $28$ to $40\,\text{s}$ Average speed $=3.75\,\text{m s}^{-1}$	Use the gradient of each section. The speeds are $54/12=4.5\,\text{m s}^{-1}$ , $(150-54)/(28-12)=96/16=6.0\,\text{m s}^{-1}$ and $(150-150)/(40-28)=0\,\text{m s}^{-1}$ . The total distance is $150\,\text{m}$ , so the average speed is $150/40=3.75\,\text{m s}^{-1}$ .	5
Total Question 1			5
02.1	$8.5\,\text{m s}^{-1}$	The instantaneous speed is the tangent gradient. Using two well-separated points on the tangent, $v=(86-18)/(11.0-3.0)=68/8.0=8.5\,\text{m s}^{-1}$ .	3
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03.1	$15\,\text{s}$	The moving times are $180/6.0=30\,\text{s}$ and $210/7.0=30\,\text{s}$ . The total distance is $390\,\text{m}$ , so the total journey time is $390/5.2=75\,\text{s}$ . The stationary time is therefore $75-30-30=15\,\text{s}$ .	4
Total Question 3			4
04.1	B catches A $65\,\text{s}$ after A starts. The catch occurs $208\,\text{m}$ from the checkpoint. B's line has the greater gradient, so B is faster.	At the catch, both walkers have travelled the same distance. If $t$ is the time after A starts, $3.2t=5.2(t-25)$ . Therefore $3.2t=5.2t-130$ , so $t=65\,\text{s}$ . The distance is $3.2\times65=208\,\text{m}$ . On the graph, B's gradient is $5.2\,\text{m s}^{-1}$ and A's is $3.2\,\text{m s}^{-1}$ , so B's line is steeper.	5
Total Question 4			5
05.1	Time for the first section $=25\,\text{s}$ and for the second section $=10\,\text{s}$ . Actual average speed $=5.71\,\text{m s}^{-1}$ . The two speeds act for unequal times, so their arithmetic mean is not total distance divided by total time.	The section times are $100/4.0=25\,\text{s}$ and $100/10=10\,\text{s}$ . The total distance is $200\,\text{m}$ and the total time is $35\,\text{s}$ , so the average speed is $200/35=5.714\ldots\,\text{m s}^{-1}$ . Averaging the two speeds directly would be valid only for equal time intervals, not these equal-distance intervals.	4
Total Question 5			4

4.5.6.1.5 · Acceleration

Tier 1 · Easy

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Question	Answers	Extra information	Mark
01.1	$1.5\,\text{m s}^{-2}$	Use $a=(v-u)/t=(15.0-3.0)/8.0=12.0/8.0=1.5\,\text{m s}^{-2}$ .	2
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02.1	First interval: $1.2\,\text{m s}^{-2}$ Second interval: $3.0\,\text{m s}^{-2}$ The second acceleration is greater.	Acceleration is change in velocity divided by time. The values are $6.0/5.0=1.2\,\text{m s}^{-2}$ and $9.0/3.0=3.0\,\text{m s}^{-2}$ , so the second interval has the greater acceleration.	3
Total Question 2			3

Tier 2 · Standard

Mark scheme for 4.5.6.1.5 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$+3.0\,\text{m s}^{-2}$ $0\,\text{m s}^{-2}$ $-4.0\,\text{m s}^{-2}$	Acceleration is the graph gradient. The first section gives $(12-0)/4.0=+3.0\,\text{m s}^{-2}$ . The horizontal section has zero gradient. The final section gives $(0-12)/3.0=-4.0\,\text{m s}^{-2}$ .	4
Total Question 1			4
02.1	The object moves at a constant velocity of $4.0\,\text{m s}^{-1}$ in the negative direction. Its acceleration is $0\,\text{m s}^{-2}$ .	A horizontal velocity–time line has zero gradient, so acceleration is zero. Its vertical coordinate is $-4.0\,\text{m s}^{-1}$ rather than zero, so the object is moving at constant velocity in the negative direction, not stationary.	3
Total Question 2			3
03.1	Time to zero velocity $=5.6\,\text{s}$ Velocity after $8.0\,\text{s}$ $=-6.0\,\text{m s}^{-1}$	Use $a=(v-u)/t$ . At zero velocity, $-2.5=(0-14)/t$ , so $t=14/2.5=5.6\,\text{s}$ . After $8.0\,\text{s}$ , $-2.5=(v-14)/8.0$ , so $v=-6.0\,\text{m s}^{-1}$ ; the object is then moving in the negative direction.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.5.6.1.5 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	$1.0\,\text{m s}^{-2}$ $12\,\text{s}$	Use $v^2-u^2=2as$ : $20.0^2-8.0^2=2a(168)$ . Thus $336=336a$ , so $a=1.0\,\text{m s}^{-2}$ . Then $a=(v-u)/t$ , so $t=(20.0-8.0)/1.0=12\,\text{s}$ .	5
Total Question 1			5
02.1	$198\,\text{m}$	Distance is the area under the velocity–time graph. The first trapezium has area $0.5\times(2.0+14)\times5.0=40\,\text{m}$ . The rectangle has area $(14-5.0)\times14=126\,\text{m}$ . The final trapezium has area $0.5\times(14+2.0)\times(18-14)=32\,\text{m}$ . Total distance $=40+126+32=198\,\text{m}$ .	5
Total Question 2			5
03.1	Acceleration $=3.0\,\text{m s}^{-2}$ Distance $=36\,\text{m}$ The reported $45\,\text{m}$ is not consistent with the calculated distance.	The acceleration is $(15.0-3.0)/4.0=3.0\,\text{m s}^{-2}$ . Using $v^2-u^2=2as$ , $s=(15.0^2-3.0^2)/(2\times3.0)=(225-9)/6=36\,\text{m}$ . The sensor's reported $45\,\text{m}$ differs from the calculated $36\,\text{m}$ by $9\,\text{m}$ , so the report is not consistent with the recorded velocities.	5
Total Question 3			5
04.1	First-stage acceleration $=3.0\,\text{m s}^{-2}$ Braking acceleration $=-3.0\,\text{m s}^{-2}$ Braking time $=6.0\,\text{s}$ Total time $=10.0\,\text{s}$	For the first stage, $a=(18-6.0)/4.0=3.0\,\text{m s}^{-2}$ . For braking, use $v^2-u^2=2as$ : $0^2-18^2=2a(54)$ , giving $a=-3.0\,\text{m s}^{-2}$ . Then $a=(v-u)/t$ gives $-3.0=(0-18)/t$ , so the braking time is $6.0\,\text{s}$ . The total time is $4.0+6.0=10.0\,\text{s}$ .	6
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05.1	Final velocity $=+8.0\,\text{m s}^{-1}$ Displacement $=+52\,\text{m}$ The velocity remains positive but decreases in magnitude, so the object is slowing down while still moving forward.	From $a=(v-u)/t$ , $-2.5=(v-18)/4.0$ , so $v=+8.0\,\text{m s}^{-1}$ . Then use $v^2-u^2=2as$ : $s=(8.0^2-18^2)/(2\times-2.5)=(64-324)/(-5.0)=+52\,\text{m}$ . As a cross-check, the mean velocity is $(18+8.0)/2=13\,\text{m s}^{-1}$ , giving $s=13\times4.0=52\,\text{m}$ . The positive final velocity and displacement show continued forward motion, while the negative acceleration reduces the velocity.	5
Total Question 5			5

4.5.6.2.1 · Newton's First Law

Tier 1 · Easy

Mark scheme for 4.5.6.2.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	$680\,\text{N}$ opposite to the motion	Constant velocity means zero resultant force. The resistive force must therefore be equal in magnitude and opposite in direction to the $680\,\text{N}$ propeller force.	2
Total Question 1			2
02.1	The puck continues north at constant speed because zero resultant force means its velocity remains constant.	Newton's First Law says a moving object keeps the same speed and direction when the resultant force is zero. The puck therefore continues with constant northward velocity rather than slowing down.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.5.6.2.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	The resistive forces total $920\,\text{N}$ backwards. The resultant force is $0\,\text{N}$ , so the van continues at constant velocity.	Steady speed in a straight line means constant velocity. By Newton's First Law the resultant force is zero, so the backward resistive forces balance the $920\,\text{N}$ engine force.	3
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02.1	The probe continues west at $3.0\,\text{m s}^{-1}$ . The forces are balanced, so the resultant force is zero and the velocity does not change.	The equal opposite forces give a resultant of $0\,\text{N}$ . By Newton's First Law, zero resultant force preserves the probe's existing velocity, so it continues west at the same speed.	3
Total Question 2			3
03.1	Resultant force $=85\,\text{N}$ west The crate begins to accelerate west.	The two original forces balance, so the stationary crate remains at rest. Removing the eastward force leaves the $85\,\text{N}$ westward force unbalanced. Its velocity therefore changes from zero by accelerating west.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.5.6.2.1 Tier 3 · Hard
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01.1	Initially the lift moves upward at constant velocity. The new resultant force is $1200\,\text{N}$ downward. The upward-moving lift decelerates.	Initially, the downward forces total $5800+400=6200\,\text{N}$ , balancing the motor force, so the lift's velocity is constant. After the change, the downward resultant is $5800+400-5000=1200\,\text{N}$ . Because this resultant opposes the upward velocity, the lift slows down.	5
Total Question 1			5
02.1	The loaded trolley has greater mass and therefore greater inertia, so it has a greater tendency to remain at rest or to continue moving at constant velocity. A greater resultant force is needed to produce the same change in velocity.	Inertia is the tendency to resist a change in the state of motion. Loading the trolley increases its mass and inertia, so changing its velocity from zero or back to zero is more difficult.	3
Total Question 2			3
03.1	A: the eastward resultant force produces an eastward acceleration. The probe therefore speeds up towards the east during A. B: zero resultant force leaves the probe moving east at constant velocity. C: the westward resultant force produces a westward acceleration. Because the probe is still moving east, this acceleration makes it slow down during C.	The eastward resultant in A gives an eastward acceleration, which increases the eastward speed. Zero resultant force in B preserves the eastward velocity. The westward resultant in C gives a westward acceleration; because the probe is still moving east, this opposing acceleration reduces its speed.	5
Total Question 3			5
04.1	Weight $=441\,\text{N}$ Initial normal contact force $=441\,\text{N}$ Normal contact force during the pull $=241\,\text{N}$ Resultant force remains $0\,\text{N}$	The weight is $W=mg=45\times9.8=441\,\text{N}$ downward. Initially the crate is at rest, so the floor provides an equal $441\,\text{N}$ normal force upward. During the pull, vertical balance gives $N+200-441=0$ , so $N=241\,\text{N}$ . The upward forces still total $441\,\text{N}$ , equal to the weight, so the resultant remains zero.	5
Total Question 4			5
05.1	Before release, the resultant force is zero and the cart has constant velocity. After release, the leftward resistance opposes the rightward motion, so the cart slows. Once stationary, the rolling resistance is zero, so there is no leftward resultant and the cart remains at rest. The student's claim that the cart reverses is incorrect.	The balanced $240\,\text{N}$ forces initially give zero resultant force. Removing the tow force leaves resistance opposing the existing motion, so the velocity decreases to zero. At rest the stated rolling resistance is also zero. With no horizontal resultant, Newton's First Law says that the stationary cart remains stationary rather than beginning to move left.	4
Total Question 5			4

4.5.6.2.2 · Newton's Second Law

Tier 1 · Easy

Mark scheme for 4.5.6.2.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	$18\,\text{N}$	Use $F=ma=12\times1.5=18\,\text{N}$ .	2
Total Question 1			2
02.1	$1.6\,\text{kg}$	Rearrange $F=ma$ to $m=F/a$ . Then $m=2.4/1.5=1.6\,\text{kg}$ .	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.5.6.2.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Move slotted masses from the cart to the hanging mass so the pulling force changes but the total moving mass stays constant. Use light gates or a motion sensor and a data logger to measure acceleration. Repeat each force setting and calculate a mean, identifying anomalies. Acceleration should be directly proportional to resultant force for constant mass.	Keep the combined mass of cart plus hanging masses unchanged. Transfer one slotted mass at a time from the cart to the hanger, increasing the driving force without changing total mass. Release the cart from the same position and obtain acceleration with light gates or a motion sensor. Repeat each setting, check anomalies and calculate a mean. A graph of acceleration against resultant force should be a straight line through the origin, showing $a\propto F$ when mass is constant.	5
Total Question 1			5
02.1	Transferring masses off the hanger changes the hanging weight, so it changes the pulling force. Keep the same hanging mass and add masses to the trolley to change the total moving mass. Measure acceleration with light gates or a motion sensor, repeat each mass setting and calculate a mean after checking anomalies.	The hanger's weight supplies the pulling force, so moving masses between hanger and trolley varies both force and mass. Leave the hanger unchanged and add known masses to the trolley. Release from the same position, measure acceleration electronically, repeat each setting and use a mean so the comparison is reliable.	5
Total Question 2			5
03.1	New acceleration $=2.0\,\text{m s}^{-2}$ The new acceleration is two thirds of the original.	The new force is $24\,\text{N}$ and the new mass is $12\,\text{kg}$ , so $a=F/m=24/12=2.0\,\text{m s}^{-2}$ . This is $2.0/3.0=2/3$ of the original acceleration.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.5.6.2.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	$2.0\,\text{m s}^{-2}$ $2500\,\text{N}$ resultant $3200\,\text{N}$ engine force	The acceleration is $a=(26.0-8.0)/9.0=2.0\,\text{m s}^{-2}$ . Hence the resultant force is $F=ma=1250\times2.0=2500\,\text{N}$ . The engine must supply the resultant plus the resistance: $2500+700=3200\,\text{N}$ .	5
Total Question 1			5
02.1	Both observations give an inertial mass of $15\,\text{kg}$ . Required resultant force $=42\,\text{N}$ .	Inertial mass is $m=F/a$ . The two observations give $18/1.2=15\,\text{kg}$ and $30/2.0=15\,\text{kg}$ , so they are consistent. For $a=2.8\,\text{m s}^{-2}$ , $F=ma=15\times2.8=42\,\text{N}$ .	5
Total Question 2			5
03.1	Each pair gives a resultant force of $2.4\,\text{N}$ . Acceleration decreases in inverse proportion to mass for a constant resultant force. A resistive force of about $0.4\,\text{N}$ accounts for the difference from the pulling weight.	Using $F=ma$ gives $1.0\times2.4=2.4\,\text{N}$ , $1.5\times1.6=2.4\,\text{N}$ and $2.0\times1.2=2.4\,\text{N}$ . The constant products show that increasing mass reduces acceleration for a constant resultant force. The pulling weight exceeds the resultant by $2.8-2.4=0.4\,\text{N}$ because friction or another resistive force opposes the motion.	6
Total Question 3			6
04.1	Mass $=2.0\,\text{kg}$ Resistive force $=3.0\,\text{N}$ Acceleration at $12.6\,\text{N}$ $=4.8\,\text{m s}^{-2}$	The extra $9.0-5.4=3.6\,\text{N}$ of applied force produces an extra $3.0-1.2=1.8\,\text{m s}^{-2}$ of acceleration, so $m=3.6/1.8=2.0\,\text{kg}$ . Using the first observation, the resistance is $5.4-(2.0\times1.2)=3.0\,\text{N}$ . At $12.6\,\text{N}$ applied, the resultant is $12.6-3.0=9.6\,\text{N}$ , so $a=9.6/2.0=4.8\,\text{m s}^{-2}$ .	6
Total Question 4			6
05.1	Each of the first three pairs gives a mass of $1.5\,\text{kg}$ . Predicted fourth acceleration $=4.4\,\text{m s}^{-2}$ The reported $2.8\,\text{m s}^{-2}$ is anomalous because it is $1.6\,\text{m s}^{-2}$ below the prediction.	With negligible friction, the applied force is the resultant force. Using $m=F/a$ , the first three masses are $2.1/1.4=1.5\,\text{kg}$ , $3.6/2.4=1.5\,\text{kg}$ and $5.1/3.4=1.5\,\text{kg}$ . Therefore the predicted acceleration is $a=F/m=6.6/1.5=4.4\,\text{m s}^{-2}$ . The reported value is $4.4-2.8=1.6\,\text{m s}^{-2}$ lower, so it is anomalous and the measurement should be repeated.	6
Total Question 5			6

4.5.6.2.3 · Newton's Third Law

Tier 1 · Easy

Mark scheme for 4.5.6.2.3 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	The nail exerts an equal-magnitude upward force on the hammer.	Reverse both the objects and the direction: hammer on nail downward pairs with nail on hammer upward. The two forces have equal magnitude.	2
Total Question 1			2
02.1	Both forces act on the parachutist, whereas a Newton's Third Law pair acts on two different interacting objects.	Weight and air resistance balance because they are equal and opposite forces on one object. A third-law pair instead contains one force on each of the two objects in a single interaction.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.5.6.2.3 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	The book pulls the Earth upward with an equal gravitational force. The table force acts on the book, whereas the third-law partner acts on the Earth.	The interaction is gravitational and the two objects are Earth and book. Earth on book pairs with book on Earth. The normal contact force is a different interaction; it balances the weight on the book but is not its third-law partner.	3
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02.1	The iron block pulls the magnet with $36\,\text{N}$ in the opposite direction. The forces act on different objects, so they cannot both appear on one object's force diagram or cancel there.	Reverse the interacting objects: magnet on block pairs with block on magnet. The forces are equal in magnitude and opposite in direction, but each belongs to a different object's force diagram.	4
Total Question 2			4
03.1	Cart B exerts $42\,\text{N}$ west on cart A. The forces act on different carts, and different cart masses can give different accelerations.	Newton's Third Law gives an equal, opposite and simultaneous force: B on A is $42\,\text{N}$ west. The forces do not act on one common mass. Since $a=F/m$ , equal force magnitudes produce different acceleration magnitudes if the cart masses differ.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.5.6.2.3 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	The water pushes the propeller forward with $4.2\,\text{kN}$ . $2.0\,\text{m s}^{-2}$ forward	By Newton's Third Law, the water exerts $4.2\,\text{kN}$ forward on the propeller. The resultant force on the boat is $4.2-1.8=2.4\,\text{kN}=2400\,\text{N}$ . Therefore $a=F/m=2400/1200=2.0\,\text{m s}^{-2}$ forward.	5
Total Question 1			5
02.1	The lamp pulls the Earth upward gravitationally with $48\,\text{N}$ . The lamp pulls the cable downward with $48\,\text{N}$ . The original weight and tension balance on the lamp; each third-law pair acts on two different objects.	Earth on lamp pairs with lamp on Earth in the gravitational interaction. Cable on lamp pairs with lamp on cable in the contact interaction. The two stated forces are equal and opposite on the same lamp, so they balance; they are not a single interaction pair.	5
Total Question 2			5
03.1	Student acceleration $=3.0\,\text{m s}^{-2}$ Platform acceleration $=0.75\,\text{m s}^{-2}$ The accelerations are in opposite directions.	The platform experiences $180\,\text{N}$ , so its acceleration is $180/240=0.75\,\text{m s}^{-2}$ . By Newton's Third Law, the student experiences an equal $180\,\text{N}$ force in the opposite direction, giving $a=180/60=3.0\,\text{m s}^{-2}$ . The paired forces point oppositely, so the accelerations do too.	5
Total Question 3			5
04.1	A pulls B with $320\,\text{N}$ right and B pulls A with $320\,\text{N}$ left. Resultant on A $=400\,\text{N}$ right; acceleration $=0.80\,\text{m s}^{-2}$ right. Resultant on B $=240\,\text{N}$ right; acceleration $=0.80\,\text{m s}^{-2}$ right.	The rope interaction gives equal and opposite forces on different vessels. On A, the resultant is $920-200-320=400\,\text{N}$ right, so $a=400/500=0.80\,\text{m s}^{-2}$ . On B, the resultant is $320-80=240\,\text{N}$ right, so $a=240/300=0.80\,\text{m s}^{-2}$ . The paired rope forces do not cancel on either individual vessel because each acts on a different object.	6
Total Question 4			6
05.1	The magnetic force on B is $0.90\,\text{N}$ west. Resistance on A $=0.30\,\text{N}$ west. Resistance on B $=0.42\,\text{N}$ east. The accelerations differ because the carts have different masses and different resultant forces, even though the magnetic interaction forces are equal.	Newton's Third Law gives a $0.90\,\text{N}$ westward magnetic force on B. For A, the eastward resultant is $ma=0.40\times1.5=0.60\,\text{N}$ , so resistance is $0.90-0.60=0.30\,\text{N}$ west. For B, the westward resultant magnitude is $0.60\times0.80=0.48\,\text{N}$ , so its eastward resistance is $0.90-0.48=0.42\,\text{N}$ . Equal interaction forces do not require equal accelerations because the total forces and masses of the carts differ.	6
Total Question 5			6

4.5.6.3.1 · Stopping distance

Tier 1 · Easy

Mark scheme for 4.5.6.3.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	$40\,\text{m}$	Add the two components: $s_{\text{stop}}=12+28=40\,\text{m}$ .	1
Total Question 1			1
02.1	Time in seconds cannot be added to distance in metres. The vehicle's speed is needed to calculate thinking distance from speed multiplied by reaction time; that distance is then added to braking distance.	Stopping distance is thinking distance plus braking distance, so both terms must be distances. Measure the vehicle speed, calculate $s_{\text{thinking}}=vt$ , then add the result in metres to $24\,\text{m}$ .	3
Total Question 2			3

Tier 2 · Standard

Mark scheme for 4.5.6.3.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Thinking distance $=9.6\,\text{m}$ Stopping distance $=28.6\,\text{m}$	During the reaction time, $s=vt=15\times0.64=9.6\,\text{m}$ . Therefore the stopping distance is $9.6+19=28.6\,\text{m}$ .	3
Total Question 1			3
02.1	Thinking distance $=12\,\text{m}$ Use $t=s/v$ $0.50\,\text{s}$	Thinking distance $=43-31=12\,\text{m}$ . During the reaction time $s=vt$ , so $t=s/v=12/24=0.50\,\text{s}$ .	3
Total Question 2			3
03.1	Thinking distance $=14\,\text{m}$ Stopping distance $=48\,\text{m}$ The car fails to stop in time by $3\,\text{m}$ .	Thinking distance is $s=vt=20\times0.70=14\,\text{m}$ . Stopping distance is $14+34=48\,\text{m}$ . Since $48-45=3\,\text{m}$ , the car fails to stop in time by $3\,\text{m}$ .	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.5.6.3.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	$47\,\text{m}$ at $20\,\text{m s}^{-1}$ $68.75\,\text{m}$ at $25\,\text{m s}^{-1}$ Increase $=21.75\,\text{m}$	At $20\,\text{m s}^{-1}$ , the thinking distance is $20\times0.75=15\,\text{m}$ , so the stopping distance is $15+32=47\,\text{m}$ . At $25\,\text{m s}^{-1}$ , the thinking distance is $18.75\,\text{m}$ , so the stopping distance is $18.75+50=68.75\,\text{m}$ . The increase is $68.75-47=21.75\,\text{m}$ .	5
Total Question 1			5
02.1	Speed $=20\,\text{m s}^{-1}$ Initial stopping distance $=42\,\text{m}$ Distracted stopping distance $=48.4\,\text{m}$ Increase $=6.4\,\text{m}$	Convert using $72\div3.6=20\,\text{m s}^{-1}$ . The initial thinking distance is $20\times0.60=12\,\text{m}$ , giving $12+30=42\,\text{m}$ . With distraction it is $20\times0.92=18.4\,\text{m}$ , giving $18.4+30=48.4\,\text{m}$ . The increase is $48.4-42=6.4\,\text{m}$ .	6
Total Question 2			6
03.1	New braking distance $=39.1\,\text{m}$ Thinking distance $=17\,\text{m}$ Stopping distance $=56.1\,\text{m}$ The car cannot stop before the obstacle.	The speed factor is $25/20=1.25$ , so the braking-distance factor is $1.25^2=1.5625$ . The new braking distance is $25\times1.5625=39.0625\,\text{m}$ . Thinking distance is $25\times0.68=17.0\,\text{m}$ , so stopping distance is $56.0625\,\text{m}$ , about $56.1\,\text{m}$ . This exceeds $55\,\text{m}$ by about $1.1\,\text{m}$ .	6
Total Question 3			6
04.1	Total stopping distance at $13\,\text{m s}^{-1}$ $=9+14=23\,\text{m}$ . At $22\,\text{m s}^{-1}$ $=15+38=53\,\text{m}$ ; at $31\,\text{m s}^{-1}$ $=21+75=96\,\text{m}$ . Between the lowest and highest speeds, braking distance grows by a factor of $75/14\approx5.4$ while thinking distance grows by only $21/9\approx2.3$ . For a similar reaction time, thinking distance rises roughly in proportion to speed, whereas the kinetic energy that the brakes must transfer rises with speed squared, so braking distance rises much more rapidly if the braking force is similar.	Adding each pair gives total stopping distances $9+14=23\,\text{m}$ , $15+38=53\,\text{m}$ and $21+75=96\,\text{m}$ . Between $13$ and $31\,\text{m s}^{-1}$ , braking distance grows by $75/14\approx5.4$ times while thinking distance grows by $21/9\approx2.3$ times. During the driver's reaction time, distance is $s=vt$ , so a similar reaction time makes thinking distance approximately proportional to speed. The car's kinetic energy is proportional to $v^2$ . With a similar braking force, transferring this greater energy requires a much greater braking distance.	5
Total Question 4			5
05.1	Thinking distance $=13.2\,\text{m}$ Dry-road stopping distance $=49.2\,\text{m}$ Wet-road stopping distance $=67.2\,\text{m}$ A $65\,\text{m}$ clear distance is insufficient by $2.2\,\text{m}$ .	Thinking distance is $24\times0.55=13.2\,\text{m}$ . The dry stopping distance is $13.2+36=49.2\,\text{m}$ . A $50\%$ increase makes the wet braking distance $36\times1.50=54\,\text{m}$ , so the wet stopping distance is $13.2+54=67.2\,\text{m}$ . This is $67.2-65=2.2\,\text{m}$ more than the clear distance.	5
Total Question 5			5

4.5.6.3.2 · Reaction time

Tier 1 · Easy

Mark scheme for 4.5.6.3.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	$0.2\,\text{s}$ to $0.9\,\text{s}$ One of tiredness, alcohol, some drugs or distraction.	Recall the AQA typical range, then name one accepted factor that delays the driver's response.	2
Total Question 1			2
02.1	Releasing without warning prevents the student from anticipating the drop. Several readings allow anomalies to be identified and a mean to be calculated.	The release should be unpredictable so the measurement represents a reaction rather than anticipation. Repeats expose variation and improve reliability when a suitable mean is calculated.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.5.6.3.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$10.56\,\text{m}$	The vehicle continues at its original speed during the reaction time, so $s=vt=22\times0.48=10.56\,\text{m}$ .	2
Total Question 1			2
02.1	Doing every no-conversation trial first confounds the condition with practice or fatigue; alternate or randomise the order. One reading is unreliable; repeat each condition several times, identify anomalies and compare means.	A fixed order can make later results differ because the participant has practised or tired, not because of conversation. Randomising or alternating the order reduces this effect. Repeating both conditions and comparing suitable means reduces the influence of random variation.	4
Total Question 2			4
03.1	Mean distance $=15\,\text{cm}$ Estimated reaction time $=0.17\,\text{s}$ This is faster than the typical $0.2$ to $0.9\,\text{s}$ range. Possible reason: anticipating the release, a systematic error in the method, or a genuinely faster-than-typical reaction.	The mean catch distance is $(14+16+15)/3=15\,\text{cm}$ . The lookup table maps this to $0.17\,\text{s}$ , which is faster than the stated typical range. Anticipation or a systematic error is a possible explanation, but the comparison alone does not prove either.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.5.6.3.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	$0.89\,\text{s}$ is anomalous. Mean without distraction $=0.30\,\text{s}$ after excluding the anomaly. Mean with distraction $=0.455\,\text{s}$ . The distraction increased reaction time by about $0.155\,\text{s}$ .	The $0.89\,\text{s}$ value is far from the other no-distraction repeats, so exclude it with justification. The remaining mean is $(0.31+0.29+0.30)/3=0.30\,\text{s}$ . The distracted mean is $(0.44+0.47+0.45+0.46)/4=0.455\,\text{s}$ . The increase is $0.455-0.300=0.155\,\text{s}$ , supporting the conclusion that the distraction slowed the response.	5
Total Question 1			5
02.1	Speed $=25\,\text{m s}^{-1}$ Paired mean result: $0.25\,\text{s}$ without and $0.37\,\text{s}$ with conversation Thinking distances: $6.25\,\text{m}$ and $9.25\,\text{m}$ Increase $=3.00\,\text{m}$ For example, only one driver was tested, so the result may not represent other drivers.	Convert $90\,\text{km h}^{-1}$ using $90\div3.6=25\,\text{m s}^{-1}$ . The means are $(0.24+0.26+0.25)/3=0.25\,\text{s}$ and $(0.39+0.35+0.37)/3=0.37\,\text{s}$ . Thus the thinking distances are $25\times0.25=6.25\,\text{m}$ and $25\times0.37=9.25\,\text{m}$ , an increase of $3.00\,\text{m}$ . Three repeats for one driver do not establish the effect for a population.	6
Total Question 2			6
03.1	Increases: $0.11$ , $0.09$ , $0.14$ and $0.11\,\text{s}$ Mean increase $=0.1125\,\text{s}$ Pairing controls for differences in participants' usual reaction times. Only four participants were tested, so the evidence may not represent the wider population.	Subtract each no-distraction time from its paired distraction time: $0.11$ , $0.09$ , $0.14$ and $0.11\,\text{s}$ . Their mean is $(0.11+0.09+0.14+0.11)/4=0.1125\,\text{s}$ . Testing each person in both conditions reduces the effect of different baseline reaction times, but a sample of four is too small to generalise confidently.	6
Total Question 3			6
04.1	Use practice block number as the independent variable and measured reaction time as the dependent variable. Use the same computer, test settings, hand and starting position throughout. Keep the number of trials and the rest interval the same in every block. Record several reaction times in each block for each participant. Identify anomalous readings and calculate a mean for each block. Test several participants and compare how their block means change, noting that fatigue could also affect later blocks.	Give each participant equal-sized blocks of trials on the same computer and stimulus, using the same hand and position. Keep rest periods fixed so later blocks are not given systematically longer recovery. Record several readings per block, identify anomalies and calculate suitable means. Repeat with several participants, plot or compare mean reaction time against block number, and interpret a consistent change while recognising fatigue as a possible competing explanation.	6
Total Question 4			6
05.1	Each original pair gives a reaction time of $0.40\,\text{s}$ . Later reaction time $=0.60\,\text{s}$ Percentage increase $=50\%$	Use $t=s/v$ . The original values are $4.8/12=0.40\,\text{s}$ , $7.2/18=0.40\,\text{s}$ and $9.6/24=0.40\,\text{s}$ , so the measurements are consistent. The later value is $10.8/18=0.60\,\text{s}$ . The increase is $0.20\,\text{s}$ , giving $(0.20/0.40)\times100=50\%$ .	5
Total Question 5			5

4.5.6.3.3 · Factors affecting braking distance 1

Tier 1 · Easy

Mark scheme for 4.5.6.3.3 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	For example, worn tyres reduce friction with the road, reducing the braking force and increasing braking distance.	Use either tyres or brakes, then link the poor condition to reduced effective friction or braking force and therefore a longer braking distance.	2
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02.1	Tiredness increases reaction time and therefore thinking distance. Ice reduces grip and therefore increases braking distance.	Separate the driver response from the braking process. Tiredness delays the driver's reaction, while ice affects tyre–road friction after the brakes are applied.	3
Total Question 2			3

Tier 2 · Standard

Mark scheme for 4.5.6.3.3 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$36\,\text{m}$	The same initial kinetic energy must be removed, so $Fs$ is the same. Halving the braking force from $6200$ to $3100\,\text{N}$ doubles the distance: $s=18\times6200/3100=36\,\text{m}$ .	3
Total Question 1			3
02.1	Worn tyres increase braking distance: by $7\,\text{m}$ on the dry road and $14\,\text{m}$ on the wet road. A wet road increases braking distance: by $14\,\text{m}$ with new tyres and $21\,\text{m}$ with worn tyres.	Compare one factor at a time because the initial speed is controlled. At fixed road condition, worn tyres give longer distances. At fixed tyre condition, the wet road gives longer distances. Quoting the paired differences supports both conclusions.	4
Total Question 2			4
03.1	$20\,\text{m s}^{-1}$	Use $s_2/s_1=(v_2/v_1)^2$ . Thus $v_2=12\sqrt{50/18}=12\sqrt{25/9}=12\times5/3=20\,\text{m s}^{-1}$ .	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.5.6.3.3 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	The values of $s/v^2$ are all about $0.062\,\text{s}^2\,\text{m}^{-1}$ . Estimated braking distance $=38.8\,\text{m}$ (about $39\,\text{m}$ ).	Calculate $s/v^2$ : $6.2/10^2=0.062$ , $13.9/15^2=0.0618$ and $24.8/20^2=0.062$ . The near-constant ratio supports $s\propto v^2$ . At $25\,\text{m s}^{-1}$ , $s=0.062\times25^2=38.75\,\text{m}$ , or about $39\,\text{m}$ .	4
Total Question 1			4
02.1	$48\,\text{m}$ is anomalous. Mean before servicing $=32\,\text{m}$ after excluding the anomaly. Mean after servicing $=22\,\text{m}$ . Servicing reduced the mean braking distance by $10\,\text{m}$ under the controlled conditions.	The $48\,\text{m}$ result is far from the other pre-service repeats, so exclude it with justification. The suitable pre-service mean is $(31+33+32)/3=32\,\text{m}$ . The post-service mean is $(22+21+23+22)/4=22\,\text{m}$ . The $10\,\text{m}$ reduction supports the conclusion that servicing improved braking because car, driver, speed and road were controlled.	5
Total Question 2			5
03.1	Release the trolley from the same point on the same ramp and verify its speed at the start of each braking surface with a light gate. Control the trolley mass and brake setting (same pad, same applied pressure). Measure from the point where braking begins to the point where the trolley stops. Repeat on each surface, identify anomalies and compare mean braking distances.	Use a fixed ramp and release point, with a light gate at the boundary to check equal initial speeds. Change only the track surface; keep the trolley mass and brake setting constant by using the same pad with the same applied pressure. Measure the stopping position from the same braking-start line. Repeat every surface condition, check anomalous readings and compare suitable means.	6
Total Question 3			6
04.1	Dry-road distance at $24\,\text{m s}^{-1}$ $=45\,\text{m}$ Wet-road factor $=1.4$ Predicted wet-road distance at $20\,\text{m s}^{-1}$ $=43.75\,\text{m}$	The speed factor from $16$ to $24\,\text{m s}^{-1}$ is $1.5$ , so the dry-road distance would be $20\times1.5^2=45\,\text{m}$ . The observed wet-road value is larger by a factor $63/45=1.4$ . At $20\,\text{m s}^{-1}$ , the dry-road prediction is $20\times(20/16)^2=31.25\,\text{m}$ . Applying the wet-road factor gives $31.25\times1.4=43.75\,\text{m}$ .	6
Total Question 4			6
05.1	Braking distance with new tyres on a dry road $=168\,000/7000=24\,\text{m}$ . Braking distance with worn tyres on a dry road $=168\,000/5600=30\,\text{m}$ . Braking distance with new tyres on a wet road $=168\,000/4200=40\,\text{m}$ . Changing only to a wet road is the more dangerous single change: the distance rises by $16\,\text{m}$ , compared with $6\,\text{m}$ for changing only to worn tyres. Each change reduces tyre–road grip, so the frictional braking force and deceleration are smaller and the car travels farther before stopping.	The brakes must transfer the whole kinetic energy store, so $Fs=E_k$ gives $s=E_k/F$ . With new tyres on a dry road, $s=168\,000/7000=24\,\text{m}$ ; with worn tyres on a dry road, $s=168\,000/5600=30\,\text{m}$ ; with new tyres on a wet road, $s=168\,000/4200=40\,\text{m}$ . Relative to $24\,\text{m}$ , worn tyres alone add $6\,\text{m}$ while a wet road alone adds $16\,\text{m}$ , so the wet road is the more dangerous single change. Each adverse change reduces grip, reducing the frictional braking force and hence the deceleration, so the braking distance grows.	5
Total Question 5			5

4.5.6.3.4 · Factors affecting braking distance 2

Tier 1 · Easy

Mark scheme for 4.5.6.3.4 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	The vehicle's kinetic energy store decreases and energy is transferred to the thermal energy stores of the brakes and surroundings.	Name the initial kinetic energy store and the thermal stores that increase because friction does work.	2
Total Question 1			2
02.1	A greater braking force produces a greater deceleration, which can overheat the brakes or cause loss of control even though the vehicle can stop in a shorter distance.	Link force to deceleration: for the same vehicle mass, a larger resultant braking force gives a larger deceleration. Excessive deceleration creates the stated safety risks, so the claim that greater force is always safer is false.	3
Total Question 2			3

Tier 2 · Standard

Mark scheme for 4.5.6.3.4 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$1.782\times10^5\,\text{J}$ $27\,\text{m}$	The kinetic energy is $E_k=\dfrac{1}{2}mv^2=0.5\times1100\times18^2=178200\,\text{J}$ . Since work done $Fs$ equals this energy change, $s=178200/6600=27\,\text{m}$ .	4
Total Question 1			4
02.1	$240000\,\text{J}$ $8000\,\text{N}$	Convert $240\,\text{kJ}=240000\,\text{J}$ . The work done by the average braking force equals the kinetic energy decrease, so $F=W/s=240000/30=8000\,\text{N}$ .	4
Total Question 2			4
03.1	Each vehicle has $135000\,\text{J}$ of kinetic energy. Each braking distance is $30\,\text{m}$ , so the distances are equal.	For A, $E_k=0.5\times1200\times15^2=135000\,\text{J}$ . For B, $E_k=0.5\times675\times20^2=135000\,\text{J}$ . Since $Fs=E_k$ , both give $s=135000/4500=30\,\text{m}$ .	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.5.6.3.4 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Braking distance $=44\,\text{m}$ Deceleration $=5.5\,\text{m s}^{-2}$ A larger deceleration can overheat the brakes or cause loss of control.	The initial kinetic energy is $0.5\times1500\times22^2=363000\,\text{J}$ . Set this equal to $Fs$ : $s=363000/8250=44\,\text{m}$ . From $F=ma$ , the deceleration magnitude is $a=8250/1500=5.5\,\text{m s}^{-2}$ . A substantially larger force creates a larger deceleration, increasing the risk of brake overheating, skidding or loss of control.	6
Total Question 1			6
02.1	Speed $=25\,\text{m s}^{-1}$ Average deceleration $=7.8125\,\text{m s}^{-2}$ (about $7.8\,\text{m s}^{-2}$ ) Average resultant force $\approx1.1\times10^4\,\text{N}$ The braking force and deceleration vary during a real stop, so values found from total time are averages.	Convert the speed using $90\div3.6=25\,\text{m s}^{-1}$ . The average deceleration magnitude is $25/3.2=7.8125\,\text{m s}^{-2}$ . Then $F=ma=1400\times7.8125=10937.5\,\text{N}$ , which is about $1.1\times10^4\,\text{N}$ . The calculation assumes a constant rate, whereas the actual resultant braking force changes during the stop.	6
Total Question 2			6
03.1	Maximum permitted force $=7200\,\text{N}$ Minimum distance at that force $=33.3\,\text{m}$ Stopping within $28\,\text{m}$ would not meet the limit.	The greatest permitted force is $F=ma=1200\times6.0=7200\,\text{N}$ . The initial kinetic energy is $0.5\times1200\times20^2=240000\,\text{J}$ . From $Fs=E_k$ , $s=240000/7200=33.3\,\text{m}$ . A $28\,\text{m}$ stop would require $240000/28=8571\,\text{N}$ and a deceleration of $8571/1200=7.14\,\text{m s}^{-2}$ , above the limit.	6
Total Question 3			6
04.1	Initial kinetic energy $=162000\,\text{J}$ Energy transferred in stage one $=72000\,\text{J}$ Remaining energy $=90000\,\text{J}$ Stage-two average braking force $=6000\,\text{N}$	The initial kinetic energy is $0.5\times1000\times18^2=162000\,\text{J}$ . In stage one, $W=Fs=4000\times18=72000\,\text{J}$ . The remaining energy is $162000-72000=90000\,\text{J}$ . Removing this over $15\,\text{m}$ requires $F=W/s=90000/15=6000\,\text{N}$ .	6
Total Question 4			6
05.1	Kinetic energy decrease $=180000\,\text{J}$ Energy transferred to the discs $=108000\,\text{J}$ Disc temperature rise $=10\,{}^{\circ}\text{C}$ The remaining $72000\,\text{J}$ is transferred mainly to thermal energy stores of the surroundings, with some energy possibly transferred by sound.	The kinetic energy decrease is $0.5\times1200\times(20^2-10^2)=180000\,\text{J}$ . The discs receive $0.60\times180000=108000\,\text{J}$ . From $\Delta E=mc\Delta\theta$ , $\Delta\theta=108000/(24\times450)=10\,{}^{\circ}\text{C}$ . The other $0.40\times180000=72000\,\text{J}$ increases thermal energy stores elsewhere, such as the tyres, road and surrounding air, with some transfer by sound where applicable.	6
Total Question 5			6

4.5.7.1 · Momentum is a property of moving objects (HT only)

Tier 1 · Easy

Mark scheme for 4.5.7.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	$2.16\,\text{kg m s}^{-1}$	Use $p=mv=0.18\times12=2.16\,\text{kg m s}^{-1}$ .	2
Total Question 1			2
02.1	X has twice the momentum magnitude of Y, and their momenta point in opposite directions.	Momentum is $p=mv$ . Equal masses make momentum magnitude proportional to speed, so doubling speed doubles momentum magnitude. Momentum has the same direction as velocity, so the directions are opposite.	3
Total Question 2			3

Tier 2 · Standard

Mark scheme for 4.5.7.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$2.16\times10^4\,\text{kg m s}^{-1}$ west	Use $p=mv=1350\times16=21600\,\text{kg m s}^{-1}$ . Momentum has the same direction as velocity, so it is westward.	2
Total Question 1			2
02.1	Mass $=0.065\,\text{kg}$ $6.0\,\text{m s}^{-1}$ east	Convert $65\,\text{g}=65/1000=0.065\,\text{kg}$ . Rearrange $p=mv$ to $v=p/m$ , giving $v=0.390/0.065=6.0\,\text{m s}^{-1}$ east.	4
Total Question 2			4
03.1	Mass $=0.80\,\text{kg}$ Momentum $=-7.2\,\text{kg m s}^{-1}$	From $p=mv$ , the gradient of a momentum–velocity graph is mass. Hence $m=5.6/7.0=0.80\,\text{kg}$ . At $v=-9.0\,\text{m s}^{-1}$ , $p=0.80\times(-9.0)=-7.2\,\text{kg m s}^{-1}$ .	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.5.7.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Vehicle A: $+12960\,\text{kg m s}^{-1}$ Vehicle B: $-11880\,\text{kg m s}^{-1}$ Vehicle A has the greater magnitude by $1080\,\text{kg m s}^{-1}$ .	For A, $p=720\times18=+12960\,\text{kg m s}^{-1}$ . West is negative, so for B, $p=1080\times(-11)=-11880\,\text{kg m s}^{-1}$ . Compare magnitudes: $12960-11880=1080\,\text{kg m s}^{-1}$ , so A's momentum magnitude is greater.	5
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02.1	Cart A: $+2.94\,\text{kg m s}^{-1}$ Cart B: $-2.94\,\text{kg m s}^{-1}$ Mass of Cart B $=1.4\,\text{kg}$	Cart A has momentum $p_A=0.84\times3.5=+2.94\,\text{kg m s}^{-1}$ . The stated opposite momentum is $p_B=-2.94\,\text{kg m s}^{-1}$ . Using magnitudes in $m=p/v$ , $m_B=2.94/2.1=1.4\,\text{kg}$ .	5
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03.1	Minimum momentum $=1.15\,\text{kg m s}^{-1}$ Maximum momentum $=1.26\,\text{kg m s}^{-1}$ The reported $1.30\,\text{kg m s}^{-1}$ is outside the possible range.	The smallest product is $0.395\times2.9=1.1455\,\text{kg m s}^{-1}$ , or $1.15\,\text{kg m s}^{-1}$ . The largest is $0.405\times3.1=1.2555\,\text{kg m s}^{-1}$ , or $1.26\,\text{kg m s}^{-1}$ . Since $1.30$ exceeds the upper bound by $0.0445\,\text{kg m s}^{-1}$ , the report is not supported by the stated measurement ranges.	5
Total Question 3			5
04.1	Cart A momentum $=+3.6\,\text{kg m s}^{-1}$ Cart B momentum $=-1.8\,\text{kg m s}^{-1}$ Cart C velocity $=-3.0\,\text{m s}^{-1}$	Using $p=mv$ , $p_A=0.75\times4.8=+3.6\,\text{kg m s}^{-1}$ and $p_B=1.20\times(-1.5)=-1.8\,\text{kg m s}^{-1}$ . For a zero total, C must have momentum $-1.8\,\text{kg m s}^{-1}$ . Therefore $v_C=p_C/m_C=-1.8/0.60=-3.0\,\text{m s}^{-1}$ .	5
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05.1	Common kinetic energy $=180000\,\text{J}$ Vehicle B speed $=15\,\text{m s}^{-1}$ Momentum of A $=18000\,\text{kg m s}^{-1}$ Momentum of B $=24000\,\text{kg m s}^{-1}$ B has the greater momentum magnitude by $6000\,\text{kg m s}^{-1}$ .	For A, $E_k=0.5\times900\times20^2=180000\,\text{J}$ . Equal kinetic energy gives $180000=0.5\times1600\times v_B^2$ , so $v_B^2=225$ and $v_B=15\,\text{m s}^{-1}$ . The momenta are $900\times20=18000\,\text{kg m s}^{-1}$ and $1600\times15=24000\,\text{kg m s}^{-1}$ . B's greater mass more than offsets its lower speed.	6
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4.5.7.2 · Conservation of momentum (HT only)

Tier 1 · Easy

Mark scheme for 4.5.7.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	$2.0\,\text{m s}^{-1}$ in the original direction	Initial momentum is $2.0\times3.0=6.0\,\text{kg m s}^{-1}$ . The joined mass is $3.0\,\text{kg}$ . Conservation gives $6.0=3.0v$ , so $v=2.0\,\text{m s}^{-1}$ .	3
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02.1	$+5.5\,\text{kg m s}^{-1}$	Conservation gives $+4.0=-1.5+p_B$ . Therefore $p_B=4.0+1.5=+5.5\,\text{kg m s}^{-1}$ .	3
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Tier 2 · Standard

Mark scheme for 4.5.7.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$3.4\,\text{m s}^{-1}$ to the right	The initial total momentum is $(0.75\times6.4)+(1.25\times1.6)=4.8+2.0=6.8\,\text{kg m s}^{-1}$ to the right. The joined mass is $0.75+1.25=2.00\,\text{kg}$ . Therefore $6.8=2.00v$ , giving $v=3.4\,\text{m s}^{-1}$ to the right.	4
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02.1	Total before $=+0.50\,\text{kg m s}^{-1}$ Total after $=+0.490\,\text{kg m s}^{-1}$ The totals differ by $0.010\,\text{kg m s}^{-1}$ , or $2\%$ . Their closeness supports conservation, but measurement uncertainties are needed to judge the difference fully.	Before the collision, $p=(0.50\times1.6)+(0.30\times-1.0)=0.80-0.30=+0.50\,\text{kg m s}^{-1}$ . Afterwards, $p=0.175+0.315=+0.490\,\text{kg m s}^{-1}$ . The difference is $0.010\,\text{kg m s}^{-1}$ , which is $(0.010/0.50)\times100=2\%$ . This is close enough to support conservation, while stated measurement uncertainties would be required to decide whether the difference is significant.	6
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03.1	$5.0\,\text{m s}^{-1}$ to the right	Take right as positive. Initial momentum is $1.5\times2.4=3.6\,\text{kg m s}^{-1}$ . The other mass is $1.5-0.60=0.90\,\text{kg}$ . Conservation gives $3.6=(0.60\times-1.5)+0.90v=-0.90+0.90v$ , so $v=5.0\,\text{m s}^{-1}$ to the right.	4
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Tier 3 · Hard

Mark scheme for 4.5.7.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	$1.68\,\text{m s}^{-1}$ opposite to the projectile	Initial total momentum is zero. Take the projectile direction as positive: $0=(0.20\times32)+(3.8v)$ . Thus $3.8v=-6.4$ , so $v=-1.684\ldots\,\text{m s}^{-1}$ . The negative sign means a recoil speed of $1.68\,\text{m s}^{-1}$ opposite to the projectile.	4
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02.1	$+2.67\,\text{m s}^{-1}$	Initial total momentum is $(0.80\times5.0)+(1.20\times-1.0)=4.0-1.2=+2.8\,\text{kg m s}^{-1}$ . Conservation gives $2.8=(0.80\times-0.50)+1.20v_B=-0.40+1.20v_B$ . Hence $1.20v_B=3.20$ and $v_B=2.666\ldots\,\text{m s}^{-1}$ , or $+2.67\,\text{m s}^{-1}$ .	5
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03.1	Momentum before and after $=2.4\,\text{kg m s}^{-1}$ Kinetic energy before $=4.8\,\text{J}$ Kinetic energy after $=2.88\,\text{J}$ Momentum is conserved, but kinetic energy decreases by $1.92\,\text{J}$ ; this energy is transferred mainly to thermal energy stores of the carts and surroundings, with some possibly transferred by sound.	Before, total momentum is $0.60\times4.0=2.4\,\text{kg m s}^{-1}$ and kinetic energy is $0.5\times0.60\times4.0^2=4.8\,\text{J}$ . After, momentum is $(0.60+0.40)\times2.4=2.4\,\text{kg m s}^{-1}$ and kinetic energy is $0.5\times1.00\times2.4^2=2.88\,\text{J}$ . Momentum is conserved, whereas $4.8-2.88=1.92\,\text{J}$ leaves the kinetic energy store and is transferred mainly to thermal energy stores of the carts and surroundings, with some transferred by sound where applicable.	6
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04.1	Third-fragment mass $=0.70\,\text{kg}$ Momentum of the first two fragments together $=+0.90\,\text{kg m s}^{-1}$ Third-fragment velocity $=-1.29\,\text{m s}^{-1}$	The third mass is $1.50-0.30-0.50=0.70\,\text{kg}$ . The first two momenta total $(0.30\times8.0)+(0.50\times-3.0)=2.4-1.5=+0.90\,\text{kg m s}^{-1}$ . Initial momentum is zero, so the third fragment must have momentum $-0.90\,\text{kg m s}^{-1}$ . Its velocity is $-0.90/0.70=-1.2857\ldots\,\text{m s}^{-1}$ , or $-1.29\,\text{m s}^{-1}$ .	5
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05.1	Initial momentum $=+20\,\text{kg m s}^{-1}$ Velocity after the first release $=+5.25\,\text{m s}^{-1}$ Platform velocity after the second release $=+7.25\,\text{m s}^{-1}$	Initially, total mass is $5.0\,\text{kg}$ and momentum is $5.0\times4.0=20\,\text{kg m s}^{-1}$ . After the first release, $20=(1.0\times-1.0)+(4.0v)$ , so $v=21/4=5.25\,\text{m s}^{-1}$ . Just before the second release, the remaining $4.0\,\text{kg}$ system has momentum $4.0\times5.25=21\,\text{kg m s}^{-1}$ . Conservation for that release gives $21=(1.0\times-0.75)+(3.0v)$ , so $v=21.75/3.0=7.25\,\text{m s}^{-1}$ .	6
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4.5.7.3 · Changes in momentum (physics only) (HT only)

Tier 1 · Easy

Mark scheme for 4.5.7.3 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	$28\,\text{N}$	The momentum change magnitude is $m\Delta v=0.35\times8.0=2.8\,\text{kg m s}^{-1}$ . Hence $F=\Delta p/\Delta t=2.8/0.10=28\,\text{N}$ .	3
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02.1	Lining B produces half the average force of lining A and is safer because it reduces the force on the cyclist's head.	For the same momentum change, $F=\Delta p/\Delta t$ . Doubling the time halves the average force, so lining B reduces the force and injury risk.	3
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Tier 2 · Standard

Mark scheme for 4.5.7.3 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$64\,\text{N}$ away from the wall	Take motion toward the wall as positive. Then $u=+12$ and $v=-8.0\,\text{m s}^{-1}$ , so $\Delta p=m(v-u)=0.16(-8.0-12)=-3.2\,\text{kg m s}^{-1}$ . Thus $F=-3.2/0.050=-64\,\text{N}$ : magnitude $64\,\text{N}$ away from the wall.	4
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02.1	Pad P: $300\,\text{N}$ Pad Q: $120\,\text{N}$ Pad Q reduces the force more.	Use $F=\Delta p/\Delta t$ . For P, $F=3.6/0.012=300\,\text{N}$ . For Q, $F=3.6/0.030=120\,\text{N}$ . The longer action time makes Q's average force smaller.	4
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03.1	$0.60\,\text{kg}$	The change in momentum is $F\Delta t=48\times0.075=3.6\,\text{kg m s}^{-1}$ . The velocity change is $4.0-(-2.0)=6.0\,\text{m s}^{-1}$ . Since $\Delta p=m\Delta v$ , $m=3.6/6.0=0.60\,\text{kg}$ .	4
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Tier 3 · Hard

Mark scheme for 4.5.7.3 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Rigid structure: $1.66\times10^5\,\text{N}$ Crumple zone: $4.16\times10^4\,\text{N}$ The crumple zone makes the stopping time four times longer, so the average force is four times smaller for the same momentum change.	The momentum change magnitude is $m\Delta v=950\times14=13300\,\text{kg m s}^{-1}$ . For the rigid structure, $F=13300/0.080=166250\,\text{N}$ . With the crumple zone, $F=13300/0.32=41562.5\,\text{N}$ . The momentum change is unchanged, but the fourfold increase in time reduces the average force to one quarter, lowering the risk of injury.	6
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02.1	Change in momentum $=-3.6\,\text{kg m s}^{-1}$ Contact time $=0.040\,\text{s}$ Padded-bat force $=-36\,\text{N}$ The padded bat reduces the average force magnitude from $90\,\text{N}$ to $36\,\text{N}$ .	The signed change is $\Delta p=m(v-u)=0.12(-12-18)=0.12(-30)=-3.6\,\text{kg m s}^{-1}$ . From $F=\Delta p/\Delta t$ , $\Delta t=\Delta p/F=(-3.6)/(-90)=0.040\,\text{s}$ . For the padded bat, $F=-3.6/0.10=-36\,\text{N}$ . Its longer contact time reduces the force magnitude by $54\,\text{N}$ .	6
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03.1	Momentum change magnitude $=4.0\,\text{kg m s}^{-1}$ Minimum contact time $=0.10\,\text{s}$ Material B Average force with B $=26.7\,\text{N}$	The object is brought to rest, so the momentum change magnitude is $\|\Delta p\|=0.50\times8.0=4.0\,\text{kg m s}^{-1}$ . The minimum allowed time is $\Delta t=\|\Delta p\|/F=4.0/40=0.10\,\text{s}$ . A is too thin because its contact time is below the minimum; B is the least thick material with a long enough contact time. Its force magnitude is $4.0/0.15=26.7\,\text{N}$ .	6
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04.1	First momentum change $=+1.8\,\text{kg m s}^{-1}$ Second momentum change $=+1.0\,\text{kg m s}^{-1}$ Final momentum $=+2.0\,\text{kg m s}^{-1}$ Final velocity $=+2.5\,\text{m s}^{-1}$	The initial momentum is $0.80\times(-1.0)=-0.80\,\text{kg m s}^{-1}$ . From $F\Delta t=\Delta p$ , the two changes are $6.0\times0.30=+1.8\,\text{kg m s}^{-1}$ and $2.0\times0.50=+1.0\,\text{kg m s}^{-1}$ . The final momentum is $-0.80+1.8+1.0=+2.0\,\text{kg m s}^{-1}$ , so $v=p/m=2.0/0.80=+2.5\,\text{m s}^{-1}$ .	5
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05.1	Water momentum change in $0.50\,\text{s}$ $=-14.4\,\text{kg m s}^{-1}$ Average force on the water $=-28.8\,\text{N}$ The water exerts an average force of $+28.8\,\text{N}$ on the wall.	For each $2.4\,\text{kg}$ portion, $\Delta p=m(v-u)=2.4(0-6.0)=-14.4\,\text{kg m s}^{-1}$ . Therefore $F=\Delta p/\Delta t=-14.4/0.50=-28.8\,\text{N}$ on the water. By Newton's Third Law, the water exerts an equal $28.8\,\text{N}$ force on the wall in the positive direction.	5
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