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AQA GCSE Physics revision notes

Forces

Section 4.5
25 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against AQA 8463 section 4.5

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

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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 6.0kg6.0\,\text{kg} and moves with a velocity of 2.5m/s2.5\,\text{m/s} east. State which of these two quantities is a vector.

  1. 1.Mass needs only a magnitude, so it is scalar.
  2. 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

ORIGINAL

Explain why a speedometer reading alone cannot give a vehicle's velocity.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A scale drawing uses 1.0cm1.0\,\text{cm} to represent 4.0N4.0\,\text{N}. Describe the arrow that represents a force of 20N20\,\text{N} acting vertically downwards.

[2 marks]

Total for this question: 2

Tier 3 · Hard

ORIGINAL

A cyclist's speedometer reads 11m/s11\,\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

Your progress and exam materials
4.5.1.2

Contact and non-contact forces

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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. 1.Friction requires touching surfaces.
  2. 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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

4.5.1.3

Gravity

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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=mgW=mg, with WW in newtons, mm in kilograms and gg in newtons per kilogram; use the value of gg supplied in the question.
  • For example, a 3.0kg3.0\,\text{kg} object where g=9.8N/kgg=9.8\,\text{N/kg} has weight 3.0×9.8=29.4N3.0\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 gg and is measured with a calibrated newtonmeter.
Worked example

A 7.5kg7.5\,\text{kg} bag is in a region where g=9.8N/kgg=9.8\,\text{N/kg}. Calculate the bag's weight.

  1. 1.Use W=mgW=mg: W=7.5×9.8=73.5NW=7.5\times9.8=73.5\,\text{N}.

Answer: 73.5N73.5\,\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=mgW=mg calculation, use mass in kilograms and the value of gg supplied in the question.

Tier 1 · Easy

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

A sample has mass 2.4kg2.4\,\text{kg}. Its weight is 23.5N23.5\,\text{N} at location A and 3.8N3.8\,\text{N} at location B. Calculate the gravitational field strength at each location.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

An explorer has mass 68kg68\,\text{kg}. A newtonmeter would show 666N666\,\text{N} for the explorer on planet P and 177N177\,\text{N} on moon Q. Determine gg at P and Q, then explain what happens to the explorer's mass during the journey.

[4 marks]

Total for this question: 4

4.5.1.4

Resultant forces

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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, 18N18\,\text{N} right and 11N11\,\text{N} left give a resultant of 1811=7N18-11=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: 16N16\,\text{N} east and 9N9\,\text{N} west. Calculate the resultant force.

  1. 1.The forces oppose, so subtract their magnitudes: 169=7N16-9=7\,\text{N}.
  2. 2.The larger force acts east, so the resultant is east.

Answer: 7N7\,\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 7N7\,\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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

A model boat is pulled forwards by 42N42\,\text{N}. Water resistance is 27N27\,\text{N} and air resistance is 6N6\,\text{N}, both backwards. Determine the resultant force on the boat.

[2 marks]

Total for this question: 2

Tier 3 · Hard

ORIGINAL

Forces on a rail cart are 85N85\,\text{N} east, 34N34\,\text{N} west and 19N19\,\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

4.5.2

Work done and energy transfer

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Work is done when a force causes a displacement, transferring energy between stores; work done against friction raises temperature.
  • Use W=FsW=Fs, where ss is the distance moved along the force's line of action; WW is in joules, FF in newtons and ss in metres.
  • For example, a 25N25\,\text{N} force moving an object 3.0m3.0\,\text{m} along its line of action does 25×3.0=75J25\times3.0=75\,\text{J} of work.
  • Do not multiply by a distance perpendicular to the force, and remember that 1J=1N m1\,\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 35N35\,\text{N} moves a box 4.0m4.0\,\text{m} horizontally. Calculate the work done by the force.

  1. 1.The movement is along the force's line of action, so W=Fs=35×4.0=140JW=Fs=35\times4.0=140\,\text{J}.

Answer: 140J140\,\text{J}

Common mistakes

  • Don't fall into the trap of using a distance perpendicular to the force in W=FsW=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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

A winch lifts a load vertically through 6.5m6.5\,\text{m} using a constant upward force of 480N480\,\text{N}. Calculate the work done and identify the energy store that increases.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A powered trolley moves 12m12\,\text{m} along a level floor. Its motor provides a forward force of 180N180\,\text{N} while friction is 140N140\,\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

4.5.3

Forces and elasticity

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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=keF=ke, measuring extension ee 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=mgW=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=160N/mk=160\,\text{N/m} extended by 0.050m0.050\,\text{m} needs F=160×0.050=8.0NF=160\times0.050=8.0\,\text{N} and stores Ee=12ke2=0.20JE_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 0.18m0.18\,\text{m} to 0.23m0.23\,\text{m}. Calculate its extension.

  1. 1.Extension is stretched length minus original length: e=0.230.18=0.050me=0.23-0.18=0.050\,\text{m}.

Answer: 0.050m0.050\,\text{m}

Common mistakes

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

Exam tip

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

Tier 1 · Easy

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

Within its linear region, a spring extends by 0.075m0.075\,\text{m} when a force of 18N18\,\text{N} is applied. Calculate the spring constant and predict the extension produced by 12N12\,\text{N}.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A spring of constant 240N/m240\,\text{N/m} is stretched by 0.18m0.18\,\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

4.5.4

Moments, levers and gears (physics only)

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A moment is the turning effect of a force about a pivot, and its size is M=FdM=Fd where dd 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, 15N15\,\text{N} acting 0.40m0.40\,\text{m} from a pivot produces a moment of 6.0N m6.0\,\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 28N28\,\text{N} acts perpendicular to a handle 0.35m0.35\,\text{m} from its pivot. Calculate the moment.

  1. 1.Use M=FdM=Fd: M=28×0.35=9.8N mM=28\times0.35=9.8\,\text{N m}.

Answer: 9.8N m9.8\,\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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

A seesaw is balanced. A child of weight 360N360\,\text{N} sits 1.5m1.5\,\text{m} to the left of the pivot. Calculate how far to the right of the pivot a child of weight 450N450\,\text{N} must sit.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A uniform 4.0m4.0\,\text{m} beam of weight 220N220\,\text{N} is supported at its left end and at a point 3.2m3.2\,\text{m} from the left end. A 300N300\,\text{N} load is placed at the right end. Calculate the upward force from the support at 3.2m3.2\,\text{m}.

[4 marks]

Total for this question: 4

4.5.5.1.1

Pressure in a fluid 1 (physics only)

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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/Ap=F/A, with normal force FF in newtons, surface area AA in square metres and pressure pp in pascals.
  • For example, a normal force of 90N90\,\text{N} on 0.030m20.030\,\text{m}^2 produces p=90÷0.030=3000Pap=90\div0.030=3000\,\text{Pa}.
  • Do not use an area in cm2\text{cm}^2 without converting it to m2\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 240N240\,\text{N}. The hatch area is 0.080m20.080\,\text{m}^2. Calculate the pressure.

  1. 1.Use p=F/Ap=F/A: p=240÷0.080=3000Pap=240\div0.080=3000\,\text{Pa}.

Answer: 3000Pa3000\,\text{Pa}

Common mistakes

  • Don't fall into the trap of using area in cm2\text{cm}^2 without converting it to m2\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

ORIGINAL

The same normal force acts on two flat surfaces, but surface BB has twice the area of surface AA. Compare the pressures on the two surfaces.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A liquid exerts a normal force of 54N54\,\text{N} on a sensor of area 30cm230\,\text{cm}^2. Calculate the pressure on the sensor.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A sealed chamber produces the same normal force of 1260N1260\,\text{N} on either of two removable panels. Panel X measures 0.30m0.30\,\text{m} by 0.20m0.20\,\text{m} and panel Y measures 0.45m0.45\,\text{m} by 0.35m0.35\,\text{m}. Calculate the pressure on each panel and determine which panel experiences the lower pressure.

[5 marks]

Total for this question: 5

4.5.5.1.2

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

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Higher tier: in a liquid, pressure increases with the height of liquid above the point and with the liquid's density. Use p=hρgp=h\rho g for pressure due to a liquid column, with hh in metres, ρ\rho in kg/m3\text{kg/m}^3 and gg in N/kg\text{N/kg}; subtract depths before using it for a pressure difference.
  • For example, in water with ρ=1000kg/m3\rho=1000\,\text{kg/m}^3 and g=9.8N/kgg=9.8\,\text{N/kg}, a 0.50m0.50\,\text{m} depth change gives Δp=0.50×1000×9.8=4900Pa\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=hρgp=h\rho g.
Pressure difference across a submerged object.
Worked example

Oil has density 820kg/m3820\,\text{kg/m}^3. Calculate the pressure due to a 1.5m1.5\,\text{m} column of the oil when g=9.8N/kgg=9.8\,\text{N/kg}.

  1. 1.Use p=hρg=1.5×820×9.8=12054Pap=h\rho g=1.5\times820\times9.8=12054\,\text{Pa}, which is 1.21×104Pa1.21\times10^4\,\text{Pa} to three significant figures.

Answer: 1.21×104Pa1.21\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=hρgp=h\rho g, use vertical depth below the liquid surface and explain upthrust using the pressure difference.

Tier 1 · Easy

ORIGINAL

Explain why a fully submerged object experiences an upward force due to the liquid pressure on it.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Two pressure sensors are 0.40m0.40\,\text{m} and 2.10m2.10\,\text{m} below the surface of a liquid of density 960kg/m3960\,\text{kg/m}^3. Calculate the pressure difference when g=9.8N/kgg=9.8\,\text{N/kg}.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A fully submerged cuboid is 0.30m0.30\,\text{m} high and has horizontal top and bottom areas of 0.012m20.012\,\text{m}^2. It is in water of density 1000kg/m31000\,\text{kg/m}^3, where g=9.8N/kgg=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 30N30\,\text{N}.

[5 marks]

Total for this question: 5

4.5.5.2

Atmospheric pressure (physics only)

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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/Ap=F/A to F=pAF=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. 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

ORIGINAL

Explain why atmospheric pressure is lower at a higher altitude.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

At high altitude, the pressure inside a sealed case is 101kPa101\,\text{kPa} and the atmospheric pressure outside is 75kPa75\,\text{kPa}. A flat lid has area 0.012m20.012\,\text{m}^2. Calculate the resultant force on the lid and state its direction.

[4 marks]

Total for this question: 4

4.5.6.1.1

Distance and displacement

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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 9m9\,\text{m} east and then 4m4\,\text{m} west gives distance 13m13\,\text{m} and displacement 5m5\,\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 14m14\,\text{m} north and then 5m5\,\text{m} south. Determine the distance and displacement.

  1. 1.Distance adds both path lengths: 14+5=19m14+5=19\,\text{m}.
  2. 2.Taking north as positive, displacement is 145=9m14-5=9\,\text{m} north.

Answer: Distance 19m19\,\text{m}; displacement 9m9\,\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

ORIGINAL

An athlete completes one 400m400\,\text{m} lap and finishes at the starting point. State the distance travelled and the displacement.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A robot moves 6.0m6.0\,\text{m} east and then 8.0m8.0\,\text{m} north. Calculate its distance. Then, using a scale of 1.0cm=2.0m1.0\,\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

Tier 3 · Hard

ORIGINAL

A survey drone flies 600m600\,\text{m} east, 250m250\,\text{m} west and then 120m120\,\text{m} north. Calculate the total distance and the net eastward displacement before the northward flight. Then use a scale of 1.0cm=50m1.0\,\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

4.5.6.1.2

Speed

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Speed is a scalar rate of change of distance; typical values are about 1.5m/s1.5\,\text{m/s} for walking, 3m/s3\,\text{m/s} for running, 6m/s6\,\text{m/s} for cycling and 330m/s330\,\text{m/s} for sound in air.
  • Typical transport-system speeds are about 131330m/s30\,\text{m/s} for a car, 55m/s55\,\text{m/s} for a train and 250m/s250\,\text{m/s} for a passenger plane.
  • Use s=vts=vt for constant speed and rearrange to v=s/tv=s/t; for non-uniform motion, average speed is total distance divided by total time.
  • For example, 450m450\,\text{m} travelled in 30s30\,\text{s} gives an average speed of 450÷30=15m/s450\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 24m24\,\text{m} in 6.0s6.0\,\text{s} at constant speed. Calculate its speed.

  1. 1.Use v=s/tv=s/t: v=24÷6.0=4.0m/sv=24\div6.0=4.0\,\text{m/s}.

Answer: 4.0m/s4.0\,\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

ORIGINAL

A runner covers 120m120\,\text{m} in 20s20\,\text{s}, rests for 10s10\,\text{s}, then covers 180m180\,\text{m} in 30s30\,\text{s}. Calculate the average speed for the whole interval.

[3 marks]

Total for this question: 3

Tier 2 · Standard

ORIGINAL

A runner covers 300m300\,\text{m} in 48s48\,\text{s}, rests for 12s12\,\text{s} and then covers another 200m200\,\text{m} in 32s32\,\text{s}. Calculate the average speed for the whole interval.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A delivery drone travels 1.8km1.8\,\text{km} to a site in 2min 20s2\,\text{min}\ 20\,\text{s}. It waits for 30s30\,\text{s}, then returns along the same route at a constant 15m/s15\,\text{m/s}. Calculate its average speed for the complete trip from departure to return.

[5 marks]

Total for this question: 5

4.5.6.1.3

Velocity

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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 60m60\,\text{m} west in 15s15\,\text{s} gives an average velocity of 4.0m/s4.0\,\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 18m/s18\,\text{m/s}. State its velocity.

  1. 1.Velocity is speed with direction, so attach the stated direction north to the magnitude 18m/s18\,\text{m/s}.

Answer: 18m/s18\,\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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

A cart moves 50m50\,\text{m} east and then 20m20\,\text{m} west in a total time of 10s10\,\text{s}. Calculate its average velocity.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A rescue vehicle travels 80m80\,\text{m} east and then 150m150\,\text{m} north in 30s30\,\text{s}. Using a scale of 1.0cm=20m1.0\,\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

4.5.6.1.4

The distance–time relationship

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A distance–time graph shows distance travelled on the vertical axis and time on the horizontal axis.
  • Its gradient is speed: v=ΔsΔtv=\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 30m30\,\text{m} at 8s8\,\text{s} to 102m102\,\text{m} at 20s20\,\text{s}. Calculate the speed on this section.

  1. 1.Distance change =10230=72m=102-30=72\,\text{m}.
  2. 2.Time change =208=12s=20-8=12\,\text{s}.
  3. 3.Gradient =7212=6.0m/s=\dfrac{72}{12}=6.0\,\text{m/s}.

Answer: 6.0m/s6.0\,\text{m/s}

Common mistakes

  • Don't fall into the trap of using 102/20102/20 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

ORIGINAL

A runner travels 156m156\,\text{m} in 48s48\,\text{s} at constant speed. Calculate the gradient of the distance–time graph.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A distance–time graph is a straight line from (0s,0m)(0\,\text{s},0\,\text{m}) to (24s,120m)(24\,\text{s},120\,\text{m}). It is then horizontal for 8s8\,\text{s}. State what happens during the horizontal section and calculate the speed during the first section.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A robot moves along a straight track. Its distance from the start is 0m0\,\text{m} at 0s0\,\text{s}, 54m54\,\text{m} at 12s12\,\text{s}, 150m150\,\text{m} at 28s28\,\text{s} and 150m150\,\text{m} at 40s40\,\text{s}. The graph joins these points with straight lines. Calculate the speed in each time interval and the average speed over all 40s40\,\text{s}.

[5 marks]

Total for this question: 5

4.5.6.1.5

Acceleration

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Acceleration is the rate at which velocity changes. Calculate it using a=vuta=\dfrac{v-u}{t}, where uu is initial velocity, vv is final velocity and tt is the time for the change; acceleration is measured in m/s2\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 v2u2=2asv^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 4.0m/s4.0\,\text{m/s} to 16.0m/s16.0\,\text{m/s} in 6.0s6.0\,\text{s}. Calculate the average acceleration.

  1. 1.Change in velocity =16.04.0=12.0m/s=16.0-4.0=12.0\,\text{m/s}.
  2. 2.a=vut=12.06.0a=\dfrac{v-u}{t}=\dfrac{12.0}{6.0}.

Answer: 2.0m/s22.0\,\text{m/s}^2

Common mistakes

  • Don't fall into the trap of dividing the final velocity by time instead of first calculating vuv-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

ORIGINAL

A scooter increases its velocity from 3.0m s13.0\,\text{m s}^{-1} to 15.0m s115.0\,\text{m s}^{-1} in 8.0s8.0\,\text{s}. Calculate its average acceleration.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A velocity–time graph rises uniformly from 00 to 12m s112\,\text{m s}^{-1} in 4.0s4.0\,\text{s}, stays horizontal for 6.0s6.0\,\text{s}, then falls uniformly to 00 in 3.0s3.0\,\text{s}. Determine the acceleration in each section.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

A train accelerates uniformly from 8.0m s18.0\,\text{m s}^{-1} to 20.0m s120.0\,\text{m s}^{-1} while travelling 168m168\,\text{m}. Calculate its acceleration and the time taken.

[5 marks]

Total for this question: 5

4.5.6.2.1

Newton's First Law

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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 760N760\,\text{N} force east. Determine the total resistive force.

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

Answer: 760N760\,\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

ORIGINAL

A boat moves at constant velocity. Its propeller provides a forward force of 680N680\,\text{N}. Determine the total resistive force.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A van travels in a straight line at a steady 18m s118\,\text{m s}^{-1}. The engine force is 920N920\,\text{N}. Explain the motion in terms of the forces and state the resultant force.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A lift moves upward. The motor force is 6200N6200\,\text{N} upward, its weight is 5800N5800\,\text{N} and friction is 400N400\,\text{N} downward. Describe its motion. The motor force then falls to 5000N5000\,\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

4.5.6.2.2

Newton's Second Law

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Newton's Second Law links resultant force, mass and acceleration: F=maF=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 m/s2\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/am=F/a.
Worked example

A 920kg920\,\text{kg} car has a driving force of 3100N3100\,\text{N} and resistive forces totalling 1260N1260\,\text{N}. Calculate its acceleration.

  1. 1.Resultant force =31001260=1840N=3100-1260=1840\,\text{N} forwards.
  2. 2.a=Fm=1840920a=\dfrac{F}{m}=\dfrac{1840}{920}.

Answer: 2.0m/s22.0\,\text{m/s}^2 forwards

Common mistakes

  • Don't fall into the trap of substituting the driving force into F=maF=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=maF=ma.

Tier 1 · Easy

ORIGINAL

A resultant force accelerates a 12kg12\,\text{kg} object at 1.5m s21.5\,\text{m s}^{-2}. Calculate the resultant force.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

A 1250kg1250\,\text{kg} car increases its velocity from 8.0m s18.0\,\text{m s}^{-1} to 26.0m s126.0\,\text{m s}^{-1} in 9.0s9.0\,\text{s}. The resistive forces total 700N700\,\text{N}. Calculate the acceleration, the resultant force and the engine force.

[5 marks]

Total for this question: 5

4.5.6.2.3

Newton's Third Law

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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 12N12\,\text{N}. State the paired force.

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

Answer: The Earth pulls the book downwards gravitationally with a force of 12N12\,\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

ORIGINAL

A hammer exerts a downward force on a nail. State the corresponding Newton's Third Law force.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

A propeller pushes water backward with a force of 4.2kN4.2\,\text{kN}. The water resistance on the boat is 1.8kN1.8\,\text{kN} backward and the boat's mass is 1200kg1200\,\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

4.5.6.3.1

Stopping distance

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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=vts=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 18m/s18\,\text{m/s}. Its driver's reaction time is 0.65s0.65\,\text{s} and its braking distance is 24m24\,\text{m}. Calculate the stopping distance.

  1. 1.Thinking distance =vt=18×0.65=11.7m=vt=18\times0.65=11.7\,\text{m}.
  2. 2.Stopping distance =11.7+24=35.7m=11.7+24=35.7\,\text{m}.

Answer: 35.7m35.7\,\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

ORIGINAL

A driver's thinking distance is 12m12\,\text{m} and the braking distance is 28m28\,\text{m}. Calculate the stopping distance.

[1 mark]

Total for this question: 1

Evidence from answers you checked

Checked automatically against the model answer once you submit.

Tier 2 · Standard

ORIGINAL

A car travels at 15m s115\,\text{m s}^{-1}. The driver's reaction time is 0.64s0.64\,\text{s} and the braking distance is 19m19\,\text{m}. Calculate the thinking distance and the stopping distance.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A delivery van has a reaction time of 0.75s0.75\,\text{s}. At 20m s120\,\text{m s}^{-1} its braking distance is 32m32\,\text{m}; at 25m s125\,\text{m s}^{-1} its braking distance is 50m50\,\text{m}. Calculate the stopping distance at each speed and the increase in stopping distance.

[5 marks]

Total for this question: 5

4.5.6.3.2

Reaction time

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Human reaction times vary, with typical values from 0.2s0.2\,\text{s} to 0.9s0.9\,\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 22m/s22\,\text{m/s} and the driver reacts in 0.48s0.48\,\text{s}. Calculate the thinking distance.

  1. 1.During the reaction time the car continues at 22m/s22\,\text{m/s}.
  2. 2.s=vt=22×0.48s=vt=22\times0.48.

Answer: 10.56m10.56\,\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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

A driver travels at 22m s122\,\text{m s}^{-1} and reacts in 0.48s0.48\,\text{s}. Calculate the thinking distance.

[2 marks]

Total for this question: 2

Tier 3 · Hard

ORIGINAL

A student measures reaction time four times without a distraction and obtains 0.310.31, 0.290.29, 0.300.30 and 0.89s0.89\,\text{s}. With a distraction the results are 0.440.44, 0.470.47, 0.450.45 and 0.46s0.46\,\text{s}. Identify the anomalous result, calculate suitable mean times and evaluate the effect of the distraction.

[5 marks]

Total for this question: 5

4.5.6.3.3

Factors affecting braking distance 1

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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 6200N6200\,\text{N} on a dry road and 3100N3100\,\text{N} on a wet road. A car stops in 18m18\,\text{m} on the dry road from the same initial speed. Estimate its wet-road braking distance.

  1. 1.The same kinetic energy must be removed, so braking work FsFs is unchanged.
  2. 2.The force halves, so the distance doubles: 18×6200310018\times\dfrac{6200}{3100}.

Answer: 36m36\,\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

ORIGINAL

Give one example of poor vehicle condition that increases braking distance and explain why it does so.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

The average braking force on a car is 6200N6200\,\text{N} on a dry road and 3100N3100\,\text{N} on a wet road. The car has the same initial speed in both tests and stops in 18m18\,\text{m} on the dry road. Estimate the wet-road braking distance.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

For one car in fixed conditions, measured braking distances are 6.2m6.2\,\text{m} at 10m s110\,\text{m s}^{-1}, 13.9m13.9\,\text{m} at 15m s115\,\text{m s}^{-1} and 24.8m24.8\,\text{m} at 20m s120\,\text{m s}^{-1}. Show that the data are consistent with braking distance being proportional to speed squared, then estimate the distance at 25m s125\,\text{m s}^{-1}.

[4 marks]

Total for this question: 4

4.5.6.3.4

Factors affecting braking distance 2

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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=FsW=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=maF=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 1100kg1100\,\text{kg} car travels at 18m/s18\,\text{m/s} and is stopped by an average braking force of 6600N6600\,\text{N}. Calculate its braking distance.

  1. 1.Ek=12mv2=0.5×1100×182=178200JE_k=\dfrac{1}{2}mv^2=0.5\times1100\times18^2=178200\,\text{J}.
  2. 2.Set braking work equal to the energy transferred: Fs=178200Fs=178200.
  3. 3.s=178200÷6600=27ms=178200\div6600=27\,\text{m}.

Answer: 27m27\,\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=FsW=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

ORIGINAL

Describe the main energy transfer when friction in a vehicle's brakes brings the vehicle to rest.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A 1100kg1100\,\text{kg} car travels at 18m s118\,\text{m s}^{-1}. Its average braking force is 6600N6600\,\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

Tier 3 · Hard

ORIGINAL

A 1500kg1500\,\text{kg} car travelling at 22m s122\,\text{m s}^{-1} is stopped by an average braking force of 8250N8250\,\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

4.5.7.1

Momentum is a property of moving objects (HT only)

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Higher tier: momentum is a property of a moving object and is calculated using p=mvp=mv.
  • Here pp is momentum in kg m/s\text{kg m/s}, mm is mass in kilograms and vv is velocity in m/s\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 1350kg1350\,\text{kg} car travels west at 16m/s16\,\text{m/s}. Calculate its momentum.

  1. 1.p=mv=1350×16=21600kg m/sp=mv=1350\times16=21600\,\text{kg m/s}.
  2. 2.Momentum has the same direction as velocity, so it is westward.

Answer: 2.16×104kg m/s2.16\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

ORIGINAL

A 0.18kg0.18\,\text{kg} ball moves at 12m s112\,\text{m s}^{-1}. Calculate its momentum.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A 1350kg1350\,\text{kg} car travels west at 16m s116\,\text{m s}^{-1}. Calculate its momentum, including direction.

[2 marks]

Total for this question: 2

Tier 3 · Hard

ORIGINAL

Vehicle A has mass 720kg720\,\text{kg} and travels east at 18m s118\,\text{m s}^{-1}. Vehicle B has mass 1080kg1080\,\text{kg} and travels west at 11m s111\,\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

4.5.7.2

Conservation of momentum (HT only)

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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=mvp=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 (m1+m2)v(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 2.0kg2.0\,\text{kg} trolley moving right at 3.0m/s3.0\,\text{m/s} hits a stationary 1.0kg1.0\,\text{kg} trolley. They stick together. Calculate their velocity.

  1. 1.Initial momentum =(2.0×3.0)+(1.0×0)=6.0kg m/s=(2.0\times3.0)+(1.0\times0)=6.0\,\text{kg m/s}.
  2. 2.Joined mass =2.0+1.0=3.0kg=2.0+1.0=3.0\,\text{kg}.
  3. 3.6.0=3.0v6.0=3.0v, so v=2.0m/sv=2.0\,\text{m/s}.

Answer: 2.0m/s2.0\,\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

ORIGINAL

A 2.0kg2.0\,\text{kg} trolley moving at 3.0m s13.0\,\text{m s}^{-1} collides with a stationary 1.0kg1.0\,\text{kg} trolley. They stick together. Calculate their common velocity.

[3 marks]

Total for this question: 3

Tier 2 · Standard

ORIGINAL

A 0.75kg0.75\,\text{kg} trolley moving right at 6.4m s16.4\,\text{m s}^{-1} catches a 1.25kg1.25\,\text{kg} trolley moving right at 1.6m s11.6\,\text{m s}^{-1}. The trolleys lock together. Determine their final velocity.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

A launcher of mass 3.8kg3.8\,\text{kg} and a 0.20kg0.20\,\text{kg} projectile are initially at rest. The projectile is fired horizontally at 32m s132\,\text{m s}^{-1}. Calculate the launcher's recoil velocity.

[4 marks]

Total for this question: 4

4.5.7.3

Changes in momentum (physics only) (HT only)

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Higher tier: force equals the rate of change of momentum, so F=ΔpΔtF=\dfrac{\Delta p}{\Delta t}.
  • For constant mass, Δp=m(vu)\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 FΔt=ΔpF\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 0.16kg0.16\,\text{kg} ball travels towards a wall at 12m/s12\,\text{m/s} and rebounds at 8.0m/s8.0\,\text{m/s}. Contact lasts 0.050s0.050\,\text{s}. Calculate the average force.

  1. 1.Take towards the wall as positive: u=+12m/su=+12\,\text{m/s} and v=8.0m/sv=-8.0\,\text{m/s}.
  2. 2.Δp=m(vu)=0.16(8.012)=3.2kg m/s\Delta p=m(v-u)=0.16(-8.0-12)=-3.2\,\text{kg m/s}.
  3. 3.F=3.20.050=64NF=\dfrac{-3.2}{0.050}=-64\,\text{N}.

Answer: 64N64\,\text{N} away from the wall

Common mistakes

  • Don't fall into the trap of using 128=4m/s12-8=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 vuv-u.

Tier 1 · Easy

ORIGINAL

A 0.35kg0.35\,\text{kg} ball moving at 8.0m s18.0\,\text{m s}^{-1} is brought to rest in 0.10s0.10\,\text{s}. Calculate the magnitude of the average force.

[3 marks]

Total for this question: 3

Tier 2 · Standard

ORIGINAL

A 0.16kg0.16\,\text{kg} ball travels toward a wall at 12m s112\,\text{m s}^{-1} and rebounds at 8.0m s18.0\,\text{m s}^{-1}. Contact lasts 0.050s0.050\,\text{s}. Calculate the magnitude and direction of the average force on the ball.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

A 950kg950\,\text{kg} vehicle travelling at 14m s114\,\text{m s}^{-1} stops in a collision. A rigid structure would stop it in 0.080s0.080\,\text{s}, while a crumple zone increases the stopping time to 0.32s0.32\,\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

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