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    Newton's laws — OCR GCSE Physics

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    Newton's laws explained

    This criterion asks you to remember Newton's second law in the form force = mass × acceleration and to use it in calculations.

    Read the full explanation

    Force is measured in newtons (N), mass in kilograms (kg) and acceleration in metres per second squared (m/s²). A resultant force of 12 N on a 3 kg trolley gives an acceleration of 12 ÷ 3 = 4 m/s². Rearrange the relationship to find mass as force ÷ acceleration, or acceleration as force ÷ mass. Remember that the force here is the resultant (net) force, so subtract any opposing force first. In a multiple-choice question, identify the two known quantities, rearrange if needed, substitute with units in kg, N and m/s², and check the answer's magnitude and unit. Convert grams to kilograms and any other units before calculating.

    recall and apply: momentum (kg m/s) = mass (kg) × velocity (m/s) (Higher tier only)

    Momentum is a vector quantity that measures how hard it is to stop a moving object. For Higher tier you recall and apply p = m × v, where momentum p is in kilogram metres per second (kg m/s), mass m is in kilograms (kg) and velocity v is in metres per second (m/s). Because velocity carries direction, momentum does too: a 2 kg ball moving at 3 m/s to the right has momentum 6 kg m/s to the right, while the same ball moving at 3 m/s to the left has momentum 6 kg m/s to the left. Convert grams to kilograms and cm/s or km/h to m/s before substituting, then multiply mass by velocity and attach the correct unit and direction.

    recall and apply: work done (J) = force (N) × distance (m) (along the line of action of the force)

    Work done measures the energy transferred when a force moves an object. You recall and apply W = F × d, where work done W is in joules (J), force F is in newtons (N) and distance d is in metres (m). The distance must be measured along the line of action of the force, so only the part of the movement in the direction of the force counts. Lifting a 20 N box 1.5 m upwards does 20 × 1.5 = 30 J of work, because the lift is along the line of the weight. Pushing a box sideways while it moves forward does no work in the forward direction if the push is at right angles to the motion. Convert centimetres to metres and kilonewtons to newtons before multiplying.

    recall and apply: power (W) = work done (J) / time (s)

    Power measures how quickly work is done, so a machine that transfers 600 J in 3 s has a power of 600 J ÷ 3 s = 200 W. The equation power (W) = work done (J) / time (s) links three quantities: work done in joules, time in seconds and power in watts, where 1 W = 1 J/s. To apply it, identify the work done and the time taken, divide, and give the answer in watts. To rearrange, multiply power by time to find work done, or divide work done by power to find time. Always convert milliseconds or minutes to seconds before dividing, and check that a larger power for the same work means a shorter time.

    recall examples of ways in which objects interact (to include: electrostatics, gravity, magnetism and by contact (including normal contact force and friction))

    Objects interact when a force acts between them, and the specification names five routes: electrostatics, gravity, magnetism, and contact through the normal contact force and through friction. A charged balloon attracting small paper pieces is an electrostatic interaction acting at a distance. The Earth pulling a falling apple is a gravitational interaction. Two magnets repelling or attracting illustrate magnetism. A book resting on a table experiences a normal contact force pushing perpendicular to the surface, while a box sliding across a floor is slowed by friction acting along the surface. Non-contact interactions (electrostatics, gravity, magnetism) act without touching; contact interactions require surfaces to press or rub together.

    describe how such examples involve interactions between pairs of objects which produce a force on each object

    Forces never come alone: they arise from an interaction between two objects, and each object in the pair experiences a force. If you push a wall, your hand pushes the wall and the wall pushes back on your hand with an equal-sized force in the opposite direction. The same pattern appears when a swimmer pushes water backwards and the water pushes the swimmer forwards, and when the Earth pulls a falling ball downwards while the ball pulls the Earth upwards. To describe an example, name both objects, state the direction of the force each one exerts on the other, and note that the two forces are equal in size and opposite in direction. The two forces act on different objects, so they never cancel each other out.

    represent forces as vectors (to include: drawing free body force diagrams to demonstrate understanding of forces acting as vectors)

    A vector has both magnitude and direction, so a force is fully described only when you give its size in newtons and the direction in which it acts. A free body force diagram isolates one object and shows every force acting on that object as an arrow: the arrow's length represents the force's magnitude and the arrowhead gives its direction. Draw each arrow starting from the object or from a labelled point on it, and label each force (for example weight, normal contact force, friction, tension, thrust). Choose a scale so lengths are comparable, and draw opposing forces along the same line. If arrows are equal and opposite the forces balance and the resultant is zero; if not, the resultant is found by vector addition. A moving object at constant velocity still has balanced forces, so no resultant arrow is needed.

    apply Newton’s first law to explain the motion of an object moving with uniform velocity and also an object where the speed and/or direction change (to include: looking at forces on one body and resultant forces and their effects (qualitative only))

    Newton's first law says an object keeps a constant velocity unless a resultant force acts on it. To apply it, isolate one body, list the forces on that body only, then combine them to judge the resultant. If the forces balance, the resultant is zero and the object continues at uniform velocity, or stays at rest. If they do not balance, a resultant force exists and the speed and/or direction change: the object accelerates, slows, or curves. For example, a car cruising at a steady 20 m/s on a straight road has driving force balanced by friction and drag, so its velocity is constant. When the driver brakes, friction exceeds the driving force, a backward resultant acts, and the car slows. Reasoning stays qualitative: name the forces, state whether they balance, and link that to the motion.

    use vector diagrams to illustrate resolution of forces, a net force (resultant force), and equilibrium situations (to include: scale drawings limited to parallel and perpendicular vectors only) (Higher tier only)

    Higher tier learners use vector diagrams to show three situations: resolving a force into components, finding a net (resultant) force, and showing equilibrium. Scale drawings are limited to parallel and perpendicular vectors, so you draw arrows to scale along two perpendicular directions. To resolve a force, draw it as the diagonal of a rectangle and read the two perpendicular components from the sides. To find a net force, add parallel vectors tip to tail and combine perpendicular ones with a right-angled triangle, measuring the resultant with a ruler and protractor. Equilibrium occurs when vectors close to form a complete shape, giving zero resultant. For example, a 5 N force at an angle resolves into a horizontal and a vertical component whose lengths you measure against the scale. Keep the scale consistent and state it on the diagram.

    describe examples of the forces acting on an isolated solid object or system (to include: examples of objects that reach terminal velocity for example skydivers and applying similar ideas to vehicles) (Higher tier only)

    Treat the object or system as isolated and list only the forces acting on it, each with its direction. For a skydiver, weight acts downwards and drag acts upwards; drag grows with speed until it equals weight, so the resultant force becomes zero and the skydiver falls at constant terminal velocity. If the skydiver opens a parachute, drag increases, the resultant force acts upwards, the skydiver decelerates and settles at a lower terminal velocity. Apply the same reasoning to a vehicle: driving force forwards, resistive forces (air resistance and friction) backwards; at constant top speed these balance. Name each force, state its direction, and explain how the resultant changes the motion.

    describe, using free body diagrams, examples where two or more forces lead to a resultant force on an object (Higher tier only)

    A free body diagram shows the object as a point or simple shape with one arrow for each force acting on it; arrow length represents magnitude and the arrowhead gives direction. When two or more forces act, combine them: forces along the same line add or subtract, so opposite forces partly cancel. If the total is not zero, a resultant force remains and the object accelerates in the direction of that resultant. For example, a box pushed with 30 N to the right while friction acts with 10 N to the left has a resultant of 20 N to the right. Read each diagram, resolve the arrows, and choose the option that correctly states the resultant's size and direction.

    describe, using free body diagrams, examples of the special case where forces balance to produce a resultant force of zero (qualitative only) (Higher tier only)

    A free body diagram shows one chosen object as a point or simple shape with an arrow for every force acting on that object only. Each arrow points in the direction of the force, and arrow length represents magnitude. When the arrows cancel — equal and opposite along each line — the resultant force is zero. This is the balanced-force special case, and it is qualitative at Higher tier: you reason about balance and direction, not about calculating a value. Example: a book resting on a table has weight acting downwards and the normal contact force from the table acting upwards; equal magnitudes give a resultant of zero, so the book stays at rest. Example: a car travelling at constant velocity along a straight level road has driving force balanced by resistive forces and weight balanced by the normal reaction.

    apply Newton’s second law in calculations relating forces, masses and accelerations

    Newton's second law links resultant force F, mass m and acceleration a by the equation F = m × a. Force is measured in newtons (N), mass in kilograms (kg) and acceleration in metres per second squared (m/s²). A resultant force of 12 N on a 3 kg trolley gives a = F ÷ m = 12 N ÷ 3 kg = 4 m/s². Rearranging gives m = F ÷ a and a = F ÷ m, so any one quantity can be found from the other two. Always use the resultant force, not a single applied force, when several forces act: subtract opposing forces first. Convert grams to kilograms by dividing by 1000, for example 500 g = 0.5 kg. Check units before substituting, then state the unit with the answer. A larger resultant force on the same mass gives a larger acceleration, and a larger mass under the same force gives a smaller acceleration.

    explain that inertia is a measure of how difficult it is to change the velocity of an object and that the inertial mass is defined as the ratio of force over acceleration (Higher tier only)

    Inertia describes how strongly an object resists any change in its velocity, whether that change is a speeding up, a slowing down or a change of direction. A loaded trolley is harder to start, stop or turn than an empty one because it has more inertia. Inertial mass puts a number on this: it is defined as the ratio of force over acceleration, so m = F ÷ a. A larger inertial mass means a smaller acceleration for the same resultant force. The definition is a ratio, so the force must be the resultant force acting on the object and the acceleration must be measured in the same direction. Units follow from the ratio: newtons divided by metres per second squared give kilograms.

    define momentum and describe examples of momentum in collisions (to include: an idea of the law of conservation of momentum in collisions) (Higher tier only)

    Momentum is a vector quantity defined as mass multiplied by velocity, p = m × v, measured in kilogram metres per second (kg m/s). Because velocity has direction, momentum has direction too: an object moving left has momentum in the opposite sense to one moving right. In a collision, the total momentum of the interacting objects before the collision equals the total momentum after it, provided no external resultant force acts on the system. This is the law of conservation of momentum. For example, two trolleys that collide and stick together share a combined momentum equal to the sum of their individual momenta before impact, allowing the combined velocity to be calculated. When objects move in opposite directions, treat one direction as positive and the other as negative before adding.

    apply formulae relating force, mass, velocity and acceleration to explain how the changes involved are inter-related (content found only in the physics separate science qualification)

    This separate-science criterion asks you to use formulae that link force, mass, velocity and acceleration, and to explain how a change in one quantity affects the others. Start from F = ma: a resultant force on a mass produces acceleration in the direction of that force, so doubling the force on a fixed mass doubles the acceleration, while doubling the mass under a fixed force halves it. Then connect acceleration to velocity change using a = (v − u) ÷ t, where u is initial velocity, v is final velocity and t is the time taken. Combining these gives F = m(v − u) ÷ t, which shows that a larger velocity change in a shorter time needs a larger force. Use consistent SI units: force in newtons (N), mass in kilograms (kg), velocity in metres per second (m/s) and acceleration in metres per second squared (m/s²).

    use the relationship between work done, force, and distance moved along the line of action of the force and describe the energy transfer involved

    This criterion asks you to calculate work done with W = F × s, where F is the force in newtons and s is the distance moved in metres along the line of action of that force, giving work done in joules (J). The distance must be measured in the same direction as the force: a force pushing a box 3 m across a floor does 60 J of work if the force is 20 N along the motion, but a force at right angles to the motion does no work because there is no movement along its line of action. Describe the energy transfer too: the work done by the pushing force transfers energy from the person to the box's kinetic energy store, and friction transfers energy by heating to the surroundings. One joule equals one newton metre, so 20 N × 3 m = 60 J.

    calculate relevant values of stored energy and energy transfers; convert between newton-metres and joules

    This criterion asks you to do two linked things. First, calculate stored energy and energy transferred in a range of situations, choosing the right relationship for the context: gravitational potential energy from mass, gravitational field strength and height; kinetic energy from mass and speed; and work done by a force from force and distance moved along the force. Second, convert between newton-metres and joules. The key idea is that one newton-metre equals one joule, because a force of 1 N moving its point of application 1 m along its own line transfers 1 J. So a work-done answer of 240 N m is 240 J, and a stored energy of 0.75 J may be written as 0.75 N m. Keep units consistent, square the speed before halving and multiplying by mass, and check whether the question wants the value in joules or newton-metres.

    explain, with reference to examples, the definition of power as the rate at which energy is transferred

    Power measures how quickly energy is transferred, not how much energy is transferred in total. The definition is that power is the rate at which energy is transferred, so a device with a larger power transfers the same energy in less time, or more energy in the same time. Use examples to make this concrete. A 2000 W kettle transfers energy to water faster than a 1000 W kettle, so it boils the same amount of water in less time. A lift motor of higher power raises a load in a shorter time. Two lamps may transfer the same total energy, but the brighter, higher-power lamp does it in less time. When explaining, name the device, state the energy transfer involved, and compare the time taken or the energy per second, linking this back to the definition of power as a rate.

    recall and apply Newton’s third law (to include: application to situations of equilibrium and non-equilibrium)

    Newton's third law says that when body A exerts a force on body B, body B exerts an equal-sized force on A in the opposite direction. The two forces act on different objects, so they never cancel each other. A book resting on a table is in equilibrium because the upward push of the table on the book balances the downward pull of gravity on the book; both of those forces act on the book. The third-law partner of the book's weight is the book's gravitational pull on the Earth, not the table's push. In a non-equilibrium case, such as a rocket accelerating, the exhaust pushes the rocket forward while the rocket pushes the exhaust backward with an equal force.

    explain why an object moving in a circle with a constant speed has a changing velocity (qualitative only) (Higher tier only)

    Velocity is a vector: it has both a magnitude (the speed) and a direction. An object moving round a circle at constant speed keeps the same magnitude of velocity but its direction of motion changes continuously, because at each instant the motion is along the tangent to the circle. Since the direction changes, the velocity changes, even though the speed does not. A changing velocity means the object is accelerating, and that acceleration is directed towards the centre of the circle, produced by a resultant force acting towards the centre. This is a qualitative explanation only: no calculation of centripetal acceleration or force is required. For example, a car going round a roundabout at a steady speed has a changing velocity because its heading keeps turning, so a resultant force towards the centre acts on it.

    Your focus

    1. State the relationship force = mass × acceleration with correct units
    2. Calculate force, mass or acceleration from the other two quantities
    3. Apply the relationship using the resultant force on an object
    Show all 63 objectives
    1. State the equation momentum = mass × velocity with correct units.
    2. Calculate momentum from a given mass and velocity.
    3. Explain why momentum has a direction.
    4. State the equation work done = force × distance with correct units.
    5. Calculate work done when a force moves an object along its line of action.
    6. Explain why a force at right angles to the motion does no work in that direction.
    7. State the equation power (W) = work done (J) / time (s) from memory.
    8. Calculate power given work done and time, giving the answer in watts.
    9. Rearrange the equation to find work done or time and convert time units to seconds.
    10. Recall the five named ways objects interact: electrostatics, gravity, magnetism, normal contact force and friction.
    11. Give a correct example of each interaction.
    12. Classify an interaction as contact or non-contact.
    13. Identify the two objects involved in a given force interaction
    14. State the direction of the force each object exerts on the other
    15. Explain why the two forces in a pair do not cancel each other
    16. Draw a free body force diagram for a named object showing each force as a labelled arrow.
    17. Explain how arrow length and arrowhead represent a force's magnitude and direction.
    18. Distinguish balanced forces from a non-zero resultant in a given diagram.
    19. Apply Newton's first law to explain an object moving with uniform velocity
    20. Apply Newton's first law to explain an object whose speed and/or direction change
    21. Identify the forces on one body and judge the resultant force qualitatively
    22. Use scale vector diagrams to resolve a force into perpendicular components
    23. Use vector diagrams to find a net (resultant) force from parallel and perpendicular vectors
    24. Use closed vector diagrams to illustrate equilibrium situations
    25. Identify the forces acting on an isolated solid object or system and state their directions.
    26. Explain how changing drag leads an object such as a skydiver to reach terminal velocity.
    27. Apply the same force-balance reasoning to a vehicle moving at constant speed.
    28. Interpret free body diagrams showing two or more forces on an object.
    29. Calculate the resultant force from forces acting along the same line.
    30. Explain how a non-zero resultant force affects the object's motion.
    31. Draw a free body diagram for a named object showing all forces acting on it
    32. Explain how equal and opposite forces give a resultant force of zero
    33. Relate a zero resultant force to an object remaining at rest or moving at constant velocity
    34. Apply F = m × a to calculate force, mass or acceleration
    35. Rearrange the equation correctly for the quantity required
    36. Use the resultant force and consistent SI units in calculations
    37. Explain that inertia is a measure of how difficult it is to change the velocity of an object.
    38. Define inertial mass as the ratio of force over acceleration.
    39. Apply m = F ÷ a to compare the acceleration of objects with different inertial masses.
    40. Define momentum and state its units.
    41. Describe examples of momentum changes in collisions.
    42. Apply conservation of momentum to calculate an unknown velocity or mass after a collision.
    43. Apply F = ma to calculate force, mass or acceleration from the other two quantities.
    44. Use a = (v − u) ÷ t to relate velocity change to time and acceleration.
    45. Explain how changing force, mass, velocity change or time affects the other quantities in these relationships.
    46. Use W = F × s to calculate work done from force and distance along the line of action.
    47. Explain why a force perpendicular to the motion does no work.
    48. Describe the energy transfers involved when work is done on an object, including heating by friction.
    49. Calculate stored energy and energy transferred in a given situation using the appropriate relationship.
    50. Convert a value between newton-metres and joules correctly.
    51. Present a calculated energy value with the correct unit and a sensible magnitude.
    52. State the definition of power as the rate at which energy is transferred.
    53. Use a named example to show how power relates to the time taken for an energy transfer.
    54. Compare two devices and explain the difference in power in terms of energy transferred per unit time.
    55. State Newton's third law in terms of equal and opposite forces acting on two different objects
    56. Distinguish third-law force pairs from balanced forces acting on a single object
    57. Apply Newton's third law to explain equilibrium and non-equilibrium situations
    58. Explain that velocity is a vector quantity with magnitude and direction.
    59. Explain why constant speed in a circle still means a changing velocity.
    60. Describe qualitatively the inward acceleration and resultant force on an object moving in a circle.

    Newton's laws exam tips

    Marking Points
    • Recalls the relationship force = mass × acceleration with force in newtons, mass in kilograms and acceleration in m/s²
    • Calculates force by multiplying mass in kilograms by acceleration in m/s²
    • Rearranges the relationship to find acceleration as force ÷ mass
    • Rearranges the relationship to find mass as force ÷ acceleration
    • Uses the resultant force, subtracting any opposing force, before applying the relationship
    • Multiplies mass in kilograms by velocity in metres per second to give momentum in kg m/s.
    • Keeps the direction of the velocity with the momentum because momentum is a vector.
    • Converts mass from grams to kilograms before substituting into p = m × v.
    • Converts a velocity given in cm/s or km/h into m/s before multiplying.
    • Rearranges p = m × v to find mass or velocity when momentum is known.
    • Multiplies force in newtons by distance in metres to give work done in joules.
    • Measures the distance along the line of action of the force, not at an angle to it.
    • Converts distance from centimetres to metres before substituting into W = F × d.
    • Converts force from kilonewtons to newtons before multiplying.
    • Rearranges W = F × d to find force or distance when work done is known.
    • Selects the relationship power = work done ÷ time and substitutes the work done in joules and the time in seconds.
    • Divides work done by time and states the result in watts, recognising that 1 W equals 1 J/s.
    • Rearranges the equation to work done = power × time when the question supplies power and time.
    • Rearranges the equation to time = work done ÷ power when the question supplies work done and power.
    • Converts a time given in minutes or milliseconds into seconds before substituting it into the equation.
    • Identifies electrostatics as an interaction between charged objects, such as a charged rod attracting small pieces of paper.
    • Identifies gravity as an interaction between masses, such as the Earth pulling a falling object downwards.
    • Identifies magnetism as an interaction between magnets or magnetic materials, such as two like poles repelling.
    • Identifies the normal contact force as a push perpendicular to a surface, such as a table supporting a book.
    • Identifies friction as a contact force opposing relative motion along surfaces, such as a box slowing on a floor.
    • Distinguishes non-contact interactions (electrostatics, gravity, magnetism) from contact interactions (normal contact force, friction).
    • Names both objects in the interaction and states the direction of the force each exerts on the other
    • States that the two forces in the pair are equal in size and opposite in direction
    • Identifies that each force in the pair acts on a different object, so the pair does not cancel
    • Applies the pair description to a named example such as a swimmer pushing water backwards while the water pushes the swimmer forwards
    • Identifies that a force vector requires both a magnitude in newtons and a direction.
    • Selects the single object under consideration and shows only the forces acting on that object.
    • Draws each force as an arrow whose length represents magnitude and whose head shows direction.
    • Labels each arrow with the name of the force, such as weight or normal contact force.
    • Recognises that equal and opposite arrows on the same line indicate balanced forces and zero resultant.
    • States that an object moving at constant velocity has balanced forces and no resultant.
    • Identifies the single body under consideration and lists only the forces acting on that body
    • Combines the forces on the body to decide whether the resultant is zero or non-zero
    • States that zero resultant force means uniform velocity or rest, in line with Newton's first law
    • Explains that a non-zero resultant force changes the speed and/or the direction of motion
    • Describes the direction of the resultant force and links it to speeding up, slowing down or curving
    • Keeps the reasoning qualitative by naming forces and balance rather than calculating values
    • Draws vectors to a stated scale with arrows showing magnitude and direction
    • Resolves a single force into two perpendicular components using a rectangle or right-angled triangle
    • Adds parallel vectors tip to tail to find their combined effect
    • Combines perpendicular vectors with a right-angled triangle to find the net force
    • Measures the resultant with a ruler and protractor and converts using the scale
    • Shows equilibrium by vectors forming a closed shape with zero resultant
    • Names the forces acting on the isolated object and gives the direction of each, for example weight downwards and drag upwards on a skydiver.
    • Explains that drag increases as speed increases until drag equals weight, giving zero resultant force and constant terminal velocity.
    • Describes the effect of opening a parachute: drag exceeds weight, resultant force acts upwards, the skydiver decelerates to a lower terminal velocity.
    • Applies the same reasoning to a vehicle, identifying driving force forwards and resistive forces backwards and explaining balance at constant speed.
    • Identifies each force arrow on the free body diagram and states its direction relative to the object.
    • Combines forces acting along the same line by adding those in one direction and subtracting those in the opposite direction.
    • States the size and direction of the resultant force when the forces do not balance.
    • Links a non-zero resultant force to acceleration of the object in the direction of that resultant.
    • Identifies the single object whose forces are being shown and draws only the forces acting on that object
    • Draws each force as an arrow whose direction shows the direction of the force
    • Represents the magnitude of each force by the length of its arrow
    • Shows weight acting vertically downwards from the object
    • Shows the normal contact force acting perpendicular to the supporting surface
    • Pairs equal and opposite forces along the same line to show that they cancel
    • States that a resultant force of zero means the object stays at rest or moves with constant velocity
    • Distinguishes balanced forces from an absence of forces acting on the object
    • Selects the equation F = m × a and substitutes the given values correctly
    • Rearranges the equation to find acceleration as force divided by mass
    • Rearranges the equation to find mass as force divided by acceleration
    • Uses the resultant force by combining opposing forces before substituting
    • Converts mass in grams to kilograms by dividing by 1000
    • States the answer with the correct unit: newtons, kilograms or metres per second squared
    • Explains that a greater resultant force on the same mass produces a greater acceleration
    • States that inertia measures how difficult it is to change an object's velocity.
    • Identifies that a greater inertial mass produces a smaller acceleration for the same resultant force.
    • Defines inertial mass as the ratio of force over acceleration, m = F ÷ a.
    • Recognises that the force in the definition is the resultant force acting on the object.
    • Applies the ratio to a numerical case, for example a resultant force of 12 N giving an acceleration of 3 m/s² implies an inertial mass of 4 kg.
    • Defines momentum as mass multiplied by velocity, p = m × v, with units of kg m/s.
    • States that momentum is a vector, so direction must be considered.
    • Applies the law of conservation of momentum: total momentum before a collision equals total momentum after, with no external resultant force.
    • Calculates total momentum before a collision by summing individual momenta, treating opposing directions with opposite signs.
    • Uses conservation of momentum to find an unknown velocity or mass after a collision, such as objects sticking together.
    • Selects the relationship F = ma and substitutes force in newtons and mass in kilograms to obtain acceleration in m/s².
    • Uses a = (v − u) ÷ t to find the change in velocity per unit time and keeps the direction of the change consistent.
    • Combines the two relationships as F = m(v − u) ÷ t to explain why a shorter stopping or contact time raises the force for the same velocity change.
    • Explains that acceleration is directly proportional to resultant force for a fixed mass and inversely proportional to mass for a fixed force.
    • Keeps all quantities in consistent SI units and states the unit of the answer with its quantity.
    • Selects W = F × s and substitutes force in newtons and distance in metres to obtain work done in joules.
    • Measures the distance along the line of action of the force and explains that a force perpendicular to the motion does no work.
    • Describes the energy transfer from the source of the force to the object's kinetic or gravitational potential energy store.
    • Identifies friction or air resistance as transferring energy by heating to the surroundings and reducing useful energy.
    • States that one joule equals one newton metre and keeps units consistent throughout the calculation.
    • Selects the relationship that matches the situation, using mass, gravitational field strength and height for gravitational potential energy, mass and speed for kinetic energy, and force and distance for work done.
    • Squares the speed before multiplying by half the mass when calculating kinetic energy, so doubling the speed quadruples the value.
    • Converts between newton-metres and joules using the equality that one newton-metre equals one joule, keeping the numerical value unchanged.
    • Keeps units consistent throughout, converting grams to kilograms and centimetres to metres before substituting into a relationship.
    • States the final value with the correct unit, choosing joules or newton-metres as the question requires.
    • States that power is the rate at which energy is transferred, linking the quantity to energy transferred per unit time.
    • Uses a named example, such as two kettles of different power, to show that a higher power transfers the same energy in less time.
    • Compares two devices transferring the same total energy and explains that the higher-power device takes less time.
    • Explains that a higher-power device transfers more energy in the same time than a lower-power device.
    • Links the example back to the definition by referring to energy transferred per second rather than total energy alone.
    • States that the two forces in a third-law pair are equal in size and opposite in direction
    • Identifies that the paired forces act on two different objects, so they do not cancel
    • Explains equilibrium by balanced forces acting on the same object, such as the table's push and the book's weight both acting on the book
    • Applies the law to a non-equilibrium case, such as a rocket and its exhaust pushing on each other with equal and opposite forces
    • States that velocity is a vector quantity with both magnitude and direction.
    • Explains that the direction of motion of the object changes continuously as it moves round the circle.
    • Concludes that a changing direction with constant speed still gives a changing velocity.
    • Links the changing velocity to an acceleration directed towards the centre of the circle, caused by a resultant force towards the centre.
    Examiner Tips
    • 💡Write the relationship, then rearrange it before substituting numbers
    • 💡Convert all masses to kilograms and check each quantity carries the correct unit
    • 💡Estimate the answer first so an arithmetic slip is easy to spot
    • 💡Write the equation p = m × v first, then list the values with their units before substituting.
    • 💡Check every unit: kg for mass, m/s for velocity, kg m/s for momentum.
    • 💡For a direction question, state the direction explicitly as well as the size of the momentum.
    • 💡Write W = F × d and record force and distance with units before calculating.
    • 💡Check that the distance is measured along the direction of the force.
    • 💡Convert cm to m and kN to N before substituting to avoid unit errors.
    • 💡Write the equation, then substitute numbers with units before calculating so the unit of the answer is clear.
    • 💡Convert every time to seconds at the start, since the equation requires seconds.
    • 💡Sanity-check the answer: doubling the time for the same work should halve the power.
    • 💡Match each named interaction to a concrete everyday example so the distinction between contact and non-contact is clear.
    • 💡Group the five interactions into non-contact (electrostatics, gravity, magnetism) and contact (normal contact force, friction) to avoid omissions.
    • 💡Read the question carefully to see whether it asks for a contact or non-contact example before choosing.
    • 💡Name both objects explicitly in every answer, for example 'the hand' and 'the wall', rather than saying 'it pushes back'.
    • 💡State the size relationship and the direction relationship separately so both are visible to the examiner.
    • 💡Explain that the forces act on different objects whenever a question asks why equal and opposite forces do not cancel.
    • 💡Read the question to identify which single object the diagram should isolate before drawing any arrow.
    • 💡Check that every arrow is labelled and that its length is consistent with the stated scale.
    • 💡For a constant-velocity situation, confirm the arrows balance rather than adding a resultant.
    • 💡Name the body first, then list each force on it with its direction before judging balance
    • 💡State explicitly whether the resultant is zero or non-zero, then link that to the described motion
    • 💡Use the words uniform velocity, resultant force and balanced to match the command of the criterion
    • 💡Write the scale on the diagram before drawing any vector
    • 💡Use a sharp pencil, ruler and protractor so measured lengths and angles are reliable
    • 💡State whether the situation shows resolution, a net force or equilibrium, then justify it from the diagram
    • 💡Start by stating which object or system you are isolating, then list only the forces on it.
    • 💡Use the phrase 'resultant force' explicitly and link each change in resultant to a change in motion.
    • 💡For the vehicle example, name both the driving force and the resistive forces and say when they balance.
    • 💡Assign a positive direction first, then treat forces in the opposite direction as negative.
    • 💡Check whether the forces are equal and opposite before concluding the object is in equilibrium.
    • 💡Always give the resultant as a size with a stated direction.
    • 💡Name the object first, then list each force acting on it before drawing any arrow
    • 💡Check each pair of arrows for equal length and exactly opposite direction along the same line
    • 💡Link the zero resultant explicitly to the motion: at rest or constant velocity, never acceleration
    • 💡Use the phrase 'resultant force is zero' rather than 'no forces' when forces cancel
    • 💡Write the equation, then rearrange it before substituting numbers to reduce errors
    • 💡Convert all masses to kilograms and check every unit before calculating
    • 💡Show the substitution line so the method is visible even if the arithmetic slips
    • 💡Sanity-check the answer: doubling the force on the same mass should double the acceleration
    • 💡Write the definition as a ratio in words and as an equation before substituting numbers.
    • 💡Check that the force is the resultant force and that acceleration is in the same direction.
    • 💡Use unit analysis: N ÷ (m/s²) gives kg, which confirms the ratio is arranged correctly.
    • 💡Write down p = m × v and list each object's mass and velocity before substituting values.
    • 💡Choose a positive direction and mark opposing velocities as negative before summing momenta.
    • 💡Check that the total momentum after the collision equals the total before as a final verification.
    • 💡Write down the formula, then list the known values with units before substituting, so a unit slip is visible.
    • 💡Check whether the question gives a velocity change and a time rather than an acceleration, and combine formulae only when needed.
    • 💡Sanity-check the size and unit of your answer against the physical situation described.
    • 💡Identify the force and the displacement along its line of action before substituting into W = F × s.
    • 💡Write the energy transfer as a chain from one store to another, naming any heating to the surroundings.
    • 💡Check the unit of the answer is the joule and that the value is sensible for the force and distance given.
    • 💡Write down the relationship you are using before substituting numbers, so the examiner can see your method.
    • 💡Convert all quantities to SI units first, then substitute, to avoid unit slips.
    • 💡Check whether the answer is required in joules or newton-metres and label it accordingly.
    • 💡Sanity-check the size of your answer: a small mass lifted a modest height gives a few joules, not thousands.
    • 💡Begin your explanation with the definition, then support it with a specific example.
    • 💡Use a comparison of two devices so the rate idea is visible, for example same energy in different times.
    • 💡Refer to energy transferred per second to make the rate explicit.
    • 💡Name the device and the energy transfer in each example rather than speaking in general terms.
    • 💡Name the two objects involved in each force pair before deciding whether the forces cancel
    • 💡Check whether the forces in a question act on the same object or on different objects
    • 💡Use the phrase 'equal in size, opposite in direction, acting on different objects' when justifying a third-law answer
    • 💡Start by defining velocity as a vector, then explain the change of direction.
    • 💡Use the word 'direction' explicitly, since constant speed alone does not change velocity.
    • 💡Keep the explanation qualitative and link the changing velocity to an inward acceleration and resultant force.
    Common Mistakes
    • Substitutes mass in grams without converting to kilograms; divide grams by 1000 first
    • Uses one applied force when two forces oppose each other; find the resultant force first
    • Divides acceleration by mass instead of force by mass when finding acceleration; use force ÷ mass
    • Omits the unit of the answer; state newtons, m/s² or kilograms as appropriate
    • Adds mass and velocity instead of multiplying them; apply p = m × v.
    • Uses mass in grams without converting; divide grams by 1000 to get kilograms.
    • Drops the direction of momentum; state the direction of the velocity with the answer.
    • Leaves a velocity in km/h; convert to m/s by dividing by 3.6 before multiplying.
    • Uses a distance measured at right angles to the force; use the distance along the line of action of the force.
    • Leaves distance in centimetres; divide by 100 to get metres.
    • Adds force and distance instead of multiplying; apply W = F × d.
    • Gives work done in newtons or joules per metre; the unit of work done is the joule.
    • Multiplies work done by time instead of dividing; divide work done by time to obtain power.
    • Leaves a time in minutes without converting; multiply minutes by 60 to obtain seconds first.
    • Reports power in joules; state power in watts because a watt is one joule per second.
    • Confuses power with energy transferred; treat power as the rate of transferring energy or doing work.
    • Treats gravity as a contact force; recognise gravity as a non-contact interaction between masses.
    • Describes the normal contact force as parallel to the surface; state that it acts perpendicular to the surface.
    • Confuses friction with the normal contact force; keep friction along the surfaces opposing motion and the normal force perpendicular to them.
    • Omits one of the named interactions; check that electrostatics, gravity, magnetism, normal contact force and friction are all covered.
    • Treats a single object as producing a force on its own; identify the second object and the force it exerts back
    • Claims the two forces in the pair cancel because they are equal and opposite; state that they act on different objects
    • Reverses the direction of one force in the pair; check each force points from one object towards the other
    • Draws arrows for forces the object exerts on other bodies; show only forces acting on the chosen object.
    • Omits the direction of a force and gives only its size; state both magnitude and direction for a vector.
    • Assumes a moving object must show a resultant arrow; recognise that constant-velocity motion has balanced forces and zero resultant.
    • Draws arrows of arbitrary length with no consistent scale; use one scale so relative magnitudes are visible.
    • Lists forces the body exerts on other objects; keep only the forces acting on the chosen body
    • Treats a moving object as necessarily having a resultant force; recognise that steady motion needs zero resultant
    • Assumes a constant speed always means zero resultant force; check whether the direction changes, as in circular motion
    • Adds forces in the same direction when they oppose; subtract opposing forces to find the resultant
    • Draws vectors without a stated scale; label the scale so lengths can be converted to newtons
    • Resolves a force along directions that are not perpendicular; use perpendicular components only as the specification limits
    • Adds perpendicular vectors by simple arithmetic; combine them geometrically with a right-angled triangle
    • Leaves a vector diagram open when forces balance; close the shape to show zero resultant in equilibrium
    • Lists forces that act on other objects or on the surroundings rather than on the isolated object; restrict the list to forces acting on the chosen object or system.
    • Claims the skydiver has no forces acting at terminal velocity; state that weight and drag are equal and opposite, so the resultant is zero.
    • Says the resultant force is zero throughout the fall; explain that it is only zero once terminal velocity is reached.
    • Treats opening a parachute as instantly stopping the skydiver; describe the upward resultant causing deceleration to a new, lower terminal velocity.
    • Adds all arrow lengths regardless of direction; subtract opposing forces to find the resultant.
    • Assumes a diagram with several arrows must show balanced forces; check whether the opposing forces are equal before deciding.
    • Gives the resultant's size but omits its direction; state both magnitude and direction.
    • Reads arrow length as the force's direction; use the arrowhead for direction and the length for magnitude.
    • Includes forces that the object exerts on other bodies; show only the forces acting on the chosen object
    • Draws weight as a horizontal arrow; draw weight vertically downwards towards the centre of the Earth
    • Treats a moving object as necessarily having a resultant force; accept that constant velocity also corresponds to zero resultant force
    • Assumes balanced forces mean no forces act; recognise that several forces can act and still cancel
    • Draws arrows of arbitrary length with no relation to magnitude; scale arrow length to the size of each force
    • Substitutes a single applied force when two opposing forces act; calculate the resultant force first
    • Uses mass in grams directly in the equation; convert to kilograms by dividing by 1000
    • Multiplies force by mass when finding acceleration; divide force by mass instead
    • Omits the unit or gives the wrong unit with the answer; state N, kg or m/s² as appropriate
    • Treats acceleration as a force; recognise that acceleration is the effect produced by a resultant force
    • Confuses inertia with weight; keep inertia as resistance to a change in velocity and weight as the force of gravity on a mass.
    • Uses total applied force rather than resultant force in the ratio; subtract opposing forces first.
    • Inverts the ratio and divides acceleration by force; write m = F ÷ a and substitute force over acceleration.
    • Treats inertia as a force with a direction; describe it as a property of the object's mass instead.
    • Uses speed instead of velocity when defining momentum; state that momentum depends on velocity and therefore has direction.
    • Adds momenta of objects moving in opposite directions without signs; assign one direction positive and the other negative.
    • Applies conservation of momentum when an external resultant force acts; state that the law holds only when no external resultant force acts on the system.
    • Omits units or uses the wrong units; give momentum in kg m/s.
    • Uses the weight formula and treats mass in kilograms as a force in newtons; multiply mass by gravitational field strength to obtain weight in newtons before applying F = ma.
    • Divides by the time in the wrong place and mixes up velocity and acceleration; divide the velocity change by time to obtain acceleration, then multiply by mass to obtain force.
    • Ignores direction and treats a deceleration as a positive acceleration without comment; state the direction of the velocity change and the direction of the resultant force.
    • Uses the total distance travelled rather than the distance along the line of action of the force; resolve the force or measure the component of displacement in the force's direction.
    • Adds force and distance instead of multiplying them; multiply the force by the distance moved along its line of action.
    • Describes energy as used up rather than transferred; state which store energy moves from and which store it moves to.
    • Multiplies mass by speed without squaring it in a kinetic energy calculation; square the speed first, then multiply by half the mass.
    • Treats newton-metres and joules as different-sized units and applies a numerical factor; recognise that one newton-metre equals one joule, so the number stays the same.
    • Substitutes a height in centimetres or a mass in grams directly into a relationship; convert to metres and kilograms before calculating.
    • Uses the total distance travelled rather than the distance moved along the line of the force when finding work done; use the displacement along the force.
    • Describes power as the total amount of energy transferred; explain instead that power is the rate, that is energy transferred per unit time.
    • Assumes a higher-power device always transfers more total energy; recognise that it transfers energy faster, so total energy depends on time as well.
    • Gives an example without linking it to the definition; state the energy transfer and the time comparison, then connect them to the rate idea.
    • Confuses power with force or with work done; keep power as the rate of energy transfer and work done as the energy transferred by a force.
    • Claims that the table's push on a resting book cancels the book's weight as a third-law pair; identify instead that the third-law partner of the book's weight is the book's pull on the Earth
    • Treats equal and opposite third-law forces as producing no motion at all; recognise instead that each force acts on a different object, so each object can accelerate
    • Describes the two forces of a pair as acting on the same object; state instead that they act on two different bodies
    • Says the velocity is constant because the speed is constant; state instead that the changing direction changes the velocity.
    • Treats speed and velocity as the same quantity; distinguish instead that velocity includes direction while speed does not.
    • Claims a force pushes the object outwards from the circle; state instead that the resultant force acts towards the centre.