GCSE Physics resources
AQA GCSE Forces
Choose one object before drawing any arrows. A force arrow identifies an interaction acting on that object, while a motion arrow describes its velocity.
Subsection scope: Scalars, vectors and collinear resultants are shared at both tiers. Free-body diagrams and resolving/vector constructions are Higher content.
Start here
Choose one object before drawing any arrows. A force arrow identifies an interaction acting on that object, while a motion arrow describes its velocity.
Before you revise
Choose an answer from memory. Your result tells you what to watch for in the guide.
Core revision guide
Choose one object before drawing any arrows. A force arrow identifies an interaction acting on that object, while a motion arrow describes its velocity.
Scope: 8463 and 8464 · Foundation and Higher
Mass, time, distance, speed and energy have magnitude only. Displacement, velocity, acceleration, momentum and force have magnitude and direction. Forces arise from interactions: weight is gravitational, a support provides a contact force and air provides drag. For opposite collinear forces, subtract the smaller from the larger and state the larger force’s direction.
Scope: 8463 and 8464 · Higher only
Isolate the chosen body, label weight vertically down and contact/tension/thrust/drag as appropriate. Arrow length represents magnitude. A book at rest on a horizontal table has equal upward contact force and downward weight. These two forces act on the book and balance, but are not a third-law pair: their reaction forces act on Earth and the table respectively.
Scope: 8463 and 8464 · Foundation and Higher
Mass is measured in kg and remains the same when the gravitational field changes. Weight W = mg is a force in N. Zero resultant force implies zero acceleration: an object can be stationary or move with constant velocity. Balanced does not mean every force is zero.
Scope: 8463 and 8464 · Higher only
For a Higher scale construction, choose 1 cm = 1 N. Draw 4 N east as a 4 cm arrow, then from its head draw 3 N north as a 3 cm arrow. The resultant runs from the first tail to the last head: measuring gives 5 cm, hence 5 N, about 37° north of east. A parallelogram built from two arrows with a common tail gives the same diagonal. To represent equilibrium, add a 5 N arrow in the opposite direction to close the triangle: a closed head-to-tail polygon has zero resultant. Resolve the 5 N arrow by replacing it with perpendicular 4 N east and 3 N north components. Components represent the same effect as the original vector; do not add the original arrow as a third force. Use a ruler and protractor rather than requiring trigonometry.
Misconception clinic
Read each belief, say what is wrong with it, then compare your correction.
Tempting idea: The book’s push on the table and the table’s push on the book.
Use instead: The book’s push acts on a different body. Both act on the chosen book and balance.
Tempting idea: 20 N right.
Use instead: Opposite forces subtract. 12 − 8 = 4 towards the larger force.
Retrieval practice
Worked examples
Apply scalars, vectors and interactions in the worked reasoning, then try the changed-context questions below.
Scope: 8463 and 8464 · Foundation and Higher
600 N forwards.
Scope: 8463 and 8464 · Higher only
Resultant 5 N at about 37° north of east; equilibrium requires 5 N in the opposite direction. Resolving this resultant gives the original 4 N and 3 N components.
Scope: 8463 and 8464 · Foundation and Higher
19.6 N.
Exam precision
Do this: For a vector construction, state the scale, measure resultant length and direction, and keep arrowheads consistent. Replace a vector with components without double-counting it.
Independent practice
Write an answer before opening the marking guidance. Then edit the exact phrase or step that would gain the next mark.
Scope: 8463 and 8464 · Foundation and Higher
Original PhysicsUK mark scheme: Point 1 (1 mark): The horizontal resultant is zero. Point 2 (1 mark): Driving and total resistive forces are equal and opposite. Award each distinct point once. Accept equivalent scientifically correct wording. No half marks.
Model answer: At constant velocity, zero resultant means driving and resistive forces balance.
Scope: 8463 and 8464 · Foundation and Higher
Original PhysicsUK mark scheme: Point 1 (1 mark): W = mg. Point 2 (1 mark): W = 2.0 × 9.8. Point 3 (1 mark): W = 19.6 N (20 N to two significant figures). Award each distinct point once. Accept equivalent scientifically correct wording. No half marks.
Model answer: W = mg. W = 2.0 × 9.8. W = 19.6 N (20 N to two significant figures).
Acceptable answers: Expected result: W = 19.6 N (20 N to two significant figures). Accept the unrounded value or a correctly rounded equivalent to two or more significant figures; accept equivalent units if converted correctly. No additional measurement interval is implied.
Marking notes: Method marks remain available after an arithmetic error. Carry an earlier numerical error into a later valid step; award that method point if the unit conversion and physics are correct. A fully correct answer with unit earns all marks unless working is expressly requested.
Scope: 8463 and 8464 · Higher only
Original PhysicsUK mark scheme: Point 1 (1 mark): Construct 3 cm east and 4 cm north head-to-tail (accept a correct scaled parallelogram). Point 2 (1 mark): Measure a 5 cm resultant: 10 N, about 53° north of east. Point 3 (1 mark): Draw an equal opposite 10 N vector closing the triangle, about 53° south of west. Point 4 (1 mark): The 6 N east and 8 N north components replace the resultant with the same effect; do not also add the original resultant as another force. Award each distinct point once. Accept equivalent scientifically correct wording. No half marks.
Model answer: Draw 3 cm east and then 4 cm north. The arrow from the first tail to the final head is 5 cm long at about 53° north of east, representing 10 N. An equal opposite 10 N arrow closes the triangle for equilibrium. The perpendicular 6 N and 8 N components replace this vector with the same effect; they are not extra forces to add alongside it.
Acceptable answers: For the prescribed ruler/protractor construction accept 9.6–10.4 N and 50–56° north of east, or equivalent bearing. The equilibrium arrow must oppose the candidate’s correctly constructed resultant; allow a carried measurement error.
Marking notes: Award the construction point independently of small reading errors. Use the stated scale. Resolution is replacement, not double-counting the vector.
Questions pupils ask
Yes. Zero resultant means zero acceleration, so straight motion at constant velocity can continue.
Useful next steps