Scalars and vectors — AQA A-Level Physics
Teaching note checked against the official specification.
Scalars are physical quantities with magnitude only, such as mass, time, and temperature, while vectors have both magnitude and direction, such as displacement, velocity, and force. In AQA A-Level Physics, you must confidently add and resolve vectors using scale diagrams, Pythagoras, and trigonometry, and apply these skills to equilibrium and motion problems.
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Scalars and vectors explained
Scalars and vectors covers 10 specification points, including Nature of scalars and vectors.
Read the full explanation
and Velocity/speed, mass, force/weight,…. Open a point below for its explanation and revision guidance.
Nature of scalars and vectors.
A scalar is a physical quantity that has magnitude only, such as mass, time, speed, distance, energy or temperature. A vector has both magnitude and direction, such as displacement, velocity, acceleration, force, momentum or electric field strength. Direction may be given as an angle, a bearing, or a sign in one dimension. Scalars add by ordinary arithmetic, but vectors must be combined by vector addition, for example using a scale diagram or resolving into perpendicular components. For instance, walking 3 m east then 4 m north gives a distance of 7 m but a displacement of 5 m at 53° north of east. Recognising whether a quantity is scalar or vector is essential before choosing the correct method for a calculation.
Examples should include: velocity/speed, mass, force/weight, acceleration, displacement/distance.
A scalar has magnitude only; a vector has magnitude and direction. Speed is how fast something moves; velocity is speed in a stated direction, so a car at 20 m s⁻¹ north has a different velocity from one at 20 m s⁻¹ south. Mass is the quantity of matter in kg; weight is the gravitational force on it in N, W = mg, acting downwards. Distance is the total path length; displacement is the straight-line change in position with direction. Acceleration is the rate of change of velocity, a vector, so slowing down or turning changes it. Force is a push or pull with direction. Classify each quantity by asking whether direction is needed to describe it fully.
Addition of vectors by calculation or scale drawing.
To add vectors, account for direction. For perpendicular vectors, resolve into components and use Pythagoras: a 3 N force east plus a 4 N force north gives a resultant of 5 N at an angle found from tan θ = 4/3. For non-perpendicular vectors, use a scale drawing: choose a scale, draw each vector tip-to-tail at the correct angle, then measure the resultant length and direction. Alternatively resolve each vector into perpendicular components, add components, then recombine. The resultant is a single vector with the same effect as the original vectors. Always state magnitude with unit and direction relative to a stated reference.
Calculations will be limited to two vectors at right angles.
This statement defines the scope of vector calculations for perpendicular vectors. When combining two vectors at 90° to each other, you draw a right-angled triangle with the vectors placed tip-to-tail. You then use Pythagoras' theorem to find the magnitude of the resultant and an inverse tangent to find its direction. For example, a 3.0 N force north and a 4.0 N force east give a resultant of 5.0 N at 37° from the 4.0 N force. Resolving a single vector into two perpendicular components is the reverse process, using sine and cosine. While equilibrium problems can involve three coplanar forces, calculations to resolve them will rely on breaking them down into two perpendicular axes or using a closed vector triangle.
Scale drawings may involve vectors at angles other than 90°.
This statement extends vector work beyond right angles: when vectors act at any angle, you can find the resultant by drawing them to scale rather than by right-angled trigonometry alone. Choose a scale such as 1 cm = 2 N, draw each vector tip-to-tail at its correct angle using a protractor, then measure the length and direction of the line from start to finish. For example, a 6.0 N force at 0° and a 5.0 N force at 60° can be drawn to scale and the resultant measured, then converted back using the scale. Calculations may also use perpendicular components, but the scale drawing itself is the method assessed here. Accuracy depends on a sharp pencil, a sensible scale and careful measurement.
Resolution of vectors into two components at right angles to each other.
Resolution replaces one vector by two perpendicular vectors whose combined effect is identical. Choose axes, usually horizontal and vertical or along and perpendicular to a surface, then project the vector onto each axis using trigonometry. For a vector F at angle θ to the x-axis, the x-component is F cos θ and the y-component is F sin θ. Check with Pythagoras: Fx² + Fy² = F². For example, a 50 N force at 30° above the horizontal gives 50 cos 30° ≈ 43 N horizontally and 50 sin 30° = 25 N vertically. The components are not extra forces; they are a mathematical replacement that simplifies adding vectors and applying Newton's laws along each axis independently.
Examples should include components of forces along and perpendicular to an inclined plane.
On an inclined plane, choose axes parallel and perpendicular to the slope rather than horizontal and vertical. For a plane at angle θ to the horizontal, the weight W of an object on the slope resolves into W sin θ down the slope and W cos θ perpendicular into the slope. The angle between the weight and the perpendicular to the slope equals θ, which is why sine acts along the slope and cosine acts into it. For example, a 20 N weight on a 25° slope gives 20 sin 25° ≈ 8.5 N down the slope and 20 cos 25° ≈ 18 N perpendicular to the slope. The perpendicular component is balanced by the normal contact force when there is no acceleration perpendicular to the surface, while the parallel component drives motion or is balanced by friction.
Problems may be solved either by the use of resolved forces or the use of a closed triangle.
When several coplanar forces act at a point, you can analyse equilibrium in two equivalent ways. Resolving splits every force into perpendicular components, usually horizontal and vertical, then applies ΣFₓ = 0 and ΣF_y = 0. The closed-triangle method draws the forces tip-to-tail; if they form a closed triangle, the resultant is zero. For example, a 50 N weight on a string pulled sideways by a horizontal force with the string at 30° to the vertical: resolving gives T cos 30° = 50 N and T sin 30° = F, so T ≈ 57.7 N and F ≈ 28.9 N. A closed triangle with the same three forces gives the same values by trigonometry. Choose whichever method suits the geometry, but show the full vector diagram or component equations so the reasoning is clear.
Conditions for equilibrium for two or three coplanar forces acting at a point.
A point is in equilibrium when the resultant force on it is zero. For two coplanar forces, this requires equal magnitudes acting in exactly opposite directions along the same line. For three coplanar forces acting at a point, equilibrium requires the vector sum to be zero, which can be tested by resolving in two perpendicular directions and applying ΣFₓ = 0 and ΣF_y = 0, or by drawing the three forces tip-to-tail to form a closed triangle. For example, a lamp of weight 30 N hangs from two strings each at 40° to the horizontal: resolving vertically gives 2T sin 40° = 30 N, so T ≈ 23.3 N. The three-force condition also means the lines of action of the forces must be concurrent if the body is to have no resultant moment.
Appreciation of the meaning of equilibrium in the context of an object at rest or moving with constant velocity.
Equilibrium means the resultant force on an object is zero, so its acceleration is zero. This occurs when the object is at rest or moving with constant velocity. For example, a car travelling at a steady 20 m s⁻¹ along a straight road has driving force balanced by drag and friction, and weight balanced by the normal contact force. A book resting on a table has weight balanced by the normal contact force. In both cases, the vector sum of all forces is zero. Equilibrium does not require the object to be stationary; constant velocity is equally valid. To test equilibrium, resolve forces in two perpendicular directions and check that each component sum is zero, or draw a closed vector triangle. If the resultant is non-zero, the object accelerates in the direction of the resultant force.
Your focus
- Define scalar and vector quantities and classify given physical quantities correctly.
- Add two perpendicular vectors to find a resultant magnitude and direction.
- Explain why direction must be considered when combining vector quantities but not scalar quantities.
Show all 31 objectives
- Classify given quantities as scalar or vector with a reason.
- Distinguish velocity from speed and displacement from distance in written definitions.
- Distinguish mass from weight and state the direction of weight.
- Add two perpendicular vectors by calculation to find magnitude and direction.
- Construct a scale drawing to add non-perpendicular vectors and measure the resultant.
- Resolve vectors into components and recombine them to find a resultant.
- Draw a right-angled vector triangle for two perpendicular vectors.
- Calculate the magnitude of a resultant using Pythagoras' theorem.
- Calculate the direction of a resultant using an inverse tangent ratio.
- Resolve a vector into two perpendicular components.
- Draw vectors to scale tip-to-tail at angles other than 90°.
- Measure the magnitude of a resultant from a scale drawing and convert it using the scale.
- Measure the direction of a resultant relative to a named reference direction.
- Resolve a given vector into two perpendicular components using the correct trigonometric ratios.
- Select appropriate perpendicular axes for a described physical situation.
- Verify resolved components by checking that their squares sum to the square of the original magnitude.
- Resolve the weight of an object on an inclined plane into components parallel and perpendicular to the plane.
- Relate the perpendicular component of weight to the normal contact force in equilibrium.
- Apply resolved components along slope axes to determine acceleration or unknown forces.
- Resolve a set of coplanar forces into perpendicular components and apply the conditions for zero resultant.
- Construct a closed vector triangle for three coplanar forces acting at a point and solve it using trigonometry.
- Compare resolved-force and closed-triangle solutions and justify which method is more efficient for a given arrangement.
- State and apply the conditions for equilibrium of two or three coplanar forces acting at a point.
- Resolve forces in two perpendicular directions and solve the resulting equations for unknown magnitudes or angles.
- Use a closed vector triangle to verify equilibrium and to calculate an unknown force or angle.
- Define equilibrium as the state in which the resultant force on an object is zero.
- Describe examples of objects at rest and objects moving with constant velocity in equilibrium.
- Use vector diagrams or resolution to demonstrate that forces are balanced in equilibrium.
Which command word is this?
Give a precise definition, including the nature of the quantity (scalar or vector) and its units if applicable. For a vector, state that it has both magnitude and direction.
Scalars and vectors exam tips
Topic Overview
Scalars and vectors are fundamental to all of physics. Scalars are quantities that have magnitude only, such as mass, time, speed, and energy. Vectors have both magnitude and direction, such as displacement, velocity, acceleration, force, and momentum. Understanding the distinction is crucial because many physical laws, like Newton's second law, are vector equations.
In AQA A-Level Physics, you will use vector addition and resolution to solve problems in mechanics, fields, and electricity. You need to be able to add vectors graphically and by calculation, resolve vectors into perpendicular components, and apply these skills to equilibrium and motion. This topic underpins much of the rest of the course, so a solid grasp is essential.
Key Concepts
- →Scalars have magnitude only; vectors have magnitude and direction. Examples of scalars: mass, time, speed, distance, energy, temperature. Examples of vectors: displacement, velocity, acceleration, force, momentum, electric field strength.
- →Vectors can be represented by arrows: length represents magnitude, direction represents direction. The resultant of two or more vectors is the single vector that has the same effect.
- →Vector addition: for perpendicular vectors, use Pythagoras' theorem to find magnitude and trigonometry to find direction. For non-perpendicular vectors, resolve into components or use the parallelogram rule.
- →Resolving vectors: any vector can be split into two perpendicular components, usually horizontal and vertical, using Fx = F cos(theta) and Fy = F sin(theta), where theta is the angle to the horizontal.
- →Equilibrium: when the resultant force on an object is zero, it is in equilibrium. This means the vector sum of all forces is zero, and the sum of components in any direction is zero.
Marking Points
- A scalar quantity has magnitude only and no direction.
- A vector quantity has both magnitude and direction.
- Examples of scalars include mass, time, speed, distance, energy and temperature.
- Examples of vectors include displacement, velocity, acceleration, force, momentum and electric field strength.
- Vectors are added by taking direction into account, for example by drawing a vector triangle or by resolving into components.
- Scalars are added using ordinary arithmetic because direction is not involved.
- In one dimension, direction can be represented by a positive or negative sign.
- States that a scalar has magnitude only, while a vector has both magnitude and direction.
- Classifies speed, mass and distance as scalars, and velocity, force/weight, acceleration and displacement as vectors.
- Explains that velocity is speed in a stated direction, so direction must be given.
- Distinguishes mass in kg from weight in N, linking weight to W = mg acting downwards.
- Distinguishes distance as total path length from displacement as straight-line change in position with direction.
- Describes acceleration as the rate of change of velocity, so it is a vector even when speed is constant but direction changes.
- Draws vectors tip-to-tail at the correct angles using a stated scale.
- Measures the resultant length and converts it using the scale to find magnitude.
- Measures the angle of the resultant relative to a stated reference direction.
- Uses Pythagoras for perpendicular vectors and trigonometry to find the resultant direction.
- Resolves vectors into perpendicular components, adds components, then recombines to find the resultant.
- States the final answer as a magnitude with unit and a direction, for example 5.0 N at 53° north of east.
- State that the two vectors must be perpendicular to form a right-angled triangle.
- Use Pythagoras' theorem to calculate the magnitude of the resultant from two perpendicular vectors.
- Use an inverse tangent ratio to calculate the angle giving the direction of the resultant relative to one of the vectors.
- Resolve a single vector into two perpendicular components using sine and cosine.
- For three coplanar forces in equilibrium, resolve them into two perpendicular axes to simplify calculations.
- Choose a scale that makes the drawing large enough to measure accurately and state it clearly.
- Draw each vector tip-to-tail at the correct angle using a protractor, with arrows showing direction.
- Measure the length of the resultant from the tail of the first vector to the tip of the last vector.
- Convert the measured length back to the physical quantity using the stated scale.
- Measure the direction of the resultant with a protractor relative to a named reference direction.
- Use perpendicular components as an alternative check when the angle between vectors is not 90°.
- State that resolution replaces a single vector by two perpendicular components with the same combined effect as the original vector.
- Select suitable perpendicular axes for the physical situation, such as horizontal and vertical or parallel and perpendicular to a slope.
- Identify the angle between the vector and the chosen axis and apply the correct trigonometric projection, for example F cos θ along the axis and F sin θ perpendicular to it.
- Check the result using Fx² + Fy² = F² and confirm that each component is smaller than the original vector unless the vector lies along an axis.
- Use resolved components to sum forces or analyse equilibrium along each axis separately, keeping directions consistent with a stated positive sense.
- Draw a clear free-body diagram showing the weight acting vertically downwards and mark the angle θ between the plane and the horizontal.
- Resolve the weight parallel to the plane as W sin θ and perpendicular to the plane as W cos θ, explaining that the angle between the weight and the perpendicular to the plane is θ.
- Identify the normal contact force as perpendicular to the surface and, in the absence of other perpendicular forces, equate it to W cos θ.
- Resolve any applied force or friction along the same parallel and perpendicular axes before combining components.
- Apply Newton's second law along each axis separately, using the parallel component to find acceleration or the balance of forces for equilibrium.
- State that a body in equilibrium has zero resultant force, so the vector sum of all forces acting at the point is zero.
- Resolve each force into two perpendicular directions, commonly horizontal and vertical, using F cos θ and F sin θ with θ measured from the chosen axis.
- Apply ΣFₓ = 0 and ΣF_y = 0 simultaneously to form solvable equations for unknown magnitudes or angles.
- Draw a closed vector triangle with forces placed tip-to-tail, ensuring each arrow direction represents the force acting on the point.
- Use sine rule, cosine rule or right-angled trigonometry on the closed triangle to determine an unknown force or angle.
- Check that the answer is physically sensible, including units, direction and relative size compared with the given forces.
- State that equilibrium at a point means the resultant force is zero, so the vector sum of the forces is zero.
- For two forces, state that they must be equal in magnitude and opposite in direction along the same straight line.
- For three coplanar forces, resolve in two perpendicular directions and set the sum of components in each direction to zero.
- Draw the three forces tip-to-tail to form a closed triangle, showing that the resultant is zero.
- Use the closed triangle with trigonometry, including the sine rule or cosine rule, to find an unknown force or angle.
- Recognise that for three non-parallel coplanar forces in equilibrium their lines of action must meet at a common point.
- State that equilibrium occurs when the resultant force on an object is zero.
- Explain that zero resultant force means zero acceleration, so the object remains at rest or continues moving with constant velocity.
- Give a concrete example, such as a car moving at constant velocity with driving force balanced by resistive forces and weight balanced by the normal contact force.
- Use vector addition or resolution to show that the sum of forces in two perpendicular directions is zero.
- Distinguish equilibrium from a stationary state by noting that constant velocity is also equilibrium.
- Apply Newton's first law to link zero resultant force to constant velocity or rest.
Examiner Tips
- 💡When asked to classify a quantity, state whether it has direction as well as magnitude, and give a brief reason.
- 💡For vector addition, always state both the magnitude and direction of the resultant, using an angle or bearing as appropriate.
- 💡Use a clear scale diagram or resolve into perpendicular components; show the direction of each component with an arrow or sign.
- 💡When asked to classify a quantity, write scalar or vector and justify it by referring to whether direction is needed.
- 💡Use correct units and directions in answers, such as 9.81 m s⁻² downwards for acceleration of free fall.
- 💡If a question gives a speed without direction, state that it is a scalar and that velocity cannot be fully stated without a direction.
- 💡Choose a scale that makes the drawing large enough to measure angles accurately, and state the scale clearly.
- 💡Show the tip-to-tail construction and label the resultant with its direction; examiners can then follow your method.
- 💡For perpendicular vectors, calculate rather than draw, and check the resultant is larger than either component but smaller than their sum.
- 💡Sketch the two vectors tip-to-tail and mark the right angle before substituting numbers.
- 💡Check that the resultant magnitude lies between the larger vector and the arithmetic sum of the two vectors.
- 💡State the scale before drawing so the examiner can follow the conversion from length to magnitude.
- 💡Use a sharp pencil and a protractor, and measure the resultant line as precisely as the scale allows.
- 💡Label the resultant with its arrow and give its direction relative to a clearly named direction.
- 💡Sketch the vector and both axes before substituting numbers; label the angle you are using so the sine or cosine choice is visible.
- 💡Keep at least three significant figures through intermediate steps and round only the final answer to the precision justified by the data.
- 💡After calculating components, perform a quick Pythagoras check to catch a swapped sine and cosine.
- 💡Write the resolved components with their directions, for example 43 N horizontally and 25 N vertically, rather than giving bare numbers.
- 💡Mark the angle θ at the slope and again where it appears between the weight and the perpendicular to the slope; this makes the sine and cosine assignment clear.
- 💡Use the limiting cases θ = 0° and θ = 90° to check that your expressions behave sensibly.
- 💡State the direction of each component, such as down the slope or perpendicular into the slope, so the physics remains clear.
- 💡When friction is present, resolve it along the slope and include its sign relative to the chosen positive direction.
- 💡Sketch the forces at the point and label every angle relative to a clearly chosen axis before writing any equation.
- 💡If the geometry is a right-angled triangle, resolving is usually quicker; if three forces form a scalene triangle, the sine rule or cosine rule may be more efficient.
- 💡Keep full calculator precision through intermediate steps and round only the final answer to an appropriate number of significant figures.
- 💡Draw a free-body diagram of the point, showing each force with its correct direction and labelling unknown magnitudes clearly.
- 💡Choose axes that simplify the algebra, for example horizontal and vertical, and mark each angle from the chosen axis before resolving.
- 💡Check the final answer by substituting back into both component equations or by confirming the vector triangle closes.
- 💡When asked to explain equilibrium, always start from the resultant force being zero and link this to zero acceleration.
- 💡Use a free-body diagram to identify all forces and show balanced pairs or resolved components.
- 💡If a question gives a velocity-time graph, constant velocity is shown by a horizontal line, which indicates equilibrium.
- 💡Avoid saying 'no forces act' when describing equilibrium; say 'no resultant force acts'.
- 💡Always draw a clear diagram for vector problems. Label all vectors with arrows and angles. This helps you visualise the problem and can earn method marks even if your calculation goes wrong.
- 💡When adding vectors, if they are not perpendicular, resolve them into components first. Do not simply add magnitudes. For equilibrium problems, set the sum of components in each direction to zero.
- 💡Pay attention to significant figures and units. AQA often awards marks for correct units and appropriate precision. In calculations, keep full precision until the final answer, then round to the required number of significant figures.
Common Mistakes
- Treating speed as a vector: speed is scalar because it has magnitude only, whereas velocity is a vector because it includes direction. Correction: use velocity when direction matters.
- Adding vector magnitudes arithmetically without considering direction: for example, adding 3 N east and 4 N north as 7 N is incorrect. Correction: use a vector triangle or components to find the resultant magnitude and direction.
- Confusing distance with displacement: distance is the total path length (scalar), while displacement is the straight-line distance in a specified direction (vector). Correction: state the direction when giving displacement.
- Treating speed and velocity as interchangeable: correct by stating that velocity requires a direction, for example 15 m s⁻¹ east.
- Calling weight a scalar because it is measured in newtons: correct by noting weight is a force and therefore has direction, acting towards the centre of the Earth.
- Saying displacement is the same as distance travelled: correct by defining displacement as the straight-line distance and direction from start to finish.
- Drawing vectors tail-to-tail and measuring between the tails: correct by drawing them tip-to-tail so the resultant runs from the start of the first vector to the tip of the last.
- Forgetting to convert a measured drawing length back using the scale: correct by multiplying the measured length by the scale factor.
- Giving only a magnitude for a vector sum: correct by also stating the direction relative to a clear reference line.
- Adding the two magnitudes arithmetically instead of combining them as perpendicular vectors; correct this by using Pythagoras' theorem so the resultant is larger than either vector but smaller than their sum.
- Using the wrong trigonometric ratio when finding the angle; correct this by identifying which sides are opposite and adjacent to the required angle before choosing tangent.
- Forgetting to state the direction of the resultant; correct this by giving the angle relative to a named vector or a compass direction.
- Drawing vectors tail-to-tail instead of tip-to-tail; correct this by placing the tail of each new vector at the tip of the previous one.
- Using a scale that makes the diagram too small to measure; correct this by choosing a scale that gives a drawing of a convenient size.
- Measuring the angle from an unstated direction; correct this by naming the reference direction, such as the horizontal or the first vector.
- Swapping sine and cosine because the wrong angle was identified: always mark the angle between the vector and the axis, then use cosine for the component along that axis and sine for the perpendicular component.
- Treating the components as additional real forces acting alongside the original vector: state clearly that the components replace the original vector for calculation purposes.
- Forgetting that a component can be negative when it points in the negative direction of the chosen axis: assign a positive sense first and include the sign in later calculations.
- Using W cos θ down the slope and W sin θ into the slope: check by imagining θ = 0°, where the parallel component must be zero and the perpendicular component must equal the full weight.
- Assuming the normal contact force always equals the weight: on a slope it equals W cos θ only when no other force acts perpendicular to the surface.
- Mixing horizontal and vertical axes with slope axes in the same calculation: choose one pair of perpendicular axes and resolve every force onto those axes consistently.
- Using the wrong trigonometric ratio for a component: the component along an axis is F cos θ when θ is measured from that axis, and F sin θ when θ is measured from the perpendicular axis. Always mark θ on the diagram before choosing.
- Drawing the closed triangle with arrows tail-to-tail rather than tip-to-tail; the corrected diagram has each force arrow starting where the previous one ends, and the triangle closes only when the resultant is zero.
- Forgetting that the two resolution equations must be solved together; substituting a value from one equation into the other, or ignoring one direction, loses the link between the unknowns.
- Assuming three forces are in equilibrium whenever they form any triangle; the corrected condition is a closed triangle with the force arrows arranged tip-to-tail so the resultant is zero.
- Treating the weight of a suspended object as acting along the string rather than vertically downwards; the corrected free-body diagram shows weight acting vertically downwards from the point.
- Balancing forces in only one direction and ignoring the perpendicular direction; the corrected method applies both ΣFₓ = 0 and ΣF_y = 0, or closes the vector triangle fully.
- Thinking that equilibrium only applies to objects at rest; correct this by stating that constant velocity also satisfies equilibrium because acceleration is zero.
- Assuming that if an object is moving, there must be a resultant force in the direction of motion; correct this by explaining that a resultant force would cause acceleration, not constant velocity.
- Confusing balanced forces with equal magnitudes only; correct this by emphasising that forces must be equal in magnitude and opposite in direction, and may need to be resolved into components.
- Forgetting to consider all forces, such as normal contact force or friction; correct this by drawing a free-body diagram and labelling every force.
- Students often think that a scalar is just a vector without direction, but scalars are fundamentally different: they combine by ordinary arithmetic, while vectors combine by vector addition. For example, adding two displacements of 3 m and 4 m can give a resultant between 1 m and 7 m depending on direction.
- Many students confuse speed and velocity, or distance and displacement. Speed is the rate of change of distance (scalar), while velocity is the rate of change of displacement (vector). A car going around a roundabout at constant speed has changing velocity because its direction changes.
- Students sometimes forget that when resolving a vector, the component adjacent to the angle uses cosine and the opposite uses sine. A common error is to swap sin and cos, leading to incorrect components.
Revision Plan
- 1Day 1-2: Learn definitions and examples of scalars and vectors. Make a table of all quantities in the AQA specification and classify them. Test yourself with flashcards.
- 2Day 3-4: Practice vector addition using scale diagrams and calculation for perpendicular vectors. Use past paper questions on displacement and velocity.
- 3Day 5-6: Master resolving vectors into components. Practice with forces at angles, including inclined planes. Watch tutorial videos and attempt textbook problems.
- 4Day 7-8: Apply vector resolution to equilibrium problems. Solve problems involving three forces in equilibrium, such as a hanging mass or a charged particle in fields.
- 5Day 9-10: Complete a full past paper section on scalars and vectors. Review mark schemes carefully, noting where you lost marks and why. Redo any incorrect questions.
Exam Question Types
- 📋Multiple choice questions testing scalar/vector classification and simple vector addition. Advice: read each option carefully; eliminate obviously wrong answers.
- 📋Calculation questions requiring resolution of vectors or finding a resultant. Advice: draw a diagram, show all steps, and include units and direction.
- 📋Structured questions on equilibrium, often involving three forces. Advice: resolve forces horizontally and vertically, set sums to zero, and solve simultaneous equations if needed.
- 📋Practical-based questions on vector addition using force boards or inclined planes. Advice: describe the method accurately, including how you ensured forces were in equilibrium.
Command Word Expectations (AQA)
Give a precise definition, including the nature of the quantity (scalar or vector) and its units if applicable. For a vector, state that it has both magnitude and direction.
Show your working, including the formula used, substitution of values, and the final answer with correct units and significant figures. For vectors, include direction.
Provide a reasoned account, linking cause and effect. Use correct terminology and refer to physical principles. For example, explain why a vector quantity changes even if magnitude is constant.
How Students Lose Marks (Examiner Pitfalls)
Step-by-Step Worked Solutions
Question: A hiker walks 5.0 km due east, then 3.0 km due north. Calculate the magnitude and direction of the resultant displacement.
- 1.Step 1: Identify given facts: displacement 1 = 5.0 km east, displacement 2 = 3.0 km north. These are perpendicular vectors.
- 2.Step 2: Apply Pythagoras' theorem to find the magnitude of the resultant: R = sqrt(5.0^2 + 3.0^2) = sqrt(25 + 9) = sqrt(34) = 5.83 km.
- 3.Step 3: Determine the direction using trigonometry: tan(theta) = opposite/adjacent = 3.0/5.0 = 0.6, so theta = arctan(0.6) = 31 degrees north of east.
- 4.Step 4: State final conclusion with units: The resultant displacement is 5.8 km at 31 degrees north of east.
Question: A force of 20 N acts at 30 degrees above the horizontal. Resolve this force into horizontal and vertical components.
- 1.Step 1: Identify given facts: force F = 20 N, angle theta = 30 degrees to the horizontal.
- 2.Step 2: Apply trigonometric resolution: horizontal component Fx = F cos(theta) = 20 cos(30) = 20 * 0.866 = 17.3 N.
- 3.Step 3: Vertical component Fy = F sin(theta) = 20 sin(30) = 20 * 0.5 = 10 N.
- 4.Step 4: State final conclusion with units: The horizontal component is 17.3 N and the vertical component is 10 N.