Vectors

Learning goals

What students should be able to explain, calculate and investigate after completing Full Vectors.

Goals marked Higher Level are required only for Higher Level students; the remaining goals are for all students.

• Higher Level: Resolve a vector into two perpendicular components.
• Higher Level: Calculate the resultant of two vectors using components (not just scale diagrams).
• Higher Level: Apply vector resolution to resultant force problems and to two-dimensional collisions.
• Higher Level: Verify the law of addition of vectors using primary and secondary data.

Key definitions

Precise terms, symbols and SI units used in Full Vectors.

Resolving a Vector into Components

Any vector can be replaced by two perpendicular components without changing its physical effect.

Resolving replaces one vector by two perpendicular components that have exactly the same combined effect. If angle θ is measured from the positive horizontal direction, the adjacent component is A cos θ and the opposite component is A sin θ. Signs must follow the chosen axes: left and downward components are commonly negative.

Finding a Resultant Using Components

To add two or more vectors that are not simply along the same line, first resolve each into x- and y-components.

Resolve every vector along the same pair of axes, then add the horizontal components and the vertical components separately. Use Pythagoras for the magnitude and inverse tangent for the angle. Check the signs on a sketch and state the direction clearly, such as north of east.

Resultant Force

When several forces act on an object, their vector sum is the resultant force.

The resultant force is the vector sum of all forces on one object. It is the resultant—not any single force—that equals ma. If the resultant is zero, acceleration is zero, but the object may be at rest or moving with constant velocity. A free-body diagram should contain only forces acting on the chosen object.

F_resultant,x is the combined effect of all force components in the horizontal direction, and F_resultant,y is the combined effect in the vertical direction. Combine opposite directions using signs, then apply Newton’s Second Law separately in each direction.

Two-Dimensional Collisions

For a two-dimensional collision, total momentum is conserved independently in the x- and y-directions.

Resolve each momentum vector into x- and y-components, then apply conservation of momentum separately in each direction. For momentum p at angle θ to the positive x-axis, p_x = p cos θ and p_y = p sin θ. Since p = mv, these can also be written as p_x = mv cos θ and p_y = mv sin θ. Use signed components. Kinetic energy is conserved only in an elastic collision and must not be assumed.

Diagrams and formulae

Key diagrams and equations for reviewing Full Vectors.

• Resolving a vector F at angle θ into perpendicular components: Fx = F cos θ and Fy = F sin θ.
• Finding a resultant using the tip-to-tail (triangle) method, which can be confirmed by calculation using components.
• A two-dimensional collision: momentum is conserved separately along each of two perpendicular directions.

Resolving a vector F at angle θ into perpendicular components: Fx = F cos θ and Fy = F sin θ.

Finding a resultant using the tip-to-tail (triangle) method, which can be confirmed by calculation using components.

A two-dimensional collision: momentum is conserved separately along each of two perpendicular directions.

• Key relationships
• Components: Fx = F cos θ, Fy = F sin θ Resultant magnitude: R = √(Rx² + Ry²)

Worked Example

A step-by-step calculation applying the central relationships in Full Vectors.

Two forces act on an object: 8 N due east, and 6 N due north. Find the resultant force.

• These two forces are already perpendicular, so: Rx = 8 N, Ry = 6 N
• R = √(8² + 6²) = √(64+36) = √100 = 10 N
• θ = tan⁻¹(6/8) = tan⁻¹(0.75) ≈ 36.9° north of east

Practical / Investigation

Experimental planning, measurement and evaluation for Full Vectors.

Experiment: Verifying the law of addition of vectors (force table / parallelogram of forces)

• Apparatus: force table (or a board with pulleys), slotted masses, string, protractor.
• Method: apply two known forces (via hanging masses over pulleys) at measured angles to an object; measure the angle and magnitude of a third force needed to keep the object in equilibrium; this third force is equal and opposite to the resultant of the first two.
• Verify that the calculated resultant (by components, or by scale diagram) matches the measured balancing force in both magnitude and direction, within experimental error.
• Sources of error: friction in pulleys; parallax error reading angles; strings not perfectly straight/frictionless.
• Improvement: use low-friction pulleys; read angles and lengths carefully at eye level.

• Arrange two forces along the same straight line using a trolley or ring, string, pulleys and known hanging masses.
• For forces in the same direction, compare the measured balancing force with F₁ + F₂. For opposite directions, compare it with the magnitude of F₁ − F₂ and record the resultant direction.
• Repeat with several force pairs. A balancing force equal and opposite to the calculated resultant supports the law of vector addition in one dimension.

• Record each hanging mass and pulley angle; convert mass to force using F = mg.
• At equilibrium, the third force is the equilibrant: equal in magnitude and opposite in direction to the resultant of the other two.
• Centre the ring without contact, minimise pulley friction and compare measured and calculated magnitude and direction.

Quick-Check Questions

Short retrieval questions covering the essential ideas in Full Vectors.

1. Write the formulas for the x- and y-components of a vector of magnitude F at angle θ.

2. How do you find the magnitude of a resultant from its components?

3. Why is a calculation-based method for finding a resultant generally preferred over a scale diagram?

4. How is conservation of momentum applied differently in a 2D collision compared to a 1D (head-on) collision?

5. What two pieces of information must always be given to fully describe a resultant vector?

Common mistakes and exam tips

Common misconceptions and exam-technique checks for Full Vectors.

• Avoid: Choosing sine or cosine from memory without reference to the stated angle. Correct idea: Use cosine for the component adjacent to the angle and sine for the opposite component.
• Avoid: Adding vector magnitudes directly when directions differ. Correct idea: Resolve vectors into common perpendicular components before adding.
• Avoid: Giving only the magnitude of a resultant vector. Correct idea: A complete vector result includes magnitude and direction relative to a stated reference.

• Draw a labelled component sketch before resolving a vector.
• Use the same x and y axes for every vector in the problem.
• State the reference direction for a resultant angle.

Exam-style questions

Five exam-style questions with worked answers on Full Vectors.

Higher Level — Required Content

Required Higher Level content for this topic.

Resolve each vector into perpendicular components, add the x-components and y-components separately, then calculate the magnitude and direction of the resultant.

• Example: 6 N east and 8 N north give a resultant √(6² + 8²) = 10 N.
• Its direction is tan⁻¹(8/6) = 53° north of east.
• In a two-dimensional collision, conserve momentum separately in the x- and y-directions.

End-of-unit revision

Formulae, definitions, experiments and mixed revision questions for Full Vectors.

• Fx = F cos θ, Fy = F sin θ
• R = √(Rx² + Ry²)
• θ = tan⁻¹(Ry/Rx)

• Component
• Resultant
• Resolving a vector

• Resolving a vector into perpendicular components
• Tip-to-tail resultant of two vectors
• 2D collision resolved into perpendicular components

• Verifying the law of addition of vectors (force table / parallelogram of forces)

• Sine/cosine mix-ups when resolving vectors
• Adding vector magnitudes directly when not along the same line
• Omitting direction from a resultant vector answer
• Not resolving into components before applying momentum conservation in 2D

1. A displacement of 12 m at 40° to the horizontal is given. Find its horizontal and vertical components.

2. Two perpendicular forces of 9 N and 12 N act on an object. Find the resultant force.

3. Explain why resolving vectors into components is useful for analysing motion on an inclined slope.

4. A ball moving east collides with a stationary ball and deflects north-east after the collision. Explain how conservation of momentum is applied to find the second ball's velocity.

1. Horizontal = 12 cos 40° ≈ 9.19 m. Vertical = 12 sin 40° ≈ 7.71 m.

2. R = √(9² + 12²) = √(81+144) = √225 = 15 N, at θ = tan⁻¹(12/9) ≈ 53.1° from the 9 N force.

3. On a slope, gravity acts straight down, which is not aligned with the direction of motion (along the slope) or perpendicular to it (into the slope); resolving weight into components along and perpendicular to the slope allows the along-slope component to be used directly in F = ma for motion along the slope, and the perpendicular component to be balanced against the normal force.

4. Resolve the momentum of both balls into components along the original direction of motion (x) and perpendicular to it (y); apply conservation of momentum separately in each direction (total x-momentum before = total x-momentum after; total y-momentum before = total y-momentum after, which is zero before the collision) to solve for the unknown velocity components of the second ball.

The table below shows which equations from this unit appear in the official State Examinations Commission Formulae and Tables book, and where.

Resolving vectors into perpendicular components and calculating resultants are identified as Higher Level content.