Scalars and Vectors
Section: Motion, Forces & Energy | Syllabus: Cambridge IGCSE Physics 0625
Scalars and Vectors Physical quantities can be classified into two types based on whether they require direction to be fully described. Scalar Quantity A quantity that has magnitude only (no direction).
Vector Quantity A quantity that has both magnitude and direction. Comparison of Scalars and Vectors Scalar Quantities Vector Quantities Distance Displacement Speed Velocity Mass Weight Time Force Energy Acceleration Temperature Momentum Gravitational field strength Electric field strength Syllabus List - Learn These Scalars (syllabus): distance, speed, time, mass, energy, temperature Vectors (syllabus): force, weight, velocity, acceleration, momentum, gravitational field strength, electric field strength Common Mistake Do not confuse speed (scalar) with velocity (vector).
Velocity includes direction! Representing Vectors Vectors are represented by arrows to show both their magnitude and direction. Length of arrow → magnitude of the vector Direction of arrow → direction of the vector FIG 1.1: Vector Representation A straight arrow representing a vector.
The length of the arrow represents the magnitude, and the arrowhead indicates the direction of the quantity. Calculating with Vectors Vectors can be added, subtracted, and resolved into components using mathematical methods.
Finding the Resultant of Two Perpendicular Vectors The syllabus requires you to find the resultant of two vectors at right angles - limited to forces or velocities only . Two methods are acceptable: Method 1: Graphical (Scale Drawing) Draw the two vectors to scale, placed head to tail .
The resultant is drawn from the tail of the first vector to the head of the second. Measure its length and direction with a ruler and protractor. FIG 1.2: Graphical Method - Head-to-Tail (for forces or velocities) A worked example with two perpendicular forces: Force A = 3 N pointing right (horizontal arrow), Force B = 4 N pointing up (vertical arrow).
Step 1: draw force A to scale (e.g. 1 cm = 1 N, so 3 cm rightward). Step 2: from the tip of A, draw force B (4 cm upward). Step 3: draw the resultant R from the tail of A to the tip of B - a diagonal arrow.
A ruler measurement shows R = 5 cm = 5 N. A protractor shows the angle θ = 53° above the horizontal. The scale is noted at the bottom: "Scale: 1 cm = 1 N". Method 2: By Calculation (Pythagoras' Theorem) For two vectors at right angles, the resultant is the hypotenuse of a right-angled triangle: Magnitude: R = √(A² + B²) Direction: θ = tan⁻¹(B / A) (angle from A towards B) FIG 1.3: Calculation Method - Right-Angled Triangle A right-angled triangle.
The horizontal side is labelled "A (e.g. horizontal force or velocity)". The vertical side is labelled "B (e.g. vertical force or velocity)". The hypotenuse is labelled "R = √(A² + B²)" with an arrowhead showing direction.
The angle θ at the bottom-left is labelled "θ = tan⁻¹(B/A)". The right-angle symbol is shown at the bottom-right corner. Important In the IGCSE syllabus, you only need to find the resultant of vectors that are at right angles (90°) to each other, and only for forces or velocities .
Always draw the right-angled triangle first before applying Pythagoras. Resolving a Vector into Components A vector can be split into horizontal and vertical components: Horizontal component: F cos θ Vertical component: F sin θ FIG 1.4: Resolving a Vector into Components Force F shown as a diagonal arrow at angle θ above the horizontal.
Two dashed lines show the components: a horizontal dashed arrow labelled "Fx = F cos θ" and a vertical dashed arrow labelled "Fy = F sin θ". Together the two components form the two shorter sides of a right-angled triangle, with the original force F as the hypotenuse.
Tip Always draw a right-angled triangle showing components before using sine or cosine! Worked Examples Example: Resolving Force Components Question: A force of 5 N acts at 60° to the horizontal. Find its horizontal and vertical components.
Solution Horizontal = 5 cos 60° = 2.5 N Vertical = 5 sin 60° = 4.33 N Example: Finding Resultant Force Question: Two perpendicular forces of 3 N and 4 N act on a point. Find the resultant force. Solution R = √(3² + 4²) = 5 N θ = tan⁻¹(4/3) = 53.1° Exam Tip For perpendicular vectors, use Pythagoras.
For non-perpendicular ones, use the parallelogram rule or trigonometry (law of cosines).
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