Scalars and vectors

1. Overview

In physics, quantities are classified based on whether their direction is necessary to describe them. This classification into scalars and vectors is fundamental because it determines the mathematical rules used for their manipulation. While scalars follow the laws of basic arithmetic, vectors require geometric and trigonometric analysis. In the Cambridge 9702 syllabus, the ability to resolve and combine vectors is a core skill; it is the mathematical foundation for understanding projectile motion, forces in equilibrium, momentum, and field theory. A vector is not merely a number with a unit; it is a physical entity that exists in a specific direction in space, and its effect is entirely dependent on that orientation.

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Key Definitions

• Scalar: A physical quantity that is completely described by its magnitude (size) and unit only. It has no directional component.
• Vector: A physical quantity that requires both a magnitude (with unit) and a direction for its description to be complete.
• Resultant: A single vector that has the same effect as two or more individual vectors acting together.
• Coplanar Vectors: Vectors that act within the same two-dimensional plane (e.g., all acting on a flat sheet of paper).
• Component: The effective part of a vector acting in a specific direction. Most commonly, a vector is resolved into two components at right angles (perpendicular) to each other.
• Equilibrium: A state where the resultant force acting on an object is zero. For three coplanar forces, this is represented geometrically by a closed triangle of forces.
• Displacement: The distance moved in a specified direction from a fixed reference point. It is the vector equivalent of distance.

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Content

3.1 Scalar and Vector Quantities

The Cambridge 9702 syllabus requires you to distinguish between scalars and vectors across all topics. A common exam question presents a list of physical quantities and asks you to categorize them.

Scalar Quantities	Vector Quantities
Distance ($d$ or $l$): The total path length traveled.	Displacement ($s$): Straight-line distance in a specific direction.
Speed ($v$): The rate of change of distance.	Velocity ($v$): The rate of change of displacement.
Mass ($m$): A measure of the amount of matter/inertia.	Weight ($W$): The gravitational force acting on a mass.
Time ($t$): The duration between events.	Acceleration ($a$): The rate of change of velocity.
Energy / Work ($E, W$): The capacity to do work.	Force ($F$): A push or pull (e.g., friction, upthrust).
Power ($P$): The rate at which work is done.	Momentum ($p$): The product of mass and velocity.
Pressure ($p$): Force per unit area.	Electric Field Strength ($E$): Force per unit positive charge.
Temperature ($T$): Measure of average kinetic energy.	Magnetic Flux Density ($B$): Strength of a magnetic field.
Density ($\rho$): Mass per unit volume.	Torque / Moment: The turning effect of a force.

The Case of Pressure and Work:

• Pressure: Even though pressure involves force (a vector), pressure at a point in a fluid acts equally in all directions. It does not have a single specific direction, thus it is a scalar.
• Work/Energy: Work is the product of force and displacement in the direction of the force. Mathematically, this is a "dot product" of two vectors, which results in a scalar. You cannot have "20 Joules North."

3.2 Addition and Subtraction of Coplanar Vectors

When multiple vectors act on a single point, we must find the resultant. The method used depends on whether the vectors are collinear (same line), perpendicular, or at an arbitrary angle.

1. Vector Addition (Finding the Resultant)

• Triangle Method (Tip-to-Tail):
  1. Draw the first vector to scale, maintaining its direction.
  2. Draw the second vector starting from the "tip" (arrowhead) of the first.
  3. The resultant is the vector drawn from the "tail" of the first to the "tip" of the second.
• Parallelogram Method:
  1. Draw both vectors starting from the same origin point.
  2. Complete a parallelogram using these two vectors as adjacent sides.
  3. The resultant is the diagonal of the parallelogram starting from the common origin.

2. Vector Subtraction Subtraction is mathematically equivalent to adding a negative vector. To find $\mathbf{A} - \mathbf{B}$:

1. Reverse the direction of vector $\mathbf{B}$ to create $-\mathbf{B}$.
2. Add $-\mathbf{B}$ to $\mathbf{A}$ using the tip-to-tail method: $\mathbf{R} = \mathbf{A} + (-\mathbf{B})$. This is vital for calculating the change in velocity ($\Delta v = v - u$) in momentum and circular motion problems.

3. Mathematical Calculation (Perpendicular Vectors) If two vectors $V_1$ and $V_2$ are at $90^\circ$ to each other:

• Magnitude: Use Pythagoras’ Theorem: $R = \sqrt{V_1^2 + V_2^2}$
• Direction: Use trigonometry: $\theta = \tan^{-1}\left(\frac{\text{Opposite}}{\text{Adjacent}}\right)$

3.3 Resolving a Vector into Two Perpendicular Components

Resolving is the process of breaking a single vector into two parts that act at right angles to each other. This is a powerful tool because perpendicular components are independent of each other. For example, a horizontal force cannot cause vertical acceleration.

For a vector $V$ at an angle $\theta$ to the horizontal:

• Horizontal Component ($V_x$): $V_x = V \cos \theta$
• Vertical Component ($V_y$): $V_y = V \sin \theta$

The "Inclined Plane" Scenario: This is a high-frequency exam context. When an object of weight $W$ sits on a slope inclined at angle $\alpha$ to the horizontal:

• The component of weight acting down the slope: $W_{\parallel} = W \sin \alpha$
• The component of weight acting perpendicular (normal) to the slope: $W_{\perp} = W \cos \alpha$ (Note: The angle between the weight vector and the line perpendicular to the slope is equal to the angle of the slope $\alpha$.)

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Worked Example 1 — Resultant of Non-Perpendicular Forces

A ship is being towed by two cables. Cable 1 exerts a force of $2500\text{ N}$ at an angle of $20^\circ$ North of East. Cable 2 exerts a force of $1500\text{ N}$ at an angle of $30^\circ$ South of East. Calculate the magnitude and direction of the resultant force.

Step 1: Resolve both forces into East ($x$) and North ($y$) components.

• Cable 1:
  • $F_{x1} = 2500 \cos(20^\circ) = 2349.2\text{ N}$
  • $F_{y1} = 2500 \sin(20^\circ) = 855.1\text{ N}$
• Cable 2:
  • $F_{x2} = 1500 \cos(30^\circ) = 1299.0\text{ N}$
  • $F_{y2} = 1500 \sin(-30^\circ) = -750.0\text{ N}$ (Negative because it is South)

Step 2: Sum the components.

• $\sum F_x = 2349.2 + 1299.0 = 3648.2\text{ N}$
• $\sum F_y = 855.1 - 750.0 = 105.1\text{ N}$

Step 3: Calculate the resultant magnitude. $R = \sqrt{(\sum F_x)^2 + (\sum F_y)^2}$ $R = \sqrt{3648.2^2 + 105.1^2} = \sqrt{13309363 + 11046}$ $R = 3649.7\text{ N}$

Step 4: Calculate the direction. $\theta = \tan^{-1}\left(\frac{105.1}{3648.2}\right) = 1.65^\circ$

Answer: Magnitude = $3650\text{ N}$, Direction = $1.7^\circ$ North of East (to 3 s.f.)

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Worked Example 2 — Change in Velocity (Vector Subtraction)

An electron enters a magnetic field traveling East at $4.0 \times 10^6\text{ m s}^{-1}$. It is deflected and leaves the field traveling North at $4.0 \times 10^6\text{ m s}^{-1}$. Determine the magnitude and direction of the change in velocity $\Delta v$.

Step 1: Identify the vectors.

• Initial velocity $\mathbf{u} = 4.0 \times 10^6\text{ m s}^{-1}$ East.
• Final velocity $\mathbf{v} = 4.0 \times 10^6\text{ m s}^{-1}$ North.

Step 2: Apply the change in velocity formula. $\Delta \mathbf{v} = \mathbf{v} - \mathbf{u} = \mathbf{v} + (-\mathbf{u})$ $-\mathbf{u}$ is a vector of $4.0 \times 10^6\text{ m s}^{-1}$ pointing West.

Step 3: Calculate the magnitude. Since North and West are perpendicular: $|\Delta v| = \sqrt{(4.0 \times 10^6)^2 + (4.0 \times 10^6)^2}$ $|\Delta v| = \sqrt{2 \times (16 \times 10^{12})} = \sqrt{32 \times 10^{12}}$ $|\Delta v| = 5.66 \times 10^6\text{ m s}^{-1}$

Step 4: Determine direction. The vector points from the start of $\mathbf{v}$ to the end of $-\mathbf{u}$ (or vice versa depending on construction). The resultant points North-West. $\theta = \tan^{-1}\left(\frac{4.0 \times 10^6}{4.0 \times 10^6}\right) = 45^\circ$

Answer: $5.7 \times 10^6\text{ m s}^{-1}$ at $45^\circ$ North of West.

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Worked Example 3 — Equilibrium on an Inclined Plane

A block of mass $5.0\text{ kg}$ is held stationary on a smooth slope inclined at $25^\circ$ to the horizontal by a string parallel to the slope. Calculate the tension $T$ in the string and the normal contact force $N$ exerted by the slope. (Take $g = 9.81\text{ m s}^{-2}$)

Step 1: Calculate the weight. $W = mg = 5.0 \times 9.81 = 49.05\text{ N}$

Step 2: Resolve the weight into components.

• Component down the slope: $W_{\parallel} = 49.05 \sin(25^\circ) = 20.73\text{ N}$
• Component perpendicular to the slope: $W_{\perp} = 49.05 \cos(25^\circ) = 44.45\text{ N}$

Step 3: Apply equilibrium conditions.

• Parallel to slope: $\sum F = 0 \implies T = W_{\parallel}$
• Perpendicular to slope: $\sum F = 0 \implies N = W_{\perp}$

Answer: Tension $T = 21\text{ N}$, Normal force $N = 44\text{ N}$ (to 2 s.f.)

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Key Equations

Equation	Description	Data Sheet?
$V_x = V \cos \theta$	Component adjacent to the angle $\theta$.	No
$V_y = V \sin \theta$	Component opposite to the angle $\theta$.	No
$R = \sqrt{V_x^2 + V_y^2}$	Magnitude of the resultant of two perpendicular vectors.	No
$\theta = \tan^{-1}\left(\frac{V_y}{V_x}\right)$	Angle of the resultant vector relative to the x-axis.	No
$\Delta \mathbf{v} = \mathbf{v} - \mathbf{u}$	Vector change in velocity (Final - Initial).	No
$W_{\parallel} = mg \sin \alpha$	Component of weight acting down a slope of angle $\alpha$.	No
$W_{\perp} = mg \cos \alpha$	Component of weight acting normal to a slope of angle $\alpha$.	No

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Common Mistakes to Avoid

• ❌ Wrong: Adding vector magnitudes directly (e.g., $3\text{ N} + 4\text{ N} = 7\text{ N}$) when they are not in the same direction.
  • ✓ Right: Use the tip-to-tail method or resolve into perpendicular components before adding.
• ❌ Wrong: Assuming the horizontal component is always $\cos$ and the vertical is always $\sin$.
  • ✓ Right: The component adjacent to the given angle uses $\cos$. If the angle is given relative to the vertical, the horizontal component will be $V \sin \theta$.
• ❌ Wrong: Forgetting to check the calculator mode.
  • ✓ Right: Ensure your calculator is in Degrees (DEG) mode. Radian mode will lead to incorrect trigonometric values in almost all AS mechanics problems.
• ❌ Wrong: Confusing "Distance" and "Displacement" in velocity calculations.
  • ✓ Right: Velocity is $\frac{\text{displacement}}{\text{time}}$. If an object returns to its starting point, its total displacement is zero, and thus its average velocity is zero, regardless of the distance traveled.
• ❌ Wrong: Incorrectly identifying the angle on an inclined plane.
  • ✓ Right: The angle between the weight vector (acting vertically down) and the normal to the slope is always equal to the angle of the slope $\alpha$.

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Exam Tips

1. The Closed Triangle Rule: If a question states that three forces are in equilibrium, their vector arrows must form a closed loop (triangle) when drawn tip-to-tail. If there is a gap, the object is accelerating.
2. Scale Drawings: If the exam specifically asks for a scale drawing:
   • Use a sharp pencil and a long ruler.
   • Choose a scale that makes the diagram large (e.g., $1\text{ cm} = 5\text{ N}$ to fill half a page).
   • State your scale clearly (e.g., "Scale: $1\text{ cm} : 10\text{ m s}^{-1}$").
3. Component Independence: Always remember that the horizontal component of motion is completely independent of the vertical component. This is the "golden rule" for solving projectile motion questions in Topic 2.
4. Significant Figures: Match your answer's significant figures to the least precise data given in the question (usually 2 or 3 s.f.).
5. Direction Description: Never just give a number for a vector. State the direction clearly: "at $35^\circ$ to the horizontal" or "on a bearing of $045^\circ$."
6. Free-Body Diagrams: Before starting any calculation involving forces, draw a simple sketch with arrows showing all forces acting on the object. This prevents sign errors during resolution.