What this chapter is about
This chapter extends your understanding of motion from one dimension to two dimensions. In Class 9 and the previous chapter, you studied objects moving along a straight line. Now you learn to describe motion when an object can move in a plane — forward-backward and left-right, or horizontal and vertical simultaneously. This requires the mathematical tool of vectors.
You will learn what vectors are, how to add and subtract them, and how to resolve them into components. The chapter then applies these ideas to two important types of two-dimensional motion: projectile motion (like a ball thrown at an angle) and uniform circular motion (like a stone whirled on a string). These form the foundation for understanding more complex motions in mechanics and beyond.
After studying this chapter, you should be able to represent physical quantities as vectors, perform vector operations graphically and algebraically, analyse the trajectory of a projectile, and describe the velocity and acceleration of an object moving in a circle at constant speed.
Key ideas
- Scalars and vectors: Scalars have only magnitude (mass, temperature, speed). Vectors have both magnitude and direction (displacement, velocity, acceleration, force). Vectors obey special rules of addition.
- Vector addition: Two vectors add by the triangle law or parallelogram law. The resultant of vectors A and B at angle θ has magnitude R = √(A² + B² + 2AB cos θ). Vector addition is commutative: A + B = B + A.
- Resolution of vectors: Any vector in a plane can be written as the sum of two perpendicular components. If vector A makes angle θ with the x-axis, then Aₓ = A cos θ and Aᵧ = A sin θ. The magnitude A = √(Aₓ² + Aᵧ²).
- Position, displacement and velocity vectors: Position vector r gives location relative to origin. Displacement is Δr = r₂ − r₁. Average velocity is Δr/Δt. Instantaneous velocity v = dr/dt is tangent to the path.
- Projectile motion: An object launched with initial velocity and moving under gravity alone follows a parabolic path. Horizontal and vertical motions are independent. Horizontal velocity stays constant; vertical velocity changes by g every second.
- Equations for projectile: For launch angle θ and initial speed u: horizontal range R = u² sin 2θ / g, maximum height H = u² sin² θ / 2g, time of flight T = 2u sin θ / g. Maximum range occurs at θ = 45°.
- Uniform circular motion: An object moving in a circle at constant speed has velocity changing direction continuously. This requires an acceleration directed towards the centre, called centripetal acceleration: aᶜ = v²/r = ω²r, where ω is angular speed.
- Angular quantities: Angular displacement θ is measured in radians. Angular velocity ω = dθ/dt. For uniform circular motion, v = ωr and the period T = 2π/ω.
Formulas and facts to remember
- Magnitude of resultant of two vectors A and B at angle θ: R = √(A² + B² + 2AB cos θ)
- Components of vector A at angle θ to x-axis: Aₓ = A cos θ, Aᵧ = A sin θ
- For projectile with initial speed u at angle θ to horizontal:
- Horizontal range: R = u² sin 2θ / g
- Maximum height: H = u² sin² θ / 2g
- Time of flight: T = 2u sin θ / g
- Trajectory equation: y = x tan θ − (g x²) / (2u² cos² θ), which is a parabola
- Centripetal acceleration for circular motion: aᶜ = v²/r = ω²r, directed towards centre
- Relation between linear and angular speed: v = ωr
- Period of circular motion: T = 2πr/v = 2π/ω
- One complete revolution = 2π radians = 360°
Worked examples
Example 1: Adding two displacement vectors
A person walks 40 m east, then 30 m north. Find the magnitude and direction of the net displacement.
Solution: Take east as positive x-direction and north as positive y-direction. First displacement: Aₓ = 40 m, Aᵧ = 0 Second displacement: Bₓ = 0, Bᵧ = 30 m
Net displacement components: Rₓ = 40 + 0 = 40 m, Rᵧ = 0 + 30 = 30 m
Magnitude: R = √(40² + 30²) = √(1600 + 900) = √2500 = 50 m
Direction: tan α = Rᵧ/Rₓ = 30/40 = 0.75, so α = 37° north of east
The net displacement is 50 m at 37° north of east.
Example 2: Projectile motion — a cricket ball
A fielder throws a ball at 20 m/s at an angle of 30° to the horizontal. Find (a) the maximum height, (b) the time of flight, and (c) the horizontal range. Take g = 10 m/s².
Solution: Given: u = 20 m/s, θ = 30°, g = 10 m/s² sin 30° = 0.5, cos 30° = 0.866, sin 60° = 0.866
(a) Maximum height: H = u² sin² θ / 2g = (20)² × (0.5)² / (2 × 10) = 400 × 0.25 / 20 = 5 m
(b) Time of flight: T = 2u sin θ / g = 2 × 20 × 0.5 / 10 = 2 s
(c) Horizontal range: R = u² sin 2θ / g = (20)² × sin 60° / 10 = 400 × 0.866 / 10 = 34.6 m
Example 3: Centripetal acceleration of a ceiling fan
A ceiling fan blade is 0.5 m long and rotates at 300 revolutions per minute. Find the centripetal acceleration of the tip of the blade.
Solution: Radius r = 0.5 m Frequency = 300 rpm = 300/60 = 5 revolutions per second
Angular speed: ω = 2π × 5 = 10π rad/s
Centripetal acceleration: aᶜ = ω²r = (10π)² × 0.5 = 100π² × 0.5 = 50π² m/s² aᶜ = 50 × 9.87 ≈ 493 m/s²
This is about 50 times g, showing why loose parts can fly off a fast-rotating fan.
Common mistakes
- Adding vector magnitudes directly without considering direction → Use the parallelogram law or component method; magnitudes add directly only when vectors are in the same direction.
- Using sin θ and cos θ interchangeably in projectile formulas → Remember: vertical component uses sin θ, horizontal component uses cos θ.
- Thinking velocity is constant in uniform circular motion → Speed is constant, but velocity changes direction continuously, so there is acceleration.
- Forgetting to convert angles to radians when using ω → Angular speed ω must be in rad/s for the formula v = ωr to give v in m/s.
- Believing an object at the highest point of its trajectory has zero acceleration → Acceleration due to gravity acts throughout the flight; at the top, only vertical velocity is zero.
Quick revision
- Vectors have magnitude and direction; add them using components or the parallelogram law.
- Projectile motion is two independent motions: constant horizontal velocity and uniformly accelerated vertical motion under gravity.
- Maximum range of a projectile occurs at 45° launch angle (for level ground).
- In uniform circular motion, speed is constant but velocity is not; centripetal acceleration aᶜ = v²/r points towards the centre.
- Always resolve vectors into perpendicular components before adding or using equations of motion.