✈️
Chapter 3: Motion in a Plane
Scalars & Vectors · Vector Operations · Projectile Motion · Relative Velocity · Circular Motion
Chapter 3
📌 3.1 Scalars and Vectors
Definition
Scalar quantity — has only magnitude. e.g. mass, speed, time, temperature, energy, distance.
Vector quantity — has both magnitude and direction. e.g. displacement, velocity, acceleration, force, momentum.
- Vectors are represented by bold letters or letters with arrow on top: A⃗ or A
- Equal vectors — same magnitude and same direction.
- Negative vector — same magnitude, opposite direction. −A⃗
- Zero (null) vector — magnitude = 0, no specific direction.
- Unit vector — magnitude = 1, used to indicate direction. â = A⃗/|A⃗|
- Unit vectors along x, y, z axes: î, ĵ, k̂. |î| = |ĵ| = |k̂| = 1.
- Position vector r⃗ = xî + yĵ + zk̂ — specifies position of a point from origin.
- Vectors can be added geometrically (Triangle/Parallelogram law) or analytically (component method).
➕ 3.2 Addition of Vectors
Triangle Law of Vector Addition
If two vectors A⃗ and B⃗ are represented by two sides of a triangle taken in order (head to tail), then the third side (closing side, taken in opposite order) represents their resultant R⃗ = A⃗ + B⃗.
Parallelogram Law of Vector Addition
If two vectors A⃗ and B⃗ are represented by two adjacent sides of a parallelogram drawn from a common point, then the diagonal of the parallelogram drawn from that point represents their resultant.
Magnitude of Resultant
R = √(A² + B² + 2AB cosθ)
where θ = angle between A⃗ and B⃗
Direction of Resultant
tan α = B sinθ / (A + B cosθ)
α = angle of R⃗ with A⃗
| Angle θ between A⃗ & B⃗ | Resultant R | Special Case |
|---|
| θ = 0° (same direction) | R = A + B | Maximum resultant |
| θ = 180° (opposite direction) | R = |A − B| | Minimum resultant |
| θ = 90° (perpendicular) | R = √(A² + B²) | Pythagoras case |
| A = B, θ = 60° | R = A√3 | Equilateral triangle case |
| A = B, θ = 120° | R = A | R equals each vector |
⭐ Properties of Vector Addition: Commutative — A⃗ + B⃗ = B⃗ + A⃗. Associative — (A⃗ + B⃗) + C⃗ = A⃗ + (B⃗ + C⃗).
🔀 3.3 Resolution of Vectors (Components)
Resolution
Breaking a vector into two or more component vectors along chosen directions. The components are its projections along the axes.
Components of a vector A⃗ making angle θ with x-axis
Aₓ = A cosθ (x-component) Aᵧ = A sinθ (y-component)
A⃗ = Aₓî + Aᵧĵ |A⃗| = √(Aₓ² + Aᵧ²) θ = tan⁻¹(Aᵧ/Aₓ)
- To add vectors analytically: Add x-components separately, y-components separately.
- R⃗ = A⃗ + B⃗ → Rₓ = Aₓ + Bₓ, Rᵧ = Aᵧ + Bᵧ
- R = √(Rₓ² + Rᵧ²), direction: θ = tan⁻¹(Rᵧ/Rₓ)
- In 3D: A⃗ = Aₓî + Aᵧĵ + Aᵤk̂, |A⃗| = √(Aₓ² + Aᵧ² + Aᵤ²)
✖️ 3.4 Dot Product & Cross Product
Dot Product (Scalar Product)
A⃗ · B⃗ = AB cosθ
Result is a SCALAR. θ = angle between them.
Cross Product (Vector Product)
|A⃗ × B⃗| = AB sinθ
Result is a VECTOR. Direction by right-hand rule.
Dot Product in components
A⃗·B⃗ = AₓBₓ + AᵧBᵧ + AᵤBᵤ
î·î = ĵ·ĵ = k̂·k̂ = 1, î·ĵ = 0
Cross Product — unit vectors
î×ĵ = k̂, ĵ×k̂ = î, k̂×î = ĵ
î×î = ĵ×ĵ = k̂×k̂ = 0
| Property | Dot Product A⃗·B⃗ | Cross Product A⃗×B⃗ |
|---|
| Result | Scalar | Vector |
| Formula | AB cosθ | AB sinθ (magnitude) |
| Commutative | A⃗·B⃗ = B⃗·A⃗ ✓ | A⃗×B⃗ = −B⃗×A⃗ ✗ |
| If A⃗ ⊥ B⃗ (θ=90°) | A⃗·B⃗ = 0 | |A⃗×B⃗| = AB (max) |
| If A⃗ ∥ B⃗ (θ=0°) | A⃗·B⃗ = AB (max) | |A⃗×B⃗| = 0 |
| Physical use | Work = F⃗·d⃗, Power = F⃗·v⃗ | Torque τ⃗ = r⃗×F⃗, L⃗ = r⃗×p⃗ |
🗺️ 3.5 Motion in a Plane — 2D Kinematics
Key Principle
Motion in 2D can be resolved into two independent 1D motions along x and y axes. x and y motions are completely independent of each other.
Position vector
r⃗ = xî + yĵ
Velocity vector
v⃗ = vₓî + vᵧĵ = dr⃗/dt
Acceleration vector
a⃗ = aₓî + aᵧĵ = dv⃗/dt
Average velocity
v⃗_av = Δr⃗/Δt
Equations of Motion in 2D (vector form)
v⃗ = u⃗ + a⃗t s⃗ = u⃗t + ½a⃗t²
x-direction: vₓ = uₓ + aₓt sₓ = uₓt + ½aₓt²
y-direction: vᵧ = uᵧ + aᵧt sᵧ = uᵧt + ½aᵧt²
🎯 3.6 Projectile Motion — Complete Notes
Definition
A projectile is any object thrown (projected) with an initial velocity and then moving under the influence of gravity alone (no air resistance). Its path is called trajectory — always a PARABOLA.
- Initial velocity u at angle θ with horizontal → uₓ = u cosθ (horizontal), uᵧ = u sinθ (vertical)
- Horizontal: No force acts → aₓ = 0 → vₓ = u cosθ = constant throughout
- Vertical: Gravity acts downward → aᵧ = −g → vᵧ changes with time
- At highest point: vᵧ = 0, only vₓ remains → velocity = u cosθ (horizontal only)
- Trajectory equation: y = x tanθ − gx²/(2u²cos²θ) — form y = Ax + Bx² → parabola
🌟 All Projectile Formulas — Must Memorise
Time of Flight: T = 2u sinθ / g
Maximum Height: H = u² sin²θ / 2g
Horizontal Range: R = u² sin2θ / g = u² × 2sinθcosθ / g
Maximum Range: R_max = u²/g (at θ = 45°)
Velocity at time t: vₓ = u cosθ, vᵧ = u sinθ − gt, |v| = √(vₓ² + vᵧ²)
| Angle θ | Time of Flight T | Max Height H | Range R |
|---|
| 30° | u/g | u²/8g | u²√3/2g |
| 45° | u√2/g | u²/4g | u²/g (maximum) |
| 60° | u√3/g | 3u²/8g | u²√3/2g |
| 90° | 2u/g | u²/2g | 0 (straight up) |
⭐ Same range for complementary angles: θ and (90°−θ) give the same range. e.g. 30° and 60° → same R. This is because sin2θ = sin2(90°−θ).
⭐ Relation between H and R: R = 4H/tanθ → at θ=45°, R = 4H. Also H_max = R_max/4 (at 45°).
📝 Proof: Trajectory is a Parabola
Derivation
Horizontal: x = uₓt = u cosθ · t → t = x/(u cosθ)
Vertical: y = uᵧt − ½gt² = u sinθ · t − ½gt²
Substitute t: y = u sinθ · [x/(u cosθ)] − ½g · [x/(u cosθ)]²
y = x tanθ − gx²/(2u²cos²θ)
This is of the form y = Ax + Bx² → equation of a parabola ✓
🚂 3.7 Relative Velocity in 2D
Definition
The velocity of object A as observed from object B is called the velocity of A relative to B.
Relative Velocity Formula
v⃗_AB = v⃗_A − v⃗_B (velocity of A w.r.t. B)
v⃗_BA = v⃗_B − v⃗_A (velocity of B w.r.t. A)
|v⃗_AB| = |v⃗_BA| but directions are opposite
| Situation | Relative Speed |
|---|
| Both moving in same direction with speeds v₁, v₂ | |v₁ − v₂| |
| Moving in opposite directions with speeds v₁, v₂ | v₁ + v₂ |
| Moving at angle θ to each other | √(v₁² + v₂² − 2v₁v₂cosθ) |
⭐ Rain-Man Problem: To protect from rain, umbrella must be tilted at angle α = tan⁻¹(v_man/v_rain) in the direction of walking. The effective relative velocity of rain w.r.t. man determines the angle.
⭐ River-Boat Problem: To cross river of width d, boat speed v_b, river current v_r. Time to cross = d/v_b (independent of current). To cross in minimum time, point boat perpendicular to river bank.
🔵 3.8 Uniform Circular Motion (UCM)
Definition
Motion of an object along a circular path with constant speed is called Uniform Circular Motion. Speed is constant but velocity direction changes continuously → acceleration exists.
Angular displacement θ
θ = arc/radius = s/r
Unit: radianAngular velocity ω
ω = dθ/dt = v/r = 2π/T = 2πn
Unit: rad/sLinear velocity v
v = rω = 2πr/T = 2πrn
Tangential directionTime period T
T = 2πr/v = 2π/ω
Time for 1 revolutionCentripetal acceleration
a_c = v²/r = ω²r = vω
Towards centreCentripetal force
F_c = mv²/r = mω²r
Towards centre
- In UCM: speed is constant but velocity direction changes → NOT in equilibrium → net force acts (centripetal).
- Centripetal acceleration is always directed towards the centre, perpendicular to velocity.
- Centripetal force is not a new type of force — it is provided by existing forces (tension, gravity, friction, normal reaction depending on situation).
- Centrifugal force — pseudo force in rotating (non-inertial) frame. Equal in magnitude, opposite in direction to centripetal force. It is NOT a real force.
| Situation | Centripetal Force provided by |
|---|
| Stone on a string (horizontal circle) | Tension T in string |
| Planet orbiting Sun | Gravitational force F = GMm/r² |
| Car on circular road (flat) | Friction between tyre and road |
| Car on banked road | Component of Normal force N sinθ |
| Electron orbiting nucleus | Electrostatic attraction (Coulomb force) |
📐 Conical Pendulum & Banking of Roads
Conical Pendulum
T = 2π√(L cosθ/g)
L = string length, θ = half-angle with vertical
Banking of Road (no friction)
tan θ = v²/rg
v = √(rg tanθ)
θ = banking angle, v = optimum speed
Max speed on flat road
v_max = √(μrg)
μ = coefficient of friction
Max speed on banked road
v_max = √[rg(tanθ+μ)/(1−μtanθ)]
With friction on banked road
📋 Chapter 3 — Complete Formula Quick Reference
| Quantity | Formula | Unit |
|---|
| Resultant of two vectors | R = √(A²+B²+2AB cosθ) | same as A, B |
| Direction of resultant | tan α = B sinθ/(A+B cosθ) | degree/radian |
| Dot product | A⃗·B⃗ = AB cosθ | scalar |
| Cross product | |A⃗×B⃗| = AB sinθ | vector |
| Time of flight | T = 2u sinθ/g | second (s) |
| Maximum height | H = u²sin²θ/2g | metre (m) |
| Horizontal range | R = u²sin2θ/g | metre (m) |
| Maximum range | R_max = u²/g at θ=45° | metre (m) |
| Trajectory equation | y = x tanθ − gx²/2u²cos²θ | — |
| Angular velocity | ω = v/r = 2π/T = 2πn | rad/s |
| Centripetal acceleration | a = v²/r = ω²r | m/s² |
| Centripetal force | F = mv²/r = mω²r | Newton (N) |
| Relative velocity | v⃗_AB = v⃗_A − v⃗_B | m/s |
| Banking angle | tan θ = v²/rg | degree |
🎯 Exam Important Questions — Chapter 3: Motion in a Plane
- State and prove the parallelogram law of vector addition. Derive expression for magnitude and direction of resultant vector. 5 marks
- Define scalar and vector quantities. Give 4 examples of each. What are unit vectors? 2 marks
- Distinguish between dot product and cross product of two vectors with two examples each. 3 marks
- What is a projectile? Prove that the trajectory of a projectile is a parabola. 4 marks
- Derive expressions for (i) Time of flight (ii) Maximum height (iii) Horizontal range of a projectile thrown at angle θ with horizontal. 5 marks
- Show that the horizontal range is maximum at 45°. Also show that two angles of projection θ and (90°−θ) give the same range. 3 marks
- A ball is projected with 40 m/s at 30° above horizontal. Find: (i) time of flight (ii) maximum height (iii) horizontal range. (g = 10 m/s²) 4 marks
- Define uniform circular motion. Show that a body in UCM has centripetal acceleration directed towards the centre. Derive a = v²/r. 4 marks
- What is centripetal force? Give 3 examples of circular motion and identify the force providing centripetal force in each. 3 marks
- Derive the expression for maximum safe speed of a car on a flat circular road. What is the effect of increasing the radius of the road? 3 marks
- Explain banking of roads. Derive expression for optimum speed on a banked road (without friction). 4 marks
- Define relative velocity. Two trains A and B move at 60 km/h and 80 km/h in the same direction. Find velocity of B relative to A and A relative to B. 2 marks
- A particle moves in a circle of radius 2 m with speed 4 m/s. Find (i) angular velocity (ii) time period (iii) centripetal acceleration. 3 marks
- A stone is tied to a string of length 1 m and whirled in a horizontal circle at 2 rev/s. Find linear speed, angular velocity and centripetal force if mass = 0.5 kg. 3 marks
- Distinguish between centripetal force and centrifugal force. Why is centrifugal force called a pseudo force? 2 marks
🌊
Chapter 10: Oscillations & Waves
SHM · Simple pendulum · Speed of sound · Standing waves · Beats · Doppler effect
Chapter 10
🔑 Key Points — SHM & Waves
- SHM: Motion where restoring force ∝ displacement and directed towards equilibrium. F = −kx. a = −ω²x.
- Displacement in SHM: x = A sin(ωt + φ). Velocity: v = Aω cos(ωt + φ). Max velocity = Aω at x = 0.
- Max acceleration = Aω² at extreme positions (x = ±A). Acceleration = 0 at x = 0.
- Energy in SHM: KE = ½mω²(A²−x²). PE = ½mω²x². Total E = ½mω²A² = constant.
- Simple pendulum: T = 2π√(L/g). Does not depend on mass or amplitude (for small angles).
- Spring-mass system: T = 2π√(m/k). ω = √(k/m).
- Wave equation: y = A sin(kx − ωt). Wave speed v = ω/k = λ/T = λf. k = 2π/λ (wave number).
- Speed of sound: v = √(B/ρ) (Newton). Laplace correction: v = √(γP/ρ). In air at 0°C: v = 332 m/s. v_t = v₀ + 0.6t.
- Standing waves: y = 2A cos(kx) sin(ωt). Nodes at kx = nπ (spacing λ/2). Antinodes at kx = (2n+1)π/2.
- Beats: n_beats = |n₁ − n₂|. Heard when two close frequencies superpose.
- Doppler effect: Apparent frequency changes when source or observer moves.
📐 Key Formulas
SHM period
T = 2π/ω = 2π√(m/k)
Pendulum period
T = 2π√(L/g)
Total energy SHM
E = ½mω²A² = ½kA²
Speed of sound
v = √(γP/ρ) = √(γRT/M)
Doppler (source moving)
n' = n(v ± v_o)/(v ∓ v_s)
⭐ Organ pipe: Open pipe — L = nλ/2, harmonics n = 1,2,3... Closed pipe — L = (2n−1)λ/4, only ODD harmonics.
⭐ Doppler sign convention: Numerator: + if observer moves towards source, − if away. Denominator: − if source moves towards observer, + if away.
🎯 Exam Important Questions — Ch 10
- Define SHM. Show that the motion of a particle attached to a spring is SHM. Derive expression for time period. 5 marks
- Derive expression for time period of a simple pendulum. Why does it not depend on mass or amplitude? 4 marks
- Show that total energy in SHM is constant. Draw graphs of KE, PE and total energy vs displacement. 4 marks
- Explain the formation of standing waves. Differentiate between nodes and antinodes. 3 marks
- Derive an expression for speed of sound in a gas. What is Laplace's correction and why is it needed? 4 marks
- What are beats? How are they produced? Give one practical application. 2 marks
- State and explain Doppler effect. Derive expression for apparent frequency when source moves towards stationary observer. 4 marks
- A tuning fork of 256 Hz produces 4 beats/s with a wire. The wire is tightened and beats drop to 2/s. Find original frequency of wire. 3 marks