📌 3.1 Scalars and Vectors
Scalar Quantity
Has only magnitude. No direction.
Examples: mass, speed, time, temperature, energy, distance, work, power, pressure.
Vector Quantity
Has both magnitude and direction. Follows laws of vector addition.
Examples: displacement, velocity, acceleration, force, momentum, torque, weight.
- Equal vectors — same magnitude AND same direction.
- Negative vector — same magnitude, opposite direction. −A⃗
- Zero (null) vector — magnitude = 0. e.g. A⃗ − A⃗ = 0⃗
- Unit vector — magnitude = 1. Used only to show direction. â = A⃗ / |A⃗|
- Standard unit vectors: î (x-axis), ĵ (y-axis), k̂ (z-axis). Each has magnitude 1.
- Position vector r⃗ — gives position of a point from origin. r⃗ = xî + yĵ + zk̂
- Vectors are written as bold letters (A) or with arrow (A⃗). Magnitude = |A⃗| = A
- A vector can be moved parallel to itself without changing it — called free vector.
⭐ Important: Distance is scalar but displacement is vector. Speed is scalar but velocity is vector. Temperature is scalar but temperature gradient is vector.
➕ 3.2 Addition of Vectors — Laws
Triangle Law of Vector Addition
If two vectors A⃗ and B⃗ are placed head to tail (in order), the third side of the triangle drawn from the tail of A⃗ to the head of B⃗ gives the resultant R⃗ = A⃗ + B⃗.
Parallelogram Law of Vector Addition
If two vectors A⃗ and B⃗ act simultaneously at a point, represented as two adjacent sides of a parallelogram drawn from that point, then the diagonal of the parallelogram from that same point represents their resultant R⃗.
🌟 Magnitude of Resultant
R = √(A² + B² + 2AB cosθ)
θ = angle between A⃗ and B⃗
Direction of Resultant (α = angle with A⃗)
tan α = B sinθ / (A + B cosθ)
| θ (angle between vectors) | Resultant R | Remarks |
|---|
| θ = 0° (same direction) | R = A + B | Maximum resultant |
| θ = 180° (opposite) | R = |A − B| | Minimum resultant |
| θ = 90° (perpendicular) | R = √(A² + B²) | Pythagoras theorem |
| A = B, θ = 60° | R = A√3 | Equilateral triangle case |
| A = B, θ = 120° | R = A | R = each vector |
| A = B, θ = 90° | R = A√2 | Direction at 45° with each |
- Commutative law: A⃗ + B⃗ = B⃗ + A⃗ (order doesn't matter)
- Associative law: (A⃗ + B⃗) + C⃗ = A⃗ + (B⃗ + C⃗)
- To subtract: A⃗ − B⃗ = A⃗ + (−B⃗). Add negative of B⃗ to A⃗.
⭐ The range of resultant: |A−B| ≤ R ≤ (A+B). Resultant can never be greater than sum or less than difference of magnitudes.
🔀 3.3 Resolution of Vectors into Components
Resolution
Splitting a single vector into two or more component vectors. The original vector is the resultant of its components. This is the reverse of addition.
Components of A⃗ at angle θ with x-axis
Aₓ = A cosθ (x-component, horizontal)
Aᵧ = A sinθ (y-component, vertical)
A⃗ = Aₓ î + Aᵧ ĵ
Magnitude from components
|A⃗| = √(Aₓ² + Aᵧ²)
Direction from components
θ = tan⁻¹(Aᵧ / Aₓ)
Adding vectors analytically
Rₓ = Aₓ + Bₓ
Rᵧ = Aᵧ + Bᵧ
Resultant magnitude & direction
R = √(Rₓ²+Rᵧ²)
θ = tan⁻¹(Rᵧ/Rₓ)
- In 3D: A⃗ = Aₓî + Aᵧĵ + Aᵤk̂ and |A⃗| = √(Aₓ² + Aᵧ² + Aᵤ²)
- Component method is more accurate than graphical method for adding vectors
- A vector along x-axis has only x-component. A vector along y-axis has only y-component.
✖️ 3.4 Dot Product and Cross Product
Dot Product (Scalar Product)
A⃗ · B⃗ = AB cosθ → result is a SCALAR
In components: A⃗·B⃗ = AₓBₓ + AᵧBᵧ + AᵤBᵤ
Cross Product (Vector Product)
|A⃗ × B⃗| = AB sinθ → result is a VECTOR
Direction: Right-hand rule (fingers curl from A⃗ to B⃗, thumb points in direction of A⃗×B⃗)
| Property | Dot Product A⃗·B⃗ | Cross Product A⃗×B⃗ |
|---|
| Nature of result | Scalar (number) | Vector |
| Formula | AB cosθ | AB sinθ (magnitude) |
| If θ = 0° (parallel) | A⃗·B⃗ = AB (maximum) | |A⃗×B⃗| = 0 |
| If θ = 90° (perpendicular) | A⃗·B⃗ = 0 | |A⃗×B⃗| = AB (maximum) |
| Commutative? | A⃗·B⃗ = B⃗·A⃗ ✅ YES | A⃗×B⃗ = −B⃗×A⃗ ❌ NO |
| A⃗·A⃗ | A² (= |A|²) | 0⃗ (zero vector) |
| î·î = ĵ·ĵ = k̂·k̂ | = 1 | = 0⃗ |
| î·ĵ = ĵ·k̂ = k̂·î | = 0 | î×ĵ=k̂, ĵ×k̂=î, k̂×î=ĵ |
| Physical examples | Work W = F⃗·d⃗
Power P = F⃗·v⃗ | Torque τ⃗ = r⃗×F⃗
Angular momentum L⃗ = r⃗×p⃗ |
⭐ How to find angle between two vectors: cosθ = (A⃗·B⃗)/(AB). If A⃗·B⃗ = 0 and neither is zero vector → vectors are perpendicular.
🗺️ 3.5 Motion in a Plane — 2D Kinematics
Key Principle — MOST IMPORTANT
Motion in a plane (2D) can be resolved into two completely independent rectilinear motions — one along X-axis and one along Y-axis. They do not affect each other.
Position vector
r⃗ = xî + yĵ
Displacement
Δr⃗ = r⃗₂ − r⃗₁
Average velocity
v⃗_av = Δr⃗/Δt
Instantaneous velocity
v⃗ = dr⃗/dt
Acceleration
a⃗ = dv⃗/dt = d²r⃗/dt²
Equations of Motion in 2D (Uniform Acceleration)
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 with an initial velocity that then moves under the influence of gravity alone (no air resistance, no other force). The path of a projectile is called its trajectory — always a parabola.
- Initial speed u at angle θ with horizontal → uₓ = u cosθ, uᵧ = u sinθ
- Horizontal (X): No force → aₓ = 0 → vₓ = u cosθ = constant always
- Vertical (Y): Gravity acts → aᵧ = −g → vᵧ changes every second
- At highest point: vᵧ = 0, velocity = u cosθ (purely horizontal)
- Trajectory is symmetric about the highest point — equal time going up and coming down
- Horizontal distance covered = Range R. Vertical maximum height = H
🌟 All Projectile Formulas — Memorise These
Velocity at time t: vₓ = u cosθ, vᵧ = u sinθ − gt
Position at time t: x = u cosθ · t, y = u sinθ · t − ½gt²
Time of Flight: T = 2u sinθ / g
Maximum Height: H = u² sin²θ / 2g
Horizontal Range: R = u² sin2θ / g
Maximum Range: R_max = u²/g (at θ = 45°)
Speed at any time: v = √(vₓ² + vᵧ²)
Proof: Trajectory is a Parabola
From horizontal: x = u cosθ · t → t = x / (u cosθ)
Substitute in vertical: y = u sinθ · [x/(u cosθ)] − ½g · [x/(u cosθ)]²
y = x tanθ − gx² / (2u²cos²θ)
This is of the form y = Ax + Bx² → This is a PARABOLA ✓
| Launch Angle θ | Time of Flight T | Max Height H | Range R |
|---|
| 0° (horizontal throw) | √(2H₀/g) (H₀ = launch height) | 0 | u√(2H₀/g) |
| 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 (same as 30°) |
| 90° (vertically up) | 2u/g | u²/2g (max height) | 0 |
⭐ Complementary Angles — Same Range: θ and (90°−θ) give the same range because sin2θ = sin(180°−2θ). e.g. 30° and 60°, 20° and 70°, 15° and 75° all give the same R.
⭐ Relation: H_max = R_max/4 (at θ=45°). Also: R = 4H/tanθ for any angle.
⭐ Horizontal Projectile (thrown horizontally from height h):
Time to fall: t = √(2h/g) | Horizontal range: R = u√(2h/g) | Final velocity: v = √(u²+2gh)
⚠️ Common Mistake: At the highest point, velocity is NOT zero. Only the vertical component vᵧ = 0. The horizontal component vₓ = u cosθ is still there.
🚂 3.7 Relative Velocity in 2D
Definition
The velocity of object A as observed by object B is called the velocity of A relative to B (v⃗_AB). Motion is always relative — it depends on the observer's frame.
Relative Velocity Formulas
v⃗_AB = v⃗_A − v⃗_B (A relative to B)
v⃗_BA = v⃗_B − v⃗_A (B relative to A)
|v⃗_AB| = |v⃗_BA| but directions are opposite
| Situation | Relative Speed |
|---|
| Moving in SAME direction (speeds v₁, v₂) | |v₁ − v₂| |
| Moving in OPPOSITE directions (speeds v₁, v₂) | v₁ + v₂ |
| Moving at angle θ to each other | √(v₁² + v₂² − 2v₁v₂cosθ) |
⭐ Rain-Man Problem: A man walks at speed v_m. Rain falls vertically at speed v_r. Relative velocity of rain w.r.t. man = v⃗_rain − v⃗_man. To avoid rain, tilt umbrella at angle α = tan⁻¹(v_m / v_r) towards direction of walking.
⭐ River-Boat Problem:
River width = d, Boat speed = v_b (⊥ to bank), River current = v_r
Time to cross = d/v_b (independent of river current!)
Drift (downstream) = v_r × (d/v_b)
To reach directly opposite: aim upstream at angle α = sin⁻¹(v_r/v_b)
🔵 3.8 Uniform Circular Motion (UCM)
Definition
Motion of a body along a circular path with constant speed. Speed is constant but velocity direction changes continuously → body is always accelerating → net force acts on it.
Angular displacement (θ)
θ = s/r
s = arc length, r = radius. Unit: radianAngular velocity (ω)
ω = dθ/dt = v/r
ω = 2π/T = 2πn
Unit: rad/sLinear speed (v)
v = rω = 2πr/T = 2πrn
Tangential directionTime period (T)
T = 2πr/v = 2π/ω
Time for 1 full revolutionFrequency (n)
n = 1/T = ω/2π
Revolutions per second (Hz)Centripetal acceleration
a_c = v²/r = ω²r = vω
Direction: towards centreCentripetal force
F_c = mv²/r = mω²r
Direction: towards centreAngular acceleration (α)
α = dω/dt = a_t/r
For non-uniform circular motion
- In UCM, speed is constant but velocity direction changes → acceleration exists → it is NOT equilibrium.
- Centripetal acceleration is always perpendicular to velocity and directed towards the centre.
- Centripetal force is not a new type of force — it is provided by existing forces (gravity, tension, friction, normal reaction).
- Centrifugal force — appears only in rotating (non-inertial) frame. It is a pseudo force, equal and opposite to centripetal force. NOT a real force.
| Example of Circular Motion | Force providing centripetal force |
|---|
| Stone tied to string, whirled in circle | Tension T in string |
| Planet orbiting the Sun | Gravitational force (Sun's gravity) |
| Car on flat circular road | Friction between tyre and road |
| Car on banked circular road | Component of normal force (N sinθ) |
| Electron orbiting nucleus (Bohr model) | Electrostatic (Coulomb) force |
| Satellite in orbit | Gravitational force of Earth |
| Rider on banked curve | Normal force component |
🛣️ Banking of Roads & Conical Pendulum
Max speed — flat road (with friction)
v_max = √(μrg)
μ = coeff. of friction, r = radiusOptimum speed — banked road (no friction)
v = √(rg tanθ)
θ = banking angleMax speed — banked road (with friction)
v_max = √[rg(tanθ+μ)/(1−μtanθ)]
Banking angle
tan θ = v²/rg
θ = angle of bankingConical pendulum period
T = 2π√(L cosθ/g)
L = string length, θ with verticalConical pendulum ω
ω = √(g/L cosθ)
angular velocity
⭐ Why are roads banked? On a banked road, the component of the normal force (N sinθ) provides the centripetal force. This reduces the dependence on friction and allows vehicles to safely take turns at higher speeds.
⭐ Death well / Circus bike: On vertical wall, friction acts upward = weight, Normal force inward = centripetal. Condition: μ ≥ g·r/v² = μ_min.
📋 Complete Formula Quick Reference — Chapter 3
| Quantity | Formula | Unit |
|---|
| Resultant of vectors | R = √(A²+B²+2AB cosθ) | Same as A, B |
| Direction of resultant | tan α = B sinθ / (A + B cosθ) | degrees / radian |
| Components | Aₓ = A cosθ, Aᵧ = A sinθ | — |
| Unit vector | â = A⃗ / |A⃗| | dimensionless |
| Dot product | A⃗·B⃗ = AB cosθ = AₓBₓ+AᵧBᵧ | scalar |
| Cross product (magnitude) | |A⃗×B⃗| = AB sinθ | vector |
| Time of Flight | T = 2u sinθ / g | second (s) |
| Maximum Height | H = u² sin²θ / 2g | metre (m) |
| Horizontal Range | R = u² sin2θ / g | metre (m) |
| Max Range (θ=45°) | R_max = u²/g | metre (m) |
| Trajectory (parabola) | y = x tanθ − gx²/2u²cos²θ | — |
| Horizontal projectile range | R = u√(2h/g) | metre (m) |
| Relative velocity | v⃗_AB = v⃗_A − v⃗_B | m/s |
| Angular velocity | ω = v/r = 2π/T = 2πn | rad/s |
| Centripetal acceleration | a_c = v²/r = ω²r | m/s² |
| Centripetal force | F = mv²/r = mω²r | Newton (N) |
| Max speed on flat road | v_max = √(μrg) | m/s |
| Banking angle | tan θ = v²/rg | degrees |
| Conical pendulum T | T = 2π√(L cosθ/g) | second (s) |
⚠️ Common Mistakes — Avoid These in Exam:
1. At highest point of projectile — velocity ≠ 0. Only vᵧ = 0. vₓ = u cosθ still exists.
2. Range formula uses sin2θ — not sinθ or sin²θ.
3. Centripetal force is NOT a new force. It is provided by existing forces.
4. Centrifugal force is a pseudo force — it does NOT exist in inertial frames.
5. Complementary angles 30° & 60° give same RANGE but different HEIGHT and TIME OF FLIGHT.
6. For river-boat problem: time to cross = d/v_b (not affected by river current v_r).
🎯 Exam Important Questions — Chapter 3: Motion in a Plane
- Define scalar and vector quantities. Give 4 examples of each. What are unit vectors? 2 marks
- State and prove the parallelogram law of vector addition. Derive expression for magnitude and direction of resultant. 5 marks
- State the triangle law of vector addition. How does it differ from parallelogram law? 2 marks
- Distinguish between dot product and cross product of two vectors. Give two physical examples of each. 3 marks
- If A⃗ = 3î + 4ĵ and B⃗ = 4î − 3ĵ. Find (i) A⃗+B⃗ (ii) A⃗−B⃗ (iii) |A⃗| (iv) A⃗·B⃗ (v) angle between them. 4 marks
- What is a projectile? Show that the path of a projectile is a parabola. 4 marks
- Derive expressions for (i) Time of flight (ii) Maximum height (iii) Horizontal range of a projectile launched at angle θ. 5 marks
- Show that the horizontal range of a projectile is maximum when the angle of projection is 45°. 3 marks
- Show that the range is the same for angles θ and (90°−θ) of projection. 3 marks
- A ball is projected at 40 m/s at 30° above horizontal. Find: (i) time of flight (ii) maximum height (iii) horizontal range. (g = 10 m/s²) 4 marks
- A stone is thrown horizontally from the top of a cliff 80 m high with velocity 20 m/s. Find (i) time to reach ground (ii) horizontal range (iii) velocity on hitting ground. (g = 10 m/s²) 4 marks
- Define uniform circular motion. Prove that centripetal acceleration = v²/r. Why is it directed towards the centre? 4 marks
- What is centripetal force? Give 4 examples identifying the force that provides centripetal force. 3 marks
- Distinguish between centripetal force and centrifugal force. Why is centrifugal force called a pseudo force? 2 marks
- Derive the expression for maximum safe speed of a car on a flat circular road. What is the role of friction? 3 marks
- Explain why roads are banked at curves. Derive expression for optimum speed on a banked road (frictionless). 4 marks
- Define relative velocity. A train A moves north at 72 km/h and train B moves south at 54 km/h. Find velocity of B relative to A. 2 marks
- Explain the river-boat problem. A boat can row at 5 m/s in still water. River flows at 3 m/s. Find (i) time to cross 100 m wide river (ii) drift. 3 marks
- A particle moves in a circle of radius 0.5 m at 3 rev/s. Find: (i) angular velocity (ii) linear speed (iii) centripetal acceleration. 3 marks
- What is a conical pendulum? Derive expression for its time period. 4 marks