Motion in a Plane Class 11 Notes | CBSE Physics Chapter 3

Motion in a Plane is Chapter 3 of CBSE Class 11 Physics — the chapter where one-dimensional kinematics grows up into two dimensions using vectors. It teaches you how to handle quantities that have both size and direction, and then uses that machinery to crack two exam favourites: projectile motion and uniform circular motion. Get comfortable here and JEE, NEET, and your board paper all become much friendlier.

By the end of these notes you will be able to add and resolve vectors, write the equations of motion in two dimensions, derive the time of flight, maximum height and range of a projectile, and handle angular velocity and centripetal acceleration in circular motion. This is a high-weightage chapter carrying roughly 6–8 marks in boards, and the vector foundation for Laws of Motion, Work-Energy, and almost all of physics.

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Table of Contents

• Key Concepts — Scalars & vectors, vector addition, resolution, projectile & circular motion
• Weightage in Board & Entrance Exams
• Important Definitions
• Solved Examples
• Important Questions for Board Exams
• Quick Revision Points

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Key Concepts

1. Scalars and Vectors

A scalar is a quantity that has only magnitude — like mass, time, speed, distance, energy, or temperature. You can add scalars with ordinary arithmetic.

A vector has both magnitude and direction, and it follows the triangle (or parallelogram) law of addition. Displacement, velocity, acceleration, force, and momentum are vectors. A vector is written in bold (A) or with an arrow, and its magnitude is written |A| or simply A.

Types of Vectors

• Equal vectors: same magnitude and same direction.
• Unit vector: a vector of magnitude 1 used to show direction; Â = A/|A|.
• Zero (null) vector: magnitude zero, arbitrary direction; result of adding a vector to its negative.
• Negative vector: same magnitude, opposite direction (−A).
• Collinear vectors: act along the same or parallel lines.

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2. Unit Vectors

A unit vector has a magnitude of exactly one and only specifies direction. The standard unit vectors along the x, y and z axes are î, ĵ, k̂ respectively.

Any vector can be written in terms of unit vectors: A = Aₓ î + A_y ĵ + A_z k̂

• î, ĵ, k̂ are mutually perpendicular (orthogonal).
• Each has magnitude 1 and no units.
• The unit vector of any vector is  = A/|A|.

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3. Addition of Vectors — Triangle Law

The triangle law states that if two vectors are represented by two sides of a triangle taken in order, their resultant is given by the third side taken in the opposite order.

[DIAGRAM: Vector A drawn head-to-tail with vector B; the closing side from the tail of A to the head of B is the resultant R = A + B.]

Vector addition is commutative (A + B = B + A) and associative (A + (B + C) = (A + B) + C).

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4. Addition of Vectors — Parallelogram Law

The parallelogram law states that if two vectors acting at a point are represented by two adjacent sides of a parallelogram, their resultant is represented by the diagonal passing through that point.

For two vectors A and B with angle θ between them, the magnitude of the resultant R is:

R = √(A² + B² + 2AB cos θ)

The direction of the resultant (angle β with A) is given by:

tan β = (B sin θ)/(A + B cos θ)

• When θ = 0°: R = A + B (maximum).
• When θ = 180°: R = |A − B| (minimum).
• When θ = 90°: R = √(A² + B²).

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5. Subtraction of Vectors

Subtracting a vector means adding its negative: A − B = A + (−B). So to subtract B, reverse its direction and add it to A using the triangle law.

Magnitude: |A − B| = √(A² + B² − 2AB cos θ)

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6. Resolution of Vectors

Splitting a vector into two or more components is called resolution. The most useful is resolving a vector into two mutually perpendicular components (rectangular components).

If a vector A makes an angle θ with the x-axis:

• x-component: Aₓ = A cos θ
• y-component: A_y = A sin θ

Then A = Aₓ î + A_y ĵ, with magnitude A = √(Aₓ² + A_y²) and direction tan θ = A_y/Aₓ.

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7. Analytical Method of Vector Addition

To add several vectors accurately, resolve each into components, add the components separately, then recombine.

If R = A + B, then:

• Rₓ = Aₓ + Bₓ
• R_y = A_y + B_y

Magnitude: R = √(Rₓ² + R_y²), and direction tan α = R_y/Rₓ. This method extends easily to any number of vectors and to three dimensions.

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8. Motion in a Plane with Constant Acceleration

In two dimensions, position, velocity and acceleration are all vectors, and motion along x and y is independent. The equations of motion apply separately to each axis.

• v = u + at → vₓ = uₓ + aₓt, v_y = u_y + a_yt
• r = r₀ + ut + ½at²
• v² = u² + 2a·s (along each axis)

Key idea: The horizontal and vertical motions are independent — this is exactly what makes projectile motion solvable.

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9. Projectile Motion

A projectile is any object thrown into the air that moves under gravity alone (air resistance neglected). Its path is a parabola. The horizontal velocity stays constant; the vertical velocity changes due to g.

For a projectile launched with speed u at angle θ to the horizontal:

• Horizontal velocity: uₓ = u cos θ (constant)
• Vertical velocity: u_y = u sin θ (decreases, then increases)

Key Projectile Equations

Quantity	Formula
Time of flight (T)	T = (2u sin θ)/g
Maximum height (H)	H = (u² sin²θ)/(2g)
Horizontal range (R)	R = (u² sin 2θ)/g
Equation of path	y = x tan θ − (gx²)/(2u²cos²θ)

Maximum range occurs at θ = 45°, giving R_max = u²/g. Two complementary angles (θ and 90° − θ) give the same range.

Horizontal Projectile

For a body projected horizontally with speed u from height h: time of fall t = √(2h/g), and horizontal range = u√(2h/g).

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10. Uniform Circular Motion

When a body moves along a circle at constant speed, it is in uniform circular motion. The speed is constant, but the velocity keeps changing direction, so the body is accelerating.

Angular Velocity

Angular velocity (ω) is the rate of change of angular displacement. It links to linear speed by v = ωr.

• ω = θ/t = 2π/T = 2πn, where T is the time period and n is frequency.
• SI unit: radian per second (rad/s).
• Linear speed: v = ωr

Centripetal Acceleration

The acceleration in uniform circular motion always points towards the centre and is called centripetal acceleration.

a_c = v²/r = ω²r = (4π²r)/T²

Important: In uniform circular motion the speed is constant, so there is no tangential acceleration — only the centre-directed centripetal acceleration that changes direction.

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Weightage in Board & Entrance Exams

Exam	Typical Weightage	Most-Tested Areas
CBSE Board (Class 11)	6–8 marks	Vector addition/resolution, projectile equations, circular motion
JEE Main / Advanced	1–3 questions	Projectile range/height, relative velocity, circular motion
NEET	2–3 questions	Resolution of vectors, projectile motion, centripetal acceleration

[TABLE: Question-type split — VSA (1 mark): scalar/vector, unit vectors, definitions; SA (2–3 marks): resolution, projectile numericals, ω–v relation; LA (5 marks): derivation of time of flight, max height and range; centripetal acceleration derivation.]

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Important Definitions

Term	Definition
Scalar	A quantity with magnitude only, e.g. mass, speed, energy
Vector	A quantity with both magnitude and direction obeying the triangle law
Unit vector	A vector of magnitude one showing direction: Â = A/|A|
Resolution of a vector	Splitting a vector into rectangular components Aₓ = A cos θ, A_y = A sin θ
Resultant vector	Single vector with the same effect as two or more vectors combined
Projectile	A body moving under gravity alone after being thrown; path is parabolic
Time of flight	Total time a projectile stays in air: T = 2u sin θ/g
Horizontal range	Horizontal distance covered by a projectile: R = u² sin 2θ/g
Angular velocity	Rate of change of angular displacement: ω = θ/t = 2π/T
Centripetal acceleration	Centre-directed acceleration in circular motion: a_c = v²/r = ω²r

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Solved Examples

Example 1

Two vectors of magnitude 3 units and 4 units act at right angles to each other. Find the magnitude of their resultant.

Answer: R = √(A² + B² + 2AB cos 90°) = √(3² + 4² + 0) = √(9 + 16) = √25 = 5 units.

Example 2

A force of 10 N makes an angle of 30° with the x-axis. Find its rectangular components.

Answer: Fₓ = F cos 30° = 10 × (√3/2) = 8.66 N; F_y = F sin 30° = 10 × 0.5 = 5 N.

Example 3

A ball is thrown with a speed of 20 m/s at an angle of 30° to the horizontal. Find the time of flight. (g = 10 m/s²)

Answer: T = (2u sin θ)/g = (2 × 20 × sin 30°)/10 = (2 × 20 × 0.5)/10 = 20/10 = 2 s.

Example 4

For the ball in Example 3, find the maximum height and horizontal range.

Answer: H = (u² sin²θ)/(2g) = (400 × 0.25)/20 = 100/20 = 5 m. R = (u² sin 2θ)/g = (400 × sin 60°)/10 = (400 × 0.866)/10 = 34.64 m.

Example 5

A particle moves in a circle of radius 2 m at a constant speed of 4 m/s. Find its centripetal acceleration.

Answer: a_c = v²/r = 4²/2 = 16/2 = 8 m/s², directed towards the centre.

Example 6

A stone tied to a string of length 0.5 m is whirled in a horizontal circle, completing 2 revolutions per second. Find its angular velocity and linear speed.

Answer: ω = 2πn = 2π × 2 = 4π ≈ 12.57 rad/s. v = ωr = 12.57 × 0.5 = 6.28 m/s.

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Important Questions for Board Exams

1-Mark Questions (VSA)

1. Define a unit vector. What is its use?
2. Can the resultant of two vectors of unequal magnitude be zero? Justify.
3. What is the angle of projection for maximum horizontal range?
4. Is uniform circular motion accelerated motion? Why?
5. Give one example each of a scalar and a vector quantity.

2–3-Mark Questions (SA)

1. State the parallelogram law of vector addition and write the expression for the magnitude of the resultant.
2. Show that the horizontal and vertical motions of a projectile are independent of each other.
3. Derive the expression for the time of flight of a projectile.
4. Define angular velocity and derive the relation v = ωr.

5-Mark Questions (LA)

1. Define projectile motion. Derive expressions for time of flight, maximum height, and horizontal range of a projectile.
2. Show that the path of a projectile is a parabola and find the angle for maximum range.
3. Derive an expression for the centripetal acceleration of a body in uniform circular motion.

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Quick Revision Points

• Scalar = magnitude only; vector = magnitude + direction (obeys triangle law)
• Unit vector  = A/|A|; standard unit vectors î, ĵ, k̂ are mutually perpendicular
• Parallelogram law: R = √(A² + B² + 2AB cos θ); max at θ = 0°, min at θ = 180°
• Resolution: Aₓ = A cos θ, A_y = A sin θ; A = √(Aₓ² + A_y²)
• Analytical addition: Rₓ = ΣAₓ, R_y = ΣA_y, R = √(Rₓ² + R_y²)
• 2D motion: x and y are independent; apply v = u + at and r = r₀ + ut + ½at² per axis
• Projectile path is a parabola; horizontal velocity u cos θ is constant
• Time of flight T = 2u sin θ/g; max height H = u²sin²θ/2g; range R = u²sin 2θ/g
• Maximum range at θ = 45°, R_max = u²/g; complementary angles give equal range
• Uniform circular motion: v = ωr; ω = 2π/T = 2πn
• Centripetal acceleration a_c = v²/r = ω²r, always towards the centre

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Next Chapter: Chapter 4 — Laws of Motion