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NCERT Physics Class 11 - Chapter 4: Motion in a Plane - Notes

CBSEClass 11Physicsसमतल में गति

Learning Objectives

  • Understand scalar and vector quantities and their operations
  • Learn resolution of vectors and vector addition
  • Study projectile motion and derive relevant equations
  • Understand uniform circular motion and centripetal acceleration
  • Apply vector concepts to two-dimensional kinematics

Key Concepts

Scalars and Vectors

Scalar: A quantity with only magnitude (e.g., mass, speed, temperature, energy).

Vector: A quantity with both magnitude and direction (e.g., displacement, velocity, force, momentum).

Unit Vector: A vector with magnitude 1, denoted by a hat symbol. â = A/|A|. The unit vectors along x, y, z axes are î, ĵ, k̂.

Vector Operations

Vector Addition (Triangle Law): Place vectors head to tail; the resultant goes from the tail of the first to the head of the last.

Parallelogram Law: If two vectors are represented as adjacent sides of a parallelogram, the diagonal gives the resultant.

Resultant magnitude: R = √(A² + B² + 2AB cos θ)

Direction: tan α = B sin θ / (A + B cos θ)

Vector Subtraction: A - B = A + (-B)

Resolution of Vectors

Any vector A in a plane can be resolved into two perpendicular components:

Aₓ = A cos θ (x-component), Aᵧ = A sin θ (y-component)

A = Aₓ î + Aᵧ ĵ, |A| = √(Aₓ² + Aᵧ²), tan θ = Aᵧ/Aₓ

Dot Product (Scalar Product)

A · B = AB cos θ = AₓBₓ + AᵧBᵧ + A_zB_z

Properties: Commutative, distributive. A · A = A². If A ⊥ B, then A · B = 0.

Cross Product (Vector Product)

A × B = AB sin θ n̂ (where n̂ is perpendicular to both A and B by right-hand rule)

Properties: Not commutative (A × B = -B × A), distributive. If A ∥ B, then A × B = 0.

î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ

Projectile Motion

A projectile is any object thrown with an initial velocity that moves under gravity alone. The path is a parabola.

For a projectile launched with velocity u at angle θ with horizontal:

  • Horizontal component: uₓ = u cos θ (remains constant)
  • Vertical component: uᵧ = u sin θ (changes due to g)
  • Time of flight: T = 2u sin θ / g
  • Maximum height: H = u² sin²θ / 2g
  • Range: R = u² sin 2θ / g
  • Maximum range at θ = 45°: R_max = u²/g
  • Equation of trajectory: y = x tan θ - gx²/(2u² cos²θ)

Complementary angles (θ and 90° - θ) give the same range.

Uniform Circular Motion

An object moving in a circle with constant speed has continuously changing velocity direction.

Angular velocity: ω = Δθ/Δt = 2π/T = 2πf (rad/s)

Linear speed: v = rω

Centripetal acceleration: a_c = v²/r = rω² (directed towards the centre)

Time period: T = 2πr/v = 2π/ω

Frequency: f = 1/T = ω/2π

Summary

Motion in a plane requires vector analysis. Vectors can be added using the triangle or parallelogram law and resolved into components. Projectile motion is a combination of uniform horizontal motion and uniformly accelerated vertical motion under gravity, resulting in a parabolic trajectory. Uniform circular motion involves constant speed but changing direction, requiring centripetal acceleration directed towards the centre.

Important Terms

  • Resultant Vector: Single vector that produces the same effect as two or more vectors combined
  • Resolution: Splitting a vector into perpendicular components
  • Projectile: Object moving under the influence of gravity alone after being launched
  • Trajectory: Path followed by a projectile (parabolic)
  • Range: Horizontal distance covered by a projectile
  • Centripetal Acceleration: Acceleration directed towards the centre in circular motion

Quick Revision

  • R = √(A² + B² + 2AB cos θ) for resultant of two vectors
  • Projectile: T = 2u sin θ/g, H = u² sin²θ/2g, R = u² sin 2θ/g
  • Maximum range at θ = 45°
  • Centripetal acceleration: a = v²/r = rω², directed toward centre
  • v = rω, T = 2π/ω
  • A · B = AB cos θ (scalar), A × B = AB sin θ n̂ (vector)
NCERT Physics Class 11 - Chapter 4: Motion in a Plane - Notes | EduMunch