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Motion in a Plane – Class 11 Physics Chapter 3 Notes & NCERT PDF

Class 11PhysicsChapter 3NCERT book: Physics Part I - Class 11

Notes and a simple summary of Chapter 3 of the NCERT Class 11 Physics book, with the official chapter PDF, flashcards, an MCQ quiz and an AI tutor for your doubts.

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Chapter 3: Motion in a Plane

Introduction

In one dimension, + and − signs handled direction. In a plane (two dimensions) or in space, direction can point anywhere, so we need vectors. This chapter first builds the language of vectors — equality, multiplication by a real number, addition and subtraction, resolution into components — and then uses it to describe motion in a plane: motion with constant acceleration, projectile motion and uniform circular motion. The same equations extend easily to three dimensions.

Key Concepts

3.2 Scalars and vectors

  • A scalar has magnitude only (distance, mass, temperature, time); it combines by ordinary algebra (e.g. a rectangle's perimeter 1.0 + 0.5 + 1.0 + 0.5 = 3.0 m).
  • A vector has magnitude and direction and obeys the triangle / parallelogram law of addition: displacement, velocity, acceleration, force. Written in bold (v) or with an arrow; its magnitude |v| = v.
  • Position vector r = OP; displacement vector PP′ joins initial and final positions and does not depend on the path — so |displacement| ≤ path length.
  • Equal vectors have the same magnitude and direction; a vector shifted parallel to itself is unchanged (free vectors).
  • 3.3 Multiplying a vector by a real number

  • λA (λ > 0) has magnitude λ|A| in the same direction; −λA points opposite. λ may carry a dimension (constant velocity × time = displacement).
  • 3.4 Addition and subtraction — graphical method

  • Head-to-tail (triangle) method and parallelogram method give the same resultant.
  • Vector addition is commutative (A + B = B + A) and associative ((A + B) + C = A + (B + C)).
  • Null vector 0: A + (−A) = 0; |0| = 0 and its direction cannot be specified. A + 0 = A, λ0 = 0, 0A = 0. A round trip P → P′ → P has a null displacement.
  • Subtraction: A − B = A + (−B).
  • 3.5 Resolution of vectors

  • Any vector A in a plane can be written as A = λa + μb along two non-collinear vectors a and b.
  • Unit vectors î, ĵ, k̂ (magnitude 1, no unit) point along x, y, z. A = |A| n̂.
  • Rectangular components: A = A_x î + A_y ĵ with A_x = A cos θ, A_y = A sin θ; A = √(A_x² + A_y²), tan θ = A_y/A_x.
  • In 3D: A_x = A cos α, A_y = A cos β, A_z = A cos γ; A = √(A_x² + A_y² + A_z²); r = x î + y ĵ + z k̂.
  • 3.6 Vector addition — analytical method

  • Add components: R_x = A_x + B_x, R_y = A_y + B_y (and R_z in 3D). Extends to any number of vectors, e.g. T = a + b − c.
  • For two vectors at angle θ: Law of cosines R² = A² + B² + 2AB cos θ; direction from Law of sines R/sin θ = A/sin β = B/sin α, or tan α = B sin θ/(A + B cos θ).
  • 3.7 Motion in a plane

  • Displacement Δr = r′ − r = Δx î + Δy ĵ.
  • Velocity v = dr/dt = v_x î + v_y ĵ, v_x = dx/dt, v_y = dy/dt; v = √(v_x² + v_y²), tan θ = v_y/v_x. Its direction is always tangent to the path.
  • Acceleration a = dv/dt = a_x î + a_y ĵ. In 2D/3D, v and a can make any angle from 0° to 180°.
  • 3.8 Motion in a plane with constant acceleration

  • v = v₀ + at and r = r₀ + v₀t + ½at², or in components x = x₀ + v₀ₓt + ½aₓt², y = y₀ + v₀ᵧt + ½aᵧt².
  • Key idea: motion in x and y can be treated independently as two simultaneous 1D motions.
  • 3.9 Projectile motion

    With launch speed v₀ at angle θ₀ and a = −g ĵ (air resistance neglected), v₀ₓ = v₀ cos θ₀ and v₀ᵧ = v₀ sin θ₀:

  • x = (v₀ cos θ₀) t; y = (v₀ sin θ₀) t − ½gt²
  • v_x = v₀ cos θ₀ (constant); v_y = v₀ sin θ₀ − gt
  • Path: y = (tan θ₀) x − g x²/[2(v₀ cos θ₀)²] — a parabola.
  • Time to maximum height t_m = v₀ sin θ₀/g; time of flight T_f = 2v₀ sin θ₀/g = 2t_m.
  • Maximum height h_m = (v₀ sin θ₀)²/2g.
  • Range R = v₀² sin 2θ₀/g; maximum range R_m = v₀²/g at θ₀ = 45°. Angles 45° ± α give equal ranges (Galileo).
  • Galileo first stated the independence of horizontal and vertical motion (Dialogue on the great world systems, 1632).
  • 3.10 Uniform circular motion

  • Constant speed on a circle; the velocity changes direction, so there is acceleration.
  • Centripetal acceleration a_c = v²/R, always towards the centre ("centre-seeking"); constant in magnitude but not a constant vector. First analysed in print by Christiaan Huygens (1673).
  • Angular speed ω = Δθ/Δt; v = Rω; a_c = ω²R.
  • Period T, frequency ν = 1/T: v = 2πR/T = 2πRν, ω = 2πν, a_c = 4π²ν²R.
  • Worked Examples

  • Umbrella in rain and wind (Example 3.1): rain 35 m s⁻¹ down, wind 12 m s⁻¹ east to west. R = √(35² + 12²) = 37 m s⁻¹; tan θ = 12/35 = 0.343 ⇒ θ ≈ 19° from vertical — tilt the umbrella towards the east.
  • Motorboat (Example 3.3): boat 25 km/h north, current 10 km/h at 60° east of south (120° between them). R = √(25² + 10² + 2·25·10·cos 120°) ≈ 22 km/h; sin φ = 10 sin 120°/21.8 ≈ 0.397 ⇒ φ ≈ 23.4° east of north.
  • Particle with r = 3.0t î + 2.0t² ĵ + 5.0 k̂ (Example 3.4): v = 3.0 î + 4.0t ĵ, a = 4.0 ĵ m s⁻². At t = 1.0 s, v = 3 î + 4 ĵ, |v| = 5.0 m s⁻¹ at ≈ 53° to the x-axis.
  • Constant acceleration from the origin (Example 3.5): v₀ = 5.0 î m/s, a = (3.0 î + 2.0 ĵ) m/s². x = 5.0t + 1.5t² = 84 m ⇒ t = 6 s; y = 1.0t² = 36 m; v = 23.0 î + 12.0 ĵ, speed ≈ 26 m s⁻¹.
  • Stone thrown horizontally off a 490 m cliff at 15 m s⁻¹ (Example 3.7): −490 = −½(9.8)t² ⇒ t = 10 s; v_y = −98 m s⁻¹, speed = √(15² + 98²) ≈ 99 m s⁻¹.
  • Cricket ball, 28 m s⁻¹ at 30° (Example 3.8): h_m = (28 sin 30°)²/(2 × 9.8) = 10.0 m; T_f = 2 × 28 × sin 30°/9.8 ≈ 2.9 s; R = 28² sin 60°/9.8 ≈ 69 m.
  • Insect in a circular groove (Example 3.9): R = 12 cm, 7 revolutions in 100 s. ω = 2π × 7/100 ≈ 0.44 rad s⁻¹; v = ωR ≈ 5.3 cm s⁻¹; a = ω²R ≈ 2.3 cm s⁻² towards the centre — not a constant vector.
  • Formula / Key-Facts Sheet

  • A_x = A cos θ, A_y = A sin θ; A = √(A_x² + A_y²); tan θ = A_y/A_x
  • R² = A² + B² + 2AB cos θ; R/sin θ = A/sin β = B/sin α
  • v = dr/dt; a = dv/dt; v = v₀ + at; r = r₀ + v₀t + ½at²
  • Projectile: T_f = 2v₀ sin θ₀/g; h_m = v₀² sin²θ₀/2g; R = v₀² sin 2θ₀/g; R_max = v₀²/g at 45°
  • Circular motion: v = Rω = 2πR/T; a_c = v²/R = ω²R = 4π²ν²R; ω = 2πν
  • Exam Focus

  • Scalar or vector? (Ex. 3.1–3.3) — e.g. work and current are scalars; impulse is the only vector among temperature, pressure, impulse, time, power, path length, energy, potential, friction coefficient and charge.
  • Vector inequalities (Ex. 3.6): |a + b| ≤ |a| + |b| etc., with equality when a and b are parallel (or antiparallel for the difference forms).
  • Displacement vs path length (Ex. 3.8–3.11): the taxi case — 23 km in 28 min gives average speed ≈ 49.3 km/h but |average velocity| for the 10 km straight distance ≈ 21.4 km/h; they differ because the path was not straight.
  • Projectile with a ceiling (Ex. 3.12): a ball at 40 m s⁻¹ under a 25 m ceiling — choose θ so that h_m = 25 m, then R = v₀² sin 2θ/g ≈ 150.5 m. Outline: sin²θ = 2gh/v₀², find θ, then R.
  • Circular motion numericals — always state the direction of a_c (towards the centre).
  • Key Terms

  • Scalar (अदिश)
  • Vector (सदिश)
  • Position vector (स्थिति सदिश)
  • Displacement vector (विस्थापन सदिश)
  • Null vector (शून्य सदिश)
  • Unit vector (एकांक सदिश)
  • Resolution of a vector (सदिश का वियोजन)
  • Resultant (परिणामी)
  • Parallelogram law (समांतर चतुर्भुज नियम)
  • Projectile (प्रक्षेप्य)
  • Time of flight (उड्डयन काल)
  • Horizontal range (क्षैतिज परास)
  • Uniform circular motion (एकसमान वृत्तीय गति)
  • Centripetal acceleration (अभिकेंद्र त्वरण)
  • Angular speed (कोणीय चाल)
  • Common Mistakes

  • Adding magnitudes of vectors directly (35 + 12 = 47) instead of using the triangle/parallelogram law (37 m s⁻¹).
  • Treating a vector's component as a vector — A_x is a number; A_x î is the vector.
  • Forgetting that the horizontal velocity of a projectile stays constant while only v_y changes.
  • Using R = v₀² sin 2θ₀/g when the landing point is not at the launch height (e.g. a stone thrown off a cliff).
  • Calling centripetal acceleration "constant" — its magnitude is constant, but its direction keeps changing.
  • Using degrees where ω must be in rad s⁻¹ in v = Rω.
  • 💡 Key Learning Points

    • ✓Vectors have magnitude and direction and add by the triangle or parallelogram law; addition is commutative and associative
    • ✓A vector resolves into rectangular components A_x = A cos θ and A_y = A sin θ, and vectors add analytically by adding components
    • ✓With constant acceleration, v = v₀ + at and r = r₀ + v₀t + ½at², and x- and y-motions are independent
    • ✓A projectile follows a parabola: T_f = 2v₀ sin θ₀/g, h_m = v₀² sin²θ₀/2g, R = v₀² sin 2θ₀/g, maximum range at 45°
    • ✓In uniform circular motion a_c = v²/R = ω²R points to the centre; v = Rω and ω = 2πν

    👨‍🏫 Teaching Tips

    • →Walk the displacement idea in the classroom: three different routes between two desks share one displacement vector
    • →Use string and a protractor on the board to add 35 and 12 m s⁻¹ perpendicular vectors, then confirm 37 m s⁻¹ with Pythagoras
    • →Give component-addition drills with î, ĵ notation before introducing the law of cosines
    • →Drop one coin and flick another horizontally off a desk at the same moment; both land together, showing independent vertical motion
    • →Plot range against angle on graph paper for a fixed v₀ to show the 45° maximum and the 45° ± α equal ranges
    • →Swing a ball on a string and release it to show the velocity is tangential, then discuss why acceleration points to the centre
    • →Insist that every projectile answer states the chosen origin and the positive x and y directions

    📋 Assessment Questions

    1. Classify volume, speed, angular velocity and displacement as scalar or vector
    2. Find the magnitude and direction of the resultant of two vectors at 120°
    3. Derive the equation of the trajectory of a projectile and show it is a parabola
    4. Show that ranges are equal for projection angles 45° + α and 45° − α
    5. Find ω, v and a for a body making 7 revolutions in 100 s on a circle of radius 12 cm

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