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NCERT SOLUTIONS

Chapter 4-Motion in a Plane

Find NCERT Solutions for Class 11 Physics Chapter 4 Motion in a Plane with step-by-step answers, vector concepts, and important numerical solutions.

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NCERT Solutions for Class 11 Physics Chapter 4 – Motion in a Plane

Vectors, projectile motion, uniform circular motion formulas, subtopics, and free PDF for Chapter 4.

About Chapter 4 – Motion in a Plane

Chapter 4, Motion in a Plane, extends kinematics from one dimension (Chapter 3) to two dimensions. This chapter is fundamental because most real-world motions – from a ball thrown at an angle to a planet orbiting the Sun – occur in a plane or in three-dimensional space. The chapter begins with a thorough treatment of vectors, covering addition, subtraction, resolution into components, and the laws of vector addition including the triangle law and parallelogram law.

A major topic is projectile motion – the motion of an object launched at an angle to the horizontal, moving under the influence of gravity alone. Students derive expressions for the time of flight, maximum height, horizontal range, and the trajectory equation (which turns out to be a parabola). The concept of splitting motion into independent horizontal (uniform) and vertical (uniformly accelerated) components is a key analytical technique that students must master.

The chapter also covers uniform circular motion – an object moving along a circular path with constant speed. Although the speed is constant, the velocity keeps changing direction, resulting in a centripetal acceleration directed toward the centre. Students learn to derive the expression for centripetal acceleration and understand the role of centripetal force. For board exams, projectile motion and circular motion are the most numerically intensive topics. JEE aspirants should pay special attention to the derivation of range and height formulas, as well as relative velocity in two dimensions, which is an extension of the concept introduced in Chapter 3.

NCERT Solutions for Class 11 Physics Chapter 4 – Free PDF Download

NCERT Solutions – Chapter 4: Motion in a Plane (All Exercises)

Download solved PDF with all exercise solutions covering vector operations, projectile motion numericals, and circular motion problems.

Important Formulas – Chapter 4: Motion in a Plane

ConceptFormulaDescription
Resultant of Two VectorsR = √(A² + B² + 2AB cosθ)Magnitude of resultant; θ is the angle between A and B
Direction of Resultanttan φ = B sinθ / (A + B cosθ)Angle φ that resultant makes with vector A
Unit Vectorâ = A⃗ / |A|Vector of magnitude 1 in direction of A
Dot ProductA⃗ · B⃗ = AB cosθScalar product; used in work calculations
Cross Product (magnitude)|A⃗ × B⃗| = AB sinθVector product; direction given by right-hand rule
Projectile – Time of FlightT = 2u sinθ / gTotal time from launch to landing (same level)
Projectile – Maximum HeightH = u² sin²θ / (2g)Maximum vertical rise above launch point
Projectile – Horizontal RangeR = u² sin 2θ / gHorizontal distance; maximum when θ = 45°
Projectile – Trajectoryy = x tanθ − gx²/(2u²cos²θ)Parabolic path equation
Centripetal Accelerationa_c = v²/r = ω²rDirected toward centre; for uniform circular motion
Angular Velocityω = Δθ/Δt = 2π/T = 2πfRate of change of angle; rad/s
Centripetal ForceF_c = mv²/r = mω²rNet inward force required for circular motion
Relative Velocity (2D)v⃗_AB = v⃗_A − v⃗_BVelocity of A relative to B in vector form

Subtopics of Chapter 4 – Motion in a Plane

4.1 Scalars and Vectors

Scalars have magnitude only (speed, mass); vectors have both magnitude and direction (velocity, force). Vectors follow different addition rules.

4.2 Vector Addition – Graphical Methods

Triangle law and parallelogram law for adding vectors. The resultant is the closing side of the triangle or diagonal of the parallelogram.

4.3 Resolution of Vectors

Any vector can be resolved into two perpendicular components: A_x = A cosθ (horizontal), A_y = A sinθ (vertical).

4.4 Vector Addition – Analytical Method

Add x-components and y-components separately. R_x = A_x + B_x; R_y = A_y + B_y; |R| = √(R_x² + R_y²).

4.5 Motion in a Plane

Position vector r⃗ = xî + yĵ. Velocity and acceleration are also vectors. Equations of motion apply separately to x and y components.

4.6 Projectile Motion

Horizontal: uniform velocity; vertical: uniform acceleration due to gravity. The path is a parabola. Key angle for max range = 45°.

4.7 Uniform Circular Motion

Speed constant but velocity direction continuously changes. Centripetal acceleration = v²/r directed toward centre of circle.

4.8 Relative Velocity in 2D

For two objects moving in a plane, relative velocity is found by vector subtraction. Important for river-boat and rain-umbrella problems.

Resource NameDescriptionBest For
NCERT SolutionsDetailed answers and explanations for NCERT textbook questions across all classes and subjects.Homework, assignments, and exam preparation
NCERT Solutions for Class 11Chapter-wise solutions for all Class 11 subjects including Physics, Chemistry, Mathematics, Biology, and English.Class 11 board exam preparation
NCERT Solutions for Class 11 PhysicsStep-by-step solutions covering all chapters such as Motion, Laws of Motion, Work Energy and Power, Thermodynamics, and Waves.Concept building and numerical problem-solving
NCERT Exemplar Class 11 PhysicsAdvanced and application-based questions designed to strengthen conceptual understanding and analytical skills.JEE, NEET, Olympiads, and higher-order practice
Physics FormulaChapter-wise collection of important formulas, equations, and derivations for quick revision.Last-minute revision and numerical practice

Quick Reference – Projectile Motion Summary

ParameterFormulaValue at Special Point
Initial horizontal velocityu_x = u cosθSame throughout flight (constant)
Initial vertical velocityu_y = u sinθZero at maximum height
Vertical velocity at time tv_y = u sinθ − gtv_y = 0 at H_max
Time to reach max heightt = u sinθ / gHalf the total time of flight
Max height Hu² sin²θ / 2gMaximum when θ = 90°
Range Ru² sin2θ / gMaximum when θ = 45°; R(θ) = R(90°−θ)
Speed at any time tv = √(u_x² + v_y²)Minimum at top = u cosθ
Angle of velocity at ttan α = v_y / u_xα = 0 at highest point

NCERT Solutions for Class 11 Physics Chapter 4: Motion in a Plane – FAQs