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CBSE Class 11 Physics★ 6-8 marks · CBSE Class 11 Physics Chapter 3; vectors and projectiles are high-yield for JEE and NEET

Motion in a Plane

Two-dimensional kinematics — vectors, projectile motion, and uniform circular motion. CBSE Class 11 Physics Chapter 3.

⏱ 15 min readMedium difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Distinguish scalars and vectors and resolve a vector into components
2Compute dot and cross products of vectors
3Apply the independence of horizontal and vertical motion to projectiles
4Derive and use time of flight, maximum height, and range
5Analyse uniform circular motion and centripetal acceleration

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Why this chapter matters

Real motion happens in two and three dimensions. Vectors are the mathematical language of physics, and projectile and circular motion are classic applications that appear throughout mechanics, gravitation, and electromagnetism.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Class 11 Physics: Motion in a Straight Line

One-dimensional kinematics and equations of motion

Motion in a Plane

'Motion in two dimensions is the bridge between one-dimensional kinematics and the real world.' — Physics Essentials

1. Chapter Overview

While Chapter 2 dealt with motion along a STRAIGHT LINE, real-world motion is usually in TWO or THREE dimensions. This chapter introduces VECTORS as the mathematical tool for handling multi-dimensional motion, then applies them to two IMPORTANT cases: PROJECTILE motion and UNIFORM CIRCULAR motion.

2. Scalars and Vectors

Property	Scalar	Vector
Definition	Only magnitude	Magnitude + Direction
Examples	Mass, speed, energy, time	Displacement, velocity, force
Addition	Simple algebra	Parallelogram/Triangle law
Multiplication	Ordinary multiplication	Dot product, Cross product

Vector Operations

Addition: Triangle law (head-to-tail), Parallelogram law
Subtraction: A — B = A + (-B)
Resolution: Breaking a vector into perpendicular components
For a vector A at angle θ with x-axis: Aₓ = Acosθ, Aᵧ = Asinθ, |A| = √(Aₓ² + Aᵧ²), θ = tan⁻¹(Aᵧ/Aₓ)

Dot and Cross Products

Dot Product: A·B = |A||B|cosθ (scalar result)
Cross Product: A × B = |A||B|sinθ n̂ (vector result, perpendicular to both A and B)

3. Motion in 2D — Position, Velocity, Acceleration

Position Vector: r = xî + yĵ
Displacement: Δr = r₂ — r₁
Average Velocity: v̄ = Δr/Δt
Instantaneous Velocity: v = dr/dt = (dx/dt)î + (dy/dt)ĵ = vₓî + vᵧĵ
Acceleration: a = dv/dt = (dvₓ/dt)î + (dvᵧ/dt)ĵ

Independence of Motion

Motion along x and y axes are INDEPENDENT of each other
This principle is the KEY to solving projectile problems

4. Projectile Motion

Key Assumptions

Only force acting is GRAVITY (air resistance ignored)
Acceleration: aₓ = 0, aᵧ = -g

Analysis

Horizontal: x = ucosθ × t (constant velocity)
Vertical: y = usinθ × t — ½gt² (constant acceleration)

Key Results

Quantity	Formula
Time of Flight (T)	2usinθ/g
Maximum Height (H)	u²sin²θ/2g
Horizontal Range (R)	u²sin2θ/g
Range is maximum at	θ = 45°
Equal ranges at	θ and (90° — θ)

Worked Problem

Q: A projectile is fired at 50 m/s at 30° to horizontal. Find T, H, R (g = 10 m/s²). Solution:

uₓ = 50cos30° = 43.3 m/s, uᵧ = 50sin30° = 25 m/s
T = 2usinθ/g = 2×25/10 = 5 s
H = u²sin²θ/2g = 625/20 = 31.25 m
R = u²sin2θ/g = 2500×sin60°/10 = 2500×0.866/10 = 216.5 m

5. Uniform Circular Motion

Angular Displacement Δθ (in radians)
Angular Velocity ω = dθ/dt (constant in UCM)
Linear Speed v = ωR
Centripetal Acceleration a_c = v²/R = ω²R (directed TOWARDS centre)
Centripetal Force F_c = mv²/R

Key Point

In UCM, speed is CONSTANT but velocity changes (direction keeps changing). Hence it is ACCELERATED motion — the acceleration is directed toward the centre.

Worked Problem

Q: A mass tied to a string rotates in a horizontal circle of radius 0.5 m with speed 4 m/s. Find centripetal acceleration. A: a_c = v²/R = 16/0.5 = 32 m/s² (more than 3g!)

6. Common Mistakes

Forgetting that velocity changes direction in projectile motion: Only the vertical component changes; horizontal is constant
Confusing range maximum at 45°: sin2θ = 1 when 2θ = 90°, so θ = 45°
Adding vectors as scalars: A + B = √(A² + B² + 2ABcosθ), NOT A + B
Believing UCM has no acceleration: Direction changes, so acceleration exists

7. CBSE Exam Focus

Resolution of vectors into components (1-mark)
Projectile — derivation of T, H, R (5-mark)
Numerical on projectile motion
Centripetal acceleration derivation
Dot and cross product properties

8. Key Formulas

Aₓ = Acosθ, Aᵧ = Asinθ
A·B = ABcosθ
|A × B| = ABsinθ
T = 2usinθ/g, H = u²sin²θ/2g, R = u²sin2θ/g
a_c = v²/r = ω²r

9. Self-Test (5+ Q&A)

Q1: A vector has components Aₓ = 3, Aᵧ = 4. Find its magnitude and direction. A: |A| = 5, θ = tan⁻¹(4/3) = 53.13°.

Q2: What is the dot product and cross product of î and ĵ? A: î·ĵ = 0 (perpendicular), î × ĵ = k̂.

Q3: For what angle of projection is the horizontal range maximum? A: 45° (sin2θ = 1 when 2θ = 90°).

Q4: A stone is thrown horizontally at 20 m/s from a height of 80 m. Find the range (g = 10 m/s²). A: Time to fall: 80 = ½×10×t² → t = 4 s. Range = 20×4 = 80 m.

Q5: A particle moves in a circle of radius 20 cm with constant speed completing 5 revolutions per second. Find speed and centripetal acceleration. A: ω = 2πf = 10π rad/s. v = ωR = 10π×0.2 = 2π m/s. a_c = ω²R = (100π²)×0.2 = 20π² m/s².

10. Conclusion

Motion in a plane introduces VECTORS — the mathematical language of physics. Projectile motion is the CLASSIC application of 2D kinematics, and uniform circular motion introduces you to the concept of centripetal force that will be essential in gravitation and electromagnetism. Master vectors NOW; they appear in every physics topic that follows.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Vector components

Ax = A cos(theta); Ay = A sin(theta); |A| = sqrt(Ax^2 + Ay^2)

Resolve along perpendicular axes.

Dot and cross products

A.B = AB cos(theta); |A x B| = AB sin(theta)

Dot product is scalar; cross product is a perpendicular vector.

Projectile results

T = 2u sin(theta)/g; H = u^2 sin^2(theta)/2g; R = u^2 sin(2theta)/g

Range is maximum at theta = 45 degrees.

Centripetal acceleration

a_c = v^2/R = omega^2 R; v = omega R

Directed toward the centre; speed constant but velocity changing.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Adding vectors like scalars

✓ Resultant = sqrt(A^2 + B^2 + 2AB cos(theta)), not A + B unless they are parallel.

WATCH OUT

✗ Thinking horizontal velocity changes in a projectile

✓ Horizontal velocity is constant (a_x = 0); only the vertical component changes due to gravity.

WATCH OUT

✗ Believing uniform circular motion is unaccelerated

✓ Speed is constant but direction (hence velocity) changes, so there is centripetal acceleration toward the centre.

WATCH OUT

✗ Forgetting range is symmetric about 45 degrees

✓ Angles theta and (90 - theta) give the same range because sin(2theta) is symmetric.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Vectors

Addition, resolution, dot and cross products

Questions

Projectile motion

Time of flight, height, range, trajectory

Questions

Circular motion

Angular velocity and centripetal acceleration

Questions

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Vectors

A vector has components Ax = 3, Ay = 4. Find its magnitude and direction.

▶ Show solution

|A| = sqrt(9 + 16) = 5. Direction theta = tan^-1(4/3) = 53.13 degrees from the x-axis.

Q2MEDIUM· Projectile

A projectile is fired at 50 m/s at 30 degrees to the horizontal. Find T, H, and R (g = 10 m/s^2).

▶ Show solution

uy = 25 m/s. T = 2uy/g = 5 s. H = u^2 sin^2(theta)/2g = 625/20 = 31.25 m. R = u^2 sin(60)/g = 2500 x 0.866/10 = 216.5 m.

Q3MEDIUM· Projectile

A stone is thrown horizontally at 20 m/s from 80 m height. Find its range (g = 10 m/s^2).

▶ Show solution

Time to fall: 80 = (1/2)(10)t^2, so t = 4 s. Range = 20 x 4 = 80 m.

Q4HARD· Circular Motion

A particle moves in a circle of radius 20 cm at 5 revolutions per second. Find its speed and centripetal acceleration.

▶ Show solution

omega = 2 pi f = 10 pi rad/s. v = omega R = 10 pi x 0.2 = 2 pi m/s. a_c = omega^2 R = (100 pi^2)(0.2) = 20 pi^2 m/s^2.

Q5EASY· Vectors

What are i.j and i x j?

▶ Show solution

i.j = 0 (perpendicular unit vectors). i x j = k (right-hand rule).

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Vectors have magnitude and direction; add by triangle/parallelogram law.
•Resolve: Ax = A cos(theta), Ay = A sin(theta).
•Dot product is scalar (AB cos theta); cross product is a perpendicular vector (AB sin theta).
•Horizontal and vertical motions of a projectile are independent.
•T = 2u sin(theta)/g; H = u^2 sin^2(theta)/2g; R = u^2 sin(2theta)/g; R max at 45 degrees.
•In UCM, speed is constant but velocity changes direction.
•Centripetal acceleration a_c = v^2/R points to the centre.

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 6-8 marks across the chapter

Question type	Marks each	Typical count	What it tests
Projectile motion	3-5	1	Derivation and numericals for T, H, R
Vectors	2-3	1	Resolution, dot and cross products
Circular motion	2-3	1	Centripetal acceleration

Prep strategy

Practise resolving vectors and product operations
Derive projectile formulae from independence of motion
Memorise that range is maximum at 45 degrees
Understand centripetal acceleration direction

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Sports and ballistics

Projectile equations predict the flight of balls, javelins, and projectiles for optimal range.

Vehicle cornering

Centripetal force concepts explain why roads are banked and how fast a car can turn safely.

Satellites and rides

Circular motion underlies satellite orbits, centrifuges, and amusement-park rides.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Resolve initial velocity into horizontal and vertical parts first
Treat the two directions independently
State the formula before substituting in projectile problems
Remember a_c always points toward the centre

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

Solve projectile motion on an inclined plane and find range up/down the incline.
Analyse non-uniform circular motion with both tangential and centripetal acceleration.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 11 Physics examHigh

JEE Main and Advanced (Vectors & Projectiles)High

NEET PhysicsHigh

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

The horizontal motion has constant velocity (x increases linearly with time) while the vertical motion has constant downward acceleration (y depends on t squared). Eliminating time between x = u cos(theta) t and y = u sin(theta) t - (1/2)g t^2 gives an equation of the form y = ax - bx^2, which is a parabola.

Acceleration is the rate of change of velocity, and velocity is a vector with both magnitude and direction. In uniform circular motion the magnitude (speed) stays constant but the direction continuously changes, so the velocity vector changes. This change produces a centripetal acceleration of magnitude v^2/R directed toward the centre.

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Units and Measurements

The backbone of experimental physics — SI units, dimensional analysis, errors in measurement, and significant figures. CBSE Class 11 Physics Chapter 1.

Motion in a Straight Line

The kinematics of one-dimensional motion — position, velocity, acceleration, equations of motion, and relative velocity. CBSE Class 11 Physics Chapter 2.

Laws of Motion

Newton's three laws of motion — the foundation of classical mechanics — along with impulse, friction, and circular motion dynamics. CBSE Class 11 Physics Chapter 4.