NCERT Plus One (Class 11) Physics • Chapter 3

Motion in a Plane

Scalars vs vectors, graphical & analytical vector addition, 2D kinematics, projectile motion & uniform circular motion.

📄 Official NCERT Textbook PDF

Topic 3.2: Scalars vs Vectors & Multiplication by Scalars

If your amma asks the temperature in Munnar, saying 16 °C tells the whole story — no arrow needed! That is a scalar. Now suppose a friend asks how to reach the boat jetty from Kochi metro station: saying "walk 500 m" leaves them completely lost unless you point 500 m West! Size PLUS direction PLUS arrow-addition rules make a vector.
SCALAR vs VECTORScalar: Magnitude + unit (mass, speed, time, work, electric current).
Vector: Magnitude + direction + triangle addition law (displacement, velocity, force).
SPECIAL VECTORSEqual vectors: Same magnitude & direction. Unit vector: Â = A/|A| (length = 1, dimensionless).
MULTIPLY BY SCALAR λλ > 0 scales length (same way); λ < 0 scales AND flips 180° opposite!
🧠 Memory Trick: A Unit Vector is just a pure "direction pointer" — divide any vector by its own length |A| to strip its size down to 1!
💡 Simple Kid-Check: Draw a vector as an arrow! Its length shows magnitude, and its arrowhead shows direction. Slide that arrow parallel to itself anywhere on the paper without tilting or stretching it — it stays the exact same free vector!
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    Scalars: Completely described by a number and unit. Examples: distance, mass, temperature, volume, speed, work, energy, pressure, and electric current.
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    Vectors & Unit Vectors: Represented in bold or with a top bar (A). Dividing A by its magnitude A creates unit vector  = A / A with magnitude 1 and no units!
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    Multiplying by Scalar λ: Multiplying by a pure number 2 doubles length; multiplying by −1.5 makes it 1.5× longer in the exact opposite direction. Multiplying velocity v by mass m creates momentum p = mv with new units!
|λ A| = |λ| · |A|  |  Â = A|A|  |  A = |A| Â
🧒 Quick Check: Vector A points 10 m North. What is −0.5 A?
Magnitude = 0.5 × 10 = 5 m, and the minus sign flips North to South → 5 m South!
Giant 1-mark trap: Electric current has BOTH magnitude (e.g. 5 A) AND direction along a wire — yet it is a SCALAR! Why? Because two wire currents of 3 A and 4 A meeting at a junction ALWAYS add by ordinary arithmetic to 7 A, NEVER by vector parallelogram addition!

Topic 3.4: Graphical Vector Addition & Subtraction

A traditional wooden vallam in Alappuzha backwaters rows straight across toward the opposite bank at 4 km/h while the river current sweeps it sideways at 3 km/h! Does the boat move at 4 + 3 = 7 km/h? Nope — draw those two arrows head-to-tail and Pythagoras instantly gives 5 km/h diagonally!
TRIANGLE LAWPlace tail of B on head of A. Resultant R goes from tail of A to head of B!
PARALLELOGRAM LAWJoin tails of A and B. Resultant R = diagonal from common origin!
SUBTRACTION RULEA − B = A + (−B) (flip B by 180°, then add!)
🧠 Memory Trick: Resultant of two vectors A and B can NEVER exceed A + B and can NEVER drop below |A − B|!
💡 Simple Kid-Check: Walking 3 blocks East and 4 blocks North lands you at the exact same spot as walking diagonally 5 blocks Northeast! Triangle law puts arrows nose-to-tail like train coaches; Parallelogram law puts both tails at the start corner and takes the main diagonal!
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    Commutative & Associative: Vector addition works in any order: A + B = B + A and (A + B) + C = A + (B + C).
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    Polygon Law (Many Vectors): Chain multiple vectors head-to-tail; if the final tip lands right back on the starting tail (closed polygon), the resultant is the null vector 0!
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    Subtraction & 2D Relative Velocity: Subtracting B from A means flipping B to −B and adding! That is why rain falling straight down (v_r) while you cycle forward (v_c) hits you at v_rc = v_r − v_c — tilted from the front, so you lean your umbrella forward!
Allowed Resultant Range:   |A − B| ≤ |A + B| ≤ A + B
🧒 Quick Check: Can three forces of 3 N, 4 N, and 9 N keep a knot in equilibrium (zero sum)?
Wait! The biggest sum of 3 and 4 is 3 + 4 = 7 N, which can NEVER cancel 9 N! (Triangle rule fails: sum of any two must be ≥ the third!)
Beware of the umbrella/rain question and resultant-range MCQ! If two forces are 6 N and 8 N, their resultant can ONLY lie between 8 − 6 = 2 N and 8 + 6 = 14 N — an option like 1 N or 16 N is impossible! And in a parallelogram, the common-tail diagonal is A + B, while the other diagonal is A − B.

Topic 3.5 – 3.6: Resolution of Vectors & Analytical Method

Picture pulling a heavy suitcase at Kochi airport using a tilted handle at 37° to the floor! Part of your pull slides the bag forward along the tiles, and part lifts it slightly upward. Breaking one slanted vector into neat perpendicular East–North shadows along x and y axes is called vector resolution!
RECTANGULAR COMPONENTSA_x = A cos θ,  A_y = A sin θ  ⇒  A = A_x î + A_y ĵ
MAGNITUDE & ANGLE|A| = √(A_x² + A_y²)  |  tan θ = A_y / A_x
PARALLELOGRAM FORMULAR = √(A² + B² + 2AB cos θ)  |  tan α = B sin θA + B cos θ
🧠 Memory Trick: Adjacent to angle θ = cos θ (Cozy with the angle!); Opposite side = sin θ! Two equal vectors P at angle θ give R = 2P cos(θ/2)!
💡 Simple Kid-Check: Drawing graphs with a ruler and protractor is slow and slightly sloppy. Instead, chop every slanted vector into flat horizontal (î) and straight vertical (ĵ) pieces! Adding vectors then becomes elementary-school arithmetic: add all î pieces together, add all ĵ pieces together!
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    Splitting into Unit Vectors: Using right-triangle trigonometry for angle θ with the +x axis: A_x = A cos θ and A_y = A sin θ. Then A + B = (A_x + B_x)î + (A_y + B_y)ĵ.
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    Law of Cosines (Analytical Parallelogram): For two vectors A and B placed tail-to-tail with angle θ between them: R = √(A² + B² + 2AB cos θ), making angle α with A via tan α = B sin θ / (A + B cos θ).
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    Special Angles Cheat Table: θ = 0° ⇒ R = A + B; θ = 90° ⇒ R = √(A² + B²); θ = 120° with equal vectors P, P ⇒ R = P; θ = 180° ⇒ R = |A − B|.
Addition: R = √(A² + B² + 2AB cos θ)  |  Subtraction: |A − B| = √(A² + B² − 2AB cos θ)
🧒 Super-Easy Example: Two equal forces of 10 N each act at 120° to each other. Find their resultant!
R = √(10² + 10² + 2(10)(10) cos 120°) = √(100 + 100 + 200(−½)) = √100 = 10 N! (Shortcut: 2P cos(60°) = 2(10)(½) = 10 N!)
Watch TWO sneaky traps! First: Never blindly assume horizontal is always cos θ — if an exam gives the angle θ measured from the vertical y-axis, then A_y = A cos θ and A_x = A sin θ! Second: In √(A² + B² + 2AB cos θ), θ MUST be the angle when arrows are tail-to-tail!

Topic 3.7 – 3.8: Motion & Constant Acceleration in a Plane

Imagine a football gliding across a rain-slick Kochi turf while a steady side breeze pushes it East! Does the side wind speed up or slow down how fast the ball reaches the North goal line? Not at all — in a 2D plane, motion along x and motion along y live completely separate lives sharing only the same clock t!
2D VECTOR KINEMATICSv = drdt = v_x î + v_y ĵ (ALWAYS tangent to path!)
CONSTANT ACCEL IN 2Dv = v₀ + at  |  r = r₀ + v₀t + ½at²
INDEPENDENCE PRINCIPLEMotion along x & y are independent 1D motions linked only by time t!
🧠 Memory Trick: Split EVERY 2D problem into two side-by-side 1D tables — an X-column (x, v_x, a_x) and a Y-column (y, v_y, a_y)!
💡 Simple Kid-Check: Stand two shadow flashlights around a glowing flying puck — one shining onto the floor (x-shadow) and one onto the side wall (y-shadow). Even when the puck swoops along a graceful curve in 2D space, each shadow simply moves like a plain Chapter-2 straight-line object!
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    Tangent Velocity Rule: As Δt → 0, displacement Δr aligns along the curve, so instantaneous velocity v is always tangent to the trajectory at every point.
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    Component Equations (Constant a):
    x(t) = x₀ + v₀ₓ t + ½ a_x t²  and  y(t) = y₀ + v₀ᵧ t + ½ a_y t²
    v_x(t) = v₀ₓ + a_x t  and  v_y(t) = v₀ᵧ + a_y t.
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    Two Key Shapes: If a is parallel to v₀, path is a straight line; if a is constant at an angle to v₀, path curves into a parabola!
🧒 Super-Easy Example: A particle starts at origin with v₀ = 5î m/s and constant acceleration a = (3î + 2ĵ) m/s². Find speed after t = 2 s!
• v_x = 5 + 3(2) = 11 m/s,  v_y = 0 + 2(2) = 4 m/s.
• Speed = √(11² + 4²) = √137 ≈ 11.7 m/s!
Classic demonstration trap: If Bullet A is dropped straight down from a cliff and Bullet B is fired horizontally at 300 m/s from the exact same height at the same instant, BOTH hit the ground at the EXACT SAME TIME! Why? Horizontal velocity v_x has zero effect on vertical gravity drop h = ½gt²!

Topic 3.9: Projectile Motion: Trajectory, Time, Height & Range

When Lionel Messi or a Kerala Santosh Trophy striker curls a long kick downfield at angle θ, gravity ONLY tugs vertically downward (a_y = −g) while nothing pushes horizontally (a_x = 0). That mismatch between steady horizontal cruising and gravity-pulled rise-and-fall carves the famous symmetric rainbow arch — a parabola!
TRAJECTORY (PARABOLA)y = (tan θ)x − g2v₀²cos²θ x²
TIME & MAX HEIGHTT = 2 v₀ sin θg  |  H = v₀² sin² θ2g
HORIZONTAL RANGER = v₀² sin 2θg  ⇒  Max at θ = 45° (R_max = v₀²/g)
🧠 Memory Trick: Complementary angles θ and (90° − θ) (like 30° & 60°) land at the EXACT SAME Range R! And at the peak, v_min = v₀ cos θ ≠ 0!
💡 Simple Kid-Check: A projectile is just TWO simple motions glued together: horizontally it glides forward at constant speed v₀ cos θ forever, while vertically it behaves exactly like a ball tossed straight up at v₀ sin θ!
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    No Horizontal Force: a_x = 0 ⇒ v_x = v₀ cos θ stays constant everywhere; x = (v₀ cos θ) t.
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    Vertical Gravity: a_y = −g ⇒ v_y = v₀ sin θ − gt; y = (v₀ sin θ)t − ½gt². Setting y = 0 gives flight time T = 2v₀ sin θ / g; setting v_y = 0 at t = T/2 gives max height H = v₀² sin²θ / (2g).
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    Range & Bonus Ratios: R = v₀ cos θ · T = v₀² sin(2θ) / g. Because sin(2(90°−θ)) = sin(180°−2θ) = sin(2θ), complementary launch angles hit identical targets! Also 4H / R = tan θ (so R = H when tan θ = 4 ⇒ θ ≈ 76°!).
At Peak: v = v₀ cos θ (horizontal),  a = g (downward),  Angle between v & a = 90°!
🧒 Super-Easy Example: A cricketer hits a ball at 20 m/s at 30° above the ground (g = 10 m/s²).
• T = 2(20)(0.5)/10 = 2 s  |  H = 400(0.25)/20 = 5 m
• R = 400 sin(60°)/10 = 40 × 0.866 = 34.6 m (same range as launching at 60°!).
Top Kerala DHSE 1-mark trap: "What is the velocity and acceleration at the highest point of a projectile?" Velocity is NOT zero! Only v_y = 0, so speed at top equals v₀ cos θ (purely horizontal), making a 90° angle with downward gravity a = g!

Topic 3.10: Uniform Circular Motion & Centripetal Acceleration

Tie a small stone to a string and whirl it in a steady horizontal circle overhead at constant speed 5 m/s. Is its acceleration zero just because the speedometer stays fixed at 5 m/s? Look at the direction — your arrow keeps spinning 360°! Turning that velocity arrow every instant demands a steady inward tug called centripetal acceleration.
ANGULAR LINKv = ω R,  ω = 2πT = 2πν (rad/s)
CENTRIPETAL ACCELa_c = v²R = ω² R = 4π² ν² R
DIRECTION RULEVelocity v: Tangent to circle.
Accel a_c: Always toward CENTRE (90° to v)!
🧠 Memory Trick: Centri-petal = Centre-seeking! In UCM, |v| and |a_c| stay constant, while vectors v and a_c rotate continuously!
💡 Simple Kid-Check: Speed is just how fast; velocity includes direction! Even at fixed speed, forcing a moving scooter around a roundabout bends its velocity vector inward toward the centre — and any change in velocity vector means non-zero acceleration!
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    Why Acceleration Points Inward: Compare position r and velocity v at two nearby points separated by tiny angle Δθ. Since v₁ ⊥ r₁ and v₂ ⊥ r₂, the velocity triangle Δv = v₂ − v₁ is geometrically similar to the position triangle, pointing straight toward the centre as Δt → 0!
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    Magnitude Derivation: By similarity, |Δv| / v = |Δr| / R ≈ (v Δt) / R, so dividing by Δt yields a_c = v² / R.
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    Angular Speed Form: Using arc length Δs = R Δθ, speed v = ω R, giving a_c = ω² R = 4π² ν² R!
v = ωR  |  a_c = v²R = ω²R  |  Frequency in Hz: ν = rpm60
🧒 Super-Easy Example: An insect trapped in a circular groove of radius 12 cm (0.12 m) completes 7 revolutions in 100 s.
• Frequency ν = 7/100 = 0.07 Hz,  ω = 2π(0.07) ≈ 0.44 rad/s.
• Linear speed v = ωR = 0.44 × 0.12 = 0.053 m/s;  a_c = ω²R = (0.44)²(0.12) ≈ 0.023 m/s²!
Watch TWO classic exam traps! (1) "In Uniform Circular Motion, acceleration is constant." — FALSE! Its magnitude v²/R is constant, but its direction keeps turning toward the centre, so the vector a_c is NOT constant! (2) Always convert rpm to rev/sec by dividing by 60 before multiplying by 2π!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 1 Marks): At the highest point of a projectile's path, is its velocity zero? Justify your answer.

No! Only the vertical component becomes zero (v_y = 0). The horizontal component remains constant at v_x = v₀ cos θ, so speed at the peak is v₀ cos θ (horizontal).

Q2 (Kerala HSE Model • 1 Marks): For a given initial speed, at what angle of projection does a projectile achieve (a) maximum horizontal range and (b) equal horizontal range to a 35° launch?

(a) Maximum range occurs at θ = 45° (sin 2θ = 1).
(b) Complementary angle 90° − 35° = 55° gives the exact same horizontal range.

Q3 (Kerala HSE Say • 2 Marks): State the Parallelogram Law of vector addition and write the expression for the magnitude and direction of the resultant of two vectors A and B inclined at angle θ.

Statement: If two vectors acting simultaneously at a point are represented in magnitude and direction by two adjacent sides of a parallelogram drawn from that point, their resultant is represented by the diagonal passing through that same common point.
Magnitude: R = √(A² + B² + 2AB cos θ)
Direction (angle α with A): tan α = (B sin θ) / (A + B cos θ).

Q4 (Kerala HSE March • 2 Marks): Rain is falling vertically at 35 m/s. A girl rides a bicycle from East to West at 12 m/s. In which direction should she hold her umbrella?

Velocity of rain relative to cyclist: v_rc = v_r − v_c.
Since cyclist moves West, −v_c points East at 12 m/s while v_r points downward at 35 m/s.
Angle θ with vertical: tan θ = 12 / 35 ≈ 0.343 ⇒ θ ≈ 19° tilted toward West (forward in direction of motion)!

Q5 (Kerala HSE March • 3 Marks): Show that the path followed by a projectile fired at an angle θ with horizontal is a parabola.

Let launch speed be v₀ at angle θ: v₀ₓ = v₀ cos θ, v₀ᵧ = v₀ sin θ, a_x = 0, a_y = −g.
1. Along x: x = (v₀ cos θ) t ⇒ t = x / (v₀ cos θ).
2. Along y: y = (v₀ sin θ) t − ½ g t².
Substitute t: y = (tan θ) x − [g / (2 v₀² cos² θ)] x².
This is of the quadratic form y = ax − bx², which is the equation of a parabola!

Q6 (Kerala HSE Model • 3 Marks): Derive expressions for (a) Time of flight, (b) Maximum height, and (c) Horizontal range of a projectile launched with speed v₀ at angle θ.

(a) Time of flight (T): At return to ground y = 0: 0 = (v₀ sin θ)T − ½gT² ⇒ T = (2 v₀ sin θ)/g.
(b) Max Height (H): At peak v_y = 0: 0 − (v₀ sin θ)² = 2(−g)H ⇒ H = (v₀² sin² θ)/(2g).
(c) Range (R): R = v₀ₓ · T = (v₀ cos θ)(2v₀ sin θ / g) = (v₀² sin 2θ)/g.

Q7 (NCERT / DHSE Say • 3 Marks): A cricket ball is thrown at a speed of 28 m/s in a direction 30° above the horizontal. Calculate (a) maximum height, (b) time taken to return to same level, and (c) horizontal distance from thrower to landing point. (g = 9.8 m/s²)

Given v₀ = 28 m/s, θ = 30°, g = 9.8 m/s²:
(a) H = (28² sin²30°)/(2×9.8) = (784 × 0.25)/19.6 = 10 m.
(b) T = (2 × 28 × sin 30°)/9.8 = 28/9.8 = 2.86 s.
(c) R = (28² sin 60°)/9.8 = 80 × 0.866 = 69.3 m.

Q8 (Kerala HSE March • 3 Marks): Define centripetal acceleration in uniform circular motion and derive a_c = v²/R.

Centripetal acceleration is the radially inward acceleration experienced by a body moving along a circular path at constant speed due to continuous change in velocity direction.
Derivation: As particle moves arc Δs = R Δθ from P to P' in tiny time Δt, v₁ ⊥ r₁ and v₂ ⊥ r₂, so angle between v₁ & v₂ is also Δθ.
By similar isosceles triangles: |Δv|/v = |Δr|/R ≈ (v Δt)/R.
Dividing by Δt as Δt → 0: a_c = v² / R = ω² R directed inward!