Exercise Solutions Available
Step-by-step answers for all textbook questions in this chapter
Chapter Introduction
Motion in a Plane extends one-dimensional kinematics into two dimensions using vectors. This chapter introduces vector algebra (addition, resolution, dot product, cross product), projectile motion (a guaranteed 5-mark topic), and uniform circular motion. It is one of the most important chapters for both the board exam and competitive exams — expect 8–10 marks. The key skill is resolving vectors into components and applying kinematic equations independently along each axis.
What this guide covers: scalars vs. vectors, vector addition (triangle, parallelogram, polygon), vector resolution, dot product, cross product, projectile motion (all formulas + derivation), uniform circular motion — all with worked numericals.
Key Definitions
- Scalar: A quantity with magnitude only (mass, speed, work, energy, temperature, distance, time).
- Vector: A quantity with magnitude and direction that obeys vector addition rules (displacement, velocity, force, momentum, acceleration).
- Unit vector: A vector of unit magnitude in the direction of a given vector. For vector A, Â = A/|A|. Standard unit vectors: î (x-axis), ĵ (y-axis), k̂ (z-axis).
- Null (zero) vector: A vector with zero magnitude and indeterminate direction. A − A = 0.
- Equal vectors: Same magnitude and same direction (can be at different positions).
- Negative vector: Same magnitude as the original but opposite direction. −A has magnitude |A|.
- Collinear vectors: Vectors along the same or parallel lines.
- Position vector: Vector from the origin to the particle's position. r = xî + yĵ.
- Displacement vector: Change in position. Δr = r<sub>2</sub> − r<sub>1</sub>.
- Resolution of vectors: Splitting a vector into components along chosen axes. A = A<sub>x</sub>î + A<sub>y</sub>ĵ, where A<sub>x</sub> = A cos θ, A<sub>y</sub> = A sin θ.
- Projectile: An object thrown into space that moves under gravity alone. Its path is a parabola.
- Uniform circular motion: Motion along a circular path at constant speed. Speed is constant but velocity changes (direction changes), so acceleration is non-zero and directed towards the centre.
Detailed Concepts
1. Scalars vs. Vectors
| Property | Scalar | Vector |
|---|---|---|
| Has magnitude? | Yes | Yes |
| Has direction? | No | Yes |
| Obeys vector addition? | No (ordinary addition) | Yes (triangle/parallelogram law) |
| Can be negative? | Yes (e.g., temperature −5°C) | Yes (opposite direction) |
| Examples | Speed, mass, energy, work, time, distance | Velocity, displacement, force, momentum, acceleration |
| Representation | By a number + unit | By arrow (length = magnitude, arrow = direction) |
| Multiplication by scalar | Product is scalar | Product is vector (same direction if scalar > 0) |
Note: Some quantities like electric current and time have direction but are scalars — they do not obey vector addition laws.
2. Vector Addition
Triangle law: If two vectors A and B are represented as two sides of a triangle in the same order, the resultant R = A + B is the third side taken in the reverse order.
Parallelogram law: If two vectors A and B are represented as adjacent sides of a parallelogram, the diagonal from the same origin gives the resultant R = A + B.
| Magnitude of resultant: | R | = √(A<sup>2</sup> + B<sup>2</sup> + 2AB cos θ), where θ is the angle between A and B. |
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Direction of resultant: tan α = (B sin θ) / (A + B cos θ), where α is the angle of R with A.
Special cases:
| θ | R | Condition | ||
|---|---|---|---|---|
| 0° (same direction) | A + B | Maximum resultant | ||
| 180° (opposite) | A − B | Minimum resultant | ||
| 90° (perpendicular) | √(A<sup>2</sup> + B<sup>2</sup>) | Pythagoras applies |
| Maximum and minimum resultant: | R | <sub>max</sub> = A + B (θ = 0°); | R | <sub>min</sub> = | A − B | (θ = 180°). The resultant of two vectors always lies between | A − B | and A + B. |
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Polygon law: For adding more than two vectors, place them head to tail; the closing side of the polygon gives the resultant. The resultant of several vectors forming a closed polygon is zero.
Vector addition in components: If A = A<sub>x</sub>î + A<sub>y</sub>ĵ and B = B<sub>x</sub>î + B<sub>y</sub>ĵ, then R = A + B = (A<sub>x</sub> + B<sub>x</sub>)î + (A<sub>y</sub> + B<sub>y</sub>)ĵ
3. Vector Multiplication
Scalar (Dot) Product: A · B = AB cos θ. Result is a scalar.
| Property | Dot Product |
|---|---|
| Result | Scalar |
| Formula | A · B = AB cos θ |
| θ = 0° | AB (maximum, parallel) |
| θ = 90° | 0 (vectors perpendicular) |
| θ = 180° | −AB (anti-parallel) |
| Commutative? | Yes: A · B = B · A |
| In components | A · B = A<sub>x</sub>B<sub>x</sub> + A<sub>y</sub>B<sub>y</sub> + A<sub>z</sub>B<sub>z</sub> |
| Geometric meaning | Projection of one vector on another |
| Uses | Finding angle between vectors, finding work (W = F · d) |
Vector (Cross) Product: A × B = AB sin θ n̂. Result is a vector perpendicular to both A and B, given by right-hand rule.
| Property | Cross Product |
|---|---|
| Result | Vector (perpendicular to both A and B) |
| Formula | A × B = AB sin θ n̂ |
| θ = 0° | 0 (vectors parallel) |
| θ = 90° | AB n̂ (maximum, perpendicular) |
| θ = 180° | 0 (anti-parallel) |
| Commutative? | No: A × B = −(B × A) |
| In components | A × B = (A<sub>y</sub>B<sub>z</sub> − A<sub>z</sub>B<sub>y</sub>)î − (A<sub>x</sub>B<sub>z</sub> − A<sub>z</sub>B<sub>x</sub>)ĵ + (A<sub>x</sub>B<sub>y</sub> − A<sub>y</sub>B<sub>x</sub>)k̂ |
| Geometric meaning | Area of parallelogram with sides A and B |
| Uses | Finding torque (τ = r × F), angular momentum (L = r × p) |
Standard results to memorise: î × î = 0, ĵ × ĵ = 0, k̂ × k̂ = 0 î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ ĵ × î = −k̂, k̂ × ĵ = −î, î × k̂ = −ĵ
4. Projectile Motion
A projectile is launched with initial velocity u at angle θ to the horizontal. Motion is analysed by resolving into horizontal and vertical components.
Initial velocity components: u<sub>x</sub> = u cos θ (constant, no horizontal acceleration); u<sub>y</sub> = u sin θ (affected by gravity).
Equations of motion at time t:
| Quantity | Horizontal | Vertical |
|---|---|---|
| Acceleration | a<sub>x</sub> = 0 | a<sub>y</sub> = −g |
| Velocity at time t | v<sub>x</sub> = u cos θ | v<sub>y</sub> = u sin θ − gt |
| Displacement at time t | x = u cos θ · t | y = u sin θ · t − ½gt<sup>2</sup> |
Key results — these formulas must be memorised:
| Quantity | Formula | Notes |
|---|---|---|
| Time of flight | T = 2u sin θ / g | Time to go up and come back to same level |
| Time to reach max height | t<sub>max</sub> = u sin θ / g | Half of total time of flight |
| Maximum height | H = u<sup>2</sup> sin<sup>2</sup> θ / (2g) | Highest point of trajectory |
| Horizontal range | R = u<sup>2</sup> sin 2θ / g | Horizontal distance covered |
| Range is maximum | At θ = 45°, R<sub>max</sub> = u<sup>2</sup> / g | sin 90° = 1 |
| Same range for θ and (90° − θ) | sin 2θ = sin(180° − 2θ) | Complementary angles give same R |
| Velocity at any time t | v = √(v<sub>x</sub><sup>2</sup> + v<sub>y</sub><sup>2</sup>) | Speed at any instant |
| Angle of velocity at time t | tan α = v<sub>y</sub>/v<sub>x</sub> | Angle with horizontal |
Equation of trajectory: Eliminate t from x and y equations: y = x tan θ − (g x<sup>2</sup>) / (2u<sup>2</sup> cos<sup>2</sup> θ)
This is a parabola (y = ax − bx<sup>2</sup> form), confirming that projectile motion is parabolic.
Derivation of equation of trajectory (important for exams): From x = u cos θ · t → t = x / (u cos θ) Substituting in y: y = u sin θ · (x / u cos θ) − ½g(x / u cos θ)<sup>2</sup> y = x tan θ − gx<sup>2</sup> / (2u<sup>2</sup> cos<sup>2</sup> θ)
Horizontal projectile (θ = 0°, launched from height h):
| Quantity | Formula |
|---|---|
| Time of flight | t = √(2h/g) |
| Horizontal range | R = u × √(2h/g) |
| Velocity on hitting ground | v = √(u<sup>2</sup> + 2gh) |
| Angle at impact | tan α = √(2gh)/u |
5. Uniform Circular Motion
A body moves in a circle of radius r with constant speed v.
- Speed: Constant (v = 2πr / T = 2πrf)
- Velocity: Changes direction continuously; always tangential to the circle
- Centripetal acceleration: a<sub>c</sub> = v<sup>2</sup>/r = ω<sup>2</sup>r = 4π<sup>2</sup>r/T<sup>2</sup>, directed towards the centre
- Angular velocity: ω = 2π/T = 2πf, where f = frequency
- Angular acceleration: α = dω/dt (zero for uniform circular motion)
- Centripetal force: F<sub>c</sub> = mv<sup>2</sup>/r = mω<sup>2</sup>r (towards centre)
Relations between linear and angular quantities:
| Linear Quantity | Angular Quantity | Relation |
|---|---|---|
| Distance s | Angle θ | s = rθ |
| Speed v | Angular speed ω | v = rω |
| Acceleration a<sub>t</sub> | Angular acceleration α | a<sub>t</sub> = rα |
| Centripetal acceleration a<sub>c</sub> | — | a<sub>c</sub> = ω<sup>2</sup>r = v<sup>2</sup>/r |
Important: In uniform circular motion, the net force is centripetal — it does not change the speed, only the direction. There is no outward "centrifugal force" in an inertial frame.
Non-uniform circular motion: When speed changes, there is both centripetal acceleration (v<sup>2</sup>/r, towards centre) and tangential acceleration (dv/dt, along the tangent). Total acceleration = √(a<sub>c</sub><sup>2</sup> + a<sub>t</sub><sup>2</sup>).
Important Numericals
Numerical 1: Projectile — Range and Height
A ball is thrown at 30 m s<sup>-1</sup> at 45° to the horizontal. Find the maximum height and range. (g = 10 m s<sup>-2</sup>)
Numerical 2: Vector Components
A vector of magnitude 10 makes 60° with the x-axis. Find its x and y components.
Numerical 3: Centripetal Acceleration
A body moves in a circle of radius 4 m with speed 8 m s<sup>-1</sup>. Find centripetal acceleration.
Numerical 4: Dot and Cross Product
Find the dot product and cross product of A = 3î + 4ĵ and B = 2î − ĵ.
| A × B = | î ĵ k̂; 3 4 0; 2 −1 0 | = î(0−0) − ĵ(0−0) + k̂(3(−1) − 4(2)) = k̂(−3 − 8) = −11k̂ |
|---|---|---|
| A × B | = 11 |
Numerical 5: Time of Flight and Angle
A projectile has a range of 100 m and maximum height of 25 m. Find the angle of projection and initial velocity. (g = 10 m s<sup>-2</sup>)
Numerical 6: Horizontal Projectile
A body is thrown horizontally from a tower 20 m high at 10 m s<sup>-1</sup>. Find the time of flight, range, and velocity on hitting the ground. (g = 10 m s<sup>-2</sup>)
NCERT Exercise Reference
| NCERT Exercise | Topic | Marks to Expect |
|---|---|---|
| 4.1–4.4 | Scalars, vectors, vector addition | 1–2 marks (definitions, comparisons) |
| 4.5–4.7 | Vector multiplication (dot, cross) | 2 marks (numerical) |
| 4.8–4.12 | Projectile motion | 3–5 marks (derivation + numerical) |
| 4.13–4.15 | Uniform circular motion | 2 marks (formula + numerical) |
Common Mistakes
1. Treating velocity and speed as the same in 2D: Speed is the magnitude of velocity; in circular motion, speed is constant but velocity (a vector) keeps changing direction. 2. Confusing dot and cross product: Dot product gives a scalar (AB cos θ); cross product gives a vector perpendicular to both (AB sin θ n̂). Do not mix their formulas or results. 3. Forgetting that horizontal velocity is constant in projectile motion: There is no horizontal acceleration (no air resistance), so v<sub>x</sub> = u cos θ at all times. 4. Applying range formula incorrectly: R = u<sup>2</sup> sin 2θ / g applies only when launch point and landing point are at the same height. If launch and land at different heights, you must solve from first principles. 5. Centripetal vs. centrifugal: In an inertial frame, only centripetal force (inward) exists. "Centrifugal force" appears only in a rotating (non-inertial) frame — it is a pseudo force. 6. Sign of g in projectile equations: Taking upward as positive, a<sub>y</sub> = −g. Be consistent throughout the problem. 7. Cross product direction: Use the right-hand rule. If fingers curl from A to B, the thumb gives the direction of A × B.
Exam Tips
- Projectile motion is a guaranteed 3–5 mark question. Practise deriving the equation of trajectory, time of flight, max height, and range from scratch.
- Vector addition often appears as a 2-mark question: know the parallelogram law formula for magnitude and direction.
- Dot product and cross product questions test whether you know the component formulas — memorise them.
- Uniform circular motion: centripetal acceleration formula a<sub>c</sub> = v<sup>2</sup>/r is frequently asked as a 1-mark question.
- For derivations, always start with resolving the initial velocity into components, then write horizontal and vertical equations separately.
- In numericals, show all steps: resolve, write the two independent sets of equations, combine to eliminate time if needed.
- Same range for complementary angles is a common MCQ — remember θ and (90° − θ) give the same range.
- Practise drawing the trajectory parabola with labelled quantities (H, R, T, θ).
Quick Revision Notes
- Vectors have magnitude + direction; obey triangle/parallelogram law of addition.
- Parallelogram result: |R| = √(A<sup>2</sup> + B<sup>2</sup> + 2AB cos θ). Max = A + B (θ = 0°), Min = |A − B| (θ = 180°).
- Resolution: A<sub>x</sub> = A cos θ, A<sub>y</sub> = A sin θ. |A| = √(A<sub>x</sub><sup>2</sup> + A<sub>y</sub><sup>2</sup>).
- Dot product: A · B = AB cos θ = A<sub>x</sub>B<sub>x</sub> + A<sub>y</sub>B<sub>y</sub>. Scalar. Zero when perpendicular.
- Cross product: |A × B| = AB sin θ. Vector perpendicular to both. Zero when parallel. A × B = −(B × A).
- Projectile: T = 2u sin θ / g; H = u<sup>2</sup> sin<sup>2</sup> θ / (2g); R = u<sup>2</sup> sin 2θ / g.
- Max range at θ = 45°; same range at θ and (90° − θ).
- Trajectory equation: y = x tan θ − gx<sup>2</sup>/(2u<sup>2</sup> cos<sup>2</sup> θ) — parabola.
- Horizontal projectile: t = √(2h/g); R = u√(2h/g); v = √(u<sup>2</sup> + 2gh).
- Circular motion: a<sub>c</sub> = v<sup>2</sup>/r = ω<sup>2</sup>r (towards centre); velocity is tangential; speed is constant.
Important One-Mark Facts for Quick Recall
1. A scalar multiplied by a vector gives a vector (same direction if scalar positive, opposite if negative). 2. Two vectors are perpendicular if their dot product is zero (A · B = 0). 3. Two vectors are parallel if their cross product is zero (A × B = 0).
5. The cross product of parallel vectors is zero; the dot product of perpendicular vectors is zero. 6. In projectile motion, the horizontal component of velocity is constant (no horizontal acceleration). 7. At the highest point of a projectile, the velocity is purely horizontal (v<sub>y</sub> = 0, v<sub>x</sub> = u cos θ). 8. The range of a projectile is the same for θ and (90° − θ) (complementary angles). 9. Maximum range occurs at θ = 45° where R<sub>max</sub> = u<sup>2</sup>/g. 10. The trajectory of a projectile is a parabola (equation: y = ax − bx<sup>2</sup>). 11. In uniform circular motion, speed is constant but velocity changes (direction changes). 12. Centripetal acceleration is always directed towards the centre; there is no outward "centrifugal force" in an inertial frame. 13. A × B = −(B × A): the cross product is not commutative. 14. The dot product is commutative: A · B = B · A. 15. In components: A · B = A<sub>x</sub>B<sub>x</sub> + A<sub>y</sub>B<sub>y</sub> + A<sub>z</sub>B<sub>z</sub> (no unit vectors in the result).
| 4. The maximum value of | A + B | = A + B (when θ = 0°); the minimum = ** | A − B | ** (when θ = 180°). |
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