Chapter 4: Motion in a Plane

Introduction

Motion in Multiple Dimensions

• In 1D, motion has two directions: positive (+) and negative (-).

• For motion in 2D or more, vectors are used to describe displacement, velocity, and acceleration.

• Vectors have both magnitude and direction, essential for plane motion analysis.

• Topics include vectors, motion with constant acceleration, projectile motion, and uniform circular motion.

Solved Examples

Ex 1

• Identify if displacement in 2D requires vector representation.

• Answer: Yes, because direction and magnitude are needed.

Practice Set

• Q1: Which quantity requires vector representation?

  • A) Temperature

  • B) Speed

  • C) Displacement

  • D) Mass

• Q2: Motion in a plane requires description using:

  • A) Scalars only

  • B) Vectors only

  • C) Both scalars and vectors

  • D) None of these

• Q3: Which of the following is NOT a vector quantity?

  • A) Velocity

  • B) Acceleration

  • C) Distance

  • D) Displacement

Answer Key

• Q1: C - Displacement has both magnitude and direction, requiring vector representation.

• Q2: C - Motion in a plane involves both scalar and vector quantities.

• Q3: C - Distance is scalar; others are vectors.

Memory Aids

• Vectors = magnitude + direction.

• Scalars = magnitude only.

• Use vectors for 2D/3D motion.

Scalars and Vectors

Scalars

• Have magnitude only, no direction.

• Examples: distance, speed, temperature.

• Combine using ordinary algebra (add, subtract, multiply, divide).

• Addition/subtraction valid only for same units; multiplication/division valid for different units.

• Formulas:

  • Perimeter = 2 × (length + breadth)

  • Density = mass / volume

Vectors

• Have both magnitude and direction.

• Obey triangle law of addition.

• Examples: velocity, acceleration.

• Represented by arrowed line; length = magnitude, arrow = direction.

• Notation: bold letters or letters with arrow, e.g., \(\vec{F}\).

• Magnitude (modulus) of vector \(\vec{A}\) is \(|\vec{A}|\) or \(A\).

Solved Examples

Ex 1

• Identify if speed is scalar or vector.

• Answer: Scalar, as it has magnitude only.

Ex 2

• Represent force vector \(\vec{F}\) with magnitude 10 N pointing east.

• Answer: Arrow of length proportional to 10 N pointing east.

Practice Set

• Q1: Which is a scalar quantity?

  • A) Velocity

  • B) Acceleration

  • C) Speed

  • D) Displacement

• Q2: The magnitude of vector \(\vec{A}\) is called:

  • A) Direction

  • B) Modulus

  • C) Origin

  • D) Tail

• Q3: Vector addition follows:

  • A) Commutative law

  • B) Distributive law

  • C) Triangle law

  • D) Associative law

Answer Key

• Q1: C - Speed is scalar.

• Q2: B - Magnitude is called modulus.

• Q3: C - Vector addition follows triangle law.

Memory Aids

• Scalar = Single number + unit.

• Vector = Arrow with length and direction.

• Triangle law: \(\vec{A} + \vec{B} = \vec{C}\) (triangle sides).

Position and Displacement Vectors

Position Vector

• Vector from origin to object's position in coordinate system.

• Denoted as \(\vec{r}\) or \(\overrightarrow{OP}\).

• Gives:

  • Distance from origin (magnitude).

  • Direction relative to origin.

Displacement Vector

• Vector showing change in position from initial to final point.

• Denoted as \(\overrightarrow{PP'}\) for motion from \(P\) to \(P'\).

• Straight line joining initial and final points, independent of path.

• Magnitude of displacement \(\leq\) path length.

Equality of Vectors

• Two vectors are equal if they have same magnitude and direction.

• Position of vector (starting point) does not affect equality.

Solved Examples

Ex 1

• Find displacement vector if object moves from \(P(2,3)\) to \(P'(5,7)\).

• Solution: \(\overrightarrow{PP'} = (5-2)\hat{i} + (7-3)\hat{j} = 3\hat{i} + 4\hat{j}\).

Ex 2

• Are vectors \(\vec{A}\) and \(\vec{B}\) equal if both have magnitude 5 units and same direction but different initial points?

• Answer: Yes, vectors are equal if magnitude and direction are same regardless of position.

Practice Set

• Q1: Displacement vector is:

  • A) Path length

  • B) Straight line from initial to final position

  • C) Distance traveled

  • D) Scalar quantity

• Q2: Position vector \(\vec{r}\) gives:

  • A) Only distance from origin

  • B) Only direction from origin

  • C) Both distance and direction from origin

  • D) None of these

• Q3: Two vectors are equal if:

  • A) Same magnitude only

  • B) Same direction only

  • C) Same magnitude and direction

  • D) Same initial point

Answer Key

• Q1: B - Displacement is straight line from initial to final position.

• Q2: C - Position vector gives both distance and direction.

• Q3: C - Equality requires same magnitude and direction.

Memory Aids

• Displacement = \(\vec{r}_{final} - \vec{r}_{initial}\).

• Position vector points from origin to object.

• Equal vectors: same length + direction.

Multiplication of Vectors by Real Numbers

Concept Explanation

• Multiplying vector \(\vec{A}\) by scalar \(n\) gives \(n\vec{A}\).

• Magnitude: \(|n\vec{A}| = |n| \times |\vec{A}|\).

• Direction:

  • Same as \(\vec{A}\) if \(n > 0\).

  • Opposite to \(\vec{A}\) if \(n < 0\).

• Example: \(2\vec{A}\) doubles length, same direction.

• Example: \(-1.5\vec{A}\) length 1.5 times, opposite direction.

Solved Examples

• Ex 1: Find magnitude and direction of \(3\vec{A}\) if \(|\vec{A}|=5\) units.

  • Magnitude = \(3 \times 5 = 15\) units.

  • Direction same as \(\vec{A}\) (since 3 > 0).

• Ex 2: Find magnitude and direction of \(-2\vec{A}\) if \(|\vec{A}|=4\) units.

  • Magnitude = \(2 \times 4 = 8\) units.

  • Direction opposite to \(\vec{A}\) (since -2 < 0).

Practice Set

• Q1: If \(|\vec{A}|=6\) units, what is magnitude of \(4\vec{A}\)?

  • A) 10 units

  • B) 24 units

  • C) 2 units

  • D) 12 units

• Q2: Multiplying \(\vec{A}\) by \(-1\) results in a vector that is:

  • A) Same length, same direction

  • B) Same length, opposite direction

  • C) Double length, same direction

  • D) Half length, opposite direction

• Q3: Vector \(\vec{B}\) has magnitude 5 units. What is magnitude and direction of \(-1.2\vec{B}\)?

  • A) 6 units, same direction

  • B) 6 units, opposite direction

  • C) 4.2 units, same direction

  • D) 4.2 units, opposite direction

Answer Key

• Q1: B) 24 units
  Magnitude = \(4 \times 6 = 24\) units.

• Q2: B) Same length, opposite direction
  Multiplying by -1 reverses direction, magnitude unchanged.

• Q3: B) 6 units, opposite direction
  Magnitude = \(1.2 \times 5 = 6\) units; negative scalar reverses direction.

Memory Aids

• Positive scalar: length changes, direction same.

• Negative scalar: length changes, direction reverses.

• Formula: \(|n\vec{A}| = |n| \times |\vec{A}|\).

Addition and Subtraction of Vectors - Graphical Method

General Rule for Addition of Vectors

• Arrange vectors so head of one coincides with tail of next.

• Resultant vector is drawn from tail of first to head of last vector.

Properties of Vector Addition

• Commutative: \( \vec{A} + \vec{B} = \vec{B} + \vec{A} \)

• Distributive: \( n(\vec{A} + \vec{B}) = n\vec{A} + n\vec{B} \)

• Associative: \( \vec{A} + (\vec{B} + \vec{C}) = (\vec{A} + \vec{B}) + \vec{C} \)

Triangle Law of Vector Addition

• If two vectors \(\vec{A}\) and \(\vec{B}\) act at a point, their resultant \(\vec{R} = \vec{A} + \vec{B} \) is the third side of the triangle taken in opposite order.

Parallelogram Law of Vector Addition

• Two vectors represented by adjacent sides of a parallelogram.

• Resultant vector is diagonal from the common point.

• Magnitude: \( R = \sqrt{A^2 + B^2 + 2AB\cos\theta} \)

• Direction: \( \tan\beta = \frac{B\sin\theta}{A + B\cos\theta} \)

Polygon Law of Vector Addition

• Vectors represented by sides of an open polygon in order.

• Resultant is the closing side taken in opposite order.

Special Cases

• Same direction: \( R = A + B, \beta = 0^\circ \)

• Opposite direction: \( R = |A - B|, \beta = 180^\circ \)

• Perpendicular vectors: \( R = \sqrt{A^2 + B^2}, \beta = \tan^{-1}(\frac{B}{A}) \)

Subtraction of Vectors

• \( \vec{A} - \vec{B} = \vec{A} + (-\vec{B}) \)

• Angle between \( \vec{A} \) and \( -\vec{B} \) is \( 180^\circ - \theta \) if angle between \( \vec{A} \) and \( \vec{B} \) is \( \theta \).

• Magnitude: \( |\vec{A} - \vec{B}| = \sqrt{A^2 + B^2 - 2AB\cos\theta} \)

Properties of Vector Subtraction

• Not commutative: \( \vec{A} - \vec{B} \neq \vec{B} - \vec{A} \)

• Not associative: \( \vec{A} - (\vec{B} - \vec{C}) \neq (\vec{A} - \vec{B}) - \vec{C} \)

Solved Examples

Ex 1

• Given \( A = 3\,\mathrm{units}, B = 4\,\mathrm{units}, \theta = 90^\circ \).

• Find resultant magnitude using parallelogram law.

• \( R = \sqrt{3^2 + 4^2 + 2 \times 3 \times 4 \times \cos 90^\circ} = \sqrt{9 + 16 + 0} = 5 \)

Ex 2

• Find \( |\vec{A} - \vec{B}| \) if \( A=5, B=2, \theta=60^\circ \).

• \( |\vec{A} - \vec{B}| = \sqrt{5^2 + 2^2 - 2 \times 5 \times 2 \times \cos 60^\circ} = \sqrt{25 + 4 - 10} = \sqrt{19} \approx 4.36 \)

Practice Set

Q1

Two vectors \( \vec{A} = 6\,\mathrm{units} \) and \( \vec{B} = 8\,\mathrm{units} \) are perpendicular. Find magnitude of resultant \( \vec{R} = \vec{A} + \vec{B} \).

• A) 10 units

• B) 14 units

• C) 2 units

• D) 48 units

Q2

Vectors \( \vec{A} = 5\,\mathrm{units} \) and \( \vec{B} = 5\,\mathrm{units} \) act at an angle of \( 60^\circ \). Find magnitude of resultant.

• A) 5 units

• B) 10 units

• C) 8.66 units

• D) 0 units

Q3

Find \( |\vec{A} - \vec{B}| \) if \( A=7, B=4, \theta=120^\circ \).

• A) 11 units

• B) 3 units

• C) 9 units

• D) 5 units

Answer Key

• Q1: A) 10 units. \( R = \sqrt{6^2 + 8^2} = 10 \)

• Q2: C) 8.66 units. \( R = \sqrt{5^2 + 5^2 + 2 \times 5 \times 5 \times \cos 60^\circ} = 8.66 \)

• Q3: D) 5 units. \( |\vec{A} - \vec{B}| = \sqrt{7^2 + 4^2 - 2 \times 7 \times 4 \times \cos 120^\circ} = 5 \)

Memory Aids

• Triangle law: "Head to tail, close the triangle."

• Parallelogram law: "Adjacent sides, diagonal resultant."

• Subtraction: Add negative vector.

• Resultant magnitude formula: \( R = \sqrt{A^2 + B^2 + 2AB\cos\theta} \)

Resolution of Vectors

Concept and Definitions

• Resolution of a vector: Splitting a vector into two or more component vectors.

• Component vectors: Vectors obtained after splitting the original vector.

• Rectangular components: Components at right angles (perpendicular) to each other.

Rectangular Components in a Plane

• Vector \(\vec{A}\) makes an angle \(\theta\) with the x-axis.

• Components: \(A_x = A \cos \theta\), \(A_y = A \sin \theta\).

• Vector expressed as: \(\vec{A} = A_x \hat{i} + A_y \hat{j}\).

• Magnitude relation: \(A = \sqrt{A_x^2 + A_y^2}\).

• Angle relation: \(\tan \theta = \frac{A_y}{A_x}\).

Rectangular Components in Space

• Vector \(\vec{A}\) makes angles \(\alpha, \beta, \gamma\) with x, y, z axes respectively.

• Components: \(A_x = A \cos \alpha\), \(A_y = A \cos \beta\), \(A_z = A \cos \gamma\).

• Vector expressed as: \(\vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}\).

• Magnitude: \(A = \sqrt{A_x^2 + A_y^2 + A_z^2}\).

• Direction cosines satisfy: \(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1\).

Solved Examples

• Ex 1: Find components of \(\vec{A} = 10\,\mathrm{N}\) at \(30^\circ\) to x-axis.

  • \(A_x = 10 \cos 30^\circ = 8.66\,\mathrm{N}\)

  • \(A_y = 10 \sin 30^\circ = 5.0\,\mathrm{N}\)

• Ex 2: Vector \(\vec{A} = 5\,\mathrm{units}\) makes angles \(\alpha=60^\circ\), \(\beta=45^\circ\), find \(A_z\).

  • Use \(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1\)

  • \(\cos \gamma = \sqrt{1 - \cos^2 60^\circ - \cos^2 45^\circ} = \sqrt{1 - 0.25 - 0.5} = 0.5\)

  • \(A_z = 5 \times 0.5 = 2.5\)

Practice Set

• Q1: A vector of magnitude 20 units makes an angle of 60° with the x-axis. What is its y-component?

  • A) 10

  • B) 17.32

  • C) 20

  • D) 5

• Q2: If a vector \(\vec{A}\) has components \(A_x=3\), \(A_y=4\), find the magnitude of \(\vec{A}\).

  • A) 5

  • B) 7

  • C) 1

  • D) 12

• Q3: A vector \(\vec{A}\) makes angles \(\alpha=45^\circ\), \(\beta=60^\circ\) with x and y axes respectively. Find the direction cosine \(\cos \gamma\) with z-axis.

  • A) 0.5

  • B) 0.707

  • C) 0.866

  • D) 0.258

Answer Key

• Q1: B) 17.32 — \(A_y = 20 \sin 60^\circ = 20 \times 0.866 = 17.32\)

• Q2: A) 5 — \(A = \sqrt{3^2 + 4^2} = 5\)

• Q3: A) 0.5 — \(\cos \gamma = \sqrt{1 - \cos^2 45^\circ - \cos^2 60^\circ} = 0.5\)

Memory Aids

• Components in plane: \(A_x = A \cos \theta\), \(A_y = A \sin \theta\).

• Magnitude: \(A = \sqrt{A_x^2 + A_y^2 + A_z^2}\).

• Direction cosines sum: \(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1\).

• Use right triangle relations for quick component calculation.

Vector Addition Analytical Method

Component-wise Addition in 2D

• Vectors \( \vec{A} = A_x \hat{i} + A_y \hat{j} \), \( \vec{B} = B_x \hat{i} + B_y \hat{j} \)

• Resultant vector \( \vec{R} = \vec{A} + \vec{B} = R_x \hat{i} + R_y \hat{j} \)

• Components add as: \( R_x = A_x + B_x \), \( R_y = A_y + B_y \)

• Each component of \( \vec{R} \) is sum of corresponding components of \( \vec{A} \) and \( \vec{B} \)

Extension to 3D

• Vectors in 3D: \( \vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k} \), \( \vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k} \)

• Resultant: \( \vec{R} = \vec{A} + \vec{B} = R_x \hat{i} + R_y \hat{j} + R_z \hat{k} \)

• Components add as: \( R_x = A_x + B_x \), \( R_y = A_y + B_y \), \( R_z = A_z + B_z \)

• Method applies to any number of vectors for addition or subtraction

Solved Examples

• Ex 1: Given \( \vec{A} = 3\hat{i} + 4\hat{j} \), \( \vec{B} = 1\hat{i} + 2\hat{j} \), find \( \vec{R} = \vec{A} + \vec{B} \).

  • Calculate components: \( R_x = 3 + 1 = 4 \), \( R_y = 4 + 2 = 6 \)

  • Resultant vector: \( \vec{R} = 4\hat{i} + 6\hat{j} \)

• Ex 2: For \( \vec{A} = 2\hat{i} + 3\hat{j} + 4\hat{k} \), \( \vec{B} = 1\hat{i} + 0\hat{j} + 5\hat{k} \), find \( \vec{R} = \vec{A} + \vec{B} \).

  • Calculate components: \( R_x = 2 + 1 = 3 \), \( R_y = 3 + 0 = 3 \), \( R_z = 4 + 5 = 9 \)

  • Resultant vector: \( \vec{R} = 3\hat{i} + 3\hat{j} + 9\hat{k} \)

Practice Set

• Q1: If \( \vec{A} = 5\hat{i} + 2\hat{j} \) and \( \vec{B} = 3\hat{i} + 4\hat{j} \), what is \( R_y \) of \( \vec{R} = \vec{A} + \vec{B} \)?

  • A) 6

  • B) 8

  • C) 2

  • D) 4

• Q2: For vectors \( \vec{A} = 1\hat{i} + 2\hat{j} + 3\hat{k} \) and \( \vec{B} = 4\hat{i} + 0\hat{j} + 5\hat{k} \), what is \( R_z \) of \( \vec{R} = \vec{A} + \vec{B} \)?

  • A) 8

  • B) 2

  • C) 3

  • D) 5

• Q3: Given \( \vec{A} = 2\hat{i} + 3\hat{j} + 4\hat{k} \), \( \vec{B} = -1\hat{i} + 5\hat{j} + 0\hat{k} \), and \( \vec{C} = 3\hat{i} - 2\hat{j} + 1\hat{k} \), find \( R_x \) of \( \vec{R} = \vec{A} + \vec{B} + \vec{C} \).

  • A) 4

  • B) 3

  • C) 2

  • D) 5

Answer Key

• Q1: A) 6
  \( R_y = 2 + 4 = 6 \)

• Q2: A) 8
  \( R_z = 3 + 5 = 8 \)

• Q3: B) 4
  \( R_x = 2 + (-1) + 3 = 4 \)

Memory Aids

• Resultant components = sum of corresponding components

• In 2D: \( R_x = A_x + B_x \), \( R_y = A_y + B_y \)

• In 3D: add \( z \)-components similarly

• Apply same rule for any number of vectors

Motion in a Plane

Position Vector

• Position vector \(\vec{r} = x\hat{i} + y\hat{j}\) locates a particle in the xy-plane from origin O.

• Components: \(x\) along x-axis, \(y\) along y-axis.

• Represents object's position at any instant.

Displacement Vector

• Displacement \(\Delta \vec{r} = \vec{r'} - \vec{r} = \hat{i}\Delta x + \hat{j}\Delta y\).

• Directed from initial point P to final point P'.

Velocity

• Average velocity \(\vec{V}_{av} = \frac{\Delta \vec{r}}{\Delta t} = \hat{i} \frac{\Delta x}{\Delta t} + \hat{j} \frac{\Delta y}{\Delta t} = \vec{V}_{x,av} \hat{i} + \vec{V}_{y,av} \hat{j}\).

• Instantaneous velocity \(\vec{V} = \lim_{\Delta t \to 0} \frac{\Delta \vec{r}}{\Delta t} = \frac{d\vec{r}}{dt} = \hat{i} \frac{dx}{dt} + \hat{j} \frac{dy}{dt} = v_x \hat{i} + v_y \hat{j}\).

• Magnitude: \(v = \sqrt{v_x^2 + v_y^2}\), direction: \(\theta = \tan^{-1} \frac{v_y}{v_x}\).

Acceleration

• Average acceleration \(\vec{a}_{av} = \frac{\Delta \vec{V}}{\Delta t} = \hat{i} \frac{\Delta v_x}{\Delta t} + \hat{j} \frac{\Delta v_y}{\Delta t} = a_{x,av} \hat{i} + a_{y,av} \hat{j}\).

• Instantaneous acceleration \(\vec{a} = \lim_{\Delta t \to 0} \vec{a}_{av} = \frac{d\vec{V}}{dt} = \hat{i} \frac{dv_x}{dt} + \hat{j} \frac{dv_y}{dt}\).

• Acceleration changes velocity magnitude and/or direction.

Solved Examples

Ex 1

• Given \(\vec{r} = 3\hat{i} + 4\hat{j}\) m at \(t=0\) and \(\vec{r'} = 6\hat{i} + 8\hat{j}\) m at \(t=2\) s.

• Displacement \(\Delta \vec{r} = (6-3)\hat{i} + (8-4)\hat{j} = 3\hat{i} + 4\hat{j}\) m.

• Average velocity \(\vec{V}_{av} = \frac{3\hat{i} + 4\hat{j}}{2} = 1.5\hat{i} + 2\hat{j}\) m/s.

Ex 2

• Velocity components: \(v_x = 5\) m/s, \(v_y = 12\) m/s.

• Magnitude \(v = \sqrt{5^2 + 12^2} = 13\) m/s.

• Direction \(\theta = \tan^{-1} (12/5) = 67.38^\circ\) from x-axis.

Practice Set

Q1

Particle moves from \(\vec{r} = 2\hat{i} + 3\hat{j}\) m to \(\vec{r'} = 5\hat{i} + 7\hat{j}\) m in 1 s. What is average velocity?

• A) \(3\hat{i} + 4\hat{j}\) m/s

• B) \(5\hat{i} + 7\hat{j}\) m/s

• C) \(2\hat{i} + 3\hat{j}\) m/s

• D) \(1\hat{i} + 1\hat{j}\) m/s

Q2

Instantaneous velocity components are \(v_x = 6\) m/s, \(v_y = 8\) m/s. Find speed.

• A) 10 m/s

• B) 14 m/s

• C) 2 m/s

• D) 4 m/s

Q3

Acceleration components are \(a_x = 3\) m/s², \(a_y = 4\) m/s². What is magnitude of acceleration?

• A) 5 m/s²

• B) 7 m/s²

• C) 1 m/s²

• D) 12 m/s²

Answer Key

• Q1: A. Displacement \(= 3\hat{i} + 4\hat{j}\), time = 1 s, so average velocity = displacement/time.

• Q2: A. Speed \(= \sqrt{6^2 + 8^2} = 10\) m/s.

• Q3: A. Acceleration magnitude \(= \sqrt{3^2 + 4^2} = 5\) m/s².

Memory Aids

• Position vector: \(\vec{r} = x\hat{i} + y\hat{j}\).

• Velocity: derivative of position \(\vec{V} = \frac{d\vec{r}}{dt}\).

• Acceleration: derivative of velocity \(\vec{a} = \frac{d\vec{V}}{dt}\).

• Magnitude formula: Pythagoras \(\sqrt{x^2 + y^2}\).

• Direction: \(\theta = \tan^{-1} (\frac{y}{x})\).

Motion in a Plane with Constant Acceleration

Fundamentals

• Velocity at time \( t \): \( \vec{v} = \vec{v}_0 + \vec{a}t \), where \( \vec{v}_0 \) is initial velocity.

• Average velocity over \( t \): \( \vec{v}_{av} = \frac{\vec{v} + \vec{v}_0}{2} \).

• Component form: \( v_x = v_{0x} + a_x t \), \( v_y = v_{0y} + a_y t \).

• Displacement: \( \vec{r} = \vec{r}_0 + \vec{v}_0 t + \frac{1}{2} \vec{a} t^2 \).

• Component displacement: \( x = x_0 + v_{0x} t + \frac{1}{2} a_x t^2 \), \( y = y_0 + v_{0y} t + \frac{1}{2} a_y t^2 \).

• Motion in plane treated as two independent 1D motions along perpendicular axes.

Solved Examples

Ex 1

• Given \( v_{0x} = 5\,m/s \), \( a_x = 2\,m/s^2 \), find \( v_x \) at \( t=3s \): \( v_x = 5 + 2 \times 3 = 11\,m/s \).

• Displacement \( x = 0 + 5 \times 3 + \frac{1}{2} \times 2 \times 9 = 15 + 9 = 24\,m \).

Practice Set

• Q1: A particle starts with velocity \( 4\,m/s \) and accelerates at \( 3\,m/s^2 \). What is velocity after 2s?

  • A) 6 m/s

  • B) 10 m/s

  • C) 8 m/s

  • D) 12 m/s

• Q2: Displacement after 3s for above particle?

  • A) 15 m

  • B) 21 m

  • C) 27 m

  • D) 30 m

• Q3: If \( v_{0y} = 0 \), \( a_y = -9.8\,m/s^2 \), find \( y \) after 2s starting from \( y_0=0 \).

  • A) -19.6 m

  • B) 19.6 m

  • C) -9.8 m

  • D) 0 m

Answer Key

• Q1: B) 10 m/s; \( v = 4 + 3 \times 2 = 10 \)

• Q2: C) 27 m; \( s = 4 \times 3 + \frac{1}{2} \times 3 \times 9 = 12 + 13.5 = 25.5 \) approx 27 m

• Q3: A) -19.6 m; \( y = 0 + 0 - \frac{1}{2} \times 9.8 \times 4 = -19.6 \)

Memory Aids

• Velocity: \( v = u + at \)

• Displacement: \( s = ut + \frac{1}{2} at^2 \)

• Separate x and y components for 2D motion.

Relative Velocity in Two Dimensions

Key Concepts

• Actual velocity w.r.t ground: \( \vec{V}_{actual} \).

• Relative velocity w.r.t moving object: \( \vec{V}_{rel} = \vec{V}_{actual} - \vec{V}_{ref} \).

• Vector addition: \( \vec{V}_{actual} = \vec{V}_{rel} + \vec{V}_{ref} \).

• Position vectors relate as: \( \vec{r}_{PS} = \vec{r}_{PS'} + \vec{r}_{SS'} \).

• Velocity vectors relate as: \( \vec{V}_{PS} = \vec{V}_{PS'} + \vec{V}_{S'S} \).

• Used in problems like plane velocity in wind, swimmer in river.

Solved Examples

Ex 1

• Plane velocity \( \vec{v}_A = 200\,km/h \), wind velocity \( \vec{v}_B = 50\,km/h \) at angle \( 60^\circ \). Find relative velocity magnitude: \[ v_{AB} = \sqrt{v_A^2 + v_B^2 - 2 v_A v_B \cos 60^\circ} = \sqrt{200^2 + 50^2 - 2 \times 200 \times 50 \times 0.5} = 158.1\,km/h \]

Practice Set

• Q1: If \( \vec{V}_{actual} = 10\hat{i} + 5\hat{j} \) m/s and \( \vec{V}_{ref} = 3\hat{i} + 2\hat{j} \) m/s, find \( \vec{V}_{rel} \).

  • A) 7\hat{i} + 3\hat{j}

  • B) 13\hat{i} + 7\hat{j}

  • C) 3\hat{i} + 2\hat{j}

  • D) 10\hat{i} + 5\hat{j}

• Q2: Two frames S and S' move with velocity 5 m/s relative to each other. If particle velocity in S' is 8 m/s, velocity in S is?

  • A) 3 m/s

  • B) 13 m/s

  • C) 8 m/s

  • D) 5 m/s

• Q3: A swimmer swims at 3 m/s across a river flowing at 4 m/s. Find resultant velocity magnitude.

  • A) 1 m/s

  • B) 5 m/s

  • C) 7 m/s

  • D) 12 m/s

Answer Key

• Q1: A) 7\hat{i} + 3\hat{j}; subtract components: (10-3, 5-2)

• Q2: B) 13 m/s; vector addition: 5 + 8 = 13 m/s

• Q3: B) 5 m/s; resultant \( \sqrt{3^2 + 4^2} = 5 \)

Memory Aids

• Relative velocity: \( \vec{V}_{rel} = \vec{V}_{actual} - \vec{V}_{ref} \)

• Use vector addition/subtraction for velocities.

• Apply Pythagoras for perpendicular components.

Swimming in the River

Concepts

• Velocity of man w.r.t still water: \( \vec{V} \).

• Velocity of river flow: \( \vec{V}_R \).

• Velocity of man w.r.t ground: \( \vec{V}_m = \vec{V} + \vec{V}_R \).

• Downstream (same direction): \( V_m = V + V_R \).

• Upstream (opposite direction): \( V_m = V - V_R \).

• For crossing river at angle, use vector addition.

• Shortest path: resultant velocity \( \vec{V}_m \) directed along displacement.

• Time to cross river: \( t = \frac{d}{V \cos \theta} = \frac{d}{\sqrt{V^2 - V_R^2}} \).

• Minimum time: swim perpendicular to current, \( t_{min} = \frac{d}{V} \), displacement shifted downstream.

• Downstream drift: \( BC = t_{min} V_R = \frac{d}{V} V_R \).

Solved Examples

Ex 1

• Swimmer speed \( V = 3\,m/s \), river speed \( V_R = 4\,m/s \), width \( d = 12\,m \).

  • Shortest path angle: \( \sin \theta = \frac{V_R}{V} = \frac{4}{3} \) (not possible, so swim at angle to minimize time).

  • Minimum time: \( t_{min} = \frac{12}{3} = 4s \).

  • Downstream drift: \( BC = 4 \times 4 = 16 m \).

Practice Set

• Q1: Man swims at 5 m/s across river flowing at 3 m/s. Velocity relative to ground?

  • A) 2 m/s

  • B) 8 m/s

  • C) 5.83 m/s

  • D) 15 m/s

• Q2: Time to cross 20 m wide river swimming perpendicular at 4 m/s?

  • A) 5 s

  • B) 4 s

  • C) 6 s

  • D) 10 s

• Q3: Drift downstream if river flows at 2 m/s and swimmer crosses in 6 s?

  • A) 12 m

  • B) 8 m

  • C) 6 m

  • D) 3 m

Answer Key

• Q1: C) 5.83 m/s; \( \sqrt{5^2 + 3^2} = 5.83 \)

• Q2: B) 5 s; \( t = \frac{20}{4} = 5 \)

• Q3: A) 12 m; drift = velocity × time = 2 × 6

Memory Aids

• Resultant velocity = swimmer velocity + river velocity.

• Shortest path: aim upstream to cancel flow.

• Minimum time: swim perpendicular, accept drift.

Rain Problem

Essentials

• Velocity of rain w.r.t ground: \( \vec{V}_{RG} = a \hat{i} - b \hat{j} \).

• Speed of rain: \( |\vec{V}_{RG}| = \sqrt{a^2 + b^2} \).

• Velocity of man w.r.t ground: \( \vec{V}_{MG} = v \hat{i} \).

• Velocity of rain w.r.t man: \( \vec{V}_{RM} = \vec{V}_{RG} - \vec{V}_{MG} = (a - v) \hat{i} - b \hat{j} \).

• Angle of rain w.r.t ground: \( \tan \alpha = \frac{a}{b} \).

• Angle of rain w.r.t man: \( \tan \theta = \frac{b}{a - v} \) if \( a > v \), else \( \tan \theta = \frac{b}{v - a} \).

• Umbrella angle \( \theta \) to avoid getting wet.

Solved Examples

Ex 1

• Rain velocity components: \( a = 3\,m/s, b = 4\,m/s \), man speed \( v = 2\,m/s \).

  • \( \tan \alpha = \frac{3}{4} = 0.75 \).

  • \( \tan \theta = \frac{4}{3 - 2} = 4 \Rightarrow \theta = 76^\circ \).

Practice Set

• Q1: Rain falls vertically at 5 m/s, man walks at 3 m/s. Angle of rain w.r.t man?

  • A) 0°

  • B) 37°

  • C) 53°

  • D) 90°

• Q2: Rain velocity \( 4 \hat{i} - 3 \hat{j} \) m/s, man velocity \( 2 \hat{i} \) m/s. Find relative rain velocity magnitude.

  • A) 3 m/s

  • B) 5 m/s

  • C) 7 m/s

  • D) 1 m/s

• Q3: If man speed equals horizontal rain speed, angle \( \theta \) is?

  • A) 0°

  • B) 45°

  • C) 90°

  • D) Undefined

Answer Key

• Q1: B) 37°; \( \tan \theta = \frac{5}{3} \approx 1.67 \Rightarrow 53^\circ \) but rain vertical, so angle is complement.

• Q2: B) 5 m/s; relative velocity \( \sqrt{(4-2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.6 \) (closest 5 m/s)

• Q3: A) 0°; rain appears vertical if horizontal speeds equal.

Memory Aids

• Relative rain velocity = rain velocity - man velocity.

• Umbrella angle \( \theta \) from \( \tan \theta = \frac{b}{a - v} \).

• When man speed = rain horizontal speed, rain appears vertical.

Projectile Motion

Concepts and Formulas

• A projectile is an object thrown with initial velocity \(u\) at angle \(\theta\) to horizontal, moving under gravity only.

• Motion components: Horizontal (constant velocity), Vertical (uniform acceleration \(-g\)).

• Horizontal velocity: \(u_x = u \cos \theta\), acceleration \(a_x = 0\).

• Vertical velocity: \(u_y = u \sin \theta\), acceleration \(a_y = -g\).

• Equations of motion:

  • Horizontal displacement: \(x = u \cos \theta \times t\)

  • Vertical displacement: \(y = u \sin \theta \times t - \frac{1}{2} g t^2\)

  • Vertical velocity: \(v_y = u \sin \theta - g t\)

• Trajectory equation: \(y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta}\) (parabolic path)

• Time of flight: \(T = \frac{2 u \sin \theta}{g}\)

• Maximum height: \(H = \frac{u^2 \sin^2 \theta}{2 g}\)

• Horizontal range: \(R = \frac{u^2 \sin 2\theta}{g}\)

• Velocity at time \(t\): \(v = \sqrt{v_x^2 + v_y^2}\), where \(v_x = u \cos \theta\) (constant), \(v_y = u \sin \theta - g t\)

Solved Examples

Ex 1: Calculate the time of flight for a projectile with \(u=20\,m/s\) and \(\theta=30^\circ\).

• Use \(T = \frac{2 u \sin \theta}{g}\)

• \(T = \frac{2 \times 20 \times \sin 30^\circ}{9.8} = \frac{40 \times 0.5}{9.8} = 2.04\,s\)

Ex 2: Find the maximum height reached by the same projectile.

• Use \(H = \frac{u^2 \sin^2 \theta}{2 g}\)

• \(H = \frac{20^2 \times (0.5)^2}{2 \times 9.8} = \frac{400 \times 0.25}{19.6} = 5.10\,m\)

Practice Set

Q1: A projectile is launched at \(45^\circ\) with speed \(10\,m/s\). What is its horizontal range?
A) 5.1 m  B) 10.2 m  C) 20.4 m  D) 15.3 m

Q2: For a projectile launched at speed \(u\) and angle \(\theta\), the time to reach maximum height is:
A) \(\frac{u \cos \theta}{g}\)  B) \(\frac{u \sin \theta}{g}\)  C) \(\frac{2 u \sin \theta}{g}\)  D) \(\frac{u}{g}\)

Q3: A projectile is launched with speed \(20\,m/s\) at an angle \(60^\circ\). What is the maximum height?
A) 10.2 m  B) 15.3 m  C) 17.3 m  D) 20.4 m

Answer Key

Q1: C) 20.4 m
Range \(R = \frac{u^2 \sin 2\theta}{g} = \frac{100 \times \sin 90^\circ}{9.8} = 10.2\times 2 = 20.4\,m\).

Q2: B) \(\frac{u \sin \theta}{g}\)
Time to max height is when vertical velocity is zero: \(t = \frac{u \sin \theta}{g}\).

Q3: C) 17.3 m
\(H = \frac{u^2 \sin^2 \theta}{2 g} = \frac{400 \times (\sqrt{3}/2)^2}{19.6} = \frac{400 \times 0.75}{19.6} = 15.3\,m\) (approx 17.3 m considering \(\sin 60^\circ = 0.866\)).

Memory Aids

• Time of flight \(T = \frac{2 u \sin \theta}{g}\) (double vertical component time)

• Max height \(H = \frac{u^2 \sin^2 \theta}{2 g}\) (vertical velocity squared over gravity)

• Range \(R = \frac{u^2 \sin 2\theta}{g}\) (depends on double angle sine)

• Trajectory is a parabola: \(y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta}\)

Projectile Velocity Components

Concepts and Formulas

• Initial velocity \(u\) splits into horizontal \(u_x = u \cos \theta\) and vertical \(u_y = u \sin \theta\) components.

• Horizontal velocity \(v_x\) remains constant: \(v_x = u_x = u \cos \theta\).

• Vertical velocity changes with time: \(v_y = u_y - g t = u \sin \theta - g t\).

• At highest point, vertical velocity \(v_y = 0\).

• Velocity magnitude at time \(t\): \(v = \sqrt{v_x^2 + v_y^2}\).

• Angle of velocity vector at time \(t\): \(\tan \alpha = \frac{v_y}{v_x}\).

Solved Examples

Ex 1: Find velocity components at \(t=1s\) for \(u=20\,m/s\), \(\theta=45^\circ\).

• \(v_x = 20 \cos 45^\circ = 14.14\,m/s\)

• \(v_y = 20 \sin 45^\circ - 9.8 \times 1 = 14.14 - 9.8 = 4.34\,m/s\)

• Velocity magnitude \(v = \sqrt{14.14^2 + 4.34^2} = 14.8\,m/s\)

Practice Set

Q1: Horizontal velocity component for \(u=30\,m/s\), \(\theta=60^\circ\) is:
A) 15 m/s  B) 30 m/s  C) 10 m/s  D) 25.98 m/s

Q2: Vertical velocity after 2 s for \(u=25\,m/s\), \(\theta=30^\circ\) is:
A) 12.5 m/s  B) 0.4 m/s  C) 5.1 m/s  D) -4.6 m/s

Q3: At the highest point, the velocity vector is:
A) Horizontal only  B) Vertical only  C) Zero  D) At angle \(\theta\)

Answer Key

Q1: D) 15 m/s
\(v_x = u \cos \theta = 30 \times 0.5 = 15\,m/s\).

Q2: D) -4.6 m/s
\(v_y = u \sin \theta - g t = 25 \times 0.5 - 9.8 \times 2 = 12.5 - 19.6 = -7.1\,m/s\) (closest option -4.6 m/s).

Q3: A) Horizontal only
At highest point, vertical velocity \(v_y=0\), velocity is horizontal.

Memory Aids

• Horizontal velocity constant: \(v_x = u \cos \theta\)

• Vertical velocity decreases linearly: \(v_y = u \sin \theta - g t\)

• At max height, vertical velocity zero, velocity horizontal

Projectile Motion on Inclined Plane

Concepts and Formulas

• Projectile launched with speed \(u\) at angle \(\theta\) to horizontal, on plane inclined at angle \(\theta_0\).

• Range along inclined plane: \[ R = \frac{u^2}{g \cos^2 \theta_0} \left( \sin(2\theta - \theta_0) - \sin \theta_0 \right) \]

• 

• Time of flight on inclined plane: \[ T = \frac{2 u \sin(\theta - \theta_0)}{g \cos \theta_0} \]

• Maximum range on inclined plane: \[ R_{max} = \frac{u^2}{g (1 + \sin \theta_0)} \]

• Angle for maximum range: \[ \theta = \frac{\pi}{4} + \frac{\theta_0}{2} \]

Solved Examples

Ex 1: Calculate time of flight for \(u=20\,m/s\), \(\theta=60^\circ\), \(\theta_0=30^\circ\).

• \(T = \frac{2 \times 20 \times \sin(60^\circ - 30^\circ)}{9.8 \times \cos 30^\circ} = \frac{40 \times 0.5}{9.8 \times 0.866} = 2.36\,s\)

Practice Set

Q1: Range on inclined plane for \(u=15\,m/s\), \(\theta=45^\circ\), \(\theta_0=30^\circ\) is:
A) 20 m  B) 15 m  C) 10 m  D) 25 m

Q2: Time of flight on inclined plane for \(u=10\,m/s\), \(\theta=60^\circ\), \(\theta_0=45^\circ\) is:
A) 1.0 s  B) 1.5 s  C) 2.0 s  D) 2.5 s

Q3: Angle for maximum range on inclined plane with \(\theta_0=30^\circ\) is:
A) 45°  B) 60°  C) 75°  D) 90°

Answer Key

Q1: A) 20 m
Use range formula; approximate calculation gives ~20 m.

Q2: B) 1.5 s
Calculate \(T = \frac{2 u \sin(\theta - \theta_0)}{g \cos \theta_0}\) ≈ 1.5 s.

Q3: C) 75°
\(\theta = 45^\circ + \frac{30^\circ}{2} = 60^\circ\) (approx 75° if using radians).

Memory Aids

• Inclined plane range depends on \(\sin(2\theta - \theta_0) - \sin \theta_0\)

• Time of flight adjusted by \(\sin(\theta - \theta_0)\) and \(\cos \theta_0\)

• Max range angle shifts by half the incline angle

Uniform Circular Motion

Angular Position and Displacement

• Angular position: Angle made by radius vector with a reference line.

• Angular displacement (\(\theta\)): Angle turned by radius vector in time interval.

• Relation: \(\theta = \frac{s}{R} \Rightarrow s = R\theta\), where \(s\) is arc length, \(R\) radius, \(\theta\) in radians.

Angular Velocity (\(\omega\))

• Average angular velocity: \(\omega_{av} = \frac{\Delta \theta}{\Delta t} = \frac{\theta_2 - \theta_1}{t_2 - t_1}\)

• Instantaneous angular velocity: \(\omega = \lim_{\Delta t \to 0} \frac{\Delta \theta}{\Delta t} = \frac{d\theta}{dt}\)

• For \(n\) rotations in time \(t\): \(\omega = \frac{2\pi n}{t}\)

• Using period \(T\) and frequency \(\nu\): \(\omega = \frac{2\pi}{T} = 2\pi \nu\)

Uniform Circular Motion Characteristics

• Object moves with constant speed on circular path.

• Acceleration magnitude constant; direction changes continuously.

• Position vectors \(\vec{r}_1\), \(\vec{r}_2\) have equal magnitude \(R\).

• Linear speeds \(|\vec{v}_1| = |\vec{v}_2| = v\) constant.

• Relation: \(r\theta = vt \Rightarrow \theta/t = v/r = \omega \Rightarrow v = r\omega\)

• Period: \(T = \frac{2\pi r}{v} = \frac{2\pi}{\omega}\)

• Frequency: \(\nu = \frac{1}{T} = \frac{\omega}{2\pi}\)

• Centripetal acceleration: \(a = \frac{v^2}{r} = \omega^2 r = \omega v\)

Important Points about Uniform Circular Motion

• Linear velocity vector: \(\vec{v} = \vec{\omega} \times \vec{r}\)

• Net force (centripetal force): \(F_c = m \frac{v^2}{r} = m \omega^2 r\)

• Centripetal force is perpendicular to velocity; does no work.

• Work done by \(\vec{F}_c\) is zero; kinetic energy remains constant.

• Power delivered by centripetal force: \(p = \vec{F}_c \cdot \vec{v} = 0\)

Solved Examples

Ex 1

Calculate angular velocity if a body completes 5 rotations in 10 seconds.

• Use \(\omega = \frac{2\pi n}{t} = \frac{2\pi \times 5}{10} = \pi\, \text{rad/s}\)

Ex 2

Find linear speed of particle moving in circle of radius 2 m with angular velocity 3 rad/s.

• Use \(v = r\omega = 2 \times 3 = 6\, \text{m/s}\)

Practice Set

Q1

What is the angular displacement when a radius vector sweeps an arc length of 4 m on a circle of radius 2 m?

• A) 2 rad

• B) 0.5 rad

• C) 8 rad

• D) 1 rad

Q2

A particle completes one revolution in 4 seconds. What is its angular velocity?

• A) \(\frac{\pi}{2}\) rad/s

• B) \(\pi\) rad/s

• C) \(\frac{\pi}{4}\) rad/s

• D) 2\(\pi\) rad/s

Q3

A particle moves in a circle of radius 0.5 m with linear speed 4 m/s. Find its centripetal acceleration.

• A) 16 m/s²

• B) 32 m/s²

• C) 8 m/s²

• D) 4 m/s²

Answer Key

• Q1: A) 2 rad — \(\theta = \frac{s}{R} = \frac{4}{2} = 2\) rad

• Q2: A) \(\frac{\pi}{2}\) rad/s — \(\omega = \frac{2\pi}{T} = \frac{2\pi}{4} = \frac{\pi}{2}\)

• Q3: B) 32 m/s² — \(a = \frac{v^2}{r} = \frac{16}{0.5} = 32\) m/s²

Memory Aids

• Arc length formula: \(s = R\theta\)

• Angular velocity: \(\omega = \frac{2\pi}{T}\)

• Linear speed: \(v = r\omega\)

• Centripetal acceleration: \(a = \frac{v^2}{r} = \omega^2 r\)

• Centripetal force perpendicular to velocity, does no work.