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Unit 2, Chapter 4: Motion in a Plane | Physics | Class 11th
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Chapter 4: Motion in a Plane

1. Introduction

While motion in a straight line provides a fundamental understanding of kinematics, most real-world motions are not one-dimensional. A car turning on a road, a thrown javelin, or a planet orbiting the sun are all examples of motion in a plane (two-dimensional motion). To describe such motion, we require a new mathematical tool that can handle both magnitude and direction simultaneously. This is the role of vectors.

2. Scalar and Vector Quantities

Scalar Quantities (Scalars)

Definition: A scalar quantity is a physical quantity that has only magnitude and no direction. It is specified by a single number and a unit.
Examples: Distance, speed, mass, time, temperature, work, energy.

Vector Quantities (Vectors)

Definition: A vector quantity is a physical quantity that has both magnitude and direction and obeys the laws of vector algebra.
Representation: A vector is represented by a letter with an arrow over it (e.g., A).
Examples: Displacement, velocity, acceleration, force, momentum, torque.

3. General Vectors and their Notations

Position and Displacement Vectors

To describe an object's motion in a plane, we use a Cartesian coordinate system (x-y plane).

Displacement Vector (Δr) = r₂ - r₁

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Unit 2, Chapter 4: Motion in a Plane | Physics | Class 11th
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4. Addition and Subtraction of Vectors

4.1 Graphical Methods of Vector Addition

(a) Triangle Law of Vector Addition (Head-to-Tail Method)

Statement:

If two vectors are represented in magnitude and direction by the two sides of a triangle taken in the same order, then their resultant is represented completely by the third side of the triangle taken in the opposite order.

A B R = A + B

(b) Parallelogram Law of Vector Addition (Tail-to-Tail Method)

Statement:

If two vectors are represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from a common point, then their resultant is represented by the diagonal of the parallelogram drawn from that same common point.

4.2 Analytical Method of Vector Addition: Detailed Derivation

We can derive a formula for the magnitude and direction of the resultant of two vectors A and B inclined at an angle `θ`.

Derivation of Resultant Magnitude and Direction

1. Geometric Construction

Consider vectors OP = A and OQ = B, with an angle θ between them. Complete the parallelogram OQSP. The diagonal OS = R is the resultant. Drop a perpendicular SN from point S to the extended line OP.

A O P B Q R S N θ α θ

2. Use Trigonometry on the Small Triangle

In the right-angled triangle PSN, the hypotenuse PS has magnitude B.
Therefore, PN = PS cosθ = Bcosθ and SN = PS sinθ = Bsinθ.

3. Apply Pythagorean Theorem to the Large Triangle

Now consider the large right-angled triangle OSN.
OS² = ON² + SN²
Since ON = OP + PN, we have:
R² = (A + Bcosθ)² + (Bsinθ)²

4. Simplify the Expression

R² = A² + B²cos²θ + 2ABcosθ + B²sin²θ
R² = A² + B²(cos²θ + sin²θ) + 2ABcosθ
Since cos²θ + sin²θ = 1, we get:
R² = A² + B² + 2ABcosθ

5. Find the Direction (Angle α)

The direction of the resultant R is the angle α it makes with vector A. In triangle OSN:
tan(α) = Perpendicular / Base = SN / ON = Bsinθ / (A + Bcosθ)

Magnitude of Resultant: R = √(A² + B² + 2ABcosθ)

Direction of Resultant: tan(α) = (B sinθ) / (A + B cosθ)

Solved Example 1: Finding the Resultant Force

Question: Two forces of 5 N and 12 N act on a particle at an angle of 90° to each other. Find the magnitude and direction of the resultant force.

Solution:
Given: F₁ = 5 N, F₂ = 12 N, θ = 90°.
Using the analytical formula for magnitude: R = √(F₁² + F₂² + 2F₁F₂cosθ)
Since cos(90°) = 0, the formula reduces to the Pythagorean theorem:
R = √(5² + 12²) = √(25 + 144) = √169 = 13 N.
To find the direction (angle α with F₁): tan(α) = (F₂ sinθ) / (F₁ + F₂ cosθ)
tan(α) = (12 × 1) / (5 + 0) = 12 / 5 = 2.4.
α = tan⁻¹(2.4) ≈ 67.4°.
Answer: The resultant force is 13 N acting at an angle of 67.4° from the 5 N force.

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Unit 2, Chapter 4: Motion in a Plane | Physics | Class 11th
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5. Resolution of a Vector and Products

5.1 Resolution of a Vector

The process of splitting a vector into two or more components is called the resolution of a vector. The most useful components are mutually perpendicular, known as **rectangular components**.

x y O A Ax Ay θ

A vector A in the x-y plane making an angle `θ` with the x-axis can be resolved as:

Horizontal Component: Ax = A cosθ

Vertical Component: Ay = A sinθ

In terms of unit vectors `î` and `ĵ`:

A = Ax î + Ay ĵ

5.2 Scalar (Dot) Product

Definition of Scalar Product:

The scalar product of two vectors A and B is a scalar quantity defined as the product of their magnitudes and the cosine of the angle `θ` between them.

A · B = AB cosθ

Example: Work done (W) is the dot product of Force (F) and displacement (s): W = F · s.

5.3 Vector (Cross) Product

Definition of Vector Product:

The vector product of two vectors A and B is a vector C whose magnitude is `AB sinθ`. The direction of C is perpendicular to the plane containing A and B, given by the Right-Hand Thumb Rule.

A × B = (AB sinθ) n̂

Example: Torque (τ) is the cross product of position vector (r) and Force (F): τ = r × F.

Solved Example 2: Work Done (Dot Product)

Question: A force `F = 2î + 3ĵ - k̂` N acts on a body and displaces it from origin to a point `s = 4î - ĵ + 2k̂` m. Calculate the work done.

Solution:
Work done `W = F · s`.
W = (2î + 3ĵ - k̂) · (4î - ĵ + 2k̂)
W = (2)(4) + (3)(-1) + (-1)(2) = 8 - 3 - 2 = 3 Joules.
Answer: The work done is 3 J.

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Unit 2, Chapter 4: Motion in a Plane | Physics | Class 11th
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6. Projectile Motion

Definition of Projectile Motion:

A projectile is any object that is thrown into space upon which the only force acting is gravity (air resistance is neglected). The curved path followed by the projectile is called its trajectory. Projectile motion is a classic example of motion in a plane with constant acceleration.

To analyze projectile motion, we resolve the motion into two independent components:

  1. Horizontal Motion: Constant velocity (acceleration ax = 0).
  2. Vertical Motion: Constant downward acceleration (ay = -g).
X Y O u ux uy θ H Range (R)

6.1 Detailed Derivations for Projectile Motion

Derivation of the Equation of Trajectory

1. Decompose Initial Velocity

Initial velocity: `u`. Angle of projection: `θ`.
Horizontal component: ux = u cosθ
Vertical component: uy = u sinθ

2. Write Position Equations

Using kinematic equations for constant acceleration (ax=0, ay=-g):
Horizontal position at time t: x = uxt = (u cosθ)t
Vertical position at time t: y = uyt + ½ayt² = (u sinθ)t - ½gt²

3. Eliminate Time (t)

From the horizontal equation, solve for t: t = x / (u cosθ)

4. Substitute t into the Vertical Equation

y = (u sinθ) [x / (u cosθ)] - ½g [x / (u cosθ)]²

5. Simplify to Final Form

y = (sinθ/cosθ)x - (g / 2u²cos²θ)x²

Trajectory Equation: y = (tanθ)x - ( g / (2u²cos²θ) )x²
This is the equation of a parabola.

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Unit 2, Chapter 4: Motion in a Plane | Physics | Class 11th
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Derivation of Time of Flight (T)

1. Condition for Time of Flight

The time of flight is the total time the projectile is in the air. This ends when the vertical displacement (sy) returns to zero.

2. Apply Vertical Motion Equation

Use sy = uyt + ½ayt². Set sy=0 and t=T.

3. Solve for T

0 = (u sinθ)T - ½gT²
½gT² = (u sinθ)T
Since T≠0, divide by T: ½gT = u sinθ

Time of Flight: T = (2u sinθ) / g

Derivation of Maximum Height (H)

1. Condition for Maximum Height

At the maximum height, the vertical component of velocity (vy) is momentarily zero.

2. Apply Velocity-Position Equation

Use vy² = uy² + 2aysy. Set vy=0 and sy=H.

3. Solve for H

0² = (u sinθ)² + 2(-g)H
0 = u²sin²θ - 2gH
2gH = u²sin²θ

Maximum Height: H = (u²sin²θ) / 2g

Derivation of Horizontal Range (R)

1. Condition for Horizontal Range

Range is the horizontal distance (sx) covered during the Time of Flight (T).

2. Apply Horizontal Motion Equation

Use sx = uxt. Set sx=R and t=T.

3. Substitute and Simplify

R = ux × T = (u cosθ) × ( (2u sinθ) / g )
R = (u²(2sinθcosθ)) / g
Using the identity `sin(2θ) = 2sinθcosθ`:

Horizontal Range: R = (u²sin(2θ)) / g

Solved Example 3: Projectile Motion

Question: A cricket ball is thrown at a speed of 28 m/s in a direction 30° above the horizontal. Calculate (a) the maximum height, (b) the time taken by the ball to return to the same level, and (c) the horizontal distance from the thrower to the point where the ball returns to the same level. (Assume g = 9.8 m/s²)

Solution:
Given: u = 28 m/s, θ = 30°, g = 9.8 m/s².
(a) Maximum Height (H): H = (u²sin²θ) / 2g
H = (28² × (sin30°)²) / (2 × 9.8) = (784 × 0.25) / 19.6 = 196 / 19.6 = 10 m.
(b) Time of Flight (T): T = (2u sinθ) / g
T = (2 × 28 × sin30°) / 9.8 = (56 × 0.5) / 9.8 = 28 / 9.8 ≈ 2.86 s.
(c) Horizontal Range (R): R = (u²sin(2θ)) / g
R = (28² × sin60°) / 9.8 = (784 × 0.866) / 9.8 ≈ 69.3 m.
Answer: H = 10 m, T ≈ 2.86 s, R ≈ 69.3 m.

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Unit 2, Chapter 4: Motion in a Plane | Physics | Class 11th
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7. Uniform Circular Motion

Definition of Uniform Circular Motion:

When an object moves along a circular path at a constant speed, its motion is called uniform circular motion (UCM). Although the speed is constant, the velocity is continuously changing because the direction of motion changes at every point. Therefore, UCM is an example of accelerated motion.

The acceleration in UCM is directed towards the center of the circle and is called centripetal acceleration.

7.1 Detailed Derivation of Centripetal Acceleration (ac)

Derivation of Centripetal Acceleration

1. Geometric Setup

Consider a particle moving in a circle of radius 'r' with constant speed 'v'. Let its position at time t be P and at time t+Δt be P'. The position vectors are r and r' and velocity vectors are v and v' respectively. Let the angle between the position vectors be Δθ.

O r P v r' P' v' Δr Δθ Velocity Triangle v v' Δv Δθ

2. Analyze Velocity Change

The change in velocity is `Δv = v' - v`. Since speed is constant, |v'| = |v| = v. The velocity vector triangle is an isosceles triangle. As the position vectors rotate by Δθ, the velocity vectors (which are always tangential) also rotate by the same angle Δθ.

3. Use Similar Triangles

The triangle formed by position vectors r, r', and displacement `Δr` is similar to the triangle formed by velocity vectors v, v', and change in velocity `Δv`. For small Δθ, we can write the ratio of corresponding sides:

|Δv| / v ≈ |Δr| / r

4. Introduce Time and take the Limit

Rearranging, |Δv| = (v/r)|Δr|. Divide by Δt:
|Δv| / Δt = (v/r) (|Δr| / Δt)
Taking the limit as Δt → 0:
lim ( |Δv| / Δt ) = a (magnitude of acceleration)
lim ( |Δr| / Δt ) = v (instantaneous speed)
So, a = (v/r) × v = v²/r.
The direction of `Δv` can be shown to point towards the center.

Centripetal Acceleration (ac) = v² / r

Solved Example 4: Centripetal Acceleration

Question: A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?

Solution:
Given: radius r = 80 cm = 0.8 m.
Frequency of revolution ν = 14 / 25 Hz.
Angular speed ω = 2πν = 2 × (22/7) × (14/25) = 88 / 25 = 3.52 rad/s.
Centripetal acceleration ac = ω²r
ac = (3.52)² × 0.8 ≈ 12.39 × 0.8 ≈ 9.9 m/s².
Answer: The acceleration is 9.9 m/s² directed along the radius towards the centre of the circle.

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Unit 2, Chapter 4: Motion in a Plane | Physics | Class 11th
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Important Questions and Answers

Conceptual Questions

Q1: Is it possible for an object to have zero speed but non-zero velocity?

A: No, this is not possible. Velocity is a vector quantity with both magnitude and direction. Speed is the magnitude of velocity. If the speed is zero, the magnitude of velocity is zero, and therefore the velocity itself must be a null vector (zero).

Q2: What is the angle between two vectors A and B if A + B = C and A² + B² = C²?

A: The magnitude of the resultant C is given by `C = √(A² + B² + 2ABcosθ)`. Squaring both sides gives `C² = A² + B² + 2ABcosθ`. Given that `C² = A² + B²`, this implies `2ABcosθ = 0`. Since A and B are non-zero magnitudes, `cosθ = 0`, so the angle `θ` is 90°.

Q3: What is the condition for the dot product and cross product of two non-zero vectors to be zero?

A: The dot product is zero when the vectors are perpendicular. The cross product is zero when the vectors are parallel or anti-parallel.

Q4: At what point in its trajectory does a projectile have its minimum speed? What is its acceleration at this point?

A: A projectile has its minimum speed at the highest point of its trajectory, where the vertical velocity component is zero. The acceleration at this point is still the acceleration due to gravity, g, acting vertically downwards.

Q5: Why is an object in uniform circular motion considered to be in accelerated motion?

A: Because its velocity is continuously changing due to a continuous change in its direction of motion, even though its speed is constant.

Q6: What is the physical significance of the unit vectors î, ĵ, and k̂?

A: Their physical significance is to provide a clear and unambiguous way to specify the direction of a vector's components in a 3D Cartesian coordinate system.

Q7: Can the resultant of two unequal vectors be zero?

A: No. The resultant can only be zero if the two vectors are equal in magnitude and opposite in direction.

Q8: For what angle of projection is the horizontal range of a projectile equal to its maximum height?

A: Setting Range (R) = Max Height (H): `(u²sin(2θ))/g = (u²sin²θ)/2g`. This simplifies to `tanθ = 4`, so `θ = tan⁻¹(4) ≈ 76°`.

Q9: What is the source of centripetal force for a planet revolving around the Sun?

A: The gravitational force of attraction exerted by the Sun on the planet.

Q10: If `A + B = A - B`, what can you conclude about vector B?

A: This simplifies to `2B = 0`, which implies that B must be a null vector.

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Unit 2, Chapter 4: Motion in a Plane | Physics | Class 11th
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Numerical Problems

Q11: Two forces of magnitudes 10 N and 6 N act on a body. The angle between the forces is 60°. Find the magnitude of the resultant force.

A: Given F₁ = 10 N, F₂ = 6 N, θ = 60°.
R = √(F₁² + F₂² + 2F₁F₂cosθ) = √(10² + 6² + 2(10)(6)cos60°)
R = √(100 + 36 + 120(1/2)) = √196 = 14 N.
Answer: 14 N.

Q12: Find the angle between two vectors `A = 2î + 3ĵ + k̂` and `B = -3î + ĵ - 2k̂`.

A: Use `A · B = ABcosθ`.
`A · B = (2)(-3) + (3)(1) + (1)(-2) = -5`.
A = √14, B = √14.
cosθ = -5 / (√14 × √14) = -5 / 14.
θ = cos⁻¹(-5/14) ≈ 110.9°.
Answer: ≈ 110.9°.

Q13: Find a unit vector perpendicular to both `A = î - 2ĵ + k̂` and `B = 3î + ĵ - 2k̂`.

A: Find `C = A × B`.
C = 3î + 5ĵ + 7k̂.
Magnitude of C = `√83`.
Unit vector `ĉ` = `C / C`.
Answer: `(1/√83)(3î + 5ĵ + 7k̂)`.

Q14: A projectile is fired with a velocity of 20 m/s at an angle of 30° with the horizontal. Find (a) the time of flight, (b) the maximum height, and (c) the horizontal range. (g = 10 m/s²)

A: Given u = 20 m/s, θ = 30°, g = 10 m/s².
(a) Time of Flight (T): (2 × 20 × sin30°) / 10 = 2 s.
(b) Max Height (H): (20² × (sin30°)²) / (2 × 10) = 5 m.
(c) Range (R): (20²sin(60°)) / 10 = 20√3 m ≈ 34.64 m.
Answer: T = 2 s, H = 5 m, R ≈ 34.64 m.

Q15: An insect in a circular groove of radius 12 cm completes 7 revolutions in 100 s. What is the centripetal acceleration?

A: r = 0.12 m. ν = 7/100 Hz. ω = 2πν = 0.14π rad/s.
ac = ω²r = (0.14π)² × 0.12 ≈ 0.023 m/s².
Answer: ≈ 0.023 m/s².

Q16: A motorboat races towards north at 25 km/h. The water current is 10 km/h at 60° east of south. Find the resultant velocity.

A: vb = 25ĵ. vc = 5√3î - 5ĵ.
V = vb + vc = 5√3î + 20ĵ.
|V| = `√((5√3)² + 20²) ≈ 21.8 km/h`.
Answer: ≈ 21.8 km/h at an angle of about 66.6° north of east.

Q17: A particle starts from origin at t=0 with velocity 5î m/s and moves in x-y plane with acceleration (3î + 2ĵ) m/s². At what time is the x-coordinate 84 m?

A: ux=5, ax=3. sx = uxt + ½axt².
84 = 5t + 1.5t² → 1.5t² + 5t - 84 = 0.
Solving the quadratic equation gives t = 6 s.
Answer: t = 6 s.

Q18: It is raining vertically at 35 m/s. A woman rides a bicycle at 12 m/s from east to west. In what direction should she hold her umbrella?

A: vr = -35ĵ. vw = -12î.
vrw = vr - vw = 12î - 35ĵ.
tanθ = 12/35 ≈ 0.343. θ ≈ 19°.
Answer: At an angle of ≈ 19° with the vertical, towards the west.

Q19: A stone on a string of length 80 cm is whirled in a horizontal circle. It makes 14 revolutions in 25 s. Find its acceleration.

A: r = 0.8 m. ν = 14/25 Hz. ω = 2πν.
ac = ω²r = (2π × 14/25)² × 0.8 ≈ 9.9 m/s².
Answer: ≈ 9.9 m/s².

Q20: An aircraft executes a horizontal loop of radius 1.00 km at 900 km/h. Compare its centripetal acceleration with g.

A: r = 1000 m. v = 900 km/h = 250 m/s.
ac = v²/r = (250)² / 1000 = 62.5 m/s².
Ratio ac / g = 62.5 / 9.8 ≈ 6.38.
Answer: The centripetal acceleration is ≈ 6.4 times g.