Class 11 Physics Chapter 3: Motion in a Plane – Notes & Questions

Class 11 Physics Chapter 3: Motion in a Plane

Motion in a Plane is the third chapter of Class 11 Physics. In the previous chapter, we studied motion along a straight line. Here, the motion becomes more interesting because an object can move in more than one direction at the same time.

This chapter introduces one of the most useful ideas in Physics: vectors. Vectors help us describe quantities such as displacement, velocity and acceleration when direction matters.

These notes are written in simple language for CBSE Class 11 Physics students and are useful for understanding concepts, revising formulas and preparing for numerical questions.

Chapter idea: Motion in a plane can usually be understood by breaking it into two perpendicular directions, such as the x-direction and y-direction.

In this chapter:

• Motion in a Plane
• Scalar and Vector Quantities
• Vectors
• Addition of Vectors
• Resolution of Vectors
• Relative Velocity
• Projectile Motion
• Uniform Circular Motion
• Quick Revision

1. What is Motion in a Plane?

When an object changes its position in two dimensions, its motion is called motion in a plane or two-dimensional motion.

A ball thrown at an angle, the motion of a stone projected from the ground and the motion of a satellite around Earth are examples where direction changes or motion cannot be described completely using only one straight line.

To study such motion, we normally use two perpendicular axes: the x-axis and the y-axis.

Simple picture in your mind: Imagine walking 3 m east and then 4 m north. Your final position is not described by one number alone. We need both horizontal and vertical information.

2. Scalar and Vector Quantities

Scalar Quantity

A scalar quantity has magnitude only. It does not need a direction for its complete description.

Examples include:

• Mass
• Time
• Distance
• Speed
• Temperature
• Energy

Vector Quantity

A vector quantity has both magnitude and direction.

Examples include:

• Displacement
• Velocity
• Acceleration
• Force
• Momentum

Scalar	Vector
Magnitude only	Magnitude and direction
Added using ordinary algebra	Requires vector rules
Example: speed	Example: velocity

3. Understanding Vectors

A vector is commonly represented by a directed line segment. The length represents its magnitude, while the arrow indicates its direction.

A vector can be written using an arrow above the symbol, for example: →A.

Magnitude of a Vector

If a vector has rectangular components A_x and A_y, its magnitude is:

|A| = √(A_x² + A_y²)

Direction of a Vector

If the vector makes an angle θ with the positive x-axis:

tan θ = A_y / A_x

Therefore:

θ = tan⁻¹(A_y/A_x)

Important: While finding the angle, always look at the signs of the x and y components to identify the correct quadrant.

4. Addition of Vectors

Two or more vectors can be added to obtain a single vector called the resultant vector.

Triangle Law of Vector Addition

If two vectors are represented by two sides of a triangle taken in order, their resultant is represented by the third side drawn from the starting point to the final point.

Parallelogram Law

If two vectors acting at a point are represented by the adjacent sides of a parallelogram, their resultant is represented by the diagonal passing through the common point.

Resultant of Two Vectors

If two vectors A and B make an angle θ with each other, the magnitude of their resultant R is:

R = √(A² + B² + 2AB cos θ)

The direction of the resultant with respect to A is given by:

tan α = B sin θ / (A + B cos θ)

5. Resolution of a Vector

Sometimes adding vectors is easier if we split one vector into perpendicular components.

Suppose a vector A makes an angle θ with the positive x-axis. Its rectangular components are:

A_x = A cos θ

A_y = A sin θ

Therefore, the vector can be represented in terms of its components as:

A = A_x î + A_y ĵ

Here î and ĵ are unit vectors along the x-axis and y-axis respectively.

6. Unit Vector

A unit vector is a vector having magnitude equal to one. It indicates direction without changing the physical meaning of the quantity.

Along the x-axis, we use î, and along the y-axis, we use ĵ.

|î| = |ĵ| = 1

7. Position Vector

The position of a particle in a plane can be represented by a position vector drawn from the origin to the position of the particle.

r = x î + y ĵ

Here x and y are the coordinates of the particle.

8. Displacement in Two Dimensions

If a particle moves from position vector r₁ to r₂, its displacement is:

Δr = r₂ − r₁

Thus, displacement is a vector quantity because it has both magnitude and direction.

9. Velocity in Two Dimensions

Average velocity is defined as displacement divided by the corresponding time interval.

v_avg = Δr / Δt

Instantaneous velocity is the rate of change of position with time.

v = dr/dt

In component form:

v = v_x î + v_y ĵ

10. Acceleration in Two Dimensions

Acceleration describes the rate at which velocity changes with time. Since velocity is a vector, acceleration can result from a change in speed, direction, or both.

a = dv/dt

In component form:

a = a_x î + a_y ĵ

Key idea: An object can have constant speed but still have acceleration if its direction of motion is continuously changing. Uniform circular motion is the classic example.

11. Relative Velocity

The velocity of one object as observed from another moving object is called relative velocity.

If A and B have velocities v_A and v_B, then the velocity of A relative to B is:

v_AB = v_A − v_B

Similarly:

v_BA = v_B − v_A

Why Relative Velocity Matters

When two vehicles move on a road, the speed at which one appears to approach the other depends on their relative velocity, not simply on either vehicle's speed alone.

12. Projectile Motion

A projectile is an object projected into the air that subsequently moves under the influence of gravity, when air resistance is neglected.

The path followed by an ideal projectile is a parabola.

Most important idea: Projectile motion can be treated as two independent motions: horizontal motion and vertical motion.

Horizontal Component of Initial Velocity

If an object is projected with initial speed u at an angle θ:

u_x = u cos θ

Vertical Component of Initial Velocity

u_y = u sin θ

Horizontal Motion

Neglecting air resistance, horizontal acceleration is zero. Therefore, horizontal velocity remains constant.

x = (u cos θ)t

Vertical Motion

Vertical acceleration is due to gravity and is directed downward.

y = (u sin θ)t − ½gt²

13. Time of Flight

For a projectile that is projected and lands at the same level, the total time of flight is:

T = 2u sin θ / g

14. Maximum Height

The maximum vertical height reached by a projectile is:

H = u² sin²θ / 2g

15. Horizontal Range

The horizontal distance travelled by the projectile before returning to the same level is called its range.

R = u² sin 2θ / g

Maximum Range

Since sin 2θ has its maximum value of 1:

R_max = u² / g

The maximum range occurs when:

θ = 45°

16. Uniform Circular Motion

When an object moves along a circular path with constant speed, its motion is called uniform circular motion.

Even though the speed remains constant, the velocity changes because the direction of motion changes continuously.

Centripetal Acceleration

The acceleration of an object moving in a circle is directed towards the centre of the circle. It is called centripetal acceleration.

a_c = v²/r

Using v = ωr:

a_c = ω²r

Here r is the radius of the circular path, v is linear speed and ω is angular speed.

Angular Speed

ω = θ/t

For one complete revolution:

ω = 2π/T

Relation Between Linear and Angular Speed

v = ωr

17. Quick Revision

Concept	Key Point
Scalar	Magnitude only.
Vector	Magnitude and direction.
Vector magnitude	√(Ax² + Ay²)
x-component	A cos θ
y-component	A sin θ
Relative velocity	vAB = vA − vB
Projectile time	T = 2u sin θ/g
Projectile height	H = u²sin²θ/2g
Projectile range	R = u²sin2θ/g
Centripetal acceleration	ac = v²/r = ω²r
Linear-angular relation	v = ωr

Class 11 Physics Chapter 3 Important Questions

Practising questions is the best way to turn vector concepts and projectile formulas into marks. Try answering these questions before checking the solutions.

Very Short Answer Questions

1. What is a scalar quantity?
2. What is a vector quantity?
3. Give two examples of vector quantities.
4. What is a unit vector?
5. What is a position vector?
6. What is meant by relative velocity?
7. What is projectile motion?
8. What is the shape of the trajectory of an ideal projectile?
9. What is the horizontal acceleration of a projectile when air resistance is neglected?
10. What is centripetal acceleration?

Short Answer Questions

1. Differentiate between scalar and vector quantities.
2. Explain the triangle law of vector addition.
3. Explain the parallelogram law of vector addition.
4. What is meant by resolution of a vector?
5. Why is projectile motion treated as two independent motions?
6. Why does a projectile have zero horizontal acceleration when air resistance is neglected?
7. Why does a projectile follow a parabolic path?
8. Why does an object moving with constant speed in a circle have acceleration?
9. Explain centripetal acceleration.
10. Derive the relation between linear speed and angular speed.

Important Conceptual Questions with Answers

Q1. Can a vector have zero magnitude?

Answer: Yes. Such a vector is called a zero or null vector. It has zero magnitude and no definite direction.

Q2. Can two vectors of unequal magnitudes have zero resultant?

Answer: No. Two vectors can have zero resultant only when their magnitudes are equal and they act in exactly opposite directions.

Q3. Why is velocity a vector but speed a scalar?

Answer: Speed describes only how fast an object moves, whereas velocity also specifies the direction of motion.

Q4. Why does a projectile have constant horizontal velocity?

Answer: In ideal projectile motion, air resistance is ignored and gravity acts vertically downward. Therefore, there is no horizontal acceleration and the horizontal velocity remains constant.

Q5. Is acceleration zero in uniform circular motion?

Answer: No. Although speed remains constant, velocity changes continuously because its direction changes. Therefore, centripetal acceleration is present.

Q6. At what angle is the range of a projectile maximum?

Answer: For projection and landing at the same level, the maximum range occurs at 45°.

Multiple Choice Questions

1. Which of the following is a vector quantity?
   A. Mass   B. Time   C. Speed   D. Velocity
   Answer: D. Velocity


2. The magnitude of a vector with components 3 and 4 is:
   A. 3   B. 4   C. 5   D. 7
   Answer: C. 5


3. A vector of magnitude A makes an angle θ with the x-axis. Its x-component is:
   A. A sin θ   B. A cos θ   C. A tan θ   D. A/ sin θ
   Answer: B. A cos θ


4. The trajectory of an ideal projectile is:
   A. Circle   B. Straight line   C. Parabola   D. Ellipse
   Answer: C. Parabola


5. At the highest point of projectile motion, the vertical component of velocity is:
   A. Maximum   B. Zero   C. Equal to g   D. Negative maximum
   Answer: B. Zero


6. The horizontal acceleration of an ideal projectile is:
   A. g   B. 2g   C. Zero   D. g/2
   Answer: C. Zero


7. The maximum range of a projectile occurs at:
   A. 30°   B. 45°   C. 60°   D. 90°
   Answer: B. 45°


8. Centripetal acceleration is directed:
   A. Away from centre   B. Along tangent   C. Towards centre   D. Vertically upward
   Answer: C. Towards centre


9. The relation between linear speed and angular speed is:
   A. v = ω/r   B. v = ωr   C. v = r/ω   D. v = ω + r
   Answer: B. v = ωr


10. If two vectors are equal in magnitude and opposite in direction, their resultant is:
    A. Maximum   B. Zero   C. Twice either vector   D. Half
    Answer: B. Zero

Numerical Problems with Solutions

Numerical 1: Magnitude of a Vector

A vector has rectangular components 6 units and 8 units. Find its magnitude.

Solution:

|A| = √(A_x² + A_y²)

|A| = √(6² + 8²) = √100 = 10 units

Answer: 10 units

Numerical 2: Vector Components

A velocity of 20 m/s makes an angle of 30° with the x-axis. Find its horizontal and vertical components.

v_x = v cos 30°

v_x = 20 × √3/2 = 10√3 m/s

v_y = v sin 30° = 20 × 1/2 = 10 m/s

Answer: Horizontal component = 10√3 m/s; Vertical component = 10 m/s.

Numerical 3: Resultant of Perpendicular Vectors

Two perpendicular vectors have magnitudes 5 N and 12 N. Find the magnitude of their resultant.

R = √(A² + B²)

R = √(5² + 12²) = √169 = 13 N

Answer: 13 N

Numerical 4: Projectile Time of Flight

A ball is projected with a speed of 20 m/s at an angle of 30° with the horizontal. Take g = 10 m/s². Find the time of flight.

T = 2u sin θ / g

T = [2 × 20 × sin 30°] / 10

T = [40 × 1/2] / 10 = 2 s

Answer: 2 s

Numerical 5: Maximum Height of a Projectile

A projectile is fired with a speed of 20 m/s at 30° to the horizontal. Find its maximum height. Take g = 10 m/s².

H = u² sin²θ / 2g

H = 20² × (1/2)² / (2 × 10)

H = 400 × 1/4 / 20 = 5 m

Answer: 5 m

Numerical 6: Horizontal Range

A projectile is fired with speed 20 m/s at 45° to the horizontal. Take g = 10 m/s². Find its range.

R = u² sin 2θ / g

R = 20² × sin 90° / 10

R = 400 / 10 = 40 m

Answer: 40 m

Numerical 7: Centripetal Acceleration

A car moves around a circular track of radius 50 m with a speed of 10 m/s. Find its centripetal acceleration.

a_c = v²/r

a_c = 10²/50 = 2 m/s²

Answer: 2 m/s²

Numerical 8: Angular Speed

An object completes one revolution around a circular path in 4 seconds. Find its angular speed.

ω = 2π/T

ω = 2π/4 = π/2 rad/s

Answer: π/2 rad/s

Numerical strategy: For projectile problems, first split the initial velocity into horizontal and vertical components. Then solve the x-motion and y-motion separately. This single habit prevents many common mistakes.

Assertion-Reason Questions

1. Assertion: An object moving with constant speed in a circular path has acceleration.

Reason: Its velocity changes continuously because its direction changes.

Answer: Both statements are true, and the reason correctly explains the assertion.

2. Assertion: The horizontal velocity of an ideal projectile remains constant.

Reason: Gravity acts vertically downward and there is no horizontal acceleration when air resistance is neglected.

Answer: Both statements are true, and the reason correctly explains the assertion.

3. Assertion: The range of a projectile is maximum at 45°.

Reason: The value of sin 2θ is maximum when 2θ = 90°.

Answer: Both statements are true, and the reason correctly explains the assertion.

Class 11 Physics Chapter 3 FAQs

1. What is Chapter 3 of Class 11 Physics?

In the current CBSE Class 11 Physics sequence, Chapter 3 is Motion in a Plane.

2. What are the main topics in Motion in a Plane?

The important topics include scalar and vector quantities, vector addition and resolution, relative velocity, projectile motion and uniform circular motion.

3. What is a vector quantity?

A vector quantity is a physical quantity that has both magnitude and direction. Displacement, velocity and acceleration are common examples.

4. What is a scalar quantity?

A scalar quantity has magnitude only. Mass, time, distance and speed are examples of scalar quantities.

5. What is the formula for the magnitude of a vector?

|A| = √(A_x² + A_y²)

6. What is projectile motion?

Projectile motion is the motion of an object projected into the air that subsequently moves under gravity, when air resistance is neglected.

7. What is the shape of a projectile's path?

For ideal projectile motion, the trajectory is a parabola.

8. What is the formula for the time of flight?

T = 2u sin θ / g

This formula applies when the projectile is projected and lands at the same level.

9. What is the formula for maximum height?

H = u² sin²θ / 2g

10. What is the formula for the range of a projectile?

R = u² sin 2θ / g

For projection and landing at the same level, the maximum range occurs at 45°.

11. What is relative velocity?

Relative velocity is the velocity of one object as observed from another object. For two objects A and B:

v_AB = v_A − v_B

12. Why is there acceleration in uniform circular motion?

Although the speed remains constant, the direction of velocity changes continuously. Since acceleration is the rate of change of velocity, the object has centripetal acceleration.

13. What is centripetal acceleration?

Centripetal acceleration is the acceleration directed towards the centre of a circular path.

a_c = v²/r = ω²r

14. What is the relation between linear and angular velocity?

v = ωr

15. What is the most important concept in projectile motion?

The most useful idea is to separate the motion into horizontal and vertical components. Horizontal motion has zero acceleration, while vertical motion has acceleration due to gravity.

Important Formula Sheet

Concept	Formula
Vector magnitude	|A| = √(Ax² + Ay²)
Vector direction	tan θ = Ay/Ax
x-component	Ax = A cos θ
y-component	Ay = A sin θ
Resultant of two vectors	R = √(A²+B²+2AB cos θ)
Relative velocity	vAB = vA − vB
Projectile horizontal velocity	u cos θ
Projectile vertical velocity	u sin θ
Time of flight	T = 2u sin θ/g
Maximum height	H = u² sin²θ/2g
Horizontal range	R = u² sin 2θ/g
Maximum range	Rmax = u²/g
Angular speed	ω = 2π/T
Linear speed	v = ωr
Centripetal acceleration	ac = v²/r = ω²r

How to Prepare Motion in a Plane for CBSE

1. Start with vectors. Do not jump directly to projectile numericals. Make sure you understand components and resultant vectors first.
2. Learn the x-y method. Whenever possible, separate a two-dimensional problem into horizontal and vertical components.
3. Remember the projectile conditions. The standard time-of-flight, maximum-height and range formulas assume the projectile lands at the same level from which it was projected.
4. Draw a small diagram. A simple diagram can show the direction of velocity, angle of projection, components and acceleration much more clearly.
5. Do not confuse speed with velocity. This becomes particularly important in circular motion.
6. Practise numerical problems. Formula memorisation alone is not enough for this chapter.

One-Minute Chapter Revision

Scalar: Magnitude only.

Vector: Magnitude + direction.

Vector components: A_x = A cos θ and A_y = A sin θ.

Relative velocity: v_AB = v_A − v_B.

Projectile: Horizontal and vertical motions are treated independently.

Time of flight: T = 2u sin θ/g.

Maximum height: H = u² sin²θ/2g.

Range: R = u² sin 2θ/g.

Maximum range: Occurs at 45° for the standard same-level projectile.

Circular motion: Speed may be constant but velocity changes.

Centripetal acceleration: a_c = v²/r.

Final Words

Motion in a Plane becomes much easier once you stop looking at it as a collection of formulas. Think of every two-dimensional motion as a combination of simpler motions along two perpendicular directions. That idea connects vectors, projectiles and circular motion.

For exam preparation, focus especially on vector components, resultant vectors, projectile motion formulas, relative velocity and centripetal acceleration. Practise enough numericals to become comfortable with signs, angles and units.

Understand the direction first, choose the formula second, and calculate last.

Note: These are independently written educational notes for learning and revision. Students should also refer to the latest CBSE syllabus and prescribed NCERT textbook for the complete course content.