Motion in a Plane Class 11 Notes PDF Download

Motion in a Plane Class 11 Notes help students understand two-dimensional motion using vectors, projectile motion, uniform circular motion, and relative velocity. This chapter is an important part of Class 11 Physics Syllabus because it connects basic one-dimensional motion with real-life motion in two dimensions.

In Motion in a Straight Line, an object moves only along one axis. In Motion in a Plane, an object can move along both the x-axis and y-axis. Examples include a football kicked in air, a stone thrown at an angle, a car moving around a circular road, rain falling while a person walks, and a boat crossing a river.

These notes are useful for CBSE Class 11 Physics, JEE Main, JEE Advanced, NEET, and school-level exam preparation.

Download Motion in a Plane Class 11 Notes PDF

Motion in a Plane is the study of motion in two dimensions where both horizontal and vertical coordinates may change with time.

This chapter mainly covers:

• Scalars and vectors
• Types of vectors
• Vector addition and subtraction
• Resolution of vectors
• Dot product and cross product
• Projectile motion
• Uniform circular motion
• Relative velocity in two dimensions
• Rain-man and river-boat problems
• Important formulas and solved examples

The chapter is formula-based, concept-based, and graphically important. Students should focus on understanding vector components and projectile motion formulas clearly.

Why Motion in a Plane Class 11 is Important

Motion in a Plane is important because most real-life motion happens in two or three dimensions, not only in a straight line.

This chapter helps students understand:

• How to represent physical quantities with direction
• How to split a vector into horizontal and vertical components
• How projectile motion is analysed
• Why the path of a projectile is a parabola
• How circular motion has acceleration even when speed is constant
• How relative velocity is used in rain and river problems

Concepts from this chapter are used later in Laws of Motion, Work Energy and Power, Rotational Motion, Gravitation, Oscillations, Waves, and Electromagnetism.

CBSE, JEE and NEET Preparation Note for Motion in a Plane

Motion in a Plane is important for both CBSE Class 11 Physics and competitive exams like IIT JEE and NEET. CBSE students should focus strongly on NCERT Solutions concepts such as scalars and vectors, vector addition, resolution of vectors, projectile motion, uniform circular motion, and basic derivations.

JEE and NEET aspirants should study the same topics in more depth and also practice advanced applications such as relative velocity in two dimensions, rain-man problems, river-boat problems, and multi-step projectile motion numericals.

Scalars and Vectors Class 11 Physics Notes

Scalars are physical quantities that have magnitude only, while vectors are physical quantities that have both magnitude and direction.

Scalars can be added using ordinary algebra. Vectors cannot always be added directly because their direction also matters.

Examples of Scalar Quantities

Examples of Vector Quantities

Difference Between Scalars and Vectors Class 11

Common Mistake

Students often think that a negative sign always means a vector. This is not true. A vector must have magnitude, direction, and must obey vector addition laws.

Types of Vectors in Motion in a Plane Class 11

Unit Vector Class 11

A unit vector is a vector whose magnitude is 1.

It is used to show direction only.

If a vector is A⃗, then the unit vector along A⃗ is:

â = A⃗ / |A⃗|

Where:

• A⃗ = vector
• |A⃗| = magnitude of vector A⃗
• â = unit vector along A⃗

The common unit vectors along coordinate axes are:

• î along x-axis
• ĵ along y-axis
• k̂ along z-axis

Zero Vector Class 11

A zero vector is a vector whose magnitude is zero and whose direction is not fixed.

Example:

If the initial and final positions of a body are the same, then displacement is zero.

Displacement = 0

So, the displacement vector is a zero vector.

Position Vector Class 11

Position vector gives the position of a point with respect to the origin.

If a point P has coordinates (x, y), then its position vector is:

r⃗ = xî + yĵ

In three dimensions:

r⃗ = xî + yĵ + zk̂

Displacement Vector Class 11

Displacement vector is the vector drawn from the initial position to the final position of an object.

If initial position vector is r⃗1 and final position vector is r⃗2, then:

Δr⃗ = r⃗2 - r⃗1

Where:

• Δr⃗ = displacement vector
• r⃗1 = initial position vector
• r⃗2 = final position vector

Vector Addition Class 11 Notes

Vector addition is the process of combining two or more vectors to get a single resultant vector.

Since vectors have direction, they cannot always be added like ordinary numbers. The direction of each vector must be considered.

If two vectors A⃗ and B⃗ are added, then the resultant vector is:

R⃗ = A⃗ + B⃗

Triangle Law of Vector Addition Class 11

Triangle law of vector addition states that if two vectors are represented by two sides of a triangle taken in the same order, then their resultant is represented by the third side of the triangle taken in the opposite order.

Explanation

If vector A⃗ is followed by vector B⃗, then the resultant vector R⃗ is drawn from the tail of A⃗ to the head of B⃗.

R⃗ = A⃗ + B⃗

Solved Example: Triangle Law of Vector Addition

Question: A student walks 3 m east and then 4 m north. Find the magnitude of displacement.

Step 1: Write the given values

Distance east = 3 m
Distance north = 4 m

These two directions are perpendicular to each other.

Step 2: Use Pythagoras theorem

R = √(32 + 42)

Step 3: Substitute values

R = √(9 + 16)

R = √25

R = 5 m

Final Answer: The displacement is 5 m.

Parallelogram Law of Vector Addition Class 11

Parallelogram law of vector addition states that if two vectors acting at a point are represented by two adjacent sides of a parallelogram, then the diagonal passing through the common point represents their resultant.

If two vectors A⃗ and B⃗ act at an angle θ, then the magnitude of the resultant vector is:

R = √(A2 + B2 + 2AB cos θ)

The direction angle α of the resultant with vector A⃗ is:

tan α = (B sin θ) / (A + B cos θ)

Special Cases of Vector Addition

Solved Example: Resultant of Two Vectors

Question: Two vectors of magnitudes 6 N and 8 N act at right angles. Find the resultant.

Step 1: Write the given values

A = 6 N
B = 8 N
θ = 90°

Step 2: Use resultant formula

R = √(A2 + B2 + 2AB cos θ)

Since cos 90° = 0,

R = √(A2 + B2)

Step 3: Substitute values

R = √(62 + 82)

R = √(36 + 64)

R = √100

R = 10 N

Final Answer: The resultant force is 10 N.

Resolution of a Vector Class 11 Notes

Resolution of a vector means splitting a vector into two or more components along chosen directions.

In Class 11 Physics, a vector is usually resolved into horizontal and vertical components.

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

Ax = A cos θ

Ay = A sin θ

The vector can be written as:

A⃗ = Axî + Ayĵ

Substituting the components:

A⃗ = A cos θ î + A sin θ ĵ

Where:

• Ax = horizontal component
• Ay = vertical component
• A = magnitude of vector
• θ = angle with x-axis

Common Mistake in Vector Resolution

Students often confuse sin θ and cos θ. The component adjacent to the angle uses cos θ, and the component opposite to the angle uses sin θ.

Solved Example: Resolution of a Vector

Question: A force of 10 N acts at an angle of 30° with the horizontal. Find its horizontal and vertical components.

Step 1: Write the given values

A = 10 N
θ = 30°

Step 2: Write component formulas

Ax = A cos θ

Ay = A sin θ

Step 3: Substitute values

Ax = 10 cos 30°

Since cos 30° = √3 / 2,

Ax = 10 × √3 / 2

Ax = 5√3 N

Now,

Ay = 10 sin 30°

Since sin 30° = 1 / 2,

Ay = 10 × 1 / 2

Ay = 5 N

Final Answer:
Horizontal component = 5√3 N
Vertical component = 5 N

Vector Multiplication Class 11 Notes

Vectors can be multiplied in two important ways:

1. Scalar product or dot product
2. Vector product or cross product

Scalar Product or Dot Product Class 11

The scalar product of two vectors gives a scalar quantity and is equal to the product of their magnitudes and the cosine of the angle between them.

Formula:

A⃗ · B⃗ = AB cos θ

In component form:

A⃗ · B⃗ = AxBx + AyBy + AzBz

Also Check: CBSE Class 11 Physics Notes

Important Cases of Dot Product

Application of Dot Product: Work Done

Work done by a force is:

W = F⃗ · s⃗

or

W = Fs cos θ

Where:

• W = work done
• F = force
• s = displacement
• θ = angle between force and displacement

Solved Example: Work Done Using Dot Product

Question: A force of 20 N moves a body through 5 m. The angle between force and displacement is 60°. Find the work done.

Step 1: Write the given values

F = 20 N
s = 5 m
θ = 60°

Step 2: Use work formula

W = Fs cos θ

Step 3: Substitute values

W = 20 × 5 × cos 60°

Since cos 60° = 1 / 2,

W = 20 × 5 × 1 / 2

W = 50 J

Final Answer: Work done = 50 J.

Vector Product or Cross Product Class 11

The vector product of two vectors gives a vector quantity and is equal to the product of their magnitudes and the sine of the angle between them.

Formula:

A⃗ × B⃗ = AB sin θ n̂

Here, n̂ is a unit vector perpendicular to the plane containing A⃗ and B⃗.

Important Cases of Cross Product

Application of Cross Product: Torque

Torque is given by:

τ⃗ = r⃗ × F⃗

Magnitude of torque:

τ = rF sin θ

Where:

• τ = torque
• r = position vector or lever arm
• F = force
• θ = angle between r⃗ and F⃗

Dot Product vs Cross Product Class 11

Projectile Motion Class 11 Notes

Projectile motion is the motion of an object thrown near Earth’s surface that moves under the influence of gravity only, neglecting air resistance.

The object thrown is called a projectile, and its path is called trajectory.

Examples of projectile motion:

• A football kicked at an angle
• A stone thrown from the ground
• A bullet fired from a gun
• Water coming out of a hose
• A ball thrown horizontally from a height

Types of Projectile Motion Class 11

Projectile motion is mainly of two types:

1. Horizontal projectile motion
2. Oblique projectile motion

Horizontal Projectile Motion

Horizontal projectile motion occurs when an object is projected horizontally from a height.

In this case:

• Initial vertical velocity is zero
• Horizontal velocity remains constant
• Vertical velocity changes due to gravity
• Path is curved

Oblique Projectile Motion

Oblique projectile motion occurs when an object is projected at an angle θ with the horizontal.

If a projectile is thrown with initial speed u at an angle θ, then the initial velocity is resolved into two components:

ux = u cos θ

uy = u sin θ

Where:

• u = initial velocity
• θ = angle of projection
• ux = horizontal component of velocity
• uy = vertical component of velocity

Projectile Motion Formulas Class 11

For a projectile projected with initial velocity u at an angle θ with the horizontal:

Why is the Trajectory of a Projectile a Parabola?

The trajectory of a projectile is a parabola because horizontal motion has constant velocity while vertical motion has constant acceleration due to gravity.

Step-by-Step Derivation

For horizontal motion:

x = u cos θ × t

So,

t = x / (u cos θ)

For vertical motion:

y = u sin θ × t - (1 / 2)gt2

Substitute the value of t:

y = u sin θ × [x / (u cos θ)] - (1 / 2)g[x / (u cos θ)]2

Simplify the first term:

u sin θ × [x / (u cos θ)] = x tan θ

Simplify the second term:

(1 / 2)g[x / (u cos θ)]2 = gx2 / (2u2 cos2θ)

Therefore:

y = x tan θ - gx2 / (2u2 cos2θ)

This equation is of the form:

y = ax - bx2

This is the equation of a parabola.

Final Conclusion: The path of a projectile is parabolic.

Time of Flight in Projectile Motion Class 11

Time of flight is the total time for which the projectile remains in air.

For a projectile projected from ground and landing at the same level:

T = (2u sin θ) / g

Derivation of Time of Flight

Vertical displacement after total flight is zero:

y = 0

Using vertical motion equation:

y = u sin θ × t - (1 / 2)gt2

Put y = 0:

0 = u sin θ × t - (1 / 2)gt2

Take t common:

t[u sin θ - (1 / 2)gt] = 0

Ignoring t = 0, we get:

u sin θ - (1 / 2)gt = 0

(1 / 2)gt = u sin θ

t = (2u sin θ) / g

Therefore:

T = (2u sin θ) / g

Maximum Height in Projectile Motion Class 11

Maximum height is the greatest vertical height reached by a projectile during its motion.

At maximum height, the vertical component of velocity becomes zero.

Formula:

H = (u2 sin2θ) / 2g

Derivation of Maximum Height

Initial vertical velocity:

uy = u sin θ

At maximum height:

vy = 0

Use:

vy2 = uy2 - 2gH

Substitute values:

0 = (u sin θ)2 - 2gH

2gH = u2 sin2θ

Therefore:

H = (u2 sin2θ) / 2g

Horizontal Range of Projectile Class 11

Horizontal range is the total horizontal distance covered by a projectile before it returns to the same vertical level.

Formula:

R = (u2 sin 2θ) / g

Derivation of Horizontal Range

Horizontal displacement is:

x = u cos θ × t

For complete time of flight:

T = (2u sin θ) / g

So range is:

R = u cos θ × T

Substitute T:

R = u cos θ × [(2u sin θ) / g]

R = (2u2 sin θ cos θ) / g

Using:

2 sin θ cos θ = sin 2θ

Therefore:

R = (u2 sin 2θ) / g

Angle of Projection for Maximum Range Class 11

For a given initial velocity, the horizontal range of a projectile is maximum when the angle of projection is 45°.

We know:

R = (u2 sin 2θ) / g

For maximum range, sin 2θ must be maximum.

Maximum value of sin 2θ is 1.

So:

sin 2θ = 1

This happens when:

2θ = 90°

Therefore:

θ = 45°

Maximum range is:

Rmax = u2 / g

Two angles θ and (90° - θ) give the same horizontal range if the initial velocity is the same.

Example:

30° and 60° give the same range.

Uniform Circular Motion Class 11 Notes

Uniform circular motion is the motion of an object along a circular path with constant speed.

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

Examples:

• Stone tied to a string and rotated
• Moon moving around Earth
• Car moving around a circular track
• Fan blade rotating
• Electron moving in a circular path in a magnetic field

Angular Displacement and Angular Velocity Class 11

Angular Displacement

Angular displacement is the angle swept by the radius vector of a particle moving in a circle.

It is usually represented by θ and measured in radians.

Angular Velocity

Angular velocity is the rate of change of angular displacement.

Formula:

ω = dθ / dt

For uniform circular motion:

ω = θ / t

Relation between linear speed and angular speed:

v = rω

Where:

• v = linear speed
• r = radius of circular path
• ω = angular velocity

Centripetal Acceleration Class 11

Centripetal acceleration is the acceleration of a body moving in a circular path and is always directed toward the centre of the circle.

Formula:

ac = v2 / r

Using:

v = rω

We get:

ac = (rω)2 / r

ac = r2ω2 / r

Therefore:

ac = rω2

So:

ac = v2 / r = rω2

What is the Angle Between Velocity and Acceleration in Uniform Circular Motion?

In uniform circular motion, velocity is tangential to the circle and centripetal acceleration is directed toward the centre. Therefore, the angle between velocity and acceleration is 90°.

Relative Velocity in 2D Class 11 Notes

Relative velocity in two dimensions is the velocity of one object as observed from another moving object, using vector subtraction.

If body A has velocity v⃗A and body B has velocity v⃗B, then velocity of A with respect to B is:

v⃗AB = v⃗A - v⃗B

Similarly:

v⃗BA = v⃗B - v⃗A

Therefore:

v⃗BA = -v⃗AB

This topic is very useful in JEE and NEET problems involving rain, river, boats, swimmers, aircraft, and moving observers.

Rain Umbrella Problem Class 11 Physics

In rain-man problems, the apparent direction of rain depends on the relative velocity of rain with respect to the moving person.

If rain falls vertically with speed vr and a man moves horizontally with speed vm, then the relative velocity of rain with respect to the man is:

v⃗rm = v⃗r - v⃗m

The umbrella must be held in the direction of the relative velocity of rain.

If θ is the angle with the vertical, then:

tan θ = vm / vr

Solved Example: Rain Umbrella Problem

Question: Rain is falling vertically with speed 10 m s-1. A man walks horizontally with speed 5 m s-1. At what angle with the vertical should he hold the umbrella?

Step 1: Write the given values

vr = 10 m s-1
vm = 5 m s-1

Step 2: Use formula

tan θ = vm / vr

Step 3: Substitute values

tan θ = 5 / 10

tan θ = 1 / 2

Step 4: Find angle

θ = tan-1(1 / 2)

θ ≈ 26.6°

Final Answer: The umbrella should be held at about 26.6° with the vertical.

River Boat Problems Relative Velocity Class 11

River-boat problems are based on relative velocity. A boat has velocity relative to water, while the river has its own flow velocity.

Let:

• vbr = velocity of boat relative to river
• vr = velocity of river
• d = width of river

Shortest Time to Cross River Class 11

For shortest time, the boat should be aimed perpendicular to the river flow.

Time taken:

t = d / vbr

Drift:

x = vrt

Substitute t:

x = vr(d / vbr)

So:

x = vrd / vbr

Shortest Path or Zero Drift in River Boat Problems

For shortest path or zero drift, the boat must be aimed upstream so that the river drift is cancelled.

Condition:

sin θ = vr / vbr

Resultant speed across the river:

v = √(vbr2 - vr2)

Time taken:

t = d / √(vbr2 - vr2)

Solved Example: Shortest Time to Cross River

Question: A river is 100 m wide. A boat can move with speed 5 m s-1 relative to water. The river flows at 3 m s-1. Find the shortest time to cross the river and the drift.

Step 1: Write the given values

d = 100 m
vbr = 5 m s-1
vr = 3 m s-1

Step 2: Use shortest time formula

t = d / vbr

Step 3: Substitute values

t = 100 / 5

t = 20 s

Step 4: Calculate drift

x = vrt

x = 3 × 20

x = 60 m

Final Answer:
Shortest time = 20 s
Drift = 60 m

Motion in a Plane Class 11 Formula Sheet

Motion in a Plane Class 11 Solved Numericals

Solved Example 1: Resultant of Two Perpendicular Vectors

Question: Two vectors of magnitudes 3 N and 4 N act at right angles to each other. Find the resultant.

Step 1: Write the given values

A = 3 N
B = 4 N
θ = 90°

Step 2: Use formula

R = √(A2 + B2 + 2AB cos θ)

Since cos 90° = 0,

R = √(A2 + B2)

Step 3: Substitute values

R = √(32 + 42)

R = √(9 + 16)

R = √25

R = 5 N

Final Answer: Resultant = 5 N

Solved Example 2: Projectile Maximum Height

Question: A ball is projected with speed 20 m s-1 at an angle of 30° with the horizontal. Find the maximum height. Take g = 10 m s-2.

Step 1: Write the given values

u = 20 m s-1
θ = 30°
g = 10 m s-2

Step 2: Use formula

H = (u2 sin2θ) / 2g

Step 3: Substitute values

H = [202 × sin230°] / (2 × 10)

Since sin 30° = 1 / 2,

sin230° = (1 / 2)2 = 1 / 4

So,

H = [400 × 1 / 4] / 20

H = 100 / 20

H = 5 m

Final Answer: Maximum height = 5 m

Solved Example 3: Horizontal Range of Projectile

Question: A projectile is fired with speed 20 m s-1 at an angle of 45°. Find its horizontal range. Take g = 10 m s-2.

Step 1: Write the given values

u = 20 m s-1
θ = 45°
g = 10 m s-2

Step 2: Use formula

R = (u2 sin 2θ) / g

Step 3: Substitute values

R = [202 × sin(2 × 45°)] / 10

R = [400 × sin 90°] / 10

Since sin 90° = 1,

R = 400 / 10

R = 40 m

Final Answer: Horizontal range = 40 m

Solved Example 4: Centripetal Acceleration

Question: A body moves in a circle of radius 2 m with speed 10 m s-1. Find centripetal acceleration.

Step 1: Write the given values

v = 10 m s-1
r = 2 m

Step 2: Use formula

ac = v2 / r

Step 3: Substitute values

ac = 102 / 2

ac = 100 / 2

ac = 50 m s-2

Final Answer: Centripetal acceleration = 50 m s-2

Solved Example 5: River Boat Shortest Path

Question: A river is 80 m wide. A boat can move at 10 m s-1 in still water. The river flows at 6 m s-1. Find the resultant speed for shortest path.

Step 1: Write the given values

vbr = 10 m s-1
vr = 6 m s-1

Step 2: Use formula for shortest path

v = √(vbr2 - vr2)

Step 3: Substitute values

v = √(102 - 62)

v = √(100 - 36)

v = √64

v = 8 m s-1

Final Answer: Resultant speed across the river = 8 m s-1

Common Mistakes in Motion in a Plane Class 11

Motion in a Plane Class 11 Important Questions

1-Mark Questions

1. Define scalar quantity.
2. Define vector quantity.
3. What is a unit vector?
4. What is a zero vector?
5. Write the formula for dot product.
6. Write the formula for cross product.
7. What is projectile motion?
8. What is time of flight?
9. Define centripetal acceleration.
10. What is relative velocity?

2-Mark Questions

1. Differentiate between scalars and vectors.
2. State triangle law of vector addition.
3. State parallelogram law of vector addition.
4. Write the components of a vector A⃗ making angle θ with x-axis.
5. Why is horizontal acceleration zero in projectile motion?
6. What is the angle between velocity and acceleration in uniform circular motion?

3-Mark Questions

1. Derive the expression for resultant of two vectors using parallelogram law.
2. Explain dot product and cross product with examples.
3. Derive the equation of trajectory of a projectile.
4. Derive the expression for time of flight.
5. Derive the expression for maximum height.
6. Derive the expression for horizontal range.

5-Mark Questions

1. Explain projectile motion and derive time of flight, maximum height, and range.
2. Explain vector addition using triangle law and parallelogram law.
3. Explain uniform circular motion and derive centripetal acceleration formula.
4. Explain relative velocity in two dimensions with rain-man and river-boat examples.
5. Compare scalar product and vector product with formulas and applications.