MOTION IN A PLANE

This article is formulated according to the 4th chapter in the NCERT Syllabus of 11th Std Physics OR 1st PUC Physics.

When a body moves in a plane (a two-dimensional motion) or in a space (a three-dimensional motion) then the position, displacement, velocity and acceleration of the body have two or three components respectively. Then we need to use Vectors to describe the concept of position, displacement, velocity and acceleration.

Scalar quantity

A physical quantity having only magnitude is called a scalar quantity. It is specified completely by a single number along with proper unit. Ex: mass, length, temperature, speed, charge, area etc. Scalars can be added, subtracted, multiplied and divided just as the ordinary numbers. They follow the rules of algebra.

Vector quantity

A Physical quantity having both magnitude and direction and obey the triangle law of addition is called Vector quantity. It is represented by a number with an appropriate unit and direction. Ex: Displacement, velocity, acceleration, force, momentum etc.

Differences between Scalar quantity and Vector quantity

Scalar Quantity	Vector Quantity
It has only magnitude	It has both magnitude and direction
They follow the rules of ordinary algebra	They follow the rules of vector algebra
These change when magnitude changes	These changes when magnitude changes or direction changes or both of them changes.
Ex: Mass, Length, Temperature, Area	Ex: Displacement, velocity, Acceleration, Force

Representation of a vector

To represent a vector we use a bold face letters or an arrow placed over a letter.

Ex:

Here O is called the initial point and P is called the terminal point. The length of the line segment OP represents the magnitude and the arrow at the endpoint indicates the direction. The magnitude of a vector is often called the absolute value and indicated by,

Classification of vectors

Parallel vectors

Two or more vectors having same direction are called parallel vectors.

Anti-parallel vectors (opposite vectors)

Vectors having opposite directions are called anti-parallel vectors (opposite vectors).

Equality of vector (Equal vectors)

Two (or more) vectors having same magnitude and direction, representing the same physical quantity are called Equal vectors.

Negative of a vector

A vector having same magnitude but having opposite direction to that of the given vector is called negative of a given vector.

Zero(Null) vector

A vector whose magnitude is zero is called Zero vector. It is represented by 0⃗⃗ and the direction is not specified.

Properties of Zero vector are, 𝐴⃗ − 𝐴⃗ = 0⃗⃗

|0⃗⃗| = 0

𝐴⃗ − 0⃗⃗ = 𝐴⃗

0⃗⃗ 𝐴⃗ = 𝐴⃗ 0⃗⃗ = 0⃗⃗

𝜆 0⃗⃗ = 0⃗⃗

Unit vector

A vector having unit magnitude is called unit vector. Its purpose is to specify a direction. Unit vector has no dimensions and unit. If 𝑎⃗ is a vector, then the unit vector in direction of 𝑎⃗ is written as 𝑎̂ (read as “a cap”) 𝑎⃗ = |𝑎⃗| 𝑎̂, Then, mathematically unit vector can be represented as, 𝑎̂ = 𝑎⃗ |𝑎⃗|

Note: The unit vectors in the positive directions of x, y and z axes are labeled as 𝑖,̂ 𝑗̂ 𝑎𝑛𝑑 𝑘̂ respectively.

Addition of vectors – Graphical method

Two vectors representing the same quantity in the same unit are added using the following rules.

(i) Triangle method of vector addition: Law of the triangle of vectors or Triangular law of vector addition:

If two vectors 𝑎⃗ and 𝑏⃗⃗ are represented by two sides of a triangle in head-to-tail form, then the closing side of the triangle taken from the tail of the first to head of the second represents the vector sum of 𝑎⃗ and 𝑏⃗.

Explanation: Consider two vectors, 𝑎⃗ = 𝐴𝐵⃗ and 𝑏⃗⃗ = 𝐵𝐶⃗ are of same nature. The triangle ABC is completed by joining A and C. According to the triangle law of addition, 𝐴𝐶⃗ = 𝑟⃗ represents the sum of 𝑎⃗ and 𝑏⃗. 𝑎⃗ + 𝑏⃗ = 𝑟⃗ or 𝐴𝐵⃗ + 𝐵𝐶⃗ = 𝐴𝐶⃗

Note: In this procedure of vector addition, vectors are arranged head-to-tail. Hence it is called the head-to-tail method.

Properties vector addition

(a) Vector addition is commutative.

(b) Vector addition is Associative.

𝑎⃗ + 𝑏⃗ = 𝑟⃗ = 𝑏⃗ + 𝑎⃗ (𝑎⃗ + 𝑏) + 𝑐⃗ = 𝑎⃗ + (𝑏⃗ + 𝑐⃗)

(ii) Parallelogram method of vector addition

Law of parallelogram of vectors or Parallelogram law of vector addition

If two vectors are represented by two adjacent sides of a parallelogram, then the diagonal drawn from the common initial point represents their vector sum.

Explanation: Vector 𝑎⃗ and 𝑏⃗ are drawn with a common initial point and parallelogram is constructed using these two vectors as two adjacent sides of a parallelogram. The diagonal originating from the common initial point is vector sum of 𝑎⃗ and 𝑏⃗.

Subtraction vectors – Graphical method

Subtraction of vectors can be defined in terms of addition of vectors. Consider two vectors 𝑎⃗ and 𝑏⃗ of same nature and another vector −𝑏⃗ which is opposite (negative) vector of 𝑏⃗ , then 𝑎⃗ + ( −𝑏⃗) = 𝑎⃗ + 𝑏⃗

Note: Subtraction vector is neither commutative nor associative.

Multiplication of a vector by real (Scalar) number OR Scalar multiplication of a Vector

The product of a vector 𝑣⃗ and a positive number (Scalar) 𝜆 gives a vector, whose magnitude is changed by a factor 𝜆 but direction is same as that of 𝑣⃗. |𝜆𝑣⃗| = 𝜆|𝑣⃗ | (𝑖𝑓 𝜆 > 0) If 𝜆 is negative, the direction of the vector 𝜆𝑣⃗ is opposite to the direction of the vector 𝑣⃗ and magnitude is – 𝜆 times |𝑣⃗ |. If the multiplying factor 𝜆 is dimensionless then 𝜆𝑣⃗ have the same dimensions as that of 𝑣⃗ and is product of dimensions if 𝜆 has dimensions.

Resolution of vectors

Splitting a given vector into a number of components is called resolution of vectors OR The process of finding the components of a given vector is called resolution the vector.

Expressions for X and Y components of a Vector

Consider a vector 𝑂𝐴⃗ = 𝑎⃗ in X-Y plane, which makes an angle 𝜃 with the positive X-axis. Draw AM and AN perpendicular to X and Y axes respectively. Let 𝑂𝑀⃗ = 𝑎⃗_𝑥 and 𝑂𝑁⃗ = 𝑎⃗_𝑦. From parallelogram law of addition, we have 𝑂𝑀⃗ + 𝑂𝑁⃗ = 𝑂𝐴⃗⃗

𝑎⃗_𝑥 + 𝑎⃗_𝑦 = 𝑎⃗ (Here 𝑎⃗_𝑥 is x – component of 𝑎⃗ and 𝑎⃗_𝑦 is y – component of 𝑎⃗) From ∆𝑂𝐴𝑀, 𝑐𝑜𝑠 𝜃 = 𝑂𝑀/𝑂𝐴 = 𝑎𝑥/𝑎

𝒂_𝒙 = 𝒂 𝐜𝐨𝐬 𝜽

and 𝑠𝑖𝑛 𝜃 = 𝐴𝑀/𝑂𝐴 = 𝑎_𝑦/𝑎

𝒂_𝒚 = 𝒂 𝐬𝐢𝐧 𝜽

Vectoricaly 𝒂⃗_𝒙 = 𝒂⃗ 𝐜𝐨𝐬 𝜽 and 𝒂⃗_𝒚 = 𝒂⃗𝐬𝐢𝐧 𝜽

Note: 𝑎⃗_𝑥 and 𝑎⃗_𝑦 being perpendicular are called rectangular components of 𝑎⃗.

Magnitude

Magnitude of 𝑎⃗ is given by |𝑎⃗| = 𝑎

Now, 𝑎_𝑥² = 𝑎²cos²𝜃 and 𝑎_𝑦² = 𝑎²sin²𝜃

Taking 𝑎_𝑥² + 𝑎_𝑦² = 𝑎²cos²𝜃 + 𝑎²sin²𝜃

𝑎_𝑥² + 𝑎_𝑦² = 𝑎²(cos²𝜃 + sin²𝜃)

𝑎_𝑥² + 𝑎_𝑦² = 𝑎²

𝒂 = √(𝒂_𝒙^𝟐 + 𝒂_𝒚^𝟐)

𝐃𝐢𝐫𝐞𝐜𝐭𝐢𝐨𝐧

By taking 𝑎_𝑦/𝑎_𝑥 = 𝑎sin𝜃/𝑎cos𝜃

𝑎_𝑦/𝑎_𝑥 = tan𝜃

𝜽 = 𝐭𝐚𝐧^−𝟏(𝒂_𝒚/𝒂_𝒙)

Note: (i) In terms of unit vectors, 𝑎⃗ = 𝑎_𝑥𝑖̂+ 𝑎_𝑦𝑗̂= 𝑎cos𝜃𝑖̂ + sin𝜃𝑗̂

where 𝑎⃗_𝑥 = 𝑎 cos 𝜃 𝑖̂

and 𝑎⃗_𝑦 = 𝑎 sin 𝜃 𝑗̂

(ii) If 𝑎⃗ is in XYZ plane and makes an angle 𝛼, 𝛽 and 𝛾 with X, Y and Z axes respectively, then 𝑎_𝑥 = 𝑎 cos 𝛼, 𝑎_𝑦 = 𝑎 cos𝛽, 𝑎_𝑧 = 𝑎 cos 𝛾

and 𝑎⃗ = 𝑎_𝑥𝑖̂+ 𝑎_𝑦𝑗̂+ 𝑎_𝑧𝑘̂

The magnitude of 𝑎⃗ is, 𝑎 = √(𝑎_𝑥² + 𝑎_𝑦² + 𝑎_𝑧²).

Find the magnitude and direction of the resultant of two vectors 𝑨⃗ and 𝑩⃗ in terms of their magnitudes and angle 𝜽 between them.

Let 𝑂𝑃⃗ and 𝑂𝑄⃗ represent the two vectors 𝐴⃗ and 𝐵⃗ making an angle 𝜃. Then using the parallelogram method of vector addition 𝑂𝑆 ⃗ represents the resultant vector 𝑅⃗.

𝑅⃗ = 𝐴⃗ + 𝐵⃗

Draw SN is normal to OP extended. In ∆𝑆𝑃𝑁, 𝑐𝑜𝑠 𝜃 = 𝑃𝑁/𝑆𝑃

𝑃𝑁 = 𝑆𝑃 cos 𝜃 = 𝐵 cos 𝜃 and 𝑠𝑖𝑛 𝜃 = 𝑆𝑁/𝑆𝑃

𝑆𝑁 = 𝑆𝑃 sin 𝜃 = 𝐵 sin 𝜃

Magnitude

From geometry, 𝑂𝑆² = 𝑂𝑁² + 𝑆𝑁²

𝑂𝑆² = (𝑂𝑃 + 𝑃𝑁)² + 𝑆𝑁²

𝑅² = (𝐴 + 𝐵 cos 𝜃)² + (𝐵 sin 𝜃)²

𝑅² = 𝐴² + (𝐵 cos 𝜃)² + 2𝐴𝐵 cos 𝜃 + 𝐵² sin²𝜃

𝑅² = 𝐴² + 𝐵² cos²𝜃 + 2𝐴𝐵 cos 𝜃 + 𝐵² sin² 𝜃

𝑅² = 𝐴² + 𝐵² (cos²𝜃 + sin² 𝜃) + 2𝐴𝐵cos𝜃

𝑅² = 𝐴² + 𝐵² + 2𝐴𝐵cos 𝜃

𝑹 = √(𝑨^𝟐 + 𝑩^𝟐 + 𝟐𝑨𝑩 𝐜𝐨𝐬𝜽)

Direction: Let 𝛼 be angle made by the resultant vector 𝑅⃗ with the vector 𝐴⃗, then tan 𝛼 = 𝑆𝑁/𝑂𝑁 = 𝑆𝑁/(𝑂𝑃 + 𝑃𝑁)

tan 𝛼 = 𝐵sin𝜃/(𝐴 + 𝐵cos𝜃)

𝜶 = 𝐭𝐚𝐧^−𝟏(𝑩𝐬𝐢𝐧𝜽/(𝑨 + 𝑩𝐜𝐨𝐬𝜽)

Limitations of Graphical method of adding vectors

(i) It is very difficult method.

(ii) It has limited accuracy.

To overcome these limitations Analytical method of addition of vectors is preferred.

Addition of vectors – Analytical method

In two Dimensions

Consider two vectors 𝑎⃗ and 𝑏⃗ in X-Y plane. If 𝑎⃗ = 𝑎_𝑥𝑖̂+ 𝑎_𝑦𝑗 ̂ and 𝑏⃗ = 𝑏_𝑥𝑖̂+ 𝑏_𝑦𝑗 ̂ then, 𝑅⃗ = 𝑎⃗ + 𝑏⃗

𝑅⃗ = (𝑎_𝑥𝑖̂+ 𝑎_𝑦𝑗̂) + (𝑏_𝑥𝑖̂+ 𝑏_𝑦𝑗̂)

𝑅⃗ = 𝑎_𝑥𝑖̂+ 𝑏_𝑥𝑖̂+ 𝑎_𝑦𝑗̂+ 𝑏_𝑦𝑗 ̂

𝑅⃗ = (𝑎_𝑥 + 𝑏_𝑥 )𝑖̂+ (𝑎_𝑦 + 𝑏_𝑦)𝑗̂

𝑹⃗ = 𝑹_𝒙𝒊̂+ 𝑹_𝒚𝒋 ̂

Where 𝑅_𝑥 = 𝑎_𝑥 + 𝑏_𝑥 and 𝑅_𝑦 = 𝑎_𝑦 + 𝑏_𝑦

In three Dimensions

If 𝑎⃗ = 𝑎_𝑥𝑖̂+ 𝑎_𝑦𝑗̂+ 𝑎_𝑧𝑘̂ and 𝑏⃗ = 𝑏_𝑥𝑖̂+ 𝑏_𝑦𝑗̂+ 𝑏_𝑧𝑘̂

then, 𝑅⃗ = 𝑎⃗ + 𝑏⃗

𝑅⃗ = (𝑎_𝑥𝑖̂+ 𝑎_𝑦𝑗̂+ 𝑎_𝑧𝑘̂) + (𝑏_𝑥𝑖̂+ 𝑏_𝑦𝑗̂+ 𝑏_𝑧𝑘̂)

𝑅⃗ = 𝑎_𝑥𝑖̂ + 𝑏_𝑥𝑖̂ + 𝑎_𝑦𝑗̂ + 𝑏_𝑦𝑗̂ + 𝑎_𝑧𝑘̂ + 𝑏_𝑧𝑘̂

𝑅⃗ = (𝑎_𝑥 + 𝑏_𝑥) 𝑖̂+ (𝑎_𝑦 + 𝑏_𝑦) 𝑗̂+ (𝑎_𝑧 + 𝑏_𝑧) 𝑘̂

𝑹⃗ = 𝑹_𝒙𝒊̂+ 𝑹_𝒚𝒋̂+ 𝑹_𝒛𝒌̂

This method can be extended to addition and subtraction of any number of vectors.

Motion in a plane

Position vector

The position vector 𝑟⃗ of a particle located in X-Y plane with reference to the origin is given by, 𝑟⃗ = 𝑥𝑖̂+ 𝑦𝑗̂

Where 𝑥 and 𝑦 are component of 𝑟⃗ along X-axis and Y-axis respectively.

Displacement

Consider a particle moves along curve. Initially it is at 𝑃₁ at time 𝑡₁ and moves to a new position 𝑃₂ at time 𝑡₂. Then the displacement is given by, ∆𝑟⃗ = 𝑟⃗₂ − 𝑟⃗₁

∆𝑟⃗ = (𝑥₂𝑖̂+ 𝑦₂𝑗̂) − (𝑥₁𝑖̂+ 𝑦₁𝑗̂)

∆𝑟⃗ = (𝑥₂𝑖̂−𝑥₁𝑖)̂ + (𝑦_2𝑗̂− 𝑦₁𝑗̂)

∆𝑟⃗ = (𝑥₂−𝑥₁) 𝑖̂ + (𝑦₂ − 𝑦₁) 𝑗̂

∆𝒓⃗ = ∆𝒙𝒊̂+ ∆𝒚𝒋̂

Where ∆𝑥 = 𝑥₂ − 𝑥₁ and ∆𝑦 = 𝑦₂ − 𝑦₁

Velocity

Average velocity

It is defined as ratio of the displacement to the time taken.

𝑣⃗̅= ∆𝑟⃗/∆𝑡 = (∆𝑥𝑖̂+ ∆𝑦𝑗̂)/∆𝑡

𝑣⃗̅= (∆𝑥/∆𝑡) 𝑖̂+ (∆𝑦/∆𝑡) 𝑗̂

𝒗⃗̅ = 𝒗̅_𝒙𝒊̂+ 𝒗̅_𝒚𝒋̂

Where 𝒗̅_𝒙 = ∆𝑥/∆𝑡 and 𝒗̅_𝒚 = ∆𝑦/∆𝑡

Direction of the average velocity is same as that of the displacement.

Instantaneous velocity (Velocity)

It is given by the limiting value of the average velocity as the time interval approaches to zero.

𝑣⃗ = lim _∆𝑡→0 ∆𝑟⃗/∆𝑡

𝒗⃗ = 𝒅𝒓⃗/𝒅𝒕

The direction of velocity at any point on the path of the object is tangential to the path at that point and in the direction of the motion. The components of the velocity 𝑣⃗ are given by, 𝑣⃗ = lim _∆𝑡→0 ∆𝑟⃗/∆𝑡

𝑣⃗ = lim _∆𝑡→0 (∆𝑥/∆𝑡) 𝑖̂+ (∆𝑦/∆𝑡) 𝑗̂

𝑣⃗ = lim _∆𝑡→0 (∆𝑥/∆𝑡) 𝑖̂ + lim _∆𝑡→0 (∆𝑦/∆𝑡) 𝑗̂

𝑣⃗ = (𝑑𝑥/𝑑𝑡) 𝑖̂+ (𝑑𝑦/𝑑𝑡) 𝑗̂

𝒗⃗ = 𝒗_𝒙𝒊̂+ 𝒗_𝒚𝒋̂

The magnitude is given by 𝒗 = √(𝒗_𝒙^𝟐 + 𝒗_𝒚^𝟐) and Direction is given by

𝜽 = 𝐭𝐚𝐧^−𝟏(𝒗_𝒚/𝒗_𝒙)

Acceleration

Average acceleration

It is defined as the change in velocity divided by time interval.

𝑎̅⃗ = ∆𝑣⃗/∆𝑡 = ∆(𝑣_𝑥𝑖̂+𝑣_𝑦𝑗̂)/∆𝑡

𝑎̅⃗ = (∆𝑣_𝑥/∆𝑡) 𝑖̂ + (∆𝑣_𝑦/∆𝑡) 𝑗̂

𝒂⃗̅⃗ = 𝒂̅_𝒙𝒊̂+ 𝒂̅_𝒚𝒋̂

Instantaneous acceleration (Acceleration)

It is the limiting value of the average acceleration as the time interval approaches zero.

𝑎⃗ = lim_∆𝑡→0 (∆𝑣⃗/∆𝑡)

𝒂⃗ = 𝒅𝒗⃗/𝒅𝒕

The Components are given by, 𝑎⃗ = lim _∆𝑡→0 (∆𝑣⃗/∆𝑡)

𝑎⃗ = lim _∆𝑡→0 ∆(𝑣_𝑥𝑖̂+ 𝑣_𝑦𝑗̂)/∆𝑡

𝑎⃗ = lim _∆𝑡→0 (∆𝑣_𝑥/∆𝑡) 𝑖̂+ lim _∆𝑡→0 (∆𝑣_𝑦/∆𝑡) 𝑗̂

𝑎⃗ = (𝑑𝑣_𝑥/𝑑𝑡) 𝑖̂+ (𝑑𝑣_𝑦/𝑑𝑡) 𝑗̂

𝒂⃗ = 𝒂_𝒙𝒊̂+ 𝒂_𝒚𝒋̂

In one dimension the direction of velocity and acceleration is same or in opposite direction but in two or three dimensions, velocity and acceleration vectors may have any angle between 0° and 180°.

Motion in a plane with constant acceleration

Consider an object moving in X-Y plane and its acceleration 𝑎⃗ is constant. Let the velocity of the object be 𝑣⃗₀ at time 𝑡 = 0 and 𝑣⃗ at time 𝑡, then

(i) 𝑣⃗ = 𝑣⃗₀ + 𝑎⃗𝑡

In terms of its components,

𝑣_𝑥 = 𝑣_0𝑥 + 𝑎_𝑥𝑡

𝑣_𝑦 = 𝑣_0𝑦 + 𝑎_𝑦𝑡

(ii) Displacement is 𝑟⃗ − 𝑟⃗₀ = 𝑣⃗₀𝑡 + (1/2)𝑎⃗𝑡²

In terms of its components,

𝑥 − 𝑥₀ = 𝑣_0𝑥𝑡 + (1/2)𝑎_𝑥𝑡²

𝑦 − 𝑦₀ = 𝑣_0𝑦𝑡 + (1/2)𝑎_𝑦𝑡²

(iii) 𝑣 ⃗² = 𝑣₀ ⃗² + 2𝑎⃗(𝑟⃗ − 𝑟⃗₀)

In terms of its components,

𝑣_𝑥² = 𝑣_0𝑥² + 2𝑎𝑥(𝑥 − 𝑥₀)

𝑣_𝑦² = 𝑣_0𝑦² + 2𝑎𝑦(𝑦 − 𝑦₀)

The motion in plane can be treated as two separate simultaneous one-dimensional motions with constant acceleration along two perpendicular directions.

Relative velocity in two Dimensions

Suppose two objects 𝐴 and 𝐵 are moving with velocities 𝑣⃗_𝐴 and 𝑣⃗_𝐵, then the velocity of the object 𝐴 relative to that of 𝐵 is, 𝑣⃗_𝐴𝐵 = 𝑣⃗_𝐴 − 𝑣⃗_𝐵.

Similarly, the velocity of the object 𝐵 relative to that of 𝐴 is, 𝑣⃗_𝐵𝐴 = 𝑣⃗_𝐵 − 𝑣⃗_𝐴 Therefore, 𝑣⃗_𝐴𝐵 = −𝑣⃗_𝐵𝐴 and |𝑣⃗_𝐴𝐵| = |𝑣⃗_𝐵𝐴|

Examples for Motion in a plane

(i) Projectile motion (Uniformly accelerated motion)

(ii) Circular motion (non-uniformly accelerated motion)

When a particle traces a curve in two-dimensional plane, the velocity of the particle changes at least in direction. Hence, a two-dimensional motion along a curve is essentially an accelerated motion. Acceleration may be uniform or non-uniform.

Projectile

A projectile is any object thrown into air or space.

Projectile motion

Motion associated with a projectile in parabolic path is called Projectile motion. Ex: A ball leaving the hand of a bowler, A stone thrown at an angle to the horizontal, an object dropped from an aeroplane in horizontal flight. The motion of projectile may be thought of as the result of two separate, simultaneously occurring components of motion. One component is along a horizontal direction with any acceleration and other along the vertical direction with constant acceleration due to gravity. It was Galileo, who first stated this independency of the horizontal and vertical components of projectile motion.

Analysis of Projectile motion

Let a projectile is projected with initial velocity 𝑣⃗₀ that makes an angle 𝜃 with x-axis. The acceleration acting on it is due to gravity and is directed vertically downwards. 𝑎_𝑥 = 0, 𝑎_𝑦 = −𝑔, hence 𝑎⃗ = −𝑔𝑗̂.

The components of initial velocity 𝑣⃗₀ are, 𝑣_0𝑥 = 𝑣₀cos𝜃, 𝑣_0𝑦 = 𝑣₀sin𝜃

The components of velocity at time 𝑡 are, 𝑣_𝑥 = 𝑣₀ cos𝜃, 𝑣_𝑦 = 𝑣₀ sin 𝜃 – 𝑔𝑡.

The components of displacements at time 𝑡 are,

(i) 𝑥 = 𝑣_0𝑥𝑡 + (1/2)𝑎_𝑥𝑡²

𝑥 = 𝑣_0𝑥𝑡 (∵ 𝑎_𝑥 = 0), 𝒙 = (𝒗_𝟎 𝐜𝐨𝐬 𝜽) 𝒕 (along X-axis)

(ii) 𝑦 = 𝑣_0𝑦𝑡 + (1/2)𝑎_𝑦𝑡², 𝒚 = (𝒗_𝟎 𝐬𝐢𝐧𝜽)𝒕 – (𝟏/𝟐)𝒈𝒕^𝟐 (along 𝑌 − axis)

Path of a projectile

The path described by the projectile is called trajectory. The trajectory is a parabola.

Expression for path of a projectile (Show that the path of a projectile is Parabola)

The displacement of the projectile along X-axis is, 𝑥 = (𝑣₀ cos 𝜃)𝑡, 𝑡 = 𝑥/𝑣₀cos 𝜃 The displacement of the projectile along Y-axis is,

𝑦 = (𝑣₀ sin 𝜃)𝑡 – (1/2)𝑔𝑡²

𝑦 = (𝑣₀ sin 𝜃) (𝑥/𝑣₀ cos 𝜃) – (1/2)𝑔(𝑥/𝑣₀cos 𝜃)²

𝑦 = (sin 𝜃/cos 𝜃)𝑥 – (1/2)(𝑔/𝑣₀² cos²𝜃)𝑥²

𝑦 = 𝑥 tan 𝜃 – (1/2)(𝑔/𝑣₀²cos²𝜃)𝑥²

𝒚 = 𝒂_𝒙 − 𝒃_𝒙^𝟐

where 𝑎 = 𝑡𝑎𝑛𝜃 and 𝑏 = (1/2)𝑔𝑣₀²𝑐𝑜𝑠²𝜃

The equation 𝑦 = 𝑎𝑥 − 𝑏𝑥² represents a parabola.

Hence the trajectory is a parabola.

Time of Flight

It is the time during which the projectile is in flight. It is denoted by 𝑇_𝑓.

Expression for Time of flight

The component of velocity along Y-axis at time t is, 𝑣_𝑦 = 𝑣₀ sin𝜃 − 𝑔𝑡

At maximum height 𝑣_𝑦 = 0 and time for maximum height, 𝑡 = 𝑡_𝑚.

0 = 𝑣₀ sin 𝜃 − 𝑔𝑡_𝑚

𝑔𝑡_𝑚 = 𝑣₀ sin𝜃

𝑡_𝑚 = 𝑣₀ sin𝜃/𝑔

Time of flight 𝑇_𝑓 = 2𝑡_𝑚 because “time of ascent = time of descent”

𝑻_𝒇 = 𝟐𝒗_𝟎𝐬𝐢𝐧𝜽/𝒈

Maximum height of a projectile

It is the maximum height reached by the projectile in time 𝑡_𝑚. It is denoted by ℎ_𝑚.

Expression for maximum height of a projectile

The displacement along Y − axis is, 𝑦 = (𝑣₀ sin𝜃)𝑡_𝑚 – (1/2)𝑔𝑡_𝑚²

ℎ_𝑚 = (𝑣₀ sin 𝜃)(𝑣₀ sin 𝜃/𝑔) – (1/2)𝑔(𝑣₀sin 𝜃/𝑔)²

ℎ_𝑚 = (𝑣₀ sin 𝜃)²/𝑔 – (1/2)𝑔((𝑣₀ sin𝜃)²/𝑔²)

ℎ_𝑚 = (𝑣₀ sin 𝜃)²/𝑔 – (1/2)(𝑣₀ sin𝜃)²𝑔

ℎ_𝑚 = (1 – (1/2))(𝑣₀ sin𝜃)²/𝑔

ℎ_𝑚 = (1/2)(𝑣₀ sin𝜃)²/𝑔

𝒉_𝒎 = (𝒗_𝟎𝐬𝐢𝐧𝜽)^𝟐/𝟐𝒈

Horizontal Range of projectile

It is the horizontal distance covered by the projectile during its flight. It is denoted by 𝑅.

Expression for Horizontal Range of projectile

Displacement along X-axis is, 𝑥 = (𝑣₀ cos𝜃)𝑡

Now 𝑥 = 𝑅 and 𝑡 = 𝑇_𝑓

𝑅 = (𝑣₀ cos𝜃)𝑇_𝑓

𝑅 = (𝑣₀ cos𝜃)(2𝑣₀ sin𝜃/𝑔)

𝑅 = (𝑣₀²/𝑔)(2cos𝜃sin𝜃)

𝑹 = (𝒗_𝟎^𝟐/𝒈)𝐬𝐢𝐧𝟐𝜽

Note:

(i) For a given speed of projection, the projectile will have a maximum range (𝑅𝑚) when sin 2𝜃 is maximum or the angle of projection is 45°. sin 2𝜃 = 1 ⟹ 2𝜃 = 90° Then angle of projection, 𝜃 = 45°.

Maximum range, 𝑅_𝑚 = (𝑣₀²/𝑔) sin2(45°)

𝑹_𝒎 = 𝒗_𝟎^𝟐/𝒈

(ii) Show that 𝑹_𝒎 = 𝟒𝒉_𝒎 when the angle of projection is 𝜽 = 𝟒𝟓°.

For 𝜃 = 45°, 𝑅_𝑚 = 𝑣₀²/𝑔 and ℎ_𝑚 = 𝑣₀²sin²𝜃/2𝑔 = 𝑣₀²sin²45°/2𝑔 = (𝑣₀²/2𝑔) × (1/2) ℎ_𝑚 = 𝑣₀²/4𝑔 = (1/4)(𝑣₀²/𝑔)

ℎ_𝑚 = (1/4)(𝑅_𝑚)

𝑹_𝒎 = 𝟒𝒉_𝒎

Since the maximum range of a projectile is equal to 4 times the maximum height reached.

Uniform circular motion

The motion of the object in a circular path at a constant speed is called uniform circular motion. Even though the object moves at a constant speed it has acceleration, because there is a continuous change in its direction of motion. Hence there is a change in its velocity from point to point.

Expression for Acceleration

Let 𝑟⃗ and 𝑟⃗′ be the position vectors and 𝑣⃗ and 𝑣⃗′ are the velocities of the object when it is at 𝑃 and 𝑄 as shown. Velocity at a point is along the tangent at that point in the direction of motion. From ∆𝐴𝐵𝐶, 𝐴𝐵⃗ + 𝐵𝐶⃗ = 𝐴𝐶⃗

𝑣⃗ + 𝐵𝐶⃗ = 𝑣⃗′

𝐵𝐶⃗ = 𝑣⃗′ − 𝑣⃗ = ∆𝑣⃗

∆𝑣⃗ is the change in velocity, which is towards the center. Since the path is circular, 𝑣⃗ and 𝑣⃗′ are perpendicular to 𝑟⃗ and 𝑟⃗′ respectively. Therefore ∆𝑣⃗ is perpendicular to ∆𝑟⃗. The average acceleration = ∆𝑣⃗/∆𝑡.

Since ∆𝑣⃗ is perpendicular to ∆𝑟⃗, 𝑎̅ is along ∆𝑣⃗ and perpendicular to ∆𝑟⃗ and directed towards the center of the circle. The Instantaneous acceleration is,

𝑎⃗ = lim _∆𝑡→0 ∆𝑣⃗/∆𝑡

Its magnitude is given by,

𝑎_𝑐 = |𝑎⃗|

𝑎_𝑐 = lim _∆𝑡→0 |∆𝑣⃗|/∆𝑡

Since the velocity vectors 𝑣⃗ and 𝑣⃗′ are always perpendicular to 𝑟⃗ and 𝑟⃗′ , the angle between 𝑣⃗ and 𝑣⃗′ is also ∆𝜃. Since ∆𝐴𝐵𝐶 and ∆𝑂𝑃𝑄 are similar.

Then, 𝐵𝐶/𝑃𝑄 = 𝐴𝐵/𝑂𝑃

|∆𝑣⃗|/|∆𝑟⃗| = 𝑣/𝑅 (∵ |𝑟⃗| = 𝑅 = 𝑟𝑎𝑑𝑖𝑢𝑠)

∆𝑣⃗ = (|∆𝑟⃗|𝑣)/𝑅

∴ 𝑎_𝑐 = lim _∆𝑡→0 (|∆𝑟⃗|𝑣/∆𝑡𝑅)

𝑎_𝑐 = (𝑣/𝑅)lim _∆𝑡→0 |∆𝑟⃗|/∆𝑡

𝑎_𝑐 = (𝑣/𝑅)𝑣

𝒂_𝒄 = 𝒗^𝟐/𝑹

This equation represents the magnitude of acceleration and is directed toward the center.

Centripetal acceleration

The acceleration, which is directed towards the center, is called centripetal acceleration. The term centripetal acceleration was termed by Newton and Centripetal comes from a Greek term that means Centre seeking towards the center.

Note:

(i) In uniform circular motion as the object moves from P to Q (in the above figure) in time ∆𝑡 the line OP turns through an angle ∆𝜃, called angular distance. But angular speed, 𝜔 = ∆𝜃/∆𝑡 If the distance traveled 𝑃𝑄 = ∆𝑠 then, Speed, 𝑣 = ∆𝑠/∆𝑡 But ∆𝑠 = 𝑅∆𝜃, where R is the radius of the trajectory.

∴ 𝑣 = 𝑅∆𝜃/∆𝑡 = 𝑅(∆𝜃/∆𝑡)

𝒗 = 𝑹𝝎 → (1)

The Centripetal acceleration 𝑎_𝑐 = 𝑣²/𝑅 = 𝑅²𝜔²/𝑅

𝒂_𝒄 = 𝑹𝝎^𝟐

(ii) Time period (T)

Time is taken by an object to make one revolution.

(iii)Frequency (𝝂)

The number of revolutions made in one second. 𝜈 = 1/𝑇.

Distance moved in time period 𝑇 = 2𝜋𝑅

Speed, 𝑣 = 2𝜋𝑅/𝑇 = 2𝜋𝑅𝜈

𝑣 = 𝑅(2𝜋𝜈) → (2)

Comparing equations (1) and (2), we get 𝝎 = 𝟐𝝅𝝂

Then Acceleration, 𝑎_𝑐 = (2𝜋𝜈)²𝑅

𝒂_𝒄 = 𝟒𝝅^𝟐𝝂^𝟐𝑹

Problems

1) A cricket ball is thrown at a speed of 28𝑚𝑠⁻¹ in a direction of 30° with the horizontal.

Calculate

(a) the maximum height

(b) the time taken by the ball to return to the same level.

(c) the distance from the thrower to the point where the ball returns to the same level.

Given 𝑣₀ = 28𝑚𝑠⁻¹, 𝜃 = 30°

(a) ℎ_𝑚 = (𝑣₀ sin 𝜃)²/2𝑔

ℎ_𝑚 = (28 × sin 30°)²/(2 × 9.8)

ℎ_𝑚 = (28 × (1/2)²/(2 × 9.8)

ℎ_𝑚 = 14²/19.6

ℎ_𝑚 = 10𝑚

(b) 𝑇_𝑓 = 2𝑣₀ sin 𝜃/𝑔

𝑇_𝑓 = (2 × 28 × sin 30°)/9.8 = (2 × 28 × (1/2))/9.8

𝑇_𝑓 = 2.85𝑠

(c) 𝑅 = (𝑣₀²/𝑔)sin2𝜃

𝑅 = (28²/9.8)sin(2 × 30°) = (28²/9.8) × 0.866

𝑅 = 69.28𝑚

2) A cricket ball projected at an angle of 30° with the horizontal takes 3 seconds to reach the ground.

Calculate

(a) the velocity of the projection.

(b) The horizontal range of the ball.

Given, = 30°, 𝑇_𝑓 = 3𝑠

(𝑎) 𝑇_𝑓 = 2𝑣₀ sin 𝜃/𝑔

3 = (2 × 𝑣₀ × sin 30°/9.8)

𝑣₀ = (3 × 9.8)/(2 × sin 30°) = (3 × 9.8)/(2 × (1/2))

𝑣₀ = 29.4𝑚𝑠⁻¹

(𝑏) 𝑅 = (𝑣₀²/𝑔)sin2𝜃

𝑅 = ((29.4)²/9.8) sin(2 × 30°) = (864.36/9.8) × 0.866

𝑅 = 76.38𝑚

3) The ceiling of a long hall is 25m high. What is the maximum horizontal distance that a ball thrown with a speed of 40𝑚𝑠⁻¹ can go without hitting the ceiling of the hall?

Given 𝑣₀ = 40𝑚𝑠⁻¹ , Height of the hall, ℎ_𝑚 = 25𝑚, 𝜃 = ?, ℎ_𝑚 = (𝑣₀ sin 𝜃)²/2𝑔

25 = (40 × sin 𝜃)²/(2 × 9.8)

25 × 2 × 9.8 = (40 × sin 𝜃)²

40 × sin 𝜃 = √(25 × 2 × 9.8)

sin 𝜃 = √(25 × 2 × 9.8)/40 = 0.5534

𝜃 = sin⁻¹(0.5534) = 33.6°

𝑅 = (𝑣₀²/𝑔)sin 2𝜃

𝑅 = (40²/9.8) sin(2 × 33.6°)

𝑅 = (1600/9.8) × 0.9219

𝑅 = 150.5𝑚

4) A cricketer can throw a ball to a maximum horizontal distance of 100m. How much high above the ground can the cricketer throw the same ball?

Given 𝑅_𝑚 = 100𝑚,

For maximum range, 𝜃 = 45°

𝑅_𝑚 = 𝑣₀²/𝑔

100 = 𝑣₀²/9.8

𝑣₀² = 100 × 9.8 = 980

𝑣₀ = 31.3𝑚𝑠⁻¹

To throw the ball vertically upwards, 𝜃 = 90°

ℎ_𝑚 = (𝑣₀ sin 𝜃)²/2𝑔

ℎ_𝑚 = (31.3 × sin 90°)²/(2 × 9.8)

ℎ_𝑚 = (31.3 × 31.3)(2 × 9.8)

ℎ_𝑚 = 49.98𝑚 ≈ 50𝑚

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

Given Radius, 𝑟 = 80𝑐𝑚 = 0.80𝑚

Frequency, 𝜈 = 14/25 𝑠⁻¹

𝜔 = 2𝜋𝜈

𝜔 = 2 × (22/7) × (14/25)

𝜔 = 3.52 𝑟𝑎𝑑/𝑠

𝑎_𝑐 = 𝑟𝜔²

𝑎_𝑐 = 0.80 × (3.52)²

𝑎_𝑐 = 9.91𝑚𝑠⁻²

6) An aircraft executes a horizontal loop of radius 1km with a speed of 900kmph. Compare its centripetal acceleration with the acceleration due to gravity.

Given Radius, 𝑟 = 1𝑘𝑚 = 1000𝑚

Speed, 𝑣 = 900𝑘𝑚𝑝ℎ = 900 × (1000/3600)𝑚𝑠⁻¹ = 250𝑚𝑠⁻¹

𝑎_𝑐 = 𝑣²/𝑟

𝑎_𝑐 = 250²/1000 = 62500/1000

𝑎_𝑐 = 62.5𝑚𝑠⁻²

Comparison: But 𝑔 = 9.8𝑚𝑠⁻²

𝑎_𝑐/𝑔 = 62.5/9.8 = 6.38

𝑎_𝑐 = 6.38𝑔

7)Rain is falling vertically with a speed of 30𝑚𝑠⁻¹. A woman rides a bicycle with a speed of 10𝑚𝑠⁻¹ in the north-south direction. What is the direction in which she should hold her umbrella?

Given Velocity of bicycle, 𝑣⃗_𝐵 = 10𝑚𝑠⁻¹

Velocity of rain, 𝑣⃗_𝑅 = 30𝑚𝑠⁻¹

Magnitude of velocity, 𝑣_𝑅𝐵 = √(𝑣_𝐵² + 𝑣_𝑅²)

𝑣_𝑅𝐵 = √(30² + 10²) = √1000

𝑣_𝑅𝐵 = 31.6𝑚𝑠⁻¹

Direction, 𝜃 = tan⁻¹(𝑣_𝐵/𝑣𝑅)

𝜃 = tan⁻¹ (10/30) = tan⁻¹(0.3333)

𝜃 = 18°26′ She should hold the umbrella at an angle 18°26′ with the vertical in south-west direction.

8) An aeroplane flying at 540kmph drops a missile towards the ground. If the height of the plane is 1000m then calculate

(a) time taken by the missile to hit the ground.

(b) Horizontal distance covered by the missile from the initial point.

Given 𝑎_𝑦 = 9.8𝑚𝑠⁻², 𝑎_𝑥 = 0, 𝑦 = 1000𝑚.

At the top, 𝑣_0𝑥 = 540𝑘𝑚𝑝ℎ = 150𝑚𝑠⁻¹, 𝑣_0𝑦 = 0

(a) Time taken can be calculated as, 𝑦 = 𝑣_0𝑦𝑡 + (1/2)𝑎_𝑦𝑡²

1000 = 0 + (1/2) × 9.8 × 𝑡²

𝑡² = (1000 × 2)/9.8

𝑡 = 14.29𝑠

(b) 𝑥 = 𝑣_0𝑥𝑡 + (1/2)𝑎_𝑥𝑡²

𝑥 = 150 × 14.29

𝑥 = 2142.86𝑚

9) Two concurrent forces 20N and 30N are acting at an angle of 60° with respect to each other. Calculate the magnitude and direction of the resultant. Given 𝐴 = 20𝑁, 𝐵 = 30𝑁, 𝜃 = 60°

(a) Magnitude, 𝑅 = √(𝐴² + 𝐵² + 2𝐴𝐵 cos𝜃)

𝑅 = √(20² + 30² + (2 × 20 × 30 × cos 60°)

𝑅 = √(400 + 900 + (2 × 600 × (1/2)) = √(1900)

𝑅 = 43.50𝑁

(b) Direction, 𝛼 = tan⁻¹(𝐵 sin𝜃/(𝐴 + 𝐵 cos 𝜃))

𝛼 = tan⁻¹(30 × sin 60°/(20 + (30 × cos 60°))

𝛼 = tan⁻¹(25.98/35) = 36°58′

The resultant force is 43.50𝑁 in the direction of 36°58′ with respect to 20𝑁.

Important Questions.

One mark.

1) What is unit vector?

2) What is Zero (null) vector?

3) Is scalar multiplied by a vector, a vector or a scalar?

4) What is the minimum number of vectors to give zero resultant?

5) When will be the resultant of two given vectors is maximum?

6) What is resolution of vector?

7) What is time of flight of a projectile?

8) At what angle range of a projectile is maximum? or when the range of a projectile does become maximum?

9) What is the relation between maximum height and maximum range of a projectile?

10) For angle of projection 30°, 𝑅 is the range of the projectile. Then write another angle of projection for which the range is same.

11) Represent the unit vector in mathematical form.

Two marks.

1) What are scalar and vector? Give example. OR distinguish between scalar and vector.

2) What is a projectile? Give an example.

3) Write the equation for the trajectory of a projectile motion. What is the nature of its trajectory?

4) State and explain parallelogram law of vector addition.

5) A unit vector is represented by 𝐴𝑖̂+ 𝐵𝑗̂+ 𝐶𝑘̂. If the value of 𝐴 and 𝐵 are 0.5 and 0.8 respectively, then find the value of 𝐶.

Three marks.

1) State and explain the triangle law of vectors addition.

2) Obtain an expression for maximum height reached by a projectile.

3) Obtain an expression for time of flight of a projectile.

4) What is resolution of vectors? Write expressions for 𝑥 and 𝑦 components (Rectangular) of a vector. or obtain the equations for rectangular components of a vector in two dimensions.

5) Derive an expression for magnitude of resultant of two concurrent vectors. or find the magnitude of the resultant of two vectors 𝐴 and 𝐵 in terms of their magnitude and angle 𝜃 between them.

6) Obtain the expression for range of a projectile.

Five marks

1) What is centripetal acceleration? Derive an expression for centripetal acceleration of a particle in uniform circular motion. or What is centripetal acceleration? Derive the expression for radial acceleration.

2) What is projectile motion? Show that trajectory of projectile is a parabola. or What is projectile motion? Derive an expression for trajectory of projectile. or show that the path of the projectile is a parabola.

Additional Problems

1) A bullet is fired at a velocity of 392𝑚𝑠⁻¹ at an angle of 30° to the horizontal. Find the maximum height attended ad time of flight.

2) A body is projected with a velocity of 50𝑚𝑠⁻¹ in a direction making an angle of 30° with the horizontal.

Find

(a) The maximum height.

(b) the time taken by the body to return to the same level

and (c) the range.

3) A ball is thrown into air with a speed of 62𝑚𝑠⁻¹ at an angle 45° with the horizontal.

Calculate

(a) The maximum height attained.

(b) the time of flight.

(c) the horizontal range.

4) A player hits a cricket ball at angle of 40° to the horizontal. If the ball moves with a velocity of 20𝑚𝑠⁻¹. Find

(a) The maximum height reached by the ball.

(b) the time of flight.

(c) the horizontal range.

Given 𝑔 = 10𝑚𝑠⁻¹.

5) A football player kicks a ball at an angle of 30° to the horizontal with an initial velocity of 15𝑚𝑠⁻¹. Assuming that the ball travels in a vertical plane. Calculate (a) The maximum height reached. (b) the time of flight and (c) the horizontal range.