 # MOTION IN A PLANE

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.

## 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,

## 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.

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 = 𝑎⃗𝑥 and 𝑂𝑁⃗ = 𝑎⃗𝑦. From parallelogram law of addition, we have 𝑂𝑀⃗ + 𝑂𝑁⃗ = 𝑂𝐴⃗⃗

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

𝒂𝒙 = 𝒂 𝐜𝐨𝐬 𝜽

and 𝑠𝑖𝑛 𝜃 = 𝐴𝑀/𝑂𝐴 = 𝑎𝑦/𝑎

𝒂𝒚 = 𝒂 𝐬𝐢𝐧 𝜽

Vectoricaly 𝒂⃗𝒙 = 𝒂⃗ 𝐜𝐨𝐬 𝜽 and 𝒂⃗𝒚 = 𝒂⃗𝐬𝐢𝐧 𝜽

Note: 𝑎⃗𝑥 and 𝑎⃗𝑦 being perpendicular are called rectangular components of 𝑎⃗.

Magnitude

Magnitude of 𝑎⃗ is given by |𝑎⃗| = 𝑎

Now, 𝑎𝑥2 = 𝑎2cos2𝜃 and 𝑎𝑦2 = 𝑎2sin2𝜃

Taking 𝑎𝑥2 + 𝑎𝑦2 = 𝑎2cos2𝜃 + 𝑎2sin2𝜃

𝑎𝑥2 + 𝑎𝑦2 = 𝑎2(cos2𝜃 + sin2𝜃)

𝑎𝑥2 + 𝑎𝑦2 = 𝑎2

𝒂 = √(𝒂𝒙𝟐 + 𝒂𝒚𝟐)

𝐃𝐢𝐫𝐞𝐜𝐭𝐢𝐨𝐧

By taking 𝑎𝑦/𝑎𝑥 = 𝑎sin𝜃/𝑎cos𝜃

𝑎𝑦/𝑎𝑥 = tan𝜃

𝜽 = 𝐭𝐚𝐧−𝟏(𝒂𝒚/𝒂𝒙)

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

where 𝑎⃗𝑥 = 𝑎 cos 𝜃 𝑖̂

and 𝑎⃗𝑦 = 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 𝑎𝑥 = 𝑎 cos 𝛼, 𝑎𝑦 = 𝑎 cos𝛽, 𝑎𝑧 = 𝑎 cos 𝛾

and 𝑎⃗ = 𝑎𝑥𝑖̂+ 𝑎𝑦𝑗̂+ 𝑎𝑧𝑘̂

The magnitude of 𝑎⃗ is, 𝑎 = √(𝑎𝑥2 + 𝑎𝑦2 + 𝑎𝑧2).

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 = √(𝑨𝟐 + 𝑩𝟐 + 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 = 𝐭𝐚𝐧𝟏(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 = 𝑎𝑥𝑖̂+ 𝑎𝑦𝑗 ̂ and 𝑏⃗ = 𝑏𝑥𝑖̂+ 𝑏𝑦𝑗 ̂ then, 𝑅⃗ = 𝑎⃗ + 𝑏⃗

𝑅⃗ = (𝑎𝑥𝑖̂+ 𝑎𝑦𝑗̂) + (𝑏𝑥𝑖̂+ 𝑏𝑦𝑗̂)

𝑅⃗ = 𝑎𝑥𝑖̂+ 𝑏𝑥𝑖̂+ 𝑎𝑦𝑗̂+ 𝑏𝑦𝑗 ̂

𝑅⃗ = (𝑎𝑥 + 𝑏𝑥 )𝑖̂+ (𝑎𝑦 + 𝑏𝑦)𝑗̂

𝑹⃗ = 𝑹𝒙𝒊̂+ 𝑹𝒚𝒋 ̂

Where 𝑅𝑥 = 𝑎𝑥 + 𝑏𝑥 and 𝑅𝑦 = 𝑎𝑦 + 𝑏𝑦

In three Dimensions

If 𝑎⃗ = 𝑎𝑥𝑖̂+ 𝑎𝑦𝑗̂+ 𝑎𝑧𝑘̂ and 𝑏⃗ = 𝑏𝑥𝑖̂+ 𝑏𝑦𝑗̂+ 𝑏𝑧𝑘̂

then, 𝑅⃗ = 𝑎⃗ + 𝑏⃗

𝑅⃗ = (𝑎𝑥𝑖̂+ 𝑎𝑦𝑗̂+ 𝑎𝑧𝑘̂) + (𝑏𝑥𝑖̂+ 𝑏𝑦𝑗̂+ 𝑏𝑧𝑘̂)

𝑅⃗ = 𝑎𝑥𝑖̂ + 𝑏𝑥𝑖̂ + 𝑎𝑦𝑗̂ + 𝑏𝑦𝑗̂ + 𝑎𝑧𝑘̂ + 𝑏𝑧𝑘̂

𝑅⃗ = (𝑎𝑥 + 𝑏𝑥) 𝑖̂+ (𝑎𝑦 + 𝑏𝑦) 𝑗̂+ (𝑎𝑧 + 𝑏𝑧) 𝑘̂

𝑹⃗ = 𝑹𝒙𝒊̂+ 𝑹𝒚𝒋̂+ 𝑹𝒛𝒌̂

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

## 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)̂ + (𝑦2𝑗̂− 𝑦1𝑗̂)

∆𝑟⃗ = (𝑥2−𝑥1) 𝑖̂ + (𝑦2 − 𝑦1) 𝑗̂

∆𝒓⃗ = ∆𝒙𝒊̂+ ∆𝒚𝒋̂

Where ∆𝑥 = 𝑥2 − 𝑥1 and ∆𝑦 = 𝑦2 − 𝑦1

## 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 = 𝒗̅𝒙𝒊̂+ 𝒗̅𝒚𝒋̂

Where 𝒗̅𝒙 = ∆𝑥/∆𝑡 and 𝒗̅𝒚 = ∆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 = 𝒗𝒙𝒊̂+ 𝒗𝒚𝒋̂

The magnitude is given by 𝒗 = √(𝒗𝒙𝟐 + 𝒗𝒚𝟐) and Direction is given by

𝜽 = 𝐭𝐚𝐧−𝟏(𝒗𝒚/𝒗𝒙)

## Average acceleration

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

𝑎̅⃗ = ∆𝑣⃗/∆𝑡 = ∆(𝑣𝑥𝑖̂+𝑣𝑦𝑗̂)/∆𝑡

𝑎̅⃗ = (∆𝑣𝑥/∆𝑡) 𝑖̂ + (∆𝑣𝑦/∆𝑡) 𝑗̂

𝒂⃗̅⃗ = 𝒂̅𝒙𝒊̂+ 𝒂̅𝒚𝒋̂

## Instantaneous acceleration (Acceleration)

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

𝑎⃗ = lim𝑡→0 (∆𝑣⃗/∆𝑡)

𝒂⃗ = 𝒅𝒗⃗/𝒅𝒕

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

𝑎⃗ = lim 𝑡→0 ∆(𝑣𝑥𝑖̂+ 𝑣𝑦𝑗̂)/∆𝑡

𝑎⃗ = lim ∆𝑡→0 (∆𝑣𝑥/∆𝑡) 𝑖̂+ lim ∆𝑡→0 (∆𝑣𝑦/∆𝑡) 𝑗̂

𝑎⃗ = (𝑑𝑣𝑥/𝑑𝑡) 𝑖̂+ (𝑑𝑣𝑦/𝑑𝑡) 𝑗̂

𝒂⃗ = 𝒂𝒙𝒊̂+ 𝒂𝒚𝒋̂

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,

𝑣𝑥 = 𝑣0𝑥 + 𝑎𝑥𝑡

𝑣𝑦 = 𝑣0𝑦 + 𝑎𝑦𝑡

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

In terms of its components,

𝑥 − 𝑥0 = 𝑣0𝑥𝑡 + (1/2)𝑎𝑥𝑡2

𝑦 − 𝑦0 = 𝑣0𝑦𝑡 + (1/2)𝑎𝑦𝑡2

(iii) 𝑣 ⃗2 = 𝑣02 + 2𝑎⃗(𝑟⃗ − 𝑟⃗0)

In terms of its components,

𝑣𝑥2 = 𝑣0𝑥2 + 2𝑎𝑥(𝑥 − 𝑥0)

𝑣𝑦2 = 𝑣0𝑦2 + 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 𝑣⃗𝐴 and 𝑣⃗𝐵, then the velocity of the object 𝐴 relative to that of 𝐵 is, 𝑣⃗𝐴𝐵 = 𝑣⃗𝐴 − 𝑣⃗𝐵.

Similarly, the velocity of the object 𝐵 relative to that of 𝐴 is, 𝑣⃗𝐵𝐴 = 𝑣⃗𝐵 − 𝑣⃗𝐴 Therefore, 𝑣⃗𝐴𝐵 = −𝑣⃗𝐵𝐴 and |𝑣⃗𝐴𝐵| = |𝑣⃗𝐵𝐴|

## 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. 𝑎𝑥 = 0, 𝑎𝑦 = −𝑔, hence 𝑎⃗ = −𝑔𝑗̂.

The components of initial velocity 𝑣⃗0 are, 𝑣0𝑥 = 𝑣0cos𝜃, 𝑣0𝑦 = 𝑣0sin𝜃

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

The components of displacements at time 𝑡 are,

(i) 𝑥 = 𝑣0𝑥𝑡 + (1/2)𝑎𝑥𝑡2

𝑥 = 𝑣0𝑥𝑡 (∵ 𝑎𝑥 = 0), 𝒙 = (𝒗𝟎 𝐜𝐨𝐬 𝜽) 𝒕 (along X-axis)

(ii) 𝑦 = 𝑣0𝑦𝑡 + (1/2)𝑎𝑦𝑡2, 𝒚 = (𝒗𝟎 𝐬𝐢𝐧𝜽)𝒕 – (𝟏/𝟐)𝒈𝒕𝟐 (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 = 𝒂𝒙 − 𝒃𝒙𝟐

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 𝑇𝑓.

## Expression for Time of flight

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

At maximum height 𝑣𝑦 = 0 and time for maximum height, 𝑡 = 𝑡𝑚.

0 = 𝑣0 sin 𝜃 − 𝑔𝑡𝑚

𝑔𝑡𝑚 = 𝑣0 sin𝜃

𝑡𝑚 = 𝑣0 sin𝜃/𝑔

Time of flight 𝑇𝑓 = 2𝑡𝑚 because “time of ascent = time of descent”

𝑻𝒇 = 𝟐𝒗𝟎𝐬𝐢𝐧𝜽/𝒈

## Maximum height of a projectile

It is the maximum height reached by the projectile in time 𝑡𝑚. It is denoted by ℎ𝑚.

## Expression for maximum height of a projectile

The displacement along Y − axis is, 𝑦 = (𝑣0 sin𝜃)𝑡𝑚 – (1/2)𝑔𝑡𝑚2

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

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

𝑚 = (𝑣0 sin 𝜃)2/𝑔 – (1/2)(𝑣0 sin𝜃)2𝑔

𝑚 = (1 – (1/2))(𝑣0 sin𝜃)2/𝑔

𝑚 = (1/2)(𝑣0 sin𝜃)2/𝑔

𝒉𝒎 = (𝒗𝟎𝐬𝐢𝐧𝜽)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 = 𝑇𝑓

𝑅 = (𝑣0 cos𝜃)𝑇𝑓

𝑅 = (𝑣0 cos𝜃)(2𝑣0 sin𝜃/𝑔)

𝑅 = (𝑣02/𝑔)(2cos𝜃sin𝜃)

𝑹 = (𝒗𝟎𝟐/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, 𝑅𝑚 = (𝑣02/𝑔) sin2(45°)

𝑹𝒎 = 𝒗𝟎𝟐/𝒈

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

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

𝑚 = (1/4)(𝑅𝑚)

𝑹𝒎 = 𝟒𝒉𝒎

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,

𝑎𝑐 = |𝑎⃗|

𝑎𝑐 = 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

∴ 𝑎𝑐 = lim 𝑡→0 (|∆𝑟⃗|𝑣/∆𝑡𝑅)

𝑎𝑐 = (𝑣/𝑅)lim 𝑡→0 |∆𝑟⃗|/∆𝑡

𝑎𝑐 = (𝑣/𝑅)𝑣

𝒂𝒄 = 𝒗𝟐/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 𝑎𝑐 = 𝑣2/𝑅 = 𝑅2𝜔2/𝑅

𝒂𝒄 = 𝑹𝝎𝟐

(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, 𝑎𝑐 = (2𝜋𝜈)2𝑅

𝒂𝒄 = 𝟒𝝅𝟐𝝂𝟐𝑹

## 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) ℎ𝑚 = (𝑣0 sin 𝜃)2/2𝑔

𝑚 = (28 × sin 30°)2/(2 × 9.8)

𝑚 = (28 × (1/2)2/(2 × 9.8)

𝑚 = 142/19.6

𝑚 = 10𝑚

(b) 𝑇𝑓 = 2𝑣0 sin 𝜃/𝑔

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

𝑇𝑓 = 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°, 𝑇𝑓 = 3𝑠

(𝑎) 𝑇𝑓 = 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, ℎ𝑚 = 25𝑚, 𝜃 = ?, ℎ𝑚 = (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 𝑅𝑚 = 100𝑚,

For maximum range, 𝜃 = 45°

𝑅𝑚 = 𝑣02/𝑔

100 = 𝑣02/9.8

𝑣02 = 100 × 9.8 = 980

𝑣0 = 31.3𝑚𝑠−1

To throw the ball vertically upwards, 𝜃 = 90°

𝑚 = (𝑣0 sin 𝜃)2/2𝑔

𝑚 = (31.3 × sin 90°)2/(2 × 9.8)

𝑚 = (31.3 × 31.3)(2 × 9.8)

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

𝑎𝑐 = 𝑟𝜔2

𝑎𝑐 = 0.80 × (3.52)2

𝑎𝑐 = 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

𝑎𝑐 = 𝑣2/𝑟

𝑎𝑐 = 2502/1000 = 62500/1000

𝑎𝑐 = 62.5𝑚𝑠−2

Comparison: But 𝑔 = 9.8𝑚𝑠−2

𝑎𝑐/𝑔 = 62.5/9.8 = 6.38

𝑎𝑐 = 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, 𝑣⃗𝐵 = 10𝑚𝑠−1

Velocity of rain, 𝑣⃗𝑅 = 30𝑚𝑠−1

Magnitude of velocity, 𝑣𝑅𝐵 = √(𝑣𝐵2 + 𝑣𝑅2)

𝑣𝑅𝐵 = √(302 + 102) = √1000

𝑣𝑅𝐵 = 31.6𝑚𝑠−1

Direction, 𝜃 = tan−1(𝑣𝐵/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 𝑎𝑦 = 9.8𝑚𝑠−2, 𝑎𝑥 = 0, 𝑦 = 1000𝑚.

At the top, 𝑣0𝑥 = 540𝑘𝑚𝑝ℎ = 150𝑚𝑠−1, 𝑣0𝑦 = 0

(a) Time taken can be calculated as, 𝑦 = 𝑣0𝑦𝑡 + (1/2)𝑎𝑦𝑡2

1000 = 0 + (1/2) × 9.8 × 𝑡2

𝑡2 = (1000 × 2)/9.8

𝑡 = 14.29𝑠

(b) 𝑥 = 𝑣0𝑥𝑡 + (1/2)𝑎𝑥𝑡2

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.

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.

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