##### Assignment for chapter 4

__SCALARS AND VECTORS__

__SCALARS AND VECTORS__

**Scalars:** A quantity which has ** only magnitude** but doesn’t have any direction is called Scalar. For example:

**mass**,

**temperature, speed, energy, density, volume etc.**

- Scalar quantities are represented as numbers and hence can be
**positive**and**negative numbers**. - Scalars are
**added**,**subtracted**and**multiplied**like**real numbers**

**Vectors: **A quantity which has *both magnitude *** and direction** is called vector. For example:

**force, displacement, velocity, acceleration.**

- If a is a
**vector**quantity , it must be written as**ā** **Magnitude of ā**is a positive vector quantity and is represented by**|ā|**

**Positon & Displacement vectors:**

To describe the position of an object moving in a plane, a vector is joined from a fixed point (known as origin) to the moving particle. This is known as *position vector.*

**Displacement vector** is the **straight line** joining **the initial** and **final positions** and does not depend on the actual path undertaken by the object between the two positions.

*OP = r (Position vector)*

*OP’ = r’ (Position vector)*

*PP’ = r’ – r (Displacement vector)*

**Equality of Vectors**

Two vectors A and B are said to be equal if, and only if, they have the **same magnitude** and the **same direction.**

**Multiplication of Vectors by Real Numbers**

- Multiplying a vector
**A**with a positive number**λ**gives a vector whose magnitude is changed by the factor λ but the direction is the same as that of A :

**|λ A| = λ |A| if λ > 0**

- Multiplying a vector A by a negative number λ gives a vector
**λA**whose direction is opposite to the direction of A and whose magnitude is**–λ times |A|** - The factor λ by which a vector A is multiplied could be a scalar having its own physical dimension. Then, the dimension of λ A is the product of the dimensions of λ and A.

**Unit Vector**

- A unit vector
**ā**can also be denoted as**â** **In every direction**there can be a**unit vector.**- The unit vector in direction of
**+ X**axis is called as**î**and the unit vector in the direction of**+Y**axis is given by**ĵ**. - Any vector in
**+X**axis direction can be written as its magnitude multiplied by**î**and in**–X**axis direction as multiplied by**– î**. - A
**line perpendicular to XY**plane and passing through origin is called**Z**axis. The unit vector along**+Z**axis is denoted by**ƙ**.

__ADDITION AND SUBTRACTION OF VECTORS — GRAPHICAL METHOD__

__ADDITION AND SUBTRACTION OF VECTORS — GRAPHICAL METHOD__

**Collinear Vectors:**

- If two vectors in
**same direction**are**added,**their**magnitudes**are**added**together. - If two vectors in
**opposite**directions are**added**, their magnitudes are**subtracted.**The resultant is in the direction of the vector whose magnitude is greater.

**Triangle Law of Addition:**

- If two vectors
**ā**and**ȳ**lie along thein consecutive order, the third side represents the sum of ā and ȳ.*two sides of a triangle*

**Parallelogram Law of Addition: **

- If two vectors lie along
as shown , the diagonal of the parallelogram through common vertex represents their sum.(or resultant)*two adjacent sides of a parallelogram*

*Note: “The vectors must originate from same vertex O.”*

- Consider two vectors
**ū**&**ȳ**of magnitudes of u and y respectively making an angle between them. - If ū and ȳ lie along the adjacent sides of sides of a parallelogram, the length of the diagonal represents the magnitude of the resultant
**R.**

- The direction of
**R**is measured by the angle between**R**and the ū vector (angle α)

sinα= ȳsinθ/ **R**

** or**

**Concept: The magnitude of resultant of two vectors ā and ȳ is minimum when they are opposite to each other and maximum when ā & ȳ are in the same direction.**

R = ā + ȳ

R_{min}= |ā| - |ȳ| (where |ā| > |ȳ|)

R_{max}= |ā| + |ȳ|

**RESOLUTION OF VECTOR**

Consider a given reference of X and Y and a given vector ā (as shown).

- We can split ā in two parts (known as components of ā) , so that one part is parallel to Y axis .

- According to parallelogram law of vectors we can see that

** ā = PA + PB**

- PA is parallel to X axis and is known as X component of ā and PB is parallel to Y axis and is known as Y component of ā .
- If magnitude of ā = |ā| = atanθ is the angle between ā and X axis :

X component of ā = a_{x} = acosθ

Y component of ā = a_{y} = asinθ

- In vector notation :

ā =PA + PB

ā = a_{x}î + a_{y}ĵ

as PA = a_{x}î , PB = a_{y}ĵ

ā = acosθ î + asinθĵ

a_{x }and a_{y} can be negative or positive according to the direction of components.

If A and θ are given, A_{x} and A_{y} can be obtained using above equations. If A_{x} and A_{y} are given, A and θ can be obtained as follows:

**A _{x}^{2} + A_{y}^{2 }= A^{2}cos^{2}θ+ A^{2}sin^{2}θ= A^{2}**

The same procedure can be used to resolve a general vector A into three components along x-, y-, and z-axes in three dimensions.

If α, β, and γ are the angles between A and the x-, y-, and z-axes, respectively Fig. 4.9(d), we have,

A_{x}= A cos α , A_{y}= A cos β , A_{z}= A cosγ

A = A_{x}î + A_{y}ĵ + A_{z}ƙ

The magnitude of A vector is given by

*Note: (a) The components can be resolved along two perpendicular axes. They may not be horizontal and vertical always.*

*Observe that the given angle always comes between given vector and its cosine component.*

**VECTOR ADDITION – ANALYTICAL METHOD**

**Problem solving technique:**

- Resolve each vector into components.
**Add**the**X components**and**Y components**separately.- If the components along an axis in
**same direction**,**magnitudes are added**(like vector addition of parallel vectors) - If the components along an
**opposite direction**, magnitudes are**subtracted.** **R**component of resultant_{x}= X**R**component of resultant_{y}= Y- The magnitude of resultant is given by ,
**R =**{\sqrt {{a}^{2}+{b}^{2}}}

Consider two vectors A and B in x-y plane with components A_{x} , A_{y} and B_{x} , B_{y}

** A = A _{x}î + A_{y}ĵ**

** B = B _{x}î + B_{y}ĵ**

**R = A + B**

**R _{x} = A_{x}î + B_{x}î R_{y} = A_{y}ĵ + B_{y}ĵ**

This method can be extended to addition and subtraction of any number of vectors. For example, if vectors a, b and c are given as

**A** = A_{x}î + A_{y}ĵ + A_{z}ƙ

**B** = B_{x}î + B_{y}ĵ + B_{z}ƙ

**C** = C_{x}î + C_{y}ĵ + C_{z}ƙ

**Then , a vector T = A + B – C**

*T _{x} = A_{x}î + B_{x}î + C_{x}î *

** T_{y} = A_{y}ĵ + B_{y}ĵ + C_{y}ĵ**

** T_{z} = A_{z}ƙ + B_{z}ƙ + C_{z}ƙ**

**Law of cosines :**

**R ^{2} = A^{2} + B^{2} + 2ABcosθ**

**Law of sines :**

__DOT PRODUCT AND CROSS PRODUCT__

__DOT PRODUCT AND CROSS PRODUCT__

**ANGLE BETWEEN TWO VECTORS**

Angle between any two vectors is taken as the angle between their direction. To find the angle between two vectors ā and ū , we should first imagine (make) their tales at same point θ: angle between ā and ū .

**Scalar (Dot) Product (ā.ū)**

The ** scalar or dot product** of two vectors

**ā**and

**ū**is defined as the product of their magnitudes and the

**cosines**of angle between them . It is represented by a

**dot (.)**between ā and ū . The product itself is a

*scalar quantity .***ā . ū = aucosθ ** 0{\leq {\theta }{\leq }{\pi }}

IMPORTANT POINTS

- If ā & ū are in the same direction
**(θ= 0**then^{0})**ā.ū = ab** - If ā & ū are in opposite direction
**(θ=180**, then^{0})**ā.ū = aucos180**^{0}= -ab - If ā is perpendicular to ū
**(θ=90**, then^{0})**ā.ū=0** **î .î = ĵ.ĵ = ƙ.ƙ = 1 , î.ĵ = ĵ.ƙ= ƙ.î = 0**- If θ is
**acute**,**ā.ū**is**positive**

**ā .ū = ū.ā***( Commutative property)***ā.(ū + ȳ) = ā.ū + ā.ȳ***(Distributive Property)***ā .ā = aacos0**^{0}= a^{2 }, a

- If
**θ is obtuse**, then**ā.ū = -ve**

**Vector or Cross Product**

The vector product of two vectors ā and ū is defined as vector whose

- Magnitude is equal to the
**product of their magnitudes**and the**sine of the angle****θ****between them.** **Direction**is**perpendicular**to the**plane containing the vectors ā and ū**and is given by**right hand thumb rule.**- It is represented by a
**cross (x)**between ā & ū

The product itself is a vector quantity

**|axu|= |ā||ū|sinθ = absinθ**

Right hand thumb Rule

If we bend the fingers of the right hand in such a way that they rotate the vector ā towards ū through the angle θ , then the htumb gives the direction of vector āxū.

Note that **ā x ū = -(ū x ā )** ,

It means their magnitudes are equal and directions are opposite

__MOTION IN A PLANE__

__MOTION IN A PLANE__

**Position Vector and Displacement**

The position vector r of a particle P located in a plane with reference to the origin of an x-y reference frame is given by

**r = xî + yĵ**

Where x and y are components of r along x-, and y- axes or simply they are

the coordinates of the object.

Displacement is given by :

∆r = r’ – r and is directed from P to P’

∆r = x’î + y’ĵ – (xî + yĵ )

= îᅀx + ĵᅀy

Where ᅀx = x’ – x , ᅀy = y’ – y

**Velocity**

The average velocity (v) of an object is the ratio of the displacement and the corresponding time interval

v = v_{x}î + v_{y}ĵ

The direction of the average velocity is the same as that of ∆r.

*As the time interval ∆t approaches zero, the average velocity approaches the velocity v.*

*The direction of v is parallel to the line tangent to the path.*

- As ∆t → 0, ∆r → 0 and is along the tangent to the path . Therefore, the direction of velocity at any point on the path of an object is tangential to the path at that point and is in the direction of motion.
- We can express
**v**in component form :

** v _{x}î = dx/dt , v_{y}ĵ = dy/dt**

**Acceleration**

The average acceleration a of an object for a time interval ∆t moving in x-y plane is the change in velocity divided by the time interval

a= **a _{x}î + a_{y}ĵ**

The acceleration (instantaneous acceleration) is the limiting value of the average acceleration as the time interval approaches zero:

The acceleration (instantaneous acceleration) is the limiting value of the average acceleration as the time interval approaches zero:

* “Note that in one dimension, the velocity and the acceleration of an object are always along the same straight line (either in the same direction or in the opposite direction). However, for motion in two or three dimensions, velocity and acceleration vectors may have any angle between 0° and 180° between them.”*

**MOTION IN A PLANE WITH CONSTANT ACCELERATION**

Rewriting the equations of motion in a plane

**1 ^{st} equation of motion :**

**v _{x} = v_{0x} + a_{x}t**

** v _{y} = v_{0y} + a_{y}t**

**2 ^{nd} Equation of motion:**

** x= x _{0} + v_{0}t + ½(a_{x}t^{2})**

** y = y _{0} + v_{0}t + ½(a_{y}t^{2})**

**3 ^{rd} Equation of motion:**

**v _{x}^{2} – u_{x}^{2} = 2a_{x}x**

** v _{y}^{2} – u_{y}^{2} = 2a_{y}y**

__RELATIVE VELOCITY IN TWO DIMENSIONS__

__RELATIVE VELOCITY IN TWO DIMENSIONS__

Suppose that two objects A and B are moving with velocities v_{A} and v_{B} (each with respect to some common frame of reference, say ground.)

A relative to that of B is:

**v _{AB} = v_{A }– v_{B}**

and similarly, the velocity of object B relative to that of A is :

**v _{BA} = v_{B} – v_{A}**

Therefore,** v _{AB} = – v_{BA}**

**And | v _{AB}|= | v_{BA}|**

__PROJECTILE MOTION__

__PROJECTILE MOTION__

A particle when given velocity at ** an arbitrary angle **made with

**surface is known as**

*horizontal***projectile.**

**Motion over a Horizontal Plane **

If a particle is projected form point O, at an angle θ from the horizontal, with initial velocity ū then they have components of ū in X and Y directions are given as

u_{x} = ucosθ

u_{y} = usinθ where ū = u_{x} î + u_{y} ĵ

**ū = ucosθî + usinθĵ**

The X axis is parallel to the horizontal . Yaxis is parallel to the vertical and ū lies in the plane X-Y. The constant acceleration ā is given by

**ā = a _{x}î +a_{y}ĵ**

where a_{x} = 0 , and a_{y} = -g (acceleration due to gravity)

Now , velocity after time t is given as

**v _{x} = u_{x}t + a_{x}t = ucosθ (as a_{x} = 0 )**

**v _{y} = u_{y}t + a_{y}t = usinθ – gt **

v= ucosθî + (usinθ-gt)ĵ

Thus the direction of v with the horizontal is given by **tan ^{-1}(v_{y}/v_{x})**

Now, coordinates of the projectile after time t is given by

**x= x _{0} + u_{x}t + ½(a_{x}t^{2})**

x = ucosθt …. (1)

**y = y _{0} + u_{y}t + ½(a_{y}t^{2})**

y = usinθ – ½(gt^{2}) …… (2)

from equation (1) and (2) eliminating t , we get

y = usinθ (x/ucosθ) – ½ (g)(x^{2}/u^{2}cos^{2}θ)

The above equation shows relation between x and y and represents the path of the projectile known as **trajectory.**

**The equation of parabola is given by:**

**y = bx ^{2} + cx^{2}** Where

**b = tanθ**= constant and c =

^{ }**– gx**

^{2}/(2u^{2}cos^{2}θ)**Time Of Flight**

*“It is the time interval during which the projectile remains in air .”*

Putting y= 0 in (2), we get

**T = 2usinθ/g**, where T = time of flight

**Range :**

*“ The horizontal range R of the projectile is the horizontal distance between the initial point and the point where the projectile is again at same horizontal level.”*

If R be the horizontal range , R = ucosθ x (2usinθ/g) = (u^{2}sin2θ)/g

Since , sin2θ = sin(Π-2θ) = sin[2(Π/2 – θ)]

Let (Π/2 –θ) = β

Sin2θ=sin2β

Hence , range is same for two angles of projection provided the angles be complimentary .

For a given velocity of projection , R is maximum when sin2θ= 1

2θ= 90° , θ= 45°

** R _{max} = u^{2}/g**

We can show that y = xtanθ[1-x/R]

**Maximum Height**

Since , v_{y}^{2} = u_{y}^{2} + 2a_{y}y

At , y = y_{max} , v_{y} = 0 , 0 = u^{2}sin^{2}θ – 2 gy_{max}

Thus the maximum height attained by the projectile is given by

**y _{max} = u^{2}sin^{2}θ/2g**

**HORIZONTAL PROJECTION FROM A GIVEN HEIGHT**

Referring to the figure , let a particle be projected with a horizontal velocity v_{0} , which remains constant along horizontal line due to the absence of any horizontal force .

Due to earth’s gravitation the particle acquires vertical velocity v_{y} at any time t and at any position P(x,y)

**v _{y} = u_{y} + gt**

Since there is no vertical component of v_{0} initially , u_{y} = 0

v_{y}= gt

and the vertical displacement is

**y = u _{y}t + ½(gt^{2})**

Again , v_{y}^{2} = u_{y}^{2} + 2gy

Putting u_{y} = 0 , we get v_{y} = {\sqrt {2gy}}

**Displacement**

Let the position vector of this point be r . Now the horizontal displacement x = v_{0}t and the vertical displacement y = ½(gt^{2}).

Since the position vector r = xî + yĵ , putting the values of x and y , we obtain

**r= v _{0}tî + ½(gt^{2})**

**Velocity**

v= v_{x}î + v_{y}ĵ

**v=v _{0}î + v_{y}ĵ**

**Range**

When y = H (height of the cliff or height of fall of projectile) , the corresponding horizontal distance (Range R) can be found by putting the values of time of fall

**t =** ** ** {\sqrt {\frac {2H} {g}}}

**R = v _{0}** {\sqrt {\frac {2H} {g}}}

**Equation of Trajectory**

The locus of the path of the particle is given as

y = ½(gt^{2}) where t= x/v_{0}

This equation represents a parabola.

__UNIFORM CIRCULAR MOTION__

__UNIFORM CIRCULAR MOTION__

As another small illustration, of motion of a particle in two dimensions let’s analyse the uniform circular motion of a particle.

In **uniform circular motion**, the particle moves in a circular path with ** constant speed**.

Lets choose the centre of circular path , as the origin of the reference frame . Point ‘P’ is an arbitrary point on the path whose position vector r = xî + yĵ

Where r , the raidus of circular path is related to x and y by the following equations

**x= rcos****θ**** , y = rsin****θ**** ****and x ^{2} + y^{2} = r^{2}**

**r = rcos****θ****î + rsin****θ****ĵ**

Now , the velocity of particle ‘P’ is given as

v= dr/dt = (dx/dt)î + (dy/dt)ĵ ={ d(rcosθ)/dt }î + {d(rsinθ)/dt} ĵ

**v = -risnθ(dθ/dt) î + rcosθ(dθ/dt) ĵ**

But dθ/dt = ω = constant (for uniform circular motion)

Thus **v**= ωr(-sinθî +cosθĵ)

Now , **v.r** = ωr{-cosθ(dθ/dt) î – sinθ(dθ/dt)ĵ }

**ā= -ω ^{2} r (-cos**

**θ**

**î + sin**

**θ**

**ĵ)**

**ā= -ω ^{2}r = ω^{2}(-r)**

which shows that ā is directed in the opposite direction of r. Thus ā is always directed towards the centre.

**Non-Uniform Circular Motion**

In a non-uniform motion, the ** speed** also

**along with the**

*changes*

*direction.*** Acceleration due to change in direction** is called

**radial / centripetal /normal acceleration**. It is denoted by

**a**

_{r}**a _{r} = ω^{2}r = v^{2}/r = ωv** , direction being towards centre .

Acceleration due to change in speed (tangential acceleration)

a= **a _{t} = dv/dt** along

**the tangent**

Thus magnitude of the net acceleration

*Note : When velocity and acceleration make an acute angle , the speed increases and when the angle is obtuse , the speed decreases.*

For detailed understanding of concepts regarding vector , motion in a plane , relative velocity , projectile motion and uniform circular motion please refer to the above PDF.

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