Assignment for chapter 4


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.


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 andȳ lie along the two sides of a triangle in consecutive order, the third side represents the sum of ā and ȳ.

 Parallelogram Law of Addition:  

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

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


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
Rmin = |ā| - |ȳ|    (where |ā| > |ȳ|)
Rmax = |ā| + |ȳ|


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 = ax = acosθ

Y component of ā = ay = asinθ

  • In vector notation :                             

ā =PA  +  PB 

ā = axî + ayĵ

as   PA = axî ,    PB = ayĵ

ā = acosθ î + asinθĵ

ax and ay can be negative or positive according to the direction of components.

    If A and θ are given, Ax and Ay can be obtained using above equations. If Ax and Ay are given, A and θ can be obtained as follows:

Ax2 + Ay2 = A2cos2θ+ A2sin2θ= A2

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,

Ax= A cos α , Ay= A cos β , Az= A cosγ

A = Axî + Ayĵ + Azƙ

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.


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.
  • Rx = X component of resultant
  • Ry = Y component of resultant
  • 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 Ax , Ay and Bx  , By

A = Axî + Ayĵ

                                                            B = Bxî + Byĵ

                                                              R = A + B

                                            Rx = Axî + Bxî               Ry = Ayĵ + Byĵ

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 = Axî + Ayĵ + Azƙ

B = Bxî + Byĵ + Bzƙ

C = Cxî + Cyĵ + Czƙ

Then , a vector T = A + B – C

                                                             Tx = Axî + Bxî + Cxî 

                                                              Ty = Ayĵ + Byĵ + Cyĵ

                                                              Tz = Azƙ + Bzƙ + Czƙ

Law of cosines :

   R2 = A2 + B2 + 2ABcosθ

Law of sines :

{\frac {R} {sin{\theta }}}{=}{\frac {A} {sin{\beta }}}{=}{\frac {B} {sin{\alpha }}}



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 }}

  • If ā & ū are in the same direction (θ= 00) then ā.ū = ab
  • If ā & ū are in opposite direction (θ=1800) , then                        ā.ū = aucos1800 = -ab
  • If ā is perpendicular to ū (θ=900) , then 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 = aacos00 = a2  , 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


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


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

v = vxî + vyĵ

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 :

                                                  vxî = dx/dt  , vyĵ =  dy/dt


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= axî + ayĵ

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.”


Rewriting the equations of motion in a plane

1st equation of motion :

vx = v0x + axt

                                                           vy = v0y + ayt

2nd Equation of motion:

                                                     x= x0 + v0t + ½(axt2)

                                                     y = y0 + v0t + ½(ayt2)

3rd Equation of motion:

vx2 – ux2 = 2axx

                                                           vy2 – uy2 = 2ayy


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

A relative to that of B is:

vAB = vA – vB

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

vBA = vB – vA

Therefore, vAB = – vBA

And | vAB|= | vBA|


A particle when given velocity at an arbitrary angle made with horizontal surface is known as 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

ux =  ucosθ

uy = usinθ where ū = ux î + uy 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 = axî +ayĵ

where ax = 0 , and ay = -g (acceleration due to gravity)

Now , velocity after time t is given as

vx =    uxt + axt = ucosθ  (as ax = 0 )

vy = uyt + ayt = usinθ – gt    

v= ucosθî + (usinθ-gt)ĵ

Thus the direction of v with the horizontal is given by tan-1(vy/vx)

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

x= x0 + uxt + ½(axt2)

x = ucosθt                                …. (1)

y = y0 + uyt + ½(ayt2)

y = usinθ – ½(gt2)                     …… (2)

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

y = usinθ  (x/ucosθ) – ½ (g)(x2/u2cos2θ)

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 = bx2 + cx2 Where b = tanθ = constant and c =– gx2 /(2u2cos2θ)

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) = (u2sin2θ)/g

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

Let (Π/2 –θ) = β


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°

                                                             Rmax = u2/g

We can show that y = xtanθ[1-x/R]

Maximum Height

Since , vy2 = uy2 + 2ayy

At , y = ymax , vy = 0 , 0 = u2sin2θ – 2 gymax

Thus the maximum height attained by the projectile is given by

ymax = u2sin2θ/2g


Referring to the figure ,  let a particle be projected with a horizontal velocity v0 , which remains constant along horizontal line due to the absence of any horizontal force .

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

vy = uy + gt

Since there is no vertical component of v0 initially , uy = 0

vy= gt

and the vertical displacement is

y = uyt + ½(gt2)

Again , vy2 = uy2 + 2gy

Putting uy = 0 , we get vy = {\sqrt {2gy}}


Let the position vector of this point be r . Now the horizontal displacement x = v0t and the vertical displacement y = ½(gt2).

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

r= v0tî + ½(gt2)


v= vxî + vyĵ

v=v0î + vyĵ


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 = v0 {\sqrt {\frac {2H} {g}}}

Equation of Trajectory

The locus of the path of the particle is given as

y = ½(gt2) where t= x/v0

This equation represents a parabola.


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= rcosθ , y = rsinθ and x2 + y2 = r2

r = rcosθî + rsinθĵ

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= -ω2 r (-cosθî + sinθĵ)

ā= -ω2r = ω2(-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 changes along with the direction.

Acceleration due to change in direction is called radial / centripetal /normal acceleration. It is denoted by ar

ar =  ω2r  = v2/r = ωv , direction being towards centre .

Acceleration due to change in speed (tangential acceleration)

a= at = 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 .

Categories: General


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