Motion 2d Aakash

SCALARS AND VECTORS Scalars

Scalars are physical quantities which are completely described by their magnitude only. For example: mass, length, time, temperature energy etc.

Vectors

Vectors are those physical quantities having both magnitude as well as direction and they obeys vector algebra (eg. parallelogram law or triangle law of vector addition). For example: displacement, velocity, acceleration, force, momentum, impulse, electric field intensity etc.

TYPES OF VECTOR Axial and Polar Vectors

Vectors, which have some starting point or point of application are called polar vectors. E.g., displacement, force but those vectors representing rotational effects and are always along the axis of rotation in accordance with right hand screw rule are axial vectors.

E.g., angular velocity, angular acceleration, angular displacement, torque etc. Unit Vector : A vector having unit magnitude is called a unit vector. Thus, unit vector of a vector V is | | ˆ V V V  = V V where | V| = V = Magnitude of V . k j

iˆ,ˆ and ˆare unit v ectors along x, y and z ax is respectiv ely..

1 | ˆ | | ˆ | | ˆ |i  j  k  .

Unit vector along a direction is unique and have no unit. Coplanar vectors are vectors lying in same plane.

I

I and are in the plane of paper so they are CoplanerJ J

3

Description of Motion in

Two and Three Dimensions

C H A P T E R

AIEEE Syllabus

Scalars and Vectors, Vector addition and Subtraction, Zero Vector, Scalar and Vector products, Unit Vector, Resolution of a Vector. Relative Velocity, Motion in a plane, Projectile Motion, Uniform Circular Motion.

THIS CHAPTER

COVERS :

 Scalar and Vector

product

 Relative Motion

 Projectile Motion

Collinear vectors are vectors having common line of action. These are of two types (i) Parallel or like (angle between them is 0°)

B A

A and are parallel vectors ( = 0°)B 

Parallel vectors having equal magnitude are known as equal vectors (ii) Antiparallel or unlike (angle between them is 180°).

C D

C and are antiparallel vectors (q = 180°)D

Antiparallel vectors of equal megnitude are known as negative vectors of each other.

Null vector (₀) : Two opposite vectors added to form a null vector it has zero magnitude. If A and B are two negative vectors then A B0 vector

VECTOR ADDITION

1. Triangle Law of Vector Addition : If a and b are two vectors represented as sides of a triangle in same order then other side cin opposite order is the resultant.

c = a+ b

a

b

2. Polygon Law : If a number of vectors are represented as sides of a polygon in same order then the side which closes the polygon in opposite order in the resultant.

a b c C = a + b + c

Vector addition obeys commutative law (ABBA) and associative law (AB)C=A(BC) 3. Parallelogram Law of Vector Addition : If two vectors having common origin are represented both in

magnitude and direction as the two adjacent sides of a parallelogram, then the diagonal which originates from the common origin represents the resultant of these two vectors. The result are listed below: (a) _RAB.    R A B (b) 2 2 1/2 ) cos 2 ( | |R  A B  AB  (c)           cos sin tan , cos sin tan A B A B A B

(d) If |A||B|x(say), then R = x 2(1cos) =

2 cos 2x  and 2    

 i.e., resultant bisect angle between A and B.

(e) If |A | |B| then  < 

(f) R_max = A + B, when  = 0 and R_min = |A – B| when  = 180°. (g) R2A2B2, if  = 90° i.e., A and B are perpendicular.. (h) If |A||B||R| then  = 120°.

(i) If R is perpendicular to A, then

B A    cos and A2 _{+ R}2 _{= B}2_.

(j) For n equal vectors acting at a point such that angle between them are equal       n 360 , the resultant is zero. VECTOR SUBTRACTION

Subtraction of vector B from vector A is simply addition of vectorB with A i.e., ABA( B )

Using parallelogram law

–B A O  ( – ) A–B B  Result : R =





Note : If |A||B|x(say), then R = x 2(1cos) =

2 sin 2x  .

RESOLUTION OF VECTORS

Any vector _V can be represented as a sum of two vectors _P and Q which are in same plane as Q

P

V   , where  and  are two real numbers. We say that _V has been resolved in two component vector P and Q along _P and _Q respectively..

Rectangular components in two dimensions : 2 2 , ˆ ˆ , _{x y x y} y x V V V i V j V V V V V       y x V

V and are rectangular component of vector in 2-dimension.

V_x = V cos  Vy Vx V  X Y O V_y = V sin  = V cos(90 – ) V_z = zero. j V i V V  cosˆ sinˆ

Rectangular Component in Three Dimension :

A vector _V is in a space which is making ,  and  with x-axis, y-axis and z-axis respectively..

z y x V V V V    Vx Vz Vy Z X Y  _  V 2 1 2 2 2 ) ( | |V  V_x V_y V_z k V j V i V V _xˆ _yˆ _zˆ V V l V V_x  coscos  x V V m V V_y  coscos  y V V n V V_z  coscos  z k v j V i V

V  cosˆ cosˆ cosˆ

l, m, n are called direction cosines of vector _V. , 1 cos cos cos2 2 2 2 2 2         m n

l sin2sin2sin22.

Unit vector along V liˆmjˆnkˆcosiˆcosˆjcoskˆ.

SCALAR AND VECTOR PRODUCTS

Scalar (dot) Product of Two Vectors : The scalar product of two vectors a and b is defined as   cos .b ab a    a b ab b a   . cos 

If a and b are perpendicular, then a.b 0   If  < 90° , then a.b0   and if a.b0   , then  > 90°. Projection of vector a on b is 2 ) . ( b b b a . a a a2  . . 1 ˆ . ˆ ˆ . ˆ ˆ . ˆ _{i  j j k k } i

Scalar product is commutative i.e., a.bb.a. Vector Product of two Vectors :

Mathematically, if  is the angle between vectors A and B , then

n AB B A  sin ˆ   …(i)

The direction of vector AB is the same as that A

A

B B

A×B A×B

(a) (b)

 

(a) Right handed screw rule : Rotate a right handed screw from vector A to B through the smaller angle between them; then the direction of motion of screw gives the direction of vector A_B (Fig. a)

(b) Right hand thumb rule : Bend the finger of the right hand in such a way that they point in the direction of rotation from vector A to B through the smaller angle between them; then the thumb points in the direction of vector AB (Fig. b)

Some Important Points :

1. The cross product of the two vectors does not obey commutative law. As discussed above

) ( ., . ) (               B B A ie A B B A A

2. The cross product follows the distributive law i.e.,

             B C A B A C A ( )

3. The cross product of a vector with itself is a NULL vector i.e., 0 ˆ 0 sin ) ( ) (       n A A A A

4. The cross product of two vectors represents the area of the parallelogram formed by them,

(Figure., shows a parallelogram PQRS whose adjacent sides PQ and PS are represented by vectors

 A and



B respectively..

Now, area of parallelogram = QP × SM = QPAB sin Because, the magnitude of vectors AB is AB sin , hence cross product of two vectors represents the area of parallelogram formed by it. It is worth noting that area vector AB acts along the perpendicular to the plane of two vectors



A and B . 5. In case of unit vectors iˆ, ˆj,kˆ, we obtain following two important properties:

(a) iˆiˆ ˆjˆj kˆkˆ(1) (1) sin0(nˆ)0

(b) iˆjˆ(1)(1) sin90(kˆ)kˆ

where, kˆ is a unit vector perpendicular to the plane of iˆand jˆ in a direction in which a right hand screw will advance, when rotated from iˆ to jˆ

Also, ˆjiˆ(1)(1)sin90(kˆ)kˆ

Similarly, jˆkˆkˆˆj iˆ and kˆiˆiˆkˆ jˆ

6. Cross product of two vectors in terms of their rectangular components : ) ˆ ˆ ˆ ( ) ˆ ˆ ˆ (A i A j A k B i B j B k B A  _x  _y  _z  _x  _y  _z   (A_yB_z A_zB_y)iˆ(A_zB_x A_xB_z) ˆj(A_xB_y A_yB_x)kˆ z y x z y x B B B A A A k j iˆ ˆ ˆ 

7. _{Multiplication of a vector A} with a real number m

If m is positive real number then a parallel vector is obtained having magnitude m times the magnitude of A If m is negative real number then antiparallel vector is obtained having magnitude m times the magnitude of A



8. If angle between A and B is , then angle between A B  



and or between _{A andB} is (180° – ).

RELATIVE MOTION IN TWO DIMENSIONS Relative velocity :

Velocity of object A w.r.t. object B is v_AB v_A v_B,v_BA v_B v_A

1. Direction of Umbrella : A person moving one straight road has to hold his umbrella opposite ot direction of relative velocity of rain. The angle  is given by

R M v v  

tan with vertical in forward direction.

vR vM  _–v M   vRM vR Umbrella

2. Closest approach : Two objects A and B having velocities v_A and

B

v at separation x are shown in figure

vA

A B

vB

x

The relative velocity of A with respect to B is given by

B A AB v v v   vA -v_B v_AB -v_B  B A v v   tan

The above situation is similar to figure given below y is the distance of closest approach.

Now, x y   sin y v = 0 A x B  (Dire ctio n of mot ion of w.r. t. ) A B vAB  y = x sin 2 2 2 tan 1 tan A B A V V xV x y      

3. Crossing a river :

v = velocity of the man in still water.

= angle of which man swims w.r.t. normal to bank such that v_x = – v sin , v_y = v cos  D B y v cos  d v v sin  A u x 

Time taken to cross the river is given by

   cos v d v d t y

Velocity along the river

  

 _{u vsin} v_x

Distance drifted along the river _ 

x v t D ) sin ( cos    u v v d D Case I :

The Minimum time to cross the river is given by

v d 

min (whencos  = 1, = 0°, u v)

Distance drifted is given by

u v d D 

Case II :

To cross the riven straight v

u v u2 2 - drift D = 0  u – v sin  = 0 v u   sin  (v > u)

Time to cross the river straight across is given by

2 2 cos _{v u} d v d t     PROJECTILE MOTION

An object moving in space under the influence of gravity is called projectile. Two important cases of interest are discussed below :

1. Horizontal projection :

A body of mass m is projected horizontally with a speed u from a height h at the moment t = 0. The path followed by it is a parabola.

g H T  2 j gT i u gH u v  22  ˆ ˆ t = t0 t = T R H t = 0 x v x-axix y-axix y

The position at any instant t₀ is given by x = ut₀ y 02 2 1 gt  y ₂ 2 2u gx 

The velocity at any instant t₀ is given by j

gt i u v₀  ˆ ₀ˆ

2. Oblique projection : A body of mass m is projected from ground with speed u at an angle  above horizontal at the moment t = 0.

It hits the ground at a horizontal distance R at the moment t = T.

t = T₂ v uy t = 0  H t = T u cos =  ux ux  R u 1. Time of flight g u g u T  2 y 2 sin 2. Maximum height g u g u H y 2 sin 2 2 2 2    3. Horizontal range g u g u u T u R _x  2 x y  sin2 2 4. Equation of trajectory;     2 2 2 cos 2 tan u gx x y or          R x x y tan 1

5. Instantaneous velocity v  u2 (gt)2 2u(gt)sin

and direction of motion is such that,

     cos tan tan u gt v sin v cos  v (a)    cos cos u

v _{[ Horizontal component is same every where]} (b) v sin = u sin – gt

(c) When _v (velocity at any instant ‘t’) is perpendicular to u (initial velocity)   = 90° –  (i)        cot ) 90 cos( cos u u v u   v (ii)   sin g u t Applications :

1. The height attained by the particle is largest when  = 90°. In this situation, time of flight is maximum and range is minimum (zero).

2. The horizontal range is same for complimentary angles like (, 90 – ) or (45 + , 45 – ). It is maximum for  = 45°.

3. When horizontal range is maximum,

4 max R H 

4. When R is range, T is time of flight and H is maximum height, then

(a) R gT 2 tan 2   (b) R H 4 tan

5. If A and B are two points at same level such that the object passes A at t = t₁ and B at T = t₂, then

h h y t = 0 x t t= 1 t t= 2 t T= u  A _B (i) 2 sin t₁ t₂ g u T     (ii) 12 2 1 t gt h 

(iii) Average velocity in the interval AB is

v_av = ucos [ vertical displacement is zero]

CIRCULAR MOTION

An object of mass m is moving on a circular track of radius r. At t = 0, it was at A. At any moment of time ‘t’, it has moved to B, such that  AOB. Let its speed at this instant be v and direction is along the tangent. In a small time dt, it moves to B such that BOBd.

The angular displacement vector is ddkˆ

The angular velocity vector is k dt d _ˆ   .

At B, the speed of the object has become v + dv. y x O A  d r B B v+dv

The tangential acceleration is dt dv a_t 

The radial (centripetal acceleration) is r r v a_c 2 2   

The angular acceleration is dt d  

Relations among various quantities. 1. _v__r 2. r ac at dt d dt r d dt v d a           3. a_c v 4. _at _r

  