Motion In a Plane LNN [Lecture Note].pdf

31
NEET 2024 LECTURE NOTE - PHYSICS [FIRST YEAR]
MOTION IN A PLANE
Physical quantities can be classified into two types. Scalar quantities and vector quantities. Those
quantities which have only magnitude are known as scalars.
Examples : Mass, distance, speed etc.
Those quantities which have both magnitude and direction but do not obey laws of vector addition are
not vectors such as electric current.
A vector is represented by straight line with an arrow head. The length of the lines gives the magnitude
of the vector and arrow head gives the direction of the vector.
The vectors are represented by boldface letters or with an arrow over simple letter. If A is a vector then
it is represented by A

and magnitude of that vector is represented by A

or A
A
Types of Vectors
1. Equal vectors : Two vectors are equal if their magnitude and direction same
2. Negative vector : A vector is said to be negative vector if the magnitude is same but direction is
opposite
3. Parallel vectors : Vectors in the same direction
4. Antiparallel vectors : Vectors in the opposite direction
5. Zero or null vector : A vector whose magnitude is zero and direction is in determinate
6. Unit vector : It is a vector of unit magnitude If A is a vector then its unit vector is denoted by A

Then
A
A
A




A A A
 
  
7. Co-initial vectors : The vectors which have the same starting point are called coinitial vectors
8. Coplanar vectors : Three or more vectors are lying in the same plane or parallel to the same plane are
known as coplanar vectors
9. Collinear vectors : Vectors which lie along the same line or parallel lines are known to be collinear
vectors

Brilliant STUDY CENTRE
32
Multiplication of a vector by a scalar
When a vector A

is multiplied by scalar K, then the resultant vector has magnitude KA, and direction
same as that of A

Addition of Vectors
1. Triangular Law of vector addition
If two vectors are arranged as the adjacent sides of a triangle, then the third side taken in the opposite
direction will represent their resultant
Then R

is the resultant of A

and B

2. Parallelogram Law of vector addition
If two vectors are arranged as the adjacent sides of a parallelogram, then the diagonal passing through
the point of intersection of these vectors represent their resultant
Here R

is the resultant of A

and B

3. Polygon Law of vector addition
If we need to find resultant of P, Q


and S

. Then join the head of P

to the tail of Q

, then head of Q

to the tail of S

. Then join tail of P

and head of S

and it will represent their resultant.
Here R

represent the resultant of P, Q


and S

.
Magnitude and direction of the resultant vector

33
   
2 2
R A Bcos Bsin
    
= 2 2 2 2 2
A 2ABcos B cos B sin
   
=  
2 2 2 2
A 2ABcos B sin cos
   
= 2 2
A 2ABcos B 1
  
 is the angle made by R

with A

then
Bsin
tan d
A Bcos


 
1 Bsin
tan
A Bcos
 
 
   
 
 
Case (i)
A

and B

in same direction
0
 
2 2
R A B 2ABcos
   
= 2 2
A B 2AB 1
  
= 2 2
A B 2AB
 
=  
2
A B A B
  
Case (ii)
A

and B

in opposite direction
o
180
 
2 2
R A B 2ABcos180
  
= 2 2
A B 2ABx 1
  

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= 2 2
A B 2AB
 
=  
2
A B

= A – B
Case (iii)
A

and B

are perpendicular
o
90
 
2 2
R A B 2ABcos90
  
= 2 2
A B 2AB 0
  
= 2 2
A B

Note:
Max value of resultant is A + B and minimum value of resultant is A – B
A B R A B
    
Subtraction of Vectors
 
A B A B
   
 
 
Two subtract two vectors, we are adding one vector with negative vector of other vector
 
2 2
A B A B 2ABcos 180
     
 
= 2 2
A B 2ABx cos
   
= 2 2
A B 2ABcos
  

35
Resolution of Vectors
Ax
cos
A
 
Ay
sin
A
 
Ax Acos
  Ay Asin
 
A Ax Ay
 
   ˆ ˆ
A Axi Ayj
 

ˆ ˆ
A Acos i Asin j
   

angle made by vector with horizontal = 
1 Ax
tan
Ay
 
1 Ay
tan
Ax
  
   
 
angle made by vector with vertical = 1

1 Ax
tan
Ay
 

36
1 1 Ax
tan
Ay
  
   
 
Scalar Product of Vectors
The dot product of two vector is scalar and it is given by
A.B A B cos
 
 
 
Where  is the angle between A

and B

B. A B A cos
 
 
 
*) The scalar product is scalar. it can be +ve, –ve or zero
If 0 90
   scalar product is positive
If 90 180
   scalar product is negative
If o
90
  , scalar product is zero
*)
2
2
A.A A A cos A A
   
    
*) Angle between two vectors is given by
A . B A B cos
 
 
 
A.B
cos
A B
 
 
 
1 A.B
cos
A B

 
 
 
 
 
 
 
*) If two vectors are perpendicular, then their dot product will be zero
*) ˆ ˆ ˆ ˆ ˆ ˆ
i.i j.j k.k 1
  
i.j j.k k.i 0
  
If x y z
ˆ ˆ ˆ
A A i A j A k
  

and x y z
ˆ ˆ ˆ
B B i B j B k
  

then x x y y z z
A.B A B A B A B
  
 
*) Component of A

along B

is given by

37
A.B
Acos
B
 
 

Vector or Cross Product
The product of two vectors are given by
ˆ
A B A B sin n
  
 
 
Where n̂ is a unit vector perpendicular to both A

and B

*) A B B A
  
 
 
*)
A B
sin
A B

 
 
 
*) A A 0
 
 
i i j j k k 0
     
*)  
ˆ ˆ ˆ ˆ ˆ
i j k j k
    
*)  
ˆ ˆ ˆ ˆ ˆ
j k i k j
    
*)  
ˆ ˆ ˆ ˆ ˆ
k i j i k
    
Projectile Motion
When a body is thrown at angle with horizontal, then its motion is governed by gravitational acceleration.
(neglect air resistance)
The body is known as projectile and its motion is known as projectile motion
Horizontal motion Vertical motion
x
u u cos
  y
u u sin
 
x
a 0
 y
a g
 
2
x x x
1
S u t a t
2
  2
y y y
1
S u t a t
2
 
Time of Flight (T)

38
at t = T, y
S 0

2
y y y
1
S u t a t
2
 
    2
1
0 u sin T g T
2
   
 
2
1
gT usin T
2
 
2usin
T
g


Range (R)
at t = T, x
S R

2
x x
1
S u t a t
2
  
 
R u cos T 0
   
2usin
R u cos
g

 
2
u sin 2
R
g


Maximum Height (H)
When y
S H
 , then y
V 0

2 2
y y y y
V U 2a S
 
 
2 2
0 u sin 2 g H
  
2 2
2gH U sin
 
2 2
U sin
H
2g


Note :
* Range will be maximum when angle of projection is 45o

39
* for angle of projection  and 90  range will be same
Velocity of projectile at any time
y
ˆ ˆ
v v i v j
  

x x x
v u a t ucos 0 ucos
      
 
y y y
v u a t usin g t usin gt
      
2 2
x y
v v v v
  

   
2 2
v u cos usin gt
   
Velocity of Projectile at any Height
x y
ˆ ˆ
v v i v j
 

2 2 2 2
x x x x
v u 2a s u cos
   
 
2 2 2 2 2 2
y y y y
v u 2a s u sin 2 g h u sin 2gh
      
2 2
x y
v v v
 
 
2 2 2 2
v u cos u sin 2gh
  
2
v u 2gh
 

40
Angle made by projectile at any time
  angle made by projectile with horizontal
y
x
v
tan
v
 
y y
x x
u a t
tan
u a t

 

 
usin g t
tan
ucos
 
 

1 usin gt
tan
u cos
  
 
   

 
Angle made by projectile at any height
y
x
v
tan
v
 
2
y y y
2
x x
u 2a s
tan
u 2a s

 
 
2 2
2 2
u sin 2gh
tan
u cos

 

2 2
u sin 2gh
tan
u cos

 

2 2
1 u sin 2gh
tan
ucos

 
 
 
 
 

 

41
Projectile projected from the top of a building (Projected Horizontally)
x
u u
 y
u 0

x
a 0
 y
a g
 
Time of flight (T)
at t = T, y
s h
 
2
y y y
1
s u t a t
2
 
  2
1
h 0 T g T
2
    
2
1
h gT
2

2
2h
T
g

2h
T
g

Range (R)
at t = T, y
s R

2
x x x
1
s u t a t
2
 
R = UT
2h
R u
g


42
Projectile Projected from the top of a building (Projected upwards)
Horizontal Motion Vertical motion
x
u u cos
  y
u usin
 
x
a 0
 y
a g
 
Time of flight (T)
at t = T, y
s h
 
2
y y y
1
s u t a t
2
 
  2
1
h usin T g T
2
    
Solving this equation ‘T’ will be obtained
Range(R)
at t = T, x
s R

2
x x x
1
s u t a t
2
 
R u cos T 0
  
R u cos T
 
Projectile Projected from the top of building (Projected downwards)
x
u u cos
  y
u usin
 

43
x
a 0
 y
a g
 
Time of flight (T)
at t = T, y
s h
 
2
y y y
1
s u t a t
2
 
  2
1
h u sin T g T
2
     
Solving this equation ‘T’ will be obtained
Range R ucos T
 
Relative Motion
Relative Velocity
Relative velocity of a particle A with respect to B is defined as the velocity with which A appears to move if
B is considered to be at rest.
A
V 

Velocity of A with respect to ground
B
V 

velocity of B with respect to ground
Then velocity of A with respect to B is given by
AB A B
v v v
 
  
Similarly velocity of B with respect to A is given by
BA B A
v v v
 
  
Note :
If two particles are moving in the same direction, then the magnitude of their relative velocity is the
difference of individual velocities
AB A B
v v v
 
Note :
If two particles are moving in the opposite direction, then magnitude of their relative velocity is the sum
of their individual velocities
AB A B
v v v
 
Relative acceleration
It is the rate at which relative velocity is changing acceleration of A with respect to B is given by
 
 
AB
AB A B
d v d
a v v
dt dt
  

  

44
= A B
d d
v v
dt dt

 
= A B
a a

 
AB A B
a a a
 
  
Note
In the above question, range of the projectile is given by
 
2
u sin 2
R
g a



Maximum height attained by the projectile is given by
 
2 2
u sin
H
2 g a



River-Crossing Problems
d Width of river x drift
t
MR
v  Velocity of man with respect to river
R
v  velocity of river
M
v  velocity of man with respect to ground
Vertical Motion Horizontal motion
Vel = MR
v sin
  Vel =  
MR R
v cos v
 
disp = d disp = x
MR
disp d
time
vel v sin
 
 disp = vel time

 
MR R
MR
d
x v cos v
v sin
  

Time taken to cross the river

45
MR
d
t
v sin


Drift  
MR R
MR
d
x v cos v
v sin
  

Resultant vel. of man 2 2
M MR R MR R
v v v 2v v cos
   
To cross the river in shortest time
MR
d
t
v sin


t  min. sin max
  o
90
 
 Time taken to cross the river
MR MR
d d
t
v sin90 v
 
Drift  
MR R R
MR MR
d d
x v cos90 v v
v sin90 d
   
Ve. of man 2 2
M MR R MR R
v v v 2v v cos 0
    = 2 2
MR R
v v

Cross the river in shortest path
Shortest path drift = 0
MR R
v cos v 0
 
MR R
v cos v
  
R
MR
v
cos
v

 
cos is negative, o
90
 
Time taken to cross the river
2 2
MR R
v v


46
2 2
MR MR R
MR
MR
d d
t
v sin v v
v
v
 
 
2 2
MR R
d
t
v v


drift = 0
Resultant velocity of man
2 2 R
M MR R MR R
MR
v
v v v 2v v
v
 

     
 
2 2 2
M MR R R
v v v 2v
  
2 2
M MR R
v v v
 
Rain Umbrella problems
R
v

- velocity of rain
M
v

- velocity of man
RM
v

- velocity of rain with respect to man
RM R M
v v v
 
  
Relative Motion between two Projectiles
Consider two projectiles projected simultaneously.
1st
projectile is seen from second projectile, for that second projectile makes at rest

47
Since acceleration is zero both horizontal and vertical component remains constant. Therefore it
looks like a straight line
Note
If 1 1 2 2
u cos u cos
  
then it will be a vertical straight line
If 1 1 2 2
u sin u sin
  
then it will be a horizontal straight line and they will collide after some time
 Condition for collision is 1 1 2 2
u sin u sin
  
Circular Motion
Angular displacement ( )
Angle traced by the position vector of a particle moving w.r.t some fixed point is called angular
displacement
  angular displacement
Frequency (  )
Number of revolutions described by particle in one second
Time period (T) : It is the time taken by particle to complete one revolution
Angular velocity ( ) : It is defined as the rate of change of angular displacement
angle traced
time taken t

  
average angular velocity
t

 

Instantaneous angular velocity
d
dt

 
Relation between linear velocity and angular velocity

48
arc
angle
radius

S
r

 
S r
  
S r
t t
 

 
d d
r
dt dt
 

v r
 
v r
 
 

Angular acceleration (  )
Rate of change of angular velocity is known as angular acceleration
t 0
w dw
lim
t dt
 

  

average angular acceleration avg
w
t

 

Tangential acceleration  
T

Tangential acceleration is in the direction of motion or opposite to motion, and this acceleration is
responsible for change in speed of the particle. Its magnitude is rate of change of speed of the particle
T
dv
a
dt
 or T
d v
a
dt


or
 
T
d r
a
dt

 ; T
a r
 
Centripetal acceleration (ac
)
In circular motion, direction of motion of the body changes due to centripetal force and acceleration
produced by this force is known as centripetal acceleration. Its direction is always towards centre of
the body magnitude of centripetal acceleration is given by
2
C
v
a
R
 or 2
C
a R
 
Types of circular motion
1) Uniform circular motion
If a particle moves with constant speed in a circle then that circular motion is known as uniform
circular motion

49
In uniform circular motion
tangential acceleration = 0
Centripetal acceleration =
2
v
R
Non uniform circular motion
If a particle moves with variable speed in circle, then that motion is known as non uniform circular
motion
In this motion, there will be both centripetal acceleration and tangential acceleration
C
a 0
 T
a 0

then net acceleration is given by
3 2
net C T
a a a
 
Equations of circular motion
0 t
    
2
0
1
t t
2
    
2 2
0 2
    
  final angular velocity
0
  initial angular velocity
  angular displacement

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