Physics
Session
Kinematics - 1
Session Opener
You fly from Delhi to Mumbai
New Delhi
Mumbai
Hyderabad
•
•
•
Then you fly from Mumbai to Hyderabad
Does it...
Session Objectives
1. Scalars and vectors
2. Definitions
3. Vector addition and vector subtraction
4. Components of vector...
Scalars And Vectors
Scalars : physical quantities that can
be completely specified by
just numbers.
Vectors : physical qua...
Vectors
OP : Position vector
uuur
Length OP meter
Directed from O to P.
Both r and θ are needed to specify P
Vectors also ...
Vectors
Component of vector
( )
1
2 2 2
1
OP OX OY
OY
tan
OX
−
= +
θ =
OX OP cos
OY OP sin
= θ
= θ
O x
y
P
θ
r
r
X
Y
Unit Vectors
O x
y
P
θ
r
r
X
Y
Unit vector : vector with magnitude
of unity
= =
= θ + θ
r
r
r
r
ˆ ˆr , r one unit
r
ˆ ˆr i...
Class Exercise
Class Exercise - 6
∧ ∧ ∧ ∧
= + = +
ur ur
A 3 i 4 j ; B 7 i 24 j .
Find such that C = B and is in
the direction of .
ur
C
u...
Vector Definitions
Null vector : vector with zero
magnitude
Equal Vectors : ˆˆa b aa bb= ⇒ =
r r
If a = b
Direction of bot...
Addition of vectors
OP PQ OQ+ =
uuur uuur uuur
(displacement O to P, then P to Q :
same as displacement O to Q)
Triangle l...
Addition of vectors
Parallelogram law
From geometry : ( )
1/2
2 2
c a b 2abcos= + + θ
bsin
tan
a bcos
θ
α =
+ θ
a b c+ =
r...
Subtraction of Vectors
Subtracting a vector from vector
Is equivalent to adding vector
b
r
a
r
( b) to a−
r r
( )d a b a b...
Class Exercise
Class Exercise - 1
The maximum resultant of 2 vectors
is 18 units. The resultant magnitude
is 12. If the resultant is
perp...
Class Exercise - 2
+ =
ur ur ur
A B R , + =
ur ur ur
A 2B P ,
ur
P
is perpendicular to . Then
ur
A
(a) A = B (b) A + B = R...
Class Exercise - 3
The components along the X and Y-axis of
are 3 m and 4 m respectively. The
components along X and Y-axi...
Class Exercise - 4
A man swims across a river that flows
at 3 m/s. The man moves in a
direction directly perpendicular to ...
Solution - 4
r
R
m
m r R . m, r , R
→ → → → → →
+ = denote
the components of velocities of man,
river and resultant during...
Multiplication of vector by a real
number
a multiplied by λ
r
b a a a= λ ⇒ λ = λ
r r r r
If is negativeλ
c a= −λ
r r
Direc...
Scalar multiplication of vectors
Scalar product (dot product)
x x y yc a.b abcos a b a b= = θ = +
r r
a.b is scalar.
a.b b...
Class Exercise
Class Exercise - 8
∧ ∧ ∧ ∧ ∧ ∧
= + = +
ur ur
A 5 i 7 j– 3k;B 2 i 2 j– ck .
If the two vectors are perpendicular
to each ot...
Class Exercise - 9
What is the angle between the two
vectors
∧ ∧ ∧ ∧ ∧ ∧
= + + = +
ur ur
A –2 i 3 j k ; B i 2 j– 4k ?
A·B ...
Class Exercise - 10
Find the component of the vector
in the direction of
the vector .
∧ ∧ ∧
= + +
ur
A 3 i 4 j 5k ∧ ∧
= +
...
Vector multiplication of vectors
Vector product (cross product)
ˆc a b is vector absin c= × = θ
r r r
( )x y y x
ˆc k a b ...
Vector multiplication of vectors
Vector product has an
orientation given by the
right-hand thumb rule.
Curl palm of your r...
Vector multiplication of vectors
A
ur
θA
ur
B
ur
C A B= ×
ur ur ur
Product of two vectors
ˆˆ ˆi j k
ˆˆ ˆj k i
ˆ ˆ ˆk i j
× =
× =
× =
Some special cases
(ii) For unit vectors
ˆ ˆˆ ˆ ˆ ˆi i 0...
Class Exercise
Class Exercise - 7
What is the torque of a force
acting
at the point
and about the origin?
∧ ∧ ∧ 
= + ÷ ÷
 
r
F 2 i ...
Class Exercise - 5
The angle between two forces of
equal magnitude acting at a point,
such that the resultant force also h...
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Kinematics-1

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Session-1 of Kinematics for Arc-IITJEE study program

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Kinematics-1

  1. 1. Physics
  2. 2. Session Kinematics - 1
  3. 3. Session Opener You fly from Delhi to Mumbai New Delhi Mumbai Hyderabad • • • Then you fly from Mumbai to Hyderabad Does it mean that you have flown from Delhi to Hyderabad ?
  4. 4. Session Objectives 1. Scalars and vectors 2. Definitions 3. Vector addition and vector subtraction 4. Components of vectors 5. Multiplication of vectors
  5. 5. Scalars And Vectors Scalars : physical quantities that can be completely specified by just numbers. Vectors : physical quantities that cannot be completely specified by their magnitude alone. They also need a ‘direction’ specification. For instance, displacement is a vector quantity.
  6. 6. Vectors OP : Position vector uuur Length OP meter Directed from O to P. Both r and θ are needed to specify P Vectors also obey the law a b b a+ = + r r r r O x y P θ r r
  7. 7. Vectors Component of vector ( ) 1 2 2 2 1 OP OX OY OY tan OX − = + θ = OX OP cos OY OP sin = θ = θ O x y P θ r r X Y
  8. 8. Unit Vectors O x y P θ r r X Y Unit vector : vector with magnitude of unity = = = θ + θ r r r r ˆ ˆr , r one unit r ˆ ˆr i r cos j r sin O x y θ ˆrˆj ˆi
  9. 9. Class Exercise
  10. 10. Class Exercise - 6 ∧ ∧ ∧ ∧ = + = + ur ur A 3 i 4 j ; B 7 i 24 j . Find such that C = B and is in the direction of . ur C ur Cur A The unit vector in the direction of , where 5 is the magnitude of . Magnitude of the vector = 25 units. Hence the value of ∧ ∧ + = ur 3 i 4 j A 5ur Aur B ∧ ∧ ∧ ∧   +  = = +     ur 3 i 4 j C 25 15 i 20 j 5 Solution :
  11. 11. Vector Definitions Null vector : vector with zero magnitude Equal Vectors : ˆˆa b aa bb= ⇒ = r r If a = b Direction of both are the same ˆˆa b=
  12. 12. Addition of vectors OP PQ OQ+ = uuur uuur uuur (displacement O to P, then P to Q : same as displacement O to Q) Triangle law of addition a b c+ = r r r O P Q b r a r c r α θ
  13. 13. Addition of vectors Parallelogram law From geometry : ( ) 1/2 2 2 c a b 2abcos= + + θ bsin tan a bcos θ α = + θ a b c+ = r r r b r a r c r αθ a b θ bsinθ bcosθ
  14. 14. Subtraction of Vectors Subtracting a vector from vector Is equivalent to adding vector b r a r ( b) to a− r r ( )d a b a b= − = + − r r r r r b r a r b− r d r
  15. 15. Class Exercise
  16. 16. Class Exercise - 1 The maximum resultant of 2 vectors is 18 units. The resultant magnitude is 12. If the resultant is perpendicular to the smaller vector , find the magnitude of the two vectors. R A B A + B = 18 [The maximum amplitude of the resultant is when they are collinear.] Also 122 + A2 = (18 – A)2 . Solving, we get A = 5 and B = 13 Solution :
  17. 17. Class Exercise - 2 + = ur ur ur A B R , + = ur ur ur A 2B P , ur P is perpendicular to . Then ur A (a) A = B (b) A + B = R (c) A = R (d) B = R Since the given triangle is a right-angled triangle B = R, where point O is the mid point of the hypotenuse. P A B B R o Solution :
  18. 18. Class Exercise - 3 The components along the X and Y-axis of are 3 m and 4 m respectively. The components along X and Y-axis of the vectors is 2 m and 6 m respectively. Find the components of along X and Y- axis? ur A + ur ur A B ur B ∧ ∧ ∧ ∧ ∧ ∧ = + + + + = + ur Let B x i y j .Then (3 x) i (4 y) j 2 i 6 j So x = –1 and y = 2. ∧ ∧ = + ur Hence B – i 2 j Solution :
  19. 19. Class Exercise - 4 A man swims across a river that flows at 3 m/s. The man moves in a direction directly perpendicular to the flow of the river at 4 m/s. If the width of the river is 100 m, then find the time taken by the man to reach the opposite bank.
  20. 20. Solution - 4 r R m m r R . m, r , R → → → → → → + = denote the components of velocities of man, river and resultant during the motion. The resultant direction in which the man moves is not along the shortest line joining the two banks. But nevertheless the component of the velocities in the Y direction is 4 m/s. Hence the man takes , i.e 25 s to cross over to the opposite bank. 100 4
  21. 21. Multiplication of vector by a real number a multiplied by λ r b a a a= λ ⇒ λ = λ r r r r If is negativeλ c a= −λ r r Direction : opposite a r
  22. 22. Scalar multiplication of vectors Scalar product (dot product) x x y yc a.b abcos a b a b= = θ = + r r a.b is scalar. a.b b.a= r r r r r r b r a r bcosθ θ
  23. 23. Class Exercise
  24. 24. Class Exercise - 8 ∧ ∧ ∧ ∧ ∧ ∧ = + = + ur ur A 5 i 7 j– 3k;B 2 i 2 j– ck . If the two vectors are perpendicular to each other, then find c. If the vectors A and B are perpendicular to each other, then × = ur ur A B 0 × = + + = =− ur ur Therefore A B 10 14 3c 0. So,c 8 Solution :
  25. 25. Class Exercise - 9 What is the angle between the two vectors ∧ ∧ ∧ ∧ ∧ ∧ = + + = + ur ur A –2 i 3 j k ; B i 2 j– 4k ? A·B = AB cosθ = –2 + 6 – 4 = 0. Hence the two vectors are perpendicular to each other. Solution :
  26. 26. Class Exercise - 10 Find the component of the vector in the direction of the vector . ∧ ∧ ∧ = + + ur A 3 i 4 j 5k ∧ ∧ = + ur B 3 i 4 j A·B = AB cosθ. This also means that A.B is the product of the magnitude of B and the magnitude of the component of A in the direction of B. Magnitude of B = 5 units. A·B = 25. Hence the magnitude of the component of A in the direction of B is 5, which can happen only if the component vector is . ∧ ∧ +3 i 4 j Alternately, if we compare the components of the two vectors along the X, Y directions, we find that they are the same. Hence the component of A in the direction of B must be same as the vector B. Solution :
  27. 27. Vector multiplication of vectors Vector product (cross product) ˆc a b is vector absin c= × = θ r r r ( )x y y x ˆc k a b a b= − r a b b a× = − × r r r r b r a r θ
  28. 28. Vector multiplication of vectors Vector product has an orientation given by the right-hand thumb rule. Curl palm of your right hand from the first vector to the second vector keeping the thumb upright. The direction of the thumb gives the direction of resultant.
  29. 29. Vector multiplication of vectors A ur θA ur B ur C A B= × ur ur ur
  30. 30. Product of two vectors ˆˆ ˆi j k ˆˆ ˆj k i ˆ ˆ ˆk i j × = × = × = Some special cases (ii) For unit vectors ˆ ˆˆ ˆ ˆ ˆi i 0 j j k k× = = × = × = = =ˆ ˆˆ ˆ ˆ ˆi . i j. j k. k 1 = = = ˆ ˆi. j 0 ˆˆj.k 0 ˆ ˆk . i 0 × = θ = ur ur A B AB sin 0(i) (for parallel vectors) = θ = ur ur A.B AB cos 0 (for perpendicular vectors)
  31. 31. Class Exercise
  32. 32. Class Exercise - 7 What is the torque of a force acting at the point and about the origin? ∧ ∧ ∧  = + ÷ ÷   r F 2 i – 3 j 4k N ∧ ∧ ∧ = + + r r 3 i 2 j 3k  τ = ×  r r r Note r F Solution : ∧ ∧ ∧ ∧ ∧ ∧ = + + = + r r r 3 i 2 j 3k,F 2 i – 3 j 4k ∧ ∧ ∧ ∧ ∧ ∧ ∧ = × = + + + = r F r f –9k– 12 j – 4k 8 i 6 j 9 i 17 i Hence the moment of the force
  33. 33. Class Exercise - 5 The angle between two forces of equal magnitude acting at a point, such that the resultant force also has the same magnitude is___. Solution : F F R=F θ 120∴ θ = o 2 2 2 2 2 R F F 2F.F cos (but R F given) F 2F (1 cos ) = + + θ = = + θ 1 1 cos 2 1 cos 2 ⇒ = + θ ⇒ θ = −
  34. 34. Thank you

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