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Physics..1st Year - FULL NOTES

This is a discussion on Physics..1st Year - FULL NOTES within the 11th forums, part of the Classes category; SCALARS AND VECTORS. Scalars and Vectors Scalars Physical quantities which can be completely specified by 1. A number which represents ...

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    Arrow Physics..1st Year - FULL NOTES

    SCALARS AND VECTORS.

    Scalars and Vectors
    Scalars
    Physical quantities which can be completely specified by
    1. A number which represents the magnitude of the quantity.
    2. An appropriate unit are called Scalars.
    Scalars quantities can be added, subtracted multiplied and divided by usual algebraic laws. Examples
    Mass, distance, volume, density, time, speed, temperature, energy, work, potential, entropy, charge etc.
    Vectors
    Physical quantities which can be completely specified by
    1. A number which represents the magnitude of the quantity.
    2. An specific direction
    are called Vectors.
    Special laws are employed for their mutual operation.
    Examples
    Displacement, force, velocity, acceleration, momentum.
    Representation of a Vector
    A straight line parallel to the direction of the given vector used to represent it. Length of the line on a certain scale specifies the magnitude of the vector. An arrow head is put at one end of the line to indicate the direction of the given vector.
    The tail end O is regarded as initial point of vector R and the head P is regarded as the terminal point of the vector R.

    Unit Vector
    A vector whose magnitude is unity (1) and directed along the direction of a given vector, is called the unit vector of the given vector.
    A unit vector is usually denoted by a letter with a cap over it. For example if r is the given vector, then r will be the unit vector in the direction of r such that
    r = r .r
    Or
    r = r / r
    unit vector = vector / magnitude of the vector
    Equal Vectors
    Two vectors having same directions, magnitude and unit are called equal vectors.
    Zero or Null Vector
    A vector having zero magnitude and whose initial and terminal points are same is called a null vector. It is usually denoted by O. The difference of two equal vectors (same vector) is represented by a null vector.
    R - R - O
    Free Vector
    A vector which can be displaced parallel to itself and applied at any point, is known as free vector. It can be specified by giving its magnitude and any two of the angles between the vector and the coordinate axes. In 3-D, it is determined by its three projections on x, y, z-axes.
    Position Vector
    A vector drawn from the origin to a distinct point in space is called position vector, since it determines the position of a point P relative to a fixed point O (origin). It is usually denoted by r. If xi, yi, zk be the x, y, z components of the position vector r, then
    r = xi + yj + zk

    Negative of a Vector
    The vector A. is called the negative of the vector A, if it has same magnitude but opposite direction as that of A. The angle between a vector and its negative vector is always of 180.
    Multiplication of a Vector by a Number
    When a vector is multiplied by a positive number the magnitude of the vector is multiplied by that number. However, direction of the vector remain same. When a vector is multiplied by a negative number, the magnitude of the vector is multiplied by that number. However, direction of a vector becomes opposite. If a vector is multiplied by zero, the result will be a null vector.
    The multiplication of a vector A by two number (m, n) is governed by the following rules.
    1. m A = A m
    2. m (n A) = (mn) A
    3. (m + n) A = mA + nA
    4. m(A + B) = mA + mB
    Division of a Vector by a Number (Non-Zero)
    If a vector A is divided by a number n, then it means it is multiplied by the reciprocal of that number i.e. 1/n. The new vector which is obtained by this division has a magnitude 1/n times of A. The direction will be same if n is positive and the direction will be opposite if n is negative.
    Resolution of a Vector Into Rectangular Components
    Definition
    Splitting up a single vector into its rectangular components is called the Resolution of a vector.
    Rectangular Components
    Components of a vector making an angle of 90 with each other are called rectangular components.
    Procedure
    Let us consider a vector F represented by OA, making an angle O with the horizontal direction.
    Draw perpendicular AB and AC from point on X and Y axes respectively. Vectors OB and OC represented by Fx and Fy are known as the rectangular components of F. From head to tail rule of vector addition.
    OA = OB + BA
    F = Fx + Fy
    To find the magnitude of Fx and Fy, consider the right angled triangle OBA.
    Fx / F = Cos ? => Fx = F cos ?
    Fy / F = sin ? => Fy = F sin ?
    Addition of Vectors by Rectangular Components
    Consider two vectors A1 and A2 making angles ?1 and ?2 with x-axis respectively as shown in figure. A1 and A2 are added by using head to tail rule to give the resultant vector A.
    The addition of two vectors A1 and A2 mentioned in the above figure, consists of following four steps.
    Step 1
    For the x-components of A, we add the x-components of A1 and A2 which are A1x and A2x. If the x-components of A is denoted by Ax then
    Ax = A1x + A2x
    Taking magnitudes only
    Ax = A1x + A2x
    Or
    Ax = A1 cos ?1 + A2 cos ?2 ................. (1)
    Step 2
    For the y-components of A, we add the y-components of A1 and A2 which are A1y and A2y. If the y-components of A is denoted by Ay then
    Ay = A1y + A2y
    Taking magnitudes only
    Ay = A1y + A2y
    Or
    Ay = A1 sin ?1 + A2 sin ?2 ................. (2)
    Step 3
    Substituting the value of Ax and Ay from equations (1) and (2) respectively in equation (3) below, we get the magnitude of the resultant A
    A = |A| = v (Ax)2 + (Ay)2 .................. (3)
    Step 4
    By applying the trigonometric ratio of tangent ? on triangle OAB, we can find the direction of the resultant vector A i.e. angle ? which A makes with the positive x-axis.
    tan ? = Ay / Ax
    ? = tan-1 [Ay / Ax]
    Here four cases arise
    (a) If Ax and Ay are both positive, then
    ? = tan-1 |Ay / Ax|
    (b) If Ax is negative and Ay is positive, then
    ? = 180 - tan-1 |Ay / Ax|
    ( c) If Ax is positive and Ay is negative, then
    ? = 360 - tan-1 |Ay / Ax|
    (d) If Ax and Ay are both negative, then
    ? = 180 + tan-1 |Ay / Ax|
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    Re: Physics..1st Year..Chapter 1

    Addition of Vectors by Law of Parallelogram
    According to the law of parallelogram of addition of vectors, if we are given two vectors. A1 and A2 starting at a common point O, represented by OA and OB respectively in figure, then their resultant is represented by OC, where OC is the diagonal of the parallelogram having OA and OB as its adjacent sides.
    If R is the resultant of A1 and A2, then
    R = A1 + A2
    Or
    OC = OA + OB
    But OB = AC
    Therefore,
    OC = OA + AC
    is the angle opposites to the resultant.
    Magnitude of the resultant can be determined by using the law of cosines.
    R = |R| = vA1(2) + A2(2) - 2 A1 A2 cos
    Direction of R can be determined by using the Law of sines.
    A1 / sin ? = A2 / sin a = R / sin
    This completely determines the resultant vector R.
    Properties of Vector Addition
    1. Commutative Law of Vector Addition (A+B = B+A)
    Consider two vectors A and B as shown in figure. From figure
    OA + AC = OC
    Or
    A + B = R .................... (1)
    And
    OB + BC = OC
    Or
    B + A = R ..................... (2)
    Since A + B and B + A, both equal to R, therefore
    A + B = B + A
    Therefore, vector addition is commutative.

    2. Associative Law octor Addition (A + B) + C = A + (B + C) f Ve
    Consider three vectors A, B and C as shown in figure. From figure using head - to - tail rule.
    OQ + QS = OS
    Or
    (A + B) + C = R
    And
    OP + PS = OS
    Or
    A + (B + C) = R
    Hence
    (A + B) + C = A + (B + C)
    Therefore, vector addition is associative.
    Product of Two Vectors
    1.
    Scalar Product (Dot Product)
    2. Vector Product (Cross Product)
    1. Scalar Product OR Dot Product
    If the product of two vectors is a scalar quantity, then the product itself is known as Scalar Product or Dot Product.
    The dot product of two vectors A and B having angle ? between them may be defined as the product of magnitudes of A and B and the cosine of the angle ?.
    A . B = |A| |B| cos ?
    A . B = A B cos ?
    Because a dot (.) is used between the vectors to write their scalar product, therefore, it is also called dot product.
    The scalar product of vector A and vector B is equal to the magnitude, A, of vector A times the projection of vector B onto the direction of A.
    If B(A) is the projection of vector B onto the direction of A, then according to the definition of dot product.
    A . B = A B(A)
    A . B = A B cos ? {since B(A) = B cos ?}
    Examples of dot product are
    W = F . d
    P = F . V
    Commutative Law for Dot Product (A.B = B.A)
    If the order of two vectors are changed then it will not affect the dot product. This law is known as commutative law for dot product.
    A . B = B . A
    if A and B are two vectors having an angle ? between then, then their dot product A.B is the product of magnitude of A, A, and the projection of vector B onto the direction of vector i.e., B(A).
    And B.A is the product of magnitude of B, B, and the projection of vector A onto the direction vector B i.e. A(B).
    To obtain the projection of a vector on the other, a perpendicular is dropped from the first vector on the second such that a right angled triangle is obtained
    In ? PQR,
    cos ? = A(B) / A => A(B) = A cos ?
    In ? ABC,
    cos ? = B(A) / B => B(A) = B cos ?
    Therefore,
    A . B = A B(A) = A B cos ?
    B . A = B A (B) = B A cos ?
    A B cos ? = B A cos ?
    A . B = B . A
    Thus scalar product is commutative.
    Distributive Law for Dot Product
    A . (B + C) = A . B + A . C
    Consider three vectors A, B and C.
    B(A) = Projection of B on A
    C(A) = Projection of C on A
    (B + C)A = Projection of (B + C) on A
    Therefore
    A . (B + C) = A [(B + C}A] {since A . B = A B(A)}
    = A [B(A) + C(A)] {since (B + C)A = B(A) + C(A)}
    = A B(A) + A C(A)
    = A . B + A . C
    Therefore,
    B(A) = B cos ? => A B(A) = A B cos ?1 = A . B
    And C(A) = C cos ? => A C(A) = A C cos ?2 = A . C
    Thus dot product obeys distributive law.
    2. Vector Product OR Cross Product
    When the product of two vectors is another vector perpendicular to the plane formed by the multiplying vectors, the product is then called vector or cross product.
    The cross product of two vector A and B having angle ? between them may be defined as "the product of magnitude of A and B and the sine of the angle ?, such that the product vector has a direction perpendicular to the plane containing A and B and points in the direction in which right handed screw advances when it is rotated from A to B through smaller angle between the positive direction of A and B".
    A x B = |A| |B| sin ? u
    Where u is the unit vector perpendicular to the plane containing A and B and points in the direction in which right handed screw advances when it is rotated from A to B through smaller angle between the positive direction of A and B.
    Examples of vector products are
    (a) The moment M of a force about a point O is defined as
    M = R x F
    Where R is a vector joining the point O to the initial point of F.
    (b) Force experienced F by an electric charge q which is moving with velocity V in a magnetic field B
    F = q (V x B)
    Physical Interpretation of Vector OR Cross Product
    Area of Parallelogram = |A x B|
    Area of Triangle = 1/2 |A x B|



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    Re: Physics..1st Year..Chapter 1

    thanks friend nice sharing

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    Re: Physics..1st Year..Chapter 1

    thank you very much .I pray to god that you get everything you wish for in your life inshallah

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    Re: Physics..1st Year..Chapter 1

    very nice thanks

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    Re: Physics..1st Year..Chapter 1

    thnx nice sharing

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    Re: Physics..1st Year..Chapter 1

    nice sharing

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    Re: Physics..1st Year..Chapter 1

    WORK, POWER AND ENERGY

    Work, Power and Energy

    Work

    Work is said to be done when a force causes a displacement to a body on which it acts.
    Work is a scalar quantity. It is the dot product of applied force F and displacement d.


    Diagram Coming Soon
    W = F . d
    W = F d cos θ .............................. (1)
    Where θ is the angle between F and d.
    Equation (1) can be written as
    W = (F cos θ) d
    i.e., work done is the product of the component of force (F cos θ) in the direction of displacement and the magnitude of displacement d.
    equation (1) can also be written as
    W = F (d cos θ)
    i.e., work done is the product of magnitude of force F and the component of the displacement (d cos θ) in the direction of force.

    Unit of Work
    M.K.S system → Joule, BTU, eV
    C.G.S system → Erg
    F.P.S system → Foot Pound
    1 BTU = 1055 joule
    1 eV = 1.60 x 10(-19)

    Important Cases
    Work can be positive or negative depending upon the angle θ between F and d.

    Case I
    When θ = 0 i.e., when F and d have same direction.
    W = F . d
    W = F d cos 0 .............. {since θ = 0}
    W = F d .......................... {since cos 0 = 1}
    Work is positive in this case.

    Case II
    When θ = 180 i.e., when F and d have opposite direction.

    W = F . d
    W = F d cos 180 ............................ {since θ = 180}
    W = - F d ......................................... {since cos 180 = -1}
    Work is negative in this case

    Case III
    When θ = 90 i.e, when F and d are mutually perpendicular.
    W = F . d
    W = F d cos 90 ............................ {since θ = 90}
    W = 0 ........................................... {since cos 90 = 0}

    Work Done Against Gravitational Force
    Consider a body of mass 'm' placed initially at a height h(i), from the surface of the earth. We displaces it to a height h(f) from the surface of the earth. Here work is done on the body of mass 'm' by displacing it to a height 'h' against the gravitational force.
    W = F . d = F d cos θ
    Here,
    F = W = m g
    d = h(r) - h(i) = h
    θ = 180
    {since mg and h are in opposite direction}
    Since,
    W = m g h cos 180
    W = m g h (-1)
    W = - m g h
    Since this work is done against gravitational force, therefore, it is stored in the body as its potential energy (F.E)
    Therefore,
    P . E = m g h


    Power

    Power is defined as the rate of doing work.
    If work ΔW is done in time Δt by a body, then its average power is given by P(av) = ΔW / Δt
    Power of an agency at a certain instant is called instantaneous power.

    Relation Between Power and Velocity
    Suppose a constant force F moves a body through a displacement Δd in time Δt, then
    P = ΔW / Δt
    P = F . Δd / Δt ..................... {since ΔW = F.Δd}
    P = F . Δd / Δt
    P = F . V ................................. {since Δd / Δt = V}
    Thus power is the dot product of force and velocity.

    Units of Power
    The unit of power in S.I system is watt.
    P = ΔW / Δt = joule / sec = watt
    1 watt is defined as the power of an agency which does work at the rate of 1 joule per second.
    Bigger Units → Mwatt = 10(6) watt
    Gwatt = 10(9) watt
    Kilowatt = 10(3) watt
    In B.E.S system, the unit of power is horse-power (hp).
    1 hp = 550 ft-lb/sec = 746 watt


    Energy

    The ability of a body to perform work is called its energy. The unit of energy in S.I system is joule.

    Kinetic Energy
    The energy possessed by a body by virtue of its motion is called it kinetic energy.
    K.E = 1/2 mv2
    m = mass,
    v = velocity

    Prove K.E = 1/2 mv2

    Proof
    Kinetic energy of a moving body is measured by the amount of work that a moving body can do against an unbalanced force before coming to rest.
    Consider a body of mas 'm' thrown upward in the gravitational field with velocity v. It comes to rest after attaining height 'h'. We are interested in finding 'h'.
    Therefore, we use
    2 a S = vf2 - vi2 ............................ (1)
    Here a = -g
    S = h = ?
    vi = v (magnitude of v)
    vf = 0
    Therefore,
    (1) => 2(-g) = (0)2 - (v)2
    -2 g h = -v2
    2 g h = v2
    h = v2/2g
    Therefore, Work done by the body due to its motion = F . d
    = F d cos θ
    Here
    F = m g
    d = h = v2 / 2g
    θ = 0
    Therefore, Work done by the body due to its motion = (mg) (v2/2g) cos0
    = mg x v2 / 2g
    = 1/2 m v2
    And we know that this work done by the body due to its motion.
    Therefore,
    K.E = 1/2 m v2

    Potential Energy
    When a body is moved against a field of force, the energy stored in it is called its potential energy.
    If a body of mass 'm' is lifted to a height 'h' by applying a force equal to its weight then its potential energy is given by
    P.E = m g h
    Potential energy is possessed by
    1. A spring when it is compressed
    2. A charge when it is moved against electrostatic force.

    Prove P.E = m g h OR Ug = m g h

    Proof
    Consider a ball of mass 'm' taken very slowly to the height 'h'. Therefore, work done by external force is
    Wex = Fex . S = Fex S cos θ .................................. (1)
    Since ball is lifted very slowly, therefore, external force in this case must be equal to the weight of the body i.e., mg.
    Therefore,
    Fex = m g
    S = h
    θ = 0 ......................... {since Fex and h have same direction}
    Therefore,
    (1) => Wex = m g h cos 0
    Wex = m g h .................................................. ................. (2)
    Work done by the gravitational force is
    Wg = Fg . S = Fg S cosθ ................................................. (3)
    Since,
    Fg = m g
    S = h
    θ = 180 ...................... {since Fg and h have opposite direction}
    Therefore,
    (3) => Wg = m g h cos 180
    Wg = m g h (-1)
    Wg = - m g h .................................................. ................. (4)
    Comparing (2) and (4), we get
    Wg = -Wex
    Or
    Wex = - Wg
    The work done on a body by an external force against the gravitational force is stored in it as its gravitational potential energy (Ug).
    Therefore,
    Ug = Wex
    Ug = - Wg .............................. {since Wex = -Wg}
    Ug = -(-m g h) ..................... {since Wg = - m g h}
    Ug = m g h ............................................... Proved

    Absolute Potential Energy
    In gravitational field, absolute potential energy of a body at a point is defined as the amount of work done in moving it from that point to a point where the gravitational field is zero.

    Determination of Absolute Potential Energy
    Consider a body of mass 'm' which is lifted from point 1 to point N in the gravitational field. The distance between 1 and N is so large that the value of g is not constant between the two points. Hence to calculate the work done against the force of gravity, the simple formula W = F .d cannot be applied.
    Therefore, in order to calculate work done from 1 to N, we divide the entire displacement into a large number of small displacement intervals of equal length Δr. The interval Δr is taken so small that the value of g remains constant during this interval.
    Diagram Coming Soon Now we calculate the work done in moving the body from point 1 to point 2. For this work the value of constant force F may be taken as the average of the forces acting at the ends of interval Δr. At point 1 force is F1 and at point 2, force is F2.


    Law of Conservation of Energy

    Statement
    Energy can neither be created nor be destroyed, however, it can be transformed from one form to another.

    Explanation
    According to this law energy may change its form within the system but we cannot get one form of energy without spending some other form of energy. A loss in one form of energy is accompanied by an increase in other forms of energy. The total energy remains constant.

    Proof
    For the verification of this law in case of mechanical energy (Kinetic and potential energy). Let us consider a body of mass 'm' placed at a point P which is at a height 'h' from the surface of the earth. We find total energy at point P, point O and point Q. Point Q is at a distance of (h-x) from the surface of earth.





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    Re: Physics..1st Year..Chapter 1

    GRAVITATION
    Gravitation

    Gravitation

    The property of all objects in the universe which carry mass, by virtue of which they attract one another, is called Gravitation.


    Centripetal Acceleration of the Moon

    Newton, after determining the centripetal acceleration of the moon, formulated the law of universal gravitation.
    Suppose that the moon is orbiting around the earth in a circular orbit.
    If V = velocity of the moon in its orbit,
    Rm = distance between the centres of earth and moon,
    T = time taken by the moon to complete one revolution around the earth.
    For determining the centripetal acceleration of the moon,. Newton applied Huygen's formula which is
    a(c) = v2 / r
    For moon, am = v2 / Rm ..................... (1)
    But v = s/t = circumference / time period = 2πRm/T
    Therefore,
    v2 = 4π2Rm2 / T2
    Therefore,
    => a(m) = (4π2Rm2/T2) x (1/Rm)
    a(m) = 4π2Rm / T2
    Put Rm = 3.84 x 10(8) m
    T = 2.36 x 10(6) sec
    Therefore,
    a(m) = 2.72 x 10(-3) m/s2

    Comparison Between 'am' AND 'g'
    Newton compared the centripetal acceleration of the moon 'am' with the gravitational acceleration 'g'.
    i.e., am / g = 1 / (60)2 ................. (1)
    If Re = radius of the earth, he found that
    Re2 / Rm2 - 1 / (60)2 ......................... (2)
    Comparing (1) and (2),
    am / g = Re2 / Rm2 ..................................... (3)
    From equation (3), Newton concluded that at any point gravitational acceleration is inversely to the square of the distance of that point from the centre of the earth. It is true of all bodies in the universe. This conclusion provided the basis for the Newton' Law of Universal Gravitation.
    @import "/extensions/GoogleAdSense/GoogleAdSense.css";

    Newton's Law of Universal Gravitation

    Consider tow bodies A and B having masses mA and mB respectively.


    Diagram Coming Soon
    Let,
    F(AB) = Force on A by B
    F(BA) = Force on B by A
    r(AB) = displacement from A to B
    r(BA) = displacement from B to A
    r(AB) = unit vector in the direction of r(AB).
    r(BA) = unit vector in the direction of r(BA).

    From a(m) / g = Re2 / Rm2, we have
    F(AB) ∞ 1 / r(BA)2 ......................... (1)
    Also,
    F(AB) ∞ m(A) ............................... (2)
    F(BA) ∞ m(B)
    According to the Newton's third law of motion
    F(AB) = F(BA) .................... (for magnitudes)
    Therefore,
    F(AB) ∞ m(B) ................................ (3)
    Combining (1), (2) and (3), we get
    F(AB) ∞ m(A)m(B) / r(BA)2
    F(AB) = G m(A)m(B) / r(BA)2 ........................... (G = 6.67 x 10(-11) N - m2 / kg2)

    Vector Form
    F(AB) = - (G m(A)m(B) / r(BA)2) r(BA)
    F(BA) = - (G m(B)m(A) / r(AB)22) r(AB)
    Negative sign indicates that gravitational force is attractive.

    Statement of the Law
    "Every body in the universe attracts every other body with a force which is directly proportional to the products of their masses and inversely proportional to the square of the distance between their centres."


    Mass and Average Density of Earth

    Let,
    M = Mass of an object placed near the surface of earth
    M(e) = Mass of earth
    R(e) = Radius of earth
    G = Acceleration due to gravity
    According to the Newton's Law of Universal Gravitation.
    F = G M Me / Re2 ............................. (1)
    But the force with which earth attracts a body towards its centre is called weight of that body.


    Diagram Coming Soon
    Therefore,
    F = W = Mg
    (1) => M g = G M Me / Re2
    g = G Me / Re2
    Me = g Re2 / G .................................. (2)
    Put
    g = 9.8 m/sec2,
    Re = 6.38 x 10(6) m,
    G = 6.67 x 10(-11) N-m2/kg2, in equation (2)
    (1) => Me = 5.98 x 10(24) kg. ................... (In S.I system)
    Me = 5.98 x 10(27) gm .................................... (In C.G.S system)
    Me = 6.6 x 10(21) tons
    For determining the average density of earth (ρ),
    Let Ve be the volume of the earth.
    We know that
    Density = mass / volume
    Therefore,
    ρ = Me / Ve ........................... [Ve = volume of earth]
    ρ = Me / (4/3 Π Re3) .............. [since Ve = 4/3 Π Re3]
    ρ = 3 Me / 4 Π Re3
    Put,
    Me = 5.98 x 10(24) kg
    & Re = 6.38 x 10(6) m
    Therefore,
    ρ = 5.52 x 10(3) kg / m3


    Mass of Sun

    Let earth is orbiting round the sun in a circular orbit with velocity V.
    Me = Mass of earth
    Ms = Mass of the sun
    R = Distance between the centres of the sun and the earth
    T = Period of revolution of earth around sun
    G = Gravitational constant
    According to the Law of Universal Gravitation
    F = G Ms Me / R2 .................................... (1)
    This force 'F' provides the earth the necessary centripetal force
    F = Me V2 / R ............................................ (2)
    (1) & (2) => Me V2 / R = G Ms Me / R2
    => Ms = V2 R / G ........................................ (3)
    V = s / Π = 2Π R / T
    => V2 = 4Π2 R2 / T2
    Therefore,
    (3) => Ms = (4Π R2 / T2) x (R / G)
    Ms = 4Π2 R3 / GT2 ........................................ (4)
    Substituting the value of
    R = 1.49 x 10(11),
    G = 6.67 x 10(-11) N-m2 / kg2,
    T = 365.3 x 24 x 60 x 60 seconds, in equation (4)
    We get
    Ms = 1.99 x 10(30) kg

    Variation of 'g' with Altitude
    Suppose earth is perfectly spherical in shape with uniform density ρ. We know that at the surface of earth
    g = G Me / Re2
    where
    G = Gravitational constant
    Me = Mass of earth
    Re = Radius of earth
    At a height 'h' above the surface of earth, gravitational acceleration is
    g = G Me / (Re + h)2
    Dividing (1) by (2)
    g / g = [G Me / Re2] x [(Re + h)2 / G Me)
    g / g = (Re + h)2 / Re2
    g / g = [Re + h) / Re]2
    g / g = [1 + h/Re]2
    g / g = [1 + h/Re]-2
    We expand R.H.S using Binomial Formula,
    (1 + x)n = 1 + nx + n(n-1) x2 / 1.2 + n(n + 1)(n-2)x3 / 1.2.3 + ...
    If h / Re < 1, then we can neglect higher powers of h / Re.
    Therefore
    g / g = 1 - 2 h / Re
    g = g (1 - 2h / Re) ................................. (3)
    Equation (3) gives the value of acceleration due to gravity at a height 'h' above the surface of earth.
    From (3), we can conclude that as the value of 'h' increases, the value of 'g' decreases.

    Variation of 'g' with Depth
    Suppose earth is perfectly spherical in shape with uniform density ρ.
    Let
    Re = Radius of earth
    Me = Mass of earth
    d = Depth (between P and Q)
    Me = Mass of earth at a depth 'd'
    At the surface of earth,
    g = G Me / Re2 ...................................... (1)
    At a depth 'd', acceleration due to gravity is
    g = G Me / (Re - d)2 ........................... (2)
    Me = ρ x Ve = ρ x (4/3) π Re3 = 4/3 π Re3 ρ
    Me = ρ x Ve = ρ x (4/3) π (Re - d)3 = 4/3 π (Re - d)3 ρ
    Ve = Volume of earth
    Substitute the value of Me in (1),
    (1) => g = (G / Re2) x (4/3) π Re3 ρ
    g = 4/3 π Re ρ G ................................. (3)
    Substitute the value of Me in (2)
    g = [G / Re - d)2] x (4/3) π (Re - d)3 ρ
    g = 4/3 π (Re - d) ρ G
    Dividing (4) by (3)
    g / g = [4/3 π (Re - d) ρ G] / [4/3 π Re ρ G]
    g / g = (Re - d) / Re
    g / g = 1 - d/Re
    g / g = g (1 - d / Re) ........................... (5)
    Equation (5) gives the value of acceleration due to gravity at a depth 'd' below the surface of earth
    From (5), we can conclude that as the value of 'd' increases, value of 'g' decreases.
    At the centre of the earth.
    d = Re => Re / d = 1
    Therefore,
    (5) => g = g (1-1)
    g = 0
    Thus at the centre of the earth, the value of gravitational acceleration is zero.


    Weightlessness in Satellites

    An apparent loss of weight experienced by a body in a spacecraft in orbit is called weightlessness.
    To discuss weightlessness in artificial satellites, let us take the example of an elevator having a block of mass ;m; suspended by a spring balance attached to the coiling of the elevator. The tension in the thread indicates the weight of the block.
    Consider following cases.

    1. When Elevator is at Rest
    T = m g

    2. When Elevator is Ascending with an Acceleration 'a'
    In this case
    T > m g
    Therefore, Net force = T - mg
    m a = T - m g
    T = m g + m a
    In this case of the block appears "heavier".

    3. When Elevator is Descending with an Acceleration 'a'
    In this case
    m g > T
    Therefore
    Net force = m g - T
    m a = m g - T
    T = m g - m a
    In this case, the body appears lighter

    4. When the Elevator is Falling Freely Under the Action of Gravity
    If the cable supporting the elevator breaks, the elevator will fall down with an acceleration equal to 'g'
    From (3)
    T = m g - m a
    But a = g
    Therefore
    T = m g - m g
    T = 0
    In this case, spring balance will read zero. This is the state of "weightlessness".
    In this case gravitation force still acts on the block due to the reason that elevator block, spring balance and string all have same acceleration when they fall freely, the weight of the block appears zero.
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    Artificial Gravity

    In artificial satellites, artificial gravity can be created by spinning the space craft about its own axis.
    Now we calculate frequency of revolution (v) of a space craft of length 2R to produce artificial gravity in it. Its time period be 'T' and velocity is V.





  10. #10
    iTT stranger Mavi's Avatar
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    Re: Physics..1st Year..Chapter 1

    TORQUE, ANGULAR, MOMENTUM AND EQUIVALENCE

    Torque, Angular, Momentum and Equilibrium

    Torque or Moment of Force

    Definition
    If a body is capable of rotating about an axis, then force applied properly on this body will rotate it about the axis (axis of rotating). This turning effect of the force about the axis of rotation is called torque.
    Torque is the physical quantity which produces angular acceleration in the body.

    Explanation
    Consider a body which can rotate about O (axis of rotation). A force F acts on point P whose position vector w.r.t O is r.


    Diagram Coming Soon
    F is resolved into F1 and F2. θ is the angle between F and extended line of r.
    The component of F which produces rotation in the body is F1.
    The magnitude of torgue (π) is the product of the magnitudes of r and F1.
    Equation (1) shows that torque is the cross-product of displacement r and force F.
    Torque → positive if directed outward from paper
    Torque → negative if directed inward from paper
    The direction of torque can be found by using Right Hand Rule and is always perpendicular to the plane containing r & F.
    Thus
    Clockwise torque → negative
    Counter-Clockwise torque → positive
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    Alternate Definition of Torque
    π = r x F
    |π| = r F sin θ
    |π| = F x r sin θ
    But r sin θ = L (momentum arm) (from figure)
    Therefore,
    |π| = F L
    Magnitude of Torque = Magnitude of force x Moment Arm

    Note
    If line of action of force passes through the axis of rotation then this force cannot produce torque.
    The unit of torque is N.m.

    Couple
    Two forces are said to constitute a couple if they have
    1. Same magnitudes
    2. Opposite directions
    3. Different lines of action
    These forces cannot produces transiatory motion, but produce rotatory motion.

    Moment (Torque) of a Couple
    Consider a couple composed of two forces F and -F acting at points A and B (on a body) respectively, having position vectors r1 & r2.
    If π1 is the torque due to force F, then
    π1 = r1 x F
    Similarly if π2 is the torque due to force - F, then
    π2 = r2 x (-F)
    The total torque due to the two forces is
    π = π1 + π2
    π = r1 x F + r2 x (-F)
    π = r1 x F - r2 x (-F)
    π = (r1 - r2) x F
    π = r x F
    where r is the displacement vector from B to A.
    The magnitude of torque is
    π = r F sin (180 - θ)
    π = r F sin θ .................... {since sin (180 - θ) = sin θ}
    Where θ is the angle between r and -F.
    π = F (r sin θ)
    But r sin θ is the perpendicular distance between the lines of action of forces F and -F is called moment arm of the couple denoted by d.
    π = Fd
    Thus
    [Mag. of the moment of a couple] = [Mag. of any of the forces forming the couple] x [Moment arm of the couple]
    Moment (torque) of a given couple is independent of the location of origin.


    Centre of Mass

    Definition
    The centre of mass of a body, or a system of particles, is a point on the body that moves in the same way that a single particle would move under the influence of the same external forces. The whole mass of the body is supposed to be concentrated at this point.

    Explanation
    During translational motion each point of a body moves in the same manner i.e., different particles of the body do not change their position w.r.t each other. Each point on the body undergoes the same displacement as any other point as time goes on. So the motion of one particle represents the motion of the whole body. But in rotating or vibrating bodies different particles move in different manners except one point called centre of mass. The centre of mass of a body or a system of particle is a point which represents the movement of the entire system. It moves in the same way that a single particle would move under the influence of same external forces.

    Centre of Mass and Centre of Gravity
    In a completely uniform gravitational field, the centre of mass and centre of gravity of an extended body coincides. But if gravitational field is not uniform, these points are different.

    Determination of the Centre of Mass
    Consider a system of particles having masses m1, m2, m3, ................. mn. Suppose x1, and z1, z2, z3 are their distances on z-axis, all measured from origin.


    Equilibrium

    A body is said to be in equilibrium if it is
    1. At rest, or
    2. Moving with uniform velocity
    A body in equilibrium possess no acceleration.

    Static Equilibrium
    The equilibrium of bodies at rest is called static equilibrium. For example,
    1. A book lying on a table
    2. A block hung from a string

    Dynamic Equilibrium
    The equilibrium of bodies moving with uniform velocity is called dynamic equilibrium. For example,
    1. The jumping of a paratrooper by a parachute is an example of uniform motion. In this case, weight is balanced by the reaction of the air on the parachute acts in the vertically upward direction.
    2. The motion of a small steel ball through a viscous liquid. Initially the ball has acceleration but after covering a certain distance, its velocity becomes uniform because weight of the ball is balanced by upward thrust and viscous force of the liquid. Therefore, ball is in dynamic equilibrium.

    Angular Momentum
    Definition
    The quantity of rotational motion in a body is called its angular momentum. Thus angular momentum plays same role in rotational motion as played by linear momentum in translational motion.
    Mathematically, angular momentum is the cross-product of position vector and the linear momentum, both measured in an inertial frame of reference.
    ρ = r x P

    Explanation
    Consider a mass 'm' rotating anti-clockwise in an inertial frame of reference. At any point, let P be the linear momentum and r be the position vector.
    ρ = r x P
    ρ = r P sinθ ........... (magnitude)
    ρ = r m V sinθ .......... {since P = m v)
    where,
    V is linear speed
    θ is the angle between r and P
    θ = 90 in circular motion (special case)
    The direction of the angular momentum can be determined by the Righ-Hand Rule.
    Also
    ρ = r m (r ω) sin θ
    ρ = m r2 ω sin θ

    Units of Angular Momentum
    The units of angular momentum in S.I system are kgm2/s or Js.
    1. ρ = r m V sin θ
    = m x kg x m/s
    = kg.m2/s

    2. ρ = r P sin θ
    = m x Ns
    = (Nm) x s
    = J.s

    Dimensions of Angular Momentum
    [ρ] = [r] [P]
    = [r] [m] [V]
    = L . M . L/T
    = L2 M T-1

    Relation Between Torque and Angular Momentum
    OR
    Prove that the rate of change of angular momentum is equal to the external torque acting on the body.

    Proof
    We know that rate of change of linear momentum is equal to the applied force.
    F = dP / dt
    Taking cross product with r on both sides, we get
    R x F = r x dP / dt
    τ = r x dP / dt ............................. {since r x P = τ}
    Now, according to the definition of angular momentum
    ρ = r x P
    Taking derivative w.r.t time, we get
    dρ / dt = d / dt (r x P)
    => dρ / dt = r x dP / dt + dr / dt x P
    => dρ / dt = τ + V x P .................. {since dr / dt = V}
    => dρ / dt = τ + V x mV
    => dρ / dt = τ + m (V x V)
    => dρ / dt = τ + 0 ................. {since V x V = 0}
    => dρ / dt = τ
    Or, Rate of change of Angular Momentum = External Torque ................. (Proved)







  11. #11
    iTT stranger Mavi's Avatar
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    Re: Physics..1st Year..Chapter 1

    MOTION IN TWO DIMENSIONS

    Motion in two Dimension Projectile Motion

    A body moving horizontally as well as vertically under the action of gravity simultaneously is called a projectile. The motion of projectile is called projectile motion. The path followed by a projectile is called its trajectory.
    Examples of projectile motion are
    1. Kicked or thrown balls
    2. Jumping animals
    3. A bomb released from a bomber plane
    4. A shell of a gun.

    Analysis of Projectile Motion
    Let us consider a body of mass m, projected an angle θ with the horizontal with a velocity V0. We made the following three assumptions.
    1. The value of g remains constant throughout the motion.
    2. The effect of air resistance is negligible.
    3. The rotation of earth does not affect the motion.

    Horizontal Motion
    Acceleration : ax = 0
    Velocity : Vx = Vox
    Displacement : X = Vox t

    Vertical Motion
    Acceleration : ay = - g
    Velocity : Vy = Voy - gt
    Displacement : Y = Voy t - 1/2 gt2

    Initial Horizontal Velocity
    Vox = Vo cos θ ...................... (1)

    Initial Vertical Velocity
    Voy = Vo sin θ ...................... (2)
    Net force W is acting on the body in downward vertical direction, therefore, vertical velocity continuously changes due to the acceleration g produced by the weight W.
    There is no net force acting on the projectile in horizontal direction, therefore, its horizontal velocity remains constant throughout the motion.

    X - Component of Velocity at Time t (Vx)
    Vx = Vox = Vo cos θ .................... (3)

    Y - Component of Velocity at Time t (Vy)
    Data for vertical motion
    Vi = Voy = Vo sin θ
    a = ay = - g
    t = t
    Vf = Vy = ?
    Using Vf = Vi + at
    Vy = Vo sin θ - gt .................... (4)

    Range of the Projectile (R)
    The total distance covered by the projectile in horizontal direction (X-axis) is called is range
    Let T be the time of flight of the projectile.
    Therefore,
    R = Vox x T .............. {since S = Vt}
    T = 2 (time taken by the projectile to reach the highest point)
    T = 2 Vo sin θ / g
    Vox = Vo cos θ
    Therefore,
    R = Vo cos θ x 2 Vo sin θ / g
    R = Vo2 (2 sin θ cos θ) / g
    R = Vo2 sin 2 θ / g .................. { since 2 sin θ cos θ = sin2 θ}
    Thus the range of the projectile depends on
    (a) The square of the initial velocity
    (b) Sine of twice the projection angle θ.
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    The Maximum Range
    For a given value of Vo, range will be maximum when sin2 θ in R = Vo2 sin2 θ / g has maximum value. Since
    0 ≤ sin2 θ ≤ 1
    Hence maximum value of sin2 θ is 1.
    Sin2 θ = 1
    2θ = sin(-1) (1)
    2θ = 90
    θ = 45
    Therefore,
    R(max) = Vo2 / g ; at θ = 45
    Hence the projectile must be launched at an angle of 45 with the horizontal to attain maximum range.

    Projectile Trajectory
    The path followed by a projectile is referred as its trajectory.
    We known that
    S = Vit + 1/2 at2
    For vertical motion
    S = Y
    a = - g
    Vi = Voy = Vo sin θ
    Therefore,
    Y = Vo sinθ t - 1/2 g t2 ....................... (1)
    Also
    X = Vox t
    X = Vo cosθ t ............ { since Vox = Vo cosθ}
    t = X / Vo cos θ

    (1) => Y = Vo sinθ (X / Vo cos θ) - 1/2 g (X / Vo cos θ)2
    Y = X tan θ - gX2 / 2Vo2 cos2 θ
    For a given value of Vo and θ, the quantities tanθ, cosθ, and g are constant, therefore, put
    a = tan θ
    b = g / Vo2 cos2θ
    Therefore
    Y = a X - 1/2 b X2
    Which shows that trajectory is parabola.


    Uniform Circular Motion

    If an object moves along a circular path with uniform speed then its motion is said to be uniform circular motion.

    Recitilinear Motion
    Displacement → R
    Velocity → V
    Acceleration → a

    Circular Motion
    Angular Displacement → θ
    Angular Velocity → ω
    Angular Acceleration → α

    Angular Displacement
    The angle through which a body moves, while moving along a circular path is called its angular displacement.
    The angular displacement is measured in degrees, revolutions and most commonly in radian.
    Diagram Coming Soon s = arc length
    r = radius of the circular path
    θ = amgular displacement

    It is obvious,
    s ∞ θ
    s = r θ
    θ = s / r = arc length / radius

    Radian
    It is the angle subtended at the centre of a circle by an arc equal in length to its radius.
    Therefore,
    When s = r
    θ = 1 radian = 57.3

    Angular Velocity
    When a body is moving along a circular path, then the angle traversed by it in a unit time is called its angular velocity.
    Diagram Coming Soon Suppose a particle P is moving anticlockwise in a circle of radius r, then its angular displacement at P(t1) is θ1 at time t1 and at P(t2) is θ2 at time t2.
    Average angular velocity = change in angular displacement / time interval
    Change in angular displacement = θ2 - θ1 = Δθ
    Time interval = t2 - t1 = Δt
    Therefore,
    ω = Δθ / Δt
    Angular velocity is usually measured in rad/sec.
    Angular velocity is a vector quantity. Its direction can be determined by using right hand rule according to which if the axis of rotation is grasped in right hand with fingers curled in the direction of rotation then the thumb indicates the direction of angular velocity.

    Angular Acceleration
    It is defined as the rate of change of angular velocity with respect to time.
    Thus, if ω1 and ω2 be the initial and final angular velocity of a rotating body, then average angular acceleration "αav" is defined as
    αav = (ω2 - ω1) / (t2 - t1) = Δω / Δt
    The units of angular acceleration are degrees/sec2, and radian/sec2.
    Instantaneous angular acceleration at any instant for a rotating body is given by
    Angular acceleration is a vector quantity. When ω is increasing, α has same direction as ω. When ω is decreasing, α has direction opposite to ω.

    Relation Between Linear Velocity And Angular Velocity
    Consider a particle P in an object in X-Y plane rotating along a circular path of radius r about an axis through O, perpendicular to the plane of figure as shown here (z-axis).
    If the particle P rotates through an angle Δθ in time Δt,
    Then according to the definition of angular displacement.
    Δθ = Δs / r
    Dividing both sides by Δt,
    Δθ / Δt = (Δs / Δt) (1/r)
    => Δs / Δt = r Δθ / Δt
    For a very small interval of time
    Δt → 0

    Alternate Method
    We know that for linear motion
    S = v t .............. (1)
    And for angular motion
    S = r θ ................. (2)
    Comparing (1) & (2), we get
    V t = r θ
    v = r θ/t
    V = r ω ........................... {since θ/t = ω}

    Relation Between Linear Acceleration And Angular Acceleration
    Suppose an object rotating about a fixed axis, changes its angular velocity by Δω in Δt. Then the change in tangential velocity, ΔVt, at the end of this interval is
    ΔVt = r Δω
    Dividing both sides by Δt, we get
    ΔVt / Δt = r Δω / Δt
    If the time interval is very small i.e., Δt → 0 then

    Alternate Method
    Linear acceleration of a body is given by
    a = (Vr - Vi) / t
    But Vr = r ω r and Vi = r ω i
    Therefore,
    a = (r ω r - r ω i) / t
    => a = r (ωr- ωi) / t
    a = r α .................................... {since (ωr = ωi) / t = ω}

    Time Period
    When an object is rotating in a circular path, the time taken by it to complete one revolution or cycle is called its time period, (T).
    We know that
    ω = Δθ / Δt OR Δt = Δθ / ω
    For one complete rotation
    Δθ = 2 π
    Δt = T
    Therefore,
    T = 2 π / ω
    If ω = 2πf ........................ {since f = frequency of revolution}
    Therefore,
    T = 2π / 2πf
    => T = 1 / f

    Tangential Velocity
    When a body is moving along a circle or circular path, the velocity of the body along the tangent of the circle is called its tangential velocity.
    Vt = r ω
    Tangential velocity is not same for every point on the circular path.

    Centripetal Acceleration
    A body moving along a circular path changes its direction at every instant. Due to this change, the velocity of the body 'V' is changing at every instant. Thus body has an acceleration which is called its centripetal acceleration. It is denoted by a(c) or a1 and always directed towards the centre of the circle. The magnitude of the centripetal acceleration a(c) is given as follows
    a(c) = V2 / r, ........................... r = radius of the circular path

    Prove That a(c) = V2 / r
    Proof
    Consider a body moving along a circular path of radius of r with a constant speed V. Suppose the body moves from a point P to a point Q in a small time Δt. Let the velocity of the body at P is V1 and at Q is V2. Let the angular displacement made in this time be ΔO .
    Since V1 and V2 are perpendicular to the radial lines at P and Q, therefore, the angle between V1 and V2 is also Δ0, Triangles OPQ and ABC are similar.
    Therefore,
    |ΔV| / |V1| = Δs / r
    Since the body is moving with constant speed
    Therefore,
    |V1| = |V2| = V
    Therefore,
    ΔV / V = Vs / r
    ΔV = (V / r) Δs
    Dividing both sides by Δt
    Therefore,
    ΔV / Δt = (V/r) (V/r) (Δs / Δt)
    taking limit Δt → 0.

    Proof That a(c) = 4π2r / T2
    Proof
    We know that
    a(c) = V2 / r
    But V = r ω
    Therefore,
    a(c) = r2 ω2 / r
    a(c) = r ω2 ...................... (1)
    But ω = Δθ / Δt
    For one complete rotation Δθ = 2π, Δt = T (Time Period)
    Therefore,
    ω = 2π / T
    (1) => a(c) = r (2π / T)2
    a(c) = 4 π2 r / T2 .................. Proved
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    Tangential Acceleration
    The acceleration possessed by a body moving along a circular path due to its changing speed during its motion is called tangential acceleration. Its direction is along the tangent of the circular path. It is denoted by a(t). If the speed is uniform (unchanging) the body do not passes tangential acceleration.
    Total Or Resultant Acceleration
    The resultant of centripetal acceleration a(c) and tangential acceleration a(t) is called total or resultant acceleration denoted by a.

    Centripetal Force
    If a body is moving along a circular path with a constant speed, a force must be acting upon it. Direction of the force is along the radius towards the centre. This force is called the centripetal force by F(c).
    F(c) = m a(c)
    F(c) = m v(2) / r ..................... {since a(c) = v2 / r}
    F(c) = mr2 ω2 r ....................... {since v = r ω}
    F(c) = mrω2





  12. #12
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    Re: Physics..1st Year..Chapter 1

    Quote Originally Posted by Mavi View Post
    Addition of Vectors by Law of Parallelogram
    According to the law of parallelogram of addition of vectors, if we are given two vectors. A1 and A2 starting at a common point O, represented by OA and OB respectively in figure, then their resultant is represented by OC, where OC is the diagonal of the parallelogram having OA and OB as its adjacent sides.
    If R is the resultant of A1 and A2, then
    R = A1 + A2
    Or
    OC = OA + OB
    But OB = AC
    Therefore,
    OC = OA + AC
    is the angle opposites to the resultant.
    Magnitude of the resultant can be determined by using the law of cosines.
    R = |R| = vA1(2) + A2(2) - 2 A1 A2 cos
    Direction of R can be determined by using the Law of sines.
    A1 / sin ? = A2 / sin a = R / sin
    This completely determines the resultant vector R.
    Properties of Vector Addition
    1. Commutative Law of Vector Addition (A+B = B+A)
    Consider two vectors A and B as shown in figure. From figure
    OA + AC = OC
    Or
    A + B = R .................... (1)
    And
    OB + BC = OC
    Or
    B + A = R ..................... (2)
    Since A + B and B + A, both equal to R, therefore
    A + B = B + A
    Therefore, vector addition is commutative.

    2. Associative Law octor Addition (A + B) + C = A + (B + C) f Ve
    Consider three vectors A, B and C as shown in figure. From figure using head - to - tail rule.
    OQ + QS = OS
    Or
    (A + B) + C = R
    And
    OP + PS = OS
    Or
    A + (B + C) = R
    Hence
    (A + B) + C = A + (B + C)
    Therefore, vector addition is associative.
    Product of Two Vectors
    1.
    Scalar Product (Dot Product)
    2. Vector Product (Cross Product)
    1. Scalar Product OR Dot Product
    If the product of two vectors is a scalar quantity, then the product itself is known as Scalar Product or Dot Product.
    The dot product of two vectors A and B having angle ? between them may be defined as the product of magnitudes of A and B and the cosine of the angle ?.
    A . B = |A| |B| cos ?
    A . B = A B cos ?
    Because a dot (.) is used between the vectors to write their scalar product, therefore, it is also called dot product.
    The scalar product of vector A and vector B is equal to the magnitude, A, of vector A times the projection of vector B onto the direction of A.
    If B(A) is the projection of vector B onto the direction of A, then according to the definition of dot product.
    A . B = A B(A)
    A . B = A B cos ? {since B(A) = B cos ?}
    Examples of dot product are
    W = F . d
    P = F . V
    Commutative Law for Dot Product (A.B = B.A)
    If the order of two vectors are changed then it will not affect the dot product. This law is known as commutative law for dot product.
    A . B = B . A
    if A and B are two vectors having an angle ? between then, then their dot product A.B is the product of magnitude of A, A, and the projection of vector B onto the direction of vector i.e., B(A).
    And B.A is the product of magnitude of B, B, and the projection of vector A onto the direction vector B i.e. A(B).
    To obtain the projection of a vector on the other, a perpendicular is dropped from the first vector on the second such that a right angled triangle is obtained
    In ? PQR,
    cos ? = A(B) / A => A(B) = A cos ?
    In ? ABC,
    cos ? = B(A) / B => B(A) = B cos ?
    Therefore,
    A . B = A B(A) = A B cos ?
    B . A = B A (B) = B A cos ?
    A B cos ? = B A cos ?
    A . B = B . A
    Thus scalar product is commutative.
    Distributive Law for Dot Product
    A . (B + C) = A . B + A . C
    Consider three vectors A, B and C.
    B(A) = Projection of B on A
    C(A) = Projection of C on A
    (B + C)A = Projection of (B + C) on A
    Therefore
    A . (B + C) = A [(B + C}A] {since A . B = A B(A)}
    = A [B(A) + C(A)] {since (B + C)A = B(A) + C(A)}
    = A B(A) + A C(A)
    = A . B + A . C
    Therefore,
    B(A) = B cos ? => A B(A) = A B cos ?1 = A . B
    And C(A) = C cos ? => A C(A) = A C cos ?2 = A . C
    Thus dot product obeys distributive law.
    2. Vector Product OR Cross Product
    When the product of two vectors is another vector perpendicular to the plane formed by the multiplying vectors, the product is then called vector or cross product.
    The cross product of two vector A and B having angle ? between them may be defined as "the product of magnitude of A and B and the sine of the angle ?, such that the product vector has a direction perpendicular to the plane containing A and B and points in the direction in which right handed screw advances when it is rotated from A to B through smaller angle between the positive direction of A and B".
    A x B = |A| |B| sin ? u
    Where u is the unit vector perpendicular to the plane containing A and B and points in the direction in which right handed screw advances when it is rotated from A to B through smaller angle between the positive direction of A and B.
    Examples of vector products are
    (a) The moment M of a force about a point O is defined as
    M = R x F
    Where R is a vector joining the point O to the initial point of F.
    (b) Force experienced F by an electric charge q which is moving with velocity V in a magnetic field B
    F = q (V x B)
    Physical Interpretation of Vector OR Cross Product
    Area of Parallelogram = |A x B|
    Area of Triangle = 1/2 |A x B|
    i like ur notes thanks..............
    mb kayani likes this.

  13. #13
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    Thumb up2 Re: Physics..1st Year - FULL NOTES

    thank u very much i like it

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    Re: Physics..1st Year - FULL NOTES

    nice sharing thanks a lot may ALLAH bless u always............keep it up

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    Re: Physics..1st Year..Chapter 1

    I realy like the notes from your site and I am happy that i have made an account on your site Thank You

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    Re: Physics..1st Year..Chapter 1

    nyc sharing

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    Re: Physics..1st Year - FULL NOTES

    very good work keep it up and upload more notes of all chapters also upload 2nd year notes

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    Re: Physics..1st Year - FULL NOTES

    Here are some more note
    Short cuts Physics ~ Entry Test Preparation and Admission Help Click here

    Physics Pocket Diary of Concepts and formulas ~ Entry Test Preparation and Admission Help Click here
    Last edited by Charlie; 14th September 2013 at 12:02 PM.

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    Re: Physics..1st Year - FULL NOTES

    Physics and chemistry are also available

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    Re: Physics..1st Year..Chapter 1

    NiCe brO...............

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