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 180º.
 

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 90º 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
? = 180º - tan-1 |Ay / Ax|
( c) If Ax is positive and Ay is negative, then
? = 360º - tan-1 |Ay / Ax|
(d) If Ax and Ay are both negative, then
? = 180º + tan-1 |Ay / Ax|

 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
ß 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 ß
Direction of R can be determined by using the Law of sines.
A1 / sin ? = A2 / sin a = R / sin ß
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|

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 = 0º i.e., when F and d have same direction.
W = F . d
W = F d cos 0º ………….. {since θ = 0º}
W = F d …………………….. {since cos 0º = 1}
Work is positive in this case.
Case II
When θ = 180º i.e., when F and d have opposite direction.
W = F . d
W = F d cos 180º ………………………. {since θ = 180º}
W = – F d ………………………………….. {since cos 180º = -1}
Work is negative in this case
Case III
When θ = 90º i.e, when F and d are mutually perpendicular.
W = F . d
W = F d cos 90º ………………………. {since θ = 90º}
W = 0 ……………………………………. {since cos 90º = 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 = 180º
{since mg and h are in opposite direction}
Since,
W = m g h cos 180º
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 = 0º
Therefore, Work done by the body due to its motion = (mg) (v2/2g) cos0º
= 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 = 0º ……………………. {since Fex and h have same direction}
Therefore,
(1) => Wex = m g h cos 0º
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 = 180º …………………. {since Fg and h have opposite direction}
Therefore,
(3) => Wg = m g h cos 180º
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.

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.
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.
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.
Diagram Coming Soon

Torque, Angular, Momentum and Equilibrium

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
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 = 90º 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)

Motion in two Dimension

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 call
ed 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.
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 = 90º
θ = 45º
Therefore,
R(max) = Vo2 / g ; at θ = 45º
Hence the projectile must be launched at an angle of 45º 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 = 57.3º
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
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

Motion

Motion
Definition

If an object continuously changes its position with respect to its surrounding, then it is said to be in state of motion.
Rectilinear Motion
The motion along a straight line is called rectilinear motion.
Velocity
Velocity may be defined as the change of displacement of a body with respect time.
Velocity = change of displacement / time
Velocity is a vector quantity and its unit in S.I system is meter per second (m/sec).
Average Velocity
Average velocity of a body is defined as the ratio of the displacement in a certain direction to the time taken for this displacement.
Suppose a body is moving along the path AC as shown in figure. At time t1, suppose the body is at P and its position w.r.t origin O is given by vector r2.
Diagram Coming Soon
Thus, displacement of the body = r2 – r1 = Δr
Time taken for this displacement – t2 – t1 = Δt
Therefore, average velocity of the body is given by
Vav = Δr / Δt
Instantaneous Velocity
It is defined as the velocity of a body at a certain instant.
V(ins) = 1im Δr / Δt
Where Δt → 0 is read as “Δt tends to zero”, which means that the time is very small.
Velocity From Distance – Time Graph
We can determine the velocity of a body by distance – time graph such that the time is taken on x-axis and distance on y-axis.
Acceleration
Acceleration of a body may be defined as the time rate of change velocity. If the velocity of a body is changing then it is said to posses acceleration.
Acceleration = change of velocity / time
If the velocity of a body is increasing, then its acceleration will be positive and if the velocity of a body is decreasing, then its acceleration will be negative. Negative acceleration is also called retardation.
Acceleration is a vector quantity and its unit in S.I system is meter per second per second. (m/sec2 OR m.sec-2)
Average Acceleration
Average acceleration is defined as the ratio of the change in velocity of a body and the time interval during which the velocity has changed.
Suppose that at any time t1 a body is at A having velocity V1. At a later time t2, it is at point B having velocity V2. Thus,
Change in Velocity = V2 – V1 = Δ V
Time during which velocity has changed = t2 – t1 = Δ t
Instantaneous Acceleration
It is defined as the acceleration of a body at a certain instant
a(ins) = lim Δ V / Δ t
where Δt → 0 is read as “Δt tends to zero”, which means that the time is very small.
Acceleration from Velocity – Time Graph
We can determine the acceleration of a body by velocity – time graph such that the time is taken on x-axis and velocity on y-axis.
Equations of Uniformly Accelerated Rectilinear Motion
There are three basic equations of motion. The equations give relations between
Vi = the initial velocity of the body moving along a straight line.
Vf = the final velocity of the body after a certain time.
t = the time taken for the change of velocity
a = uniform acceleration in the direction of initial velocity.
S = distance covered by the body.
Equations are
1. Vf = Vi + a t
2. S = V i t + 1/2 a t2
3. 2 a S = V f2 – V i 2
Motion Under Gravity
The force of attraction exerted by the earth on a body is called gravity or pull of earth. The acceleration due to gravity is produced in a freely falling body by the force of gravity. Equations for motion under gravity are
1. Vf = Vi + g t
2. S = V i t + 1/2 g t2
3. 2 g S = Vr2 – Vi2
where g = 9.8 m / s2 in S.I system and is called acceleration due to gravity.
Law of Motion
Isaac Newton studied motion of bodies and formulated three famous laws of motion in his famous book “Mathematical Principles of Natural Philosophy” in 1687. These laws are called Newton’s Laws of Motion.
Newton’s First Law of Motion
Statement
A body in state of rest will remain at rest and a body in state of motion continues to move with uniform velocity unless acted upon by an unbalanced force.
Explanation
This law consists of two parts. According to first part a body at rest will remain at rest will remain at rest unless some external unbalanced force acts on it. It is obvious from our daily life experience. We observe that a book lying on a table will remain there unless somebody moves it by applying certain force. According to the second part of this law a body in state of uniform motion continuous to do so unless it is acted upon by some unbalanced force.
This part of the law seems to be false from our daily life experience. We observe that when a ball is rolled in a floor, after covering certain distance, it stops. Newton gave reason for this stoppage that force of gravity friction of the floor and air resistance are responsible of this stoppage which are, of course, external forces. If these forces are not present, the bodies, one set into motion, will continue to move for ever.
Qualitative Definition of Net Force
The first law of motion gives the qualitative definition of the net force. (Force is an agent which changes or tends to change the state of rest or of uniform motion of a body).
First Law as Law of Inertia
Newton’s first law of motion is also called the Law of inertia. Inertia is the property of matter by virtue of which is preserves its state of rest or of uniform motion. Inertia of a body directly related to its mass.
Newton’s Second Law of Motion
Statement
If a certain unbalanced force acts upon a body, it produces acceleration in its own direction. The magnitude of acceleration is directly proportional to the magnitude of the force and inversely proportional to the mass of the body.
Mathematical Form
According to this law
f ∞ a
F = m a → Equation of second law
Where ‘F’ is the unbalanced force acting on the body of mass ‘m’ and produces an acceleration ‘a’ in it.
From equation
1 N = 1 kg x 1 m/sec2
Hence one newton is that unbalanced force which produces an acceleration of 1 m/sec2 in a body of mass 1 kg.
Vector Form
Equation of Newton’s second law can be written in vector form as
F = m a
Where F is the vector sum of all the forces acting on the body.
Newton’s Third Law of Motion
Statement
To every action there is always an equal and opposite reaction.
Explanation
For example, if a body A exerts force on body B (F(A) on B) in the opposite direction. This force is called reaction. Then according to third law of motion.
Examples
1. When a gun is fired, the bullet flies out in forward direction. As a reaction of this action, the gun reacts in backward direction.
2. A boatman, when he wants to put his boat in water pushes the bank with his oar, The reaction of the bank pushes the boat in forward direction.
3. While walking on the ground, as an action, we push the ground in the backward direction. As a reaction ground pushes us in the forward direction.
4. In flying a kite, the string is given a downward jerk and is then released. Thereupon the reaction of the air pushes the kite upward and makes it rise higher.
Tension in a String
Consider a body of weight W supported by a person with the help of a string. A force is experienced by the hand as well as by the body. This force is known as Tension. At B the hand experiences a downward force. So the direction of force at point B is downward. But at point A direction of the force is upward.
These forces at point A and B are tensions. Its magnitude in both cases is same but the direction is opposite. At point A,
Tension = T = W = mg
Momentum of a Body
The momentum of a body is the quantity of motion in it. It depends on two things
1. The mass of the object moving (m),
2. The velocity with which it is moving (V).
Momentum is the product of mass and velocity. It is denoted by P.
P = m V
Momentum is a vector quantity an its direction is the same as that of the velocity.
Unit of Momentum
Momentum = mass x velocity
= kg x m/s
= kg x m/s x s/s
= kg x m/s2 x s
since kg. m/s2 is newton (N)
momentum = N-s
Hence the S.I unit of momentum is N-s.
Unbalanced or Net Force is equal to the Rate of Change of Momentum
i.e., F = (mVf = mVi) / t
Proof
Consider a body of mass ‘m’ moving with a velocity Vl. A net force F acts on it for a time ‘t’. Its velocity then becomes Vf.
Therefore
Initial momentum of the body = m Vi
Final momentum of the body = m Vf
Time interval = t
Unbalanced force = F
Therefore
Rate of change of momentum = (m Vf – m Vi) / t ………………….. (1)
But
(Vf – Vi) / t = a
Therefore,
Rate of change of momentum = m a = F ………………… (2)
Substituting the value of rate of change of momentum from equation (2) in equation (1), we get
F = (m Vf – m Vi) / t ……………………….. Proved
Law of Conservation of Momentum
Isolated System
When a number of bodies are such that they exert force upon one another and no external agency exerts a force on them, then they are said to constitute and isolated system.
Statement of the Law
The total momentum of an isolated system of bodies remains constant.
OR
If there is no external force applied to a system, then the total momentum of that system remains constant.
Elastic Collision
An elastic collision is that in which the momentum of the system as well as the kinetic energy of the system before and after collision, remains constant. Thus for an elastic collision.
If P momentum and K.E is kinetic energy.
P(before collision) = P(after collision)
K.E(before collision) = K.E(after collision)
Inelastic Collision
An inelastic collision is that in which the momentum of the system before and after the collision remains constant but the kinetic energy before and after the collision changes.
Thus for an inelastic collision
P(before collision) = P(after collision)
Elastic Collision in one Dimension
Consider two smooth non rotating spheres moving along the line joining their centres with velocities U1 and U2. U1 is greater than U2, therefore the spheres of mass m1 makes elastic collision with the sphere of mass m2. After collision, suppose their velocities become V1 and V2 but their direction of motion is along same line as before.
Friction
When two bodies are in contact, one upon the other and a force is applied to the upper body to make it move over the surface of the lower body, an opposing force is set up in the plane of the contract which resists the motion. This force is the force of friction or simply friction.
The force of friction always acts parallel to the surface of contact and opposite to the direction of motion.
Definition
When one body is at rest in contact with another, the friction is called Static Friction.
When one body is just on the point of sliding over the other, the friction is called Limiting Friction.
When one body is actually sliding over the other, the friction is called Dynamic Friction.
Coefficient of Friction (μ)
The ratio of limiting friction ‘F’ to the normal reaction ‘R’ acting between two surfaces in contact is called the coefficient of friction (μ).
μ = F / R
Or
F = μ R
Fluid Friction
Stoke found that bodies moving through fluids (liquids and gases) experiences a retarding force fluid friction or viscous drag. If the moving bodies are spheres then fluid friction F is given by
F = 6 π η r v
Where η is the coefficient of viscosity,
Where r is the radius of the sphere,
Where v is velocity pf the sphere.
Terminal Velocity
When the fluid friction is equal to the downward force acting on the sphere, the sphere attains a uniform velocity. This velocity is called Terminal velocity.
The Inclined Plane
A plane which makes certain angle θ with the horizontal is called an inclined plane.
Diagram Coming Soon
Consider a block of mass ‘m’ placed on an inclined plane making certain angle θ with the horizontal. The forces acting on the block are
1. W, weight of the block acting vertically downward.
2. R, reaction of the plane acting perpendicular to the plane
3. f, force of friction which opposes the motion of the block which is moving downward.
Diagram Coming Soon
Now we take x-axis along the plane and y-axis perpendicular to the plane. We resolve W into its rectangular components.
Therefore,
Component of W along x-axis = W sin θ
And
Component of W along y-axis = W cos θ
1. If the Block is at Rest
According to the first condition of equilibrium
Σ Fx = 0
Therefore,
f – W sin θ = 0
Or
f = W sin θ
Also,
Σ Fy = 0
Therefore,
R – W cos θ = 0
Or
R = W cos θ
2. If the Block Slides Down the Inclined Plane with an Acceleration
Therefore,
W sin θ > f
Net force = F = W sin θ – f
Since F = m a and W = m g
Therefore,
m a = m g sin θ – f
3. When force of Friction is Negligible
Then f ≈ 0
Therefore,
equation (3) => m a = m g sin ≈ – 0
=> m a = m g sin ≈
or a = g sin ≈ …………. (4)
Particular Cases
Case A : If the Smooth Plane is Horizontal Then 0 = 0º
Therefore,
Equation (4) => a = g sin 0º
=> a = g x 0
=> a = 0
Case B : If the Smooth Plane is Vertical Then θ = 90º
Therefore,
Equation (4) => a = g sin 90º
=> a = g x 1
=> a = g
This is the case of a freely falling body.