First attempt to measure probability was done by Italian mathematician Galileo. The classical theory of probability is developed by French mathematicians B. Pascal and P. Fermat while solving a number of problems posed by a French nobleman and gambler Chevalier-de-mere. The famous ‘Problen of points’ posed by him to Pascal is “Two person play a game of a chance. The person who first gains the number of points wins the stake. They stopped before the game is completed. How is the stake is to be divided on the basis of the number of points each has won?” The two mathematicians doing research and study made a lengthy correspondence between them. This correspondence itself formed the basis of the classical theory of probability.

Pioneers of Theory of Probability:

• B. Pascal: Classical theory of probability
• Galileo: First attempt to measure the probability
• J.Bernoulli: Tretise on Probability
• De-Moivre: Doctrine of Chances
• T. Bayes: Inverse Probability
• P.S.Laplace: Theoric analique des probabilities
• Tsechebyscheff: Founder of Russian School of Statistics
• Liapounoff: Central limit theorem
• Knitchine: Law of large numbers
• A. Kolmogorov: Axionatic approach to probability

Terminology of Probability:

Experiment:

The operation which can produce some well-defined outcome is called an experiment. There are two types of experiments;

Deterministic experiments: 

Those experiments which when repeated under identical conditions produce the same result or outcome are called deterministic experiments. Experiments of science and engineering fall under this category.

Random Experiment:

If an experiment has more than one possible results which are known in advance and it is not possible to predict which one is going to occur, then the experiment is called a random experiment.

Examples: Tossing a fair coin, drawing a card from a well-shuffled pack of cards.

Outcome:

The result of a random experiment is called an outcome.

Sample Space:

A sample space of an experiment is the set of all possible distinct outcomes of the experiment and it is denoted by ‘S’.

Example: A sample space for different events are as follows

• A fair coin is tossed:

S  = { H, T}

∴ Total number of outcomes = 2   ∴ n (S) =  2

• Two fair coins are tossed or a Fair coin is tossed twice:

S =  { H, T} × { H, T}  =  {HH, HT, TH, TT}

∴ Total number of outcomes =  4  = 2²   ∴n (S) =  4

• Three fair coins are tossed or a Fair coin is tossed thrice: 

S =  { H, T} × { H, T} × { H, T}  =  {HH, HT, TH, TT} × { H, T} =  {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

∴ Total number of outcomes =  8  = 2³  ∴ n (S) =  8

Notes:

• An element of a sample space is called a sample point.
• If the number of elements in a sample space is finite then the sample space is called a finite sample space.
• If n coins are tossed then the number of outcomes is 2n.
• A fair coin is tossed twice is equivalent to two fair coins are tossed.
• A fair coin is tossed three times is equivalent to three fair coins are tossed.
• A fair coin is tossed ‘n’ times is equivalent to ‘n’ fair coins are
• A word fair is equivalent to unbiased.

• A fair dice is tossed:

S  = {1,2,3,4,5,6}    ∴ n (S) =  6.

• Two fair dice are tossed

S = {1,2,3,4,5,6} ×  {1,2,3,4,5,6}

∴ n (S) =  36

Notes:

• The sum of the two numbers on two dice is called the score on two dice.
• The minimum score on two dice is 2 and the maximum score on two dice is 12.
• The cases favourable to a particular score can be read along the diagonal of that score.

Event:

Any subset of a sample space is called an event. An event is denoted by a capital letter.

Example – 1:

A = an event, that a card selected from a pack of cards is a Diamond.

A Diamond card can be obtained in   13C₁ = 3 ways

Number of cases favourable to the event A = 13 ∴ n(A) = 13

Example – 2:

B = an event, that 2 cards selected are consisting of one King and other a Queen.

A King can be obtained in 4C₁ = 4 ways.

After the selection of a King in any ‘one of four ways,

the selection of a queen can be made in 4C1= 4 ways.

∴ n(B) = 4 x 4   =  16

Example – 3:

C = an event, that a die shows a number greater than 3

∴ C = { 4, 5, 6}    ∴n(C) = 3

Example – 4:

D = an event that the score on two dice is 4

∴ D = {(1, 3), (2, 2), (3, 1)}    ∴ n(D) = 3

Example – 5:

E = an event that the score on two dice is a prime number  i.e. 2,3,5,7,11

∴ E = {(1, 1), (1, 2), (2, 1), (1, 4), (2, 3), (3, 2). (4, 1), (1, 6), (2, 5), (3, 4),

(4, 3), (5, 2), (6, 1), (5, 6), (6, 5)}

∴ n(E) = 15

Probability:

If A is an event of sample space S’ then the probability of event A denoted by P(A) is defined as

P(A) = n(A)/n(S)

Important Relations and Their Proofs:

• Prove that P(Φ) = 0

Proof :

Since Φ is an impossible event. Hence Φ is an empty set.

∴  n(Φ) = 0

∴ P(Φ) = n(Φ)/n(S) = 0/n(S) = 0

∴ P(Φ) = 0  is proved

• Prove that P(S) = 1

Proof :

Since S is a certain event.

∴  n(Φ) = n(S)

∴ P(S) = n(S)/n(S) = 1

∴ P(S) = 1  is proved

• Prove that 0 < P(A)  < 1

Proof :

If A is an event of sample space S. Then we have

0 <  n(A) < n (S)

Dividing by n(S)

0/n(S) <  n(A)/n(S) < n (S)/n(S)

∴ 0   <   P(A)  < 1       (proved)

• Prove that P(A) = 1 – P(A’)

Proof :

If S contains n sample points and A contains m sample points. Then A’ will contain (n – m) sample points.

Where A’ is a complement of the set A.

∴ A U A’ = S  and A ∩ A’ = Φ

∴  n(A) + n(A’) = n(S)

Dividing by n(S)

∴  n(A)/n(S) + n(A’)/n(S) = n(S)/n(S)

∴  P(A)  +   P(A’)  =  1

∴  P(A) = 1 – P(A’)  is proved

Explanation of  the Phrases :

odds in favour of an event A  and (ii) odds against an event A

If x cases are favourable to an event A and y cases are not favourable to the event A then we say odds in favour of A are x : y  OR odds against A are y : x, Then P(A) = x/(x + y).

In the next article, we shall study to write the sample space for a given experiment.

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In this article, we shall study to solve numerical problems to calculate the magnitude of force, momentum, and change in momentum.

Example – 01:

The speed of a tortoise and hare are 1 m/s and 3 m/s respectively. The mass of the hare is 5 kg while that of tortoise is 20 kg. Which of the two has greater momentum?

Solution:

The momentum of hare = Mass of hare x Speed of hare= 5 x 3 = 15 kg m/s

The momentum of tortoise = Mass of tortoise x Speed of tortoise= 20 x 1 = 20 kg m/s

Thus the momentum of tortoise is more than that of the hare.

Example – 02:

What is the magnitude of force exerted by a horse in pulling a cart of mass 600 kg and accelerating at the rate of 1.2 m/s²?

Given: mass of cart = m = 600 kg, acceleration = a = 1.2 m/s².

To Find: F =?

Solution:

By Newton’s second law of motion, magnitude of force

F = ma = 600  x 1.2

∴ F = 720 N

Ans: The required force is 720 N.

Example – 03:

An object of mass 10 kg is moving with the initial velocity of 10 m/s. A constant force acts on it for 4 s giving it a speed of 2 m/s in opposite direction. Find the acceleration and the force.

Given: mass of object = m = 10 kg, initial velocity = u = 10 m/s, Final velocity = v = – 2m/s (negative since it is in opposite direction), time for which force is acting = t = 4 s

To Find: acceleration = a =? Force acting = F =?

Solution:

a = (v – u)/t = (-2 – 10)/4 = -12/4 = -3 m/s²

The negative sign indicated retardation.

By newton’s second law of motion we have

F = ma = 10 x (-3) = -30 N

Negative sign indicated retarding force.

Ans: The acceleration = 3 m/s² and Force = -30 N

Example – 04:

A constant force of 2 N acts on a body for 5 seconds to change its velocity. Calculate the change in the momentum.

Given: Force acting = F = 2 N, time for which the force is acting = t = 5s.

To Find:  Change in momentum =?

Solution:

Change in momentum = F x t = 2  x 5  = 10 kg m/s.

Ans: the change in momentum is 10 kg m/s.

Example – 05:

An empty truck of mass 1000 kg is moving at a speed of 36 km/hr. It is loaded with 500 kg of material on its way and again moves at the same speed. Will the momentum of the truck remain the same after loading? if not, find the momentum of the truck after loading.

Given: Mass of a truck = 1000 kg, Mass of Load = 500 kg, Speed of the vehicle = v = 36 km/hr = 36 x 5/18 = 10 m/s.

To Find: Momentum of truck = p =?

Solution:

The momentum of a body depends on its mass. In this case, the truck is loaded on the way, hence its momentum should change.

Total Mass = m = 1000 + 500 = 1500 kg

New momentum = Total Mass x Velocity = 1500 X 10 = 15000 kg m/s

Ans: The momentum of the truck after loading is 15000 kg m/s.

Example – 06:

A railway wagon of mass 1000 kg is pulled with a force of 10000 N. What is the acceleration?

Given : Force applied = F = 10000 N, mass of wagon = m = 1000 kg.

To Find:  acceleration = a =?,

Solution:

By Newton’s second law of motion, magnitude of force

F = ma

∴ 10000 = 1000 x a

∴ a = 10000/1000 = 10 m/s².

Ans: The acceleration is 10 m/s².

Example – 07:

A car of mass 1000 kg is moving at a certain speed when a constant braking force 1000 N acts on it for 5 s and speed of the car reduced to half the original speed. Find the further time required to stop the car, if the same constant force acts on it.

Given: mass of car = m = 1000 kg, Force acting = F = 1000 N,time taken =  t = 5 s, Final speed (v) = 1/2 Initial speed (u) = u/2

To find: t =? when v = 0

Solution:

By Newton’s second law of motion, magnitude of force

F = ma

∴1000 = 1000 x a

∴ a = 1000/1000 = 1 m/s².

By the first equation of motion

∴ a = (v – u)/t  = (u/2 – u)/5 = (-u/2)/5 = (-u/10)

∴ a = -u/10 = – 1

∴ u = 10 m/s

We have v = u + at

∴ 0 = 10 + (-1)t

∴ -10 = – t

∴ t = 10 s

Ans: The further time required to stop the car is 10 s

Example – 08:

Find the magnitude of the force applied to a block of mass 5 kg at rest, if it moves 36 m, in first 3 seconds. Neglect the force of friction.

Given: mass of block = m = 5 kg, Initial velocity = u = 0, Distance traveled = s = 36 m, time taken = t = 3 s,

To find: F =?

Solution:

s = ut + 1/2 at²

∴   36 = (0)(3) + 1/2 a(3)²

∴   36 = 1/2 a(9)

∴   72 = a (9)

∴ a = 72/9 = 8 m/s².

By Newton’s second law of motion, the magnitude of force

F = ma = 5 x 8 = 40 N

Ans: The force applied = 40 N

Example – 09:

Two spheres of mass 10 g and 100 g each fall on the two pans of a table balance from a height of 40 cm and 10 cm respectively. If both are brought to rest in 0.1 seconds. Determine the force exerted by each sphere on the pans.

Solution:

For the first sphere: 

Given : m₁ = 10 g = 0.01 kg, h = – 40 cm  = – 0.4 m,  t = 0.1 s, g = – 9.8 m/s²

v² = u² + 2gh

∴  v² = (0)² + 2(-9.8)(-0.4)

∴  v² =  7.84

∴  v = 2.8 m/s

F₁ = m₁a

F₁ = m₁ (v – u)/t  = 0.01 x (2.8 – 0)/0.1 = 0.28 N

For the second sphere: 

Given : m₁ = 100 g = 0.1 kg, h = – 10 cm  = – 0.1 m,  t = 0.1 s, g = – 9.8 m/s²

v² = u² + 2gh

∴ v² = (0)² + 2(-9.8)(-0.1)

∴ v² =  1.96

∴ v = 1.4 m/s

F₂ = m₂a

F₂ = m₂ (v – u)/t  = 0.1 x (1.4 – 0)/0.1 = 1.4 N

Ans: The force exerted by the first sphere is 0.28 N and that by second sphere is 1.4 N

Example – 10:

Calculate the density of a cubical ice block of side 50 cm. If a force of 1125 N applied to it produces an acceleration of 10 m/s² in it. Neglect the force of friction. Assume the ice block remains in solid state without melting.

Given: Force applied = F = 1125 N, acceleration = a = 10 ms-1, Side of a block = 50 cm = 0.5 m

To Find:  Density = ρ = ?,

Solution:

By Newton’s second law of motion

F = m.a

∴ 1125 = m x 10

∴ m = 1125/10 = 112.5 kg

Ans: The density of ice block is 900 kg/m³.

Example – 11:

A ball of mass 50 g at rest is hit by a bat and the ball covers a distance of 400 m, in 2 seconds. If the ball was in contact with it for 0.1 s, find the magnitude of the force acting on it. Assuming no other force acts on a ball after it is hit by the bat.

Given: mass of ball = m = 50 g = 0.05 kg,  time of contact = 0.1 s, distance covered  s= 400 m, time taken to cover the distance = t = 2 s.

To find: Force acting =F =?

Solution:

No other force acts on a ball after it is hit by the bat. Thus it is in uniform motion after hit.

v = s/t = 400/2 = 200 m/s

Now Force, F = ma

F = m (v – u)/t  = 0.05 x (200 – 0)/0.1 = 100 N

Ans: The force acting = 100 N

Example – 12:

A force of 500 N acts on a body of mass 1000 kg and the body is brought to rest within a distance of 64 m. Find the initial velocity and time taken by the body to come to rest.

Given: mass of body = m = 1000 kg, Force acting = F = 500 N, Final velocity = v = 0 ms-1, distance traveled = s = 64m.

To find: initial velocity =u =? time taken = t = ?

Solution:

We have     F = ma

∴   500 = 1000 x a

∴ a = 500/1000 = 0.5 m/s²

As the body is brought to rest a = – 0.5 m/s²

v² = u² + 2as

∴ (0)² = u² + 2(-0.5)(64)

∴ (0)² = u² – (64)

∴ u² = 64

∴ u = 8 m/s

By first equation of motion

v = u + at

∴ 0 = 8 + (- 0.5) x t

∴ 8 = – 0.5 x t

∴ t = 8/0.5 = 16 s

Ans: Initial velocity = 8 m/s, time taken to come to rest = 16 s

Example – 13:

A car of mass 1000 kg is moving uniformly with 10 m/s. If the engine of the car develops an extra linear momentum of 1000 kg m/s. Calculate the new velocity with which the car runs.

Solution:

Given : mass of car = m = 100 kg, initial velocity = u = 10 m/s, Extra momentum = 1000 kg m/s,

To find: Final velocity = v = ?

Initial momentum = p₁ = mu = 1000 x 10 = 10000 kg m/s

Final momentum = p₂ = 10000 + 1000 = 11000 kg ms-1.

Now, Final momentum = p₂ = mv

∴ 11000 = 1000 x v

∴ v = 11000/1000 = 11m/s

Ans: New Velocity = 11 m/s

Previous Topic: Applications of Newton’s Laws

Next Topic: Law of Conservation of Momentum

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In this article, we shall discuss the concept of balanced forces and unbalanced forces.

Kinematics: The branch of physics (mechanics) which deals with the motion of the bodies without considering the forces causing it is called kinematics.

Dynamics: The branch of physics (mechanics) which deals with the motion of the bodies and the forces causing it is called dynamics.

Concept of the Resultant Force:

Two forces T₁ and T₂ are applied by two tug boats on a ship at an angle to each other. As a result, the ship accelerates and moves in the direction as shown, as if it were under the influence of a single force F. The single force which acts on a body to produce the same effect in it as is done by all the forces collectively is called a resultant force.

Balanced Force:

When a number of forces acting on a body do not cause any change in the state of rest or of uniform motion in a straight line, then the resultant force acting on the body is said to be zero and the forces are said to be balanced forces. From this explanation, we can conclude that a state of rest or of uniform motion in a straight line, does not mean no force acts on a body. Actually a number of forces may be acting on it but they are balanced forces. Thus balanced forces do not change the state of motion of a body, but they may change the shape of the body.

Unbalanced Force:

When the resultant of all the forces acting on a body is not zero, then the forces are said to be unbalanced forces. Unbalanced forces cause the change of state of motion of a body.

Aristotle (384 BC – 322 BC):

Aristotle was a Greek philosopher, a student of Plato and teacher of Alexander the Great. His writings cover many subjects, including poetry, physics, metaphysics, theater, music, logic, rhetoric, linguistics, politics, ethics,  government, biology, and zoology. Together with Plato and Socrates (Plato’s teacher), Aristotle is one of the most important founding figures in Western philosophy. Aristotle’s writings were the first to create a comprehensive system of Western philosophy, encompassing morality and aesthetics, logic and science, politics and metaphysics.

Aristotle’s Concept of motion:

Aristotle proposed that for uniform motion a constant force is required. Thus to move a body at rest we have to apply force and to make the body to be in continued motion we have to apply constant pressure. When this constant force ceases to act, the body stops. This concept was accepted and believed for about 2000 years.

Galileo Galilei (15 February 1564 – 8 January 1642):

Galileo Galilei was an Italian physicist, mathematician, astronomer, and philosopher who played a major role in the Scientific Revolution. His achievements include improvements to the telescope and consequent astronomical observations and support for Copernicanism. Galileo has been called the “father of modern observational astronomy”. Stephen Hawking says, “Galileo, perhaps more than any other single person, was responsible for the birth of modern science.

The motion of uniformly accelerated objects, taught in nearly all high school and introductory college physics courses, was studied by Galileo as the subject of kinematics. His contributions to observational astronomy include the telescopic confirmation of the phases of Venus, the discovery of the four largest satellites of Jupiter (named the Galilean moons in his honour), and the observation and analysis of sunspots. Galileo also worked in applied science and technology, inventing an improved military compass and other instruments.

Before Galileo, it was assumed that if a heavy body and a light body are dropped from the same height simultaneously, then the heavier body will strike the ground first. Galileo by his experiments at leaning tower Pisa proved that irrespective of their masses, both the body hit the ground simultaneously. He proposed that all the bodies are attracted to the earth by constant acceleration. His work is a base for Newton’s law of motion and Newton’s law of gravitation.

The geocentric concept of the universe was accepted at that time i.e. all planets and stars revolve around the earth. Copernicus proposed heliocentric theory i.e. sun at the centre and the earth and other planets revolve around the sun. Copernicus was punished to death for proposing theory opposite to religious beliefs. Galileo was punished by the Catholic Church for supporting Copernicus heliocentric theory and was forbidden from teaching and holding any talks. Ban on him was lifted when he apologized to the church.

Galileo’s Concept of Motion:

Galileo proposed that “For uniform motion, no force is required. Force is only required to change the state of motion”. He proved it with his Marble-Tray Experiments.

Galileo’s Experiments:

For his experiment Galileo used double inclined plane one sloping down, one sloping up. He released the marble from the sloping down plane he found that the velocity of the marble increases then the marble moves up sloping up the inclined plane. When moving up the velocity of the marble decreases. It reaches the same height on sloping up the inclined plane from which it is released from sloping down the inclined plane.

Now, he decreased the angle of sloping up the inclined plane and released the marble from the same height. He observed that the marble reaches the same height covering a longer distance.

Then he made sloping up the plane completely horizontal and released the marble from the same height. He observed that marble travels larger distances and he said that the marble will be in motion continuously (will cover infinite distance) as it will never reach that height.

When the surface of the second plane is rough, the ball would cover less distance. Thus, the conclusion of the experiment was that for uniform motion no force is required. This was the basis of Newton’s laws of motion.

Note: Actually the marble cannot move forever because it stops due to the friction between it and the surface.

Thus by this experiment, Galileo proved that for uniform motion no unbalanced forces are required.

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Next Topic: Newton’s First Law of Motion

Science > Physics > Force > Concept of Balanced and Unbalanced Forces

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