Complete Notes on 9th Physics by Asif Rasheed

XSAS

2013 IDEAL SCIENCE ACADEMY

(CONTACT # 0344 -78 46 394 )

9TH CLASS PHYSICS NOTES

1) PHYSICAL QUANTITIES &

MEASURMENTS 2) KINEMATICS 3) DYNAMICS NOTES OF PHYSICS 9TH CLASS ENGLISH MEDIUM COMPLETE FIRST THREE

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Ideal Science Academy

CHAPTER PROBLEMS,SHORT AND LONG QUESTIONS

ASIF RASHEED BS HONS PHYSICS

Asif Rasheed BS (HONS) Physics # 0344 78 46 394

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Unit No 1

(1) Define Physics? (2) Ans: The branch of science which deals with the study of matter

and energy and their mutual relationship is called Physics.

(2) Write names of the branches of Physics?

Ans: (1) Mechanics (2) Heat & Thermodynamics (3) Sound (4) Light

(5) Electromagnetism (6) Atomic and Molecular Physics (7) Nuclear Physics (8) Plasma Physics (9) Solid Physics

Why do we study physics?

We study physics to under stand the laws of nature and how nature effects the human action.

(3) Define Plasma?

Ans: The state of matter at a very high temperature comprising the ions and electrons is called plasma.

(4) Name the branches of physics overlapping the other branches of science?

Ans: (1) Astrophysics (2) Geophysics (3) Biophysics

(7) Who studied the freely falling bodies?

Ans: Galileo studied the freely falling bodies

BRANCHES OF PHYSICS: Mechanics:

It is the study of motion of objects, the causes and effect.

Heat:

It deals with the nature of heat, modes of transfer and effects of heat.


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Sound:

It deals with the physical aspects of sound waves, their production, properties and applications.

Light (Optics):

It is the study of physical aspects of light, its properties, working and use of optical instruments.

Electricity and magnetism:

It is the study of the changes at rest and in motion, their effects and their relationships with magnetism.

Atomic physics:

It is the study of the structure and properties of atom.

Nuclear physics:

It deals with the properties and behaviour of nuclei and the particles within the nuclei.

Plasma physics:

It is the study of production, properties of the ionic state of matter – The fourth state of matter.

Geophysics:

It is the study of the internal structure of the earth.

System international units:

A world-wide system of measurements is known as system international units (SI). In SI, the units of seven base quantities are meter, kilogram, second, ampere, Kelvin, candela and mole.

vernier callipers:

An instrument used to measure small lengths such as internal and external diameter or length of a cylinder etc is called vernier calipers.

Least count of vernier calliper is 1/10 cm or 0.1 cm which is also called vernier constant.

Screw gauge:


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A screw gauge is used to measure small lengths such as diameter of a wire, thickness of a metal sheet etc.

The least count of micrometer screw gauge is 0.01 mm.

Physical balance

Physical balance is a modified type of a beam balance used to measure small masses by comparison with greater accuracy.

MEASUREMENTS

Why a standard unit is need to measure a quantity correctly. Ans: Various units have been in use in different times in different parts of the world. The fast means of communication systems have changed the world into a global village. Due to this reason an international system of units for mutual business became essential. The eleventh general conference of weights and measures recommended that all the countries of world should adopted a system of same kind of standard units, consisting of seven base units known as international system of units (SI) and derived units.

Q: What is meant by base and derived units? Give three examples of derived units and explain how they are derived from base units.Ans: Base Units: The units of base quantities are called base units.Examples: Kilogram (kg), meter (m), second (s), Ampere (A)Derived Units: The units of derived quantities which are derived from base units are called derived units. Examples: Unit of Area: m2.Unit of Volume: m3

Unit of Density: Kg m-3 Unit of speed: meter per second (ms-1), Unit of weight: Newton


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Unit of force: Newton,Unit of Pressure: Pascal

Q : How they are derived from base units:These units are obtained by multiplication, division or both of base units.Unit of Area: length x breadthUnit of length x unit of breadthMeter x meterm x m: m2

Unit of Volume: length x breadth x heightUnit of length x unit of breadth x unit of heightMeter x meter x meterm x m x m: m3

Unit of Density: MassVolumeUnit of massUnit of volume Q: What is the number of base units in System International (SI)?Ans. There are seven base units which are given belowNumber Physical Quantity Unit Symbol1. Length Meter (m2. Mass Kilogram(kg3. Time Seconds (s4. Electric current Ampere( A5. Temperature Kelvin( K6. Intensity of light Candela(cd7. Amount of substance(Mole mol

Q: Where multiples and sub-multiples of units are used? Describe some standard prefaces which are internationally used.

Ans. Multiples and sub-multiples of units are used to make very large and


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very small mathematical calculations easier. The multiples and sub-multiples are obtained by multiplying or dividing with ten or power of tens.

The terms used internationally for the multiples and sub-multiples for different units are called prefixes.

Q: What is the use of vernier calipers?Q1.7 What is meant by its vernier constant?Ans. A vernier calliper can be used to measure lengths accurately up to one tenth of a millimetre.Least Count/Vernier ConstantThe minimum length which can be measured accurately with the help of vernier scale or vernier calipers is called least count. Least count of vernier calliper is 1/10 mm or 0.1 cm which is also called vernier constant.

Q: Explain the statement “A micrometer screw gauge measures more accurately than a vernier caliper”?Ans. Least Count: The minimum length which can be measured accurately is called least count of any measuring device.

The least count of micrometer screw gauge is 0.01 mm and that of Vernier calliper is 0.1. cm. So a micrometer screw gauge measures more accurately than a vernier calliper.

Significant figures:

All accurately known digits and the first doubtful digit in an expression are called significant figures. It reflects the precision of a measured value of a physical quantity.

RULES TO FIND THE SIGNIFICANT FIGURE IN A MEASUREMENT:

(i) Digits other than zero are always significant.

27 have 2 significant digits.


(ii) Zeroes between significant digits are also significant.


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(iii) Final zero or zero after decimal are significant.

275.00 have 5 significant digits.

(iv) Zero used for spacing the decimal points is not significant.

Here zero is place holders only.

0.03 has 1 significant digit.

0.027 has 2 significant digits.

(v) In whole numbers that end in 1 or more zero without a decimal point. These zeroes may or may not be significant. In such cases, it is not clear which zero serve to locate the position value and which are actually parts of measurements. In such a case, express the quantity using scientific notation to find the significant zero.

APPLICATION OF PHYSICS

Physics play an important role in our daily life. We hardly find a device where physics is not involved. Consider pulleys make it easy to left heavy loads. Electricity is not used only to get light and heat but also mechanical energy that derives fans electric motors ctc.Consider the means of transportation such as car and aeroplanes domestic appliances such as air conditioners refrigerators vacuum cleaners washing machine and micro wave ovens etc.Similarly the means of communication such as radio T V telephone are the result of application of physics. These devices made our lives much comfortable and easier than the past.

BOOK EXERCISE QUESTIONS AND ANSWERS.

Q: 1.5Estimate your age in seconds?Ans : My age is sixteen years. Its value in seconds 16 x 365 x 24 x 60 x 60 = 504576000 seconds.

Q1.6: What role SI units have played in the progress of science?


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Ans : SI units are very easy to use because their addition, multiplication and division is very easy . These can be written in terms of multiples of ten.1.7 SEE ABOVE.Q1.8: what do you under stand by zero error of measuring instruments?Ans: When the zero of virnier scale is not coinciding with the zero of main scale,”then instrument has zero error”Q1.9: why is the use of zero error in a measuring instrument?Ans: By the use of zero error the observation taken can be correct, to get correct observationsQ1.10: What is stop watch? What is the least count of mechanical stop watch you have use in laboratory?Ans: The stop watch is used to measure small intervals of time. Its least count is about 0.1 seconds.Q1.11: We need to measure extremely small interval of times?Ans: Small time interval are measured o calculate instantaneous time rate of change of variable.Q1.12: What is meant by significant figures of a measurement?Ans: All the accurately known digits and the first doubtful digit in an expression are known as significant figures.Q1.13: How is precision is related to the significant figures in a measured quantity?Ans: More is the number of significant figures, when the measuring instrument used has small value of its least count. The small value of least count the large is the value of precision. For example reading taken by screw gauge has more precision than reading taken by meter rod or verneir scale.

CHAPTER: 1 Physical quantities and measurement (Problems)

P1.1) Express the following quantities using prefixes.Solution:(a) 5000gAs 1000g = 1kg so 5000/1000=5kg Ans


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(b)2000000wAs 106 = megaSo 2000000 w2 x 10 6 =2MW(c) 52 x 10-10 x kg since 1kg =1000g or103g=52 x 10-10 x 103g= 52 x 10-10+3g= 52 x 10-7g= 5.2 x 101 x 10-7g= 5.2 x 10-7+1g= 5.2 x 10-6g =5.2ug Ans(d) 225 x 10-8s =2.25 x 102 x 10-8s= 2.25 x 102-8s=2.25 x 10-6s= 2.25 us ANS

P1.2) How do prefixes micro, nano and pico related to each other.As we know that,Micro =10-6

Nano = 10-9

Pico = 10-12

1 p= 1/1000n1 p= 1/1000000 µ1 n= 1/1000 µ1 n = 1000 p1 µ = 1000n1 µ = 1000000p

P1.3) Your hair grow at the rate of 1mm per day find their growth rate in nms-1.As milli = 10-3

Nano = 10-9

1m = 10-6 n OR 1m = 1000000nBy multiplying ‘’m’’ on both sides


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1mm = 106 nm OR 1mm = 1000000nm As we know thatOne day = 24 hoursOne hour = 60 minutes One minutes = 60sSoOne day = 24 x60 x6= 86400 sSo the growth rate in nms-1 is= 1000000nm/86400s= 11.57nms-1 Ans

P1.4) Rewrite the followings in standard form:Solution:

(a) 1168 x 10−27

Solution:11.68 x 10−27

1.168 x 103 x 10−27

1.168 x 10 3-27

1.168 x 10−24 ANS(b) 32 x 105

Solution:3.2 x 101 x 105

3.2 x 10 1+5

3.2 x 106

(c) 725 x 10−25

7.25 x 10−5 x 103g7.25 x 102 x 102 x 10 -5+3

7.25 x 10 2-2g7.25 x 100

7.25g ANS(d) 0.02 x 10−8

Solution:0.02 x 10-2 x 10-8

2 x 10-2-8 (as we know that powers are added up)2 x 10-10 ANS


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P1.5) Write the following quantities in scientific notation:Solution:(a) 6400km6.4x103km ANS(b) 380000km3.8x105 ANS(c) 300000000ms-1

3.108ms-1

(d) Seconds in a day:24x60x60s=86400s8.64x104s ANS

P1.6) Question on book:As the zero of Vernier scale is on right so zero error will be positive and if its 4th division is conceding with the main scale then the zero error=0.01x4=0.04Zero error= +0.04cmAnd zero correction= -0.04cm

P1.7) A screw gauge has 50 divisions on its circular scale. The pitch of the screw gauge is 0.5mm. What is its least count?Solution:Least count= pitch of screw gauge/no. of divisions in circular0.5/50=0.01mm0.01x10-3m1x10-5m1x10-5x100cm1x10 -3cm0.001cm ANS

P1.8) Which of the following quantities have three significant figures?Solution:

(a) 3.006m(b) 5.05x10-21kg(c) 0.00309kg(d) 301.0s


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P1.9) What are the significant figures in the following measurements?

(a) 1.009m (It carry all of them 4)(b) 0.00450kg

0.00450It has 3 significant figures

(c) 1.66x10-27kg1.66x10-27kgIt has 3 significant figures.

(d) 2001sIt has 4 significant figures.

P.10) A chocolate wrapper is 6.7cm long, 5.4cm wide. Calculate its area up to reasonable number of significant figures.Solution:Area= length x width =6.7cm x 5.4cm = 36.18cm2

Area in significant figure= 36cm2

Unit # 2 Kinematics

Define Mechanics and its types.The branch of physics, which deals with the study of motion of bodies, is

called Mechanics. It has two types:

I) Kinematics

ii) Dynamics

Define Types of Mechanics. Kinematics:

It is study of motion of bodies without reference of force and mass.

Dynamics:


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It is study of motion of bodies with reference of force and mass.

Q2.3 (i) difference between rest and motion?

Define Rest If a body does not change its position with respect to some observers then it

is said to be in the state of rest.

Define Motion If a body is changing its position with respect of some observers then it is

said to be in the state of motion.

Name the types of motiona. Translatory Motion

b. Linear motion

c. Circular motion

d. Random motion

e. Rotatory Motion

f. Vibratory Motion

What is the motion butterfly? Executed by Flight of butterfly is irregular motion. Therefore its motion is called random motion.What is type of motion of free falling bodies?

Freely falling bodies move downward in straight direction under the force

of gravity. Therefore their motion is called linear motion.

What is the type of motion of a man moving in circular track?His motion is circulatory motion.

Q2.3 (iii)What is the difference between distance and

displacement?

Define Distance The path between two points is called distance. It is scalar quantity.

Define Displacement


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The shortest distance between two points is called displacement. It is a

vector quantity.

Q2.3 (VI) what is the difference between Scalar and Vector? OR

What are Scalar and Vector Quantities? Scalars are those quantities which are described by a number with

suitable unit without direction.

Vectors are those quantities which can be described by a number with

suitable unit and with direction.

Q2.3 (iv) What is the difference between Velocity and Speed?

Define Velocity (Part of 2.4 The distance covered by a body in a unit time in a particular direction is

called velocity.”

OR

The rate of change of displacement is called velocity. It is denoted by v. It

is a vector quantity.

1. Positive Acceleration

If the velocity continuously increases then the acceleration will be

positive.

2. Negative acceleration

If the velocity continuously decreases then the acceleration will be

negative.

Define Uniform Speed If a body covers an equal distance in equal interval of time in a particular

direction, the body is said to be uniform Velocity.

Define variable velocity


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If a body does not cover an equal distance in equal interval of time in a

particular direction, the body is said to be in variable velocity.

Define relative velocity When two bodies are in motion then the velocity of one body relative to

other is called relative velocity.

Define Instantaneous velocity The velocity of a body at any instance of time is called instantaneous

velocity.

Define Average velocity Average velocity of a body can be obtained by dividing the total

displacement with total time taken.

Vav = Displacement/Time = dt

Can a body moving with certain velocity in the direction of East can have acceleration in the direction of West?

Ans:

Yes, if the velocity of the body decreases, then it will have

acceleration in the opposite direction, that is, in the direction of west.

Does speedometer of a car measures its velocity?It measures only speed but not velocity. Part of 2.4 Define Acceleration.

The rate of change of velocity is called acceleration. It is denoted by “a”.

It is a vector quantity.

Define Uniform Acceleration If velocity of a body is changing equally in equal intervals of times then

its acceleration will be uniform.

Define Variable Acceleration If velocity of a body is not changing equally in equal intervals of times

then its acceleration will be variable.


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Define Average Acceleration

The average acceleration can be obtained by dividing total change in

velocity with total time taken.

aav = Change in Velocity/Total Time

=(Vf – Vi)t

Define Gravitational Acceleration The acceleration of freely falling bodies is called gravitational

acceleration. It is denoted by g. Its value

is 10 meter per second per second (10 ms-2)

2. A body is thrown vertically upward. What is gravitational acceleration?Ans : It is 10 meter per second per second (-10 ms-2)

3. What is acceleration of a body moving with uniform velocity?Ans : The acceleration will be 0.

4. What consideration should be kept in mind while using equation of motion for free falling bodies?• Initial velocity should be taken as zero.• Acceleration will be taken as (g) instead of (a)

Part of 2.4 Define Speed The distance covered in unit time is called speed.

Speed = Distance/Time

v = S/t

The unit of speed is meter per second (ms-1) or m/s

5. Q#2.7 /Can a body moving at a constant speed have acceleration?Yes, if it is moving in circular path, it can have acceleration.

6. A body is moving with uniform velocity, what will be its acceleration?Its acceleration will be zero.


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7. A body is moving with a uniform speed. Will its velocity be

uniform?Yes, if it moves in straight line and does not change its direction.

8. Can a body moving with a certain velocity in direction of East, have Acceleration in the direction of West?Yes, if its velocity will decrease, it has acceleration in the direction of

west.9. Does speedometer a car measure its velocity?

No, it only measures the speed.10. Why a stone and a piece of paper when dropped from the same

height, reach the ground at the same time.Because both have same gravitational acceleration.

11. What type of change will occur in three equations of motion under the action of gravity?Acceleration (a) will be replaced with gravitational acceleration (g) in all

equations.And distance (s) will be changed in to height (h)

Describe the different types of motion in detail?Motions of bodies are of three types:

Q2.3 (v) What is the difference between linear and random Motion?

Q2.3 (ii) What is the difference between Circular and

Translatory Motion?

i) Translatory Motion: -

A motion in which each particle of a body has exactly

same motion is called Translatory Motion. It may be of many

kinds for examples: -

a. Linear Motion: If a body moves in straight line its

motion is called linear motion. e.g. motion of free falling bodies, a

man walking on a straight path


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b. Circular Motion: If a body moves in a circle its motion

is called circular motion. e.g. motion of stone attached to

thread and whirled.

c. Random Motion: If a body moves in irregular manner

its motion is called random motion. e.g. motion of butterfly.

ii) Rotatory Motion: -

Motion is said to Rotatory, when the object rotates on its own axis.

Examples: Rotatory motion of a planets on its axis, wheels of a

vehicles, spinning top, ceiling fan etc.

iii) Vibratory Motion: -

When a body moves to and fro about a point and repeats its motion

then its motion is called vibratory motion. e.g. Motion of simple Pendulum,

motion of tuning fork, A ruler. Place one inch of it on a desk, and the other

11 off the desk. Flick the end off the desk and watch it vibrate.

Q2.10 How can a vector quantity be represented graphically?

Ans: When a vector is represented graphically, its magnitude is represented by the length of straight line and its direction is represented by the direction

of the arrow head . Here is an example

:

Q2.11 Why vectors quantities can not be added and subtracted like scalar quantities?


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Ans ; In addition of vector quantities, not their magnitude but their direction

also involved therefore vector can not be added like the addition of scalar.

Q2.12: How are vectors quantities are important to us in our daily

life?

Ans: in our daily life vectors quantities are completely explained only when

their direction are also considered Q2.13 Derive equation of motion for

uniformly accelerated rectilinear motion?

Three equations of motion are three equations of motion under the action of

gravity are

Vf = Vi + at Vf=Vi+gt

S = Vi t + 12 at2 h=Vit+

12gt2

2aS = Vf 2 – Vi 22gh = Vf 2 – Vi 2

Q2.14 : Sketch the velocity time graph for the motion of the body?

Motion Graphs

For body moving at constant velocity:

The graph of straight line parallel to the X axis shows that the body is

moving with constant velocity

a)


http://physics.tutorvista.com/motion/motion-graphs.html

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b)Q 2.7 Distance time graph for a body start from rest

uniform velocity-- time graph

Derivation of Equation of Motion (Graphically)First Equation of Motion

Consider an object moving with a uniform velocity u in a straight line. Let it

be, given a uniform acceleration at time, t = 0 when its initial velocity is V i.

As a result of the acceleration, its velocity increases to Vf (final velocity) in

time t and s is the distance covered by the object in time t. The figure shows

the velocity-time graph of the motion of the object.

Slope of the Vf - t graph gives the acceleration of the moving object.

Thus, acceleration = slope = AB = BC/AC

Where BD=Vf, CD=Vi, AC= OD=t BC=BD-CD


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(Average,acceleration(aav)= ChangeinVelocity/time)

Slope=AB= Vf - Vi/ t

a = Vf - Vi/ t

Vf – Vi = at

Vf = Vi + at ........................................................(1)

Second Equation of Motion

Let Vi be the initial velocity of an object and

'a' the acceleration produced in the body.

The distance travelled s in time t is given by

the area enclosed by the velocity-time graph

for the time interval 0 to t.

Distance travelled S = area of the

trapezium ABDO

Total aera of OABD= area of rectangle

ACDO + area of Î”ABC

= OD x OA + 12 (BC x AC)

Where, OA=VI, OD=t, BC=BD-CD, AC=t, BD=Vf, CD=Vi

= t x Vi + 12 (Vf - Vi ) x t

= Vi t + 12 (Vf - Vi ) x t

(vf = Vi + at I eqn of motion; vf - vi =

at)


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S = Vi t +

12at x t

S = Vit + 12at2.

Third Equation of MotionLet 'vi' be the initial velocity of an object and a be the acceleration produced

in the body. The distance travelled’s’ in time’t’ is given by the area enclosed

by the speed (v) - t graph.

S= area of the trapezium OABD.

= (12) (OA + BD) x AC where OA=Vi, BD= Vf, AC=OD=t

= (12) (Vi + Vf)t ....(1)

But we know that a =( Vf - Vi )t

Or t = (Vf - Vi )a

Substituting the value of t in eq. (1) we get,

s = (12) (Vi + Vf)( Vf − Vi)a = (

12) (Vf + Vi)( Vf − Vi)a

2as = (Vf + Vi) ( Vf - Vi )

(Vf + Vi)( Vf - Vi) = 2as [using the identity a2 - b2 = (a + b) (a -

b)]

v2 - Vi 2 = 2as........... Third Equation of Motion

Acceleration Due To Gravity Or Free Falling Objects

“Galileo was the first scientist to observe that, neglecting the effect of air

resistance, all bodies in free-fall close to the Earth’s surface accelerate

vertically downwards with the same acceleration: namely 9.8 m/s2″


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Example

If a ball is thrown vertically upward, it rises to a particular height and then

falls back to the ground. However this is due to the attraction of the earth

which pulls the object towards the ground”

Characteristic Of Free Falling Bodies

1. When a body is thrown vertically upward, its velocity continuously

decreases and become zero at a particular height During this motion the

value of acceleration is negative and Vf is equal to zero

(a = -9.8m/s2 , Vf = 0).

2. When a body falls back to the ground , its velocity continuously increases

and become maximum at a particular height During this motion the value of

acceleration is positive and Vi is equal to zero (a = 9.8m/s2 , Vi = 0).

3. Acceleration due to gravity is denoted by a and its value is 9.8m/s2 .

4. Equation of motion for the free-falling bodies be written as,

Vf = Vi + gt

h = Vi t + 12 gt2

2gh = Vf 2 – Vi 2

CHAPTER: 2 KINEMATICS

P2.1) A train moves with a uniform Velocity of 36kmh−1 for 10S. Find the distance travelled by it. Solution:(Velocity) V= 36 kmh−1

= 36x1000/60x60= 36000/3600= 10ms−1

(Time) t= 10ms−1

(Distance) S= ?Formula: S= Vavx t = (10) x (10)


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S= 100 m ANS

P2.2) A train starts from nest. It moves through 1km in 100S with uniform acceleration. What will be its speed at the end of 100S?Solution:(Distance) S= 1km =1000 m(Time) t= 100 S

(Velocity)Vi= 0 m/sVf=?By using formula:

S= vit+ (12) a t 2

1000= 0(t)+ 12 a (100)2

1000= 12 a (10000)

2x1000/10000=aA= 0.2 ms−2 ANSNow by using first equation of motion:

Vf= v ;+atVf= 0=(0.2)(100)Vf= 20 ms−1 ANS

P2.3) A car has a Velocity of 10m/s. At accelerate at 0.2 m s2 for half minute. Find the distance travelled during this time and the final Velocity of the car.Solution:(Initial Velocity) Vi= 10m/s(Acceleration) a= 0.2m/s2

(time) t= 12 minutes= 30s

(final velocity) Vf= ? S=?By using 1st equation of motion:Vf= Vi+atVf= (10)+(0.2)(30)Vf= 10+6Vf= 16ms−1


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By using 3rd equation of motion to find s:2aS= Vf 2-V;2

2aS= Vf 2-V;2

2(0.2)= (16 )2-(10 )2

0.4 S= 256-100

S= (1560.4

)

S= 390m ANS

P2.4) A tennis ball is hit vertically upward with a Velocity of 30m/s. it takes 3s to reach the highest point. Calculate the maximum highest reached by the ball. How long it will take to return to the ground?Solution:(Initial Velocity) Vi= 30m/s(Time) t1= 3s(Height) S=?Time required returning to the ground t2=?

g = -10m/s2

The value of g will be negative because the ball will be decelerating.Now by using the 2nd equation of the motion:

S= vit+ (12) (10)(3 )2

= 90+(-5)(9)= 90-45Height S= 45m ANS

P 2.6) A train starts from the nest with an acceleration of 0.5ms−2. Find its speed in kmh−1 when it has moved through 100m.Solution:Initial Velocity Vi= 0Acceleration a= 0.5ms−2

Distance s= 100mFinal Velocity Vf=?To find the final Velocity we have to find the time. By using 2nd equation of motion: S= vit+1/2a t 2

By putting the values:


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100= (0)t+1/2(0.5)(t 2 )100= ½(0.5)t 2

100= 0.25t 2

t 2= 1000.25=400

t 2= 400Taking square root on both sides:√ t2=√400T= 20sNow for Vf, we have formula:Vf= Vi+atVf= 0=(0.5)(20)Vf= 10ms−1

Now to convert 10m/s into km/h, we will multiply it with 3600 nad divide it by 1000.

So, Vf= 10 x 36001000

Vf= 36 kmh−1 ANS

P2.8) A cricket ball is hit vertically upward and returns to the ground 6s later. Calculate:

(i) Maximum height reached by the ball(ii) Initial Velocity of the ball

Solution:Acceleration g = -10m/s2

Time t= 6sTime for upward= t1= 6/2= 3s

Height= s?= Initial Velocity= Vi?

Final Velocity= Vf= 0By using 1st equation of motion:Vf= vi+gtO= vi+(-10)(3)O= vi-30

30=viVi=30ms−1

By using 3rd equation of motion:


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2aS =V f 2-v i2

2x(-10) x s= 0 – (30)-20s= -900

20s= 900S= 900/20S= 45m

ANS

P2.9) When brakes are applied the speed of train decreases from the 96km/h to 48km/h. In 800m how much distance will it cover before coming to rest? (Assume the retardation is constant)Solution:The situation can be divided into two parts. The parts 1 data is as follows:Initial Velocity Vi= 96 km/h = 96x1000/3600= 26.66m/sFinal Velocity Vf= 48km/h = 48x1000/3600= 13.33m/sDistance s= 800mAcceleration a=?By using 3rd equation of motion:2aS= Vf 2- V;2

2a(800)= (13.33 )2-(26.66 )2

1600s= -533.35a= -533.35/1600a= -0.33m/s1

vi= 48km/h= 13.33m/svf= 0 m/ss=?Again by using 3rd equation of motion:2aS= vf 2 – vi2

2(-0.3)s= (0 )2- (13.33 )2

-0.6s= -177.688S= -177.688/-0.6S= 266.53m

P2.10) In problem 2.9 find the time taken by the train to stop after the application of brakes.


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29


Solution:Initial Velocity Vi= 96km/h= 26.667m/sFinal Velocity Vf= 0m/sAcceleration a= -0.33m/s2

Time t=?Formula Vf= vi+at O= 26.677+(-0.33)t-26.66= -033tT= -26.667/-0.33T= 80.80Vf-vi=at0-26.677/-0.3344=tT=80s ANS

P2.11) A car moves with uniform Velocity of 5s it comes to rest in the next 10s/ Find deceleration and total distance covered by the car?Solution:Initial Velocity Vi= 40ms−1

Time t = 10sFinal Velocity Vf= 0Retardation a=?Total distance S=?As we know that:

a= (vf−vit

)

a= 0-40/40a= -4 ms−2 ANS Distance travelled in 1st five seconds.S1= vxt = 40x5S1= 200mAverage Velocity for next 10 seconds. Vav= 40+0/2= 20ms−1

S2= Vav*t = 20x10s2= 200m


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30


Total distance S= S1 + S2

= 200+200 = 400m ANS

Chapter # 3 Dynamics

1. FORCE

The Force is an agent which produces or tends to produce a motion in a

body or it stops or tends to stop the motion of a body. In simple words we

can also say that force is an agent which changes or tends to change the

state of an object.

UNIT

The unit of a Force in M.K.S System is Newton

2. MASS

The quantity of matter contained in a body is called mass. It is a scalar

quantity.

FORMULA

F = ma

m = F/a

UNIT

The unit of mass in M.K.S System is Kilogram (kg).

3. WEIGHT

The force with which earth attracts other bodies towards its centre is called

weight. It is a vector quantity.

FORMULA; W = mg


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31


UNIT

The unit of weight in M.K.S System is Newton (N).

INERTIA

Definition

“Inertia is the tendency of a body to resist a change in its state.”

Examples

Cover a glass with a post card and place a coin on it. Now strike the post

card swiftly with the nail of your finger. If the stroke has been made

correctly, the postcard will be thrown away and the coin will drop in the

glass.

If a moving bus stops suddenly, the passenger standing in it feels a jerk in

the forward direction. As a result he may fall. It is due to the fact that the

lower part of the standing passengers comes to rest as the bus stops. But

the upper portion remains in motion due to inertia.

Difference between Mass and Weight

Mass

1. The quantity of matter present in a body is called mass.

2. The mass of a body remains constant everywhere and does not change

by change in altitude.

3. It is a scalar quantity.

4. Mass can be determined by a physical balance.

Weight

1. The force with which the earth attracts a body towards its centre is called

the weight of the body.

2. The weight of a body is not constant. It is changed by altitude.


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32


3. Weight is always directed towards the center of the earth. So it is a

vector quantity.

4. Weight can be determined by only a spring balance.

MOMENTUM

“The quantity or quality of motion is called momentum and it is denoted by

P”

MATHEMATICAL DEFINITION

“It is the product of mass and velocity.”

MATHEMATICAL REPRESENTATION

P = mV

where:

p is the momentum

m is the mass

v the velocity

LAW OF MOTIONS

Newton formulated three laws of motion in his book.

NEWTON FIRST LAW OF MOTIONS

Newton’s first law of motion is also known as the Law of Inertia.

STATEMENT

“Every body continues its state of rest or uniform motion in a straight line

until it is acted upon by an external or unbalance force to change its state of

rest or uniform motion”.

EXPLANATION

This law consists of a two parts


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33


(a) When body is at rest

(b) When body is moving with uniform velocity

(a). When a body is at rest

Newton’s Law states that when a body is at rest, it continues its rest unless

we apply a force on it. When we apply a force, it changes its state of rest

and starts moving along a straight line.

(b) When body is moving with a uniform velocity

Newton’s Law states that when a body is moving, it moves in a straight line

with uniform velocity, but when we apply an opposite force, it changes its

state of motion and come to rest.

Examples

If a bus suddenly starts moving, the passengers standing in the bus will fall

in the backward direction. It is due to the reason that the lower part of the

passengers which is in contract with the floor of the bus is carried forward

by the motion of the bus, but the upper part of the body remains at rest due

to inertia and so the passengers fall in backward direction.

If a bus suddenly stops moving, the passengers standing in the bus will fall

in the forward direction. It is due to the reason that the lower part of the

passengers which is in contract with the floor of the bus is stopped with the

bus, but the upper part of the body remains moving due to inertia and so

the passengers fall in forward direction.

SECOND LAW OF MOTION

STATEMENT“When a force acts on an object it produces an acceleration which is directly proportion to the amount of the force and inversely proportional to the product of mass”


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EXPLANATION

When we push a body with greater force then its velocity increases and

change of velocity takes place in the direction of the force. If we apply a

certain force F on a mass m, then it moves with certain velocity in the

direction of the force. If the force becomes twice then its velocity will also

increase two times. In this way if we go on increasing the fore there will be

increase in velocity, which will increase the acceleration.

DERIVATION

According to the Newton`s Second law of motion when a force acts on an

object it produces an acceleration which is directly proportion to the amount

of the force.

a ∝ F

and inversely proportional to the product of mass

a ∝ 1m

Combining both.

a ∝ Fm

A = constant F/m

a = k Fm

If the Value of constant K is 1

so,

a = Fm


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35


or

F = ma

THIRD LAW OF MOTION

Statement:

“To every action there is always an equal and opposite reaction ”

EXPLANATION

According to Newton’s Law of Motion, we have:

F(action) = – F(reaction)

The negative (-) sign indicates that the two forces are parallel but in the

opposite direction. If we consider one of the interacting objects as A and the

other as B, then according to the third law of motion:

F(AB) = – F(BA)

F(AB) represents the force exerted on A and F(BA) is the force exerted

on B.

Examples

We walk on the ground, we push the ground backward and as a

reaction the ground pushes us forward. Due to this reason we are able to

move on the ground.

If a book is placed on the table, it exerts some force on the table, which is

equal to the weight of the book. The table as a reaction pushes the book

upward. This is the reason that the book is stationary on the table and it

does not fall down.

FRICTION


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36


Definition

The force, which resists the motion of one surface on another surface, is

known as

friction.

Explanation

Suppose a wooden block is placed on a table and a spring balance is

attached on it. If we apply a very small force of magnitude F by pulling the

spring gradually and increase it, we observe that the block does not move

until the applied force has reached a critical value. If F is less then critical

value, the block does not move. According to Newton’s Third Law of motion

an opposite force balance the force. This opposing force is known as the

force of friction or friction.

CausesofFriction

If we see the surface of material bodies through microscope, we observe

that they are not smooth. Even the most polished surfaces are uneven.

When one surface is placed over another, the elevations of one get

interlocked with the depression of the other. Thus they oppose relative

motion. The opposition is known as friction.

Factors on which Friction Depends

The force of friction depends upon the following factors:

1. Normal Reaction (R)

Force of friction is directly proportional to normal reaction (R), which act


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37


upon the body in upward direction against the weight of the body sliding on

the surface.

2. Nature of Surfaces

Force of friction also depends upon the nature of the two surfaces. It is

denoted as u and has constant values for every surface. It is different for

the two surfaces in contact.

Coefficient Of Friction

The coefficient of friction is a number which represents the friction between

two surfaces. Between two equal surfaces, the coefficient of friction will be

the same. The symbol usually used for the coefficient of

friction is Greek letter µ, where 0 ≤ µ ≤ 1 .

The maximum frictional force (when a body is sliding) is

equal to the coefficient of friction × the normal reaction force.

F = µ R

Where µ is the coefficient of friction and R is the normal reaction force.

This frictional force, F, will act parallel to the surfaces in contact and in a

direction to oppose the motion that is taking/ trying to take place.

Advantages of Friction

1. We could not walk without the friction between our shoes and the ground.

As we try to step forward, we push our foot backward. Friction holds our

shoe to the ground, allowing you to walk.

2. Writing with a pencil requires friction. We could not hold a pencil in our

hand without friction.


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3. A nail stays in wood due to friction

4. Nut and bold hold due to friction

DISADVANTAGES OF FRICTION

1. In any type of vehicle–such as a car, boat or airplane–excess friction

means that extra fuel must be used to power the vehicle. In other words,

fuel or energy is wasted because of the friction.

2. The Law of Conservation of Energy states that the amount of energy

remains constant. Thus, the energy that is “lost” to friction in trying to move

an object is really turned to heat energy. The friction of parts rubbing

together creates heat.

3. Due to the friction a machine has less efficiency less than 100%.

4. Due to friction machine catch fire.

Laws of Friction

Statement

The value of limiting friction increases proportionally with the increase in

normal reaction. Hence, liming friction F(s) is directly proportional to the

normal reaction.

F(s) < R (Here < represents the sign of proportionality don’t’ write it in the

examination paper.)

=> Fs = µ R ……….. (i)

u = F(s)/R

u is the constant of proportionality, which depends upon the nature of the

surfaces of the two surfaces in contact. It is known as the coefficient of

friction. It is only a number without any unit. We know that the normal

reaction is directly proportional to the weight of the block, therefore,


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39


R = W = mg

Substituting the value of R in equation (i)

=> Fs = µ mg

Rolling Friction

When a body rolls over a surface, the force of friction is called rolling

friction. Rolling friction is much less than the sliding friction. This is because

the surfaces in contact are very much less.

LONG QUESTIONSQuestion: Explain the Law of Conservation of Momentum?This law states that

“When two or more bodies collide with one another the total momentum of

the

system remains the same, provided no external force acts upon them.”

Explanation: -The law of conservation of momentum is a fundamental law of nature, and it

states that the total momentum of an isolated system of objects (which has

no interactions with external agents) is constant. One of the consequences

of this is that the of any centre of mass system of objects will always

continue with the same velocity unless acted on by a force outside the

system.

Consider two balls of masses m1 and m2. They are initially moving with

velocities u1and u2 in same direction on a straight line. If u1 > u2, then the

balls will collide. Let their velocities becomes v1 and v2 after collision.

Total momentum of balls before collision = m1u1 + m2u2

Total momentum of balls after collision = m1v1 + m2v2


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40


According to Law of conservation of momentum

Total Momentum before collision = Total Momentum after collision

m1u1 + m2u2 = m1v1 + m2v2

Rockets and jet engines also work on the same principle. In these machines, hot gases produced by burning of fuel rush out with large momentum. The machines gain an equal and opposite momentum. This enables them to move with very high velocities.

Question: Define friction and describe the types of friction ?

Question: What is force of friction? How friction can be reduced?Friction: -The force, which resists the motion of one surface on another surface, is

known as

friction.

Methods to reduce friction: -i) Sliding parts should be highly polished to reduce friction.

ii) Friction of liquids is less than solids. Therefore oil or grease is applied

between the parts of machinery.

iii) Rolling friction is less then sliding friction. Therefore sliding friction

should be converted to rolling friction by using ball bearings.

iv) Front side of vehicles, aeroplanes and ships are shaped wedge like and

pointed so that minimum friction is offered by air.

Centripetal ForceDefinition“The force that causes an object to move along a curve (or a curved path) is called centripetal force.”Mathematical Expression

We know that the magnitude of centripetal acceleration of a body in a

uniform circular motions is directly proportional to the square of velocity


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41


and inversely proportional to the radius of the path Therefore,

a(c) ∝v2

a(c) ∝ 1/r

Combining both the equations:

a(c) ∝ v2/r

From Newton’s Second Law of Motion:

F = ma

=> F(c) = mv2/r

Where,

Fc = Centripetal Force m = Mass of object

v = Velocity of object r = Radius of the curved path

Banking of the roads Factors on which Fc Depends:

Fc depends upon the following factors:

Increase in the mass will increases Fc.

It increases with the square of velocity.

It decreases with the increase in radius of the curved path.

When a car takes

Examples

The centripetal force required by natural planets to move constantly round

a circle is provided by the gravitational force of the sun.

If a stone tied to a string is whirled in a circle, the required centripetal force

is supplied to it by our hand. As a reaction the stone exerts an equal force

which is felt by our hand.


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42


The pilot while turning his aero plane tilts one wing in the upward direction

so that the air pressure may provide the required suitable Fc.

Centrifugal ForceDefinition

“A force supposed to act outward on a body moving in a curve is known as

centrifugal force.”

Explanation

Centrifugal force is actually a reaction to the centripetal force. It is a well-

known fact that Fc is directed towards the centre of the circle, so the

centrifugal force, which is a force of reaction, is directed away from the

centre of the circle or the curved path.

According to Newton’s third law of motion action and reaction do

not act on the same body, so the centrifugal force does not act on

the body moving round a circle, but it acts on the body that

provides Fc.

Examples

If a stone is tied to one end of a string and it is moved round a circle, then

the force exerted on the string on outward direction is called centrifugal

force.

The aeroplane moving in a circle exerts force in a direction opposite to the

pressure of air.

When a train rounds a curve, the centrifugal force is also exerted on the

track.


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Vertical motion of two bodies attached to the ends of a string that

passes over a frictionless pulley

Consider two bodies A and B of masses m1 and m2 respectively, let m1 is

greater than m2. the bodies are attached to the opposite ends of an

inextensible string. The string passes over a frictionless pulley. the body A

being heavier must be moving downward with some acceleration. Let this

acceleration be a. At the same time, the body B attached to the other end

of the string moves up with the same acceleration a. As the pulley is

frictionless, hence tension will be the same throughout the string. Let the

tension in the string be T.

Since the body A moves downwards, hence its weight m1g is greater than

the tension T in the string.

Net force acting on body A=m1g-T

According to the Newton’s law of motion:

m1g-T = m1a ..... ...... 1)

As body B moves upwards, hence its weight m2g is less than the tension T

in the string.

Net force acting on body B = T – m2g

According to Newton law

T- m2g = m2a ...... ..... 2)

Adding Eq.1and Eq 2 , we get acceleration a.


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a= (

m1−m2m1+m2 )g

Divide Eq. 2 by Eq.1, to find tension T in the string.

T = (2m1m2m1+m2

)g ..... ..... 3)

The above arrangement is also known as Atwood machine. It can be used to

find the acceleration g due to gravity using , g = ( m1+m2m1−m2

)

a

OR


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45


Motion of two bodies attached to the ends of a string that passes

over a frictionless pulley such that one body moves vertically and

the other moves on A smooth horizontal surface

Consider two bodies A and B masses m1 and m2 respectively attached to the

ends of an inextensible string as shown in the figure above.Let the body A

moves downwards with an acceleration a. Since the string is inextensible,

therefore, body B also moves over the horizontal surface with the same

acceleration a. As the pulley is frictionless hence tension T will be the same

throughout the string.

Since body A moves downwards, therefore, its weight m1g is greater than

the tension in T in the string.

Net force acting on the body A = m1g – T

According to the Newton’s second law of motion:

m1g – T = m1a ... ... ... (1)

The forces acting on the body are:

I. Weight m2g of the body B acting downwards.

II. Reaction R of the horizontal surface acting on the body B in the

upward direction.

III. Tension in the string pulling the body B horizontally on the smooth

surface.


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As body B has no vertical motion, hence resultant of vertical forces (m2g

and R) must be zero.

Thus the net force acting on the body B is T.

According to the Newton’s second law of motion:

T = m2a ... ... ... (2)

Adding eqs. 1 and 2, we get acceleration a as:

m 1g –T + T =m1a+ m2a

m 1g = a(m1+m2)

a = m1

m1+m 2 g ... (3)

Putting the value of a in equations 2 to get tension T as:

T = m1m2m1+m 2 g ... ... ... (4)

FORCE AND THE MOMENTUM:

Consider a body of mass m moving with initial velocity Vi. Let a force F acts on the body which produces an acceleration a in it. This changes the velocity of the body. Let its final velocity after time t becomes Vf. if Pi and Pf be the initial momentum and the final momentum related to the body related to the initial and the final velocity respectively then:

Pi = mvi and Pf = mvf

Changes in momentum = Final momentum – initial momentumOr Pf – Pi = mvf - mvi

Thus the rate of change in momentum given by:p−pt

= mv−mvt

= m vf−vit


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Since

vf−vit is the rate of change of velocity equal to the acceleration a

produced by the force F.pf−pit

= m a

According to Newton’s second law of motion:F = ma

Or pf−pit = F

Equation also defines the force and states Newton’s second aw of motion as:“When a force acts on a body, it produces acceleration in the body and will be equal to the rate of change of momentum of the body.”SI unit of momentum defined by equation is Newton-second (Ns) which is the same as kmgs-1.

3.4. What is the law of Inertia?Ans: (Inertia is the resistance of any physical object to a change in its state of motion or rest, or the tendency of an object to resist any change in its motion including a change in direction).3.5. Why is it dangerous the roof of a bus to travel on?Ans : The friction or drag force due to air acting on the upper part of the body of a person standing on the roof of running bus tries to turn over which is dangerous while the lower part of body remains at rest w.r.t roof of the bus3.6. Why does a passenger move outward when a bus takes turn?Ans: When does bus take a turn the passenger sitting inside experienced centrifugal force and moves out wards

3.7. How can you relate a force with the change of momentum of a body?

Ans : By using 2nd of motion we can write F = ma here a= vf−vit by

putting this value we can get

F = m vf−vit

F = mvf−mvi

t but F = final momentum−initialmomentum

timeForce = time rate of change of momentum


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48


3.8. What will be the tension in the rope that is pulled from the end by two opposite forces 100N each?Ans: When two forces of 100 N each applied on a string then resultant tension is equal to 100. 3.9. Action and reaction are always equal and opposite. Then how does a body move?Ans: Action and reaction force equal in magnitude but opposite in direction. These do not act upon the same body. Action force is applied on one body, which give reactional force acting on other body. Both of these do not neutralized this is the result of motion. 3.10. A horse pulls the cart. If the action and reaction are equal and opposite then the how does the cart move?Ans: The horse applies action force by feet on the road the reaction is given by road on horse due to which the cart tied to the horse also move.3.11. What is the law of conservation of momentum?Ans : “When two or more bodies collide with one another the total

momentum of the

system remains the same, provided no external force acts upon them.”

According to Law of conservation of momentum

Total Momentum before collision = Total Momentum after collision

m1u1 + m2u2 = m1v1 + m2v2

3.12. Why is the law of conservation of momentum important?Ans : by using law of conservation of momentum it is possible to calculate, force , velocity, acceleration of a body. Most of elementary particles were discovered by this law.3.13. When a gun is fired, it recoils. Why?Ans : according law of conservation of momentum the momentum gained by fired bullet is neutralized by equal and opposite momentum given to the gun recoils back.3.14. Describe two situations in which force of friction is needed.Ans: The friction between walking person and surface of earth is necessary for walking.


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To stop a moving vehicle force of friction between tyre and road is required if this is reduced by putting oil on the road then it would be impossible to stop a vehicles. 3.15. How does oiling the moving parts of a machine lower the friction?Ans : by oiling the various parts of a matching of friction is reduced which increase its efficiency3.16. Describe ways to reduce friction.Ans: Methods to reduce friction: -i) Sliding parts should be highly polished to reduce friction.

ii) Friction of liquids is less than solids. Therefore oil or grease is applied

between the parts of machinery.

iii) Rolling friction is less then sliding friction. Therefore sliding friction

should be converted to rolling friction by using ball bearings.

iv) Front side of vehicles, aeroplanes and ships are shaped wedge like and

pointed so that minimum friction is offered by air.

3.17. Why rolling friction is lower than sliding friction?Ans: the interlocking between ups and down of two surface need not be ruptured in case of rolling while in case of sliding these are to be ruptured and result to increase in friction .3.18. What do you know about the following? (i) Tension in a stringAns: Tension in string is to neutralize applied force on the string this prevents it from moving.

(iii) Limiting force of frictionAns: The force of friction on the body at rest lying in a rough surface is called force of static friction its value increases with the increase of applied force .The maximum possible value of static friction if applied force made grater then it body starts moving is called limiting force of static friction

(iv) Braking forceAns: the bracking force is the force between brake pushing and wheel of vehicle. It is help to stop the wheels.

(v) Skidding of vehicles


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Ans: when a force of friction between tyre and road is small then applying brakes tyre slide over the road .it is called skidding of vehicles

(vi) SeatbeltsAns: The seat belts provide opposition against falling ahead when vehicle is stopped suddenly

(vii) Banking of roadsAns: the outer edge of road is made higher to provide reactional force on tyre which prevents it from slopping. It is called banking of road.

(viii) Cream separatorAns: in a cream separator milk is rotated and lighter particles of cream come at the axis of rotation and are separated from milk and collected through the pipe.3.19. What would happen if all friction suddenly disappears?Ans: when the frictional force suddenly disappears the motion of the object would never be stopped.3.20. Why the spinner of a washing machine is made to spin at a high speed?Ans: at very high spinning speed the water and dirt particles are separate from cloths to clean them inside washing machine.

Chapter No 3

Dynamics

P 3.1) A force of 20 N moves a body with an acceleration of 20ms-2

what is its mass?

Solution:

Force F= 20 N Acceleration a= 20ms-2 Mass m=?

Formula F= ma

m= Fa


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20/2 = 10 kg Ans

P 3.2) Weight is 147 N what is its mass?

Solution:

Weight w=147 N

Acceleration g = 10ms-2

Mass m=?

Formula W = mg m = Wg

147/10 = 14.7 kg Ans

P3.3) How much force is needed to prevent a body of mass 10kg

from falling?

Solution:

Mass m=10kg

Force F=a

The force needed to prevent the body from falling is equal to the weight of

the body

F = W

W=mg

F=mg

F= 10 x 10= 100 N Ans


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P3.4) Find the acceleration produce by a force of 100 N in a mass of

50 kg?

Solution:

Acceleration a =?

Force F = 100 N

Mass m = 50kg

Formula:

F= ma a= Fm

a=100/50 = 2ms-2

P3.5) A body has weight 20N how much force is required to move it

vertically upward with an acceleration of 20ms-2?

Solution:

Weight W = 20N

Acceleration a = 20ms-2

Force F=?

To find out force we have to first calculate the Mass of the body

To find out the mass to use W= mg m = w/g m= 20/10 m = 2kg

So he net force will take the body upward will be

Net force F = W

F= ma

W = mg

The g will be negative because body is moving upward so

W = -mg

Net force F = F-W


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ma – m(-g )

ma + mg

m(a +g)

2 (2+10)

Force F= 24N Ans

P3.6) Two masses 52kg and 48kg are attached to the end of the

string that passes over a frictionless pulley. Find the tension in the

string and acceleration in the body? When the masses are moving

vertically.

Solution:

m1 = 52kg

m2 = 58kg

T =?

a =?

First we find tension in the string

Formula:

T = (2m1m2m1+m2

)g

T = 2(52) (48) (10)/ 52+48

T = 499.2N

T = 500N approximately Ans

Now we will find the acceleration

Formula:

a = ( m1−m2m1+m2 )g


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a = (52-48) (10)/ 52+48

a= 4x10/100

a= 40/100

a= 0.4ms-2 Ans

P3.7) Two masses 26kg and 24kg are attached to the end of a string

which passes over a frictionless pulley. 26kg is lying over a smooth

horigalal table .24kg mass is moving vertically downward. Find the

tension in string and acceleration in bodies.

Solution:

m1 = 24kg m2 = 26kg T =? a =?

Formula:

T = (2m1m2m1+m2

)g

T = (24) (26) (10)/24+26

T = 124.8N

T = 125N

Formula: a = ( m1−m2m1+m2 )g

= 24x10 / 24+ 26

a = 408ms-2 Ans

P3.8) How much time is required to change 22 Ns momentum by a

force of 20 N?

Solution:

(Initial momentum) Pi = 22Ns


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Pf = 0Ns F = 20N t =?

Formula:

Pf−Pit = F

F = Pf−Pit

= 0-22/ 20

t = -1.1s

As time cannot be negative to

t = 1.1s ANS

P3.9) How much is the force of friction between a wooden block of

mass 5 kg and the horizontal marble floor? The coefficient of

friction between the wood and marble is 0.6

Solution:

Fr =? m= 5 kg u= 0.6 Formula Fr = UF

F = mg

5 x 10 = 50 N

Fr =UF

Fr = 0.6 x 50 = 30 N ANS


Documents

Complete Notes on 9th Physics by Asif Rasheed