Chapter Two Class 9th

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CHP # 2

KINEMATICS

KINEMATICS The branch of physics which deals with the study of motion of material objects without discussing its causes (forces) is called kinematics. Q.1 A.

Define motion and rest with example? Motion When a body changes its position with respect to its surrounding, we say that the body is in motion. For example if a body "B" changes its position with respect to another body "A", we say that body "B" is in motion. Rest When a body does not change its position with respect to its surrounding, we say that the body is at rest. For example if a body "B" does not change its position with respect to another body "A", we say that body "B" is at rest.

Q.2 A. (1)

Write types of motion with examples? There are three types of motion which are stated as; Translatory motion A type of motion in which each and every particle of the body move in the same manner as that of every other particle. For examples (i) Motion of the cars (ii) Motion of the flying birds (iii) Motion of the falling objects Rotatory motion A type of motion in which a body rotate about a fixed point and the distance of the body at any time is constant (equal) to the fixed point. For examples (i) Motion of the wheel of a cycle (ii) Motion of the hands of a clock (iii) Motion of the wings of a fan. Vibratory motion The To and fro motion of an object around a fix point known as mean or equilibrium position is called Vibratory motion. For examples (i) Motion of a swing. (ii) Motion of a simple pendulum. (iii) Motion of mass attached to a spring.

(2)

(3)

Q.3 A.

Differentiate between distance and displacement? Distance The length cover by a body on different paths is called distance. It is a scalar physical quantity. It is denoted by "S". Its SI unit is meter (m).

Displacement The shortest distance between two points is called displacement. It is a vector physical quantity. It is denoted by "S". Its SI unit is meter (m). Q.4 A.

Define speed and its derivatives? Speed The distance covered by a body in unit time is called speed. It is denoted by "V". It is a scalar physical quantity. Mathematically S Dis tan ce  Speed  t time Its SI unit is meter/second (m/s). Derivatives of the speed are;

(i)

Average speed It can be defined as "the total distance covered divided by total time taken is called average speed". Mathematically S = t Instantaneous speed It can be defined as "the time rate of change distance covered by a body is called Instantaneous speed". Mathematically S Vins  t Uniform speed It can be defined as "if a body covers equal distance in equal interval of time, then the speed of the body is called uniform speed".

(ii)

(iii)

Q.5 A.

Define Velocity and its derivatives? Velocity The displacement covered by a body in unit time is called Velocity. It is denoted by "V". It is a vector physical quantity. Mathematically Displacement Velocity  time

  S V  t

Its SI unit is meter/second (m/s).

(i)

Derivatives of the Velocity are; Average Velocity It can be defined as "the total displacement covered divided by total time taken is called average Velocity". Mathematically



 S = t (ii)

Uniform Velocity It can be defined as "if a body covers equal displacement in equal interval of time, then the Velocity of the body is called uniform Velocity". OR If the speed as well as direction of the body does not change in a time interval, then the Velocity of the body is called uniform Velocity.

(iii)

Variable Velocity It can be defined as "If the speed or direction or both of the body changed in a time interval, then the Velocity of the body is called variable Velocity.

Q.6 A.

Define Acceleration and its derivatives? Acceleration It can be defined as "the time rate of change of Velocity is called acceleration". It is dented by "a". It is a vector physical quantity. Mathematically

 V f  Vi  V a = t t

(i)

(ii)

(iii)

Its SI unit is meter per second per second (m/s2). Derivatives of the acceleration are; Positive acceleration It can be defined as "if the magnitude of Velocity increases with time, then acceleration is called positive acceleration". Negative acceleration It can be defined as "if the magnitude of Velocity decreases with time, then acceleration is called negative acceleration". Average acceleration It can be defined as "the change in Velocity divided by time interval is called average acceleration". Mathematically

 V  a  t

Q.7 A.

Define scalars and vectors with examples? Scalars Those physical quantities which are completely specified from its magnitude (number+proper unit) are called scalars. For examples Length, mass, time, speed, area, volume, etc.

Vectors Those physical quantities which are completely specified from its magnitude (number+proper unit) as well as direction are called vectors. For examples Displacement, Velocity, acceleration, Force, Momentum, torque, etc. Q.8 A. (i)

How a vector can be represented? A vector can be represented by the following two methods. Symbolic representation In this method a vector can be represented by a letter. An arrow head is placed above or below the letter. OR the letter is written as bold face For example a vector A can be stated as; 

A (ii)

A,

,



A

Graphical representation Graphically a vector can be represented by a straight line having an arrow head in the direction of the vector. This process is completed in the following four steps: 1. Select a suitable scale. 2. Draw NEWS or coordinate system. 3. Draw a 5cm representative line on the NEWS. 4. Draw the required line according to the direction. Example Al–Khalid tank is moving with a velocity of 100Km/h towards north east. Draw representative line of its velocity? Solution N 1. Scale 20Km/h = 1cm 100Km/h = 5cm W E 2. Draw NEWS system

S 3.

Representative line on the NEWS. N

5cm

45o W

E O S

4.

Required line 5cm 45o O

E

Q.9 A.

What is graph? How a graph can be drawn. Graph It is a method to show the relationship between two physical quantities. For example Distance-time graph shows that how speed or Velocity of a body changes. A graph is drawn on a graph paper. Graph paper contains horizontal and vertical lines of equal distances. We use Rectangular coordinate system for graph, which consist of two mutually perpendicular lines XOX  and YOY  . As shown below in figure;

Q.10 A. (i) (ii) (iii)

How we can determine the slope of a graph? The slope of a graph can be determined as follow; Take two points P1 ( x1 , y1 ) and P2 ( x2 , y 2 ) on the graph. Draw perpendiculars on the x-axis and y-axis from both points. Calculate the difference in x and y coordinates as; x  x2  x1 and y  y 2  y1 Divide y by x ; this is the slope of the graph. Mathematically y  y1 y Slope = = 2 x x 2  x1

(iv)

Q.11 What is Distance-time graph? A. Distance- time graph A graph which shows the relationship between distance and time is called Distance-time graph. The distance traveled is taken on the vertical axis (y-axis) while the time is taken on the horizontal axis (x-axis). The slope of the graph denotes speed or Velocity. Mathematically Slope =

y  y1 y = 2 x x 2  x1

Speed or Velocity =

y  y1 y = 2 x x 2  x1

Slope Distance (m)

O Time (sec)

Q.12 What is Speed-time graph? A. Speed- time graph A graph which shows the relationship between speed and time is called speedtime graph. The speed is taken on the vertical axis (y-axis) while the time is taken on the horizontal axis (x-axis). The slope of the graph denotes acceleration. Mathematically

y  y1 y Slope = = 2 x x 2  x1 Acceleration =

y  y1 y = 2 x x 2  x1

Slope Speed (m/s)

O Time (sec)

Q.13 How the distance traveled can be determined from speed-time graph? A. The distance traveled can be calculated from the speed-time graph by the area enclosed the graph. For example Consider a body moves with uniform speed, its initial speed is OA. After a time interval OC the speed of the body becomes BC. The graph can be drawn as; Mathematical form As we know that B S C V  t S  Vt it is clear from figure, that V = OA, t = OC Speed (m/s) S = OA x OC S = Width x Length O A Time (s)

S = Area of rectangle OABC S = Area of the graph Q.14 Prove first equation of motion ( V f  Vi  at ) by graph? A. Statement Consider the speed-time graph in which the initial speed of the body is "OA". After a time interval "OC" the speed of the body changes uniformly and becomes "BC". The slope of the graph "AB" shows acceleration "a". Figure Y B a

A Vi O

D t

t

C

Mathematical proof It is clear from figure that BC  OA  BD V f  Vi  BD ------------ (1) To find BD, we have Slope of graph AB =

BD AD

BD t at  BD a

Put the value of BD in equation (1)

V f  Vi  at

Vf

X

Q.15 Prove second equation of motion ( S  Vi t  A.

1 2 at ) by graph? 2

Statement Consider the speed-time graph in which the initial speed of the body is "OA". After a time interval "OC" the speed of the body changes uniformly and becomes "BC". The slope of the graph "AB" shows acceleration "a". The distance traveled by the speed–time graph can be calculated as; Distance traveled = Area enclosed by the graph Figure Y B a

at

A

D t

Vi O

X

C

t Mathematical proof It is clear from figure that S = Area of OABC But S = Area of rectangle OADC + Area of triangle ABD 1 S = OA  OC + (AD  BD) 2 1 S  Vi  t  (t )(at ) 2

S  Vi t 

1 2 at 2

Q.16 Prove third equation of motion A.

( 2aS  V f  Vi ) 2

2

by graph?

Statement Consider the speed-time graph in which the initial speed of the body is "OA". After a time interval "OC" the speed of the body changes uniformly and becomes "BC". The slope of the graph "AB" shows acceleration "a".

Figure

Y B a

A

D

Vf

t

Vi

O C t Mathematical proof It is clear from figure that S = Area of OABC S = Area of Trapezium OABC We know that

X

Area of Trapezium = (sum of parallel sides of Trapezium) 

AD 2 OC S  (Vi  V f )  2 t S  (Vi  V f )         ( A) 2 But from first equation of motion; V f  Vi  at S  (OA  BC ) 

t 

V f  Vi

a Put the value of “t” in equation (A); We get Vf  Vi S = ( Vi  Vf )  ( ) 2a

2aS  V f

2

 Vi

2

height 2

Q.17 What is meant by gravitational acceleration? A. Gravitational acceleration An Italian scientist Galileo drop several objects fro the famous Leaning tower at the same time. He concluded that all the objects reached at the same time towards the earth surface. He stated that earth attract every object itself with the same acceleration called gravitational acceleration. It is denoted by “g”. Its value is 9  8m / s 2 at the surface of the earth. Equations of motion for motion under gravity are; 

V f  Vi  gt



1 S  Vi t  gt 2 2

2 gh  V f  Vi 2



2

CONCEPTUALQUESTIONS (1)

The figure given shows the speed time graph for a pendulum. Write down (a) The maximum speed. (b) The time at which the maximum speed occurs. (Ans) (a) The maximum speed is 12m/s. (b) The time at which maximum speed occur is 0 3Sec . (2) Can a body at rest be regarded in a state of motion? Give example (Ans) Yes a body at rest can be regarded in a state of motion. For example A person sitting in a train is at rest with respect to an observer in the train but this person is in motion for the observer outside the train. (3)

Is the distance covered by the body may be greater than the magnitude of displacement? (Ans) Yes the distance covered by a body may be greater than the displacement. For example When an object is moving on a circular path its distance covered is greater than the displacement at any instant of time. (4) Is it possible that displacement is zero but not the distance? (Ans) Yes it is possible that displacement is zero but not the distance. When an object moves in a circle and complete one round trip its displacement become zero while the distance covered is not zero.

(5) Under what condition displacement is equal to the distance? (Ans) The displacement and distance covered is equal if and only if the body is moving in a straight line. (6) Can a body have acceleration with zero velocity? (Ans) A body has no acceleration with zero velocity. Because acceleration is produce when the velocity of the body is changed. Mathematically

  V a t 0 a t a  0m / s 2

(7) Can the speed of a body be negative? (Ans) No the speed of the body can not be zero. Because it is a scalar quantity and scalar quantities can not be negative. (8)

Is it possible that velocity of an object be in a direction other than the direction of acceleration? (Ans) Yes the velocity of a body can be in a direction other than acceleration. For example When a body moves in a circle, the direction of linear velocity is tangent to the circle while the acceleration is directed towards the centre of the circle.

a

(9)

Is the kinematics equation S  Vi t 

V

1 2 at true if acceleration is not 2

constant? (Ans) No the kinematics equation S  Vi t  constant.

1 2 at is not true if acceleration is not 2

(10) By giving an example prove that rest and motion are relative terms. (Ans) Motion and rest are relative. For example two persons are setting in a bus. They are at rest with respect to each other. But due to the motion of the bus, they are in motion with respect to their external surrounding. (11) Give an example of an accelerated body moving with a uniform speed. (Ans) The motion of a body in a circle with uniform speed have an acceleration due to change in the direction of velocity called centripetal acceleration. V

a V V

V (12) Is unit Kmh-1s-1 is same as Kms-1h-1 explain? (Ans) The unit Kmh-1s-1 is not the same as Kms-1h-1 . Because in unit Kmh-1s-1, Kmh-1 is the unit of velocity the change of which is given in second. But in the unit Kms-1h-1, Kms-1 is the unit of velocity the change of which is in hour. (13) If bus is traveling eastward, can its acceleration be westward? Explain (Ans) Yes a bus traveling eastward its acceleration can be westward. Explaination A bus is traveling eastward, if the velocity of the bus decreasing continuously, then the deceleration is produce which will be westward. (14) If an object is stationary is its acceleration necessary zero? (Ans) Yes if an object is stationary its acceleration is zero. Because the speed of the body is zero and direction is unchanged. (15)

When the velocity time graph is a straight line parallel to time axis, what can you say about its acceleration. (Ans) The acceleration of the body is zero. Because the graph shows that velocity is constant. When the velocity of the body is constant its acceleration will be zero. (16)

A ball is thrown vertically upward with an initial speed of 5m/s. What will its speed be when it returns to its starting point. (in the absence of air resistance).

(Ans) The body returns to its starting point with the same speed of 5m/s in the absence of air resistance.

(1)

(2)

NUMERICAL PROBLEMS A bus travel 15Km towards west makes u-turn back travel a further distance of 10Km, find (a) Distance traveled b) Its displacement Given data S 1 = 15Km S 2 = 10Km a) Distance, S =? S = S 1+ S 2 S = 15+10 S =25Km  b) Displacement, S =?    S = S 2-S 1  S = 15 -10  S = 5Km ---- towards east of starting point. A race car travels around a circular track, covering a distance of 850m in 25s before stopping at point from where it started. Determine the average Velocity of the car during this period of time. Given data Distance, S = 850m  Displacement, S = 0m S=850m Time, t = 25s   S  0m Average velocity, V  =? We know that   t  25sec S

V    V   V 

t 0  25  0m / sec

(3)

A truck moving at a speed of 20m/s begins to slow at constant rate of 3m/s2, find how far it goes before stopping? Given data

Vi  20m / sec

a  3m / sec2 V f  0m / sec S ? We know that

2aS  V f  Vi 2

V f  Vi 2

S

2

2

2a 0  (20) 2  400 S   66  67m 2(3) 6 2

(4)

The speed of a bus is reducing uniformly from 15m/s to 7m/s. while traveling a distance of 90m. (a) Find the acceleration (b) How much further distance will the bus travel before coming to rest, provided the acceleration remains constant? Given data

Vi  15m / sec V f  7 m / sec

(a)

S  90m a? We know that

2aS  V f  Vi 2

V f  Vi 2

a

2

2

2S (7) 2  (15) 2 a 2  90 49  225  176 a  180 180 a  0  977m / sec2

(b)

Vi  7m / sec V f  0m / sec a  0  977m / sec2 S ? We know that

2aS  V f  Vi 2

V f  Vi 2

S

2

2

2a ( 0) 2  ( 7 ) 2 S 2(0  977)  49 S  1  954 S  25  07m / sec (5)

Brakes are applied to a train traveling at 72Km/h after passing over 200m its velocity is reduced to 36Km/h at the same rate of retardation, how much further will it go before it is brought to rest? Given data

72  1000m  20m / sec 3600sec 36  1000m V f  36Km / h   10m / sec 3600sec S  200m Re tardation  a  ? S ? When Vi  10m / sec Vi  72Km / h 

V f  0m / sec To find the retardation first

2aS  V f  Vi 2

V f  Vi 2

a

2

2

2S (10) 2  (20) 2 a 2  200 100  400  300 a   0  75m / s 2 400 400 Now to find the further distance traveled

2aS  V f S 

Vf

2

2

 Vi

 Vi

2

2

2a 0 2 (10) 2 S  2( 0  75)  100 S   1  50 S  66  67m (6)

A motor cyclist is moving on a road with an acceleration of 3m/s2, how much time will it require to change the velocity from 10m/s to 20m/s? Given data

a  3m / sec2 Vi  10m / sec V f  20m / sec t ? We know that

V f  Vi  at V f  Vi  at t 

V f  Vi

a 20  10 t  3 10 t   3  33 sec 3 (7)

A cyclist starts from rest and moves with uniform acceleration of 0.2m/s 2 after 2 minutes, find the velocity of the cyclist and distance covered. Given data

Vi  0m / sec

a  0  2m / sec2 t  2 min  2  60 sec  120sec Vf  ? S ? We know that

V f  Vi  at

V f  0  (0  2)(120) V f  0  24  0  24m / sec We also know that

1 2 at 2 1 S  0  120  (0  2)(120) 2 2 1 S  (0  2)(14400) 2 S  1440m S  Vi t 

(8)

A body is thrown vertically upward with a speed of 20m/s. How high will it rise? (take downward g=10m/s2) Given data

Vi  20m / sec

g  10m / sec2 V f  0m / sec h? We know that

2 gh  V f  Vi 2

V f  Vi 2

h

2

2g

0 2  ( 20) 2 h 2( 10)  400 h  20 h  20m

2

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