Science and Technology Standard Nine PDF

 The Coordination Committee formed by GR No. Abhyas - 2116/(Pra.Kra.43/16) SD - 4 Dated 25.4.2016 has given approval to prescribe this textbook in its meeting held on 3.3.2017 SCIENCE...

 The Coordination Committee formed by GR No. Abhyas - 2116/(Pra.Kra.43/16) SD - 4 Dated 25.4.2016 has given approval to prescribe this textbook in its meeting held on 3.3.2017 SCIENCE AND TECHNOLOGY STANDARD NINE Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune. The digital textbook can be obtained through DIKSHA APP on a smartphone by using the Q. R. Code given on title page of the textbook and useful audio-visual teaching-learning material of the relevant lesson will be available through the Q. R. Code given in each lesson of this textbook. First Edition : 2017 © Maharashtra State Bureau of Textbook Production Reprint : 2022 and Curriculum Research, Pune - 411 004. The Maharashtra State Bureau of Textbook Production and Curriculum Research reserves all rights relating to the book. No part of this book should be reproduced without the written permission of the Director, Maharashtra State Bureau of Textbook Production and Curriculum Research, ‘Balbharati’, Senapati Bapat Marg, Pune 411004. Science Subject Committee : Cover and illustrations : Dr Chandrashekhar Murumkar, Chairman Shri. Vivekanand Shivshankar Patil Ashna Advani Dr Dilip Sadashiv Jog, Member Typesetting : Dr Abhay Jere, Member DTP Section, Textbook Bureau, Pune Dr Sulabha Nitin Vidhate, Member Smt. Mrinalini Desai, Member Shri. Gajanan Suryawanshi, Member Coordination : Rajiv Arun Patole Shri. Sudhir Yadavrao Kamble, Member Special Officer for Science Smt. Dipali Dhananjay Bhale, Member Shri. Rajiv Arun Patole, Member-Secretary Translation : Dr Sushma Dilip Jog Science Study Group : Dr Pushpa Khare Dr Prabhakar Nagnath Kshirsagar Shri. Sandeep Chordiya Dr Shaikh Mohammed Waquioddin H. Dr Vishnu Vaze Scrutiny : Dr Gayatri Gorakhnath Choukade Smt. Mrinalini Desai Dr Ajay Digambar Mahajan Smt. Shweta Thakur Coordination : Smt. Pushpalata Gawande Dhanavanti Hardikar Shri. Rajesh Vamanrao Roman Academic Secretary for Languages Shri. Hemant Achyut Lagvankar Shri. Nagesh Bhimsevak Telgote Santosh Pawar Smt. Dipti Chandansingh Bisht Assistant Special Officer, English Shri. Vishwas Bhave Shri. Prashant Panditrao Kolse Paper : 70 GSM Creamwove Shri. Sukumar Shrenik Navale Shri. Dayashankar Vishnu Vaidya Print Order : Smt. Kanchan Rajendra Sorate Printer : Smt. Anjali Khadke Smt. Manisha Rajendra Dahivelkar Smt. Jyoti Medpilwar Production : Shri. Shankar Bhikan Rajput Sachchitanand Aphale Shri. Mohammed Atique Abdul Shaikh Chief Production Officer Shri. Manoj Rahangdale Rajendra Vispute Smt. Jyoti Damodar Karane Production Officer Invitees : Publisher : Dr Sushma Dilip Jog Vivek Uttam Gosavi Dr Pushpa Khare Controller Dr Jaydeep Sali Maharashtra State Textbook Bureau, Shri. Sandeep Popatlal Chordiya Prabhadevi, Mumbai - 400 025. Shri. Sachin Ashok Bartakke  Preface Dear students, Welcome to Std IX. We have great pleasure in offering to you this Science and Technology textbook based on the new syllabus. From the Primary level till today, you have studied Science from various textbooks. From Std IX onwards, you will be able to study the fundamental concepts of Science and Technology from a different point of view through the medium of the different branches of Science. The basic purpose of this textbook can be said to be ‘Understand and explain to others’ the science and technology that relates to our everyday lives. While studying the concepts, principles and theories in Science, do make the effort to understand their connection with day to day affairs. While studying from this textbook, use the sections ‘Can you recall?’ and ‘Can you tell?’ for revision. You will learn Science through the many activities given under the titles such as ‘Observe and discuss.’ and ‘Try this’ or ‘Let's try this’. Activities like ‘Use your brain power!’, ‘Research’, ‘Think about it.' will stimulate your power of thinking. Many experiments have been included in the textbook. Carry out these experiments yourself, following the given procedure and making your own observations. Ask your teachers, parents or classmates for help whenever you need it. Interesting information which reveals the science underlying the events we commonly observe, and the technology developed on its basis, has been explained in this textbook through several activities. In this world of speed and technology, you have already become familiar with computers and mobile phones. While studying the textbook, make full and proper use of the devices of information communication technology, which will make your studies so much easier. While carrying out the given activities and experiments, take all precautions with regard to handling apparatus, chemicals, etc. and encourage others to take the same precautions. It is expected that while carrying out activities or observation involving plants and animals, you will also make efforts towards conservation of the environment. You must, of course, take all care to avoid doing any harm or causing injury to any plants or animals. Do tell us about the parts that you like as well as about the difficulties that you face as you read and study and understand this textbook. Our best wishes for your academic progress. Pune (Dr Sunil Magar) Date : 28 April 2017 Director Akshaya Tritiya Maharashtra State Bureau of Textbook Indian Solar Year : Production and Curriculum Research, Pune 8 Vaishakh 1939  For teachers l The real objective of science education is to learn to be able to think about events that are happening around us, logically and with discretion. l In view of the age group of Std IX students, it would be appropriate now, in the process of science education, to give freedom and scope to the students’ own curiosity about the events of the world, their propensity to go looking for the causes behind them and to their own initiative and capacity to take the lead. l As experimentation is necessary to learn the skills of observation, logic, estimation, comparison and application of available data, which form a part of science education, deliberate efforts must be made to develop these skills while dealing with laboratory experiments given in the textbook. All observations that the students have noted should be accepted, and then they should be helped to achieve the expected results. l These two years in middle school lay the foundation of higher education in Science. Hence, it is our duty and responsibility to enrich and enhance the students’ interest in science. You all will of course always actively pursue the objective of developing their creativity and imbuing them with a scientific temper. l You can use ‘Let’s recall’ to review the previous knowledge required for a lesson and ‘Can you tell?’ to introduce a topic by eliciting all the knowledge that the students already have about it from their own reading or experience. You may of course use any of your own activities or questions for this purpose. Activities given under ‘Try this’ and ‘Let’s try this’ help to explain the content of the lesson. The former are for students to do themselves and the latter are those that you are expected to demonstrate. ‘Use your brain power!’ is meant for application of previous knowledge as well as the new lesson, and ‘Always remember’ gives important suggestions/information or values. ‘Research’, ‘Find out’, ‘Do you know?’, ‘An introduction to scientists’ and ‘Institutes at work’ are meant to give some information about the world outside the textbook and to develop the habit of doing independent reference work to obtain additional information. l This textbook is not meant for reading and explaining in the classroom but guiding students to learn the methods of gaining knowledge by carrying out the given activities. An informal atmosphere in the classroom is required to achieve the aims of this textbook. All students should be encouraged to participate in discussions, experiments and activities. Special efforts should be made to organise presentations or report-reading in the class based on students’ activities and projects, besides observing of Science Day and other relevant occasions/ days. l The science and technology content of the textbook has been complemented with Information Communication Technology. These activities are to be conducted under your guidance along with the learning of various new scientific concepts. Front and back covers : Pictures of various activities, experiments and concepts in the book. DISCLAIMER Note : All attempts have been made to contact copy righters (©) but we have not heard from them. We will be pleased to acknowledge the copy right holder (s) in our next edition if we learn from them.  Competency Statements Standard IX The living world Energy 1. To distinguish between the life processes of 1. To explain the inter-relationship between plants and animals. work and energy and to identify the type 2. To use the knowledge about chemical control of work done in everyday instances. in living organisms and based on that to 2. To explain the logic involved in examples understand/ explain events in day to day life. of work, energy and power from everyday 3. To distinguish between the different types of life, and to solve numerical problems. tissue based on their exact structure. 3. To explain the importance of various 4. To explain the importance/use of micro- sound-related concepts in everyday life organisms in the production of antibiotics. and solve various related problems. 5. To explain the cause and effect relationship 4. To draw a diagram of a SONAR station and explain it. between micro-organisms and the various life 5. To explain the sound-related functions of processes of living organisms. the human ear with the help of a diagram. 6. To explain the various diseases caused by 6. To identify the different types of mirrors, harmful micro-organisms and their remedies to give a scientific explanation of the and to be able to take care of one’s own health images formed by them and to draw the and that of society. related ray diagrams. 7. To classify plants scientifically. 7. To find the number of multiple images by 8. To draw a diagram of the excretory system experiment. and nervous system of humans correctly and 8. To find out the scientific reasons for the explain their importance in our life. use of the different types of mirrors in our 9. To explain the importance of the hormones daily life. secreted by the endocrine glands in the physical development of the body and to explain their Substances causal relationship with problems like over- 1. To describe the form, structure and excitability and over-emotionality. shape of the substances of the universe and explain the science behind it. 2. To verify the laws of chemical combination, conservation of mass Diet and nutrition and constant proportions and to make 1. To explain what is tissue culture and its use inferences based on them. in agriculture and related occupations and to 3. To state the meaning of the concepts of give information about processes involved in molecular mass and mole, to recognise it. and write some molecular formulae as 2. To explain convincingly the importance also to explain them. of agriculture-related occupations for 4. To classify substances of everyday use development of society. with the help of indicators and to explain 3. To analyse the inter-relationship between the their uses on the basis of experiments. food chain and the energy pyramid. 5. To verify by experiments the effects 4. To find out the reasons for the changes in the of acids and alkalis on metals and various cycles in nature. non-metals 5. To analyse the information about the factors 6. To eradicate superstitions and rigid that endanger individual and social health customs prevailing in society with the and suggest the remedies. help of relationship between indicators, acids and bases. 6. To change one’s lifestyle by taking into 7. To produce natural indicators. account the effects of the various diseases 8. To demonstrate the effectiveness of the and disorders. chemical substances. Natural resources and disaster management Motion, Force and Machines 1. To explain the application/ relevance of 1. To write equations of motion and use them to solve modern science and technology in the numerical problems. work of the meteorological department. 2. To obtain formulae by drawing graphs involving 2. To classify garbage generated in the house quantities such as displacement, distance, time and and in the surroundings. velocity. 3. To produce manure from garbage or reuse 3. To verify the cause and effect relationships the waste materials. between motion and laws of motion as seen in 4. To undertake activities related to various events of everyday life. cleanliness of surroundings and motivate others to do the same. The Universe 5. To compile information about how the 1. To use the telescope for sky watching. disaster management systems function 2. To explain the contribution of astronomy and and make a presentation based on it. Thus modern technology to human progress. to be able to overcome crisis situations in 3. To distinguish between the different types of one’s own life. telescopes. Information Communication Technology (ICT) 1. To state the fundamental changes that computer technology has brought about in society and the fields of economics, science, industry, etc. 2. To use the computer to search for information in order to solve various problems. 3. To use the computer to explain to scientific concepts. 4. To identify problems arising in using computer software and to resolve them. 5. To use different ways of processing the information obtained by using the computer. Contents S.No. Name of Lesson Page No. 1. Laws of Motion......................................................................................................... 1 2. Work and Energy..................................................................................................... 18 3. Current Electricity................................................................................................... 30 4. Measurement of Matter........................................................................................... 46 5. Acids, Bases and Salts............................................................................................ 58 6. Classification of Plants........................................................................................... 75 7. Energy Flow in an Ecosystem................................................................................. 81 8. Useful and Harmful Microbes................................................................................. 88 9. Environmental Management................................................................................... 96 10. Information Communication Technology (ICT)....................................................... 108 11. Reflection of Light................................................................................................. 115 12. Study of Sound...................................................................................................... 128 13. Carbon : An important element............................................................................. 138 14. Substances in Common Use.................................................................................. 150 15. Life Processes in Living Organisms...................................................................... 163 16. Heredity and Variation.......................................................................................... 179 17. Introduction to Biotechnology............................................................................... 194 18. Observing Space : Telescopes.............................................................................. 209  1. 1. Laws गतीचेofनियम Motion Ø Motion Ø Distance and displacement Ø Acceleration Ø Newton’s laws of motion and related equations. Motion of an object In which of the following examples can you sense motion? Can you tell? How will you explain presence and absence of motion? 1. The flight of a bird 2. A stationary train 3. Leaves flying through air 4. A stone lying on a hill We see the motion of several objects every day. Sometimes we cannot see the motion of an object directly, as in the case of a breeze. Can you list other examples of motion, besides those given here? Think about it. 1. You are travelling in a bus. Is the person sitting next to you in motion? 2. What do you take into consideration to decide if an object is moving or not? You have learnt that motion is a relative concept. If the position of an object is changing with respect to its surroundings, then it is said to be in motion. Otherwise, it is said to be at rest. Displacement and distance A B Let’s try this (a) School 1. Measure the distance between points A and B in different ways as shown in 1200 m m 00 figure 1.1(a) 13 Path B 2. Now measure the distance along the dotted line. Which distance is correct 500 m Sheetal and Path A according to you and why? Prashant’s house Sangeeta’s house (b) Think about it. 1.1 Location of the school and houses A. Sheetal first went to her friend Sangeeta’s house on her way to school (see figure 1.1b). B. Prashant went straight from home to school. Both are walking with the same speed. Who will take less time to reach the school and why? In the above example, is there a difference between the actual distance and the distance travelled? What is it? 1  ‘Distance’ is the length of the actual path travelled by an object in motion while going from one point to another, whereas displacement is the minimum distance between the starting and finishing points. Use your brain power! 1. Every morning, Swaralee walks round the edge of a circular field having a radius of 100 m. As shown in (a) figure 1.2 (a), if she starts from the point A and takes one round, how much distance has she walked and what is her displacement? A. 2. If a car, starting from point P, goes to point Q (see figure 1.2 b) and then returns to point P, how much distance has it travelled and what is its displacement? (b) ® ® 1.2 Distance and displacement P 360 metres Q Even if the displacement of an object is zero, the actual distance traversed by it may not be zero. Speed and velocity 1. What are vectors and scalars? Can you recall? 2. Which of the quantities distance, speed, velocity, time and displacement are scalars and which are vectors? Total distance travelled Speed = Always remember Time required 1. The units of speed and velocity are the same. In the SI system, the unit The distance travelled in one direction is m/s while in the CGS system, it by an object in unit time is called its is cm/s. velocity. Here, unit time can be one second, one minute, one hour, etc. If large units are 2. Speed is related to distance while used, one year can also be used as a unit of velocity is related to the time. displacement. The displacement that occurs in unit 3. If the motion is along a straight time is called velocity. line, the values of speed and velocity are the same, otherwise they can be different. Displacement Velocity = Velocity is the displacement that Time occurs in unit time. 2  In the first example (on page 1), the straight line distance between the houses of Sheetal and Sangeeta is 500 m and that between Sangeeta’s house and school is 1200 m. Also, the straight line distance between Sheetal’s house and school is 1300 m. Suppose Sheetal takes 5 minutes to reach Sangeeta’s house and then 24 minutes to reach school from there, Then, Distance 500 m Sheetal’s speed along path A = = = 100 m/minute Time 5 minute Distance 1200 m Sheetal’s speed along path B = = = 50 m/minute Time 24 minute Sheetal’s average speed = Total distance 1700 m = 58.6 m/minute Total time 29 minute Displacement 1300 m Sheetal’s velocity = = Time 29 minute Sheetal’s velocity = 44.83 m/minute Effect of speed and direction on velocity Sachin is travelling on a motorbike. Explain what will happen in the following events during Sachin’s ride (see figure 1.3). 1. What will be the effect on the velocity of the motorcycle if its speed increases or decreases, but its direction remains unchanged? 2. In case of a turning on the road, will the velocity and speed be same? If Sachin changes the direction of the motorcycle, keeping its speed constant, what will be the effect on the velocity? 3. If, on a turning, Sachin changes the direction as well as the speed of the motorcycle, what will be the effect on its velocity? It is clear from the above that velocity depends on speed as well as direction and that velocity changes by 1. changing the speed while keeping the direction same 2. changing the direction while keeping the speed same 3. changing the speed as well as the direction. 1.3 Effect on velocity Always remember The first scientist to measure speed as the distance /time was Galileo. The speed of sound in dry air is 343.2 m/s while the speed of light is about 3 × 108 m/s. The speed of revolution of the earth around the sun is about 29770 m/s. 3 Uniform and non-uniform linear motion Amar, Akbar and Anthony are travelling in different cars with different velocities. The distances covered by them during different time intervals are given in the following table. Time in the Distance covered by Distance covered by Distance covered by clock Amar in km Akbar in km Anthony in km 5.00 0 0 0 5.30 20 18 14 6.00 40 36 28 6.30 60 42 42 7.00 80 70 56 7.30 100 95 70 8.00 120 120 84 Use your brain power ! If an object covers unequal distances 1. What is the time interval between the notings of distances made in equal time by Amar, Akbar and Anthony? intervals, it is said 2. Who has covered equal distances in equal time intervals? to be moving with 3. Are all the distances covered by Akbar in the fixed time intervals non-uniform speed. the same? For example, the 4. Considering the distances covered by Amar, Akbar and Anthony motion of a vehicle in fixed time intervals, what can you say about their speeds? being driven If an object covers equal distances in equal time intervals, it througth heavy is said to be moving with uniform speed. traffic. Acceleration 1. Take a 1m long plastic tube and cut it lengthwise into two halves. Let’s try this. 2. Take one of the channel shaped pieces. Place one of its ends on the ground and hold the other at some height from the ground as shown in figure 1.4. 3. Take a small ball and release it from the upper end of the channel. Channel 4. Observe the velocity of the ball as it rolls down Ball along the channel. 5. Is its velocity the same at all points? 6. Observe how the velocity changes as it moves from the top, through the middle and to the 1.4 Change in velocity bottom. 4  You must have all played on a slide in a park. You know that while sliding down, the velocity is less at the top, Use your brain power! it increases in the middle and becomes zero towards the 1. If the velocity end. The rate of change of velocity is called acceleration changes by equal Change in velocity amounts in equal time Acceleration = Time. intervals, the object is If the initial velocity is ‘u’ and in time ‘t’ it changes to said to be in uniform the final velocity ‘v’, acceleration. 2. If the velocity Final velocity – Initial velocity (v-u) changes by unequal Acceleration = a = \a= Time t amounts in equal time If the velocity of an object changes during a certain time intervals, the object is period, then it is said to have accelerated motion. An object in say to be non-uniform motion can have two types of acceleration. acceleration. 1. When an object is at rest in the beginning of its motion, what is its initial velocity? 2. When an object comes to rest at the end of its motion, what is its final velocity? Positive, negative and zero acceleration An object can have positive or negative acceleration. When the velocity of an object increases, the acceleration is positive. In this case, the acceleration is in the direction of velocity. When the velocity of an object decreases with time, it has negative acceleration. Negative acceleration is also called deceleration. Its direction is opposite to the direction of velocity. If the velocity of the object does not change with time, it has zero acceleration. Distance-time graph for uniform motion The following table shows the distances covered by a car in fixed time intervals. Draw a graph of distance against time taking ‘time’ along the X-axis and ‘distance’ along the Y-axis in figure 1.5. Y Scale : X axis 1 cm = Time Distance 140 Y axis 1 cm = (sec- (metres) onds) 120 100 Distance (metres) 0 0 80 10 15 20 30 60 30 45 40 40 60 20 50 75 X 60 90 0 10 20 30 40 50 60 70 80 70 105 1.5 Distance - time graph Time (seconds) 5  An object in uniform motion Use your brain power ! covers equal distances in equal time intervals. Thus, the graph between In the distance-time graph above, what distance and time is a straight line. does the slope of the straight line indicate? Distance-time graph for non-uniform motion The following table shows the distances covered by a bus in equal time intervals Draw a graph of distance against time taking the time along the X-axis and distance along the Y-axis in figure 1.6. Does the graph show a direct proportionality between distance and time? Time Distance (second) (metre) 0 0 5 7 10 12 15 20 20 30 25 41 30 50 35 58 1.6 Distance - time graph Here, the distance changes non-uniformly with time. Thus, the bus is having non-uniform motion. What difference do you see in the distance-time Use your brain power ! graphs for uniform and non-uniform motion? Velocity-time graph for Y uniform velocity Scale : X axis 1 cm = 1 hr A train is moving with a 70 Y axis 1 cm = 10 km/hr uniform velocity of 60 km/hour A B for 5 hours. The velocity-time 60 graph for this uniform motion is 50 shown in figure 1.7. Velocity (km/hr) 1. With the help of the graph, how 40 will you determine the distance 30 covered by the train between 2 and 4 hours? 20 2. Is there a relation between the 10 distance covered by the train C t1 D t2 between 2 and 4 hours and the X 0 1 2 3 4 5 6 area of a particular quadrangle in the graph? What is the Time (hours) acceleration of the train? 1.7 Velocity - time graph 6 Velocity-time graph for uniform acceleration The changes in the velocity of a car in specific time intervals are given in the following table. Y Scale : X axis 1 cm = 5 second Time Velocity Y axis 1 cm = 8 m/second 56 (seconds) (m/s) 0 0 48 5 8 40 Velocity (m/s) B 10 16 32 15 24 24 20 32 A 25 40 16 E 30 48 8 35 56 D C X 0 5 10 15 20 25 30 35 40 Time (seconds) 1.8 Velocity - time graph The velocity-time graph in figure 1.8 shows that, 1. The velocity changes by equal amounts in equal time intervals. Thus, this is uniform acceleration in accelerated motion. How much does the velocity change every 5 minutes? 2. For all uniformly accelerated motions, the velocity-time graph is a straight line. 3. For non-uniformly accelerated motions, the velocity-time graph may have any shape depending on how the acceleration changes with time. From the graph in figure 1.8, we can determine the distance covered by the car between the 10th and the 20th seconds as we did in the case of the train in the previous example. The difference is that the velocity of the car is not constant (unlike that of the train) but is continuously changing because of uniform acceleration. In such a case, we have to use the average velocity of the car in the given time interval to determine the distance covered in that interval. 32+16 = 24 m/s From the graph, the average velocity of the car = 2 Multiplying this by the time interval, i.e. 10 seconds gives us the distance covered by the car. Distance covered = 24 m/s x 10 s = 240 m Check that, similar to the example of the train, the distance covered is given by the area of quadrangle ABCD. A( ABCD ) = A ( AECD ) + A ( ABE ) Equations of motion using graphical method Newton studied motion of an object and gave a set of three equations of motion. These relate the displacement, velocity, acceleration and time of an object moving along a straight line. 7  Suppose an object is in motion along a straight line with initial velocity ‘u’. It attains a final velocity ‘v’ in time ‘t’ due to acceleration ‘a’ its desplacement is ‘s’. The three equations of motion can be written as v = u + at This is the relation between velocity and time. 1 2 s = ut + at This is the relation between displacement and time. 2 v2 = u2 + 2as This is the relation between displacement and velocity. Let us try to obtain these equations by the graphical method. Equation describing the relation between velocity and time Figure 1.9 shows the change in velocity with time of a uniformly accelerated object. The object starts from the point D in the graph with velocity u. Its velocity keeps increasing and after time t, it reaches the point B on the graph. The initial velocity of the object = u = OD The final velocity of the object = v = OC Time = t = OE B v Change in velocity C Acceleration (a) = Time (Final velocity – Initial velocity) Velocity = Time (OC - OD) = t u A \ CD = at …… (i) (OC - OD = CD) D Draw a line parallel to Y axis passing through B. This will cross the X axis in E. Draw a line t parallel to X-axis passing through D. This will O Time E cross the line BE at A. 1.9 Velocity - time graph From the graph.... BE = AB + AE \ v = CD + OD.....(AB = CD and AE = OD) \ v = at + u............(from i ) \ v = u + at This is the first equation of motion. Equation describing the relation between displacement and time Let us suppose that an object in uniform acceleration ‘a’ and it has covered the distance ‘s’ within time ‘t’. From the graph in figure 1.9, the distance covered by the object during time ‘t’ is given by the area of quadrangle DOEB. \ s = area of quadrangle DOEB = area (rectangle DOEA) + area of triangle (DAB) 1 \ s = (AE × OE ) + ( × [AB × DA]) 2 8 But, AE = u, OE = t and (OE = DA = t) AB = at ---( AB = CD ) --- from (i) 1 \s=u×t+ × at × t 2 1 2 \ Newton’s second equation of motion is s = ut + at 2 Equation describing the relation between displacement and velocity We have seen that from the graph in figure 1.9 we can determine the distance covered by the object in time t from the area of the quadrangle DOEB. DOEB is a trapezium. So we can use the formula for its area. \ s = area of trapezium DOEB \ s = 1 × sum of lengths of parallel sides × distance between the parallel sides 2 \ s = 1 × (OD + BE ) × OE But, OD = u, BE = v and OE = t 2 1 \ s = 2 × ( u + v) × t ------ (ii) (v-u) But, a = t Always remember \ t= (v-u) ---------------(iii) a The velocity of an accelerated object changes with time. Change 1 (v-u) \ s= × (u + v)× in the velocity can be due to a 2 a change in direction or magnitude (v+u) ( v-u) of the velocity or both. \ s= 2a \ 2 as =(v+u) ( v-u) = v2-u2 \ v2= u2 + 2as this is Newton’s third equation of motion. Uniform circular motion Let’s try this Observe the tip of the second hand of a clock. What can you say about its velocity and speed? The speed of the tip of a clock is constant, but the direction of its displacement and therefore, its velocity is constantly changing. As the tip is moving along a circular path, its motion is called uniform circular motion. Can you give more examples of such motion? 9  Try out and think about it 1. Draw a rectangular path as shown figure 1.10 2. Place the tip of your pencil on the middle of any side of the square path and trace the path. 3. Note how many times you have to change the direction while tracing the complete path. 4. Now repeat this action for a pentagonal, hexagonal, octagonal path and note the number of times you have to change direction. 5. If you increase the number of sides of the polygon and make it infinite, how many times will you have to change the direction? What will be the shape of the path? This shows that as we increase the number of sides, we have to keep changing direction more and more times. And when we increase the number of sides to infinity, the polygon becomes a circle. 1.10 Changes in direction When an object is moving with a constant If an object, moving along a circular speed along a circular path, the change in path of radius ‘r’, takes time ‘t’ to velocity is only due to the change in direction. come back to its starting position, Hence, it is accelerated motion. When an object its speed can be determined using moves with constant speed along a circular the formula given below : path, the motion is called uniform circular motion, e.g. the motion of a stone in a sling or Speed = Circumference Time that of any point on a bicycle wheel when they are in uniform motion. 2 pr r =radius of the circle v= t Research Find out more examples of circular motion in day to day life. Determining the direction of velocity in uniform circular motion. Take a circular disc and put a five rupee coin at a point Let’s try this along its edge. Make it move around its axis by putting a pin through it. When the Coin disc is moved at higher speed, the coin will be thrown off as shown in figure 1.11. Note the direction in which it is thrown off. Repeat the Disc action placing the coin at different points along the edge of the circle and observe the direction in which the 1.11 The coin and the disc coin is thrown off. 10  The coin will be thrown off in the direction of the tangent which is perpendicular to the radius of disc. Thus, the direction in which it gets thrown off depends on its position at the moment of getting thrown off. It means that, as the coin moves along a circular path the direction of its motion is changing at every point. Solved examples Example 1: An athlete is running on a circular track. He runs a distance of 400 m in 25 s before returning to his original position. What is his average speed and velocity? Given : Total distance travelled = 400 m Total displacement = 0, as he returns to his original position. Total time = 25 seconds. Average speed = ?, Average velocity = ? Total distance covered 400 Average speed = = = 16 m/s Total time taken 25 Total displacement 0 Average velocity = = 25 = 0 m/s Total time taken Example 2: An aeroplane taxies on the runway for 30 s with an acceleration of 3.2 m/s2 before taking off. How much distance would it have covered on the runway? Given : a= 3.2 m/s2, t = 30 s, u = o , s = ? 1 1 s = ut + at2 = 0 × 30 + × 3.2 × 302 = 1440 m. 2 2 Example 3: A kangaroo can jump 2.5 m Example 4 : A motorboat starts from rest vertically. What must be the initial velocity and moves with uniform acceleration. If it of the kangaroo? attains the velocity of 15 m/s in 5 s, Given : calculate the acceleration and the distance a = 9.8 m/s2 s = 2.5 m travelled in that time. v=0 Given : u=? Initial velocity, u = 0, v2= u2 + 2as (0)2 = u2 + 2 × (-9.8) (2.5) Negative sign final velocity, v = 15 m/s, time, t = 5 s. is used as the acceleration is in the direc- Acceleration = a = ? tion opposite to that of velocity. From the first equation of motion 0 = u2 - 49 u2 = 49 v-u 15-0 2 a = = = 3 m/s u = 7 m/s t 5 11 From the second equation of motion, the distance covered will be 1 2 s = ut + at 2 1 s = 0 × 5 + 3 × 52 2 = 0 + 75 = 37.5 m 2 Newton’s laws of motion What could be the reason for the following? 1. A static object does not move without the application of a force. 2. The force which is sufficient to lift a book from a table is not sufficient to lift the table. 3. Fruits on a tree fall down when its branches are shaken. 4. An electric fan keeps on rotating for some time even after it is switched off. If we look for reasons for the above, we realize that objects have some inertia. We have learnt that inertia is related to the mass of the object. Newton’s first law of motion describes this very property and is therefore also called the law of inertia. Newton’s first law of motion Fill a glass with sand. Keep a piece of cardboard on it. Let’s try this Keep a five rupee coin on the cardboard. Now strike the card- board hard using your fingers. Observe what happens. Balanced and unbalanced force You must have played tug-of-war. So long as the forces applied by both the sides are equal, i.e. balanced, the centre of the rope is static in spite of the applied forces. On the other hand, when the applied forces become unequal, i.e. unbalanced, a net force gets applied in the direction of the greater force and the centre of the rope shifts in that direction. ‘An object continues to remain at rest or in a state of uniform motion along a straight line unless an external unbalanced force acts on it.’ When an object is at rest or in uniform motion along a straight line, it does not mean that no force is acting on it. Actually there are a number of forces acting on it, but they cancel one another so that the net force is zero. Newton’s first law explains the phenomenon of inertia, i.e. the inability of an object to change its state of motion on its own. It also explains the unbalanced forces which cause a change in the state of an object at rest or in uniform motion. All instances of inertia are examples of Newton’s first law of Motion. 12 Newton’s second law of motion A. 1. Ask your friend to drop one plastic and one rubber Let’s try this ball from the same height. 2. You catch the balls. Which ball was easier to catch and why? B. 1. Ask your friend to throw a ball towards you at slow speed. Try to catch it. 2. Now ask your friend to throw the same ball at high speed towards you. Try to catch it. Which ball could you catch with greater ease? Why? The effect of one object Momentum has magnitude as well as striking another object depends direction. Its direction is the same as that of both on the mass of the former velocity. In SI system, the unit of momentum is object and its velocity. This kg m/s, while in CGS system, it is g cm/s. means that the effect of the force If an unbalanced force applied on an object causes depends on a property related to a change in the velocity of the object, then it also both mass and velocity of the causes a change in its momentum. The force striking object. This property necessary to cause a change in the momentum of was termed ‘momentum’ by an object depends upon the rate of change of Newton. momentum. Momentum (P) : Momentum is the product of mass and velocity of an object. P = m v. Momentum is a vector quantity. ‘The rate of change of momentum is proportional to the applied force and the change of momentum occurs in the direction of the force.’ Suppose an object of mass m has an initial velocity u. When a force F is applied in the direction of its velocity for time t, its velocity becomes v. \ The initial momentum of the object = mu, Its final momentum after time t = mv Change in momentum \ Rate of change of momentum = Time mv- mu = m (v- u) = ma \ Rate of change of momentum = t t According to Newton’s second law of motion, the rate of change of momentum is proportional to the applied force. \ ma a F \ F = k ma (k = Constant of proportionality and its value is 1). F= m×a 13  Consider two objects having different masses which are initially at rest. The In SI system, the unit of force is initial momentum for both is zero. Suppose newton. a force ‘F’ acts for time ‘t’ on both objects. Newton (N) : The force necessary to cause The lighter object starts moving faster than an acceleration of 1 m/s2 in an object of the heavier object. However, from the mass 1 kg is called 1 newton. above formula, we know that the rate of 1 N = 1 kg × 1 m/s2 change of momentum i.e. ‘F’ in both In CGS system the unit of force is a dyne. objects is same and the total change in their Dyne: The force necessary to cause an momentum will also be same i.e. ‘Ft’. acceleration of 1 cm/s2 in an object of Thus, if the same force is applied on mass 1 gm is called 1 dyne. different objects, the change in momentum is the same. 1 dyne = 1 g × 1 cm/s2 Use your brain power ! Why is there a thick bed of sand for a high jumper to fall on after his jump? Newton’s third law of motion 1. Take a plastic boat and make a hole at its rear end. Let’s try this 2. Inflate a balloon and fix it on the hole in the boat. Release the boat in water. What happens to the boat as the air in the balloon escapes slowly? Why? We have learnt about force and its effect on an object through Newton’s first and second laws of motion. ‘However, in nature force cannot act alone.’ Force is a reciprocal action between two objects. Forces are always applied in pairs. When one object applies a force on another object, the latter object also simultaneously applies a force on the former object. The forces between two objects are always equal and opposite. This idea is expressed in Newton’s third law of motion. The force applied by the first object is called action force while the force applied by the second object on the first is called reaction force. ‘Every action force has an equal and opposite reaction force which acts simultaneously.’ 1. Action and reaction are terms that Use your brain power ! express force. 1. While hitting a ball with a bat, the speed 2. These forces act in pairs. One force of the bat decreases. cannot exist by itself. 2. A gun recoils i.e. moves backwards 3. Action and reaction forces act when a bullet is fired. simultaneously. 3. Mechanism of firing of a rocket. 4. Action and reaction forces act on How will you explain these with the different objects. They do not act help of Newton’s third law of motion? on the same object and hence cannot cancel each other’s effect. 14 Law of conservation of momentum Suppose an object A has mass m1 and its initial velocity is u1. An object B has mass m2 and initial velocity u2. According to the formula for momentum, the initial momentum of A is m1 u1 and that of B is m2u2. Suppose these two objects collide. Let the force on A due to B be F1. This force will cause acceleration in A and its velocity will become v1. \ Momentum of A after collision = m1 v1 According to Newton’s third law of motion, A also exerts an equal force on B but in the opposite direction. This will cause a change in the momentum of B. If its velocity after collision is v2, The momentum of B after collision = m2v2. If F2 is the force that acts on object B, F2 = -F1 \ \ m2 a2 = - m1 a1………… F= ma (v2-u2) (v -u ) \ m2 × = - m1 × 1 1 ……… \ a = (v-u) t t t \ m2 (v2 - u2 )= - m1 (v1-u1) \ m2 v2 - m2 u2 = - m1 v1+ m1 u1 \ (m2 v2 + m1 v1 )= (m1 u1+m2 u2) The magnitude of total final momentum = the magnitude of total initial momentum. Thus, if no external force is acting on two objects, then their total initial momentum is equal to their total final momentum. This statement is true for any number of objects. ‘When no external force acts on two interacting objects, their total momentum remains constant. It does not change.’ This is a corollary to Newton’s third law of motion. The momentum is unchanged after the collision. The momentum gets redistributed between the colliding objects. The momentum of one of the objects decreases while that of the other increases. Thus, we can also state this corollary as follows. ‘When two objects collide, the total momentum before collision is equal to the total momentum after collision.’ In order to understand this principle, let us consider the example of a bullet fired from a gun. When a bullet of mass m1 is fired from a gun of mass m2, its velocity becomes v1, and its momentum becomes m1 v1. Before firing the bullet, both the gun and the bullet are at rest and hence the total initial momentum is zero. According to the above law, the total final momentum also has to be zero. Thus, the forward moving bullet causes the gun to move backward after firing. This backward motion of the gun is called its recoil. The velocity of recoil, v2 , is such that, m1 m1 v1 + m2 v2 = 0 or v2 = - m2 × v1 15  As the mass of the gun is much higher than the mass of the bullet, the velocity of the gun is much smaller than the velocity of the bullet. The magnitude of the momentum of the bullet and that of the gun are equal and their directions are opposite. Thus, the total momentum is constant. Total momentum is also constant during the launch of a rocket. Solved examples Example 1: The mass of a cannon is 500 kg and it recoils with a speed of 0.25 m/s. What is the momentum of the cannon? Given : mass of the cannon = 500 kg, recoil speed = 0.25 m/s Momentum = ? Momentum = m × v = 500 x 0.25 = 125 kg m/s Example 2: 2 balls have masses of 50 gm and 100 gm respectively and they are moving along the same line in the same direction with velocities of 3 m/s and 1.5 m/s respectively. They collide with each other and after the collision, the first ball moves with a velocity of 2.5 m/s. Calculate the velocity of the other ball after collision. Given : The mass of first ball = m1 = 50 g = 0.05 kg, mass of the second ball = m2= 100 g = 0.1 kg Initial velocity of the first ball = u1 = 3 m/s, Initial velocity of the second ball = u2 = 1.5 m/s Final velocity of the first ball = v1 = 2.5 m/s ,Final velocity of the second ball = v2 = ? According to the law of conservation of momentum, total initial momentum = Total final momentum. m1u1 + m2u2 = m1v1 + m2v2 (0.05 × 3) + (0.1 × 1.5) = (0.05 × 2.5) + (0.1 × v2) \ (0.15)+(0.15) = 0.125 + 0.1v2 \ 0.3 = 0.125 + 0.1 v2 0.175 \ 0.1v2 = 0.3 - 0.125 \v2 = = 1.75 m/s 0.1 Exercises 1. Match the first column with appropriate entries in the second and third columns and remake the table. S. No. Column 1 Column 2 Column 3 1 Negative The velocity of the ob- A car, initially at rest acceleration ject remains constant reaches a velocity of 50 km/hr in 10 seconds 2 Positive The velocity of A vehicle is moving with a acceleration the object decreases velocity of 25 m/s 3 Zero The velocity of the A vehicle moving with the velocity of acceleration object increases 10 m/s, stops after 5 seconds. 2. Clarify the differences A. Distance and displacement B. Uniform and non-uniform motion. 16 3. Complete the following table. u (m/s) a (m/s2) t (sec) v = u + at (m/s) 2 4 3 - - 5 2 20 u (m/s) a (m/s2) t (sec) 1 2 s = ut + at (m) 2 5 12 3 - 7 - 4 92 u (m/s) a (m/s2) s (m) v 2= u2 + 2as (m/s)2 4 3 - 8 - 5 8.4 10 4. Complete the sentences and explain 7. Solve the following examples. them. a) An object moves 18 m in the first a. The minimum distance between 3 s, 22 m in the next 3 s and 14 m in the start and finish points of the the last 3 s. What is its average motion of an object is called the speed? (Ans: 6 m/s) ……….. of the object. b) An object of mass 16 kg is moving b. Deceleration is -------- acceleration with an acceleration of 3 m/s2. c. When an object is in uniform Calculate the applied force. If the circular motion, its ……… same force is applied on an object of changes at every point. mass 24 kg, how much will be the d. During collision......….. remains acceleration? (Ans: 48 N, 2 m/s2) constant. c) A bullet having a mass of 10 g and e. The working of a rocket depends moving with a speed of 1.5 m/s, on Newton’s ….…. law of motion. penetrates a thick wooden plank of 5. Give scientific reasons. mass 900 g. The plank was initially a. When an object falls freely to the at rest. The bullet gets embedded in ground, its acceleration is uniform. the plank and both move together. b. Even though the magnitudes of Determine their velocity. action force and reaction force are (Ans: 0.15 m/s) equal and their directions are d) A person swims 100 m in the first opposite, their effects do not get 40 s, 80 m in the next 40 s and 45 m cancelled. in the last 20 s. What is the average c. It is easier to stop a tennis ball as speed? (Ans: 2.25 m/s2) compared to a cricket ball, when both are travelling with the same Project: velocity. Obtain information about d. The velocity of an object at rest is commonly used gadgets or devices considered to be uniform. which are based on the principles 6. Take 5 examples from your of Newton’s laws of motion. surroundings and give explanation based on Newtons laws of motion. ²²² 17  2. Work and 1. गतीचे Energy नियम Ø Work Ø Energy Ø Mechanical energy Ø Law of conservation of energy Ø Free fall Observe 2.1 Various activities 1. In which of the pictures above has work been done? Can you tell? 2. From scientific point of view, when do we say that no work was done? Generally, any mental or physical activity is referred to as work. When we walk or run, the energy in our body is used to do the necessary work. We say that a girl who is studying is working or performing work. But that is mental work. In physics, we deal with physical work. Work has a special meaning in physics. ‘Work is said to be done when a force applied on an object causes displacement of the object.’ You have already learnt that the work done by a force acting on an object is the product of the magnitude of the force and the displacement of the object in the direction of the force. Thus, Work = force × displacement Use your brain power ! What are different types of You have learnt how to Can you recall? forces and their examples? calculate the work done on an object when the displacement Minakshee wants to displace a wooden is in the direction of the applied block from point A to point B along the surface force. But if the displacement of a table as shown in figure 2.2A. She has used is not in the direction of the force F for the purpose. Has all the energy she applied force, how do we spent been used to produce acceleration in the calculate the amount of work block? Which forces have been overcome using done? that energy? 18  Observe and discuss. B A B A C 2.2 Displacement of an object You must have seen the events depicted in the pictures B and C above. When a child pulls a toy with the help of a string, the direction of the force is different from that of displacement. Similarly, when a large vehicle tows a small one, the directions of force and the displacements are different. In both these cases, the direction of force makes an angle with the direction of displacement. Let us see how to calculate work done in such cases. When a child pulls a toy cart, force is applied along the direction of the string while the cart is pulled along the horizontal surface. In this case, in order to calculate the amount of work done, we have to convert the applied force into the force acting along the direction of displacement. Let F be the applied force and F1 be its component in the direction of displacement. Let s be the displacement. The amount of work done is given by W = F1.s................................. (1) The force F is applied in the direction of the string i. e. at an angle with the horizontal. Let q be the angle that the string makes with the horizontal. We can determine the component F1, of this force F, which acts in the horizontal direction by means of trigonometry. (see figure2.3) F cos q = base / hypotenuse cos q = F1 / F F1 = F cos q Thus, the work done by F1 is )q W = F cos q s F1 W = F s cos q Enter your conclusions about the work 2.3 Force used for the displacement done for the specific values of q in the following table. q cos q W = F s cos q Conclusion 00 1 W=Fs 900 0 0 1800 -1 W = -F s 19  Unit of work Work = Force × Displacement In SI system, the unit of force is newton (N) and the unit of displacement is metre (m). Thus, the unit of force is newton-metre. This is called joule. 1 Joule : If a force of 1 newton displaces an object through 1 metre in the direction of the force, the amount of work done on the object is 1 joule. \ 1 joule = 1 newton ´ 1 metre 1J=1N ´1m In CGS system, the unit of force is dyne and that of displacement is centimeter (cm). Thus, the unit of work done is dyne-centimetre. This is called an erg. 1 erg : If a force of 1 dyne displaces an object through 1 centimetre in the direction of the force, the amount of work done is 1 erg. 1 erg = 1 dyne ´ 1 cm Relationship between joule and erg We know that, 1 newton = 105 dyne and 1 m = 102 cm Work = force ´ displacement 1 joule = 1 newton ´ 1 m 1 joule = 105 dyne ´ 102 cm = 107 dyne cm 1 joule = 107 erg Positive, negative and zero work Think before you answer Discuss the directions of force and of displacement in each of the following cases. 1. Pushing a stalled vehicle 2. Catching the ball which your friend has thrown towards you. 3. Tying a stone to one end of a string and swinging it round and round by the other end of the string. 4. Walking up and down a staircase; climbing a tree. 5. Stopping a moving car by applying brakes. You will notice that in some of the above examples, the direction of the force and displacement are the same. In some other cases, these directions are opposite to each other, while in some cases, they are perpendicular to each other. In these cases, the work done by the force is as follows. 1. When the force and the displacement are in the same direction (q = 00 ), the work done by the force is positive. 2. When the force and the displacement are in opposite directions (q = 1800), the work done by the force is negative. 3. When the applied force does not cause any displacement or when the force and the displacement are perpendicular to each other (q =900), the work done by the force is zero. 20  Try this Take a plastic cup and make a hole in the Figure A Cup centre of its bottom. Take a long thread, double it and pass it through the hole. Tie a thick enough knot at the end so that the knot will not pass through the hole, taking care that the two loose thread ends are below the bottom of the cup. Tie a nut each to the two ends as shown in figure 2.4. Now nut do the following. As shown in figure ‘A’, put the cup on a table, Cup keep one of the nuts in the cup and let the thread carrying the other nut hang down along the side Figure B of the table. What happens? As shown in figure ‘B’, when the cup is thread sliding along the table, stop it by putting a ruler in its path. nut As shown in figure ‘C’, keep the cup at the centre of the table and leave the two nuts hanging on opposite sides of the table. Cup Figure C Questions: 1. Figure A - Why does the cup get pulled? 2. Figure B - What is the relation between the displacement of the cup and the force applied thread through the ruler? 3. In figure C - Why doesn’t the cup get displaced? nut nut 4. What is the type of work done in figures A, B and C? In the three actions above, what is the relationship between the applied force and the 2.4 Positive, negative and zero work displacement? Suppose an artificial satellite is moving around the earth in a circular orbit. As the gravitational force acting on the satellite (along the radius of the circle) and its displacement (along the tangent to the circle) are perpendicular to each other, the work done by the gravitational force is zero Institutes at work The National Physical Laboratory, New Delhi, was conceptualized in 1943. It functions under the Council of Scientific and Industrial Research. Its conducts basic research in the various branches of physics and helps various industries and institutes engaged in developmental work. Its main objective is to establish national standards of various physical quantities. 21  Solved examples Example 1: Calculate the work done to Example 2 : Pravin has applied a force of take an object of mass 20 kg to a height of 100 N on an object, at an angle of 600 to 10 m. the horizontal. The object gets displaced in (g = 9.8 m/s2) the horizontal direction and 400 J work is done. What is the displacement of the Given: m = 20 kg; s = 10 m object? g = 9.8 m/s 2 (cos 600 = 1 ) \ F = m.g 2 Given : = 20 ´ (-9.8) q = 600 (The negative sign is taken because the displacement is opposite to the direction F = 100 N of the force.) W = 400 J , s = ? F = -196 N W = F s Cos q \W=Fs 400 = 100 ´ s ´ 12 = -196 ´ 10 400 = 1 ´ s W = -1960 J 100 2 (The negative sign appears because 4´2=s the direction of force is opposite to the \ s = 8 m direction of displacement so that the work The object will be displaced through 8 m. done is negative.) Energy Why does it happen? 1. If a pot having a plant is kept in the dark, the plant languishes. 2. On increasing the volume of a music system or TV beyond a limit, the vessels in the house start vibrating. 3. Collecting sunlight on a paper with the help of a convex lens burns the paper. The capacity of a body to perform work is called its energy. The units of work and energy are the same. The unit in SI system is joule while that in cgs system is erg. You have learnt that energy exists in various forms like mechanical, heat, light, sound, electro-magnetic, chemical, nuclear and solar. In this chapter, we are going to study two forms of mechanical energy, namely, potential energy and kinetic energy. Kinetic energy What will happen in the following cases? 1. A fast cricket ball strikes the stumps. 2. The striker hits a coin on the carom board. 3. One marble strikes another in a game of marbles. From the above examples we understand that when a moving object strikes a stationary object, the stationary object moves. Thus, the moving object has some energy, part or all of which it shares with the stationary object, thereby setting it in motion. ‘The energy which an object has because of its motion is called its kinetic energy’. The work done by a force to displace a stationary object through a distance s is the kinetic energy gained by the object. Kinetic energy = work done on the object \ K.E. = F × s 22 Expression for kinetic energy : Suppose a stationary object of mass m moves because of an applied force. Let u be its initial velocity (here u = 0). Let the applied force be F. This generates an acceleration a in the object, and, after time t, the velocity of the object becomes equal to v. The displacement during this time is s. The work done on the object, W = F. s W=F×s According to Newton’s second law of motion, F = ma -------- (1) Similarly, using Newton’s second equation of motion s = ut + 1 at22 However, as initial velocity is zero, u = 0. s=0+ 1 at2 2 s= 1 at2 ------(2) 2 \ W = ma ´ 1 at2 ------ using equations (1) and (2) 2 W = 12 m(at)2 -------(3) Using Newton’s first equation of motion.............. v = u + at \ v = 0 + at \ v = at \ v2 = (at)2 ------(4) \ W = 1 mv2 ------- using equations (3) and (4) 2 The kinetic energy gained by an object is the amount of work done on the object. \ K. E. = W \ K. E. = 12 mv2 Example : A stone having a mass of 250 gm is falling from a height. How much kinetic energy does it have at the moment when its velocity is 2 m/s? Given : m = 250 g m = 0.25 kg v = 2 m/s K.E. = 12 mv2 = 12 × 0.25 × (2)2 = 0.5 J The mass of a moving body is doubled, how Use your brain power ! many times will the kinetic energy increase? 23 Potential energy 1. An arrow is released from a stretched bow. Try this 2. Water kept at a height flows through a pipe into the tap below. 3. A compressed spring is released. Which words describe the state of the object in the above examples? Where did the energy required to cause the motion of objects come from? If the objects were not brought in those states, would they have moved? ‘The energy stored in an object because of its specific state or position is called its potential energy.’ 1.Hold a chalk at a height of 5 cm from the floor and release it. 2. Now stand up straight and then release the chalk. 3. Is there a difference in the results of the two activities? If so, why? Expression for potential energy To carry an object of mass ‘m’ to a height ‘h’ above the earth’s surface, a force equal to ‘mg’ has to be used against the direction of the gravitational force. The amount of work done can be calculated as follows. Work = force x displacement W = mg × h \ W = mgh \ The amount of potential energy stored in the object because of its displacement P.E. = mgh (W = P.E.) \ Displacement to height h causes energy equal to mgh to be stored in the object. Example : 500 kg water is stored in the overhead tank of a 10 m high building. Calculate the amount of potential energy stored in the water. Given : h = 10 m, m = 500 kg g = 9.8 m/s2 \ P.E. = mgh = 10 × 9.8 ×500 P.E. = 49000 J Ajay and Atul have been asked to determine the potential energy of a ball of Atul mass m kept on a table as shown in the figure. What answers will they get? Will Ajay h2 they be different? What do you conclude from this? h1 Potential energy is relative. The heights of the ball with respect to Ajay and Atul are different. So the potential energy with respect to them will be different. 24 Transformation of energy Can you tell? Which are the different forms of energy? Which type of energy is used in each of the following processes? 1. A stretched piece of rubber 2. Fast moving car 3. The whistling of a cooker due to steam 4. The crackers exploded in Diwali 5. A fan running on electricity 6. Drawing out pieces of iron from garbage, using a magnet 7. Breaking of a glass window pane because of a loud noise. Energy can be transformed from one type to another. For example, the exploding firecrackers convert the chemical energy stored in them into light, sound and heat energy. ® Engine/Fan ® Heater Mechanical energy ® ell ® Electrical energy ® Heat energy ® cc Generator ® Thermo-couple tri Lo ec ud -el ® sp oto p ® ea am ph ® ke ll / M r cl ic ce tri ro lar ph ec So on El e Sound Light energy energy Primary cell Secondary cell Chemical energy 2.5 Tranformation of energy Observe the above diagram (figure 2.5) and discuss how tranformation of energy takes place, giving example of each. Law of conservation of energy ‘Energy can neither be created nor destroyed. It can be converted from one form into another. Thus, the total amount of energy in the universe remains constant’. Make two pendulums of the same length with the help of thread Try this and two nuts. Tie another thread in the horizontal position. Tie the two pendulums to this horizontal thread thread in such a way that they will not hit each other while swinging. Now swing one of the pendulums and thread thread observe. What do you see? You will see that as the speed of oscillation of the pendulum slowly decreases, the second pendulum nut nut which was initially stationary, begins to swing. Thus, one pendulum transfers its energy to the other. 2.6 Coupled oscillators 25 Free fall If we release an object from a height, it gets pulled towards the earth because of the gravitational force. An object falling solely under the influence of gravitational force is said to be in free fall or to be falling freely. Let us look at the kinetic and potential energies of an object of mass m, falling freely from height h, when the object is at different heights As shown in the figure, the point A is at a height h A from the ground. Let the point B be at a distance x, vertically below A. Let the point C be on the ground directly below A and B. Let us calculate the energies of the object at A, B and C. x 1. When the object is stationary at A, its initial velocity is u = 0 h \ K.E. = 1 mass x velocity2 B 2 = 1 mu2 2 K.E. = 0 h-x P.E. = mgh \ Total energy = K.E. + P.E. = 0 + mgh Total Energy = mgh.--- (1) C 2. Let the velocity of the object be vB when 2.7 Free fall it reaches point B, having fallen through a 3. Let the velocity of the object be vC when distance x. it reaches the ground, near point C. u = 0, s = x, a = g u = 0, s = h, a = g v = u + 2as 2 2 v2 = u2 + 2as vB2 = 0 + 2gx vc2 = 0 + 2gh vB2 = 2gx \ K.E. = 1 mvC2 = 1 m(2gh) 2 2 \ K.E. = 12 mvB2 = 12 m(2gx) K.E. = mgh K.E. = mgx The height of the object from the Height of the object when at B = h-x ground at point C is \ P.E. = mg (h-x) h=0 P.E. = mgh - mgx \ P.E. = mgh = 0 \ Total Energy T.E. = K.E. + P.E. \ T.E. = K.E. + P.E = mgx + mgh - mgx T.E. = mgh ------(3) \ T.E. = mgh -------(2) From equations (1), (2) and (3) we see that the total energy of the object is the same at the three points A, B and C. 26  Thus, every object has potential energy when it is at a height above the ground and it keeps getting converted to kinetic energy as the object falls towards the ground. On reaching the ground (point C), all the potential energy gets converted to kinetic energy. But at any point during the fall the total energy remains constant. i.e., T.E. = P.E. + K.E. T.E. at A = mgh + 0 = mgh T.E. at B = mgx + mg (h - x) = mgh T.E. at C = 0 + mgh = mgh Power Think before you answer 1. Can your father climb stairs as fast as you can? 2. Will you fill the overhead water tank with the help of a bucket or an electric motor? 3. Suppose Rajashree, Yash and Ranjeet have to reach the top of a small hill. Rajashree went by car, Yash went cycling while Ranjeet went walking. If all of them choose the same path, who will reach first and who will reach last? In the above examples, the work done is the same in each example but the time taken to perform the work is different for each person or each method. The fast or slow rate of the work done is expressed in terms of power. ‘Power is the rate at which work is done.’ If W amount of work is done in time t then, Work W An introduction to scientists Power = P= Time t In SI system the unit of work is J, so the unit of power is J/s. This is called watt 1 watt = 1 joule / 1 second In the industrial sector the unit used to measure power is called ‘horse power.’ 1 horse power = 746 watt. The unit of energy for commercial use is kilo watt The steam engine was hour. invented in 1781 by the 1000 joule work performed in 1 second is 1 Scottish scientist James kilowatt power. Watt (1736 – 1819). This 1 kW h = 1 kW ´ 1hr invention brought about an = 1000 W ´ 3600 s industrial revolution. The = 3600000 J unit of power is called Watt in his honour. James Watt 1 kW h = 3.6 ´ 106 J was the first to use the term Electricity used for domestic purposes is measured ‘horse-power’. in units of kilowatt hour. 1 kW h = 1 unit 27  Solved problems Example 1 : Swaralee takes 20 s to carry a bag weighing 20 kg to a height of 5 m. Example 2 : A 25 W electric bulb is used How much power has she used? for 10 hours every day. How much electric- Given : m = 20 kg, h = 5 m, t = 40 s ity does it consume each day? \ The force which has to be applied Given : by Swaralee, P = 25, W = 0.025 kW F = mg = 20 ´ 9.8 \ Energy consumed = power ´ time F = 196 N = 0.025 ´ 10 Work done by Swaralee to carry the Energy = 0.25 kW hr bag to a height of 5 m, W = F s = 196 ´ 5 = 980 J Websites for more information : Work 980 www.physicscatalyst.com \ power = (P) = t = 40 www.tryscience.org P = 24.5 W Exercises 1. Write detailed answers? a. For work to be performed, energy a. Explain the difference between must be …. potential energy and kinetic energy. (i) transferred from one place to b. Derive the formula for the kinetic another (ii) concentrated energy of an object of mass m, (iii) transformed from one type to moving with velocity v. another (iv) destroyed c. Prove that the kinetic energy of a b. Joule is the unit of … freely falling object on reaching (i) force (ii) work (iii) power the ground is nothing but the (iv) energy transformation of its initial potential c. Which of the forces involved in energy. dragging a heavy object on a d. Determine the amount of work smooth, horizontal surface, have done when an object is displaced at the same magnitude? an angle of 300 with respect to the (i) the horizontal applied force (ii) direction of the applied force. gravitational force (iii) reaction e. If an object has 0 momentum, does force in vertical direction (iv) force it have kinetic energy? Explain your of friction answer. d. Power is a measure of the...…. f. Why is the work done on an object (i) the rapidity with which work is moving with uniform circular done (ii) amount of energy required motion zero? to perform the work (iii) The 2. Choose one or more correct alterna- slowness with which work is performed (iv) length of time tives. 28  e. While dragging or lifting an object, Questions negative work is done by 1. At the moment of releasing the (i) the applied force (ii) gravitational balls, which energy do the balls force (iii) frictional force have? (iv) reaction force 2. As the balls roll down which energy 3. Rewrite the following sentences is converted into which other form using proper alternative. of energy? a. The potential energy of your body is 3. Why do the balls cover the same least when you are ….. distance on rolling down? (i) sitting on a chair (ii) sitting on 4. What is the form of the eventual the ground (iii) sleeping on the total energy of the balls? ground (iv) standing on the ground 5. Which law related to energy does b. The total energy of an object falling the above activity demonstrate ? freely towards the ground … Explain. (i) decreases (ii) remains unchanged 5. Solve the following examples. (iii) increases (iv) increases in the a. An electric pump has 2 kW power. beginning and then decreases How much water will the pump lift c. If we increase the velocity of a car every minute to a height of 10 m? moving on a flat surface to four (Ans : 1224.5 kg) times its original speed, its potential b. If a 1200 W electric iron is used energy ….