NCERT Textbook Class 9

Palaemon Aranea(Spider) 7.5.8 ECHINODERMATA (Prawn) Palamnaeus In Greek, echinos means hedgehog (spiny Pariplaneta (Scorpion) mammal), and derma means skin. Thus, these (Cockroach) are spiny skinned organisms. These are Butterfly exclusively free-living marine animals. They Scolopendra are triploblastic and have a coelomic cavity. (Centipede) They also have a peculiar water-driven tube system that they use for moving around. They have hard calcium carbonate structures that they use as a skeleton. Examples are sea-stars and sea urchins (see Fig. 7.19). Musca (House fly) Fig. 7.17: Arthropoda 7.5.7 MOLLUSCA Antedon Holothuria (feather star) (sea cucumber) In the animals of this group, there is bilateral symmetry. The coelomic cavity is reduced. There is little segmentation. They have an open circulatory system and kidney-like organs for excretion. There is a foot that is used for moving around. Examples are snails and mussels (see Fig. 7.18). Echinus (sea urchin) Asterias (sea-star) Fig. 7.19: Echinodermata Chiton 7.5.9 PROTOCHORDATA Octopus These animals are bilaterally symmetrical, triploblastic and have a coelom. In addition, Unio they show a new feature of body design, Pila namely a notochord, at least at some stages during their lives. The notochord is a long Fig. 7.18: Mollusca rod-like support structure (chord=string) that runs along the back of the animal separating the nervous tissue from the gut. It provides a place for muscles to attach for ease of movement. Protochordates may not have a proper notochord present at all stages in their lives or for the entire length of the animal. Protochordates are marine animals. Examples are Balanoglossus, Herdmania and Amphioxus (see Fig. 7.20). DIVERSITY IN LIVING ORGANISMS 91

Anus Proboscis scaleless. They are ectoparasites or borers of other vertebrates. Petromyzon (Lamprey) and Posthepatic Collarette Myxine (Hagfish) are examples. region Collar 7.5.10 (ii) PISCES Branchial region Gill pores These are fish. They are exclusively aquatic Dorsally animals. Their skin is covered with scales/ curved plates. They obtain oxygen dissolved in water genital wings by using gills. The body is streamlined, and a muscular tail is used for movement. They are Middosrsal cold-blooded and their hearts have only two ridge chambers, unlike the four that humans have. Hepatic caeca They lay eggs. We can think of many kinds of fish, some with skeletons made entirely of Hepatic region cartilage, such as sharks, and some with a skeleton made of both bone and cartilage, such as tuna or rohu [see examples in Figs. 7.22 (a) and 7.22 (b)]. Fig. 7.20: Protochordata: Balanoglossus Synchiropus splendidus Caulophyryne jordani (Mandarin fish) (Angler fish) 7.5.10 VERTEBRATA Pterois volitans These animals have a true vertebral column (Lion fish) and internal skeleton, allowing a completely different distribution of muscle attachment Eye points to be used for movement. Spiracle Vertebrates are bilaterally symmetrical, triploblastic, coelomic and segmented, with Pelvic fin complex differentiation of body tissues and Dorsal fin organs. All chordates possess the following features: Tail (i) have a notochord Caudal fin Sting ray Tail (ii) have a dorsal nerve cord Electric ray (Torpedo) Dorsal fin (iii) are triploblastic Eye (iv) have paired gill pouches (v) are coelomate. Vertebrates are grouped into six classes. 7.5.10 (i) CYCLOSTOMATA Cyclostomes are jawless vertebrates. They are characterised by having an elongated eel-like body, circular mouth, slimy skin and are Fig. 7.21: A jawless vertebrate: Petromyzon Mouth Gills Pectoral Pelvic 92 fin fin Scoliodon (Dog fish) Fig. 7.22 (a): Pisces SCIENCE

Eye Head 7.5.10 (iv) REPTILIA Nostril These animals are cold-blooded, have scales Mouth Pectoral and breathe through lungs. While most of fin them have a three-chambered heart, crocodiles Caudal Pelvic Pectoral Mouth have four heart chambers. They lay eggs with fin fin fin Brood Dorsal tough coverings and do not need to lay their fin eggs in water, unlike amphibians. Snakes, pouch turtles, lizards and crocodiles fall in this category (see Fig. 7.24). Labeo rohita (Rohu) Tail Male Hippocampus (Sea horse) Wing like pectoral Scales Turtle Pelvic fin Tail Exocoetus (Flying fish) Chameleon Anabas (Climbing perch) King Cobra Fig. 7.22 (b): Pisces 7.5.10 (iii) AMPHIBIA These animals differ from the fish in the lack of scales, in having mucus glands in the skin, and a three-chambered heart. Respiration is through either gills or lungs. They lay eggs. These animals are found both in water and on land. Frogs, toads and salamanders are some examples (see Fig. 7.23). House wall lizard (Hemidactylus) Salamander Flying lizard (Draco) Toad Fig. 7.24: Reptilia Rana tigrina Hyla (Tree frog) 7.5.10 (v) AVES (Common frog) These are warm-blooded animals and have a Fig. 7.23: Amphibia four-chambered heart. They lay eggs. There is an outside covering of feathers, and two forelimbs are modified for flight. They breathe through lungs. All birds fall in this category (see Fig. 7.25 for examples). DIVERSITY IN LIVING ORGANISMS 93

Whale White Stork Rat (Ciconia ciconia) Human Bat Cat Ostrich (Struthio camelus) Male Tufted Duck Fig. 7.26: Mammalia (Aythya fuligula) Sparrow Q uestions 1. How do poriferan animals differ Pigeon from coelenterate animals? 2. How do annelid animals differ from arthropods? 3. What are the differences between amphibians and reptiles? 4. What are the differences between animals belonging to the Aves group and those in the mammalia group? Crow Carolus Linnaeus (Karl von Linne) was born in Fig. 7.25: Aves (birds) Sweden and was a doctor by professsion. He was 7.5.10 (vi) MAMMALIA interested in the study of plants. At the age of 22, he Mammals are warm-blooded animals with published his first paper on four-chambered hearts. They have mammary plants. While serving as a glands for the production of milk to nourish personal physician of a Carolus Linnaeus their young. Their skin has hairs as well as wealthy government (1707-1778) sweat and oil glands. Most mammals familiar official, he studied the diversity of plants to us produce live young ones. However, a in his employer’s garden. Later, he few of them, like the platypus and the echidna published 14 papers and also brought out lay eggs, and some, like kangaroos give birth the famous book Systema Naturae from to very poorly developed young ones. Some which all fundamental taxonomical examples are shown in Fig. 7.26. researches have taken off. His system of classification was a simple scheme for The scheme of classification of animals is arranging plants so as to be able to identify shown in Fig. 7.27. them again. 94 SCIENCE

Fig. 7.27: Classification of animals DIVERSITY IN LIVING ORGANISMS 95

7.6 Nomenclature the result of the process of classification which puts it along with the organisms it is Why is there a need for systematic naming of most related to. But when we actually name living organisms? the species, we do not list out the whole hierarchy of groups it belongs to. Instead, we Activity ______________ 7.3 limit ourselves to writing the name of the genus and species of that particular organism. World • Find out the names of the following over, it has been agreed that both these names will be used in Latin forms. animals and plants in as many Certain conventions are followed while languages as you can: writing the scientific names: 1. Tiger 2. Peacock 3. Ant 1. The name of the genus begins with a capital letter. 4. Neem 5. Lotus 6. Potato 2. The name of the species begins with a As you might be able to appreciate, it small letter. would be difficult for people speaking or writing in different languages to know when 3. When printed, the scientific name is they are talking about the same organism. This given in italics. problem was resolved by agreeing upon a ‘scientific’ name for organisms in the same 4. When written by hand, the genus name manner that chemical symbols and formulae and the species name have to be for various substances are used the world over. underlined separately. The scientific name for an organism is thus unique and can be used to identify it Activity ______________ 7.4 anywhere in the world. • Find out the scientific names of any The system of scientific naming or five common animals and plants. Do nomenclature we use today was introduced these names have anything in by Carolus Linnaeus in the eighteenth common with the names you normally century. The scientific name of an organism is use to identify them? What you have learnt • Classification helps us in exploring the diversity of life forms. • The major characteristics considered for classifying all organisms into five major kingdoms are: (a) whether they are made of prokaryotic or eukaryotic cells (b) whether the cells are living singly or organised into multi-cellular and thus complex organisms (c) whether the cells have a cell-wall and whether they prepare their own food. • All living organisms are divided on the above bases into five kingdoms, namely Monera, Protista, Fungi, Plantae and Animalia. • The classification of life forms is related to their evolution. 96 SCIENCE

• Plantae and Animalia are further divided into subdivisions on the basis of increasing complexity of body organisation. • Plants are divided into five groups: Thallophytes, Bryophytes, Pteridophytes, Gymnosperms and Angiosperms. • Animals are divided into ten groups: Porifera, Coelenterata, Platyhelminthes, Nematoda, Annelida, Arthropoda, Mollusca, Echinodermata, Protochordata and Vertebrata. • The binomial nomenclature makes for a uniform way of identification of the vast diversity of life around us. • The binomial nomenclature is made up of two words – a generic name and a specific name. Exercises 1. What are the advantages of classifying organisms? 2. How would you choose between two characteristics to be used for developing a hierarchy in classification? 3. Explain the basis for grouping organisms into five kingdoms. 4. What are the major divisions in the Plantae? What is the basis for these divisions? 5. How are the criteria for deciding divisions in plants different from the criteria for deciding the subgroups among animals? 6. Explain how animals in Vertebrata are classified into further subgroups. DIVERSITY IN LIVING ORGANISMS 97

C 8hapter MOTION In everyday life, we see some objects at rest Activity ______________ 8.1 and others in motion. Birds fly, fish swim, blood flows through veins and arteries, and • Discuss whether the walls of your cars move. Atoms, molecules, planets, stars classroom are at rest or in motion. and galaxies are all in motion. We often perceive an object to be in motion when its Activity ______________ 8.2 position changes with time. However, there are situations where the motion is inferred • Have you ever experienced that the through indirect evidences. For example, we train in which you are sitting appears infer the motion of air by observing the to move while it is at rest? movement of dust and the movement of leaves and branches of trees. What causes the • Discuss and share your experience. phenomena of sunrise, sunset and changing of seasons? Is it due to the motion of the Think and Act earth? If it is true, why don’t we directly perceive the motion of the earth? We sometimes are endangered by the motion of objects around us, especially An object may appear to be moving for if that motion is erratic and one person and stationary for some other. For uncontrolled as observed in a flooded the passengers in a moving bus, the roadside river, a hurricane or a tsunami. On the trees appear to be moving backwards. A other hand, controlled motion can be a person standing on the road–side perceives service to human beings such as in the the bus alongwith the passengers as moving. generation of hydro-electric power. Do However, a passenger inside the bus sees his you feel the necessity to study the fellow passengers to be at rest. What do these erratic motion of some objects and observations indicate? learn to control them? Most motions are complex. Some objects 8.1 Describing Motion may move in a straight line, others may take a circular path. Some may rotate and a few We describe the location of an object by others may vibrate. There may be situations specifying a reference point. Let us involving a combination of these. In this understand this by an example. Let us chapter, we shall first learn to describe the assume that a school in a village is 2 km north motion of objects along a straight line. We of the railway station. We have specified the shall also learn to express such motions position of the school with respect to the through simple equations and graphs. Later, railway station. In this example, the railway we shall discuss ways of describing circular station is the reference point. We could have motion. also chosen other reference points according to our convenience. Therefore, to describe the position of an object we need to specify a reference point called the origin. 2018-19

8.1.1 MOTION ALONG A STRAIGHT LINE while the magnitude of displacement = 35 km. Thus, the magnitude of displacement (35 km) The simplest type of motion is the motion is not equal to the path length (85 km). along a straight line. We shall first learn to Further, we will notice that the magnitude of describe this by an example. Consider the the displacement for a course of motion may motion of an object moving along a straight be zero but the corresponding distance path. The object starts its journey from O covered is not zero. If we consider the object which is treated as its reference point to travel back to O, the final position concides (Fig. 8.1). Let A, B and C represent the with the initial position, and therefore, the position of the object at different instants. At displacement is zero. However, the distance first, the object moves through C and B and covered in this journey is OA + AO = 60 km + reaches A. Then it moves back along the same 60 km = 120 km. Thus, two different physical path and reaches C through B. quantities — the distance and the Fig. 8.1: Positions of an object on a straight line path The total path length covered by the object displacement, are used to describe the overall is OA + AC, that is 60 km + 35 km = 95 km. motion of an object and to locate its final This is the distance covered by the object. To position with reference to its initial position describe distance we need to specify only the at a given time. numerical value and not the direction of motion. There are certain quantities which Activity ______________ 8.3 are described by specifying only their numerical values. The numerical value of a • Take a metre scale and a long rope. physical quantity is its magnitude. From this • Walk from one corner of a basket-ball example, can you find out the distance of the final position C of the object from the initial court to its oppposite corner along its position O? This difference will give you the sides. numerical value of the displacement of the • Measure the distance covered by you object from O to C through A. The shortest and magnitude of the displacement. distance measured from the initial to the final • What difference would you notice position of an object is known as the between the two in this case? displacement. Activity ______________ 8.4 Can the magnitude of the displacement be equal to the distance travelled by an • Automobiles are fitted with a device object? Consider the example given in that shows the distance travelled. Such (Fig. 8.1). For motion of the object from O to a device is known as an odometer. A A, the distance covered is 60 km and the car is driven from Bhubaneshwar to magnitude of displacement is also 60 km. New Delhi. The difference between the During its motion from O to A and back to B, final reading and the initial reading of the distance covered = 60 km + 25 km = 85 km the odometer is 1850 km. • Find the magnitude of the displacement between Bhubaneshwar and New Delhi by using the Road Map of India. MOTION 99

Q uestions Table 8.1 1. An object has moved through a distance. Can it have zero Time Distance Distance displacement? If yes, support travelled by travelled by your answer with an example. object A in m object B in m 2. A farmer moves along the boundary of a square field of side 9:30 am 10 12 10 m in 40 s. What will be the 9:45 am 20 19 magnitude of displacement of the 10:00 am 30 23 farmer at the end of 2 minutes 20 10:15 am 40 35 seconds from his initial position? 10:30 am 50 37 3. Which of the following is true for 10:45 am 60 41 displacement? 11:00 am 70 44 (a) It cannot be zero. (b) Its magnitude is greater than 8.2 Measuring the Rate of Motion the distance travelled by the object. 8.1.2 UNIFORM MOTION AND NON- (a) UNIFORM MOTION Consider an object moving along a straight line. Let it travel 5 m in the first second, 5 m more in the next second, 5 m in the third second and 5 m in the fourth second. In this case, the object covers 5 m in each second. As the object covers equal distances in equal intervals of time, it is said to be in uniform motion. The time interval in this motion should be small. In our day-to-day life, we come across motions where objects cover unequal distances in equal intervals of time, for example, when a car is moving on a crowded street or a person is jogging in a park. These are some instances of non-uniform motion. Activity ______________ 8.5 • The data regarding the motion of two (b) different objects A and B are given in Fig. 8.2 Table 8.1. • Examine them carefully and state whether the motion of the objects is uniform or non-uniform. 100 SCIENCE

Look at the situations given in Fig. 8.2. If Total distance travelled the bowling speed is 143 km h–1 in Fig. 8.2(a) Average speed = Total time taken what does it mean? What do you understand from the signboard in Fig. 8.2(b)? 32 m = 6 s = 5.33 m s–1 Different objects may take different Therefore, the average speed of the object amounts of time to cover a given distance. is 5.33 m s–1. Some of them move fast and some move 8.2.1 SPEED WITH DIRECTION slowly. The rate at which objects move can The rate of motion of an object can be more be different. Also, different objects can move comprehensive if we specify its direction of at the same rate. One of the ways of motion along with its speed. The quantity that measuring the rate of motion of an object is specifies both these aspects is called velocity. to find out the distance travelled by the object Velocity is the speed of an object moving in a in unit time. This quantity is referred to as definite direction. The velocity of an object speed. The SI unit of speed is metre per can be uniform or variable. It can be changed second. This is represented by the symbol by changing the object’s speed, direction of m s–1 or m/s. The other units of speed include motion or both. When an object is moving centimetre per second (cm s–1) and kilometre along a straight line at a variable speed, we per hour (km h–1). To specify the speed of an can express the magnitude of its rate of object, we require only its magnitude. The motion in terms of average velocity. It is speed of an object need not be constant. In calculated in the same way as we calculate most cases, objects will be in non-uniform average speed. motion. Therefore, we describe the rate of motion of such objects in terms of their In case the velocity of the object is average speed. The average speed of an object changing at a uniform rate, then average is obtained by dividing the total distance velocity is given by the arithmetic mean of travelled by the total time taken. That is, initial velocity and final velocity for a given period of time. That is, Total distance travelled average speed = Total time taken If an object travels a distance s in time t then its speed v is, s average velocity = initial velocity + final velocity v= t (8.1) 2 Let us understand this by an example. A u+v Mathematically, vav = 2 car travels a distance of 100 km in 2 h. Its (8.2) average speed is 50 km h–1. The car might not have travelled at 50 km h–1 all the time. where vav is the average velocity, u is the initial velocity and v is the final velocity of the object. Sometimes it might have travelled faster and sometimes slower than this. Speed and velocity have the same units, that is, m s–1 or m/s. Example 8.1 An object travels 16 m in 4 s Activity ______________ 8.6 and then another 16 m in 2 s. What is the average speed of the object? • Measure the time it takes you to walk from your house to your bus stop or Solution: the school. If you consider that your average walking speed is 4 km h–1, Total distance travelled by the object = estimate the distance of the bus stop 16 m + 16 m = 32 m or school from your house. Total time taken = 4 s + 2 s = 6 s MOTION 101

Activity ______________ 8.7 km 1000 m 1h = 50 × × • At a time when it is cloudy, there may be frequent thunder and lightning. The h 1km 3600s sound of thunder takes some time to reach you after you see the lightning. = 13.9 m s–1 The average speed of the car is • Can you answer why this happens? 50 km h–1 or 13.9 m s–1. • Measure this time interval using a Example 8.3 Usha swims in a 90 m long digital wrist watch or a stop watch. pool. She covers 180 m in one minute • Calculate the distance of the nearest by swimming from one end to the other and back along the same straight path. point of lightning. (Speed of sound in Find the average speed and average air = 346 m s-1.) velocity of Usha. Q uestions Solution: 1. Distinguish between speed and Total distance covered by Usha in 1 min velocity. is 180 m. 2. Under what condition(s) is the Displacement of Usha in 1 min = 0 m magnitude of average velocity of an object equal to its average Total distance covered speed? Average speed = Total time taken 3. What does the odometer of an automobile measure? 180 m = 180 m × 1 min 4. What does the path of an object = 1min 1min 60 s look like when it is in uniform motion? = 3 m s-1 5. During an experiment, a signal from a spaceship reached the Displacement ground station in five minutes. Average velocity = Total time taken What was the distance of the spaceship from the ground 0m station? The signal travels at the = 60 s speed of light, that is, 3 × 108 m s–1. = 0 m s–1 The average speed of Usha is 3 m s–1 Example 8.2 The odometer of a car reads and her average velocity is 0 m s–1. 2000 km at the start of a trip and 2400 km at the end of the trip. If the 8.3 Rate of Change of Velocity trip took 8 h, calculate the average speed of the car in km h–1 and m s–1. During uniform motion of an object along a straight line, the velocity remains constant Solution: with time. In this case, the change in velocity of the object for any time interval is zero. Distance covered by the car, However, in non-uniform motion, velocity s = 2400 km – 2000 km = 400 km varies with time. It has different values at Time elapsed, t = 8 h different instants and at different points of Average speed of the car is, the path. Thus, the change in velocity of the object during any time interval is not zero. vav = s 400 km Can we now express the change in velocity of = an object? t 8h = 50 km h–1 102 SCIENCE

To answer such a question, we have to attain a velocity of 6 m s–1 in 30 s. Then introduce another physical quantity called he applies brakes such that the velocity acceleration, which is a measure of the of the bicycle comes down to 4 m s-1 in change in the velocity of an object per unit the next 5 s. Calculate the acceleration time. That is, of the bicycle in both the cases. change in velocity Solution: acceleration = In the first case: time taken initial velocity, u = 0 ; final velocity, v = 6 m s–1 ; If the velocity of an object changes from time, t = 30 s . an initial value u to the final value v in time t, From Eq. (8.3), we have the acceleration a is, v–u v–u (8.3) a= a= t t Substituting the given values of u,v and This kind of motion is known as t in the above equation, we get accelerated motion. The acceleration is taken to be positive if it is in the direction of velocity ( )6 m s–1 – 0 m s–1 and negative when it is opposite to the direction of velocity. The SI unit of a= acceleration is m s–2 . 30 s If an object travels in a straight line and = 0.2 m s–2 its velocity increases or decreases by equal In the second case: amounts in equal intervals of time, then the initial velocity, u = 6 m s–1; acceleration of the object is said to be final velocity, v = 4 m s–1; uniform. The motion of a freely falling body time, t = 5 s. is an example of uniformly accelerated motion. On the other hand, an object can ( )4 m s–1 – 6 m s–1 travel with non-uniform acceleration if its velocity changes at a non-uniform rate. For Then, a = example, if a car travelling along a straight 5s road increases its speed by unequal amounts in equal intervals of time, then the car is said = –0.4 m s–2 . to be moving with non-uniform acceleration. The acceleration of the bicycle in the first case is 0.2 m s–2 and in the second Activity ______________ 8.8 case, it is –0.4 m s–2. • In your everyday life you come across Q uestions a range of motions in which 1. When will you say a body is in (a) acceleration is in the direction of (i) uniform acceleration? (ii) non- motion, uniform acceleration? (b) acceleration is against the 2. A bus decreases its speed from direction of motion, 80 km h–1 to 60 km h–1 in 5 s. (c) acceleration is uniform, Find the acceleration of the bus. (d) acceleration is non-uniform. 3. A train starting from a railway station and moving with uniform • Can you identify one example each acceleration attains a speed for the above type of motion? 40 km h–1 in 10 minutes. Find its acceleration. Example 8.4 Starting from a stationary position, Rahul paddles his bicycle to MOTION 103

8.4 Graphical Representation of distance travelled by the object is directly Motion proportional to time taken. Thus, for uniform speed, a graph of distance travelled against Graphs provide a convenient method to present basic information about a variety of time is a straight line, as shown in Fig. 8.3. events. For example, in the telecast of a The portion OB of the graph shows that the one-day cricket match, vertical bar graphs show the run rate of a team in each over. As distance is increasing at a uniform rate. Note you have studied in mathematics, a straight that, you can also use the term uniform line graph helps in solving a linear equation having two variables. velocity in place of uniform speed if you take the magnitude of displacement equal to the To describe the motion of an object, we can use line graphs. In this case, line graphs distance travelled by the object along the show dependence of one physical quantity, y-axis. such as distance or velocity, on another quantity, such as time. We can use the distance-time graph to determine the speed of an object. To do so, 8.4.1 DISTANCE–TIME GRAPHS consider a small part AB of the distance-time The change in the position of an object with graph shown in Fig 8.3. Draw a line parallel time can be represented on the distance-time graph adopting a convenient scale of choice. to the x-axis from point A and another line In this graph, time is taken along the x–axis parallel to the y-axis from point B. These two and distance is taken along the y-axis. Distance-time graphs can be employed under lines meet each other at point C to form a various conditions where objects move with triangle ABC. Now, on the graph, AC denotes uniform speed, non-uniform speed, remain at rest etc. the time interval (t2 – t1) while BC corresponds to the distance (s2 – s1). We can see from the graph that as the object moves from the point A to B, it covers a distance (s2 – s1) in time (t2 – t1). The speed, v of the object, therefore can be represented as v= s2 – s1 (8.4) t 2 – t 1 We can also plot the distance-time graph for accelerated motion. Table 8.2 shows the distance travelled by a car in a time interval of two seconds. Fig. 8.3: Distance-time graph of an object moving Table 8.2: Distance travelled by a with uniform speed car at regular time intervals We know that when an object travels equal Time in seconds Distance in metres distances in equal intervals of time, it moves with uniform speed. This shows that the 00 21 104 44 69 8 16 10 25 12 36 SCIENCE

Fig. 8.4: Distance-time graph for a car moving with and the velocity is represented along the non-uniform speed y-axis. If the object moves at uniform velocity, The distance-time graph for the motion the height of its velocity-time graph will not of the car is shown in Fig. 8.4. Note that the shape of this graph is different from the earlier change with time (Fig. 8.5). It will be a straight distance-time graph (Fig. 8.3) for uniform line parallel to the x-axis. Fig. 8.5 shows the motion. The nature of this graph shows non- linear variation of the distance travelled by velocity-time graph for a car moving with the car with time. Thus, the graph shown in uniform velocity of 40 km h–1. Fig 8.4 represents motion with non-uniform speed. We know that the product of velocity and time give displacement of an object moving 8.4.2 VELOCITY-TIME GRAPHS with uniform velocity. The area enclosed by The variation in velocity with time for an velocity-time graph and the time axis will be object moving in a straight line can be represented by a velocity-time graph. In this equal to the magnitude of the displacement. graph, time is represented along the x-axis To know the distance moved by the car Fig. 8.5: Velocity-time graph for uniform motion of between time t1 and t2 using Fig. 8.5, draw a car perpendiculars from the points corresponding to the time t1 and t2 on the graph. The velocity of 40 km h–1 is represented by the height AC or BD and the time (t2 – t1) is represented by the length AB. So, the distance s moved by the car in time (t2 – t1) can be expressed as s = AC × CD = [(40 km h–1) × (t2 – t1) h] = 40 (t2– t1) km = area of the rectangle ABDC (shaded in Fig. 8.5). We can also study about unifor mly accelerated motion by plotting its velocity– time graph. Consider a car being driven along a straight road for testing its engine. Suppose a person sitting next to the driver records its velocity after every 5 seconds by noting the reading of the speedometer of the car. The velocity of the car, in km h–1 as well as in m s–1, at different instants of time is shown in table 8.3. Table 8.3: Velocity of a car at regular instants of time Time Velocity of the car (s) (km h–1) (m s–1) 0 5 0 0 10 2.5 9 15 5.0 18 20 7.5 27 25 10.0 36 30 12.5 45 15.0 54 MOTION 105

In this case, the velocity-time graph for the motion of the car is shown in Fig. 8.6. The nature of the graph shows that velocity changes by equal amounts in equal intervals of time. Thus, for all uniformly accelerated motion, the velocity-time graph is a straight line. Velocity (km h–1) Fig. 8.6: Velocity-time graph for a car moving with Fig. 8.7: Velocity-time graphs of an object in non- uniform accelerations. uniformly accelerated motion. You can also determine the distance Fig. 8.7(a) shows a velocity-time graph moved by the car from its velocity-time graph. that represents the motion of an object whose The area under the velocity-time graph gives velocity is decreasing with time while the distance (magnitude of displacement) Fig. 8.7 (b) shows the velocity-time graph moved by the car in a given interval of time. representing the non-uniform variation of If the car would have been moving with velocity of the object with time. Try to interpret uniform velocity, the distance travelled by it these graphs. would be represented by the area ABCD under the graph (Fig. 8.6). Since the Activity ______________ 8.9 magnitude of the velocity of the car is changing due to acceleration, the distance s • The times of arrival and departure of a travelled by the car will be given by the area train at three stations A, B and C and ABCDE under the velocity-time graph the distance of stations B and C from (Fig. 8.6). station A are given in table 8.4. That is, Table 8.4: Distances of stations B s = area ABCDE and C from A and times of arrival = area of the rectangle ABCD + area of and departure of the train the triangle ADE Station Distance Time of Time of 1 from A arrival departure = AB × BC + 2 (AD × DE) (km) (hours) (hours) In the case of non-uniformly accelerated motion, velocity-time graphs can have any A 0 08:00 08:15 shape. B 120 11:15 11:30 C 180 13:00 13:15 • Plot and interpret the distance-time graph for the train assuming that its motion between any two stations is uniform. 106 SCIENCE

Activity _____________ 8.10 8.5 Equations of Motion by Graphical Method • Feroz and his sister Sania go to school on their bicycles. Both of them start at When an object moves along a straight line the same time from their home but take different times to reach the school with uniform acceleration, it is possible to although they follow the same route. Table 8.5 shows the distance travelled relate its velocity, acceleration during motion by them in different times and the distance covered by it in a certain Table 8.5: Distance covered by Feroz and Sania at different times on time interval by a set of equations known as their bicycles the equations of motion. For convenience, a set of three such equations are given below: v = u + at (8.5) s = ut + ½ at2 (8.6) 2 a s = v2 – u2 (8.7) where u is the initial velocity of the object Time Distance Distance which moves with uniform acceleration a for travelled travelled by Feroz by Sania time t, v is the final velocity, and s is the (km) (km) distance travelled by the object in time t. Eq. (8.5) describes the velocity-time relation and Eq. (8.6) represents the position-time 8:00 am 0 0 relation. Eq. (8.7), which represents the 8:05 am 1.0 0.8 8:10 am 1.9 1.6 relation between the position and the velocity, 8:15 am 2.8 2.3 8:20 am 3.6 3.0 can be obtained from Eqs. (8.5) and (8.6) by 8:25 am 3.6 – eliminating t. These three equations can be derived by graphical method. • Plot the distance-time graph for their 8.5.1 EQUATION FOR VELOCITY-TIME motions on the same scale and interpret. RELATION Consider the velocity-time graph of an object that moves under uniform acceleration as Q uestions Fig. 8.8: Velocity-time graph to obtain the equations 1. What is the nature of the of motion distance-time graphs for uniform and non-uniform motion of an 107 object? 2. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis? 3. What can you say about the motion of an object if its speed- time graph is a straight line parallel to the time axis? 4. What is the quantity which is measured by the area occupied below the velocity-time graph? MOTION

shown in Fig. 8.8 (similar to Fig. 8.6, but now 1 = OA × OC + 2 (AD × BD) with u ≠ 0). From this graph, you can see that (8.10) initial velocity of the object is u (at point A) Substituting OA = u, OC = AD = t and BD = at, and then it increases to v (at point B) in time we get t. The velocity changes at a uniform rate a. In Fig. 8.8, the perpendicular lines BC and 1 (t ×at ) 2 BE are drawn from point B on the time and s= u × t + the velocity axes respectively, so that the initial velocity is represented by OA, the final or s = u t + 1 a t 2 2 velocity is represented by BC and the time interval t is represented by OC. BD = BC – CD, represents the change in velocity in time 8.5.3 EQUATION FOR POSITION–VELOCITY interval t. RELATION Let us draw AD parallel to OC. From the graph, we observe that From the velocity-time graph shown in Fig. 8.8, the distance s travelled by the object BC = BD + DC = BD + OA in time t, moving under uniform acceleration a is given by the area enclosed within the Substituting BC = v and OA = u, trapezium OABC under the graph. That is, we get v = BD + u s = area of the trapezium OABC or BD = v – u (8.8) From the velocity-time graph (Fig. 8.8), the acceleration of the object is given by Change in velocity (OA + BC) ×OC a = time taken = BD = BD 2 = AD OC Substituting OA = u, BC = v and OC = t, we get Substituting OC = t, we get (u + v)t (8.11) BD s= a= t 2 or BD = at From the velocity-time relation (Eq. 8.6), Using Eqs. (8.8) and (8.9) we get v = u + at (8.9) we get (v – u ) (8.12) t= a 8.5.2 EQUATION FOR POSITION-TIME Using Eqs. (8.11) and (8.12) we have RELATION (v + u ) × (v - u ) Let us consider that the object has travelled a distance s in time t under uniform s= acceleration a. In Fig. 8.8, the distance 2a travelled by the object is obtained by the area enclosed within OABC under the velocity-time or 2 a s = v2 – u2 graph AB. Example 8.5 A train starting from rest Thus, the distance s travelled by the object attains a velocity of 72 km h–1 in is given by 5 minutes. Assuming that the acceleration is uniform, find (i) the s = area OABC (which is a trapezium) acceleration and (ii) the distance = area of the rectangle OADC + area of travelled by the train for attaining this the triangle ABD velocity. 108 SCIENCE

Solution: (ii) From Eq. (8.6) we have We have been given s = ut + 1at 2 u = 0 ; v = 72 km h–1 = 20 m s-1 and 2 t = 5 minutes = 300 s. (i) From Eq. (8.5) we know that 1 = 5 m s–1 × 5 s + 2 × 1 m s–2 × (5 s)2 (v – u ) = 25 m + 12.5 m a= = 37.5 m t The acceleration of the car is 1 m s–2 and the distance covered is 37.5 m. 20 m s–1 – 0 m s–1 = Example 8.7 The brakes applied to a car produce an acceleration of 6 m s-2 in 300 s the opposite direction to the motion. If = 1 m s–2 the car takes 2 s to stop after the application of brakes, calculate the 15 distance it travels during this time. (ii) From Eq. (8.7) we have Solution: 2 a s = v2 – u2 = v2 – 0 Thus, We have been given a = – 6 m s–2 ; t = 2 s and v = 0 m s–1. v2 From Eq. (8.5) we know that s= v = u + at 2a 0 = u + (– 6 m s–2) × 2 s or u = 12 m s–1 . (20 m s–1 )2 From Eq. (8.6) we get = 1 2 ×(1/15) m s–2 s=ut+ 2 at2 = 3000 m 1 = 3 km = (12 m s–1 ) × (2 s) + 2 (–6 m s–2 ) (2 s)2 = 24 m – 12 m 1 = 12 m The acceleration of the train is 15 m s– 2 Thus, the car will move 12 m before it and the distance travelled is 3 km. stops after the application of brakes. Can you now appreciate why drivers Example 8.6 A car accelerates uniformly are cautioned to maintain some from 18 km h–1 to 36 km h–1 in 5 s. distance between vehicles while Calculate (i) the acceleration and (ii) the travelling on the road? distance covered by the car in that time. Q uestions Solution: 1. A bus starting from rest moves with a uniform acceleration of We are given that 0.1 m s-2 for 2 minutes. Find (a) u = 18 km h–1 = 5 m s–1 the speed acquired, (b) the v = 36 km h–1 = 10 m s–1 and distance travelled. t = 5s. (i) From Eq. (8.5) we have v–u a= t 10 m s-1 – 5 m s-1 = 5s = 1 m s–2 MOTION 109

2. A train is travelling at a speed Let us consider an example of the motion of 90 km h–1. Brakes are applied of a body along a closed path. Fig 8.9 (a) shows so as to produce a uniform the path of an athlete along a rectangular acceleration of – 0.5 m s-2. Find track ABCD. Let us assume that the athlete how far the train will go before it runs at a uniform speed on the straight parts is brought to rest. AB, BC, CD and DA of the track. In order to keep himself on track, he quickly changes 3. A trolley, while going down an his speed at the corners. How many times inclined plane, has an will the athlete have to change his direction acceleration of 2 cm s-2. What will of motion, while he completes one round? It be its velocity 3 s after the start? is clear that to move in a rectangular track once, he has to change his direction of motion 4. A racing car has a unifor m four times. acceleration of 4 m s-2. What distance will it cover in 10 s after Now, suppose instead of a rectangular start? track, the athlete is running along a hexagonal shaped path ABCDEF, as shown 5. A stone is thrown in a vertically in Fig. 8.9(b). In this situation, the athlete upward direction with a velocity will have to change his direction six times of 5 m s-1. If the acceleration of the while he completes one round. What if the stone during its motion is 10 m s–2 track was not a hexagon but a regular in the downward direction, what octagon, with eight equal sides as shown by will be the height attained by the ABCDEFGH in Fig. 8.9(c)? It is observed that stone and how much time will it as the number of sides of the track increases take to reach there? the athelete has to take turns more and more often. What would happen to the shape of 8.6 Uniform Circular Motion the track as we go on increasing the number of sides indefinitely? If you do this you will When the velocity of an object changes, we notice that the shape of the track approaches say that the object is accelerating. The change the shape of a circle and the length of each of in the velocity could be due to change in its the sides will decrease to a point. If the athlete magnitude or the direction of the motion or moves with a velocity of constant magnitude both. Can you think of an example when an along the circular path, the only change in object does not change its magnitude of his velocity is due to the change in the velocity but only its direction of motion? direction of motion. The motion of the athlete moving along a circular path is, therefore, an (a) Rectangular track (b) Hexagonal track example of an accelerated motion. We know that the circumference of a circle of radius r is given by 2πr . If the athlete takes t seconds to go once around the circular path of radius r, the speed v is given by 2 πr (8.13) v= t (c) Octagonal shaped track (d) A circular track When an object moves in a circular path with uniform speed, its motion is called Fig. 8.9: The motion of an athlete along closed tracks uniform circular motion. of different shapes. 110 SCIENCE

Activity _____________ 8.11 If you carefully note, on being released the stone moves along a straight line • Take a piece of thread and tie a small tangential to the circular path. This is piece of stone at one of its ends. Move because once the stone is released, it the stone to describe a circular path continues to move along the direction it has with constant speed by holding the been moving at that instant. This shows that thread at the other end, as shown in the direction of motion changed at every point Fig. 8.10. when the stone was moving along the circular path. Fig. 8.10: A stone describing a circular path with a velocity of constant magnitude. When an athlete throws a hammer or a discus in a sports meet, he/she holds the • Now, let the stone go by releasing the hammer or the discus in his/her hand and thread. gives it a circular motion by rotating his/her own body. Once released in the desired • Can you tell the direction in which the direction, the hammer or discus moves in the stone moves after it is released? direction in which it was moving at the time it was released, just like the piece of stone in • By repeating the activity for a few times the activity described above. There are many and releasing the stone at different more familiar examples of objects moving positions of the circular path, check under uniform circular motion, such as the whether the direction in which the motion of the moon and the earth, a satellite stone moves remains the same or not. in a circular orbit around the earth, a cyclist on a circular track at constant speed and so on. What you have learnt • Motion is a change of position; it can be described in terms of the distance moved or the displacement. • The motion of an object could be uniform or non-uniform depending on whether its velocity is constant or changing. • The speed of an object is the distance covered per unit time, and velocity is the displacement per unit time. • The acceleration of an object is the change in velocity per unit time. • Uniform and non-uniform motions of objects can be shown through graphs. • The motion of an object moving at uniform acceleration can be described with the help of the following equations, namely v = u + at s = ut + ½ at2 2as = v2 – u2 MOTION 111

where u is initial velocity of the object, which moves with uniform acceleration a for time t, v is its final velocity and s is the distance it travelled in time t. • If an object moves in a circular path with uniform speed, its motion is called uniform circular motion. Exercises 1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s? 2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C? 3. Abdul, while driving to school, computes the average speed for his trip to be 20 km h–1. On his return trip along the same route, there is less traffic and the average speed is 30 km h–1. What is the average speed for Abdul’s trip? 4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s–2 for 8.0 s. How far does the boat travel during this time? 5. A driver of a car travelling at 52 km h–1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h–1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied? 6. Fig 8.11 shows the distance-time graph of three objects A,B and C. Study the graph and answer the following questions: Fig. 8.11 112 SCIENCE

(a) Which of the three is travelling the fastest? (b) Are all three ever at the same point on the road? (c) How far has C travelled when B passes A? (d) How far has B travelled by the time it passes C? 7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s-2, with what velocity will it strike the ground? After what time will it strike the ground? 8. The speed-time graph for a car is shown is Fig. 8.12. Fig. 8.12 (a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period. (b) Which part of the graph represents uniform motion of the car? 9. State which of the following situations are possible and give an example for each of these: (a) an object with a constant acceleration but with zero velocity (b) an object moving with an acceleration but with uniform speed. (c) an object moving in a certain direction with an acceleration in the perpendicular direction. 10. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth. MOTION 113

C 9hapter FORCE AND LAWS OF MOTION In the previous chapter, we described the In our everyday life we observe that some motion of an object along a straight line in effort is required to put a stationary object terms of its position, velocity and acceleration. into motion or to stop a moving object. We We saw that such a motion can be uniform ordinarily experience this as a muscular effort or non-uniform. We have not yet discovered and say that we must push or hit or pull on what causes the motion. Why does the speed an object to change its state of motion. The of an object change with time? Do all motions concept of force is based on this push, hit or require a cause? If so, what is the nature of pull. Let us now ponder about a ‘force’. What this cause? In this chapter we shall make an is it? In fact, no one has seen, tasted or felt a attempt to quench all such curiosities. force. However, we always see or feel the effect of a force. It can only be explained by For many centuries, the problem of describing what happens when a force is motion and its causes had puzzled scientists applied to an object. Pushing, hitting and and philosophers. A ball on the ground, when pulling of objects are all ways of bringing given a small hit, does not move forever. Such objects in motion (Fig. 9.1). They move observations suggest that rest is the “natural because we make a force act on them. state” of an object. This remained the belief until Galileo Galilei and Isaac Newton From your studies in earlier classes, you developed an entirely different approach to are also familiar with the fact that a force can understand motion. be used to change the magnitude of velocity of an object (that is, to make the object move faster or slower) or to change its direction of motion. We also know that a force can change the shape and size of objects (Fig. 9.2). (a) The trolley moves along the (b) The drawer is pulled. direction we push it. (a) (c) The hockey stick hits the ball forward (b) Fig. 9.1: Pushing, pulling, or hitting objects change Fig. 9.2: (a) A spring expands on application of force; their state of motion. (b) A spherical rubber ball becomes oblong as we apply force on it. 2018-19

9.1 Balanced and Unbalanced box with a small force, the box does not move Forces because of friction acting in a direction opposite to the push [Fig. 9.4(a)]. This friction Fig. 9.3 shows a wooden block on a horizontal force arises between two surfaces in contact; table. Two strings X and Y are tied to the two in this case, between the bottom of the box opposite faces of the block as shown. If we and floor’s rough surface. It balances the apply a force by pulling the string X, the block pushing force and therefore the box does not begins to move to the right. Similarly, if we move. In Fig. 9.4(b), the children push the pull the string Y, the block moves to the left. box harder but the box still does not move. But, if the block is pulled from both the sides This is because the friction force still balances with equal forces, the block will not move. the pushing force. If the children push the Such forces are called balanced forces and box harder still, the pushing force becomes do not change the state of rest or of motion of bigger than the friction force [Fig. 9.4(c)]. an object. Now, let us consider a situation in There is an unbalanced force. So the box which two opposite forces of different starts moving. magnitudes pull the block. In this case, the block would begin to move in the direction of What happens when we ride a bicycle? the greater force. Thus, the two forces are When we stop pedalling, the bicycle begins not balanced and the unbalanced force acts to slow down. This is again because of the in the direction the block moves. This friction forces acting opposite to the direction suggests that an unbalanced force acting on of motion. In order to keep the bicycle moving, an object brings it in motion. we have to start pedalling again. It thus appears that an object maintains its motion Fig. 9.3: Two forces acting on a wooden block under the continuous application of an unbalanced force. However, it is quite What happens when some children try to incorrect. An object moves with a uniform push a box on a rough floor? If they push the velocity when the forces (pushing force and frictional force) acting on the object are balanced and there is no net external force on it. If an unbalanced force is applied on the object, there will be a change either in its speed or in the direction of its motion. Thus, to accelerate the motion of an object, an unbalanced force is required. And the change in its speed (or in the direction of motion) would continue as long as this unbalanced force is applied. However, if this force is (a) (b) (c) FORCE AND LAWS OF MOTION Fig. 9.4 115

removed completely, the object would continue to move with the velocity it has acquired till then. 9.2 First Law of Motion Fig. 9.5: (a) the downward motion; (b) the upward motion of a marble on an inclined plane; By observing the motion of objects on an and (c) on a double inclined plane. inclined plane Galileo deduced that objects move with a constant speed when no force Newton further studied Galileo’s ideas on acts on them. He observed that when a marble force and motion and presented three rolls down an inclined plane, its velocity fundamental laws that govern the motion of increases [Fig. 9.5(a)]. In the next chapter, objects. These three laws are known as you will learn that the marble falls under the Newton’s laws of motion. The first law of unbalanced force of gravity as it rolls down motion is stated as: and attains a definite velocity by the time it reaches the bottom. Its velocity decreases An object remains in a state of rest or of when it climbs up as shown in Fig. 9.5(b). uniform motion in a straight line unless Fig. 9.5(c) shows a marble resting on an ideal compelled to change that state by an applied frictionless plane inclined on both sides. force. Galileo argued that when the marble is released from left, it would roll down the slope In other words, all objects resist a change and go up on the opposite side to the same in their state of motion. In a qualitative way, height from which it was released. If the the tendency of undisturbed objects to stay inclinations of the planes on both sides are at rest or to keep moving with the same equal then the marble will climb the same velocity is called inertia. This is why, the first distance that it covered while rolling down. If law of motion is also known as the law of the angle of inclination of the right-side plane inertia. were gradually decreased, then the marble would travel further distances till it reaches Certain experiences that we come across the original height. If the right-side plane were while travelling in a motorcar can be ultimately made horizontal (that is, the slope explained on the basis of the law of inertia. is reduced to zero), the marble would continue We tend to remain at rest with respect to the to travel forever trying to reach the same seat until the driver applies a braking force height that it was released from. The to stop the motorcar. With the application of unbalanced forces on the marble in this case brakes, the car slows down but our body are zero. It thus suggests that an unbalanced tends to continue in the same state of motion (external) force is required to change the because of its inertia. A sudden application motion of the marble but no net force is of brakes may thus cause injury to us by needed to sustain the uniform motion of the marble. In practical situations it is difficult to achieve a zero unbalanced force. This is because of the presence of the frictional force acting opposite to the direction of motion. Thus, in practice the marble stops after travelling some distance. The effect of the frictional force may be minimised by using a smooth marble and a smooth plane and providing a lubricant on top of the planes. 116 SCIENCE

Galileo Galilei was born impact or collision with the panels in front. Safety belts are worn to prevent such on 15 February 1564 in accidents. Safety belts exert a force on our body to make the forward motion slower. An Pisa, Italy. Galileo, right opposite experience is encountered when we are standing in a bus and the bus begins to from his childhood, had move suddenly. Now we tend to fall backwards. This is because the sudden start interest in mathematics of the bus brings motion to the bus as well as to our feet in contact with the floor of the and natural philosophy. bus. But the rest of our body opposes this motion because of its inertia. But his father When a motorcar makes a sharp turn at Vincenzo Galilei wanted a high speed, we tend to get thrown to one side. This can again be explained on the basis him to become a medical of the law of inertia. We tend to continue in our straight-line motion. When an doctor. Accordingly, Galileo Galilei unbalanced force is applied by the engine to Galileo enrolled himself (1564 – 1642) change the direction of motion of the for a medical degree at the motorcar, we slip to one side of the seat due to the inertia of our body. University of Pisa in 1581 which he never The fact that a body will remain at rest completed because of his real interest in unless acted upon by an unbalanced force can be illustrated through the following mathematics. In 1586, he wrote his first activities: scientific book ‘The Little Balance [La Activity ______________ 9.1 Balancitta]’, in which he described • Make a pile of similar carom coins on a table, as shown in Fig. 9.6. Archimedes’ method of finding the relative • Attempt a sharp horizontal hit at the densities (or specific gravities) of substances bottom of the pile using another carom coin or the striker. If the hit is strong using a balance. In 1589, in his series of enough, the bottom coin moves out quickly. Once the lowest coin is essays – De Motu, he presented his theories removed, the inertia of the other coins makes them ‘fall’ vertically on the table. about falling objects using an inclined plane Fig. 9.6: Only the carom coin at the bottom of a to slow down the rate of descent. pile is removed when a fast moving carom coin (or striker) hits it. In 1592, he was appointed professor of mathematics at the University of Padua in the Republic of Venice. Here he continued his observations on the theory of motion and through his study of inclined planes and the pendulum, formulated the correct law for uniformly accelerated objects that the distance the object moves is proportional to the square of the time taken. Galileo was also a remarkable craftsman. He developed a series of telescopes whose optical performance was much better than that of other telescopes available during those days. Around 1640, he designed the first pendulum clock. In his book ‘Starry Messenger’ on his astronomical discoveries, Galileo claimed to have seen mountains on the moon, the milky way made up of tiny stars, and four small bodies orbiting Jupiter. In his books ‘Discourse on Floating Bodies’ and ‘Letters on the Sunspots’, he disclosed his observations of sunspots. Using his own telescopes and through his observations on Saturn and Venus, Galileo argued that all the planets must orbit the Sun and not the earth, contrary to what was believed at that time. FORCE AND LAWS OF MOTION 117

Activity ______________ 9.2 five-rupees coin if we use a one-rupee coin, we find that a lesser force is required to perform • Set a five-rupee coin on a stiff card the activity. A force that is just enough to covering an empty glass tumbler cause a small cart to pick up a large velocity standing on a table as shown in will produce a negligible change in the motion Fig. 9.7. of a train. This is because, in comparison to the cart the train has a much lesser tendency • Give the card a sharp horizontal flick to change its state of motion. Accordingly, we with a finger. If we do it fast then the say that the train has more inertia than the card shoots away, allowing the coin to cart. Clearly, heavier or more massive objects fall vertically into the glass tumbler due offer larger inertia. Quantitatively, the inertia to its inertia. of an object is measured by its mass. We may thus relate inertia and mass as follows: • The inertia of the coin tries to maintain Inertia is the natural tendency of an object to its state of rest even when the card resist a change in its state of motion or of flows off. rest. The mass of an object is a measure of its inertia. Fig. 9.7: When the card is flicked with the finger the coin placed over it falls in the Q uestions tumbler. 1. Which of the following has more inertia: (a) a rubber ball and a Activity ______________ 9.3 stone of the same size? (b) a bicycle and a train? (c) a five- • Place a water-filled tumbler on a tray. rupees coin and a one-rupee coin? • Hold the tray and turn around as fast 2. In the following example, try to identify the number of times the as you can. velocity of the ball changes: • We observe that the water spills. Why? “A football player kicks a football to another player of his team who Observe that a groove is provided in a kicks the football towards the saucer for placing the tea cup. It prevents goal. The goalkeeper of the the cup from toppling over in case of sudden opposite team collects the football jerks. and kicks it towards a player of his own team”. 9.3 Inertia and Mass Also identify the agent supplying the force in each case. All the examples and activities given so far 3. Explain why some of the leaves illustrate that there is a resistance offered by may get detached from a tree if an object to change its state of motion. If it is we vigorously shake its branch. at rest it tends to remain at rest; if it is moving 4. Why do you fall in the forward it tends to keep moving. This property of an direction when a moving bus object is called its inertia. Do all bodies have brakes to a stop and fall the same inertia? We know that it is easier to backwards when it accelerates push an empty box than a box full of books. from rest? Similarly, if we kick a football it flies away. But if we kick a stone of the same size with 9.4 Second Law of Motion equal force, it hardly moves. We may, in fact, get an injury in our foot while doing so! The first law of motion indicates that when Similarly, in activity 9.2, instead of a an unbalanced external force acts on an 118 SCIENCE

object, its velocity changes, that is, the object change the momentum of an object depends on the time rate at which the momentum is gets an acceleration. We would now like to changed. study how the acceleration of an object The second law of motion states that the rate of change of momentum of an object is depends on the force applied to it and how proportional to the applied unbalanced force in the direction of force. we measure a force. Let us recount some observations from our everyday life. During the game of table tennis if the ball hits a player it does not hurt him. On the other hand, when a fast moving cricket ball hits a spectator, it 9.4.1 MATHEMATICAL FORMULATION OF may hurt him. A truck at rest does not require SECOND LAW OF MOTION any attention when parked along a roadside. But a moving truck, even at speeds as low as Suppose an object of mass, m is moving along a straight line with an initial velocity, u. It is 5 m s–1, may kill a person standing in its path. uniformly accelerated to velocity, v in time, t A small mass, such as a bullet may kill a by the application of a constant force, F person when fired from a gun. These throughout the time, t. The initial and final momentum of the object will be, p1 = mu and observations suggest that the impact p2 = mv respectively. produced by the objects depends on their mass and velocity. Similarly, if an object is to be accelerated, we know that a greater force The change in momentum ∝ p2 – p1 ∝ mv – mu is required to give a greater velocity. In other ∝ m × (v – u). words, there appears to exist some quantity of importance that combines the object’s mass and its velocity. One such property The rate of change of momentum ∝ m × (v − u ) t called momentum was introduced by Newton. Or, the applied force, The momentum, p of an object is defined as the product of its mass, m and velocity, v. F ∝ m × (v − u ) t That is, p = mv (9.1) Momentum has both direction and F = km × (v − u ) (9.2) magnitude. Its direction is the same as that t (9.3) of velocity, v. The SI unit of momentum is kilogram-metre per second (kg m s-1). Since = kma the application of an unbalanced force brings a change in the velocity of the object, it is Here a [ = (v – u)/t ] is the acceleration, therefore clear that a force also produces a which is the rate of change of velocity. The change of momentum. quantity, k is a constant of proportionality. The SI units of mass and acceleration are kg Let us consider a situation in which a car and m s-2 respectively. The unit of force is so with a dead battery is to be pushed along a chosen that the value of the constant, k straight road to give it a speed of 1 m s-1, becomes one. For this, one unit of force is which is sufficient to start its engine. If one defined as the amount that produces an or two persons give a sudden push acceleration of 1 m s-2 in an object of 1 kg (unbalanced force) to it, it hardly starts. But mass. That is, a continuous push over some time results in a gradual acceleration of the car to this speed. 1 unit of force = k × (1 kg) × (1 m s-2). It means that the change of momentum of the car is not only determined by the Thus, the value of k becomes 1. From Eq. (9.3) magnitude of the force but also by the time during which the force is exerted. It may then F = ma (9.4) also be concluded that the force necessary to The unit of force is kg m s-2 or newton, which has the symbol N. The second law of FORCE AND LAWS OF MOTION 119

motion gives us a method to measure the The first law of motion can be force acting on an object as a product of its mathematically stated from the mathematical mass and acceleration. expression for the second law of motion. Eq. (9.4) is The second law of motion is often seen in action in our everyday life. Have you noticed F = ma that while catching a fast moving cricket ball, a fielder in the ground gradually pulls his m (v − u ) (9.5) hands backwards with the moving ball? In or F = t doing so, the fielder increases the time during or Ft = mv – mu which the high velocity of the moving ball decreases to zero. Thus, the acceleration of That is, when F = 0, v = u for whatever time, the ball is decreased and therefore the impact t is taken. This means that the object will of catching the fast moving ball (Fig. 9.8) is continue moving with uniform velocity, u also reduced. If the ball is stopped suddenly throughout the time, t. If u is zero then v will then its high velocity decreases to zero in a also be zero. That is, the object will remain very short interval of time. Thus, the rate of at rest. change of momentum of the ball will be large. Therefore, a large force would have to be Example 9.1 A constant force acts on an applied for holding the catch that may hurt object of mass 5 kg for a duration of the palm of the fielder. In a high jump athletic 2 s. It increases the object’s velocity event, the athletes are made to fall either on from 3 m s–1 to 7 m s-1. Find the a cushioned bed or on a sand bed. This is to magnitude of the applied force. Now, if increase the time of the athlete’s fall to stop the force was applied for a duration of after making the jump. This decreases the 5 s, what would be the final velocity of rate of change of momentum and hence the the object? force. Try to ponder how a karate player breaks a slab of ice with a single blow. Solution: Fig. 9.8: A fielder pulls his hands gradually with the We have been given that u = 3 m s–1 moving ball while holding a catch. and v = 7 m s-1, t = 2 s and m = 5 kg. From Eq. (9.5) we have, m (v − u ) F= t Substitution of values in this relation gives F = 5 kg (7 m s-1 – 3 m s-1)/2 s = 10 N. Now, if this force is applied for a duration of 5 s (t = 5 s), then the final velocity can be calculated by rewriting Eq. (9.5) as v = u + Ft m On substituting the values of u, F, m and t, we get the final velocity, v = 13 m s-1. 120 SCIENCE

Example 9.2 Which would require a Solution: greater force –– accelerating a 2 kg mass From Eq. (9.4) we have m1 = F/a1; and at 5 m s–2 or a 4 kg mass at 2 m s-2 ? m2 = F/a2. Here, a1 = 10 m s-2; a2 = 20 m s-2 and F = 5 N. Solution: Thus, m1 = 5 N/10 m s-2 = 0.50 kg; and m2 = 5 N/20 m s-2 = 0.25 kg. From Eq. (9.4), we have F = ma. If the two masses were tied together, Here we have m1 = 2 kg; a1 = 5 m s-2 the total mass, m would be and m2 = 4 kg; a2 = 2 m s-2. m = 0.50 kg + 0.25 kg = 0.75 kg. Thus, F1 = m1a1 = 2 kg × 5 m s-2 = 10 N; The acceleration, a produced in the and F2 = m2a2 = 4 kg × 2 m s-2 = 8 N. combined mass by the 5 N force would ⇒ F1 > F2. be, a = F/m = 5 N/0.75 kg = 6.67 m s-2. Thus, accelerating a 2 kg mass at 5 m s-2 would require a greater force. Example 9.5 The velocity-time graph of a ball of mass 20 g moving along a Example 9.3 A motorcar is moving with a straight line on a long table is given in velocity of 108 km/h and it takes 4 s to Fig. 9.9. stop after the brakes are applied. Calculate the force exerted by the Fig. 9.9 brakes on the motorcar if its mass along with the passengers is 1000 kg. How much force does the table exert on the ball to bring it to rest? Solution: Solution: The initial velocity of the motorcar The initial velocity of the ball is 20 cm s-1. u = 108 km/h Due to the frictional force exerted by the table, the velocity of the ball decreases = 108 × 1000 m/(60 × 60 s) down to zero in 10 s. Thus, u = 20 cm s–1; = 30 m s-1 v = 0 cm s-1 and t = 10 s. Since the and the final velocity of the motorcar velocity-time graph is a straight line, it is v = 0 m s-1. clear that the ball moves with a constant The total mass of the motorcar along acceleration. The acceleration a is with its passengers = 1000 kg and the v −u time taken to stop the motorcar, t = 4 s. a= From Eq. (9.5) we have the magnitude t of the force (F) applied by the brakes as = (0 cm s-1 – 20 cm s-1)/10 s m(v – u)/t. = –2 cm s-2 = –0.02 m s-2. On substituting the values, we get F = 1000 kg × (0 – 30) m s-1/4 s 121 = – 7500 kg m s-2 or – 7500 N. The negative sign tells us that the force exerted by the brakes is opposite to the direction of motion of the motorcar. Example 9.4 A force of 5 N gives a mass m1, an acceleration of 10 m s–2 and a mass m2, an acceleration of 20 m s-2. What acceleration would it give if both the masses were tied together? FORCE AND LAWS OF MOTION

The force exerted on the ball F is, Fig. 9.10: Action and reaction forces are equal and F = ma = (20/1000) kg × (– 0.02 m s-2) opposite. = – 0.0004 N. Suppose you are standing at rest and The negative sign implies that the intend to start walking on a road. You must frictional force exerted by the table is accelerate, and this requires a force in opposite to the direction of motion of accordance with the second law of motion. the ball. Which is this force? Is it the muscular effort you exert on the road? Is it in the direction 9.5 Third Law of Motion we intend to move? No, you push the road below backwards. The road exerts an equal The first two laws of motion tell us how an and opposite force on your feet to make you applied force changes the motion and provide move forward. us with a method of determining the force. The third law of motion states that when one It is important to note that even though object exerts a force on another object, the the action and reaction forces are always second object instantaneously exerts a force equal in magnitude, these forces may not back on the first. These two forces are always produce accelerations of equal magnitudes. equal in magnitude but opposite in direction. This is because each force acts on a different These forces act on different objects and never object that may have a different mass. on the same object. In the game of football sometimes we, while looking at the football When a gun is fired, it exerts a forward and trying to kick it with a greater force, force on the bullet. The bullet exerts an equal collide with a player of the opposite team. and opposite force on the gun. This results in Both feel hurt because each applies a force the recoil of the gun (Fig. 9.11). Since the gun to the other. In other words, there is a pair of has a much greater mass than the bullet, the forces and not just one force. The two acceleration of the gun is much less than the opposing forces are also known as action and acceleration of the bullet. The third law of reaction forces. motion can also be illustrated when a sailor jumps out of a rowing boat. As the sailor Let us consider two spring balances jumps forward, the force on the boat moves it connected together as shown in Fig. 9.10. The backwards (Fig. 9.12). fixed end of balance B is attached with a rigid support, like a wall. When a force is applied Fig. 9.11: A forward force on the bullet and recoil of through the free end of spring balance A, it is the gun. observed that both the spring balances show the same readings on their scales. It means that the force exerted by spring balance A on balance B is equal but opposite in direction to the force exerted by the balance B on balance A. Any of these two forces can be called as action and the other as reaction. This gives us an alternative statement of the third law of motion i.e., to every action there is an equal and opposite reaction. However, it must be remembered that the action and reaction always act on two different objects, simultaneously. 122 SCIENCE

Fig. 9.12: As the sailor jumps in forward direction, The cart shown in this activity can be the boat moves backwards. constructed by using a 12 mm or 18 mm thick plywood board of about 50 cm × 100 cm with Activity ______________ 9.4 two pairs of hard ball-bearing wheels (skate wheels are good to use). Skateboards are not • Request two children to stand on two as effective because it is difficult to maintain separate carts as shown in Fig. 9.13. straight-line motion. • Give them a bag full of sand or some 9.6 Conservation of Momentum other heavy object. Ask them to play a game of catch with the bag. Suppose two objects (two balls A and B, say) of masses mA and mB are travelling in the same • Does each of them experience an direction along a straight line at different instantaneous force as a result of velocities uA and uB, respectively [Fig. 9.14(a)]. throwing the sand bag? And there are no other external unbalanced forces acting on them. Let uA > uB and the • You can paint a white line on two balls collide with each other as shown in cartwheels to observe the motion of the Fig. 9.14(b). During collision which lasts for two carts when the children throw the a time t, the ball A exerts a force FAB on ball B bag towards each other. and the ball B exerts a force FBA on ball A. Suppose vA and vB are the velocities of the two balls A and B after the collision, respectively [Fig. 9.14(c)]. Fig. 9.14: Conservation of momentum in collision of two balls. From Eq. (9.1), the momenta (plural of momentum) of ball A before and after the collision are mAuA and mAvA, respectively. The rate of change of its momentum (or FAB) during the collision will be mA (v A −uA) . t Fig. 9.13 Similarly, the rate of change of momentum of ball B (= FBA) during the collision will be Now, place two children on one cart and one on another cart. The second law of motion mB (v B − u B ) . can be seen, as this arrangement would show t different accelerations for the same force. According to the third law of motion, the force FAB exerted by ball A on ball B FORCE AND LAWS OF MOTION 123

and the force FBA exerted by the ball B on ball Activity ______________ 9.6 A must be equal and opposite to each other. • Take a test tube of good quality glass Therefore, material and put a small amount of water in it. Place a stop cork at the FAB = – FBA (9.6) mouth of it. or mA (v A −uA) = – mB (v B − u B ) . • Now suspend the test tube horizontally t t by two strings or wires as shown in Fig. 9.16. This gives, • Heat the test tube with a burner until mAuA + mBuB = mAvA + mBvB (9.7) water vaporises and the cork blows out. Since (mAuA + mBuB) is the total momentum • Observe that the test tube recoils in of the two balls A and B before the collision the direction opposite to the direction and (mAvA + mBvB) is their total momentum of the cork. after the collision, from Eq. (9.7) we observe that the total momentum of the two balls Fig. 9.16 remains unchanged or conserved provided no other external force acts. • Also, observe the difference in the velocity the cork appears to have and As a result of this ideal collision that of the recoiling test tube. experiment, we say that the sum of momenta of the two objects before collision is equal to Example 9.6 A bullet of mass 20 g is the sum of momenta after the collision horizontally fired with a velocity provided there is no external unbalanced 150 m s-1 from a pistol of mass 2 kg. force acting on them. This is known as the What is the recoil velocity of the pistol? law of conservation of momentum. This statement can alternatively be given as the Solution: total momentum of the two objects is We have the mass of bullet, unchanged or conserved by the collision. m1 = 20 g (= 0.02 kg) and the mass of the pistol, m2 = 2 kg; initial velocities of Activity ______________ 9.5 the bullet (u1) and pistol (u2) = 0, respectively. The final velocity of the • Take a big rubber balloon and inflate bullet, v1 = + 150 m s-1. The direction it fully. Tie its neck using a thread. Also of bullet is taken from left to right using adhesive tape, fix a straw on the (positive, by convention, Fig. 9.17). Let surface of this balloon. v be the recoil velocity of the pistol. • Pass a thread through the straw and SCIENCE hold one end of the thread in your hand or fix it on the wall. • Ask your friend to hold the other end of the thread or fix it on a wall at some distance. This arrangement is shown in Fig. 9.15. • Now remove the thread tied on the neck of balloon. Let the air escape from the mouth of the balloon. • Observe the direction in which the straw moves. Fig. 9.15 124

Total momenta of the pistol and bullet Example 9.7 A girl of mass 40 kg jumps before the fire, when the gun is at rest with a horizontal velocity of 5 m s-1 onto a stationary cart with frictionless = (2 + 0.02) kg × 0 m s–1 wheels. The mass of the cart is 3 kg. = 0 kg m s–1 What is her velocity as the cart starts Total momenta of the pistol and bullet moving? Assume that there is no after it is fired external unbalanced force working in = 0.02 kg × (+ 150 m s–1) the horizontal direction. + 2 kg × v m s–1 Solution: = (3 + 2v) kg m s–1 According to the law of conservation of Let v be the velocity of the girl on the momentum cart as the cart starts moving. Total momenta after the fire = Total The total momenta of the girl and cart momenta before the fire before the interaction 3 + 2v = 0 ⇒ v = − 1.5 m s–1. = 40 kg × 5 m s–1 + 3 kg × 0 m s–1 Negative sign indicates that the = 200 kg m s–1. direction in which the pistol would recoil is opposite to that of bullet, that Total momenta after the interaction is, right to left. = (40 + 3) kg × v m s–1 Fig. 9.17: Recoil of a pistol = 43 v kg m s–1. According to the law of conservation of momentum, the total momentum is conserved during the interaction. That is, 43 v = 200 ⇒ v = 200/43 = + 4.65 m s–1. The girl on cart would move with a velocity of 4.65 m s–1 in the direction in which the girl jumped (Fig. 9.18). (a) (b) FORCE AND LAWS OF MOTION Fig. 9.18: The girl jumps onto the cart. 125

Example 9.8 Two hockey players of If v is the velocity of the two entangled opposite teams, while trying to hit a players after the collision, the total hockey ball on the ground collide and momentum then immediately become entangled. One has a mass of 60 kg and was moving = (m1 + m2) × v with a velocity 5.0 m s–1 while the other = (60 + 55) kg × v m s–1 has a mass of 55 kg and was moving = 115 × v kg m s–1. faster with a velocity 6.0 m s–1 towards Equating the momenta of the system the first player. In which direction and before and after collision, in accordance with what velocity will they move after with the law of conservation of they become entangled? Assume that momentum, we get the frictional force acting between the v = – 30/115 feet of the two players and ground is negligible. = – 0.26 m s–1. Thus, the two entangled players would Solution: move with velocity 0.26 m s–1 from right to left, that is, in the direction the second player was moving before the collision. Fig. 9.19: A collision of two hockey players: (a) before collision and (b) after collision. Let the first player be moving from left Q uestions to right. By convention left to right is 1. If action is always equal to the taken as the positive direction and thus reaction, explain how a horse can right to left is the negative direction (Fig. pull a cart. 9.19). If symbols m and u represent the 2. Explain, why is it difficult for a mass and initial velocity of the two fireman to hold a hose, which players, respectively. Subscripts 1 and ejects large amounts of water at 2 in these physical quantities refer to a high velocity. the two hockey players. Thus, 3. From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an m1 = 60 kg; u1 = + 5 m s-1; and initial velocity of 35 m s–1. m2 = 55 kg; u2 = – 6 m s-1. Calculate the initial recoil velocity The total momentum of the two players of the rifle. before the collision = 60 kg × (+ 5 m s-1) + 55 kg × (– 6 m s-1) = – 30 kg m s-1 126 SCIENCE

4. Two objects of masses 100 g and They collide and after the 200 g are moving along the same collision, the first object moves at line and direction with velocities a velocity of 1.67 m s–1. Determine of 2 m s–1 and 1 m s–1, respectively. the velocity of the second object. CONSERVATION LAWS All conservation laws such as conservation of momentum, energy, angular momentum, charge etc. are considered to be fundamental laws in physics. These are based on observations and experiments. It is important to remember that a conservation law cannot be proved. It can be verified, or disproved, by experiments. An experiment whose result is in conformity with the law verifies or substantiates the law; it does not prove the law. On the other hand, a single experiment whose result goes against the law is enough to disprove it. The law of conservation of momentum has been deduced from large number of observations and experiments. This law was formulated nearly three centuries ago. It is interesting to note that not a single situation has been realised so far, which contradicts this law. Several experiences of every-day life can be explained on the basis of the law of conservation of momentum. What you have learnt • First law of motion: An object continues to be in a state of rest or of uniform motion along a straight line unless acted upon • by an unbalanced force. • • The natural tendency of objects to resist a change in their state • of rest or of uniform motion is called inertia. • The mass of an object is a measure of its inertia. Its SI unit is kilogram (kg). • Force of friction always opposes motion of objects. • • Second law of motion: The rate of change of momentum of an object is proportional to the applied unbalanced force in the FORCE AND LAWS OF MOTION direction of the force. The SI unit of force is kg m s–2. This is also known as newton and represented by the symbol N. A force of one newton produces an acceleration of 1 m s–2 on an object of mass 1 kg. The momentum of an object is the product of its mass and velocity and has the same direction as that of the velocity. Its SI unit is kg m s–1. Third law of motion: To every action, there is an equal and opposite reaction and they act on two different bodies. In an isolated system (where there is no external force), the total momentum remains conserved. 127

Exercises 1. An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason. 2. When a carpet is beaten with a stick, dust comes out of it. Explain. 3. Why is it advised to tie any luggage kept on the roof of a bus with a rope? 4. A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because (a) the batsman did not hit the ball hard enough. (b) velocity is proportional to the force exerted on the ball. (c) there is a force on the ball opposing the motion. (d) there is no unbalanced force on the ball, so the ball would want to come to rest. 5. A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 tonnes (Hint: 1 tonne = 1000 kg.) 6. A stone of 1 kg is thrown with a velocity of 20 m s–1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice? 7. A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate: (a) the net accelerating force and (b) the acceleration of the train. 8. An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 m s–2? 9. What is the momentum of an object of mass m, moving with a velocity v? (a) (mv)2 (b) mv2 (c) ½ mv2 (d) mv 10. Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet? 11. Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 m s-1 before the collision during which they 128 SCIENCE

stick together. What will be the velocity of the combined object after collision? 12. According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move. 13. A hockey ball of mass 200 g travelling at 10 m s–1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 m s–1. Calculate the magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick. 14. A bullet of mass 10 g travelling horizontally with a velocity of 150 m s–1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet. 15. An object of mass 1 kg travelling in a straight line with a velocity of 10 m s–1 collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object. 16. An object of mass 100 kg is accelerated uniformly from a velocity of 5 m s–1 to 8 m s–1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object. 17. Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions. 18. How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 m s–2. FORCE AND LAWS OF MOTION 129

Additional Exercises A1. The following is the distance-time table of an object in motion: Time in seconds Distance in metres 00 11 28 3 27 4 64 5 125 6 216 7 343 (a) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero? (b) What do you infer about the forces acting on the object? A2. Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 m s-2. With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort.) A3. A hammer of mass 500 g, moving at 50 m s-1, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer? A4. A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required. 130 SCIENCE

C 10hapter GRAVITATION In Chapters 8 and 9, we have learnt about Let us try to understand the motion of the motion of objects and force as the cause the moon by recalling activity 8.11. of motion. We have learnt that a force is needed to change the speed or the direction Activity _____________ 10.1 of motion of an object. We always observe that an object dropped from a height falls towards • Take a piece of thread. the earth. We know that all the planets go • Tie a small stone at one end. Hold the around the Sun. The moon goes around the earth. In all these cases, there must be some other end of the thread and whirl it force acting on the objects, the planets and round, as shown in Fig. 10.1. on the moon. Isaac Newton could grasp that • Note the motion of the stone. the same force is responsible for all these. • Release the thread. This force is called the gravitational force. • Again, note the direction of motion of the stone. In this chapter we shall learn about gravitation and the universal law of gravitation. We shall discuss the motion of objects under the influence of gravitational force on the earth. We shall study how the weight of a body varies from place to place. We shall also discuss the conditions for objects to float in liquids. 10.1 Gravitation Fig. 10.1: A stone describing a circular path with a velocity of constant magnitude. We know that the moon goes around the earth. An object when thrown upwards, Before the thread is released, the stone reaches a certain height and then falls moves in a circular path with a certain speed downwards. It is said that when Newton was and changes direction at every point. The sitting under a tree, an apple fell on him. The change in direction involves change in velocity fall of the apple made Newton start thinking. or acceleration. The force that causes this He thought that: if the earth can attract an acceleration and keeps the body moving along apple, can it not attract the moon? Is the force the circular path is acting towards the centre. the same in both cases? He conjectured that This force is called the centripetal (meaning the same type of force is responsible in both ‘centre-seeking’) force. In the absence of this the cases. He argued that at each point of its orbit, the moon falls towards the earth, instead of going off in a straight line. So, it must be attracted by the earth. But we do not really see the moon falling towards the earth. 2018-19

force, the stone flies off along a straight line. 10.1.1 UNIVERSAL LAW OF GRAVITATION This straight line will be a tangent to the circular path. Every object in the universe attracts every other object with a force which is proportional Tangent to a circle to the product of their masses and inversely proportional to the square of the distance between them. The force is along the line joining the centres of two objects. More to know A straight line that meets the circle Mm at one and only one point is called a F=G tangent to the circle. Straight line ABC is a tangent to the circle at d2 point B. Fig. 10.2: The gravitational force between two The motion of the moon around the earth uniform objects is directed along the line is due to the centripetal force. The centripetal force is provided by the force of attraction of joining their centres. the earth. If there were no such force, the moon would pursue a uniform straight line Let two objects A and B of masses M and motion. m lie at a distance d from each other as shown It is seen that a falling apple is attracted towards the earth. Does the apple attract the in Fig. 10.2. Let the force of attraction between earth? If so, we do not see the earth moving towards an apple. Why? two objects be F. According to the universal According to the third law of motion, the law of gravitation, the force between two apple does attract the earth. But according to the second law of motion, for a given force, objects is directly proportional to the product acceleration is inversely proportional to the mass of an object [Eq. (9.4)]. The mass of an of their masses. That is, apple is negligibly small compared to that of the earth. So, we do not see the earth moving F∝M × m (10.1) towards the apple. Extend the same argument for why the earth does not move towards the And the force between two objects is inversely moon. proportional to the square of the distance In our solar system, all the planets go around the Sun. By arguing the same way, between them, that is, we can say that there exists a force between the Sun and the planets. From the above facts 1 (10.2) Newton concluded that not only does the F∝ earth attract an apple and the moon, but all objects in the universe attract each other. This d2 force of attraction between objects is called the gravitational force. Combining Eqs. (10.1) and (10.2), we get M ×m (10.3) F ∝ d2 M×m (10.4) or, F = G d 2 where G is the constant of proportionality and is called the universal gravitation constant. By multiplying crosswise, Eq. (10.4) gives F×d2=GM×m 132 SCIENCE

Isaac Newton was born How did Newton guess the inverse-square rule? in Woolsthorpe near Grantham, England. He is generally There has always been a great interest regarded as the most in the motion of planets. By the 16th original and century, a lot of data on the motion of influential theorist in planets had been collected by many the history of science. astronomers. Based on these data He was born in a poor Johannes Kepler derived three laws, farming family. But he which govern the motion of planets. was not good at These are called Kepler’s laws. These are: farming. He was sent 1. The orbit of a planet is an ellipse with Isaac Newton to study at Cambridge the Sun at one of the foci, as shown in (1642 – 1727) University in 1661. In 1665 a plague broke the figure given below. In this figure O is the position of the Sun. out in Cambridge and so Newton took a year 2. The line joining the planet and the Sun off. It was during this year that the incident of sweep equal areas in equal intervals the apple falling on him is said to have of time. Thus, if the time of travel from occurred. This incident prompted Newton to A to B is the same as that from C to D, explore the possibility of connecting gravity then the areas OAB and OCD are with the force that kept the moon in its orbit. equal. This led him to the universal law of 3. The cube of the mean distance of a gravitation. It is remarkable that many great planet from the Sun is proportional to scientists before him knew of gravity but failed the square of its orbital period T. Or, to realise it. r3/T2 = constant. Newton formulated the well-known laws of It is important to note that Kepler motion. He worked on theories of light and could not give a theory to explain colour. Hedesignedanastronomicaltelescope the motion of planets. It was Newton to carry out astronomical observations. who showed that the cause of the Newton was also a great mathematician. He planetary motion is the gravitational invented a new branch of mathematics, called force that the Sun exerts on them. Newton calculus. He used it to prove that for objects used the third law C of Kepler to D outside a sphere of uniform density, the sphere behaves as if the whole of its mass is calculate the concentrated at its centre. Newton gravitational force O B of attraction. The A transformed the structure of physical gravitational force science with his three laws of motion and the universal law of gravitation. As the keystone of the earth is of the scientific revolution of the seventeenth weakened by distance. A simple argument century, Newton’s work combined the goes like this. We can assume that the contributions of Copernicus, Kepler, Galileo, planetary orbits are circular. Suppose the and others into a new powerful synthesis. orbital velocity is v and the radius of the It is remarkable that though the orbit is r. Then the force acting on an gravitational theory could not be verified at orbiting planet is given by F ∝ v2/r. that time, there was hardly any doubt about If T denotes the period, then v = 2πr/T, its correctness. This is because Newton based so that v2 ∝ r2/T2. his theory on sound scientific reasoning and We can rewrite this as v2 ∝ (1/r) × backed it with mathematics. This made the ( r3/T2). Since r3/T2 is constant by Kepler’s theory simple and elegant. These qualities are third law, we have v2 ∝ 1/r. Combining now recognised as essential requirements of a this with F ∝ v2/ r, we get, F ∝ 1/ r2. good scientific theory. GRAVITATION 133

F d2 (10.5) From Eq. (10.4), the force exerted by the or G = earth on the moon is M×m M ×m F =G d2 The SI unit of G can be obtained by substituting the units of force, distance and 6.7 ×10−11 N m2 kg-2 × 6 ×1024 kg × 7.4 ×1022 kg mass in Eq. (10.5) as N m2 kg–2. = The value of G was found out by (3.84 ×108 m)2 Henry Cavendish (1731 – 1810) by using a sensitive balance. The accepted value of G is = 2.02 × 1020 N. 6.673 × 10–11 N m2 kg–2. Thus, the force exerted by the earth on We know that there exists a force of the moon is 2.02 × 1020 N. attraction between any two objects. Compute the value of this force between you and your Q uestions friend sitting closeby. Conclude how you do 1. State the universal law of not experience this force! gravitation. 2. Write the formula to find the More to know The law is universal in the sense that magnitude of the gravitational it is applicable to all bodies, whether force between the earth and an the bodies are big or small, whether object on the surface of the earth. they are celestial or terrestrial. 10.1.2 IMPORTANCE OF THE UNIVERSAL Inverse-square LAW OF GRAVITATION Saying that F is inversely proportional to the square of d The universal law of gravitation successfully means, for example, that if d gets explained several phenomena which were bigger by a factor of 6, F becomes believed to be unconnected: 1 (i) the force that binds us to the earth; 36 times smaller. (ii) the motion of the moon around the Example 10.1 The mass of the earth is earth; 6 × 1024 kg and that of the moon is (iii) the motion of planets around the Sun; 7.4 × 1022 kg. If the distance between the earth and the moon is 3.84×105 km, and calculate the force exerted by the earth on (iv) the tides due to the moon and the Sun. the moon. (Take G = 6.7 × 10–11 N m2 kg-2) 10.2 Free Fall Solution: Let us try to understand the meaning of free The mass of the earth, M = 6 × 1024 kg fall by performing this activity. The mass of the moon, Activity _____________ 10.2 m = 7.4 × 1022 kg The distance between the earth and the • Take a stone. moon, • Throw it upwards. • It reaches a certain height and then it d = 3.84 × 105 km = 3.84 × 105 × 1000 m starts falling down. = 3.84 × 108 m We have learnt that the earth attracts G = 6.7 × 10–11 N m2 kg–2 objects towards it. This is due to the gravitational force. Whenever objects fall towards the earth under this force alone, we say that the objects are in free fall. Is there any 134 SCIENCE

change in the velocity of falling objects? While calculations, we can take g to be more or less constant on or near the earth. But for objects falling, there is no change in the direction of far from the earth, the acceleration due to gravitational force of earth is given by motion of the objects. But due to the earth’s Eq. (10.7). attraction, there will be a change in the 10.2.1 TO CALCULATE THE VALUE OF g magnitude of the velocity. Any change in To calculate the value of g, we should put the values of G, M and R in Eq. (10.9), velocity involves acceleration. Whenever an namely, universal gravitational constant, G = 6.7 × 10–11 N m2 kg-2, mass of the earth, object falls towards the earth, an acceleration M = 6 × 1024 kg, and radius of the earth, R = 6.4 × 106 m. is involved. This acceleration is due to the M earth’s gravitational force. Therefore, this g=G acceleration is called the acceleration due to R2 the gravitational force of the earth (or 6.7 ×10-11 N m2 kg-2 × 6 ×1024 kg = (6.4 ×106 m)2 acceleration due to gravity). It is denoted by = 9.8 m s–2. g. The unit of g is the same as that of Thus, the value of acceleration due to gravity acceleration, that is, m s–2. of the earth, g = 9.8 m s–2. We know from the second law of motion 10.2.2 MOTION OF OBJECTS UNDER THE that force is the product of mass and INFLUENCE OF GRAVITATIONAL acceleration. Let the mass of the stone in FORCE OF THE EARTH activity 10.2 be m. We already know that there Let us do an activity to understand whether all objects hollow or solid, big or small, will is acceleration involved in falling objects due fall from a height at the same rate. to the gravitational force and is denoted by g. Activity _____________ 10.3 Therefore the magnitude of the gravitational • Take a sheet of paper and a stone. Drop them simultaneously from the first floor force F will be equal to the product of mass of a building. Observe whether both of them reach the ground simultaneously. and acceleration due to the gravitational • We see that paper reaches the ground force, that is, little later than the stone. This happens because of air resistance. The air offers F=mg (10.6) resistance due to friction to the motion of the falling objects. The resistance From Eqs. (10.4) and (10.6) we have offered by air to the paper is more than the resistance offered to the stone. If M ×m we do the experiment in a glass jar from mg =G which air has been sucked out, the paper and the stone would fall at the d2 same rate. M g=G or d2 (10.7) where M is the mass of the earth, and d is the distance between the object and the earth. Let an object be on or near the surface of the earth. The distance d in Eq. (10.7) will be equal to R, the radius of the earth. Thus, for objects on or near the surface of the earth, M ×m (10.8) mg = G R2 M (10.9) g=G R2 The earth is not a perfect sphere. As the radius of the earth increases from the poles to the equator, the value of g becomes greater at the poles than at the equator. For most GRAVITATION 135

We know that an object experiences u +v (ii) average speed = 2 acceleration during free fall. From Eq. (10.9), = (0 m s–1+ 5 m s–1)/2 this acceleration experienced by an object is = 2.5 m s–1 (iii) distance travelled, s = ½ a t2 independent of its mass. This means that all = ½ × 10 m s–2 × (0.5 s)2 = ½ × 10 m s–2 × 0.25 s2 objects hollow or solid, big or small, should = 1.25 m Thus, fall at the same rate. According to a story, (i) its speed on striking the ground Galileo dropped different objects from the top = 5 m s–1 (ii) its average speed during the 0.5 s of the Leaning Tower of Pisa in Italy to prove = 2.5 m s–1 the same. (iii) height of the ledge from the ground As g is constant near the earth, all the = 1.25 m. equations for the uniformly accelerated Example 10.3 An object is thrown vertically upwards and rises to a height motion of objects become valid with of 10 m. Calculate (i) the velocity with which the object was thrown upwards acceleration a replaced by g (see section 8.5). and (ii) the time taken by the object to reach the highest point. The equations are: Solution: v = u + at (10.10) Distance travelled, s = 10 m 1 (10.11) Final velocity, v = 0 m s–1 s = ut + 2 at2 (10.12) Acceleration due to gravity, g = 9.8 m s–2 v2 = u2 + 2as Acceleration of the object, a = –9.8 m s–2 where u and v are the initial and final (upward motion) velocities and s is the distance covered in (i) v 2 = u2 + 2a s time, t. 0 = u 2 + 2 × (–9.8 m s–2) × 10 m In applying these equations, we will take –u 2 = –2 × 9.8 × 10 m2 s–2 acceleration, a to be positive when it is in the direction of the velocity, that is, in the u = 196 m s-1 direction of motion. The acceleration, a will u = 14 m s-1 be taken as negative when it opposes the (ii) v = u + a t motion. 0 = 14 m s–1 – 9.8 m s–2 × t t = 1.43 s. Example 10.2 A car falls off a ledge and Thus, drops to the ground in 0.5 s. Let (i) Initial velocity, u = 14 m s–1, and g = 10 m s–2 (for simplifying the (ii) Time taken, t = 1.43 s. calculations). (i) What is its speed on striking the Q uestions ground? 1. What do you mean by free fall? (ii) What is its average speed during the 2. What do you mean by acceleration 0.5 s? due to gravity? (iii) How high is the ledge from the ground? Solution: Time, t = ½ second Initial velocity, u = 0 m s–1 Acceleration due to gravity, g = 10 m s–2 Acceleration of the car, a = + 10 m s–2 (downward) (i) speed v = a t v = 10 m s–2 × 0.5 s = 5 m s–1 136 SCIENCE

10.3 Mass attracts the object. In the same way, the weight We have learnt in the previous chapter that the of an object on the moon is the force with mass of an object is the measure of its inertia (section 9.3). We have also learnt that greater which the moon attracts that object. The mass the mass, the greater is the inertia. It remains the same whether the object is on the earth, of the moon is less than that of the earth. Due the moon or even in outer space. Thus, the mass of an object is constant and does not to this the moon exerts lesser force of attraction change from place to place. on objects. Let the mass of an object be m. Let its weight on the moon be Wm. Let the mass of the moon be Mm and its radius be Rm. By applying the universal law of gravitation, the weight of the object on the 10.4 Weight moon will be We know that the earth attracts every object Wm = G Mm ×m (10.16) Rm2 with a certain force and this force depends on Let the weight of the same object on the the mass (m) of the object and the acceleration earth be We. The mass of the earth is M and its radius is R. due to the gravity (g). The weight of an object is the force with which it is attracted towards the earth. Table 10.1 We know that Celestial Mass (kg) Radius (m) body F = m × a, (10.13) 6.37 × 106 1.74 × 106 that is, F = m × g. (10.14) Earth 5.98 × 1024 Moon 7.36 × 1022 The force of attraction of the earth on an object is known as the weight of the object. It is denoted by W. Substituting the same in Eq. (10.14), we have From Eqs. (10.9) and (10.15) we have, W=m×g (10.15) As the weight of an object is the force with M ×m (10.17) We = G R2 which it is attracted towards the earth, the SI unit of weight is the same as that of force, that Substituting the values from Table 10.1 in Eqs. (10.16) and (10.17), we get is, newton (N). The weight is a force acting vertically downwards; it has both magnitude and direction. =G 7.36 × 1022 kg × m 1.74 × 106 m 2 We have learnt that the value of g is ( )Wm constant at a given place. Therefore at a given place, the weight of an object is directly Wm = 2.431 × 1010 G × m (10.18a) and We =1.474 × 1011G × m (10.18b) proportional to the mass, say m, of the object, that is, W ∝ m. It is due to this reason that at a given place, we can use the weight of an Dividing Eq. (10.18a) by Eq. (10.18b), we get object as a measure of its mass. The mass of Wm 2.431 × 1010 We 1.474 × 1011 an object remains the same everywhere, that = is, on the earth and on any planet whereas its weight depends on its location because g Wm = 0.165 ≈ 1 We 6 depends on location. or (10.19) 10.4.1 WEIGHT OF AN OBJECT ON Weight of the object on the moon = 1 Weight of the object on the earth 6 THE MOON We have learnt that the weight of an object on Weight of the object on the moon the earth is the force with which the earth = (1/6) × its weight on the earth. GRAVITATION 137

Example 10.4 Mass of an object is 10 kg. of the net force in a particular direction (thrust) What is its weight on the earth? and the force per unit area (pressure) acting on the object concerned. Solution: Let us try to understand the meanings of Mass, m = 10 kg thrust and pressure by considering the Acceleration due to gravity, g = 9.8 m s–2 following situations: Situation 1: You wish to fix a poster on a W=m×g bulletin board, as shown in Fig 10.3. To do W = 10 kg × 9.8 m s-2 = 98 N this task you will have to press drawing pins Thus, the weight of the object is 98 N. with your thumb. You apply a force on the surface area of the head of the pin. This force Example 10.5 An object weighs 10 N when is directed perpendicular to the surface area measured on the surface of the earth. of the board. This force acts on a smaller area What would be its weight when at the tip of the pin. measured on the surface of the moon? Solution: We know, Weight of object on the moon = (1/6) × its weight on the earth. That is, Wm = We = 10 N. 6 6 = 1.67 N. Thus, the weight of object on the surface of the moon would be 1.67 N. Q uestions 1. What are the differences between the mass of an object and its weight? 2. Why is the weight of an object on 1 the moon 6 th its weight on the earth? 10.5 Thrust and Pressure Fig. 10.3: To fix a poster, drawing pins are pressed with the thumb perpendicular to the board. Have you ever wondered why a camel can run in a desert easily? Why an army tank weighing Situation 2: You stand on loose sand. Your more than a thousand tonne rests upon a feet go deep into the sand. Now, lie down on continuous chain? Why a truck or a motorbus the sand. You will find that your body will not has much wider tyres? Why cutting tools have go that deep in the sand. In both cases the sharp edges? In order to address these force exerted on the sand is the weight of your questions and understand the phenomena body. involved, it helps to introduce the concepts 138 SCIENCE

You have learnt that weight is the force by the wooden block on the table top if acting vertically downwards. Here the force it is made to lie on the table top with its is acting perpendicular to the surface of the sides of dimensions (a) 20 cm × 10 cm sand. The force acting on an object and (b) 40 cm × 20 cm. perpendicular to the surface is called thrust. Solution: When you stand on loose sand, the force, that is, the weight of your body is acting on The mass of the wooden block = 5 kg an area equal to area of your feet. When you The dimensions lie down, the same force acts on an area equal to the contact area of your whole body, which = 40 cm × 20 cm × 10 cm is larger than the area of your feet. Thus, the Here, the weight of the wooden block effects of forces of the same magnitude on applies a thrust on the table top. different areas are different. In the above That is, cases, thrust is the same. But effects are different. Therefore the effect of thrust Thrust = F = m × g depends on the area on which it acts. = 5 kg × 9.8 m s–2 = 49 N The effect of thrust on sand is larger while standing than while lying. The thrust on unit Area of a side = length × breadth area is called pressure. Thus, = 20 cm × 10 cm = 200 cm2 = 0.02 m2 Pressure = thrust (10.20) area From Eq. (10.20), Substituting the SI unit of thrust and area in 49 N Pressure = 0.02 m2 Eq. (10.20), we get the SI unit of pressure as = 2450 N m-2. N/m2 or N m–2. When the block lies on its side of dimensions 40 cm × 20 cm, it exerts In honour of scientist Blaise Pascal, the the same thrust. Area= length × breadth SI unit of pressure is called pascal, denoted = 40 cm × 20 cm as Pa. = 800 cm2 = 0.08 m2 From Eq. (10.20), Let us consider a numerical example to 49 N understand the effects of thrust acting on Pressure = 0.08 m2 different areas. = 612.5 N m–2 The pressure exerted by the side 20 cm Example 10.6 A block of wood is kept on a × 10 cm is 2450 N m–2 and by the side tabletop. The mass of wooden block is 5 40 cm × 20 cm is 612.5 N m–2. kg and its dimensions are 40 cm × 20 cm × 10 cm. Find the pressure exerted Fig. 10.4 Thus, the same force acting on a smaller area exerts a larger pressure, and a smaller pressure on a larger area. This is the reason why a nail has a pointed tip, knives have sharp edges and buildings have wide foundations. 10.5.1 PRESSURE IN FLUIDS All liquids and gases are fluids. A solid exerts pressure on a surface due to its weight. Similarly, fluids have weight, and they also GRAVITATION 139

exert pressure on the base and walls of the water on the bottle is greater than its weight. container in which they are enclosed. Pressure Therefore it rises up when released. exerted in any confined mass of fluid is transmitted undiminished in all directions. To keep the bottle completely immersed, the upward force on the bottle due to water 10.5.2 BUOYANCY must be balanced. This can be achieved by an externally applied force acting downwards. Have you ever had a swim in a pool and felt This force must at least be equal to the lighter? Have you ever drawn water from a difference between the upward force and the well and felt that the bucket of water is heavier weight of the bottle. when it is out of the water? Have you ever wondered why a ship made of iron and steel The upward force exerted by the water on does not sink in sea water, but while the same the bottle is known as upthrust or buoyant amount of iron and steel in the form of a sheet force. In fact, all objects experience a force of would sink? These questions can be answered buoyancy when they are immersed in a fluid. by taking buoyancy in consideration. Let us The magnitude of this buoyant force depends understand the meaning of buoyancy by on the density of the fluid. doing an activity. 10.5.3 WHY OBJECTS FLOAT OR SINK WHEN Activity _____________ 10.4 PLACED ON THE SURFACE OF WATER? • Take an empty plastic bottle. Close the Let us do the following activities to arrive at mouth of the bottle with an airtight an answer for the above question. stopper. Put it in a bucket filled with water. You see that the bottle floats. Activity _____________ 10.5 • Push the bottle into the water. You feel • Take a beaker filled with water. an upward push. Try to push it further • Take an iron nail and place it on the down. You will find it difficult to push deeper and deeper. This indicates that surface of the water. water exerts a force on the bottle in the • Observe what happens. upward direction. The upward force exerted by the water goes on increasing The nail sinks. The force due to the as the bottle is pushed deeper till it is gravitational attraction of the earth on the completely immersed. iron nail pulls it downwards. There is an upthrust of water on the nail, which pushes • Now, release the bottle. It bounces it upwards. But the downward force acting back to the surface. on the nail is greater than the upthrust of water on the nail. So it sinks (Fig. 10.5). • Does the force due to the gravitational attraction of the earth act on this Fig. 10.5: An iron nail sinks and a cork floats when bottle? If so, why doesn’t the bottle stay placed on the surface of water. immersed in water after it is released? How can you immerse the bottle in water? The force due to the gravitational attraction of the earth acts on the bottle in the downward direction. So the bottle is pulled downwards. But the water exerts an upward force on the bottle. Thus, the bottle is pushed upwards. We have learnt that weight of an object is the force due to gravitational attraction of the earth. When the bottle is immersed, the upward force exerted by the 140 SCIENCE