PHYSICS BOOK PART-1

MOTION IN A PLANE 87 strokes normal to the river current? How far down the river does he go when he reaches the other bank ? 4.14 In a harbour, wind is blowing at the speed of 72 km/h and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction. If the boat starts moving at a speed of 51 km/h to the north, what is the direction of the flag on the mast of the boat ? 4.15 The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m s-1 can go without hitting the ceiling of the hall ? 4.16 A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball ? 4.17 A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone ? 4.18 An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity. 4.19 Read each statement below carefully and state, with reasons, if it is true or false : (a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre (b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point (c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector 4.20 The position of a particle is given by r = 3.0t ˆi − 2.0t 2 ˆj + 4.0 kˆ m where t is in seconds and the coefficients have the proper units for r to be in metres. (a) Find the v and a of the particle? (b) What is the magnitude and direction of velocity of the particle at t = 2.0 s ? 4.21 A particle starts from the origin at t = 0 s with a velocity of 10.0 j m/s and moves in ( )the x-y plane with a constant acceleration of 8.0i + 2.0 j m s-2. (a) At what time is the x- coordinate of the particle 16 m? What is the y-coordinate of the particle at that time? (b) What is the speed of the particle at the time ? 4.22 i and j are unit vectors along x- and y- axis respectively. What is the magnitude and direction of the vectors i + j, and i − j ? What are the components of a vector A= 2 i + 3j along the directions of i + j and i − j? [You may use graphical method] 4.23 For any arbitrary motion in space, which of the following relations are true : (a) vaverage = (1/2) (v (t1) + v (t2)) (b) v average = [r(t2) - r(t1) ] /(t2 – t1) (c) v (t) = v (0) + a t (d) r (t) = r (0) + v (0) t + (1/2) a t2 (e) a average =[ v (t2) - v (t1 )] /( t2 – t1) (The ‘average’ stands for average of the quantity over the time interval t to t2) 1 4.24 Read each statement below carefully and state, with reasons and examples, if it is true or false : A scalar quantity is one that (a) is conserved in a process (b) can never take negative values (c) must be dimensionless (d) does not vary from one point to another in space (e) has the same value for observers with different orientations of axes. 4.25 An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is 30°, what is the speed of the aircraft ? 2020-21

88 PHYSICS Additional Exercises 4.26 A vector has magnitude and direction. Does it have a location in space ? Can it vary with time ? Will two equal vectors a and b at different locations in space necessarily have identical physical effects ? Give examples in support of your answer. 4.27 A vector has both magnitude and direction. Does it mean that anything that has magnitude and direction is necessarily a vector ? The rotation of a body can be specified by the direction of the axis of rotation, and the angle of rotation about the axis. Does that make any rotation a vector ? 4.28 Can you associate vectors with (a) the length of a wire bent into a loop, (b) a plane area, (c) a sphere ? Explain. 4.29 A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away ? Assume the muzzle speed to be fixed, and neglect air resistance. 4.30 A fighter plane flying horizontally at an altitude of 1.5 km with speed 720 km/h passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed 600 m s-1 to hit the plane ? At what minimum altitude should the pilot fly the plane to avoid being hit ? (Take g = 10 m s-2 ). 4.31 A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ? 4.32 (a) Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by θ (t ) = tan-1 v0y − gt  vox (b) Shows that the projection angle θ0 for a projectile launched from the origin is given by θ0 = tan-1 4hm  R  where the symbols have their usual meaning. 2020-21

CHAPTER FIVE LAWS OF MOTION 5.1 Introduction 5.1 INTRODUCTION 5.2 Aristotle’s fallacy 5.3 The law of inertia In the preceding Chapter, our concern was to describe the 5.4 Newton’s first law of motion motion of a particle in space quantitatively. We saw that 5.5 Newton’s second law of uniform motion needs the concept of velocity alone whereas non-uniform motion requires the concept of acceleration in motion addition. So far, we have not asked the question as to what governs the motion of bodies. In this chapter, we turn to this 5.6 Newton’s third law of motion basic question. 5.7 Conservation of momentum 5.8 Equilibrium of a particle Let us first guess the answer based on our common 5.9 Common forces in mechanics experience. To move a football at rest, someone must kick it. 5.10 Circular motion To throw a stone upwards, one has to give it an upward 5.11 Solving problems in push. A breeze causes the branches of a tree to swing; a strong wind can even move heavy objects. A boat moves in a mechanics flowing river without anyone rowing it. Clearly, some external agency is needed to provide force to move a body from rest. Summary Likewise, an external force is needed also to retard or stop Points to ponder motion. You can stop a ball rolling down an inclined plane by Exercises applying a force against the direction of its motion. Additional exercises In these examples, the external agency of force (hands, wind, stream, etc) is in contact with the object. This is not always necessary. A stone released from the top of a building accelerates downward due to the gravitational pull of the earth. A bar magnet can attract an iron nail from a distance. This shows that external agencies (e.g. gravitational and magnetic forces ) can exert force on a body even from a distance. In short, a force is required to put a stationary body in motion or stop a moving body, and some external agency is needed to provide this force. The external agency may or may not be in contact with the body. So far so good. But what if a body is moving uniformly (e.g. a skater moving straight with constant speed on a horizontal ice slab) ? Is an external force required to keep a body in uniform motion? 2020-21

90 PHYSICS 5.2 ARISTOTLE’S FALLACY true law of nature for forces and motion, one has to imagine a world in which uniform motion is The question posed above appears to be simple. possible with no frictional forces opposing. This However, it took ages to answer it. Indeed, the is what Galileo did. correct answer to this question given by Galileo 5.3 THE LAW OF INERTIA in the seventeenth century was the foundation Galileo studied motion of objects on an inclined of Newtonian mechanics, which signalled the plane. Objects (i) moving down an inclined plane birth of modern science. accelerate, while those (ii) moving up retard. (iii) Motion on a horizontal plane is an The Greek thinker, Aristotle (384 B.C– 322 intermediate situation. Galileo concluded that B.C.), held the view that if a body is moving, an object moving on a frictionless horizontal something external is required to keep it moving. plane must neither have acceleration nor According to this view, for example, an arrow retardation, i.e. it should move with constant shot from a bow keeps flying since the air behind velocity (Fig. 5.1(a)). the arrow keeps pushing it. The view was part of an elaborate framework of ideas developed by (i) (ii) (iii) Aristotle on the motion of bodies in the universe. Fig. 5.1(a) Most of the Aristotelian ideas on motion are now known to be wrong and need not concern us. Another experiment by Galileo leading to the For our purpose here, the Aristotelian law of same conclusion involves a double inclined plane. motion may be phrased thus: An external force A ball released from rest on one of the planes rolls is required to keep a body in motion. down and climbs up the other. If the planes are smooth, the final height of the ball is nearly the Aristotelian law of motion is flawed, as we shall same as the initial height (a little less but never see. However, it is a natural view that anyone greater). In the ideal situation, when friction is would hold from common experience. Even a absent, the final height of the ball is the same small child playing with a simple (non-electric) as its initial height. toy-car on a floor knows intuitively that it needs to constantly drag the string attached to the toy- If the slope of the second plane is decreased car with some force to keep it going. If it releases and the experiment repeated, the ball will still the string, it comes to rest. This experience is reach the same height, but in doing so, it will common to most terrestrial motion. External travel a longer distance. In the limiting case, when forces seem to be needed to keep bodies in the slope of the second plane is zero (i.e. is a motion. Left to themselves, all bodies eventually horizontal) the ball travels an infinite distance. come to rest. In other words, its motion never ceases. This is, of course, an idealised situation (Fig. 5.1(b)). What is the flaw in Aristotle’s argument? The answer is: a moving toy car comes to rest because Fig. 5.1(b) The law of inertia was inferred by Galileo the external force of friction on the car by the floor from observations of motion of a ball on a opposes its motion. To counter this force, the child double inclined plane. has to apply an external force on the car in the direction of motion. When the car is in uniform motion, there is no net external force acting on it: the force by the child cancels the force ( friction) by the floor. The corollary is: if there were no friction, the child would not be required to apply any force to keep the toy car in uniform motion. The opposing forces such as friction (solids) and viscous forces (for fluids) are always present in the natural world. This explains why forces by external agencies are necessary to overcome the frictional forces to keep bodies in uniform motion. Now we understand where Aristotle went wrong. He coded this practical experience in the form of a basic argument. To get at the 2020-21

LAWS OF MOTION 91 In practice, the ball does come to a stop after accomplished almost single-handedly by Isaac moving a finite distance on the horizontal plane, Newton, one of the greatest scientists of all times. because of the opposing force of friction which can never be totally eliminated. However, if there Newton built on Galileo’s ideas and laid the were no friction, the ball would continue to move foundation of mechanics in terms of three laws with a constant velocity on the horizontal plane. of motion that go by his name. Galileo’s law of inertia was his starting point which he Galileo thus, arrived at a new insight on formulated as the first law of motion: motion that had eluded Aristotle and those who followed him. The state of rest and the state of Every body continues to be in its state uniform linear motion (motion with constant of rest or of uniform motion in a straight velocity) are equivalent. In both cases, there is line unless compelled by some external force to act otherwise. Ideas on Motion in Ancient Indian Science Ancient Indian thinkers had arrived at an elaborate system of ideas on motion. Force, the cause of motion, was thought to be of different kinds : force due to continuous pressure (nodan), as the force of wind on a sailing vessel; impact (abhighat), as when a potter’s rod strikes the wheel; persistent tendency (sanskara) to move in a straight line(vega) or restoration of shape in an elastic body; transmitted force by a string, rod, etc. The notion of (vega) in the Vaisesika theory of motion perhaps comes closest to the concept of inertia. Vega, the tendency to move in a straight line, was thought to be opposed by contact with objects including atmosphere, a parallel to the ideas of friction and air resistance. It was correctly summarised that the different kinds of motion (translational, rotational and vibrational) of an extended body arise from only the translational motion of its constituent particles. A falling leaf in the wind may have downward motion as a whole (patan) and also rotational and vibrational motion (bhraman, spandan), but each particle of the leaf at an instant only has a definite (small) displacement. There was considerable focus in Indian thought on measurement of motion and units of length and time. It was known that the position of a particle in space can be indicated by distance measured along three axes. Bhaskara (1150 A.D.) had introduced the concept of ‘instantaneous motion’ (tatkaliki gati), which anticipated the modern notion of instantaneous velocity using Differential Calculus. The difference between a wave and a current (of water) was clearly understood; a current is a motion of particles of water under gravity and fluidity while a wave results from the transmission of vibrations of water particles. no net force acting on the body. It is incorrect to The state of rest or uniform linear motion both assume that a net force is needed to keep a body imply zero acceleration. The first law of motion can, in uniform motion. To maintain a body in therefore, be simply expressed as: uniform motion, we need to apply an external If the net external force on a body is zero, its force to ecounter the frictional force, so that acceleration is zero. Acceleration can be non the two forces sum up to zero net external zero only if there is a net external force on force. the body. To summarise, if the net external force is zero, Two kinds of situations are encountered in the a body at rest continues to remain at rest and a application of this law in practice. In some body in motion continues to move with a uniform examples, we know that the net external force velocity. This property of the body is called on the object is zero. In that case we can inertia. Inertia means ‘resistance to change’. conclude that the acceleration of the object is A body does not change its state of rest or zero. For example, a spaceship out in uniform motion, unless an external force interstellar space, far from all other objects and compels it to change that state. with all its rockets turned off, has no net external force acting on it. Its acceleration, 5.4 NEWTON’S FIRST LAW OF MOTION according to the first law, must be zero. If it is in motion, it must continue to move with a Galileo’s simple, but revolutionary ideas uniform velocity. dethroned Aristotelian mechanics. A new mechanics had to be developed. This task was 2020-21

92 PHYSICS Galileo Galilei (1564 - 1642) Galileo Galilei, born in Pisa, Italy in 1564 was a key figure in the scientific revolution in Europe about four centuries ago. Galileo proposed the concept of acceleration. From experiments on motion of bodies on inclined planes or falling freely, he contradicted the Aristotelian notion that a force was required to keep a body in motion, and that heavier bodies fall faster than lighter bodies under gravity. He thus arrived at the law of inertia that was the starting point of the subsequent epochal work of Isaac Newton. Galileo’s discoveries in astronomy were equally revolutionary. In 1609, he designed his own telescope (invented earlier in Holland) and used it to make a number of startling observations : mountains and depressions on the surface of the moon; dark spots on the sun; the moons of Jupiter and the phases of Venus. He concluded that the Milky Way derived its luminosity because of a large number of stars not visible to the naked eye. In his masterpiece of scientific reasoning : Dialogue on the Two Chief World Systems, Galileo advocated the heliocentric theory of the solar system proposed by Copernicus, which eventually got universal acceptance. With Galileo came a turning point in the very method of scientific inquiry. Science was no longer merely observations of nature and inferences from them. Science meant devising and doing experiments to verify or refute theories. Science meant measurement of quantities and a search for mathematical relations between them. Not undeservedly, many regard Galileo as the father of modern science. More often, however, we do not know all the normal force R must be equal and opposite to the forces to begin with. In that case, if we know weight W ”. that an object is unaccelerated (i.e. it is either at rest or in uniform linear motion), we can infer Fig. 5.2 (a) a book at rest on the table, and (b) a car from the first law that the net external force on moving with uniform velocity. The net force the object must be zero. Gravity is everywhere. is zero in each case. For terrestrial phenomena, in particular, every object experiences gravitational force due to the Consider the motion of a car starting from earth. Also objects in motion generally experience rest, picking up speed and then moving on a friction, viscous drag, etc. If then, on earth, an smooth straight road with uniform speed (Fig. object is at rest or in uniform linear motion, it is (5.2(b)). When the car is stationary, there is no not because there are no forces acting on it, but net force acting on it. During pick-up, it because the various external forces cancel out accelerates. This must happen due to a net i.e. add up to zero net external force. external force. Note, it has to be an external force. The acceleration of the car cannot be accounted Consider a book at rest on a horizontal surface for by any internal force. This might sound Fig. (5.2(a)). It is subject to two external forces : surprising, but it is true. The only conceivable the force due to gravity (i.e. its weight W) acting external force along the road is the force of downward and the upward force on the book by friction. It is the frictional force that accelerates the table, the normal force R . R is a self-adjusting the car as a whole. (You will learn about friction force. This is an example of the kind of situation in section 5.9). When the car moves with mentioned above. The forces are not quite known constant velocity, there is no net external force. fully but the state of motion is known. We observe the book to be at rest. Therefore, we conclude from the first law that the magnitude of R equals that of W. A statement often encountered is : “Since W = R, forces cancel and, therefore, the book is at rest”. This is incorrect reasoning. The correct statement is : “Since the book is observed to be at rest, the net external force on it must be zero, according to the first law. This implies that the 2020-21

LAWS OF MOTION 93 The property of inertia contained in the First It relates the net external force to the law is evident in many situations. Suppose we acceleration of the body. are standing in a stationary bus and the driver starts the bus suddenly. We get thrown Momentum backward with a jerk. Why ? Our feet are in touch Momentum of a body is defined to be the product with the floor. If there were no friction, we would of its mass m and velocity v, and is denoted remain where we were, while the floor of the bus by p: would simply slip forward under our feet and the back of the bus would hit us. However, p=mv (5.1) fortunately, there is some friction between the feet and the floor. If the start is not too sudden, Momentum is clearly a vector quantity. The i.e. if the acceleration is moderate, the frictional following common experiences indicate the force would be enough to accelerate our feet importance of this quantity for considering the along with the bus. But our body is not strictly effect of force on motion. a rigid body. It is deformable, i.e. it allows some relative displacement between different parts. • Suppose a light-weight vehicle (say a small What this means is that while our feet go with the bus, the rest of the body remains where it is car) and a heavy weight vehicle (say a loaded due to inertia. Relative to the bus, therefore, we truck) are parked on a horizontal road. We all are thrown backward. As soon as that happens, know that a much greater force is needed to however, the muscular forces on the rest of the push the truck than the car to bring them to body (by the feet) come into play to move the body the same speed in same time. Similarly, a along with the bus. A similar thing happens greater opposing force is needed to stop a when the bus suddenly stops. Our feet stop due heavy body than a light body in the same time, to the friction which does not allow relative if they are moving with the same speed. motion between the feet and the floor of the bus. But the rest of the body continues to move • If two stones, one light and the other heavy, forward due to inertia. We are thrown forward. The restoring muscular forces again come into are dropped from the top of a building, a play and bring the body to rest. person on the ground will find it easier to catch the light stone than the heavy stone. The Example 5.1 An astronaut accidentally mass of a body is thus an important gets separated out of his small spaceship parameter that determines the effect of force accelerating in inter stellar space at a on its motion. constant rate of 100 m s–2. What is the acceleration of the astronaut the instant after • Speed is another important parameter to he is outside the spaceship ? (Assume that there are no nearby stars to exert consider. A bullet fired by a gun can easily gravitational force on him.) pierce human tissue before it stops, resulting in casualty. The same bullet fired with Answer Since there are no nearby stars to exert moderate speed will not cause much damage. gravitational force on him and the small Thus for a given mass, the greater the speed, spaceship exerts negligible gravitational the greater is the opposing force needed to stop attraction on him, the net force acting on the the body in a certain time. Taken together, astronaut, once he is out of the spaceship, is the product of mass and velocity, that is zero. By the first law of motion the acceleration momentum, is evidently a relevant variable of the astronaut is zero. of motion. The greater the change in the momentum in a given time, the greater is the 5.5 NEWTON’S SECOND LAW OF MOTION force that needs to be applied. The first law refers to the simple case when the • A seasoned cricketer catches a cricket ball net external force on a body is zero. The second law of motion refers to the general situation when coming in with great speed far more easily there is a net external force acting on the body. than a novice, who can hurt his hands in the act. One reason is that the cricketer allows a longer time for his hands to stop the ball. As you may have noticed, he draws in the hands backward in the act of catching the ball (Fig. 5.3). The novice, on the other hand, keeps his hands fixed and tries to catch the ball almost instantly. He needs to provide a much greater force to stop the ball instantly, and 2020-21

94 PHYSICS this hurts.The conclusion is clear: force not This force is provided by our hand through only depends on the change in momentum, the string. Experience suggests that our hand but also on how fast the change is brought needs to exert a greater force if the stone is about. The same change in momentum rotated at greater speed or in a circle of brought about in a shorter time needs a smaller radius, or both. This corresponds to greater applied force. In short, the greater the greater acceleration or equivalently a greater rate of change of momentum, the greater is rate of change in momentum vector. This the force. suggests that the greater the rate of change in momentum vector the greater is the force applied. Fig. 5.3 Force not only depends on the change in momentum but also on how fast the change is brought about. A seasoned cricketer draws Fig. 5.4 Force is necessary for changing the direction in his hands during a catch, allowing greater of momentum, even if its magnitude is time for the ball to stop and hence requires a constant. We can feel this while rotating a smaller force. stone in a horizontal circle with uniform speed by means of a string. • Observations confirm that the product of These qualitative observations lead to the mass and velocity (i.e. momentum) is basic to second law of motion expressed by Newton as the effect of force on motion. Suppose a fixed follows : force is applied for a certain interval of time on two bodies of different masses, initially at The rate of change of momentum of a body is rest, the lighter body picks up a greater speed directly proportional to the applied force and than the heavier body. However, at the end of takes place in the direction in which the force the time interval, observations show that each acts. body acquires the same momentum. Thus the same force for the same time causes Thus, if under the action of a force F for time the same change in momentum for interval ∆t, the velocity of a body of mass m different bodies. This is a crucial clue to the changes from v to v + ∆v i.e. its initial momentum second law of motion. p = m v changes by ∆p = m ∆v . According to the • In the preceding observations, the vector Second Law, character of momentum has not been evident. F ∝ ∆p or F = k ∆p In the examples so far, momentum and change ∆t ∆t in momentum both have the same direction. But this is not always the case. Suppose a where k is a constant of proportionality. Taking stone is rotated with uniform speed in a horizontal plane by means of a string, the the limit ∆t → 0, the term ∆p becomes the magnitude of momentum is fixed, but its ∆t direction changes (Fig. 5.4). A force is needed to cause this change in momentum vector. derivative or differential co-efficient of p with respect to t, denoted by dp . Thus dt 2020-21

LAWS OF MOTION 95 F = k dp (5.2) on the particle and a stands for acceleration dt of the particle. It turns out, however, that the law in the same form applies to a rigid body or, For a body of fixed mass m, even more generally, to a system of particles. In that case, F refers to the total external force dp = d (m v) = m dv = m a (5.3) on the system and a refers to the acceleration dt dt dt of the system as a whole. More precisely, a is the acceleration of the centre of mass of the i.e the Second Law can also be written as system about which we shall study in detail in chapter 7. Any internal forces in the system F = kma (5.4) are not to be included in F. which shows that force is proportional to the product of mass m and acceleration a. The unit of force has not been defined so far. In fact, we use Eq. (5.4) to define the unit of force. We, therefore, have the liberty to choose any constant value for k. For simplicity, we choose k = 1. The second law then is dp (5.5) F = dt = m a In SI unit force is one that causes an acceleration Fig. 5.5 Acceleration at an instant is determined by of 1 m s-2 to a mass of 1 kg. This unit is known as the force at that instant. The moment after a newton : 1 N = 1 kg m s-2. stone is dropped out of an accelerated train, it has no horizontal acceleration or force, if Let us note at this stage some important points air resistance is neglected. The stone carries about the second law : no memory of its acceleration with the train a moment ago. 1. In the second law, F = 0 implies a = 0. The second law is obviously consistent with the first law. 2. The second law of motion is a vector law. It is 4. The second law of motion is a local relation equivalent to three equations, one for each which means that force F at a point in space component of the vectors : (location of the particle) at a certain instant of time is related to a at that point at that Fx = dp x = ma x instant. Acceleration here and now is dt determined by the force here and now, not by any history of the motion of the particle Fy = dpy = ma y (See Fig. 5.5). dt Fz = dp z =m a z (5.6) dt This means that if a force is not parallel to Example 5.2 A bullet of mass 0.04 kg the velocity of the body, but makes some angle moving with a speed of 90 m s–1 enters a with it, it changes only the component of heavy wooden block and is stopped after a velocity along the direction of force. The distance of 60 cm. What is the average component of velocity normal to the force resistive force exerted by the block on the remains unchanged. For example, in the bullet? motion of a projectile under the vertical gravitational force, the horizontal component Answer The retardation ‘a’ of the bullet of velocity remains unchanged (Fig. 5.5). (assumed constant) is given by 3. The second law of motion given by Eq. (5.5) is a = –u2 = – 90 × 90 m s−2 = – 6750 m s−2 applicable to a single point particle. The force 2s 2 × 0.6 F in the law stands for the net external force 2020-21

96 PHYSICS The retarding force, by the second law of ordinary forces. Newtonian mechanics has no motion, is such distinction. Impulsive force is like any other force – except that it is large and acts for a short = 0.04 kg × 6750 m s-2 = 270 N time. The actual resistive force, and therefore, Example 5.4 A batsman hits back a ball retardation of the bullet may not be uniform. The straight in the direction of the bowler without answer therefore, only indicates the average changing its initial speed of 12 m s–1. resistive force. If the mass of the ball is 0.15 kg, determine the impulse imparted to the ball. (Assume Example 5.3 The motion of a particle of linear motion of the ball) mass m is described by y = ut + 1 gt 2 . Find Answer Change in momentum 2 = 0.15 × 12–(–0.15×12) the force acting on the particle. = 3.6 N s, Answer We know Impulse = 3.6 N s, in the direction from the batsman to the bowler. y =ut + 1 gt 2 2 This is an example where the force on the ball by the batsman and the time of contact of the Now, ball and the bat are difficult to know, but the impulse is readily calculated. v = dy = u + gt dt 5.6 NEWTON’S THIRD LAW OF MOTION acceleration, a = dv =g The second law relates the external force on a dt body to its acceleration. What is the origin of the external force on the body ? What agency Then the force is given by Eq. (5.5) provides the external force ? The simple answer in Newtonian mechanics is that the external F = ma = mg force on a body always arises due to some other body. Consider a pair of bodies A and B. B gives Thus the given equation describes the motion rise to an external force on A. A natural question is: Does A in turn give rise to an external force of a particle under acceleration due to gravity on B ? In some examples, the answer seems clear. If you press a coiled spring, the spring is and y is the position coordinate in the direction compressed by the force of your hand. The compressed spring in turn exerts a force on your of g. hand and you can feel it. But what if the bodies are not in contact ? The earth pulls a stone Impulse downwards due to gravity. Does the stone exert We sometimes encounter examples where a large a force on the earth ? The answer is not obvious force acts for a very short duration producing a since we hardly see the effect of the stone on the finite change in momentum of the body. For earth. The answer according to Newton is: Yes, example, when a ball hits a wall and bounces the stone does exert an equal and opposite force back, the force on the ball by the wall acts for a on the earth. We do not notice it since the earth very short time when the two are in contact, yet is very massive and the effect of a small force on the force is large enough to reverse the momentum its motion is negligible. of the ball. Often, in these situations, the force and the time duration are difficult to ascertain Thus, according to Newtonian mechanics, separately. However, the product of force and time, force never occurs singly in nature. Force is the which is the change in momentum of the body mutual interaction between two bodies. Forces remains a measurable quantity. This product is called impulse: Impulse = Force × time duration (5.7) = Change in momentum A large force acting for a short time to produce a finite change in momentum is called an impulsive force. In the history of science, impulsive forces were put in a conceptually different category from 2020-21

LAWS OF MOTION 97 always occur in pairs. Further, the mutual forces comes before reaction i.e action is the cause between two bodies are always equal and and reaction the effect. There is no cause- opposite. This idea was expressed by Newton in effect relation implied in the third law. The the form of the third law of motion. force on A by B and the force on B by A act at the same instant. By the same reasoning, To every action, there is always an equal and any one of them may be called action and the opposite reaction. other reaction. 3. Action and reaction forces act on different Newton’s wording of the third law is so crisp and bodies, not on the same body. Consider a pair beautiful that it has become a part of common of bodies A and B. According to the third law, language. For the same reason perhaps, misconceptions about the third law abound. Let FAB = – FBA (5.8) us note some important points about the third law, particularly in regard to the usage of the (force on A by B) = – (force on B by A) terms : action and reaction. 1. The terms action and reaction in the third law Thus if we are considering the motion of any one body (A or B), only one of the two forces is mean nothing else but ‘force’. Using different relevant. It is an error to add up the two forces terms for the same physical concept and claim that the net force is zero. can sometimes be confusing. A simple and clear way of stating the third law is as However, if you are considering the system follows : of two bodies as a whole, FAB and FBA are internal forces of the system (A + B). They add Forces always occur in pairs. Force on a up to give a null force. Internal forces in a body A by B is equal and opposite to the body or a system of particles thus cancel away force on the body B by A. in pairs. This is an important fact that enables the second law to be applicable to a 2. The terms action and reaction in the third law body or a system of particles (See Chapter 7). may give a wrong impression that action Isaac Newton (1642 – 1727) Isaac Newton was born in Woolsthorpe, England in 1642, the year Galileo died. His extraordinary mathematical ability and mechanical aptitude remained hidden from others in his school life. In 1662, he went to Cambridge for undergraduate studies. A plague epidemic in 1665 forced the university town to close and Newton had to return to his mother’s farm. There in two years of solitude, his dormant creativity blossomed in a deluge of fundamental discoveries in mathematics and physics : binomial theorem for negative and fractional exponents, the beginning of calculus, the inverse square law of gravitation, the spectrum of white light, and so on. Returning to Cambridge, he pursued his investigations in optics and devised a reflecting telescope. In 1684, encouraged by his friend Edmund Halley, Newton embarked on writing what was to be one of the greatest scientific works ever published : The Principia Mathematica. In it, he enunciated the three laws of motion and the universal law of gravitation, which explained all the three Kepler’s laws of planetary motion. The book was packed with a host of path-breaking achievements : basic principles of fluid mechanics, mathematics of wave motion, calculation of masses of the earth, the sun and other planets, explanation of the precession of equinoxes, theory of tides, etc. In 1704, Newton brought out another masterpiece Opticks that summarized his work on light and colour. The scientific revolution triggered by Copernicus and steered vigorously ahead by Kepler and Galileo was brought to a grand completion by Newton. Newtonian mechanics unified terrestrial and celestial phenomena. The same mathematical equation governed the fall of an apple to the ground and the motion of the moon around the earth. The age of reason had dawned. 2020-21

98 PHYSICS Example 5.5 Two identical billiard balls magnitude of force cannot be ascertained since strike a rigid wall with the same speed but the small time taken for the collision has not at different angles, and get reflected without been specified in the problem. any change in speed, as shown in Fig. 5.6. What is (i) the direction of the force on the Case (b) wall due to each ball? (ii) the ratio of the magnitudes of impulses imparted to the ( ) ( ),px initial = m u cos 30 balls by the wall ? p =y initial − m u sin 30 Fig. 5.6 ( ) ( )px final = – m u cos 30 , p =y final − m u sin 30 Answer An instinctive answer to (i) might be Note, while px changes sign after collision, py that the force on the wall in case (a) is normal to does not. Therefore, the wall, while that in case (b) is inclined at 30° to the normal. This answer is wrong. The force x-component of impulse = –2 m u cos 30° on the wall is normal to the wall in both cases. y-component of impulse = 0 How to find the force on the wall? The trick is The direction of impulse (and force) is the same to consider the force (or impulse) on the ball as in (a) and is normal to the wall along the due to the wall using the second law, and then negative x direction. As before, using Newton’s use the third law to answer (i). Let u be the speed third law, the force on the wall due to the ball is of each ball before and after collision with the normal to the wall along the positive x direction. wall, and m the mass of each ball. Choose the x and y axes as shown in the figure, and consider The ratio of the magnitudes of the impulses the change in momentum of the ball in each imparted to the balls in (a) and (b) is case : ( )2m u/ 2m u cos 30 = 2 ≈ 1.2 Case (a) ( )p = 0y initial 3 ( )py final = 0 ( )px initial = mu 5.7 CONSERVATION OF MOMENTUM ( )px final = −mu The second and third laws of motion lead to an important consequence: the law of Impulse is the change in momentum vector. conservation of momentum. Take a familiar Therefore, example. A bullet is fired from a gun. If the force on the bullet by the gun is F, the force on the gun x-component of impulse = – 2 m u by the bullet is – F, according to the third law. y-component of impulse = 0 The two forces act for a common interval of time Impulse and force are in the same direction. ∆t. According to the second law, F ∆t is the Clearly, from above, the force on the ball due to change in momentum of the bullet and – F ∆t is the wall is normal to the wall, along the negative the change in momentum of the gun. Since x-direction. Using Newton’s third law of motion, initially, both are at rest, the change in the force on the wall due to the ball is normal to momentum equals the final momentum for each. the wall along the positive x-direction. The Thus if pb is the momentum of the bullet after firing and pg is the recoil momentum of the gun, pg = – pb i.e. pb + pg = 0. That is, the total momentum of the (bullet + gun) system is conserved. Thus in an isolated system (i.e. a system with no external force), mutual forces between pairs of particles in the system can cause momentum change in individual particles, but since the mutual forces for each pair are equal and opposite, the momentum changes cancel in pairs and the total momentum remains unchanged. This fact is known as the law of conservation of momentum : 2020-21

LAWS OF MOTION 99 The total momentum of an isolated system Fig. 5.7 Equilibrium under concurrent forces. of interacting particles is conserved. An important example of the application of the law of conservation of momentum is the collision of two bodies. Consider two bodies A and B, with initial momenta pA and pB. The bodies collide, get apart, with final momenta p′A and p′B respectively. By the Second Law FAB ∆t = p′A − pA and In other words, the resultant of any two forces say F1 and F2, obtained by the parallelogram FBA∆t = p′B − pB law of forces must be equal and opposite to the third force, F3. As seen in Fig. 5.7, the three (where we have taken a common interval of time forces in equilibrium can be represented by the for both forces i.e. the time for which the two sides of a triangle with the vector arrows taken bodies are in contact.) in the same sense. The result can be generalised to any number of forces. A particle Since FAB = −FBA by the third law, is in equilibrium under the action of forces F1, F2,... Fn if they can be represented by the sides ( )p′A − pA = − p′B − pB of a closed n-sided polygon with arrows directed in the same sense. i.e. p′A + p′B = p A + pB (5.9) which shows that the total final momentum of Equation (5.11) implies that the isolated system equals its initial momentum. F1x + F2x + F3x = 0 Notice that this is true whether the collision is F1y + F2y + F3y = 0 elastic or inelastic. In elastic collisions, there is a second condition that the total initial kinetic F1z + F2z + F3z = 0 (5.12) energy of the system equals the total final kinetic where F1x, F1y and F1z are the components of F1 along x, y and z directions respectively. energy (See Chapter 6). 5.8 EQUILIBRIUM OF A PARTICLE Example 5.6 See Fig. 5.8. A mass of 6 kg is suspended by a rope of length 2 m Equilibrium of a particle in mechanics refers to from the ceiling. A force of 50 N in the horizontal direction is applied at the mid- the situation when the net external force on the point P of the rope, as shown. What is the angle the rope makes with the vertical in particle is zero*. According to the first law, this equilibrium ? (Take g = 10 m s-2). Neglect the mass of the rope. means that, the particle is either at rest or in uniform motion. If two forces F1 and F2, act on a particle, equilibrium requires F1 = − F2 (5.10) i.e. the two forces on the particle must be equal and opposite. Equilibrium under three concurrent forces F1, F2 and F3 requires that the vector sum of the three forces is zero. F1 + F2 + F3 = 0 (5.11) (a) (b) (c) Fig. 5.8 * Equilibrium of a body requires not only translational equilibrium (zero net external force) but also rotational equilibrium (zero net external torque), as we shall see in Chapter 7. 2020-21

100 PHYSICS Answer Figures 5.8(b) and 5.8(c) are known as other types of supports), there are mutual free-body diagrams. Figure 5.8(b) is the free-body contact forces (for each pair of bodies) satisfying diagram of W and Fig. 5.8(c) is the free-body the third law. The component of contact force diagram of point P. normal to the surfaces in contact is called normal reaction. The component parallel to the Consider the equilibrium of the weight W. surfaces in contact is called friction. Contact Clearly,T2 = 6 × 10 = 60 N. forces arise also when solids are in contact with fluids. For example, for a solid immersed in a Consider the equilibrium of the point P under fluid, there is an upward bouyant force equal to the action of three forces - the tensions T1 and the weight of the fluid displaced. The viscous T2, and the horizontal force 50 N. The horizontal force, air resistance, etc are also examples of and vertical components of the resultant force contact forces (Fig. 5.9). must vanish separately : Two other common forces are tension in a T1 cos θ = T2 = 60 N string and the force due to spring. When a spring T1 sin θ = 50 N is compressed or extended by an external force, which gives that a restoring force is generated. This force is usually proportional to the compression or Note the answer does not depend on the length elongation (for small displacements). The spring of the rope (assumed massless) nor on the point force F is written as F = – k x where x is the at which the horizontal force is applied. displacement and k is the force constant. The negative sign denotes that the force is opposite 5.9 COMMON FORCES IN MECHANICS to the displacement from the unstretched state. For an inextensible string, the force constant is In mechanics, we encounter several kinds of very high. The restoring force in a string is called forces. The gravitational force is, of course, tension. It is customary to use a constant tension pervasive. Every object on the earth experiences T throughout the string. This assumption is true the force of gravity due to the earth. Gravity also for a string of negligible mass. governs the motion of celestial bodies. The gravitational force can act at a distance without In Chapter 1, we learnt that there are four the need of any intervening medium. fundamental forces in nature. Of these, the weak and strong forces appear in domains that do not All the other forces common in mechanics are concern us here. Only the gravitational and electrical forces are relevant in the context of contact forces*. As the name suggests, a contact mechanics. The different contact forces of mechanics mentioned above fundamentally arise force on an object arises due to contact with some from electrical forces. This may seem surprising other object: solid or fluid. When bodies are in contact (e.g. a book resting on a table, a system of rigid bodies connected by rods, hinges and Fig. 5.9 Some examples of contact forces in mechanics. * We are not considering, for simplicity, charged and magnetic bodies. For these, besides gravity, there are electrical and magnetic non-contact forces. 2020-21

LAWS OF MOTION 101 since we are talking of uncharged and non- exist by itself. When there is no applied force, magnetic bodies in mechanics. At the microscopic there is no static friction. It comes into play the level, all bodies are made of charged constituents moment there is an applied force. As the applied (nuclei and electrons) and the various contact force F increases, fs also increases, remaining forces arising due to elasticity of bodies, molecular equal and opposite to the applied force (up to a collisions and impacts, etc. can ultimately be certain limit), keeping the body at rest. Hence, it traced to the electrical forces between the charged is called static friction. Static friction opposes constituents of different bodies. The detailed impending motion. The term impending motion microscopic origin of these forces is, however, means motion that would take place (but does complex and not useful for handling problems in not actually take place) under the applied force, mechanics at the macroscopic scale. This is why if friction were absent. they are treated as different types of forces with their characteristic properties determined We know from experience that as the applied empirically. force exceeds a certain limit, the body begins to 5.9.1 Friction move. It is found experimentally that the limiting Let us return to the example of a body of mass m ( )value of static friction fs max is independent of at rest on a horizontal table. The force of gravity (mg) is cancelled by the normal reaction force the area of contact and varies with the normal (N) of the table. Now suppose a force F is applied horizontally to the body. We know from force(N) approximately as : experience that a small applied force may not be enough to move the body. But if the applied ( )fs max = µsN (5.13) force F were the only external force on the body, it must move with acceleration F/m, however where µs is a constant of proportionality small. Clearly, the body remains at rest because depending only on the nature of the surfaces in some other force comes into play in the horizontal direction and opposes the applied contact. The constant µs is called the coefficient force F, resulting in zero net force on the body. of static friction. The law of static friction may This force fs parallel to the surface of the body in contact with the table is known as frictional thus be written as force, or simply friction (Fig. 5.10(a)). The subscript stands for static friction to distinguish fs ≤ µs N (5.14) it from kinetic friction fk that we consider later (Fig. 5.10(b)). Note that static friction does not ( )If the applied force F exceeds fs max the body Fig. 5.10 Static and sliding friction: (a) Impending begins to slide on the surface. It is found motion of the body is opposed by static friction. When external force exceeds the experimentally that when relative motion has maximum limit of static friction, the body begins to move. (b) Once the body is in started, the frictional force decreases from the motion, it is subject to sliding or kinetic friction which opposes relative motion between the ( )static maximum value fs max . Frictional force two surfaces in contact. Kinetic friction is usually less than the maximum value of static that opposes relative motion between surfaces friction. in contact is called kinetic or sliding friction and is denoted by fk . Kinetic friction, like static friction, is found to be independent of the area of contact. Further, it is nearly independent of the velocity. It satisfies a law similar to that for static friction: fk = µk N (5.15) where µk′ the coefficient of kinetic friction, depends only on the surfaces in contact. As mentioned above, experiments show that µk is less than µs. When relative motion has begun, the acceleration of the body according to the second law is ( F – fk )/m. For a body moving with constant velocity, F = fk. If the applied force on the body is removed, its acceleration is – fk /m and it eventually comes to a stop. The laws of friction given above do not have the status of fundamental laws like those for gravitational, electric and magnetic forces. They are empirical relations that are only 2020-21

102 PHYSICS approximately true. Yet they are very useful in Answer The forces acting on a block of mass m practical calculations in mechanics. at rest on an inclined plane are (i) the weight mg acting vertically downwards (ii) the normal Thus, when two bodies are in contact, each force N of the plane on the block, and (iii) the experiences a contact force by the other. Friction, static frictional force fs opposing the impending by definition, is the component of the contact force motion. In equilibrium, the resultant of these parallel to the surfaces in contact, which opposes forces must be zero. Resolving the weight mg impending or actual relative motion between the along the two directions shown, we have two surfaces. Note that it is not motion, but relative motion that the frictional force opposes. m g sin θ = fs , m g cos θ = N Consider a box lying in the compartment of a train As θ increases, the self-adjusting frictional force that is accelerating. If the box is stationary fs increases until at θ = θmax, fs achieves its relative to the train, it is in fact accelerating along with the train. What forces cause the acceleration ( )maximum value, fs max = µs N. of the box? Clearly, the only conceivable force in the horizontal direction is the force of friction. If Therefore, there were no friction, the floor of the train would tan θmax = µs or θmax = tan–1 µs slip by and the box would remain at its initial position due to inertia (and hit the back side of When θ becomes just a little more than θmax , the train). This impending relative motion is there is a small net force on the block and it opposed by the static friction fs. Static friction begins to slide. Note that θmax depends only on provides the same acceleration to the box as that µs and is independent of the mass of the block. of the train, keeping it stationary relative to the For θmax = 15°, train. µs = tan 15° Example 5.7 Determine the maximum = 0.27 acceleration of the train in which a box lying on its floor will remain stationary, Example 5.9 What is the acceleration of given that the co-efficient of static friction the block and trolley system shown in a between the box and the train’s floor is Fig. 5.12(a), if the coefficient of kinetic friction 0.15. between the trolley and the surface is 0.04? What is the tension in the string? (Take g = Answer Since the acceleration of the box is due 10 m s-2). Neglect the mass of the string. to the static friction, (a) ma = fs ≤ µs N = µs m g i.e. a ≤ µs g ∴ amax = µs g = 0.15 x 10 m s–2 = 1.5 m s–2 Example 5.8 See Fig. 5.11. A mass of 4 kg rests on a horizontal plane. The plane is gradually inclined until at an angle θ = 15° with the horizontal, the mass just begins to slide. What is the coefficient of static friction between the block and the surface ? Fig. 5.11 (b) (c) Fig. 5.12 2020-21

LAWS OF MOTION 103 Answer As the string is inextensible, and the is the reason why discovery of the wheel has been a major milestone in human history. pully is smooth, the 3 kg block and the 20 kg Rolling friction again has a complex origin, trolley both have same magnitude of though somewhat different from that of static and sliding friction. During rolling, the surfaces acceleration. Applying second law to motion of in contact get momentarily deformed a little, and this results in a finite area (not a point) of the the block (Fig. 5.12(b)), body being in contact with the surface. The net effect is that the component of the contact force 30 – T = 3a parallel to the surface opposes motion. Apply the second law to motion of the trolley (Fig. We often regard friction as something undesirable. In many situations, like in a 5.12(c)), machine with different moving parts, friction does have a negative role. It opposes relative T – fk = 20 a. motion and thereby dissipates power in the form Now fk = µk N, of heat, etc. Lubricants are a way of reducing kinetic friction in a machine. Another way is to Here µk = 0.04, use ball bearings between two moving parts of a machine [Fig. 5.13(a)]. Since the rolling friction N = 20 x 10 between ball bearings and the surfaces in contact is very small, power dissipation is = 200 N. reduced. A thin cushion of air maintained between solid surfaces in relative motion is Thus the equation for the motion of the trolley is another effective way of reducing friction (Fig. 5.13(a)). T – 0.04 x 200 = 20 a Or T – 8 = 20a. In many practical situations, however, friction These equations give a = 2 2 m s –2 = 0.96 m s-2 is critically needed. Kinetic friction that dissipates power is nevertheless important for and T = 27.1 N. 23 quickly stopping relative motion. It is made use of by brakes in machines and automobiles. Rolling friction Similarly, static friction is important in daily life. We are able to walk because of friction. It A body like a ring or a sphere rolling without is impossible for a car to move on a very slippery slipping over a horizontal plane will suffer no road. On an ordinary road, the friction between friction, in principle. At every instant, there is the tyres and the road provides the necessary just one point of contact between the body and external force to accelerate the car. the plane and this point has no motion relative to the plane. In this ideal situation, kinetic or static friction is zero and the body should continue to roll with constant velocity. We know, in practice, this will not happen and some resistance to motion (rolling friction) does occur, i.e. to keep the body rolling, some applied force is needed. For the same weight, rolling friction is much smaller (even by 2 or 3 orders of magnitude) than static or sliding friction. This Fig. 5.13 Some ways of reducing friction. (a) Ball bearings placed between moving parts of a machine. (b) Compressed cushion of air between surfaces in relative motion. 2020-21

104 PHYSICS 5.10 CIRCULAR MOTION is the static friction that provides the centripetal acceleration. Static friction opposes the We have seen in Chapter 4 that acceleration of impending motion of the car moving away from the circle. Using equation (5.14) & (5.16) we get a body moving in a circle of radius R with uniform the result speed v is v2/R directed towards the centre. According to the second law, the force f providing c this acceleration is : = mv2 f R ≤ µsN mv 2 fc = R (5.16) µsRN m v2 ≤ = µsRg [∵ N = mg] where m is the mass of the body. This force which is independent of the mass of the car. directed forwards the centre is called the This shows that for a given value of µs and R, centripetal force. For a stone rotated in a circle there is a maximum speed of circular motion of by a string, the centripetal force is provided by the car possible, namely the tension in the string. The centripetal force for motion of a planet around the sun is the vmax = µs Rg (5.18) (a) (b) Fig. 5.14 Circular motion of a car on (a) a level road, (b) a banked road. gravitational force on the planet due to the sun. Motion of a car on a banked road For a car taking a circular turn on a horizontal road, the centripetal force is the force of friction. We can reduce the contribution of friction to the circular motion of the car if the road is banked The circular motion of a car on a flat and (Fig. 5.14(b)). Since there is no acceleration along banked road give interesting application of the the vertical direction, the net force along this laws of motion. direction must be zero. Hence, Motion of a car on a level road N cos θ = mg + f sin θ (5.19a) Three forces act on the car (Fig. 5.14(a): The centripetal force is provided by the horizontal components of N and f. (i) The weight of the car, mg (ii) Normal reaction, N (iii) Frictional force, f mv 2 N sin θ + f cos θ = R As there is no acceleration in the vertical (5.19b) direction N – mg = 0 But f ≤ µs N N = mg (5.17) Thus to obtain vmax we put f = µs N . The centripetal force required for circular motion Then Eqs. (5.19a) and (5.19b) become is along the surface of the road, and is provided by the component of the contact force between road and the car tyres along the surface. This by definition is the frictional force. Note that it N cos θ = mg + µs N sin θ (5.20a) 2020-21

LAWS OF MOTION 105 N sin θ + µs N cos θ = mv2/R (5.20b) Example 5.11 A circular racetrack of From Eq. (5.20a), we obtain radius 300 m is banked at an angle of 15°. If the coefficient of friction between the N = mg wheels of a race-car and the road is 0.2, cosθ – µssinθ what is the (a) optimum speed of the race- car to avoid wear and tear on its tyres, and Substituting value of N in Eq. (5.20b), we get (b) maximum permissible speed to avoid slipping ? mg (sinθ + µs )cosθ = mvm2 ax cosθ – µs sinθ R Answer On a banked road, the horizontal component of the normal force and the frictional 1 force contribute to provide centripetal force to  2 keep the car moving on a circular turn without or v max =  Rg µs + tanθ  (5.21) slipping. At the optimum speed, the normal  reaction’s component is enough to provide the  1 – µs tanθ needed centripetal force, and the frictional force is not needed. The optimum speed vo is given by Comparing this with Eq. (5.18) we see that Eq. (5.22): maximum possible speed of a car on a banked road is greater than that on a flat road. vO = (R g tan θ)1/2 Here R = 300 m, θ = 15°, g = 9.8 m s-2; we For µs = 0 in Eq. (5.21 ), (5.22) have vo = ( R g tan θ ) ½ vO = 28.1 m s-1. At this speed, frictional force is not needed at all The maximum permissible speed vmax is given by Eq. (5.21): to provide the necessary centripetal force. 5.11 SOLVING PROBLEMS IN MECHANICS Driving at this speed on a banked road will cause The three laws of motion that you have learnt in little wear and tear of the tyres. The same this chapter are the foundation of mechanics. You should now be able to handle a large variety equation also tells you that for v < vo, frictional of problems in mechanics. A typical problem in force will be up the slope and that a car can be mechanics usually does not merely involve a single body under the action of given forces. parked only if tan θ ≤ µs. More often, we will need to consider an assembly of different bodies exerting forces on each other. Example 5.10 A cyclist speeding at Besides, each body in the assembly experiences 18 km/h on a level road takes a sharp the force of gravity. When trying to solve a circular turn of radius 3 m without reducing problem of this type, it is useful to remember the speed. The co-efficient of static friction the fact that we can choose any part of the between the tyres and the road is 0.1. Will assembly and apply the laws of motion to that the cyclist slip while taking the turn? part provided we include all forces on the chosen part due to the remaining parts of the assembly. Answer On an unbanked road, frictional force We may call the chosen part of the assembly as alone can provide the centripetal force needed the system and the remaining part of the to keep the cyclist moving on a circular turn assembly (plus any other agencies of forces) as without slipping. If the speed is too large, or if the environment. We have followed the same the turn is too sharp (i.e. of too small a radius) or both, the frictional force is not sufficient to provide the necessary centripetal force, and the cyclist slips. The condition for the cyclist not to slip is given by Eq. (5.18) : v2 ≤ µs R g Now, R = 3 m, g = 9.8 m s-2, µs = 0.1. That is, µs R g = 2.94 m2 s-2. v = 18 km/h = 5 m s-1; i.e., v2 = 25 m2 s-2. The condition is not obeyed. The cyclist will slip while taking the circular turn. 2020-21

106 PHYSICS method in solved examples. To handle a typical the net force on the block must be zero i.e., problem in mechanics systematically, one R = 20 N. Using third law the action of the should use the following steps : block (i.e. the force exerted on the floor by (i) Draw a diagram showing schematically the the block) is equal to 20 N and directed vertically downwards. various parts of the assembly of bodies, the (b) The system (block + cylinder) accelerates links, supports, etc. downwards with 0.1 m s-2. The free-body (ii) Choose a convenient part of the assembly diagram of the system shows two forces on as one system. the system : the force of gravity due to the (iii) Draw a separate diagram which shows this earth (270 N); and the normal force R ′ by the system and all the forces on the system by floor. Note, the free-body diagram of the the remaining part of the assembly. Include system does not show the internal forces also the forces on the system by other between the block and the cylinder. Applying agencies. Do not include the forces on the the second law to the system, environment by the system. A diagram of this type is known as ‘a free-body diagram’. 270 – R′ = 27 × 0.1N (Note this does not imply that the system ie. R′ = 267.3 N under consideration is without a net force). (iv) In a free-body diagram, include information Fig. 5.15 about forces (their magnitudes and directions) that are either given or you are By the third law, the action of the system on sure of (e.g., the direction of tension in a the floor is equal to 267.3 N vertically downward. string along its length). The rest should be Action-reaction pairs treated as unknowns to be determined using For (a): (i) the force of gravity (20 N) on the block laws of motion. (v) If necessary, follow the same procedure for by the earth (say, action); the force of another choice of the system. In doing so, gravity on the earth by the block employ Newton’s third law. That is, if in the (reaction) equal to 20 N directed free-body diagram of A, the force on A due to upwards (not shown in the figure). B is shown as F, then in the free-body (ii) the force on the floor by the block diagram of B, the force on B due to A should (action); the force on the block by the be shown as –F. floor (reaction). For (b): (i) the force of gravity (270 N) on the The following example illustrates the above system by the earth (say, action); the procedure : force of gravity on the earth by the system (reaction), equal to 270 N, Example 5.12 See Fig. 5.15. A wooden block of mass 2 kg rests on a soft horizontal floor. When an iron cylinder of mass 25 kg is placed on top of the block, the floor yields steadily and the block and the cylinder together go down with an acceleration of 0.1 m s–2. What is the action of the block on the floor (a) before and (b) after the floor yields ? Take g = 10 m s–2. Identify the action-reaction pairs in the problem. Answer (a) The block is at rest on the floor. Its free-body diagram shows two forces on the block, the force of gravitational attraction by the earth equal to 2 × 10 = 20 N; and the normal force R of the floor on the block. By the First Law, 2020-21

LAWS OF MOTION 107 directed upwards (not shown in the gravity on the mass in (a) or (b) and the normal figure). force on the mass by the floor are not action- (ii) the force on the floor by the system reaction pairs. These forces happen to be equal (action); the force on the system by the and opposite for (a) since the mass is at rest. floor (reaction). In addition, for (b), the They are not so for case (b), as seen already. force on the block by the cylinder and The weight of the system is 270 N, while the the force on the cylinder by the block normal force R′ is 267.3 N. also constitute an action-reaction pair. The practice of drawing free-body diagrams is The important thing to remember is that an of great help in solving problems in mechanics. action-reaction pair consists of mutual forces It allows you to clearly define your system and which are always equal and opposite between consider all forces on the system due to objects two bodies. Two forces on the same body which that are not part of the system itself. A number happen to be equal and opposite can never of exercises in this and subsequent chapters will constitute an action-reaction pair. The force of help you cultivate this practice. SUMMARY 1. Aristotle’s view that a force is necessary to keep a body in uniform motion is wrong. A force is necessary in practice to counter the opposing force of friction. 2. Galileo extrapolated simple observations on motion of bodies on inclined planes, and arrived at the law of inertia. Newton’s first law of motion is the same law rephrased thus: “Everybody continues to be in its state of rest or of uniform motion in a straight line, unless compelled by some external force to act otherwise”. In simple terms, the First Law is “If external force on a body is zero, its acceleration is zero”. 3. Momentum (p ) of a body is the product of its mass (m) and velocity (v) : p = mv 4. Newton’s second law of motion : The rate of change of momentum of a body is proportional to the applied force and takes place in the direction in which the force acts. Thus F = k dp = k m a dt where F is the net external force on the body and a its acceleration. We set the constant of proportionality k = 1 in SI units. Then dp F = = ma dt The SI unit of force is newton : 1 N = 1 kg m s-2. (a) The second law is consistent with the First Law (F = 0 implies a = 0) (b) It is a vector equation (c) It is applicable to a particle, and also to a body or a system of particles, provided F is the total external force on the system and a is the acceleration of the system as a whole. (d) F at a point at a certain instant determines a at the same point at that instant. That is the Second Law is a local law; a at an instant does not depend on the history of motion. 5. Impulse is the product of force and time which equals change in momentum. The notion of impulse is useful when a large force acts for a short time to produce a measurable change in momentum. Since the time of action of the force is very short, one can assume that there is no appreciable change in the position of the body during the action of the impulsive force. 6. Newton’s third law of motion: To every action, there is always an equal and opposite reaction 2020-21

108 PHYSICS In simple terms, the law can be stated thus : Forces in nature always occur between pairs of bodies. Force on a body A by body B is equal and opposite to the force on the body B by A. Action and reaction forces are simultaneous forces. There is no cause-effect relation between action and reaction. Any of the two mutual forces can be called action and the other reaction. Action and reaction act on different bodies and so they cannot be cancelled out. The internal action and reaction forces between different parts of a body do, however, sum to zero. 7. Law of Conservation of Momentum The total momentum of an isolated system of particles is conserved. The law follows from the second and third law of motion. 8. Friction Frictional force opposes (impending or actual) relative motion between two surfaces in contact. It is the component of the contact force along the common tangent to the surface in contact. Static friction fs opposes impending relative motion; kinetic friction fk opposes actual relative motion. They are independent of the area of contact and satisfy the following approximate laws : ( )fs ≤ fs max = µsR f k = µk R µs (co-efficient of static friction) and µk (co-efficient of kinetic friction) are constants characteristic of the pair of surfaces in contact. It is found experimentally that µk is less than µs . POINTS TO PONDER 1. Force is not always in the direction of motion. Depending on the situation, F may be along v, opposite to v, normal to v or may make some other angle with v. In every case, it is parallel to acceleration. 2. If v = 0 at an instant, i.e. if a body is momentarily at rest, it does not mean that force or acceleration are necessarily zero at that instant. For example, when a ball thrown upward reaches its maximum height, v = 0 but the force continues to be its weight mg and the acceleration is not zero but g. 3. Force on a body at a given time is determined by the situation at the location of the body at that time. Force is not ‘carried’ by the body from its earlier history of motion. The moment after a stone is released out of an accelerated train, there is no horizontal force (or acceleration) on the stone, if the effects of the surrounding air are neglected. The stone then has only the vertical force of gravity. 4. In the second law of motion F = m a, F stands for the net force due to all material agencies external to the body. a is the effect of the force. ma should not be regarded as yet another force, besides F. 2020-21

LAWS OF MOTION 109 5. The centripetal force should not be regarded as yet another kind of force. It is simply a name given to the force that provides inward radial acceleration to a body in circular motion. We should always look for some material force like tension, gravitational force, electrical force, friction, etc as the centripetal force in any circular motion. 6. Static friction is a self-adjusting force up to its limit µs N (fs ≤ µs N). Do not put fs= µs N without being sure that the maximum value of static friction is coming into play. 7. The familiar equation mg = R for a body on a table is true only if the body is in equilibrium. The two forces mg and R can be different (e.g. a body in an accelerated lift). The equality of mg and R has no connection with the third law. 8. The terms ‘action’ and ‘reaction’ in the third Law of Motion simply stand for simultaneous mutual forces between a pair of bodies. Unlike their meaning in ordinary language, action does not precede or cause reaction. Action and reaction act on different bodies. 9. The different terms like ‘friction’, ‘normal reaction’ ‘tension’, ‘air resistance’, ‘viscous drag’, ‘thrust’, ‘buoyancy’, ‘weight’, ‘centripetal force’ all stand for ‘force’ in different contexts. For clarity, every force and its equivalent terms encountered in mechanics should be reduced to the phrase ‘force on A by B’. 10. For applying the second law of motion, there is no conceptual distinction between inanimate and animate objects. An animate object such as a human also requires an external force to accelerate. For example, without the external force of friction, we cannot walk on the ground. 11. The objective concept of force in physics should not be confused with the subjective concept of the ‘feeling of force’. On a merry-go-around, all parts of our body are subject to an inward force, but we have a feeling of being pushed outward – the direction of impending motion. EXERCISES (For simplicity in numerical calculations, take g = 10 m s-2) 5.1 Give the magnitude and direction of the net force acting on (a) a drop of rain falling down with a constant speed, (b) a cork of mass 10 g floating on water, (c) a kite skillfully held stationary in the sky, (d) a car moving with a constant velocity of 30 km/h on a rough road, (e) a high-speed electron in space far from all material objects, and free of electric and magnetic fields. 5.2 A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble, (a) during its upward motion, (b) during its downward motion, (c) at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction? Ignore air resistance. 5.3 Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg, (a) just after it is dropped from the window of a stationary train, (b) just after it is dropped from the window of a train running at a constant velocity of 36 km/h, (c ) just after it is dropped from the window of a train accelerating with 1 m s-2, (d) lying on the floor of a train which is accelerating with 1 m s-2, the stone being at rest relative to the train. Neglect air resistance throughout. 2020-21

110 PHYSICS 5.4 One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is : (i) T, (ii) T − mv2 , (iii) T + mv2 , (iv) 0 l l T is the tension in the string. [Choose the correct alternative]. 5.5 A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 m s-1. How long does the body take to stop ? 5.6 A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s-1 to 3.5 m s-1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force ? 5.7 A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body. 5.8 The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle ? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg. 5.9 A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 m s-2. Calculate the initial thrust (force) of the blast. 5.10 A body of mass 0.40 kg moving initially with a constant speed of 10 m s-1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its position at t = –5 s, 25 s, 100 s. 5.11 A truck starts from rest and accelerates uniformly at 2.0 m s-2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t = 11s ? (Neglect air resistance.) 5.12 A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 m s-1. What is the trajectory of the bob if the string is cut when the bob is (a) at one of its extreme positions, (b) at its mean position. 5.13 A man of mass 70 kg stands on a weighing scale in a lift which is moving (a) upwards with a uniform speed of 10 m s-1, (b) downwards with a uniform acceleration of 5 m s-2, (c) upwards with a uniform acceleration of 5 m s-2. What would be the readings on the scale in each case? (d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity ? 5.14 Figure 5.16 shows the position-time graph of a particle of mass 4 kg. What is the (a) force on the particle for t < 0, t > 4 s, 0 < t < 4 s? (b) impulse at t = 0 and t = 4 s ? (Consider one-dimensional motion only). Fig. 5.16 5.15 Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force F = 600 N is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case? 2020-21

LAWS OF MOTION 111 5.16 Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released. 5.17 A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions. 5.18 Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 m s-1 collide and rebound with the same speed. What is the impulse imparted to each ball due to the other ? 5.19 A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m s-1, what is the recoil speed of the gun ? 5.20 A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball ? (Mass of the ball is 0.15 kg.) 5.21 A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string ? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N ? 5.22 If, in Exercise 5.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks : (a) the stone moves radially outwards, (b) the stone flies off tangentially from the instant the string breaks, (c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle ? 5.23 Explain why (a) a horse cannot pull a cart and run in empty space, (b) passengers are thrown forward from their seats when a speeding bus stops suddenly, (c) it is easier to pull a lawn mower than to push it, (d) a cricketer moves his hands backwards while holding a catch. Additional Exercises 5.24 Figure 5.17 shows the position-time graph of a body of mass 0.04 kg. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the body ? What is the magnitude of each impulse ? Fig. 5.17 5.25 Figure 5.18 shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 m s-2. What is the net force on the man? If the coefficient of static friction between the man’s shoes and the belt is 0.2, up to what acceleration of the belt can the man continue to be stationary relative to the belt ? (Mass of the man = 65 kg.) Fig. 5.18 2020-21

112 PHYSICS 5.26 A stone of mass m tied to the end of a string revolves in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are : [Choose the correct alternative] Lowest Point Highest Point (a) mg – T1 mg + T2 (b) mg + T1 mg – T2 (c) mg + T1 – (m v 2 ) / R mg – T2 + (m v 2 ) / R 1 1 (d) mg – T1 – (m v 2 ) / R mg + T2 + (m v 2 ) / R 1 1 T1 and v1 denote the tension and speed at the lowest point. T2 and v2 denote corresponding values at the highest point. 5.27 A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s-2. The crew and the passengers weigh 300 kg. Give the magnitude and direction of the (a) force on the floor by the crew and passengers, (b) action of the rotor of the helicopter on the surrounding air, (c) force on the helicopter due to the surrounding air. 5.28 A stream of water flowing horizontally with a speed of 15 m s-1 gushes out of a tube of cross-sectional area 10-2 m2, and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound ? 5.29 Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m. Give the magnitude and direction of (a) the force on the 7th coin (counted from the bottom) due to all the coins on its top, (b) the force on the 7th coin by the eighth coin, (c) the reaction of the 6th coin on the 7th coin. 5.30 An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15°. What is the radius of the loop ? 5.31 A train runs along an unbanked circular track of radius 30 m at a speed of 54 km/h. The mass of the train is 106 kg. What provides the centripetal force required for this purpose — The engine or the rails ? What is the angle of banking required to prevent wearing out of the rail ? 5.32 A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in Fig. 5.19. What is the action on the floor by the man in the two cases ? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding ? Fig. 5.19 2020-21

LAWS OF MOTION 113 5.33 A monkey of mass 40 kg climbs on a rope (Fig. 5.20) which Fig. 5.20 can stand a maximum tension of 600 N. In which of the Fig. 5.21 following cases will the rope break: the monkey (a) climbs up with an acceleration of 6 m s-2 (b) climbs down with an acceleration of 4 m s-2 (c) climbs up with a uniform speed of 5 m s-1 (d) falls down the rope nearly freely under gravity? (Ignore the mass of the rope). 5.34 Two bodies A and B of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wall (Fig. 5.21). The coefficient of friction between the bodies and the table is 0.15. A force of 200 N is applied horizontally to A. What are (a) the reaction of the partition (b) the action-reaction forces between A and B ? What happens when the wall is removed? Does the answer to (b) change, when the bodies are in motion? Ignore the difference between µs and µk. 5.35 A block of mass 15 kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 m s-2 for 20 s and then moves with uniform velocity. Discuss the motion of the block as viewed by (a) a stationary observer on the ground, (b) an observer moving with the trolley. 5.36 The rear side of a truck is open and a box of 40 kg Fig. 5.22 mass is placed 5 m away from the open end as shown in Fig. 5.22. The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 m s-2. At what distance from the starting point does the box fall off the truck? (Ignore the size of the box). 5.37 A disc revolves with a speed of 1 rev/min, and has a radius of 15 cm. Two coins are 33 3 placed at 4 cm and 14 cm away from the centre of the record. If the co-efficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record ? 5.38 You may have seen in a circus a motorcyclist driving in vertical loops inside a ‘death- well’ (a hollow spherical chamber with holes, so the spectators can watch from outside). Explain clearly why the motorcyclist does not drop down when he is at the uppermost point, with no support from below. What is the minimum speed required at the uppermost position to perform a vertical loop if the radius of the chamber is 25 m ? 5.39 A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis with 200 rev/min. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed ? 5.40 A thin circular loop of radius R rotates about its vertical diameter with an angular frequency ω. Show that a small bead on the wire loop remains at its lowermost point for ω ≤ g / R . What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for ω = 2g / R ? Neglect friction. 2020-21

CHAPTER SIX WORK, ENERGY AND POWER 6.1 Introduction 6.1 INTRODUCTION 6.2 Notions of work and kinetic The terms ‘work’, ‘energy’ and ‘power’ are frequently used in everyday language. A farmer ploughing the field, a energy : The work-energy construction worker carrying bricks, a student studying for theorem a competitive examination, an artist painting a beautiful landscape, all are said to be working. In physics, however, 6.3 Work the word ‘Work’ covers a definite and precise meaning. 6.4 Kinetic energy Somebody who has the capacity to work for 14-16 hours a 6.5 Work done by a variable day is said to have a large stamina or energy. We admire a long distance runner for her stamina or energy. Energy is force thus our capacity to do work. In Physics too, the term ‘energy’ is related to work in this sense, but as said above the term 6.6 The work-energy theorem for ‘work’ itself is defined much more precisely. The word ‘power’ is used in everyday life with different shades of meaning. In a variable force karate or boxing we talk of ‘powerful’ punches. These are delivered at a great speed. This shade of meaning is close to 6.7 The concept of potential the meaning of the word ‘power’ used in physics. We shall find that there is at best a loose correlation between the energy physical definitions and the physiological pictures these terms generate in our minds. The aim of this chapter is to 6.8 The conservation of develop an understanding of these three physical quantities. Before we proceed to this task, we need to develop a mechanical energy mathematical prerequisite, namely the scalar product of two vectors. 6.9 The potential energy of a 6.1.1 The Scalar Product spring We have learnt about vectors and their use in Chapter 4. 6.10 Various forms of energy : the Physical quantities like displacement, velocity, acceleration, force etc. are vectors. We have also learnt how vectors are law of conservation of energy added or subtracted. We now need to know how vectors are multiplied. There are two ways of multiplying vectors which 6.11 Power we shall come across : one way known as the scalar product 6.12 Collisions gives a scalar from two vectors and the other known as the vector product produces a new vector from two vectors. We Summary shall look at the vector product in Chapter 7. Here we take Points to ponder up the scalar product of two vectors. The scalar product or Exercises dot product of any two vectors A and B, denoted as A.B (read Additional exercises Appendix 6.1 2020-21

WORK, ENERGY AND POWER 115 A dot B) is defined as (6.1a) A = Ax i + Ay j + Azk A.B = A B cos θ where θ is the angle between the two vectors as B = Bx i + By j + Bzk shown in Fig. 6.1(a). Since A, B and cos θ are their scalar product is scalars, the dot product of A and B is a scalar quantity. Each vector, A and B, has a direction ( ) ( )A.B = Ax ˆi + Ay ˆj + Az kˆ . Bxˆi + By ˆj + Bz kˆ but their scalar product does not have a = Ax Bx + Ay By + Az Bz (6.1b) direction. From the definition of scalar product and From Eq. (6.1a), we have (Eq. 6.1b) we have : A.B = A (B cos θ ) ( i ) A.A = Ax Ax + Ay Ay + Az Az = B (A cos θ ) Or, A2 = Ax2 + Ay2 + A 2 (6.1c) z Geometrically, B cos θ is the projection of B onto A in Fig.6.1 (b) and A cos θ is the projection of A since A.A = |A ||A| cos 0 = A2. onto B in Fig. 6.1 (c). So, A.B is the product of the magnitude of A and the component of B along (ii) A.B = 0, if A and B are perpendicular. A. Alternatively, it is the product of the magnitude of B and the component of A along B. Example 6.1 Find the angle between force F = (3 ˆi + 4 ˆj - 5 kˆ ) unit and displacement Equation (6.1a) shows that the scalar product d = (5 ˆi + 4 ˆj + 3 kˆ ) unit. Also find the follows the commutative law : projection of F on d. A.B = B.A Answer F.d = Fx dx + Fydy + Fzdz Scalar product obeys the distributive law: = 3 (5) + 4 (4) + (– 5) (3) = 16 unit A. (B + C) = A.B + A.C Hence F.d = F d cosθ = 16 unit Further, A. (λ B) = λ (A.B) where λ is a real number. Now F.F = F 2 = Fx2 + Fy2 + Fz2 and d.d = 9 + 16 + 25 The proofs of the above equations are left to = 50 unit you as an exercise. = d2 = dx2 + dy2 + dz2 For unit vectors i, j, k we have = 25 + 16 + 9 i⋅i = j⋅ j= k⋅k =1 = 50 unit i⋅ j = j⋅k = k⋅i = 0 ∴ cos θ = 16 = 16 = 0.32 , Given two vectors 50 50 50 θ = cos–1 0.32 Fig. 6.1 (a) The scalar product of two vectors A and B is a scalar : A.B = A B cos θ. (b) B cos θ is the projection of B onto A. (c) A cos θ is the projection of A onto B. 2020-21

116 PHYSICS 6.2 NOTIONS OF WORK AND KINETIC to be proportional to the speed of the drop ENERGY: THE WORK-ENERGY THEOREM but is otherwise undetermined. Consider a drop of mass 1.00 g falling from a height The following relation for rectilinear motion under 1.00 km. It hits the ground with a speed of 50.0 m s-1. (a) What is the work done by the constant acceleration a has been encountered gravitational force ? What is the work done by the unknown resistive force? in Chapter 3, v2 − u2 = 2 as (6.2) where u and v are the initial and final speeds Answer (a) The change in kinetic energy of the and s the distance traversed. Multiplying both drop is sides by m/2, we have 1 mv2 − 1 mu 2 = mas = Fs (6.2a) ∆K = 1 m v2 − 0 22 2 where the last step follows from Newton’s Second = 1 × 10-3 × 50 × 50 Law. We can generalise Eq. (6.2) to three 2 dimensions by employing vectors = 1.25 J v2 − u2 = 2 a.d where we have assumed that the drop is initially at rest. Here a and d are acceleration and displacement vectors of the object respectively. Assuming that g is a constant with a value Once again multiplying both sides by m/2 , we obtain 10 m/s2, the work done by the gravitational force is, 1 mv2 − 1 mu 2 = m a.d = F.d (6.2b) Wg = mgh ×10 ×103 22 = 10-3 The above equation provides a motivation for = 10.0 J the definitions of work and kinetic energy. The (b) From the work-energy theorem left side of the equation is the difference in the ∆K = Wg + Wr quantity ‘half the mass times the square of the where Wr is the work done by the resistive force on the raindrop. Thus speed’ from its initial value to its final value. We Wr = ∆K − Wg call each of these quantities the ‘kinetic energy’, = 1.25 −10 = − 8.75 J denoted by K. The right side is a product of the displacement and the component of the force along the displacement. This quantity is called ‘work’ and is denoted by W. Eq. (6.2b) is then is negative. Kf − Ki = W (6.3) 6.3 WORK where Ki and Kf are respectively the initial and As seen earlier, work is related to force and the final kinetic energies of the object. Work refers displacement over which it acts. Consider a to the force and the displacement over which it constant force F acting on an object of mass m. acts. Work is done by a force on the body over The object undergoes a displacement d in the a certain displacement. positive x-direction as shown in Fig. 6.2. Equation (6.2) is also a special case of the work-energy (WE) theorem : The change in kinetic energy of a particle is equal to the work done on it by the net force. We shall generalise the above derivation to a varying force in a later section. Example 6.2 It is well known that a Fig. 6.2 An object undergoes a displacement d raindrop falls under the influence of the under the influence of the force F. downward gravitational force and the opposing resistive force. The latter is known 2020-21

WORK, ENERGY AND POWER 117 Table 6.1 Alternative Units of Work/Energy in J The work done by the force is defined to be the product of component of the force in the direction of the displacement and the magnitude of this displacement. Thus W = (F cos θ )d = F.d (6.4) We see that if there is no displacement, there Example 6.3 A cyclist comes to a skidding is no work done even if the force is large. Thus, stop in 10 m. During this process, the force when you push hard against a rigid brick wall, on the cycle due to the road is 200 N and the force you exert on the wall does no work. Yet is directly opposed to the motion. (a) How your muscles are alternatively contracting and much work does the road do on the cycle ? relaxing and internal energy is being used up (b) How much work does the cycle do on and you do get tired. Thus, the meaning of work the road ? in physics is different from its usage in everyday language. No work is done if : Answer Work done on the cycle by the road is the work done by the stopping (frictional) force (i) the displacement is zero as seen in the on the cycle due to the road. (a) The stopping force and the displacement make example above. A weightlifter holding a 150 an angle of 180o (π rad) with each other. kg mass steadily on his shoulder for 30 s Thus, work done by the road, Wr = Fd cosθ does no work on the load during this time. = 200 × 10 × cos π (ii) the force is zero. A block moving on a smooth = – 2000 J horizontal table is not acted upon by a It is this negative work that brings the cycle to a halt in accordance with WE theorem. horizontal force (since there is no friction), but (b) From Newton’s Third Law an equal and opposite force acts on the road due to the may undergo a large displacement. cycle. Its magnitude is 200 N. However, the road undergoes no displacement. Thus, (iii) the force and displacement are mutually work done by cycle on the road is zero. perpendicular. This is so since, for θ = π/2 rad (= 90o), cos (π/2) = 0. For the block moving on The lesson of Example 6.3 is that though the force on a body A exerted by the body B is always a smooth horizontal table, the gravitational equal and opposite to that on B by A (Newton’s Third Law); the work done on A by B is not force mg does no work since it acts at right necessarily equal and opposite to the work done on B by A. angles to the displacement. If we assume that 6.4 KINETIC ENERGY the moon’s orbits around the earth is As noted earlier, if an object of mass m has perfectly circular then the earth’s velocity v, its kinetic energy K is gravitational force does no work. The moon’s instantaneous displacement is tangential while the earth’s force is radially inwards and θ = π/2. Work can be both positive and negative. If θ is between 0o and 90o, cos θ in Eq. (6.4) is positive. If θ is between 90o and 180o, cos θ is negative. In many examples the frictional force opposes displacement and θ = 180o. Then the work done by friction is negative (cos 180o = –1). From Eq. (6.4) it is clear that work and energy K = 1 m v.v = 1 mv 2 (6.5) have the same dimensions, [ML2T–2]. The SI unit 2 2 of these is joule (J), named after the famous British physicist James Prescott Joule (1811-1869). Since Kinetic energy is a scalar quantity. The kinetic work and energy are so widely used as physical energy of an object is a measure of the work an concepts, alternative units abound and some of these are listed in Table 6.1. 2020-21

118 PHYSICS Table 6.2 Typical kinetic energies (K) object can do by the virtue of its motion. This This is illustrated in Fig. 6.3(a). Adding notion has been intuitively known for a long time. successive rectangular areas in Fig. 6.3(a) we The kinetic energy of a fast flowing stream get the total work done as has been used to grind corn. Sailing ships employ the kinetic energy of the wind. Table ∑W ≅ x f F (x )∆x (6.6) 6.2 lists the kinetic energies for various objects. xi Example 6.4 In a ballistics demonstration where the summation is from the initial position a police officer fires a bullet of mass 50.0 g with speed 200 m s-1 (see Table 6.2) on soft xi to the final position xf. plywood of thickness 2.00 cm. The bullet emerges with only 10% of its initial kinetic If the displacements are allowed to approach energy. What is the emergent speed of the zero, then the number of terms in the sum bullet ? increases without limit, but the sum approaches a definite value equal to the area under the curve in Fig. 6.3(b). Then the work done is lim xf F (x )∆x lim ∆x → 0 Answer The initial kinetic energy of the bullet ∑W = is mv2/2 = 1000 J. It has a final kinetic energy xi of 0.1×1000 = 100 J. If vf is the emergent speed of the bullet, xf (6.7) 1 2 = ∫ F (x ) dx 2 f xi mv = 100 J where ‘lim’ stands for the limit of the sum when ∆x tends to zero. Thus, for a varying force vf = 2 ×100 J the work done can be expressed as a definite 0.05 kg integral of force over displacement (see also = 63.2 m s–1 Appendix 3.1). The speed is reduced by approximately 68% (not 90%). 6.5 WORK DONE BY A VARIABLE FORCE Fig. 6.3(a) A constant force is rare. It is the variable force, which is more commonly encountered. Fig. 6.3 is a plot of a varying force in one dimension. If the displacement ∆x is small, we can take the force F (x) as approximately constant and the work done is then ∆W =F (x) ∆x 2020-21

WORK, ENERGY AND POWER 119 The work done by the frictional force is Wf → area of the rectangle AGHI Wf = (−50) × 20 = − 1000 J The area on the negative side of the force axis has a negative sign. Fig. 6.3 (a) The shaded rectangle represents the 6.6 THE WORK-ENERGY THEOREM FOR A work done by the varying force F(x), over VARIABLE FORCE the small displacement ∆x, ∆W = F(x) ∆x. (b) adding the areas of all the rectangles we We are now familiar with the concepts of work find that for ∆x → 0, the area under the curve and kinetic energy to prove the work-energy is exactly equal to the work done by F(x). theorem for a variable force. We confine ourselves to one dimension. The time rate of Example 6.5 A woman pushes a trunk on change of kinetic energy is a railway platform which has a rough surface. She applies a force of 100 N over a dK = d  1 m v2  distance of 10 m. Thereafter, she gets dt dt 2 progressively tired and her applied force reduces linearly with distance to 50 N. The = m dv v total distance through which the trunk has dt been moved is 20 m. Plot the force applied by the woman and the frictional force, which =F v (from Newton’s Second Law) is 50 N versus displacement. Calculate the work done by the two forces over 20 m. = F dx dt Answer Thus dK = Fdx Integrating from the initial position (x i ) to final position ( x f ), we have Kf xf ∫ dK = ∫ Fdx Ki xi where, Ki and K f are the initial and final kinetic energies corresponding to x i and x f. Fig. 6.4 Plot of the force F applied by the woman and xf (6.8a) the opposing frictional force f versus displacement. ∫or K f − Ki = Fdx xi From Eq. (6.7), it follows that The plot of the applied force is shown in Fig. Kf − Ki = W (6.8b) 6.4. At x = 20 m, F = 50 N (≠ 0). We are given that the frictional force f is |f|= 50 N. It opposes Thus, the WE theorem is proved for a variable motion and acts in a direction opposite to F. It is therefore, shown on the negative side of the force. force axis. While the WE theorem is useful in a variety of The work done by the woman is problems, it does not, in general, incorporate the complete dynamical information of Newton’s second law. It is an integral form of Newton’s WF → area of the rectangle ABCD + area of second law. Newton’s second law is a relation the trapezium CEID between acceleration and force at any instant of = × + 1 (100 + 50) × time. Work-energy theorem involves an integral 2 WF 100 10 10 over an interval of time. In this sense, the temporal = 1000 + 750 (time) information contained in the statement of = 1750 J Newton’s second law is ‘integrated over’ and is 2020-21

120 PHYSICS not available explicitly. Another observation is that are like ‘compressed springs’. They possess a Newton’s second law for two or three dimensions large amount of potential energy. An earthquake is in vector form whereas the work-energy results when these fault lines readjust. Thus, theorem is in scalar form. In the scalar form, potential energy is the ‘stored energy’ by virtue information with respect to directions contained of the position or configuration of a body. The in Newton’s second law is not present. body left to itself releases this stored energy in the form of kinetic energy. Let us make our notion Example 6.6 A block of mass m = 1 kg, of potential energy more concrete. moving on a horizontal surface with speed vi = 2 m s–1 enters a rough patch ranging The gravitational force on a ball of mass m is from x = 0.10 m to x = 2.01 m. The retarding mg . g may be treated as a constant near the earth force Fr on the block in this range is inversely surface. By ‘near’ we imply that the height h of proportional to x over this range, the ball above the earth’s surface is very small compared to the earth’s radius RE (h <<RE) so that Fr = −k for 0.1 < x < 2.01 m we can ignore the variation of g near the earth’s x surface*. In what follows we have taken the = 0 for x < 0.1m and x > 2.01 m upward direction to be positive. Let us raise the where k = 0.5 J. What is the final kinetic ball up to a height h. The work done by the external agency against the gravitational force is mgh. This energy and speed vf of the block as it work gets stored as potential energy. crosses this patch ? Gravitational potential energy of an object, as a function of the height h, is denoted by V(h) and it Answer From Eq. (6.8a) is the negative of work done by the gravitational force in raising the object to that height. ∫ ( )K f −k 2.01 x dx V (h) = mgh If h is taken as a variable, it is easily seen that = Ki + the gravitational force F equals the negative of the derivative of V(h) with respect to h. Thus, 0.1 F = − d V(h) = −m g ( )=1 dh 2 mvi2 −k ln x 2.01 0.1 The negative sign indicates that the gravitational force is downward. When released, = 1 mvi2 −k ln (2.01/0.1) the ball comes down with an increasing speed. 2 Just before it hits the ground, its speed is given by the kinematic relation, = 2 − 0.5 ln (20.1) v2 = 2gh = 2 − 1.5 = 0.5 J This equation can be written as v f = 2K f /m = 1 m s−1 1 2 m v2 = m g h Here, note that ln is a symbol for the natural logarithm to the base e and not the logarithm to which shows that the gravitational potential the base 10 [ln X = loge X = 2.303 log10 X]. energy of the object at height h, when the object is released, manifests itself as kinetic energy of 6.7 THE CONCEPT OF POTENTIAL ENERGY the object on reaching the ground. The word potential suggests possibility or Physically, the notion of potential energy is capacity for action. The term potential energy applicable only to the class of forces where work brings to one’s mind ‘stored’ energy. A stretched done against the force gets ‘stored up’ as energy. bow-string possesses potential energy. When it When external constraints are removed, it is released, the arrow flies off at a great speed. manifests itself as kinetic energy. Mathematically, The earth’s crust is not uniform, but has (for simplicity, in one dimension) the potential discontinuities and dislocations that are called fault lines. These fault lines in the earth’s crust * The variation of g with height is discussed in Chapter 8 on Gravitation. 2020-21

WORK, ENERGY AND POWER 121 energy V(x) is defined if the force F(x) can be which means that K + V, the sum of the kinetic written as and potential energies of the body is a constant. F (x ) = − dV Over the whole path, xi to xf, this means that dx Ki + V(xi ) = Kf + V(xf ) (6.11) This implies that The quantity K +V(x), is called the total xf Vf mechanical energy of the system. Individually ∫ F(x)dx = − ∫ dV = Vi − V f the kinetic energy K and the potential energy xi Vi V(x) may vary from point to point, but the sum is a constant. The aptness of the term The work done by a conservative force such as ‘conservative force’ is now clear. gravity depends on the initial and final positions only. In the previous chapter we have worked Let us consider some of the definitions of a on examples dealing with inclined planes. If an object of mass m is released from rest, from the conservative force. top of a smooth (frictionless) inclined plane of height h, its speed at the bottom A force F(x) is conservative if it can be derived is 2gh irrespective of the angle of inclination. from a scalar quantity V(x) by the relation given by Eq. (6.9). The three-dimensional Thus, at the bottom of the inclined plane it generalisation requires the use of a vector derivative, which is outside the scope of this acquires a kinetic energy, mgh. If the work done book. The work done by the conservative force or the kinetic energy did depend on other factors depends only on the end points. This can be seen from the relation, such as the velocity or the particular path taken W = Kf – Ki = V (xi ) – V(xf ) by the object, the force would be called non- which depends on the end points. A third definition states that the work done conservative. by this force in a closed path is zero. This is once again apparent from Eq. (6.11) since The dimensions of potential energy are xi = xf . [ML2T –2] and the unit is joule (J), the same as Thus, the principle of conservation of total mechanical energy can be stated as kinetic energy or work. To reiterate, the change in potential energy, for a conservative force, ∆V is equal to the negative of the work done by the force (6.9) The total mechanical energy of a system is ∆V = − F(x) ∆x conserved if the forces, doing work on it, are conservative. In the example of the falling ball considered in The above discussion can be made more this section we saw how potential energy was concrete by considering the example of the gravitational force once again and that of the converted to kinetic energy. This hints at an spring force in the next section. Fig. 6.5 depicts a ball of mass m being dropped from a cliff of important principle of conservation in mechanics, height H. which we now proceed to examine. 6.8 THE CONSERVATION OF MECHANICAL ENERGY For simplicity we demonstrate this important principle for one-dimensional motion. Suppose that a body undergoes displacement ∆x under the action of a conservative force F. Then from the WE theorem we have, ∆K = F(x) ∆x If the force is conservative, the potential energy function V(x) can be defined such that − ∆V = F(x) ∆x The above equations imply that (6.10) Fig. 6.5 The conversion of potential energy to kinetic ∆K + ∆V = 0 energy for a ball of mass m dropped from a ∆(K + V ) = 0 height H. 2020-21

122 PHYSICS The total mechanical energies E0, Eh, and EH Answer (i) There are two external forces on of the ball at the indicated heights zero (ground the bob : gravity and the tension (T ) in the level), h and H, are string. The latter does no work since the displacement of the bob is always normal to the EH = mgH (6.11 a) string. The potential energy of the bob is thus associated with the gravitational force only. The Eh = mgh + 1 mvh2 (6.11 b) total mechanical energy E of the system is 2 (6.11 c) conserved. We take the potential energy of the E0 = (1/2) mvf2 system to be zero at the lowest point A. Thus, at A : The constant force is a special case of a spatially dependent force F(x). Hence, the mechanical energy is conserved. Thus EH = E0 = 1 mv02 2 mgH = 1 mv 2 E (6.12) 2 f or, v f = 2gH [Newton’s Second Law] a result that was obtained in section 3.7 for a where TA is the tension in the string at A. At the freely falling body. highest point C, the string slackens, as the Further, tension in the string (TC) becomes zero. Thus, at C EH = Eh which implies, v 2 = 2g(H − h) (6.11 d) E = 1 mvc2 + 2mgL (6.13) h 2 and is a familiar result from kinematics. At the height H, the energy is purely potential. mg = mvc2 L It is partially converted to kinetic at height h and [Newton’s Second Law] (6.14) is fully kinetic at ground level. This illustrates the conservation of mechanical energy. where vC is the speed at C. From Eqs. (6.13) and (6.14) Example 6.7 A bob of mass m is suspended by a light string of length L . It is imparted a E = 5 mgL horizontal velocity vo at the lowest point A 2 such that it completes a semi-circular trajectory in the vertical plane with the string Equating this to the energy at A becoming slack only on reaching the topmost point, C. This is shown in Fig. 6.6. Obtain an 5 mgL = m v02 expression for (i) vo; (ii) the speeds at points 2 2 B and C; (iii) the ratio of the kinetic energies (KB/KC) at B and C. Comment on the nature or, v0 = 5gL of the trajectory of the bob after it reaches the point C. (ii) It is clear from Eq. (6.14) vC = gL At B, the energy is E = 1 mv 2 + mgL 2 B Equating this to the energy at A and employing the result from (i), namely v02 = 5gL , 1 mv 2 + mgL = 1 mv02 2 B 2 Fig. 6.6 = 5m g L 2 2020-21

WORK, ENERGY AND POWER 123 ∴ vB = 3gL W = +k xm2 (6.16) 2 (iii) The ratio of the kinetic energies at B and C is : KB = 1mv 2 =3 2 B KC 12mvC2 1 At point C, the string becomes slack and the velocity of the bob is horizontal and to the left. If the connecting string is cut at this instant, the bob will execute a projectile motion with horizontal projection akin to a rock kicked horizontally from the edge of a cliff. Otherwise the bob will continue on its circular path and complete the revolution. 6.9 THE POTENTIAL ENERGY OF A SPRING The spring force is an example of a variable force which is conservative. Fig. 6.7 shows a block attached to a spring and resting on a smooth horizontal surface. The other end of the spring is attached to a rigid wall. The spring is light and may be treated as massless. In an ideal spring, the spring force Fs is proportional to x where x is the displacement of the block from the equilibrium position. The displacement could be either positive [Fig. 6.7(b)] or negative [Fig. 6.7(c)]. This force law for the spring is called Hooke’s law and is mathematically stated as Fs = − kx Fig. 6.7 Illustration of the spring force with a block The constant k is called the spring constant. Its attached to the free end of the spring. unit is N m-1. The spring is said to be stiff if k is large and soft if k is small. (a) The spring force Fs is zero when the displacement x from the equilibrium position Suppose that we pull the block outwards as in Fig. 6.7(b). If the extension is xm, the work done by the spring force is is zero. (b) For the stretched spring x > 0 xm xm and Fs < 0 (c) For the compressed spring x < 0 and Fs > 0.(d) The plot of Fs versus x. ∫Ws = Fs dx = − ∫ kx dx The area of the shaded triangle represents 0 0 2 the work done by the spring force. Due to the m = − k x (6.15) opposing signs of Fs and x, this work done is 2 negative, Ws = −kx 2 / 2. m This expression may also be obtained by considering the area of the triangle as in The same is true when the spring is Fig. 6.7(d). Note that the work done by the compressed with a displacement xc (< 0). The external pulling force F is positive since it overcomes the spring force. spring force does work Ws = − kxc2 /2 while the 2020-21

124 PHYSICS external force F does work + kx 2 / 2. If the block and vice versa, however, the total mechanical c energy remains constant. This is graphically depicted in Fig. 6.8. is moved from an initial displacement xi to a final displacement xf , the work done by the spring force Ws is xf = k x 2 − k x 2 (6.17) i f Ws = − ∫ k x dx xi 22 Thus the work done by the spring force depends only on the end points. Specifically, if the block is pulled from xi and allowed to return to xi ; xi = k x 2 − k x 2 i i Ws = − ∫ k x dx xi 22 = 0 (6.18) Fig. 6.8 Parabolic plots of the potential energy V and The work done by the spring force in a cyclic kinetic energy K of a block attached to a process is zero. We have explicitly demonstrated spring obeying Hooke’s law. The two plots that the spring force (i) is position dependent are complementary, one decreasing as the only as first stated by Hooke, (Fs = − kx); (ii) other increases. The total mechanical does work which only depends on the initial and energy E = K + V remains constant. final positions, e.g. Eq. (6.17). Thus, the spring force is a conservative force. Example 6.8 To simulate car accidents, auto manufacturers study the collisions of moving We define the potential energy V(x) of the spring cars with mounted springs of different spring to be zero when block and spring system is in the constants. Consider a typical simulation with equilibrium position. For an extension (or a car of mass 1000 kg moving with a speed compression) x the above analysis suggests that 18.0 km/h on a smooth road and colliding with a horizontally mounted spring of spring V(x) = kx 2 (6.19) constant 6.25 × 103 N m–1. What is the 2 maximum compression of the spring ? You may easily verify that − dV/dx = − k x, the spring force. If the block of mass m in Fig. 6.7 is extended to xm and released from rest, then its Answer At maximum compression the kinetic total mechanical energy at any arbitrary point x, energy of the car is converted entirely into the potential energy of the spring. where x lies between – xm and + xm, will be given by The kinetic energy of the moving car is 1k x 2 =1k x 2 + 1m v2 2 m 2 2 K = 1 mv2 2 where we have invoked the conservation of mechanical energy. This suggests that the speed = 1 ×103 ×5 ×5 2 and the kinetic energy will be maximum at the equilibrium position, x = 0, i.e., K = 1.25 × 104 J 1m vm2 = 1k x 2 where we have converted 18 km h–1 to 5 m s–1 [It is 2 2 m useful to remember that 36 km h–1 = 10 m s–1]. At maximum compression xm, the potential where vm is the maximum speed. energy V of the spring is equal to the kinetic energy K of the moving car from the principle of or vm = k xm conservation of mechanical energy. m Note that k/m has the dimensions of [T-2] and V = 1k x 2 our equation is dimensionally correct. The 2 m kinetic energy gets converted to potential energy 2020-21

WORK, ENERGY AND POWER 125 = 1.25 × 104 J We obtain xm = 2.00 m We note that we have idealised the situation. The spring is considered to be massless. The surface has been considered to possess negligible friction. We conclude this section by making a few Fig. 6.9 The forces acting on the car. remarks on conservative forces. (i) Information on time is absent from the above ∆K = Kf − Ki = 0 − 1m v2 discussions. In the example considered 2 above, we can calculate the compression, but The work done by the net force is not the time over which the compression occurs. A solution of Newton’s Second Law W = − 1 kx 2 − µm g xm for this system is required for temporal 2 m information. Equating we have (ii) Not all forces are conservative. Friction, for example, is a non-conservative force. The 1 m v2 = 1k x 2 + µm g xm principle of conservation of energy will have 2 2 m to be modified in this case. This is illustrated in Example 6.9. Now µmg = 0.5 × 103 × 10 = 5 × 103 N (taking (iii) The zero of the potential energy is arbitrary. g =10.0 m s-2). After rearranging the above It is set according to convenience. For the spring force we took V(x) = 0, at x = 0, i.e. the equation we obtain the following quadratic unstretched spring had zero potential energy. For the constant gravitational force equation in the unknown xm. mg, we took V = 0 on the earth’s surface. In a later chapter we shall see that for the force k x 2 + 2µm g xm −m v2 =0 due to the universal law of gravitation, the m zero is best defined at an infinite distance from the gravitational source. However, once where we take the positive square root since the zero of the potential energy is fixed in a given discussion, it must be consistently xm is positive. Putting in numerical values we adhered to throughout the discussion. You obtain cannot change horses in midstream ! xm = 1.35 m which, as expected, is less than the result in Example 6.8. If the two forces on the body consist of a Example 6.9 Consider Example 6.8 taking conservative force Fc and a non-conservative the coefficient of friction, µ, to be 0.5 and force Fnc , the conservation of mechanical energy calculate the maximum compression of the formula will have to be modified. By the WE spring. theorem (Fc+ Fnc ) ∆x = ∆K Answer In presence of friction, both the spring Fc ∆x = − ∆V force and the frictional force act so as to oppose But ∆(K + V) = Fnc ∆x the compression of the spring as shown in Hence, ∆E = Fnc ∆x Fig. 6.9. where E is the total mechanical energy. Over We invoke the work-energy theorem, rather than the conservation of mechanical energy. the path this assumes the form The change in kinetic energy is Ef − Ei = Wnc where Wnc is the total work done by the non-conservative forces over the path. Note that 2020-21

126 PHYSICS unlike the conservative force, Wnc depends on Chemical energy arises from the fact that the the particular path i to f. molecules participating in the chemical reaction have different binding energies. A stable chemical 6.10 VARIOUS FORMS OF ENERGY : THE LAW compound has less energy than the separated parts. OF CONSERVATION OF ENERGY A chemical reaction is basically a rearrangement of atoms. If the total energy of the reactants is more In the previous section we have discussed than the products of the reaction, heat is released mechanical energy. We have seen that it can be and the reaction is said to be an exothermic classified into two distinct categories : one based reaction. If the reverse is true, heat is absorbed and on motion, namely kinetic energy; the other on the reaction is endothermic. Coal consists of configuration (position), namely potential energy. carbon and a kilogram of it when burnt releases Energy comes in many a forms which transform about 3 × 107 J of energy. into one another in ways which may not often be clear to us. Chemical energy is associated with the forces that give rise to the stability of substances. These 6.10.1 Heat forces bind atoms into molecules, molecules into polymeric chains, etc. The chemical energy We have seen that the frictional force is not a conservative force. However, work is associated arising from the combustion of coal, cooking gas, with the force of friction, Example 6.5. A block of wood and petroleum is indispensable to our daily mass m sliding on a rough horizontal surface existence. with speed v0 comes to a halt over a distance x0. The work done by the force of kinetic friction f 6.10.3 Electrical Energy over x0 is –f x0. By the work-energy theorem The flow of electrical current causes bulbs to m vo2/2 = f x0. If we confine our scope to glow, fans to rotate and bells to ring. There are mechanics, we would say that the kinetic energy laws governing the attraction and repulsion of of the block is ‘lost’ due to the frictional force. charges and currents, which we shall learn On examination of the block and the table we later. Energy is associated with an electric would detect a slight increase in their current. An urban Indian household consumes temperatures. The work done by friction is not about 200 J of energy per second on an average. ‘lost’, but is transferred as heat energy. This raises the internal energy of the block and the 6.10.4 The Equivalence of Mass and Energy table. In winter, in order to feel warm, we generate heat by vigorously rubbing our palms Till the end of the nineteenth century, physicists together. We shall see later that the internal believed that in every physical and chemical energy is associated with the ceaseless, often process, the mass of an isolated system is random, motion of molecules. A quantitative idea conserved. Matter might change its phase, e.g. of the transfer of heat energy is obtained by glacial ice could melt into a gushing stream, but noting that 1 kg of water releases about 42000 J matter is neither created nor destroyed; Albert of energy when it cools by10 °C. Einstein (1879-1955) however, showed that mass and energy are equivalent and are related by 6.10.2 Chemical Energy the relation One of the greatest technical achievements of E = m c2 (6.20) humankind occurred when we discovered how to ignite and control fire. We learnt to rub two where c, the speed of light in vacuum is flint stones together (mechanical energy), got them to heat up and to ignite a heap of dry leaves approximately 3 ×108 m s–1. Thus, a staggering (chemical energy), which then provided sustained warmth. A matchstick ignites into a amount of energy is associated with a mere bright flame when struck against a specially prepared chemical surface. The lighted kilogram of matter matchstick, when applied to a firecracker, results in a spectacular display of sound and E = 1× (3 ×108)2 J = 9 ×1016 J. light. This is equivalent to the annual electrical output of a large (3000 MW) power generating station. 6.10.5 Nuclear Energy The most destructive weapons made by man, the fission and fusion bombs are manifestations of 2020-21

WORK, ENERGY AND POWER 127 Table 6.3 Approximate energy associated with various phenomena the above equivalence of mass and energy [Eq. Example 6.10 Examine Tables 6.1-6.3 (6.20)]. On the other hand the explanation of the and express (a) The energy required to life-nourishing energy output of the sun is also break one bond in DNA in eV; (b) The based on the above equation. In this case kinetic energy of an air molecule (10—21 J) effectively four light hydrogen nuclei fuse to form in eV; (c) The daily intake of a human adult a helium nucleus whose mass is less than the in kilocalories. sum of the masses of the reactants. This mass difference, called the mass defect ∆m is the Answer (a) Energy required to break one bond source of energy (∆m)c2. In fission, a heavy of DNA is nucleus like uranium 235 U, is split by a neutron 10−20 J 92 1.6 ×10−19 J/eV ~ 0.06 eV into lighter nuclei. Once again the final mass is less than the initial mass and the mass difference translates into energy, which can be tapped to Note 0.1 eV = 100 meV (100 millielectron volt). provide electrical energy as in nuclear power plants (controlled nuclear fission) or can be (b) The kinetic energy of an air molecule is employed in making nuclear weapons 10−21 J 1.6 ×10−19 J/eV (uncontrolled nuclear fission). Strictly, the energy ~ 0.0062 eV ∆E released in a chemical reaction can also be related to the mass defect ∆m = ∆E/c2. However, This is the same as 6.2 meV. for a chemical reaction, this mass defect is much (c) The average human consumption in a day is smaller than for a nuclear reaction. Table 6.3 107 ×103 lists the total energies for a variety of events and J ~ 2400 kcal J/kcal phenomena. 4.2 2020-21

128 PHYSICS We point out a common misconception created Pav = W by newspapers and magazines. They mention t food values in calories and urge us to restrict diet intake to below 2400 calories. What they The instantaneous power is defined as the should be saying is kilocalories (kcal) and not limiting value of the average power as time calories. A person consuming 2400 calories a interval approaches zero, day will soon starve to death! 1 food calorie is 1 kcal. P = dW (6.21) dt 6.10.6 The Principle of Conservation of Energy The work dW done by a force F for a displacement dr is dW = F.dr. The instantaneous power can We have seen that the total mechanical energy also be expressed as of the system is conserved if the forces doing work on it are conservative. If some of the forces P = F. dr involved are non-conservative, part of the dt mechanical energy may get transformed into other forms such as heat, light and sound. = F.v (6.22) However, the total energy of an isolated system does not change, as long as one accounts for all where v is the instantaneous velocity when the forms of energy. Energy may be transformed from force is F. one form to another but the total energy of an isolated system remains constant. Energy can Power, like work and energy, is a scalar neither be created, nor destroyed. quantity. Its dimensions are [ML2T–3]. In the SI, its unit is called a watt (W). The watt is 1 J s–1. Since the universe as a whole may be viewed The unit of power is named after James Watt, as an isolated system, the total energy of the one of the innovators of the steam engine in the universe is constant. If one part of the universe eighteenth century. loses energy, another part must gain an equal amount of energy. There is another unit of power, namely the horse-power (hp) The principle of conservation of energy cannot be proved. However, no violation of this principle 1 hp = 746 W has been observed. The concept of conservation This unit is still used to describe the output of and transformation of energy into various forms automobiles, motorbikes, etc. links together various branches of physics, chemistry and life sciences. It provides a We encounter the unit watt when we buy unifying, enduring element in our scientific electrical goods such as bulbs, heaters and pursuits. From engineering point of view all refrigerators. A 100 watt bulb which is on for 10 electronic, communication and mechanical hours uses 1 kilowatt hour (kWh) of energy. devices rely on some forms of energy transformation. 100 (watt) × 10 (hour) = 1000 watt hour 6.11 POWER =1 kilowatt hour (kWh) = 103 (W) × 3600 (s) Often it is interesting to know not only the work = 3.6 × 106 J done on an object, but also the rate at which this work is done. We say a person is physically Our electricity bills carry the energy fit if he not only climbs four floors of a building consumption in units of kWh. Note that kWh is but climbs them fast. Power is defined as the a unit of energy and not of power. time rate at which work is done or energy is transferred. Example 6.11 An elevator can carry a maximum load of 1800 kg (elevator + The average power of a force is defined as the passengers) is moving up with a constant ratio of the work, W, to the total time t taken speed of 2 m s–1. The frictional force opposing the motion is 4000 N. Determine the minimum power delivered by the motor to the elevator in watts as well as in horse power. 2020-21

WORK, ENERGY AND POWER 129 Answer The downward force on the elevator is by the second particle. F21 is likewise the force F = m g + F = (1800 × 10) + 4000 = 22000 N exerted on the second particle by the first particle. Now from Newton’s third law, F12 = − F21. This f implies The motor must supply enough power to balance ∆p1 + ∆p2 = 0 this force. Hence, P = F. v = 22000 × 2 = 44000 W = 59 hp The above conclusion is true even though the forces vary in a complex fashion during the 6.12 COLLISIONS collision time ∆t. Since the third law is true at every instant, the total impulse on the first object In physics we study motion (change in position). is equal and opposite to that on the second. At the same time, we try to discover physical quantities, which do not change in a physical On the other hand, the total kinetic energy of process. The laws of momentum and energy the system is not necessarily conserved. The conservation are typical examples. In this impact and deformation during collision may section we shall apply these laws to a commonly generate heat and sound. Part of the initial kinetic encountered phenomena, namely collisions. energy is transformed into other forms of energy. Several games such as billiards, marbles or A useful way to visualise the deformation during carrom involve collisions.We shall study the collision is in terms of a ‘compressed spring’. If collision of two masses in an idealised form. the ‘spring’ connecting the two masses regains its original shape without loss in energy, then Consider two masses m1 and m2. The particle the initial kinetic energy is equal to the final m1 is moving with speed v1i , the subscript ‘i’ kinetic energy but the kinetic energy during the implying initial. We can cosider m2 to be at rest. collision time ∆t is not constant. Such a collision No loss of generality is involved in making such is called an elastic collision. On the other hand a selection. In this situation the mass m1 the deformation may not be relieved and the two collides with the stationary mass m2 and this bodies could move together after the collision. A is depicted in Fig. 6.10. collision in which the two particles move together after the collision is called a completely inelastic collision. The intermediate case where the deformation is partly relieved and some of the initial kinetic energy is lost is more common and is appropriately called an inelastic collision. Fig. 6.10 Collisionofmass m1,withastationarymassm2. 6.12.2 Collisions in One Dimension The masses m1 and m2 fly-off in different Consider first a completely inelastic collision directions. We shall see that there are in one dimension. Then, in Fig. 6.10, relationships, which connect the masses, the velocities and the angles. θ1 = θ2 = 0 m1v1i = (m1+m2)vf (momentum conservation) 6.12.1 Elastic and Inelastic Collisions vf = m1 v1i (6.23) m1 + m2 In all collisions the total linear momentum is The loss in kinetic energy on collision is conserved; the initial momentum of the system is equal to the final momentum of the system. ∆K = 1 m1v12i − 1 (m1 + m 2 )v 2 One can argue this as follows. When two objects 2 2 f collide, the mutual impulsive forces acting over the collision time ∆t cause a change in their = 1 m1v12i −1 m12 v12i [using Eq. (6.23)] respective momenta : 2 2 m1 + m2 ∆∆pp12 = F12 ∆t = 1 m1v12i  − m1  = F21 ∆t 2 1 m1 + m2  where F12 is the force exerted on the first particle 2020-21

130 PHYSICS An experiment on head-on collision In performing an experiment on collision on a horizontal surface, we face three difficulties. One, there will be friction and bodies will not travel with uniform velocities. Two, if two bodies of different sizes collide on a table, it would be difficult to arrange them for a head-on collision unless their centres of mass are at the same height above the surface. Three, it will be fairly difficult to measure velocities of the two bodies just before and just after collision. By performing this experiment in a vertical direction, all the three difficulties vanish. Take two balls, one of which is heavier (basketball/football/volleyball) and the other lighter (tennis ball/rubber ball/table tennis ball). First take only the heavier ball and drop it vertically from some height, say 1 m. Note to which it rises. This gives the velocities near the floor or ground, just before and just after the bounce (by using v2 = 2gh ). Hence you will get the coefficient of restitution. Now take the big ball and a small ball and hold them in your hands one over the other, with the heavier ball below the lighter one, as shown here. Drop them together, taking care that they remain together while falling, and see what happens. You will find that the heavier ball rises less than when it was dropped alone, while the lighter one shoots up to about 3 m. With practice, you will be able to hold the ball properly so that the lighter ball rises vertically up and does not fly sideways. This is head-on collision. You can try to find the best combination of balls which gives you the best effect. You can measure the masses on a standard balance. We leave it to you to think how you can determine the initial and final velocities of the balls. 1 m1m 2 v1 f = (m1 − m 2 ) v1i (6.27) 2 m1 + m2 m1 + m 2 = v12i which is a positive quantity as expected. and v2 f = 2m1v1i (6.28) m1 + m2 Consider next an elastic collision. Using the Thus, the ‘unknowns’ {v1f, v2f} are obtained in terms of the ‘knowns’ {m1, m2, v1i}. Special cases above nomenclature with θ = θ = 0, the of our analysis are interesting. 1 2 momentum and kinetic energy conservation equations are Case I : If the two masses are equal m1v1i = m1v1f + m2v2f (6.24) v1f = 0 v2f = v1i m1v12i = m1v12f + m 2v22 f (6.25) The first mass comes to rest and pushes off the From Eqs. (6.24) and (6.25) it follows that, second mass with its initial speed on collision. m1v1i (v2 f − v1i ) = m1v1 f (v2 f − v1 f ) Case II : If one mass dominates, e.g. m2 > > m1 or, v2 f (v1i − v1 f ) = v12i − v12f v1f ~ − v1i v2f ~ 0 The heavier mass is undisturbed while the lighter mass reverses its velocity. = (v1i − v1 f )(v1i + v1 f ) (6.26) Example 6.12 Slowing down of neutrons: Hence, ∴ v2 f = v1i + v1 f In a nuclear reactor a neutron of high speed (typically 107 m s–1) must be slowed Substituting this in Eq. (6.24), we obtain 2020-21

WORK, ENERGY AND POWER 131 to 103 m s–1 so that it can have a high 6.12.3 Collisions in Two Dimensions probability of interacting with isotope 235 U Fig. 6.10 also depicts the collision of a moving 92 mass m1 with the stationary mass m2. Linear and causing it to fission. Show that a momentum is conserved in such a collision. neutron can lose most of its kinetic energy Since momentum is a vector this implies three in an elastic collision with a light nuclei equations for the three directions {x, y, z}. like deuterium or carbon which has a mass Consider the plane determined by the final of only a few times the neutron mass. The velocity directions of m1 and m2 and choose it to be the x-y plane. The conservation of the material making up the light nuclei, usually z-component of the linear momentum implies heavy water (D2O) or graphite, is called a moderator. that the entire collision is in the x-y plane. The Answer The initial kinetic energy of the neutron x- and y-component equations are is m1v1i = m1v1f cos θ 1 + m2v2f cos θ 2 (6.29) (6.30) K1i = 1 m1v12i 0 = m1v1f sin θ − m2v2f sin θ 2 1 2 while its final kinetic energy from Eq. (6.27) One knows {m1, m2, v1i} in most situations. There are thus four unknowns {v1f , v2f , θ and θ2}, and 1 only two equations. If θ 1 = θ 2 = 0, we regain K1 f = 1 m1v12f = 1 m1  m1 − m2 2 v12i 2 2  m1 + m2  Eq. (6.24) for one dimensional collision. If, further the collision is elastic, The fractional kinetic energy lost is 1 m1v1i 2 = 1 m1v1 2 + 1 m2v2 2 (6.31) 2 2 2 f f f1 = K1 f =  m1 − m2 2 We obtain an additional equation. That still K1i  m1 + m2  leaves us one equation short. At least one of while the fractional kinetic energy gained by the the tfohuerpurnobknleomwntos,bseaysoθl1v,ambules.tFboermeaxdaemkpnleo,wθn1 moderating nuclei K2f /K1i is for f2 = 1 − f1 (elastic collision) can be determined by moving a detector in an ( )= 4m1m2 angular fashion from the x to the y axis. Given {m1, m2, v1i , θ1} we can determine {v1f , v2f , θ2} m1 + m2 2 from Eqs. (6.29)-(6.31). One can also verify this result by substituting Example 6.13 Consider the collision from Eq. (6.28). depicted in Fig. 6.10 to be between two For deuterium m2 = 2m1 and we obtain f1 = 1/9 while f2 = 8/9. Almost 90% of the billiard balls with equal masses m1 = m2. neutron’s energy is transferred to deuterium. For The first ball is called the cue while the carbon f1 = 71.6% and f2 = 28.4%. In practice, however, this number is smaller since head-on second ball is called the target. The collisions are rare. billiard player wants to ‘sink’ the target ball in a corner pocket, which is at an angle θ = 37°. Assume that the collision 2 is elastic and that friction and rotational If the initial velocities and final velocities of motion are not important. Obtain θ 1. both the bodies are along the same straight line, then it is called a one-dimensional collision, or Answer From momentum conservation, since head-on collision. In the case of small spherical the masses are equal bodies, this is possible if the direction of travel of body 1 passes through the centre of body 2 v1i = v1f + v2f which is at rest. In general, the collision is two- dimensional, where the initial velocities and the ( ) ( )or v 1i 2= v1 f + v2 f ⋅ v1 f + v2 f final velocities lie in a plane. = v1 f 2 + v2 f 2 + 2v1 f .v2 f 2020-21

132 PHYSICS { }( )= v1 f 2 + v2 f 2 + 2v1 f v2 f cos θ1 + 37° (6.32) The matter simplifies greatly if we consider spherical masses with smooth surfaces, and Since the collision is elastic and m1 = m2 it follows assume that collision takes place only when the from conservation of kinetic energy that bodies touch each other. This is what happens in the games of marbles, carrom and billiards. v1i 2 = v1 f 2 + v2 f 2 (6.33) In our everyday world, collisions take place only Comparing Eqs. (6.32) and (6.33), we get when two bodies touch each other. But consider a comet coming from far distances to the sun, or cos (θ1 + 37°) = 0 alpha particle coming towards a nucleus and going away in some direction. Here we have to or θ + 37° = 90° deal with forces involving action at a distance. 1 Such an event is called scattering. The velocities and directions in which the two particles go away Thus, θ = 53° depend on their initial velocities as well as the 1 type of interaction between them, their masses, shapes and sizes. This proves the following result : when two equal masses undergo a glancing elastic collision with one of them at rest, after the collision, they will move at right angles to each other. SUMMARY 1. The work-energy theorem states that the change in kinetic energy of a body is the work done by the net force on the body. Kf - Ki = Wnet 2. A force is conservative if (i) work done by it on an object is path independent and depends only on the end points {xi, xj}, or (ii) the work done by the force is zero for an arbitrary closed path taken by the object such that it returns to its initial position. 3. For a conservative force in one dimension, we may define a potential energy function V(x) such that F (x ) = − dV (x ) dx xf or Vi − V f = ∫ F (x ) dx xi 4. The principle of conservation of mechanical energy states that the total mechanical energy of a body remains constant if the only forces that act on the body are conservative. 5. The gravitational potential energy of a particle of mass m at a height x about the earth’s surface is V(x) = m g x where the variation of g with height is ignored. 6. The elastic potential energy of a spring of force constant k and extension x is V (x) = 1 k x 2 2 7. The scalar or dot product of two vectors A and B is written as A.B and is a scalar quantity given by : A.B = AB cos θ, where θ is the angle between A and B. It can be positive, negative or zero depending upon the value of θ. The scalar product of two vectors can be interpreted as the product of magnitude of one vector and component of the other vector along the first vector. For unit vectors : ˆi ⋅ ˆi = ˆj ⋅ ˆj = kˆ ⋅ kˆ = 1 and ˆi ⋅ ˆj = ˆj ⋅ kˆ = kˆ ⋅ ˆi = 0 Scalar products obey the commutative and the distributive laws. 2020-21

WORK, ENERGY AND POWER 133 POINTS TO PONDER 1. The phrase ‘calculate the work done’ is incomplete. We should refer (or imply clearly by context) to the work done by a specific force or a group of forces on a given body over a certain displacement. 2. Work done is a scalar quantity. It can be positive or negative unlike mass and kinetic energy which are positive scalar quantities. The work done by the friction or viscous force on a moving body is negative. 3. For two bodies, the sum of the mutual forces exerted between them is zero from Newton’s Third Law, F12 + F21 = 0 But the sum of the work done by the two forces need not always cancel, i.e. W12 + W21 ≠ 0 However, it may sometimes be true. 4. The work done by a force can be calculated sometimes even if the exact nature of the force is not known. This is clear from Example 6.2 where the WE theorem is used in such a situation. 5. The WE theorem is not independent of Newton’s Second Law. The WE theorem may be viewed as a scalar form of the Second Law. The principle of conservation of mechanical energy may be viewed as a consequence of the WE theorem for conservative forces. 6. The WE theorem holds in all inertial frames. It can also be extended to non- inertial frames provided we include the pseudoforces in the calculation of the net force acting on the body under consideration. 7. The potential energy of a body subjected to a conservative force is always undetermined upto a constant. For example, the point where the potential energy is zero is a matter of choice. For the gravitational potential energy mgh, the zero of the potential energy is chosen to be the ground. For the spring potential energy kx2/2 , the zero of the potential energy is the equilibrium position of the oscillating mass. 8. Every force encountered in mechanics does not have an associated potential energy. For example, work done by friction over a closed path is not zero and no potential energy can be associated with friction. 9. During a collision : (a) the total linear momentum is conserved at each instant of the collision ; (b) the kinetic energy conservation (even if the collision is elastic) applies after the collision is over and does not hold at every instant of the collision. In fact the two colliding objects are deformed and may be momentarily at rest with respect to each other. 2020-21

134 PHYSICS EXERCISES 6.1 The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative: (a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket. (b) work done by gravitational force in the above case, (c) work done by friction on a body sliding down an inclined plane, (d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity, (e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest. 6.2 A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the (a) work done by the applied force in 10 s, (b) work done by friction in 10 s, (c) work done by the net force on the body in 10 s, (d) change in kinetic energy of the body in 10 s, and interpret your results. 6.3 Given in Fig. 6.11 are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant. Fig. 6.11 2020-21

WORK, ENERGY AND POWER 135 6.4 The potential energy function for a Fig. 6.12 particle executing linear simple harmonic motion is given by V(x) = kx2/2, where k is the force constant of the oscillator. For k = 0.5 N m-1, the graph of V(x) versus x is shown in Fig. 6.12. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2 m. 6.5 Answer the following : (a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere? (b) Comets move around the sun in highly elliptical orbits. The gravitational force on the Fig. 6.13 comet due to the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why ? (c) An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth ? (d) In Fig. 6.13(i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig. 6.13(ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater ? 6.6 Underline the correct alternative : (a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered. (b) Work done by a body against friction always results in a loss of its kinetic/potential energy. (c) The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces on the system. (d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies. 6.7 State if each of the following statements is true or false. Give reasons for your answer. (a) In an elastic collision of two bodies, the momentum and energy of each body is conserved. (b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present. (c) Work done in the motion of a body over a closed loop is zero for every force in nature. (d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system. 6.8 Answer carefully, with reasons : (a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact) ? (b) Is the total linear momentum conserved during the short time of an elastic collision of two balls ? 2020-21

136 PHYSICS (c) What are the answers to (a) and (b) for an inelastic collision ? (d) If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic ? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy). 6.9 A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to (i) t1/2 (ii) t (iii) t3/2 (iv) t2 6.10 A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to (i) t1/2 (ii) t (iii) t3/2 (iv) t2 6.11 A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by F = −ˆi + 2 ˆj + 3 kˆ N where ˆi, ˆj, kˆ are unit vectors along the x-, y- and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis ? 6.12 An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton ? Obtain the ratio of their speeds. (electron mass = 9.11×10-31 kg, proton mass = 1.67×10–27 kg, 1 eV = 1.60 ×10–19 J). 6.13 A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey ? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 m s–1 ? 6.14 A molecule in a gas container hits a horizontal wall with speed 200 m s–1 and angle 30° with the normal, and rebounds with the same speed. Is momentum conserved in the collision ? Is the collision elastic or inelastic ? 6.15 A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m3 in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump ? 6.16 Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed V. If the collision is elastic, which of the following (Fig. 6.14) is a possible result after collision ? Fig. 6.14 2020-21