4.6 NEWTON’S THIRD LAW OF MOTION 2. The terms action and reaction in the third law The second law relates the external force on a may give a wrong impression that action body to its acceleration. What is the origin of the comes before reaction i.e action is the cause external force on the body ? What agency and reaction the effect. There is no cause- provides the external force ? The simple answer effect relation implied in the third law. The in Newtonian mechanics is that the external force on A by B and the force on B by A act force on a body always arises due to some other at the same instant. By the same reasoning, body. Consider a pair of bodies A and B. B gives any one of them may be called action and the rise to an external force on A. A natural question other reaction. is: Does A in turn give rise to an external force 3. Action and reaction forces act on different on B ? In some examples, the answer seems bodies, not on the same body. Consider a pair clear. If you press a coiled spring, the spring is of bodies A and B. According to the third law, compressed by the force of your hand. The FAB = – FBA (4.8) compressed spring in turn exerts a force on your (force on A by B) = – (force on B by A)hand and you can feel it. But what if the bodies are not in contact ? The earth pulls a stone Thus if we are considering the motion of any downwards due to gravity. Does the stone exert one body (A or B), only one of the two forces is a force on the earth ? The answer is not obvious relevant. It is an error to add up the two forces since we hardly see the effect of the stone on the and claim that the net force is zero. earth. The answer according to Newton is: Yes, However, if you are considering the system the stone does exert an equal and opposite force of two bodies as a whole, FAB and FBA are on the earth. We do not notice it since the earth internal forces of the system (A + B). They add is very massive and the effect of a small force on up to give a null force. Internal forces in a its motion is negligible. body or a system of particles thus cancel away Reprint 2025-26 LAWS OF MOTION 57 in pairs. This is an important fact that the force on the wall due to the ball is normal to enables the second law to be applicable to a the wall along the positive x-direction. The body or a system of particles (See Chapter 6). magnitude of force cannot be ascertained since the small time taken for the collision has not ⊳ Example 4.5 Two identical billiard balls been specified in the problem. strike a rigid wall with the same speed but Case (b) at different angles, and get reflected without = − m u sin 30 any change in speed, as shown in Fig. 4.6. ( p x ) initial = m u cos 30 , ( p y ) initial What is (i) the direction of the force on the wall due to each ball? (ii) the ratio of the = − m u sin 30 p x ) final = – m u cos 30 , ( p y ) final magnitudes of impulses imparted to the ( balls by the wall ? Note, while px changes sign after collision, py does not. Therefore, x-component of impulse = –2 m u cos 30° y-component of impulse = 0 The direction of impulse (and force) is the same as in (a) and is normal to the wall along the negative x direction. As before, using Newton’s third law, the force on the wall due to the ball is normal to the wall along the positive x direction. The ratio of the magnitudes of the impulses Fig. 4.6 imparted to the balls in (a) and (b) is 2 Answer An instinctive answer to (i) might be 2 m u/ 2 m u cos30 = ≈ 1.2 ⊳ ( ) 3 that the force on the wall in case (a) is normal to the wall, while that in case (b) is inclined at 30° 4.7 CONSERVATION OF MOMENTUM to the normal. This answer is wrong. The force on the wall is normal to the wall in both cases. The second and third laws of motion lead to How to find the force on the wall? The trick is an important consequence: the law of to consider the force (or impulse) on the ball conservation of momentum. Take a familiar due to the wall using the second law, and then example. A bullet is fired from a gun. If the force use the third law to answer (i). Let u be the speed on the bullet by the gun is F, the force on the gun of each ball before and after collision with the by the bullet is – F, according to the third law. wall, and m the mass of each ball. Choose the x The two forces act for a common interval of time and y axes as shown in the figure, and consider ∆t. According to the second law, F ∆t is the change the change in momentum of the ball in each in momentum of the bullet and – F ∆t is the case : change in momentum of the gun. Since initially, both are at rest, the change in momentum equals Case (a) the final momentum for each. Thus if pb is the initial = mu initial = 0 momentum of the bullet after firing and pg is the ( p x ) ( p y ) recoil momentum of the gun, pg = – pb i.e. pb + pg = 0 = 0. That is, the total momentum of the (bullet + ( p x )final = −mu ( p y )final gun) system is conserved. Thus in an isolated system (i.e. a system withImpulse is the change in momentum vector. no external force), mutual forces between pairsTherefore, of particles in the system can cause momentum x-component of impulse = – 2 m u change in individual particles, but since the y-component of impulse = 0 mutual forces for each pair are equal and Impulse and force are in the same direction. opposite, the momentum changes cancel in pairs Clearly, from above, the force on the ball due to and the total momentum remains unchanged. the wall is normal to the wall, along the negative This fact is known as the law of conservation x-direction. Using Newton’s third law of motion, of momentum : Reprint 2025-26 58 PHYSICS The total momentum of an isolated system 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 Fig. 4.7 Equilibrium under concurrent forces. respectively. By the Second Law In other words, the resultant of any two forces say F1 and F2, obtained by the parallelogram FAB ∆=t p ′A − p A and law of forces must be equal and opposite to the FBA ∆=t p ′B − p B third force, F3. As seen in Fig. 4.7, the three forces in equilibrium can be represented by the (where we have taken a common interval of time sides of a triangle with the vector arrows taken for both forces i.e. the time for which the two in the same sense. The result can be bodies are in contact.) generalised to any number of forces. A particle is in equilibrium under the action of forces F1,Since F AB = − FBA by the third law, F2,... Fn if they can be represented by the sides of a closed n-sided polygon with arrows directed p ′A − p A = − ( p ′B − p B ) in the same sense. i.e. p ′A + p ′B = p A + p B (4.9) Equation (4.11) implies that which shows that the total final momentum of F1x + F2x + F3x = 0 the isolated system equals its initial momentum. F1y + F2y + F3y = 0 Notice that this is true whether the collision is F1z + F2z + F3z = 0 (4.12) elastic or inelastic. In elastic collisions, there is where F1x, F1y and F1z are the components of F1a second condition that the total initial kinetic along x, y and z directions respectively. energy of the system equals the total final kinetic ⊳ energy (See Chapter 5). Example 4.6 See Fig. 4.8. A mass of 6 kg is suspended by a rope of length 2 m from the ceiling. A force of 50 N in the 4.8 EQUILIBRIUM OF A PARTICLE horizontal direction is applied at the mid- point P of the rope, as shown. What is theEquilibrium of a particle in mechanics refers to angle the rope makes with the vertical inthe situation when the net external force on the equilibrium ? (Take g = 10 m s-2). Neglectparticle is zero.* According to the first law, this 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 (4.10) i.e. the two forces on the particle must be equal and opposite. Equilibrium under three concurrent forces F1, F2 and F3 requires that (a) (b) (c) the vector sum of the three forces is zero. Fig. 4.8 F1 + F2 + F3 = 0 (4.11) * 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 6. Reprint 2025-26 LAWS OF MOTION 59 Answer Figures 4.8(b) and 4.8(c) are known as other types of supports), there are mutual free-body diagrams. Figure 4.8(b) is the free-body contact forces (for each pair of bodies) satisfying diagram of W and Fig. 4.8(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 Consider the equilibrium of the weight W. normal reaction. The component parallel to the Clearly,T2 = 6 × 10 = 60 N. surfaces in contact is called friction. Contact forces arise also when solids are in contact with Consider the equilibrium of the point P under fluids. For example, for a solid immersed in a the action of three forces - the tensions T1 and fluid, there is an upward bouyant force equal toT2, and the horizontal force 50 N. The horizontal the weight of the fluid displaced. The viscousand vertical components of the resultant force must vanish separately : force, air resistance, etc are also examples of contact forces (Fig. 4.9). T1 cos θ = T2 = 60 N Two other common forces are tension in a T1 sin θ = 50 N string and the force due to spring. When a spring which gives that is compressed or extended by an external force, 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 4.9 COMMON FORCES IN MECHANICS to the displacement from the unstretched state. In mechanics, we encounter several kinds of For an inextensible string, the force constant is forces. The gravitational force is, of course, very high. The restoring force in a string is called pervasive. Every object on the earth experiences tension. It is customary to use a constant tension the force of gravity due to the earth. Gravity also T throughout the string. This assumption is true governs the motion of celestial bodies. The for a string of negligible mass. gravitational force can act at a distance without We learnt that there are four fundamental the need of any intervening medium. forces in nature. Of these, the weak and strong All the other forces common in mechanics are forces appear in domains that do not concern contact forces.* As the name suggests, a contact us here. Only the gravitational and electrical force on an object arises due to contact with some forces are relevant in the context of mechanics. other object: solid or fluid. When bodies are in The different contact forces of mechanics contact (e.g. a book resting on a table, a system mentioned above fundamentally arise from of rigid bodies connected by rods, hinges and electrical forces. This may seem surprising Fig. 4.9 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. Reprint 2025-26 60 PHYSICS 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 certain limit), keeping the body at rest. Hence, itcollisions and impacts, etc. can ultimately be is called static friction. Static friction opposestraced to the electrical forces between the charged impending motion. The term impending motionconstituents of different bodies. The detailed means motion that would take place (but does microscopic origin of these forces is, however, not actually take place) under the applied force, complex and not useful for handling problems in if friction were absent. mechanics at the macroscopic scale. This is why We know from experience that as the applied they are treated as different types of forces with force exceeds a certain limit, the body begins to their characteristic properties determined move. It is found experimentally that the limiting empirically. value of static friction ( sf )max is independent of 4.9.1 Friction the area of contact and varies with the normal Let us return to the example of a body of mass m force(N) approximately as : at rest on a horizontal table. The force of gravity = µs N (4.13) ( f s )max (mg) is cancelled by the normal reaction force where µs is a constant of proportionality(N) of the table. Now suppose a force F is applied depending only on the nature of the surfaces in horizontally to the body. We know from contact. The constant µs is called the coefficientexperience that a small applied force may not of static friction. The law of static friction may be enough to move the body. But if the applied thus be written as force F were the only external force on the body, fs ≤ µs N (4.14) it must move with acceleration F/m, however small. Clearly, the body remains at rest because If the applied force F exceeds ( sf )max the body some other force comes into play in the begins to slide on the surface. It is found experi- horizontal direction and opposes the applied mentally that when relative motion has started, force F, resulting in zero net force on the body. the frictional force decreases from the static This force fs parallel to the surface of the body in maximum value ( sf )max . Frictional force thatcontact with the table is known as frictional opposes relative motion between surfaces inforce, or simply friction (Fig. 4.10(a)). The contact is called kinetic or sliding friction and issubscript stands for static friction to distinguish denoted by fk . Kinetic friction, like static fric-it from kinetic friction fk that we consider later tion, is found to be independent of the area of (Fig. 4.10(b)). Note that static friction does not contact. Further, it is nearly independent of the velocity. It satisfies a law similar to that for static friction: f k = µk N (4.15) where µk′ the coefficient of kinetic friction, depends only on the surfaces in contact. As mentioned above, experiments show that µk is Fig. 4.10 Static and sliding friction: (a) Impending less than µs. When relative motion has begun, motion of the body is opposed by static the acceleration of the body according to the friction. When external force exceeds the second law is ( F – fk )/m. For a body moving with maximum limit of static friction, the body constant velocity, F = fk. If the applied force on begins to move. (b) Once the body is in the body is removed, its acceleration is – fk /m motion, it is subject to sliding or kinetic friction and it eventually comes to a stop. which opposes relative motion between the The laws of friction given above do not have two surfaces in contact. Kinetic friction is the status of fundamental laws like those for usually less than the maximum value of static gravitational, electric and magnetic forces. They friction. are empirical relations that are only Reprint 2025-26 LAWS OF MOTION 61 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 Thus, when two bodies are in contact, each mg acting vertically downwards (ii) the normal experiences a contact force by the other. Friction, force N of the plane on the block, and (iii) the by definition, is the component of the contact force static frictional force fs opposing the impending parallel to the surfaces in contact, which opposes motion. In equilibrium, the resultant of these impending or actual relative motion between the forces must be zero. Resolving the weight mg two surfaces. Note that it is not motion, but along the two directions shown, we have 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 itsrelative to the train, it is in fact accelerating along with the train. What forces cause the acceleration maximum value, ( sf )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 µsslip 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 4.7 Determine the maximum = 0.27 ⊳ acceleration of the train in which a box ⊳ Example 4.9 What is the acceleration of lying on its floor will remain stationary, the block and trolley system shown in a given that the co-efficient of static friction Fig. 4.12(a), if the coefficient of kinetic friction between the box and the train’s floor is between the trolley and the surface is 0.04? 0.15. 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, 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 4.8 See Fig. 4.11. 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 ? (a) (b) (c) Fig. 4.11 Fig. 4.12 Reprint 2025-26 62 PHYSICS Answer As the string is inextensible, and the is the reason why discovery of the wheel has pully is smooth, the 3 kg block and the 20 kg been a major milestone in human history. trolley both have same magnitude of Rolling friction again has a complex origin, acceleration. Applying second law to motion of though somewhat different from that of static the block (Fig. 4.12(b)), and sliding friction. During rolling, the surfaces in contact get momentarily deformed a little, and 30 – T = 3a this results in a finite area (not a point) of theApply the second law to motion of the trolley (Fig. body being in contact with the surface. The net4.12(c)), effect is that the component of the contact force T – fk = 20 a. parallel to the surface opposes motion. Now fk = µk N, We often regard friction as something Here µk = 0.04, undesirable. In many situations, like in a N = 20 x 10 machine with different moving parts, friction = 200 N. does have a negative role. It opposes relative Thus the equation for the motion of the trolley is motion and thereby dissipates power in the form T – 0.04 x 200 = 20 a Or T – 8 = 20a. of heat, etc. Lubricants are a way of reducing kinetic friction in a machine. Another way is to 22 –2 These equations give a = m s = 0.96 m s-2 use ball bearings between two moving parts of a 23 and T = 27.1 N. ⊳ machine [Fig. 4.13(a)]. Since the rolling friction between ball bearings and the surfaces in Rolling friction contact is very small, power dissipation is reduced. A thin cushion of air maintainedA body like a ring or a sphere rolling without between solid surfaces in relative motion isslipping over a horizontal plane will suffer no another effective way of reducing friction friction, in principle. At every instant, there is (Fig. 4.13(a)). just one point of contact between the body and In many practical situations, however, friction the plane and this point has no motion relative is critically needed. Kinetic friction that to the plane. In this ideal situation, kinetic or dissipates power is nevertheless important for static friction is zero and the body should quickly stopping relative motion. It is made use continue to roll with constant velocity. We know, of by brakes in machines and automobiles. in practice, this will not happen and some Similarly, static friction is important in daily resistance to motion (rolling friction) does occur, life. We are able to walk because of friction. It i.e. to keep the body rolling, some applied force is impossible for a car to move on a very slippery is needed. For the same weight, rolling friction road. On an ordinary road, the friction between is much smaller (even by 2 or 3 orders of the tyres and the road provides the necessary magnitude) than static or sliding friction. This external force to accelerate the car. Fig. 4.13 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. Reprint 2025-26 LAWS OF MOTION 63 4.10 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 a body moving in a circle of radius R with uniform the circle. Using equation (4.14) & (4.16) we get speed v is v2/R directed towards the centre. the result According to the second law, the force f providing this acceleration is : c mv 2 f = ≤ µs N 2 mv R f c = (4.16) R 2 µs RN v ≤ = µs Rg [∵N = mg] where m is the mass of the body. This force m directed forwards the centre is called the which is independent of the mass of the car. centripetal force. For a stone rotated in a circle This shows that for a given value of µs and R, by a string, the centripetal force is provided by there is a maximum speed of circular motion of the tension in the string. The centripetal force the car possible, namely for motion of a planet around the sun is the v max = µs Rg (4.18) (a) (b) Fig. 4.14 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 We can reduce the contribution of friction to the road, the centripetal force is the force of friction. circular motion of the car if the road is banked The circular motion of a car on a flat and (Fig. 4.14(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 (4.19a) Three forces act on the car (Fig. 4.14(a): The centripetal force is provided by the horizontal(i) The weight of the car, mg components of N and f.(ii) Normal reaction, N (iii) Frictional force, f 2 mv As there is no acceleration in the vertical N sin θ + f cos θ = (4.19b) R direction N – mg = 0 But f ≤ µsN N = mg (4.17) Thus to obtain vmax we putThe centripetal force required for circular motion is along the surface of the road, and is provided f = µs N . by the component of the contact force between Then Eqs. (4.19a) and (4.19b) become road and the car tyres along the surface. This by definition is the frictional force. Note that it N cos θ = mg + µsN sin θ (4.20a) Reprint 2025-26 64 PHYSICS N sin θ + µsN cos θ = mv2/R (4.20b) ⊳ Example 4.11 A circular racetrack of From Eq. (4.20a), we obtain radius 300 m is banked at an angle of 15°. mg If the coefficient of friction between the N = wheels of a race-car and the road is 0.2, cosθ – µs sinθ what is the (a) optimum speed of the race- Substituting value of N in Eq. (4.20b), we get car to avoid wear and tear on its tyres, and 2 (b) maximum permissible speed to avoidmg ( sinθ+ µs cosθ) mv max = slipping ? cosθ– µs sinθ R 1 Answer On a banked road, the horizontal 2 component of the normal force and the frictional  µs + tanθ  or v max =  Rg  (4.21) force contribute to provide centripetal force to  1 – µs tanθ keep the car moving on a circular turn without Comparing this with Eq. (4.18) we see that slipping. At the optimum speed, the normal maximum possible speed of a car on a banked reaction’s component is enough to provide the road is greater than that on a flat road. needed centripetal force, and the frictional force is not needed. The optimum speed vo is given by For µs = 0 in Eq. (4.21 ), Eq. (4.22): vo = ( R g tan θ ) ½ (4.22) vO = (R g tan θ)1/2 At this speed, frictional force is not needed at all Here R = 300 m, θ = 15°, g = 9.8 m s-2; we to provide the necessary centripetal force. have Driving at this speed on a banked road will cause vO = 28.1 m s-1. little wear and tear of the tyres. The same The maximum permissible speed vmax is given by equation also tells you that for v < vo, frictional Eq. (4.21): force will be up the slope and that a car can be parked only if tan θ ≤ µs. ⊳ ⊳ Example 4.10 A cyclist speeding at 18 km/h on a level road takes a sharp circular turn of radius 3 m without reducing the speed. The co-efficient of static friction 4.11 SOLVING PROBLEMS IN MECHANICS between the tyres and the road is 0.1. Will The three laws of motion that you have learnt in the cyclist slip while taking the turn? this chapter are the foundation of mechanics. You should now be able to handle a large variety Answer On an unbanked road, frictional force of problems in mechanics. A typical problem in alone can provide the centripetal force needed mechanics usually does not merely involve a to keep the cyclist moving on a circular turn single body under the action of given forces. without slipping. If the speed is too large, or if More often, we will need to consider an assembly the turn is too sharp (i.e. of too small a radius) of different bodies exerting forces on each other. or both, the frictional force is not sufficient to Besides, each body in the assembly experiences provide the necessary centripetal force, and the the force of gravity. When trying to solve a cyclist slips. The condition for the cyclist not to problem of this type, it is useful to remember slip is given by Eq. (4.18) : the fact that we can choose any part of the assembly and apply the laws of motion to that R g v2 ≤ µs part provided we include all forces on the chosen Now, R = 3 m, g = 9.8 m s-2, µs = 0.1. That is, part due to the remaining parts of the assembly. R g = 2.94 m2 s-2. v = 18 km/h = 5 m s-1; i.e., We may call the chosen part of the assembly asµs v2 = 25 m2 s-2. The condition is not obeyed. the system and the remaining part of the The cyclist will slip while taking the assembly (plus any other agencies of forces) as circular turn. ⊳ the environment. We have followed the same Reprint 2025-26 LAWS OF MOTION 65 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 various parts of the assembly of bodies, the vertically downwards. links, supports, etc. (b) The system (block + cylinder) accelerates (ii) Choose a convenient part of the assembly downwards with 0.1 m s-2. The free-body as one system. diagram of the system shows two forces on (iii) Draw a separate diagram which shows this the system : the force of gravity due to the system and all the forces on the system by earth (270 N); and the normal force R′ by the the remaining part of the assembly. Include floor. Note, the free-body diagram of the also the forces on the system by other system does not show the internal forces agencies. Do not include the forces on the between the block and the cylinder. Applying environment by the system. A diagram of the second law to the system, 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 about forces (their magnitudes and directions) that are either given or you are sure of (e.g., the direction of tension in a string along its length). The rest should be treated as unknowns to be determined using laws of motion. (v) If necessary, follow the same procedure for another choice of the system. In doing so, employ Newton’s third law. That is, if in the free-body diagram of A, the force on A due to B is shown as F, then in the free-body diagram of B, the force on B due to A should be shown as –F. The following example illustrates the above procedure : ⊳ Example 4.12 See Fig. 4.15. A wooden Fig. 4.15 block of mass 2 kg rests on a soft horizontal By the third law, the action of the system on floor. When an iron cylinder of mass 25 kg the floor is equal to 267.3 N vertically downward. is placed on top of the block, the floor yields steadily and the block and the cylinder Action-reaction pairs together go down with an acceleration of For (a): (i) the force of gravity (20 N) on the block 0.1 m s–2. What is the action of the block by the earth (say, action); the force of on the floor (a) before and (b) after the floor gravity on the earth by the block yields ? Take g = 10 m s–2. Identify the (reaction) equal to 20 N directed action-reaction pairs in the problem. upwards (not shown in the figure). (ii) the force on the floor by the block Answer (action); the force on the block by the (a) The block is at rest on the floor. Its free-body floor (reaction). diagram shows two forces on the block, the For (b): (i) the force of gravity (270 N) on the force of gravitational attraction by the earth system by the earth (say, action); the equal to 2 × 10 = 20 N; and the normal force force of gravity on the earth by the R of the floor on the block. By the First Law, system (reaction), equal to 270 N, Reprint 2025-26 66 PHYSICS 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 = m v 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 d p F = k = k m a d t 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. 4. 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 Reprint 2025-26 LAWS OF MOTION 67 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 : = µs R f s ≤ ( f s )max f = µ R k k 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. Reprint 2025-26 68 PHYSICS 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) 4.1 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. 4.2 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. 4.3 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. Reprint 2025-26 LAWS OF MOTION 69 4.4 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 : mv2 mv 2 (i) T, (ii) T − , (iii) T + , (iv) 0 l l T is the tension in the string. [Choose the correct alternative]. 4.5 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 ? 4.6 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 ? 4.7 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. 4.8 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. 4.9 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. 4.10 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. 4.11 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.) 4.12 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. 4.13 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 ? 4.14 Figure 4.16 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. 4.16 4.15 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? Reprint 2025-26 70 PHYSICS 4.16 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. 4.17 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. 4.18 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 ? 4.19 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 ? 4.20 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.) 4.21 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 ? 4.22 If, in Exercise 4.21, 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 ? 4.23 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. Reprint 2025-26 CHAPTER FIVE WORK, ENERGY AND POWER 5.1 INTRODUCTION The terms ‘work’, ‘energy’ and ‘power’ are frequently used in everyday language. A farmer ploughing the field, a 5.1 Introduction construction worker carrying bricks, a student studying for a competitive examination, an artist painting a beautiful

3.6 VECTOR ADDITION – ANALYTICAL then, a vector T = a + b – c has components : METHOD T x = a x + b x − c x Although the graphical method of adding vectors Ty = a y + b y − c y (3.23b) helps us in visualising the vectors and the T z = a z + b z − c z .resultant vector, it is sometimes tedious and has limited accuracy. It is much easier to add vectors ⊳ Example 3.2 Find the magnitude andby combining their respective components. direction of the resultant of two vectors AConsider two vectors A and B in x-y plane with and B in terms of their magnitudes and components Ax, Ay and Bx, By : angle θ between them. ɵ ɵ A = A x i + A y j (3.18) * Note that angles α, β, and γ are angles in space. They are between pairs of lines, which are not coplanar. Reprint 2025-26 34 PHYSICS ⊳ Example 3.3 A motorboat is racing towards north at 25 km/h and the water current in that region is 10 km/h in the direction of 60° east of south. Find the resultant velocity of the boat. Answer The vector vb representing the velocity of the motorboat and the vector vc representing Fig. 3.10 the water current are shown in Fig. 3.11 in Answer Let OP and OQ represent the two vectors directions specified by the problem. Using the A and B making an angle θ (Fig. 3.10). Then, parallelogram method of addition, the resultant using the parallelogram method of vector R is obtained in the direction shown in the addition, OS represents the resultant vector R : figure. R = A + B SN is normal to OP and PM is normal to OS. From the geometry of the figure, OS2 = ON2 + SN2 but ON = OP + PN = A + B cos θ SN = B sin θ OS2 = (A + B cos θ)2 + (B sin θ)2 or, R2 = A2 + B2 + 2AB cos θ R = A 2 + B 2 + 2AB cosθ (3.24a) In ∆ OSN, SN = OS sinα = R sinα, and in ∆ PSN, SN = PS sin θ = B sin θ Therefore, R sin α = B sin θ R B or, = (3.24b) sin θ sin α Fig. 3.11 Similarly, PM = A sin α = B sin β We can obtain the magnitude of R using the Law A B or, = (3.24c) of cosine : sin β sin α Combining Eqs. (3.24b) and (3.24c), we get R = v 2b + v c2 + 2v bv c cos120 o R A B = = (3.24d) = 25 2 + 10 2 + 2 × 25 × 10 ( -1/2 ) ≅ 22 km/h sin θ sin β sin α To obtain the direction, we apply the Law of sines Using Eq. (3.24d), we get: R vc v c B = sin θ or, sin φ = sin α = sin θ (3.24e) sin θ sin φ R R where R is given by Eq. (3.24a). 10 × sin120 10 3 = = ≅ 0.397 SN B sin θ 21.8 2 × 21.8or, tan α= = (3.24f) OP + PN A + B cos θ φ ≅ 23.4 ⊳ Equation (3.24a) gives the magnitude of the 3.7 MOTION IN A PLANEresultant and Eqs. (3.24e) and (3.24f) its direction. Equation (3.24a) is known as the Law of cosines In this section we shall see how to describe and Eq. (3.24d) as the Law of sines. ⊳ motion in two dimensions using vectors. Reprint 2025-26 MOTION IN A PLANE 35 3.7.1 Position Vector and Displacement Suppose a particle moves along the curve shown The position vector r of a particle P located in a by the thick line and is at P at time t and P′ at plane with reference to the origin of an x-y time t′ [Fig. 3.12(b)]. Then, the displacement is : reference frame (Fig. 3.12) is given by ∆r = r′ – r (3.25) ɵ ɵ and is directed from P to P′. r = x i + y j We can write Eq. (3.25) in a component form: where x and y are components of r along x-, and y- axes or simply they are the coordinates of ɵ ɵ ɵ ɵ ∆r = x' i + y' j − x i + y j ( ) ( )the object. ɵ ɵ = i ∆ x + j ∆ y where ∆x = x ′ – x, ∆y = y′ – y (3.26) Velocity v The average velocity ( ) of an object is the ratio of the displacement and the corresponding time interval : ɵ ɵ ∆ r ∆ x i + ∆y j ɵ ∆ x ɵ ∆ y v = = = i + j (3.27) ∆ t ∆ t ∆ t ∆ t Or, v = v x ˆi + v y j (a) ∆ r Since v = , the direction of the average velocity ∆t is the same as that of ∆r (Fig. 3.12). The velocity (instantaneous velocity) is given by the limiting value of the average velocity as the time interval approaches zero : ∆ r d r v = lim = (3.28) ∆ t → 0 ∆ t d t The meaning of the limiting process can be easily understood with the help of Fig 3.13(a) to (d). In these figures, the thick line represents the path of an object, which is at P at time t. P1, P2 and (b) P3 represent the positions of the object after Fig. 3.12 (a) Position vector r. (b) Displacement ∆r and times ∆t1,∆t2, and ∆t3. ∆r1, ∆r2, and ∆r3 are the average velocity v of a particle. displacements of the object in times ∆t1, ∆t2, and Fig. 3.13 As the time interval ∆t approaches zero, the average velocity approaches the velocity v. The direction of v is parallel to the line tangent to the path. Reprint 2025-26 36 PHYSICS ∆t3, respectively. The direction of the average velocity v is shown in figures (a), (b) and (c) for three decreasing values of ∆t, i.e. ∆t1,∆t2, and ∆t3, (∆t1 > ∆t2 > ∆t3). As ∆t → 0, ∆r → 0 and is along the tangent to the path [Fig. 3.13(d)]. Therefore, the direction of velocity at any point on the path of an object is tangential to the path at that point and is in the direction of motion. We can express v in a component form : Fig. 3.14 The components vx and vy of velocity v and the angle θ it makes with x-axis. Note that dr vx = v cos θ, vy = v sin θ. v = dt The acceleration (instantaneous acceleration)  ∆x ɵ ∆y ɵ  is the limiting value of the average acceleration = lim  i + j  (3.29) ∆t → 0 ∆t ∆t  as the time interval approaches zero : ∆ v ∆x ∆y ɵ ɵ a = lim (3. 32a) + = i lim j lim ∆t → 0 ∆t ∆t → 0 ∆t ∆ t → 0 ∆ t ɵ ɵ ɵ dx ɵ dy ɵ ɵ Since ∆ v = ∆ v x i + ∆ v y j, we have + jOr, v = i = v x i + v y j . dt dt ɵ ∆ v x ɵ ∆ v y dx dy a = i lim + j lim ∆ t → 0 ∆ t ∆ t → 0 ∆ twhere v x = dt , v y = dt (3.30a) ɵ ɵ So, if the expressions for the coordinates x and Or, a = a x i + a y j (3.32b) y are known as functions of time, we can use d v y d v xthese equations to find vx and vy. (3.32c)* where, a x = , a y = The magnitude of v is then d t d t 2 2 As in the case of velocity, we can understand v = v x + v y (3.30b) graphically the limiting process used in defining and the direction of v is given by the angle θ : acceleration on a graph showing the path of the object’s motion. This is shown in Figs. 3.15(a) to  v y  (d). P represents the position of the object at v y −1 tanθ = , θ = tan     (3.30c) time t and P1, P2, P3 positions after time ∆t1, ∆t2,  v x  v x ∆t3, respectively (∆t 1> ∆t2>∆t3). The velocity vectors vx, vy and angle θ are shown in Fig. 3.14 for a at points P, P1, P2, P3 are also shown in Figs. 3.15 velocity vector v at point p. (a), (b) and (c). In each case of ∆t, ∆v is obtained using the triangle law of vector addition. ByAcceleration definition, the direction of average acceleration The average acceleration a of an object for a is the same as that of ∆v. We see that as ∆t time interval ∆t moving in x-y plane is the change decreases, the direction of ∆v changes and in velocity divided by the time interval : consequently, the direction of the acceleration ɵ ɵ changes. Finally, in the limit ∆t g0 [Fig. 3.15(d)], j v x i + v y ∆ v ∆ ( ∆ v y ɵ ) ∆ v x ɵ a = = = i + j (3.31a) the average acceleration becomes the ∆ t ∆ t ∆ t ∆ t instantaneous acceleration and has the direction ɵ ɵ as shown. Or, a = a x i + a y j . (3.31b) * In terms of x and y, ax and ay can be expressed as Reprint 2025-26 MOTION IN A PLANE 37 x (m) Fig. 3.15 The average acceleration for three time intervals (a) ∆t1, (b) ∆t2, and (c) ∆t3, (∆t1> ∆t2> ∆t3). (d) In the limit ∆t g0, the average acceleration becomes the acceleration. Note that in one dimension, the velocity and the acceleration of an object are always along -1  v y  − 1  4  ° θ = tan   = tan   ≅ 53 with x-axis. the same straight line (either in the same  v x   3  direction or in the opposite direction). ⊳ However, for motion in two or three dimensions, velocity and acceleration vectors may have any angle between 0° and 180° 3.8 MOTION IN A PLANE WITH CONSTANT between them. ACCELERATION ⊳ Suppose that an object is moving in x-y plane Example 3.4 The position of a particle is and its acceleration a is constant. Over an given by interval of time, the average acceleration will r = 3.0t ˆi + 2 .0t 2 ˆj + 5 .0 kˆ equal this constant value. Now, let the velocity where t is in seconds and the of the object be v0 at time t = 0 and v at time t. coefficients have the proper units for r to Then, by definition be in metres. (a) Find v(t) and a(t) of the v − v 0 v − v 0 particle. (b) Find the magnitude and a = = t − 0 t direction of v(t) at t = 1.0 s. Or, v = v 0 + a t (3.33a) Answer In terms of components : v x = v ox + a x t dr d ɵ ɵ ɵ v ( t ) = = 3.0 t i + 2.0t2 j + 5.0 k ( ) v y = v oy + a y t (3.33b) d t dt ɵ ɵ = 3.0 i + 4.0t j Let us now find how the position r changes with d v time. We follow the method used in the one- a ( t ) = = +4.0ɵj dt dimensional case. Let ro and r be the position a = 4.0 m s–2 along y- direction vectors of the particle at time 0 and t and let the velocities at these instants be vo and v. Then,At t = 1.0 s, v = 3.0ˆi + 4.0 ˆj over this time interval t, the average velocity is 2 2 -1 (vo + v)/2. The displacement is the average It’s magnitude is v = 3 + 4 = 5.0 m s velocity multiplied by the time interval : and direction is Reprint 2025-26 38 PHYSICS ˆ  v + v 0   ( v 0 + at ) + v 0  = 5.0 t + 1.5 t 2 i + 1.0 t 2 ˆj ( ) t t = r − r0 = 2  2    2 Therefore, x (t ) = 5.0 t + 1.5 t 1 2 y (t ) = +1.0 t 2 at = v 0 t + 2 Given x (t) = 84 m, t = ? 1 2 5.0 t + 1.5 t 2 = 84 ⇒ t = 6 s Or, r = r0 + v 0t + at (3.34a) At t = 6 s, y = 1.0 (6)2 = 36.0 m 2 d r 2.0 t ˆjIt can be easily verified that the derivative of Now, the velocity v = = ( 5.0 + 3.0 t ) ˆi + d t d r ɵ ɵEq. (3.34a), i.e. gives Eq.(3.33a) and it also At t = 6 s, v = 23.0 i + 12.0 j d t satisfies the condition that at t=0, r = ro. speed = v = 232 + 122 ≅ 26 m s−1 . ⊳Equation (3.34a) can be written in component form as 3.9 PROJECTILE MOTION 1 2 x = x 0 + v ox t + a x t As an application of the ideas developed in the 2 previous sections, we consider the motion of a 1 2 projectile. An object that is in flight after being y = y 0 + v oy t + a y t (3.34b) thrown or projected is called a projectile. Such 2 a projectile might be a football, a cricket ball, a One immediate interpretation of Eq.(3.34b) is that baseball or any other object. The motion of a the motions in x- and y-directions can be treated projectile may be thought of as the result of two independently of each other. That is, motion in separate, simultaneously occurring components a plane (two-dimensions) can be treated as two of motions. One component is along a horizontal separate simultaneous one-dimensional direction without any acceleration and the other motions with constant acceleration along two along the vertical direction with constant perpendicular directions. This is an important acceleration due to the force of gravity. It was result and is useful in analysing motion of objects Galileo who first stated this independency of the in two dimensions. A similar result holds for three horizontal and the vertical components ofdimensions. The choice of perpendicular projectile motion in his Dialogue on the greatdirections is convenient in many physical world systems (1632).situations, as we shall see in section 3.9 for In our discussion, we shall assume that theprojectile motion. air resistance has negligible effect on the motion ⊳ Example 3.5 A particle starts from origin of the projectile. Suppose that the projectile is launched with velocity vo that makes an angle at t = 0 with a velocity 5.0 î m/s and moves in x-y plane under action of a force which θo with the x-axis as shown in Fig. 3.16. produces a constant acceleration of (3.0iɵ+2.0jɵ) m/s 2. (a) What is the After the object has been projected, the acceleration acting on it is that due to gravity y-coordinate of the particle at the instant which is directed vertically downward: its x-coordinate is 84 m ? (b) What is the ɵ speed of the particle at this time ? a = −g j Or, ax = 0, ay = – g (3.35) Answer From Eq. (3.34a) for r0 = 0, the position The components of initial velocity vo are : of the particle is given by 1 at 2 vox = vo cos θo r (t ) = v 0 t + 2 2 voy= vo sin θo (3.36) t 3.0ˆi + 2.0 ˆj = 5.0 ˆi t + (1/2 )( ) Reprint 2025-26 MOTION IN A PLANE 39 Now, since g, θo and vo are constants, Eq. (3.39) is of the form y = a x + b x2, in which a and b are constants. This is the equation of a parabola, i.e. the path of the projectile is a parabola (Fig. 3.17). Fig 3.16 Motion of an object projected with velocity vo at angle θ0. If we take the initial position to be the origin of the reference frame as shown in Fig. 3.16, we have : xo = 0, yo = 0 Then, Eq.(3.34b) becomes : x = vox t = (vo cos θo ) t and y = (vo sin θo ) t – ( ½ )g t2 (3.37) Fig. 3.17 The path of a projectile is a parabola. The components of velocity at time t can be Time of maximum height obtained using Eq.(3.33b) : How much time does the projectile take to reach the vx = vox = vo cos θo maximum height ? Let this time be denoted by tm. Since at this point, vy= 0, we have from Eq. (3.38): vy = vo sin θo – g t (3.38) vy = vo sinθo – g tm = 0 Equation (3.37) gives the x-, and y-coordinates Or, tm = vo sinθo /g (3.40a) of the position of a projectile at time t in terms of The total time Tf during which the projectile is two parameters — initial speed vo and projection in flight can be obtained by putting y = 0 in angle θo. Notice that the choice of mutually Eq. (3.37). We get : perpendicular x-, and y-directions for the analysis of the projectile motion has resulted in Tf = 2 (vo sin θo )/g (3.40b) a simplification. One of the components of Tf is known as the time of flight of the projectile. velocity, i.e. x-component remains constant We note that Tf = 2 tm , which is expected throughout the motion and only the because of the symmetry of the parabolic path. y- component changes, like an object in free fall Maximum height of a projectile in vertical direction. This is shown graphically The maximum height hm reached by theat few instants in Fig. 3.17. Note that at the point projectile can be calculated by substituting of maximum height, vy= 0 and therefore, t = tm in Eq. (3.37) : -1 v y 2 = oθ = tan    g  v 0 sinθ0 v 0 sinθ0 v x y = h m = ( v 0 sinθ0 )  −   Equation of path of a projectile  g  2  g  What is the shape of the path followed by the ( v 0 sinθ0 ) 2 projectile? This can be seen by eliminating the Or, h m = (3.41) 2gtime between the expressions for x and y as given in Eq. (3.37). We obtain: Horizontal range of a projectile g 2 The horizontal distance travelled by a projectile from x y = ( tan θo ) x − (3.39) its initial position (x = y = 0) to the position where it 2 (v o cosθo )2 passes y = 0 during its fall is called the horizontal Reprint 2025-26 40 PHYSICS range, R. It is the distance travelled during the time y (t) = yo + voy t +(1/2) ay t2 of flight Tf . Therefore, the range R is Here, xo = yo = 0, voy = 0, ay = –g = –9.8 m s-2, R = (vo cos θo) (Tf ) vox = 15 m s-1. The stone hits the ground when y(t) = – 490 m. =(vo cos θo) (2 vo sin θo)/g – 490 m = –(1/2)(9.8) t2. 2 sin 2θ0 This gives t =10 s. v 0 Or, R = (3.42a) The velocity components are vx = vox and g vy = voy – g t Equation (3.42a) shows that for a given so that when the stone hits the ground : projection velocity vo , R is maximum when sin vox = 15 m s–1 2θ0 is maximum, i.e., when θ0 = 450. voy = 0 – 9.8 × 10 = – 98 m s–1 The maximum horizontal range is, therefore, Therefore, the speed of the stone is 2 v 0 v 2x + v y2 = 15 2 + 98 2 = 99 m s −1 ⊳ R m = (3.42b) g ⊳ ⊳ Example 3.6 Galileo, in his book Two new Example 3.8 A cricket ball is thrown at a sciences, stated that “for elevations which speed of 28 m s–1 in a direction 30° above exceed or fall short of 45° by equal amounts, the horizontal. Calculate (a) the maximum the ranges are equal”. Prove this statement. height, (b) the time taken by the ball to return to the same level, and (c) the distance from the thrower to the point Answer For a projectile launched with velocity where the ball returns to the same level. vo at an angle θo , the range is given by v 02 sin2θ0 R = Answer (a) The maximum height is given by g 2 2 sin sin 30 ( 28 ( v 0 θo ) ° ) m = m =Now, for angles, (45° + α) and ( 45° – α), 2θo is h 2 2 g ( 9.8 )(90° + 2α) and ( 90° – 2α) , respectively. The values of sin (90° + 2α) and sin (90° – 2α) are 14 × 14 = = 10.0 mthe same, equal to that of cos 2α. Therefore, 2 × 9.8 ranges are equal for elevations which exceed or fall short of 45° by equal amounts α. ⊳ (b) The time taken to return to the same level is Tf = (2 vo sin θo )/g = (2× 28 × sin 30° )/9.8 ⊳ = 28/9.8 s = 2.9 s Example 3.7 A hiker stands on the edge (c) The distance from the thrower to the point of a cliff 490 m above the ground and where the ball returns to the same level is throws a stone horizontally with an initial speed 2 of 15 m s-1. Neglecting air resistance, v o sin2θo 28 × 28 × sin60o ( ) = = 69 m ⊳ find the time taken by the stone to reach R = g 9.8 the ground, and the speed with which it hits the ground. (Take g = 9.8 m s-2 ). 3.10 UNIFORM CIRCULAR MOTION Answer We choose the origin of the x-,and y- When an object follows a circular path at a axis at the edge of the cliff and t = 0 s at the constant speed, the motion of the object is called instant the stone is thrown. Choose the positive uniform circular motion. The word “uniform” direction of x-axis to be along the initial velocity refers to the speed, which is uniform (constant) and the positive direction of y-axis to be the throughout the motion. Suppose an object is vertically upward direction. The x-, and y- moving with uniform speed v in a circle of radius components of the motion can be treated R as shown in Fig. 3.18. Since the velocity of the independently. The equations of motion are : object is changing continuously in direction, the x (t) = xo + vox t object undergoes acceleration. Let us find the magnitude and the direction of this acceleration. Reprint 2025-26 MOTION IN A PLANE 41 Fig. 3.18 Velocity and acceleration of an object in uniform circular motion. The time interval ∆t decreases from (a) to (c) where it is zero. The acceleration is directed, at each point of the path, towards the centre of the circle. Let r and r′ be the position vectors and v and r′ be ∆θ. Since the velocity vectors v and v′ are v′ the velocities of the object when it is at point P always perpendicular to the position vectors, the and P′ as shown in Fig. 3.18(a). By definition, angle between them is also ∆θ . Therefore, the velocity at a point is along the tangent at that triangle CPP′ formed by the position vectors and point in the direction of motion. The velocity the triangle GHI formed by the velocity vectors vectors v and v′ are as shown in Fig. 3.18(a1). v, v′ and ∆v are similar (Fig. 3.18a). Therefore, ∆v is obtained in Fig. 3.18 (a2) using the triangle the ratio of the base-length to side-length for law of vector addition. Since the path is circular, one of the triangles is equal to that of the other v is perpendicular to r and so is v′ to r′. triangle. That is : Therefore, ∆v is perpendicular to ∆r. Since  ∆v  ∆ v ∆ r average acceleration is along ∆v  a =  , the =  ∆t  v R average acceleration a is perpendicular to ∆r. If we place ∆v on the line that bisects the angle ∆ r Or, ∆ v = v between r and r′, we see that it is directed towards R the centre of the circle. Figure 3.18(b) shows the Therefore, same quantities for smaller time interval. ∆v and ∆ v v ∆ r v ∆r hence a is again directed towards the centre. a = lim = lim = lim t ∆ t → 0 R ∆ t R ∆ t → 0 ∆ tIn Fig. 3.18(c), ∆tŽ 0 and the average ∆ t → 0 ∆ acceleration becomes the instantaneous If ∆t is small, ∆θ will also be small and then arc acceleration. It is directed towards the centre*. PP′ can be approximately taken to be|∆r|: Thus, we find that the acceleration of an object ∆ r ≅v∆ t in uniform circular motion is always directed ∆r towards the centre of the circle. Let us now find ≅ v the magnitude of the acceleration. ∆ t The magnitude of a is, by definition, given by ∆ r lim = v ∆v Or, ∆ t → 0 ∆ t a = lim ∆ t → 0 ∆t Let the angle between position vectors r and Therefore, the centripetal acceleration ac is : * In the limit ∆tŽ0, ∆r becomes perpendicular to r. In this limit ∆v→ 0 and is consequently also perpendicular to V. Therefore, the acceleration is directed towards the centre, at each point of the circular path. Reprint 2025-26 42 PHYSICS 2 2 2  v  v R ω 2  R ac = = = ω a c =  R v = v2/R (3.43) R R Thus, the acceleration of an object moving with 2 a c = ω R (3.46) speed v in a circle of radius R has a magnitude v 2/R and is always directed towards the centre. The time taken by an object to make one revolution This is why this acceleration is called centripetal is known as its time period T and the number ofacceleration (a term proposed by Newton). A revolution made in one second is called itsthorough analysis of centripetal acceleration was frequency ν (=1/T). However, during this timefirst published in 1673 by the Dutch scientist the distance moved by the object is s = 2πR.Christiaan Huygens (1629-1695) but it was probably known to Newton also some years earlier. Therefore, v = 2πR/T =2πRν (3.47) “Centripetal” comes from a Greek term which means In terms of frequency ν, we have ‘centre-seeking’. Since v and R are constant, the ω = 2πν magnitude of the centripetal acceleration is also v = 2πRν constant. However, the direction changes — ac = 4π2 ν2R (3.48)pointing always towards the centre. Therefore, a ⊳centripetal acceleration is not a constant vector. Example 3.9 An insect trapped in a We have another way of describing the circular groove of radius 12 cm moves along velocity and the acceleration of an object in the groove steadily and completes 7 uniform circular motion. As the object moves revolutions in 100 s. (a) What is the from P to P′ in time ∆t (= t′ – t), the line CP angular speed, and the linear speed of the (Fig. 3.18) turns through an angle ∆θ as shown motion? (b) Is the acceleration vector a in the figure. ∆θ is called angular distance. We constant vector ? What is its magnitude ? define the angular speed ω (Greek letter omega) as the time rate of change of angular Answer This is an example of uniform circular displacement : motion. Here R = 12 cm. The angular speed ω is ∆θ given by ω = ∆t (3.44) ω = 2π/T = 2π × 7/100 = 0.44 rad/s The linear speed v is :Now, if the distance travelled by the object during the time ∆t is ∆s, i.e. PP′ is ∆s, then : v =ω R = 0.44 s-1 × 12 cm = 5.3 cm s-1 ∆ s The direction of velocity v is along the tangent v = ∆t to the circle at every point. The acceleration is directed towards the centre of the circle. Sincebut ∆s = R ∆θ. Therefore : this direction changes continuously, ∆θ v = R = R ω acceleration here is not a constant vector. ∆ t However, the magnitude of acceleration is v = R ω (3.45) constant: a = ω2 R = (0.44 s–1)2 (12 cm)We can express centripetal acceleration ac in terms of angular speed : = 2.3 cm s-2 ⊳ Reprint 2025-26 MOTION IN A PLANE 43 SUMMARY 1. Scalar quantities are quantities with magnitudes only. Examples are distance, speed, mass and temperature. 2. Vector quantities are quantities with magnitude and direction both. Examples are displacement, velocity and acceleration. They obey special rules of vector algebra. 3. A vector A multiplied by a real number λ is also a vector, whose magnitude is λ times the magnitude of the vector A and whose direction is the same or opposite depending upon whether λ is positive or negative. 4. Two vectors A and B may be added graphically using head-to-tail method or parallelogram method. 5. Vector addition is commutative : A + B = B + A It also obeys the associative law : (A + B) + C = A + (B + C) 6. A null or zero vector is a vector with zero magnitude. Since the magnitude is zero, we don’t have to specify its direction. It has the properties : A + 0 = A λ0 = 0 0 A = 0 7. The subtraction of vector B from A is defined as the sum of A and –B : A – B = A+ (–B) 8. A vector A can be resolved into component along two given vectors a and b lying in the same plane : A = λ a + µ b where λ and µ are real numbers. 9. A unit vector associated with a vector A has magnitude 1 and is along the vector A: A ˆn = A ɵ ɵ ɵ The unit vectors i, j, k are vectors of unit magnitude and point in the direction of the x-, y-, and z-axes, respectively in a right-handed coordinate system. 10. A vector A can be expressed as ɵ ɵ A = A x i + Ay j where Ax, Ay are its components along x-, and y -axes. If vector A makes an angle θ A y 2 2 with the x-axis, then Ax = A cos θ, Ay=A sin θ and A = A = A x + A y , tanθ = . A x 11. Vectors can be conveniently added using analytical method. If sum of two vectors A and B, that lie in x-y plane, is R, then : ɵ R = R x i + Ry ɵ,j where, Rx = Ax + Bx, and Ry = Ay + By ɵ ɵ 12. The position vector of an object in x-y plane is given by r = x i + y j and the displacement from position r to position r’ is given by ∆r = r′− r ɵ ɵ = ( x ′ − x ) i + (y ′ − y ) j ɵ ɵ = ∆x i + ∆y j 13. If an object undergoes a displacement ∆r in time ∆t, its average velocity is given by ∆ r v = . The velocity of an object at time t is the limiting value of the average velocity ∆t as ∆t tends to zero : Reprint 2025-26 44 PHYSICS ∆ r d r v = lim = . It can be written in unit vector notation as : ∆ t →0 ∆ t dt ɵ ɵ ɵ dx dy dz = y v = v x i + v y j + v z k where v x = dt , v dt , v z = dt When position of an object is plotted on a coordinate system, v is always tangent to the curve representing the path of the object. 14. If the velocity of an object changes from v to v′in time ∆t, then its average acceleration v − v' ∆ v is given by: a = = ∆ t ∆t The acceleration a at any time t is the limiting value of a as ∆t Ž0 : ∆ v dv a = lim = ∆ t → 0 ∆ t dt ɵ ɵ ɵ In component form, we have : a = a x i + a y j + a z k dvy dv x dv z where, a x = , a y = , a z = dt dt dt 15. If an object is moving in a plane with constant acceleration a = a = a x2 + a y2 and its position vector at time t = 0 is ro, then at any other time t, it will be at a point given by: 1 2 r = ro + v o t + at 2 and its velocity is given by : v = vo + a t where vo is the velocity at time t = 0 In component form : 1 2 x = x o + v ox t + a x t 2 1 2 y = yo + v oy t + a y t 2 v x = v ox + a x t v y = v oy + a y t Motion in a plane can be treated as superposition of two separate simultaneous one- dimensional motions along two perpendicular directions 16. An object that is in flight after being projected is called a projectile. If an object is projected with initial velocity vo making an angle θo with x-axis and if we assume its initial position to coincide with the origin of the coordinate system, then the position and velocity of the projectile at time t are given by : x = (vo cos θo) t y = (vo sin θo) t − (1/2) g t2 vx = vox = vo cos θo vy = vo sin θo − g t The path of a projectile is parabolic and is given by : 2 gx y = ( tanθ0 ) x – 2 2 (v o cosθo ) The maximum height that a projectile attains is : Reprint 2025-26 MOTION IN A PLANE 45 (v o sinq o )2 h m = 2g The time taken to reach this height is : v o sinθo t m = g The horizontal distance travelled by a projectile from its initial position to the position it passes y = 0 during its fall is called the range, R of the projectile. It is : v o2 R = sin2θo g 17. When an object follows a circular path at constant speed, the motion of the object is called uniform circular motion. The magnitude of its acceleration is ac = v2 /R. The direction of ac is always towards the centre of the circle. The angular speed ω, is the rate of change of angular distance. It is related to velocity v by v = ω R. The acceleration is ac = ω 2R. If T is the time period of revolution of the object in circular motion and ν is its frequency, we have ω = 2π ν, v = 2πνR, ac = 4π2ν2R Reprint 2025-26 46 PHYSICS POINTS TO PONDER 1. The path length traversed by an object between two points is, in general, not the same as the magnitude of displacement. The displacement depends only on the end points; the path length (as the name implies) depends on the actual path. The two quantities are equal only if the object does not change its direction during the course of motion. In all other cases, the path length is greater than the magnitude of displacement. 2. In view of point 1 above, the average speed of an object is greater than or equal to the magnitude of the average velocity over a given time interval. The two are equal only if the path length is equal to the magnitude of displacement. 3. The vector equations (3.33a) and (3.34a) do not involve any choice of axes. Of course, you can always resolve them along any two independent axes. 4. The kinematic equations for uniform acceleration do not apply to the case of uniform circular motion since in this case the magnitude of acceleration is constant but its direction is changing. 5. An object subjected to two velocities v1 and v2 has a resultant velocity v = v1 + v2. Take care to distinguish it from velocity of object 1 relative to velocity of object 2 : v12= v1 − v2. Here v1 and v2 are velocities with reference to some common reference frame. 6. The resultant acceleration of an object in circular motion is towards the centre only if the speed is constant. 7. The shape of the trajectory of the motion of an object is not determined by the acceleration alone but also depends on the initial conditions of motion ( initial position and initial velocity). For example, the trajectory of an object moving under the same acceleration due to gravity can be a straight line or a parabola depending on the initial conditions. EXERCISES 3.1 State, for each of the following physical quantities, if it is a scalar or a vector : volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity. 3.2 Pick out the two scalar quantities in the following list : force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity. 3.3 Pick out the only vector quantity in the following list : Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge. 3.4 State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful : (a) adding any two scalars, (b) adding a scalar to a vector of the same dimensions , (c) multiplying any vector by any scalar, (d) multiplying any two scalars, (e) adding any two vectors, (f) adding a component of a vector to the same vector. 3.5 Read each statement below carefully and state with reasons, if it is true or false : (a) The magnitude of a vector is always a scalar, (b) each component of a vector is always a scalar, (c) the total path length is always equal to the magnitude of the displacement vector of a particle. (d) the average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of average velocity of the particle over the same interval of time, (e) Three vectors not lying in a plane can never add up to give a null vector. 3.6 Establish the following vector inequalities geometrically or otherwise : (a) |a+b| < |a| + |b| (b) |a+b| > ||a| −|b|| Reprint 2025-26 MOTION IN A PLANE 47 (c) |a−b| < |a| + |b| (d) |a−b| > ||a| − |b|| When does the equality sign above apply? 3.7 Given a + b + c + d = 0, which of the following statements are correct : Q (a) a, b, c, and d must each be a null vector, (b) The magnitude of (a + c) equals the magnitude of ( b + d), (c) The magnitude of a can never be greater than the sum of the magnitudes of b, c, and d, (d) b + c must lie in the plane of a and d if a and d are not collinear, and in the line of a and d, if they are collinear ? 3.8 Three girls skating on a circular ice ground of radius 200 m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in Fig. 3.19. What is the magnitude of the displacement vector for each ? For Fig. 3.19 which girl is this equal to the actual length of path skate ? 3.9 A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference, and returns to the centre along QO as shown in Fig. 3.20. If the round trip takes 10 min, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist ? Fig. 3.20 3.10 On an open ground, a motorist follows a track that turns to his left by an angle of 600 after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case. 3.11 A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 min. What is (a) the average speed of the taxi, (b) the magnitude of average velocity ? Are the two equal ? 3.12 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 ? 3.13 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 ? Reprint 2025-26 48 PHYSICS 3.14 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 ? 3.15 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. 3.16 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 3.17 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 ? 3.18 A particle starts from the origin at t = 0 s with a velocity of 10.0 jɵ m/s and moves in ɵ ɵ the x-y plane with a constant acceleration of 8.0 i + 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 ? 3.19 ɵi and ɵj are unit vectors along x- and y- axis respectively. What is the magnitude ɵ ɵ ɵ ɵ and direction of the vectors i + j , and i − j ? What are the components of a vector ɵ ɵ ɵ ɵ ɵ ɵ A= 2 i + 3 j along the directions of i + j and i −?j [You may use graphical method] 3.20 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 t1 to t2) 3.21 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. 3.22 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 a part is 30°, wat is the speed of the aircraft ? Reprint 2025-26 CHAPTER FOUR LAWS OF MOTION 4.1 INTRODUCTION In the preceding Chapter, our concern was to describe the motion of a particle in space quantitatively. We saw that 4.1 Introduction uniform motion needs the concept of velocity alone whereas 4.2 Aristotle’s fallacy non-uniform motion requires the concept of acceleration in 4.3 The law of inertia addition. So far, we have not asked the question as to what governs the motion of bodies. In this chapter, we turn to this4.4 Newton’s first law of motion basic question.4.5 Newton’s second law of Let us first guess the answer based on our common motion experience. To move a football at rest, someone must kick it.4.6 Newton’s third law of motion To throw a stone upwards, one has to give it an upward

4.7 Conservation of momentum push. A breeze causes the branches of a tree to swing; a 4.8 Equilibrium of a particle strong wind can even move heavy objects. A boat moves in a 4.9 Common forces in mechanics flowing river without anyone rowing it. Clearly, some external 4.10 Circular motion agency is needed to provide force to move a body from rest. 4.11 Solving problems in Likewise, an external force is needed also to retard or stop mechanics motion. You can stop a ball rolling down an inclined plane by applying a force against the direction of its motion. Summary In these examples, the external agency of force (hands, Points to ponder wind, stream, etc) is in contact with the object. This is not Exercises 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? Reprint 2025-26 50 PHYSICS 4.2 ARISTOTLE’S FALLACY true law of nature for forces and motion, one has to imagine a world in which uniform motion isThe question posed above appears to be simple. possible with no frictional forces opposing. ThisHowever, it took ages to answer it. Indeed, the is what Galileo did.correct answer to this question given by Galileo in the seventeenth century was the foundation 4.3 THE LAW OF INERTIA of Newtonian mechanics, which signalled the Galileo studied motion of objects on an inclined birth of modern science. plane. Objects (i) moving down an inclined plane The Greek thinker, Aristotle (384 B.C– 322 accelerate, while those (ii) moving up retard. B.C.), held the view that if a body is moving, (iii) Motion on a horizontal plane is an interme- something external is required to keep it moving. diate situation. Galileo concluded that an object According to this view, for example, an arrow moving on a frictionless horizontal plane must shot from a bow keeps flying since the air behind neither have acceleration nor retardation, i.e. it the arrow keeps pushing it. The view was part of should move with constant velocity (Fig. 4.1(a)).an elaborate framework of ideas developed by Aristotle on the motion of bodies in the universe. Most of the Aristotelian ideas on motion are now known to be wrong and need not concern us. For our purpose here, the Aristotelian law of motion may be phrased thus: An external force (i) (ii) (iii)is required to keep a body in motion. Fig. 4.1(a) Aristotelian law of motion is flawed, as we shall Another experiment by Galileo leading to thesee. However, it is a natural view that anyone same conclusion involves a double inclined plane.would hold from common experience. Even a A ball released from rest on one of the planes rollssmall child playing with a simple (non-electric) down and climbs up the other. If the planes are toy-car on a floor knows intuitively that it needs smooth, the final height of the ball is nearly the to constantly drag the string attached to the toy- same as the initial height (a little less but never car with some force to keep it going. If it releases greater). In the ideal situation, when friction is the string, it comes to rest. This experience is absent, the final height of the ball is the same common to most terrestrial motion. External as its initial height. forces seem to be needed to keep bodies in If the slope of the second plane is decreased motion. Left to themselves, all bodies eventually and the experiment repeated, the ball will still come to rest. reach the same height, but in doing so, it will What is the flaw in Aristotle’s argument? The travel a longer distance. In the limiting case, when answer is: a moving toy car comes to rest because the slope of the second plane is zero (i.e. is a the external force of friction on the car by the floor horizontal) the ball travels an infinite distance. opposes its motion. To counter this force, the child In other words, its motion never ceases. This is, has to apply an external force on the car in the of course, an idealised situation (Fig. 4.1(b)). 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 Fig. 4.1(b) The law of inertia was inferred by Galileo went wrong. He coded this practical experience from observations of motion of a ball on a in the form of a basic argument. To get at the double inclined plane. Reprint 2025-26 LAWS OF MOTION 51 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 Newton built on Galileo’s ideas and laid the can never be totally eliminated. However, if there foundation of mechanics in terms of three laws were no friction, the ball would continue to move of motion that go by his name. Galileo’s law of with a constant velocity on the horizontal plane. inertia was his starting point which he formu- Galileo thus, arrived at a new insight on lated as the first law of motion: motion that had eluded Aristotle and those who Every body continues to be in its state followed him. The state of rest and the state of of rest or of uniform motion in a straight uniform linear motion (motion with constant line unless compelled by some external velocity) are equivalent. In both cases, there is 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 thea body at rest continues to remain at rest and a application of this law in practice. In somebody in motion continues to move with a uniform examples, we know that the net external forcevelocity. This property of the body is called on the object is zero. In that case we caninertia. Inertia means ‘resistance to change’. A body does not change its state of rest or conclude that the acceleration of the object is uniform motion, unless an external force zero. For example, a spaceship out in compels it to change that state. interstellar space, far from all other objects and with all its rockets turned off, has no net 4.4 NEWTON’S FIRST LAW OF MOTION external force acting on it. Its acceleration, Galileo’s simple, but revolutionary ideas according to the first law, must be zero. If it is dethroned Aristotelian mechanics. A new in motion, it must continue to move with a mechanics had to be developed. This task was uniform velocity. Reprint 2025-26 52 PHYSICS More often, however, we do not know all the The acceleration of the car cannot be accounted forces to begin with. In that case, if we know for by any internal force. This might sound that an object is unaccelerated (i.e. it is either surprising, but it is true. The only conceivable at rest or in uniform linear motion), we can infer external force along the road is the force of from the first law that the net external force on friction. It is the frictional force that accelerates the object must be zero. Gravity is everywhere. the car as a whole. (You will learn about friction For terrestrial phenomena, in particular, every in section 4.9). When the car moves with object experiences gravitational force due to the constant velocity, there is no net external force. earth. Also objects in motion generally experience The property of inertia contained in the First friction, viscous drag, etc. If then, on earth, an law is evident in many situations. Suppose we object is at rest or in uniform linear motion, it is are standing in a stationary bus and the driver not because there are no forces acting on it, but starts the bus suddenly. We get thrown because the various external forces cancel out backward with a jerk. Why ? Our feet are in touch i.e. add up to zero net external force. with the floor. If there were no friction, we would Consider a book at rest on a horizontal surface remain where we were, while the floor of the bus Fig. (4.2(a)). It is subject to two external forces : would simply slip forward under our feet and the the force due to gravity (i.e. its weight W) acting back of the bus would hit us. However, downward and the upward force on the book by fortunately, there is some friction between the the table, the normal force R . R is a self-adjusting feet and the floor. If the start is not too sudden, force. This is an example of the kind of situation i.e. if the acceleration is moderate, the frictional mentioned above. The forces are not quite known force would be enough to accelerate our feet fully but the state of motion is known. We observe along with the bus. But our body is not strictly the book to be at rest. Therefore, we conclude a rigid body. It is deformable, i.e. it allows some from the first law that the magnitude of R equals relative displacement between different parts. that of W. A statement often encountered is : What this means is that while our feet go with “Since W = R, forces cancel and, therefore, the book the bus, the rest of the body remains where it is is at rest”. This is incorrect reasoning. The correct due to inertia. Relative to the bus, therefore, we statement is : “Since the book is observed to be at are thrown backward. As soon as that happens, rest, the net external force on it must be zero, however, the muscular forces on the rest of the according to the first law. This implies that the body (by the feet) come into play to move the body normal force R must be equal and opposite to the along with the bus. A similar thing happens weight W ”. when the bus suddenly stops. Our feet stop due to the friction which does not allow relative motion between the feet and the floor of the bus. But the rest of the body continues to move forward due to inertia. We are thrown forward. The restoring muscular forces again come into play and bring the body to rest. ⊳ Example 4.1 An astronaut accidentally gets separated out of his small spaceship accelerating in inter stellar space at a Fig. 4.2 (a) a book at rest on the table, and (b) a car constant rate of 100 m s–2. What is the moving with uniform velocity. The net force acceleration of the astronaut the instant after is zero in each case. he is outside the spaceship ? (Assume that Consider the motion of a car starting from there are no nearby stars to exert rest, picking up speed and then moving on a gravitational force on him.) smooth straight road with uniform speed (Fig. Answer Since there are no nearby stars to exert(4.2(b)). When the car is stationary, there is no gravitational force on him and the smallnet force acting on it. During pick-up, it spaceship exerts negligible gravitationalaccelerates. This must happen due to a net attraction on him, the net force acting on theexternal force. Note, it has to be an external force. Reprint 2025-26 LAWS OF MOTION 53 astronaut, once he is out of the spaceship, is act. One reason is that the cricketer allows a zero. By the first law of motion the acceleration longer time for his hands to stop the ball. As of the astronaut is zero. ⊳ you may have noticed, he draws in the hands backward in the act of catching the ball4.5 NEWTON’S SECOND LAW OF MOTION (Fig. 4.3). The novice, on the other hand, The first law refers to the simple case when the keeps his hands fixed and tries to catch the net external force on a body is zero. The second ball almost instantly. He needs to provide a law of motion refers to the general situation when much greater force to stop the ball instantly, there is a net external force acting on the body. and this hurts. The conclusion is clear: force It relates the net external force to the not only depends on the change in momentum, acceleration of the body. but also on how fast the change is brought Momentum about. The same change in momentum Momentum of a body is defined to be the product brought about in a shorter time needs a of its mass m and velocity v, and is denoted greater applied force. In short, the greater the by p: rate of change of momentum, the greater is p = m v (4.1) the force. Momentum is clearly a vector quantity. The following common experiences indicate the importance of this quantity for considering the effect of force on motion. • Suppose a light-weight vehicle (say a small car) and a heavy weight vehicle (say a loaded truck) are parked on a horizontal road. We all know that a much greater force is needed to push the truck than the car to bring them to the same speed in same time. Similarly, a greater opposing force is needed to stop a heavy body than a light body in the same time, if they are moving with the same speed. • If two stones, one light and the other heavy, are dropped from the top of a building, a Fig. 4.3 Force not only depends on the change in person on the ground will find it easier to catch momentum but also on how fast the change is brought about. A seasoned cricketer draws the light stone than the heavy stone. The in his hands during a catch, allowing greater mass of a body is thus an important time for the ball to stop and hence requires a parameter that determines the effect of force smaller force. on its motion. • Speed is another important parameter to consider. A bullet fired by a gun can easily • Observations confirm that the product of pierce human tissue before it stops, resulting mass and velocity (i.e. momentum) is basic to in casualty. The same bullet fired with the effect of force on motion. Suppose a fixed moderate speed will not cause much damage. force is applied for a certain interval of time Thus for a given mass, the greater the speed, on two bodies of different masses, initially at the greater is the opposing force needed to stop rest, the lighter body picks up a greater speed the body in a certain time. Taken together, than the heavier body. However, at the end of the product of mass and velocity, that is the time interval, observations show that each momentum, is evidently a relevant variable body acquires the same momentum. Thus of motion. The greater the change in the the same force for the same time causes momentum in a given time, the greater is the the same change in momentum for force that needs to be applied. • A seasoned cricketer catches a cricket ball different bodies. This is a crucial clue to the second law of motion. coming in with great speed far more easily • In the preceding observations, the vector than a novice, who can hurt his hands in the Reprint 2025-26 54 PHYSICS character of momentum has not been evident. ∆p ∆p In the examples so far, momentum and change F ∝ or F = k ∆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 ∆p the limit ∆t → 0, the term becomes the horizontal plane by means of a string, the ∆t magnitude of momentum is fixed, but its derivative or differential co-efficient of p with direction changes (Fig. 4.4). A force is needed d p to cause this change in momentum vector. respect to t, denoted by . Thus dt This force is provided by our hand through the string. Experience suggests that our hand d p F = k (4.2) needs to exert a greater force if the stone is d t rotated at greater speed or in a circle of For a body of fixed mass m, smaller radius, or both. This corresponds to greater acceleration or equivalently a greater d p d d v = (m v ) = m = m a (4.3) rate of change in momentum vector. This d t d t d t suggests that the greater the rate of change i.e the Second Law can also be written as in momentum vector the greater is the force F = k m a (4.4) applied. 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. (4.4) 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 F = = m a (4.5) dt In SI unit force is one that causes an acceleration of 1 m s-2 to a mass of 1 kg. This unit is known as newton : 1 N = 1 kg m s-2. Let us note at this stage some important points Fig. 4.4 Force is necessary for changing the direction about the second law : of momentum, even if its magnitude is constant. We can feel this while rotating a 1. In the second law, F = 0 implies a = 0. The second stone in a horizontal circle with uniform speed law is obviously consistent with the first law. by means of a string. 2. The second law of motion is a vector law. It is These qualitative observations lead to the equivalent to three equations, one for each second law of motion expressed by Newton as component of the vectors : follows : d p x F x = = ma xThe rate of change of momentum of a body is d t directly proportional to the applied force and d p ytakes place in the direction in which the force F y = = ma y acts. d t dp zThus, if under the action of a force F for time F z = =m a z (4.6) dtinterval ∆t, the velocity of a body of mass m changes from v to v + ∆v i.e. its initial momentum This means that if a force is not parallel to the velocity of the body, but makes some anglep = m v changes by ∆ p = m ∆v . According to the with it, it changes only the component of Second Law, velocity along the direction of force. The Reprint 2025-26 LAWS OF MOTION 55 component of velocity normal to the force Answer The retardation ‘a’ of the bullet remains unchanged. For example, in the (assumed constant) is given by motion of a projectile under the vertical – 90 × 90 – u 2 gravitational force, the horizontal component m s −2 = – 6750 m s −2 a = = of velocity remains unchanged (Fig. 4.5). 2s 2 × 0.6 3. The second law of motion given by Eq. (4.5) is The retarding force, by the second law of applicable to a single point particle. The force motion, is F in the law stands for the net external force = 0.04 kg × 6750 m s-2 = 270 N on the particle and a stands for acceleration of the particle. It turns out, however, that the The actual resistive force, and therefore, law in the same form applies to a rigid body or, retardation of the bullet may not be uniform. The answer therefore, only indicates the average even more generally, to a system of particles. resistive force. ⊳ In that case, F refers to the total external force ⊳ on the system and a refers to the acceleration Example 4.3 The motion of a particle of of the system as a whole. More precisely, a is 1 2 the acceleration of the centre of mass of the mass m is described by y = ut + gt . Find 2 system about which we shall study in detail in the force acting on the particle. Chapter 6. Any internal forces in the system are not to be included in F. Answer We know 1 2 y = ut + gt 2 Now, d y v = = u + gt d t dv acceleration, a = = g d t Fig. 4.5 Acceleration at an instant is determined by Then the force is given by Eq. (4.5) the force at that instant. The moment after a F = ma = mg stone is dropped out of an accelerated train, Thus the given equation describes the motion it has no horizontal acceleration or force, if of a particle under acceleration due to gravity air resistance is neglected. The stone carries no memory of its acceleration with the train and y is the position coordinate in the direction a moment ago. of g. ⊳ 4. The second law of motion is a local relation Impulse which means that force F at a point in space We sometimes encounter examples where a large (location of the particle) at a certain instant force acts for a very short duration producing a of time is related to a at that point at that finite change in momentum of the body. For instant. Acceleration here and now is example, when a ball hits a wall and bounces determined by the force here and now, not by back, the force on the ball by the wall acts for a any history of the motion of the particle very short time when the two are in contact, yet the force is large enough to reverse the momentum (See Fig. 4.5). of the ball. Often, in these situations, the force ⊳ and the time duration are difficult to ascertain Example 4.2 A bullet of mass 0.04 kg separately. However, the product of force and time, moving with a speed of 90 m s–1 enters a which is the change in momentum of the body heavy wooden block and is stopped after a remains a measurable quantity. This product is distance of 60 cm. What is the average called impulse: resistive force exerted by the block on the bullet? Impulse = Force × time duration = Change in momentum (4.7) Reprint 2025-26 56 PHYSICS A large force acting for a short time to produce a Thus, according to Newtonian mechanics, finite change in momentum is called an impulsive force never occurs singly in nature. Force is the force. In the history of science, impulsive forces were mutual interaction between two bodies. Forces put in a conceptually different category from always occur in pairs. Further, the mutual forces ordinary forces. Newtonian mechanics has no such between two bodies are always equal and distinction. Impulsive force is like any other force – opposite. This idea was expressed by Newton in except that it is large and acts for a short time. the form of the third law of motion. ⊳ To every action, there is always an equal and Example 4.4 A batsman hits back a ball opposite reaction. straight in the direction of the bowler without changing its initial speed of 12 m s–1. Newton’s wording of the third law is so crisp and If the mass of the ball is 0.15 kg, determine beautiful that it has become a part of common the impulse imparted to the ball. (Assume language. For the same reason perhaps, linear motion of the ball) misconceptions about the third law abound. Let us note some important points about the third law, particularly in regard to the usage of theAnswer Change in momentum terms : action and reaction. = 0.15 × 12–(–0.15×12) 1. The terms action and reaction in the third law = 3.6 N s, mean nothing else but ‘force’. Using different Impulse = 3.6 N s, terms for the same physical concept in the direction from the batsman to the bowler. can sometimes be confusing. A simple and clear way of stating the third law is as This is an example where the force on the ball follows :by the batsman and the time of contact of the ball and the bat are difficult to know, but the Forces always occur in pairs. Force on a impulse is readily calculated. ⊳ body A by B is equal and opposite to the force on the body B by A.