5.2 NOTIONS OF WORK AND KINETIC to be proportional to the speed of the drop ENERGY: THE WORK-ENERGY THEOREM but is otherwise undetermined. Consider The following relation for rectilinear motion under a drop of mass 1.00 g falling from a height constant acceleration a has been encountered 1.00 km. It hits the ground with a speed of in Chapter 3, 50.0 m s-1. (a) What is the work done by the v2 − u2 = 2 as (5.2) gravitational force ? What is the work done where u and v are the initial and final speeds by the unknown resistive force? and s the distance traversed. Multiplying both Answer (a) The change in kinetic energy of the sides by m/2, we have drop is 1 2 1 2 1 2 mv − mu = mas = Fs (5.2a) ∆ K = m v − 0 2 2 2 where the last step follows from Newton’s Second 1 -3 = × 10 × 50 × 50 Law. We can generalise Eq. (5.2) to three 2 dimensions by employing vectors = 1.25 J v2 − u2 = 2 a.d where we have assumed that the drop is initially at rest. Here a and d are acceleration and displacement Assuming that g is a constant with a value vectors of the object respectively. 10 m/s2, the work done by the gravitational force Once again multiplying both sides by m/2 , we obtain is, 1 1 mv 2 − mu 2 = m a.d = F.d (5.2b) Wg = mgh 2 2 = 10-3 ×10 ×103 The above equation provides a motivation for = 10.0 J the definitions of work and kinetic energy. The (b) From the work-energy theorem left side of the equation is the difference in the quantity ‘half the mass times the square of the ∆ K = W g + W r speed’ from its initial value to its final value. We where Wr is the work done by the resistive force call each of these quantities the ‘kinetic energy’, on the raindrop. Thus denoted by K. The right side is a product of the Wr = ∆K − Wg displacement and the component of the force = 1.25 −10 along the displacement. This quantity is called = − 8.75 J ‘work’ and is denoted by W. Eq. (5.2b) is then is negative. ⊳ Kf − Ki = W (5.3) 5.3 WORK where Ki and Kf are respectively the initial and As seen earlier, work is related to force and the final kinetic energies of the object. Work refers displacement over which it acts. Consider a to the force and the displacement over which it constant force F acting on an object of mass m. acts. Work is done by a force on the body over The object undergoes a displacement d in the a certain displacement. positive x-direction as shown in Fig. 5.2. Equation (5.2) is also a special case of the work-energy (WE) theorem : The change in kinetic energy of a particle is equal to the work done on it by the net force. We shall generalise the above derivation to a varying force in a later section. ⊳ Example 5.2 It is well known that a raindrop falls under the influence of the Fig. 5.2 An object undergoes a displacement d downward gravitational force and the under the influence of the force F. opposing resistive force. The latter is known Reprint 2025-26 74 PHYSICS The work done by the force is defined to be Table 5.1 Alternative Units of Work/Energy in J the product of component of the force in the direction of the displacement and the magnitude of this displacement. Thus W = (F cos θ)d = F.d (5.4) We see that if there is no displacement, there is no work done even if the force is large. Thus, when you push hard against a rigid brick wall, ⊳ the force you exert on the wall does no work. Yet Example 5.3 A cyclist comes to a skidding your muscles are alternatively contracting and stop in 10 m. During this process, the force relaxing and internal energy is being used up on the cycle due to the road is 200 N and and you do get tired. Thus, the meaning of work is directly opposed to the motion. (a) How in physics is different from its usage in everyday much work does the road do on the cycle ? language. (b) How much work does the cycle do on the road ? No work is done if : (i) the displacement is zero as seen in the example above. A weightlifter holding a 150 Answer Work done on the cycle by the road is kg mass steadily on his shoulder for 30 s the work done by the stopping (frictional) force does no work on the load during this time. on the cycle due to the road. (ii) the force is zero. A block moving on a smooth (a) The stopping force and the displacement make horizontal table is not acted upon by a an angle of 180o (π rad) with each other. horizontal force (since there is no friction), but Thus, work done by the road, may undergo a large displacement. Wr = Fd cosθ (iii) the force and displacement are mutually = 200 × 10 × cos π perpendicular. This is so since, for θ= π/2 rad = – 2000 J (= 90o), cos (π/2) = 0. For the block moving on It is this negative work that brings the cycle a smooth horizontal table, the gravitational to a halt in accordance with WE theorem. force mg does no work since it acts at right (b) From Newton’s Third Law an equal and angles to the displacement. If we assume that opposite force acts on the road due to the the moon’s orbits around the earth is cycle. Its magnitude is 200 N. However, the perfectly circular then the earth’s road undergoes no displacement. Thus, gravitational force does no work. The moon’s work done by cycle on the road is zero. ⊳ instantaneous displacement is tangential while the earth’s force is radially inwards and The lesson of Example 5.3 is that though the θ = π/2. force on a body A exerted by the body B is always Work can be both positive and negative. If θ is equal and opposite to that on B by A (Newton’s between 0o and 90o, cos θ in Eq. (5.4) is positive. Third Law); the work done on A by B is not If θ is between 90o and 180o, cos θ is negative. necessarily equal and opposite to the work done In many examples the frictional force opposes on B by A. displacement and θ = 180o. Then the work done 5.4 KINETIC ENERGY by friction is negative (cos 180o = –1). As noted earlier, if an object of mass m has From Eq. (5.4) it is clear that work and energy velocity v, its kinetic energy K ishave the same dimensions, [ML2T–2]. The SI unit of these is joule (J), named after the famous British 1 1 2physicist James Prescott Joule (1811-1869). Since K = m v. v = mv (5.5) 2 2work and energy are so widely used as physical concepts, alternative units abound and some of Kinetic energy is a scalar quantity. The kinetic these are listed in Table 5.1. energy of an object is a measure of the work an Reprint 2025-26 WORK, ENERGY AND POWER 75 Table 5.2 Typical kinetic energies (K) object can do by the virtue of its motion. This This is illustrated in Fig. 5.3(a). Adding notion has been intuitively known for a long time. successive rectangular areas in Fig. 5.3(a) we The kinetic energy of a fast flowing stream get the total work done as has been used to grind corn. Sailing x f ships employ the kinetic energy of the wind. Table W ≅ F (x )∆x (5.6) ∑

5.10 POWER Often it is interesting to know not only the work u Example 5.10 An elevator can carry a done on an object, but also the rate at which maximum load of 1800 kg (elevator + this work is done. We say a person is physically passengers) is moving up with a constant fit if he not only climbs four floors of a building speed of 2 m s–1. The frictional force opposing but climbs them fast. Power is defined as the the motion is 4000 N. Determine the time rate at which work is done or energy is minimum power delivered by the motor to transferred. the elevator in watts as well as in horse The average power of a force is defined as the power. ratio of the work, W, to the total time t taken Answer The downward force on the elevator is W Pav = F = m g + = (1800 × 10) + 4000 = 22000 N t Ff The motor must supply enough power to balanceThe instantaneous power is defined as the this force. Hence,limiting value of the average power as time interval approaches zero, P = F. v = 22000 × 2 = 44000 W = 59 hp ⊳ d W 5.11 COLLISIONS P = (5.20) d t In physics we study motion (change in position). The work dW done by a force F for a displacement At the same time, we try to discover physical dr is dW = F.dr. The instantaneous power can quantities, which do not change in a physical also be expressed as process. The laws of momentum and energy conservation are typical examples. In this d r P = F. section we shall apply these laws to a commonly d t encountered phenomena, namely collisions. Several games such as billiards, marbles or = F.v (5.21) carrom involve collisions.We shall study the where v is the instantaneous velocity when the collision of two masses in an idealised form. force is F. Consider two masses m1 and m2. The particle Power, like work and energy, is a scalar m1 is moving with speed v1i , the subscript ‘i’ quantity. Its dimensions are [ML2T–3]. In the SI, implying initial. We can cosider m2 to be at rest. its unit is called a watt (W). The watt is 1 J s–1. No loss of generality is involved in making such The unit of power is named after James Watt, a selection. In this situation the mass m1 one of the innovators of the steam engine in the collides with the stationary mass m2 and this eighteenth century. is depicted in Fig. 5.10. There is another unit of power, namely the horse-power (hp) 1 hp = 746 W This unit is still used to describe the output of automobiles, motorbikes, etc. We encounter the unit watt when we buy electrical goods such as bulbs, heaters and refrigerators. A 100 watt bulb which is on for 10 hours uses 1 kilowatt hour (kWh) of energy. 100 (watt) × 10 (hour) Fig. 5.10 Collision of mass m1, with a stationary mass m2. = 1000 watt hour The masses m1 and m2 fly-off in different =1 kilowatt hour (kWh) directions. We shall see that there are = 103 (W) × 3600 (s) relationships, which connect the masses, the = 3.6 × 106 J velocities and the angles. Reprint 2025-26 84 PHYSICS 5.11.1 Elastic and Inelastic Collisions The loss in kinetic energy on collision is In all collisions the total linear momentum is 1 2 1 2 conserved; the initial momentum of the system ∆ K = m 1 v1i − (m 1 + m 2 )v f 2 2 is equal to the final momentum of the system. One can argue this as follows. When two objects 2 1 2 1 m 1 2 collide, the mutual impulsive forces acting over = m 1v1i − v1i [using Eq. (5.22)] the collision time ∆t cause a change in their 2 2 m 1 + m 2 respective momenta : 1 2  m 1  ∆p1 = F12 ∆t = 2 m 1v1i 1 − m 1 + m 2  ∆p2 = F21 ∆t where F12 is the force exerted on the first particle 1 m 1m 2 2by the second particle. F21 is likewise the force = v1i 2 m 1 + m 2exerted on the second particle by the first particle. Now from Newton’s third law, F12 = − F21. This which is a positive quantity as expected.implies ∆p1 + ∆p2 = 0 Consider next an elastic collision. Using the above nomenclature with θ1 = θ2 = 0, the The above conclusion is true even though the momentum and kinetic energy conservation forces vary in a complex fashion during the equations are collision time ∆t. Since the third law is true at every instant, the total impulse on the first object m1v1i = m1v1f + m2v2f (5.23) is equal and opposite to that on the second. 2 2 2 m 1v1i = m 1v1 f + m 2 v 2 f (5.24) On the other hand, the total kinetic energy of the system is not necessarily conserved. The From Eqs. (5.23) and (5.24) it follows that, impact and deformation during collision may generate heat and sound. Part of the initial kinetic m 1v1i (v 2 f − v1i ) = m 1v1 f (v 2 f − v1 f ) energy is transformed into other forms of energy. A useful way to visualise the deformation during or, v 2 f (v1i − v1 f ) = v12i − v12f collision is in terms of a ‘compressed spring’. If the ‘spring’ connecting the two masses regains = (v1i − v1 f )(v 1i + v 1 f ) its original shape without loss in energy, then the initial kinetic energy is equal to the final Hence, ∴ v 2 f = v1i + v1 f (5.25) kinetic energy but the kinetic energy during the Substituting this in Eq. (5.23), we obtaincollision time ∆t is not constant. Such a collision is called an elastic collision. On the other hand (m 1 − m 2 ) v1 f = v1i (5.26)the deformation may not be relieved and the two m 1 + m 2 bodies could move together after the collision. A 2m 1v1icollision in which the two particles move together and v 2 f = (5.27) m 1 + m 2after the collision is called a completely inelastic collision. The intermediate case where the Thus, the ‘unknowns’ {v1f, v2f} are obtained in deformation is partly relieved and some of the terms of the ‘knowns’ {m1, m2, v1i}. Special cases initial kinetic energy is lost is more common and of our analysis are interesting. is appropriately called an inelastic collision. Case I : If the two masses are equal 5.11.2 Collisions in One Dimension v1f = 0 Consider first a completely inelastic collision v2f = v1i in one dimension. Then, in Fig. 5.10, The first mass comes to rest and pushes off the θ 1 = θ 2 = 0 second mass with its initial speed on collision. Case II : If one mass dominates, e.g. m2 > > m1 m1v1i = (m1+m2)vf (momentum conservation) v1f ~ − v1i v2f ~ 0 m 1 The heavier mass is undisturbed while the v f = v1i (5.22) m 1 + m 2 lighter mass reverses its velocity. Reprint 2025-26 WORK, ENERGY AND POWER 85 ⊳ dimensional, where the initial velocities and the Example 5.11 Slowing down of neutrons: final velocities lie in a plane. In a nuclear reactor a neutron of high speed (typically 107 m s–1) must be slowed 5.11.3 Collisions in Two Dimensions to 103 m s–1 so that it can have a high Fig. 5.10 also depicts the collision of a moving probability of interacting with isotope 235U92 mass m1 with the stationary mass m2. Linear momentum is conserved in such a collision. and causing it to fission. Show that a neutron can lose most of its kinetic energy Since momentum is a vector this implies three in an elastic collision with a light nuclei equations for the three directions {x, y, z}. like deuterium or carbon which has a mass Consider the plane determined by the final of only a few times the neutron mass. The velocity directions of m1 and m2 and choose it to material making up the light nuclei, usually be the x-y plane. The conservation of the heavy water (D2O) or graphite, is called a z-component of the linear momentum implies moderator. that the entire collision is in the x-y plane. The x- and y-component equations are Answer The initial kinetic energy of the neutron m1v1i = m1v1f cos θ1 + m2v2f cos θ2 (5.28)is 0 = m1v1f sin θ1 − m2v2f sin θ2 (5.29) 1 2 K 1i = m 1v1i 2 One knows {m1, m2, v1i} in most situations. There are thus four unknowns {v1f, v2f , θ1 and θ2}, andwhile its final kinetic energy from Eq. (5.26) only two equations. If θ 1 = θ 2 = 0, we regain 1 2 1  m 1 − m 2  2 2 Eq. (5.23) for one dimensional collision. K 1 f = m 1v1 f = m 1 v1i 2 2  m 1 + m 2  If, further the collision is elastic, 1 2 1 2 1 2 m1v1i = m1v1 f + m 2v 2 f (5.30) The fractional kinetic energy lost is 2 2 2 2 We obtain an additional equation. That still K 1 f  m 1 − m 2  f 1 = = leaves us one equation short. At least one of K 1i  m 1 + m 2  the four unknowns, say θ1, must be made known while the fractional kinetic energy gained by the for the problem to be solvable. For example, θ1 moderating nuclei K2f /K1i is can be determined by moving a detector in an angular fashion from the x to the y axis. Given f2 = 1 − f1 (elastic collision) {m1, m2, v1i , θ1} we can determine {v1f , v2f, θ2} 4m 1m 2 from Eqs. (5.28)-(5.30). = 2 ⊳ (m1 + m 2 ) Example 5.12 Consider the collision depicted in Fig. 5.10 to be between two One can also verify this result by substituting billiard balls with equal masses m1 = m2.from Eq. (5.27). The first ball is called the cue while the For deuterium m2 = 2m1 and we obtain second ball is called the target. The f1 = 1/9 while f2 = 8/9. Almost 90% of the billiard player wants to ‘sink’ the target neutron’s energy is transferred to deuterium. For ball in a corner pocket, which is at an carbon f1 = 71.6% and f2 = 28.4%. In practice, angle θ2 = 37°. Assume that the collision ishowever, this number is smaller since head-on elastic and that friction and rotational collisions are rare. ⊳ motion are not important. Obtain θ1. If the initial velocities and final velocities of Answer From momentum conservation, since both the bodies are along the same straight line, the masses are equal then it is called a one-dimensional collision, or head-on collision. In the case of small spherical v1i = v 1f + v 2f bodies, this is possible if the direction of travel of body 1 passes through the centre of body 2 ⋅ or v 1i 2 = ( v1 f + v 2 f ) ( v1 f + v 2 f ) which is at rest. In general, the collision is two- = v1 f 2 + v 2 f 2 + 2 v1 f .v 2 f Reprint 2025-26 86 PHYSICS 2 2 The matter simplifies greatly if we consider= (5.31) { v1 f + v 2 f + 2v1 f v 2 f cos (θ1 + 37 ° ) } spherical masses with smooth surfaces, and assume that collision takes place only when the Since the collision is elastic and m1 = m2 it follows bodies touch each other. This is what happensfrom conservation of kinetic energy that in the games of marbles, carrom and billiards. v1i 2 = v1 f 2 + v 2 f 2 (5.32) In our everyday world, collisions take place only when two bodies touch each other. But considerComparing Eqs. (5.31) and (5.32), we get a comet coming from far distances to the sun, or cos (θ1 + 37°) = 0 alpha particle coming towards a nucleus and or θ1 + 37° = 90° going away in some direction. Here we have to deal with forces involving action at a distance. Thus, θ1 = 53° Such an event is called scattering. The velocities This proves the following result : when two equal and directions in which the two particles go away masses undergo a glancing elastic collision with depend on their initial velocities as well as the one of them at rest, after the collision, they will type of interaction between them, their masses, move at right angles to each other. ⊳ shapes and sizes. SUMMARY 1. The work-energy theorem states that the change in kinetic energy of a body is the work done by the net force on the body. Kf - Ki = Wnet 2. A force is conservative if (i) work done by it on an object is path independent and depends only on the end points {xi, xj}, or (ii) the work done by the force is zero for an arbitrary closed path taken by the object such that it returns to its initial position. 3. For a conservative force in one dimension, we may define a potential energy function V(x) such that d V ( x ) F ( x ) = − d x x f F ( x ) d x or Vi − V f = ∫ x i 4. The principle of conservation of mechanical energy states that the total mechanical energy of a body remains constant if the only forces that act on the body are conservative. 5. The gravitational potential energy of a particle of mass m at a height x about the earth’s surface is V(x) = m g x where the variation of g with height is ignored. 5. The elastic potential energy of a spring of force constant k and extension x is 1 2 V ( x ) = k x 2 7. The scalar or dot product of two vectors A and B is written as A.B and is a scalar quantity given by :A.B = AB cos θ, where θ is the angle between A and B. It can be positive, negative or zero depending upon the value of θ. The scalar product of two vectors can be interpreted as the product of magnitude of one vector and component of the other vector along the first vector. For unit vectors : ˆi ⋅ ˆi = ˆj ⋅ ˆj = kˆ ⋅ kˆ = 1 and ˆi ⋅ ˆj = ˆj ⋅ kˆ = kˆ ⋅ ˆi = 0 Scalar products obey the commutative and the distributive laws. Reprint 2025-26 WORK, ENERGY AND POWER 87 POINTS TO PONDER 1. The phrase ‘calculate the work done’ is incomplete. We should refer (or imply clearly by context) to the work done by a specific force or a group of forces on a given body over a certain displacement. 2. Work done is a scalar quantity. It can be positive or negative unlike mass and kinetic energy which are positive scalar quantities. The work done by the friction or viscous force on a moving body is negative. 3. For two bodies, the sum of the mutual forces exerted between them is zero from Newton’s Third Law, F12 + F21 = 0 But the sum of the work done by the two forces need not always cancel, i.e. W12 + W21 ≠ 0 However, it may sometimes be true. 4. The work done by a force can be calculated sometimes even if the exact nature of the force is not known. This is clear from Example 5.2 where the WE theorem is used in such a situation. 5. The WE theorem is not independent of Newton’s Second Law. The WE theorem may be viewed as a scalar form of the Second Law. The principle of conservation of mechanical energy may be viewed as a consequence of the WE theorem for conservative forces. 5. The WE theorem holds in all inertial frames. It can also be extended to non- inertial frames provided we include the pseudoforces in the calculation of the net force acting on the body under consideration. 7. The potential energy of a body subjected to a conservative force is always undetermined upto a constant. For example, the point where the potential energy is zero is a matter of choice. For the gravitational potential energy mgh, the zero of the potential energy is chosen to be the ground. For the spring potential energy kx2/2 , the zero of the potential energy is the equilibrium position of the oscillating mass. 8. Every force encountered in mechanics does not have an associated potential energy. For example, work done by friction over a closed path is not zero and no potential energy can be associated with friction. 9. During a collision : (a) the total linear momentum is conserved at each instant of the collision ; (b) the kinetic energy conservation (even if the collision is elastic) applies after the collision is over and does not hold at every instant of the collision. In fact the two colliding objects are deformed and may be momentarily at rest with respect to each other. Reprint 2025-26 88 PHYSICS EXERCISES 5.1 The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative: (a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket. (b) work done by gravitational force in the above case, (c) work done by friction on a body sliding down an inclined plane, (d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity, (e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest. 5.2 A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the (a) work done by the applied force in 10 s, (b) work done by friction in 10 s, (c) work done by the net force on the body in 10 s, (d) change in kinetic energy of the body in 10 s, and interpret your results. 5.3 Given in Fig. 5.11 are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant. Fig. 5.11 Reprint 2025-26 WORK, ENERGY AND POWER 89 5.4 The potential energy function for a particle executing linear simple harmonic motion is given by V(x) = kx2/2, where k is the force constant of the oscillator. For k = 0.5 N m-1, the graph of V(x) versus x is shown in Fig. 5.12. Show that a particle of total energy 1 J moving under this Fig. 5.12 potential must ‘turn back’ when it reaches x = ± 2 m. 5.5 Answer the following : (a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere? (b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not Fig. 5.13 normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why ? (c) An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth ? (d) In Fig. 5.13(i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig. 5.13(ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater ? 5.6 Underline the correct alternative : (a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered. (b) Work done by a body against friction always results in a loss of its kinetic/potential energy. (c) The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces on the system. (d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies. 5.7 State if each of the following statements is true or false. Give reasons for your answer. (a) In an elastic collision of two bodies, the momentum and energy of each body is conserved. (b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present. (c) Work done in the motion of a body over a closed loop is zero for every force in nature. (d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system. 5.8 Answer carefully, with reasons : (a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact) ? (b) Is the total linear momentum conserved during the short time of an elastic collision of two balls ? Reprint 2025-26 90 PHYSICS (c) What are the answers to (a) and (b) for an inelastic collision ? (d) If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic ? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy). 5.9 A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to (i) t1/2 (ii) t (iii) t3/2 (iv) t2 5.10 A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to (i) t1/2 (ii) t (iii) t3/2 (iv) t2 5.11 A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by F = −ˆi + 2 ˆj + 3 kˆ N where ˆ,i ˆ,j kˆ are unit vectors along the x-, y- and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis ? 5.12 An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton ? Obtain the ratio of their speeds. (electron mass = 9.11×10-31 kg, proton mass = 1.67×10–27 kg, 1 eV = 1.60 ×10–19 J). 5.13 A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey ? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 m s–1 ? 5.14 A molecule in a gas container hits a horizontal wall with speed 200 m s–1 and angle 30° with the normal, and rebounds with the same speed. Is momentum conserved in the collision ? Is the collision elastic or inelastic ? 5.15 A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m3 in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump ? 5.16 Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed V. If the collision is elastic, which of the following (Fig. 5.14) is a possible result after collision ? Fig. 5.14 Reprint 2025-26 WORK, ENERGY AND POWER 91 5.17 The bob A of a pendulum released from 30o to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 5.15. How high does the bob A rise after the collision ? Neglect the size of the bobs and assume the collision to be elastic. 5.18 The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance ? 5.19 A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a Fig. 5.15 frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.05 kg s–1. What is the speed of the trolley after the entire sand bag is empty ? 5.20 A body of mass 0.5 kg travels in a straight line with velocity v =a x3/2 where a = 5 m–1/2 s–1. What is the work done by the net force during its displacement from x = 0 to x = 2 m ? 5.21 The blades of a windmill sweep out a circle of area A. (a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t ? (b) What is the kinetic energy of the air ? (c) Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that A = 30 m2, v = 36 km/h and the density of air is 1.2 kg m–3. What is the electrical power produced ? 5.22 A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated. (a) How much work does she do against the gravitational force ? (b) Fat supplies 3.8 × 107J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up? 5.23 A family uses 8 kW of power. (a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW? (b) Compare this area to that of the roof of a typical house. Reprint 2025-26 CHAPTER SIX SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 6.1 INTRODUCTION In the earlier chapters we primarily considered the motion of a single particle. (A particle is ideally represented as a 6.1 Introduction point mass having no size.) We applied the results of our 6.2 Centre of mass study even to the motion of bodies of finite size, assuming 6.3 Motion of centre of mass that motion of such bodies can be described in terms of the

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