8.2 STRESS AND STRAIN forces are applied parallel to the cross-sectional When forces are applied on a body in such a area of the cylinder, as shown in Fig. 8.1(b), manner that the body is still in static equilibrium, there is relative displacement between the it is deformed to a small or large extent depending opposite faces of the cylinder. The restoring force upon the nature of the material of the body and per unit area developed due to the applied the magnitude of the deforming force. The tangential force is known as tangential or deformation may not be noticeable visually in shearing stress. many materials but it is there. When a body is As a result of applied tangential force, there subjected to a deforming force, a restoring force is a relative displacement ∆x between opposite is developed in the body. This restoring force is faces of the cylinder as shown in the Fig. 8.1(b). equal in magnitude but opposite in direction to The strain so produced is known as shearing the applied force. The restoring force per unit area strain and it is defined as the ratio of relative is known as stress. If F is the force applied normal displacement of the faces ∆x to the length of the to the cross–section and A is the area of cross cylinder L. section of the body, ∆x Magnitude of the stress = F/A (8.1) Shearing strain = = tan θ (8.3) L The SI unit of stress is N m–2 or pascal (Pa) and its dimensional formula is [ ML–1T–2 ]. where θ is the angular displacement of the There are three ways in which a solid may cylinder from the vertical (original position of the change its dimensions when an external force cylinder). Usually θ is very small, tan θ acts on it. These are shown in Fig. 8.1. In is nearly equal to angle θ, (if θ = 10°, for Fig.8.1(a), a cylinder is stretched by two equal example, there is only 1% difference between θ forces applied normal to its cross-sectional area. and tan θ). The restoring force per unit area in this case is It can also be visualised, when a book is called tensile stress. If the cylinder is pressed with the hand and pushed horizontally, compressed under the action of applied forces, as shown in Fig. 8.2 (c). the restoring force per unit area is known as Thus, shearing strain = tan θ ≈ θ (8.4) compressive stress. Tensile or compressive In Fig. 8.1 (d), a solid sphere placed in the fluid stress can also be termed as longitudinal stress. under high pressure is compressed uniformly on In both the cases, there is a change in the all sides. The force applied by the fluid acts in length of the cylinder. The change in the length perpendicular direction at each point of the ∆L to the original length L of the body (cylinder surface and the body is said to be under in this case) is known as longitudinal strain. hydraulic compression. This leads to decrease (a) (b) (c) (d) Fig. 8.1 (a) A cylindrical body under tensile stress elongates by ∆L (b) Shearing stress on a cylinder deforming it by an angle θ(c) A body subjected to shearing stress (d) A solid body under a stress normal to the surface at every point (hydraulic stress). The volumetric strain is ∆V/V, but there is no change in shape. Reprint 2025-26 MECHANICAL PROPERTIES OF SOLIDS 169 in its volume without any change of its compression and shear stress may also be geometrical shape. obtained. The stress-strain curves vary from The body develops internal restoring forces material to material. These curves help us to that are equal and opposite to the forces applied understand how a given material deforms with by the fluid (the body restores its original shape increasing loads. From the graph, we can see and size when taken out from the fluid). The that in the region between O to A, the curve is internal restoring force per unit area in this case linear. In this region, Hooke’s law is obeyed. is known as hydraulic stress and in magnitude The body regains its original dimensions when is equal to the hydraulic pressure (applied force the applied force is removed. In this region, the per unit area). solid behaves as an elastic body. The strain produced by a hydraulic pressure is called volume strain and is defined as the ratio of change in volume (∆V) to the original volume (V). ∆V Volume strain = (8.5) V Since the strain is a ratio of change in dimension to the original dimension, it has no units or dimensional formula.

5.8 THE CONSERVATION OF MECHANICAL a ball of mass m being dropped from a cliff of ENERGY height H. For simplicity we demonstrate this important principle for one-dimensional motion. Suppose that a body undergoes displacement ∆x under the action of a conservative force F. Then from the WE theorem we have, ∆K = F(x) ∆x If the force is conservative, the potential energy function V(x) can be defined such that − ∆V = F(x) ∆x The above equations imply that Fig. 5.5 The conversion of potential energy to kinetic ∆K + ∆V = 0 energy for a ball of mass m dropped from a ∆(K + V ) = 0 (5.10) height H. Reprint 2025-26 WORK, ENERGY AND POWER 79 The total mechanical energies E0, Eh, and EH Answer (i) There are two external forces on of the ball at the indicated heights zero (ground the bob : gravity and the tension (T ) in the level), h and H, are string. The latter does no work since the displacement of the bob is always normal to the EH = mgH (5.11 a) string. The potential energy of the bob is thus 1 2 E h = mgh + mv h (5.11 b) associated with the gravitational force only. The 2 2 total mechanical energy E of the system is E0 = (1/2) mvf (5.11 c) conserved. We take the potential energy of the The constant force is a special case of a spatially system to be zero at the lowest point A. Thus, dependent force F(x). Hence, the mechanical at A : energy is conserved. Thus EH = E0 1 2 1 2 E = mv0 (5.12) or, mgH = mv f 2 2 v f = 2 gH [Newton’s Second Law] a result that was obtained in section 5.7 for a where TA is the tension in the string at A. At thefreely falling body. highest point C, the string slackens, as the Further, tension in the string (TC) becomes zero. EH = Eh Thus, at Cwhich implies, 2 1 2 v h = 2 g(H − h) (5.11 d) E = mv c + 2mgL (5.13) 2 and is a familiar result from kinematics. At the height H, the energy is purely potential. mvc2It is partially converted to kinetic at height h and mg = [Newton’s Second Law] (5.14) L is fully kinetic at ground level. This illustrates the conservation of mechanical energy. where vC is the speed at C. From Eqs. (5.13) ⊳ and (5.14) Example 5.7 A bob of mass m is suspended 5 by a light string of length L . It is imparted a E = mgL horizontal velocity vo at the lowest point A 2 such that it completes a semi-circular Equating this to the energy at A trajectory in the vertical plane with the string 5 m 2 becoming slack only on reaching the topmost mgL = v 0 point, C. This is shown in Fig. 5.6. Obtain an 2 2 expression for (i) vo; (ii) the speeds at points or, v 0 = 5 gL B and C; (iii) the ratio of the kinetic energies (ii) It is clear from Eq. (5.14) (KB/KC) at B and C. Comment on the nature of the trajectory of the bob after it reaches vC = gL the point C. At B, the energy is 1 2 E = mv B + mgL 2 Equating this to the energy at A and employing the result from (i), namely v 02 = 5 gL , 1 2 1 2 mv B + mgL = mv 0 2 2 5 = m g L Fig. 5.6 2 Reprint 2025-26 80 PHYSICS ∴ vB = 3 gL k x m2 W = + (5.16) 2 (iii) The ratio of the kinetic energies at B and C is : 1 2 mv B K B 2 3 = = 2 1 K C 1 mvC 2 At point C, the string becomes slack and the velocity of the bob is horizontal and to the left. If the connecting string is cut at this instant, the bob will execute a projectile motion with horizontal projection akin to a rock kicked horizontally from the edge of a cliff. Otherwise the bob will continue on its circular path and complete the revolution. ⊳

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