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. ⊳

6.12 ANGULAR MOMENTUM IN CASE OF For computing the total angular momentum ROTATION ABOUT A FIXED AXIS of the whole rigid body, we add up the contribution of each particle of the body. We have studied in section 6.7, the angular momentum of a system of particles. We already Thus know from there that the time rate of total We denote by L ⊥ and L z the components of angular momentum of a system of particles L respectively perpendicular to the z-axis andabout a point is equal to the total external torque along the z-axis;on the system taken about the same point. When OC i × m i v i (6.42a)the total external torque is zero, the total angular L ⊥= ∑ momentum of the system is conserved. where mi and vi are respectively the mass and We now wish to study the angular momentum the velocity of the ith particle and Ci is the centrein the special case of rotation about a fixed axis. of the circle described by the particle; The general expression for the total angular momentum of the system of n particles is N and ˆ (6.42b) L = =∑i 1 ri × p i (6.25b) or L z = Iωk We first consider the angular momentum of The last step follows since the perpendicular a typical particle of the rotating rigid body. We distance of the ith particle from the axis is ri; andthen sum up the contributions of individual by definition the moment of inertia of the body particles to get L of the whole body. m i ri2 . For a typical particle l = r × p. As seen in the about the axis of rotation is I =∑ last section r = OP = OC + CP [Fig. 6.17(b)]. With Note L = L z + L ⊥ (6.42c)p = m v , l = ( OC × m v ) + ( CP × m v ) The rigid bodies which we have mainly considered in this chapter are symmetric about The magnitude of the linear velocity v of the the axis of rotation, i.e. the axis of rotation is particle at P is given by v = ωr⊥ where r⊥ is the one of their symmetry axes. For such bodies, for length of CP or the perpendicular distance of P a given OCi, for every particle which has a from the axis of rotation. Further, v is tangential velocity vi , there is another particle of velocity at P to the circle which the particle describes. –vi located diametrically opposite on the circle Using the right-hand rule one can check that with centre Ci described by the particle. TogetherCP × v is parallel to the fixed axis. The unit vector along the fixed axis (chosen as the z-axis) such pairs will contribute zero to L ⊥ and as a is ˆk . Hence result for symmetric bodies L ⊥ is zero, and CP × m v = r⊥ (mv ) kˆ hence z = Iωkˆ (6.42d) = mr⊥2ω kˆ (since υ = ωr⊥ ) L = L Similarly, we can check that OC × v is For bodies, which are not symmetric about perpendicular to the fixed axis. Let us denote the axis of rotation, L is not equal to Lz and hence the part of l along the fixed axis (i.e. the z-axis) L does not lie along the axis of rotation. by lz, then Referring to Table 6.1, can you tell in which l z = CP × m v = mr⊥2ωkˆ cases L = Lz will not apply? Let us differentiate Eq. (6.42b). Since ˆk is a and l = l z + OC × m v fixed (constant) vector, we get We note that lz is parallel to the fixed axis, ˆbut l is not. In general, for a particle, the angular I ω) k d ( L z ) =  d (  d t momentum l is not along the axis of rotation, d t i.e. for a particle, l and ω are not necessarily Now, Eq. (6.28b) states parallel. Compare this with the corresponding dL fact in translation. For a particle, p and v are = τ dtalways parallel to each other. Reprint 2025-26 122 PHYSICS As we have seen in the last section, only We have already derived this equation using those components of the external torques which the work - kinetic energy route. are along the axis of rotation, need to be taken into account, when we discuss rotation about a 6.12.1 Conservation of angular momentum fixed axis. This means we can take τ = τkˆ . We are now in a position to revisit the principle of conservation of angular momentum in the Since L = L z + L ⊥ and the direction of Lz (vector context of rotation about a fixed axis. From Eq. ˆk ) is fixed, it follows that for rotation about a (6.43c), if the external torque is zero, fixed axis, Lz = Iω = constant (6.44) For symmetric bodies, from Eq. (6.42d), Lz d L z = τkˆ (6.43a) may be replaced by L .(L and Lz are respectively d t the magnitudes of L and Lz.) This then is the required form, for fixed axis d L rotation, of Eq. (6.29a), which expresses theand ⊥= 0 (6.43b) dt general law of conservation of angular momentum Thus, for rotation about a fixed axis, the of a system of particles. Eq. (6.44) applies to many component of angular momentum perpendicular situations that we come across in daily life. You may do this experiment with your friend. Sit on a to the fixed axis is constant. As L z = Iωkˆ , we swivel chair (a chair with a seat, free to rotate get from Eq. (6.43a), about a pivot) with your arms folded and feet not resting on, i.e., away from, the ground. Ask your d ( Iω) = τ (6.43c) friend to rotate the chair rapidly. While the chair d t is rotating with considerable angular speed If the moment of inertia I does not change with stretch your arms horizontally. What happens? time, Your angular speed is reduced. If you bring back d dω your arms closer to your body, the angular speed ( Iω) = I = Iα increases again. This is a situation where thed t d t principle of conservation of angular momentumand we get from Eq. (6.43c), is applicable. If friction in the rotational τ = I α (6.41) Fig 6.32 (a) A demonstration of conservation of Fig 6.32 (b) An acrobat employing the principle of angular momentum. A girl sits on a conservation of angular momentum in swivel chair and stretches her arms/ her performance. brings her arms closer to the body. Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 123 mechanism is neglected, there is no external A circus acrobat and a diver take advantage torque about the axis of rotation of the chair and of this principle. Also, skaters and classical, hence Iω is constant. Stretching the arms Indian or western, dancers performing a increases I about the axis of rotation, resulting in pirouette (a spinning about a tip–top) on the toes decreasing the angular speed ω. Bringing the of one foot display ‘mastery’ over this principle. arms closer to the body has the opposite effect. Can you explain? SUMMARY 1. Ideally, a rigid body is one for which the distances between different particles of the body do not change, even though there are forces on them. 2. A rigid body fixed at one point or along a line can have only rotational motion. A rigid body not fixed in some way can have either pure translational motion or a combination of translational and rotational motions. 3. In rotation about a fixed axis, every particle of the rigid body moves in a circle which lies in a plane perpendicular to the axis and has its centre on the axis. Every Point in the rotating rigid body has the same angular velocity at any instant of time. 4. In pure translation, every particle of the body moves with the same velocity at any instant of time. 5. Angular velocity is a vector. Its magnitude is ω = dθ/dt and it is directed along the axis of rotation. For rotation about a fixed axis, this vector ω has a fixed direction. 6. The vector or cross product of two vector a and b is a vector written as a×b. The magnitude of this vector is absinθ and its direction is given by the right handed screw or the right hand rule. 7. The linear velocity of a particle of a rigid body rotating about a fixed axis is given by v = ω × r, where r is the position vector of the particle with respect to an origin along the fixed axis. The relation applies even to more general rotation of a rigid body with one point fixed. In that case r is the position vector of the particle with respect to the fixed point taken as the origin. 8. The centre of mass of a system of n particles is defined as the point whose position vector is ri ∑m i R = M 9. Velocity of the centre of mass of a system of particles is given by V = P/M, where P is the linear momentum of the system. The centre of mass moves as if all the mass of the system is concentrated at this point and all the external forces act at it. If the total external force on the system is zero, then the total linear momentum of the system is constant. 10. The angular momentum of a system of n particles about the origin is n L = ri × pi i =∑1 The torque or moment of force on a system of n particles about the origin is τ = ∑ri × Fi 1 The force Fi acting on the ith particle includes the external as well as internal forces. Assuming Newton’s third law of motion and that forces between any two particles act along the line joining the particles, we can show τint = 0 and Reprint 2025-26 124 PHYSICS dL = τ ext dt 11. A rigid body is in mechanical equilibrium if (1) it is in translational equilibrium, i.e., the total external force on it is zero : Fi = 0 , ∑ and (2) it is in rotational equilibrium, i.e. the total external torque on it is zero : Fi = 0 . ∑ τi = ∑ri × 12. The centre of gravity of an extended body is that point where the total gravitational torque on the body is zero. 13. The moment of intertia of a rigid body about an axis is defined by the formula I m i ri2 =∑ where ri is the perpendicular distance of the ith point of the body from the axis. The 1 2 kinetic energy of rotation is K = Iω . 2 POINTS TO PONDER 1. To determine the motion of the centre of mass of a system no knowledge of internal forces of the system is required. For this purpose we need to know only the external forces on the body. 2. Separating the motion of a system of particles as the motion of the centre of mass, (i.e., the translational motion of the system) and motion about (i.e. relative to) the centre of mass of the system is a useful technique in dynamics of a system of particles. One example of this technique is separating the kinetic energy of a system of particles K as the kinetic energy of the system about its centre of mass K′ and the kinetic energy of the centre of mass MV2/2, K = K′ + MV2/2 3. Newton’s Second Law for finite sized bodies (or systems of particles) is based in Newton’s Second Law and also Newton’s Third Law for particles. 4. To establish that the time rate of change of the total angular momentum of a system of particles is the total external torque in the system, we need not only Newton’s second law for particles, but also Newton’s third law with the provision that the forces between any two particles act along the line joining the particles. 5. The vanishing of the total external force and the vanishing of the total external torque are independent conditions. We can have one without the other. In a couple, total external force is zero, but total torque is non-zero. 6. The total torque on a system is independent of the origin if the total external force is zero. 7. The centre of gravity of a body coincides with its centre of mass only if the gravitational field does not vary from one part of the body to the other. Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 125 8. The angular momentum L and the angular velocity ω are not necessarily parallel vectors. However, for the simpler situations discussed in this chapter when rotation is about a fixed axis which is an axis of symmetry of the rigid body, the relation L = Iω holds good, where I is the moment of the inertia of the body about the rotation axis. EXERCISES 6.1 Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body ? 6.2 In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10-10 m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus. 6.3 A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system ? 6.4 Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b. 6.5 Show that a.(b × c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors , a, b and c. 6.6 Find the components along the x, y, z axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components px, py and pz. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component. 6.7 Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the angular momentum vector of the two particle system is the same whatever be the point about which the angular momentum is taken. 6.8 A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.6.33. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end. Fig. 6.33 6.9 A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel. Reprint 2025-26 126 PHYSICS 6.10 Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time. 6.11 A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s-1. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis? 6.12 (a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value ? Assume that the turntable rotates without friction. (b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy? 6.13 A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N ? What is the linear acceleration of the rope ? Assume that there is no slipping. 6.14 To maintain a rotor at a uniform angular speed of 200 rad s-1, an engine needs to transmit a torque of 180 N m. What is the power required by the engine ? (Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient. 6.15 From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body. 6.16 A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick? 6.17 The oxygen molecule has a mass of 5.30 × 10-26 kg and a moment of inertia of 1.94 ×10-46 kg m2 about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule. Reprint 2025-26 CHAPTER SEVEN GRAVITATION 7.1 INTRODUCTION Early in our lives, we become aware of the tendency of all material objects to be attracted towards the earth. Anything

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