B. B. 6yX08I4e8. B. JI. KpU8IleHIC08. r. fl. MRKuwe8, B. n. llJaAbH08 CBOPHHK 3A)lAq no 3JIEMEHTAPHOA 0 seconds after the first from a height of h2 = 11 em. After a certain time the velocities of the balls coincide in magnitude and direction. Find the time 't and the interval during which the velocities of the two balls remain the same. The balls do not collide. 31. How long will a body freely falling without any initial velocity pass the n-th centimetre of its path? 32. Two bodies are thrown one after the other with the same velocities Vo from a high tower. The first body is thrown vertically upward, and the second one vertically downward after the time 't'. Determine the velocities of the bodies with respect to each other and the distance between them at the moment of time t >T. 33. At the initial moment three points A, Band C are on a horizontal straight line at equal distances from one another. Point A begins to move vertically upward with a constant vee locity v and point C vertically downward without any initial 14 PROBLEMS velocity at a constant acceleration a. How should point B move vertically for all the three points to constantly be on one straight line? The points begin to move simultaneously. 34. Two trucks tow a third one by means of a pulley tied to it (Fig. 10). The accelerations of the trucks are a1 and a2 • Find the acceleration as of the truck being towed. 35. A lift moves with an acceleration a. A passenger in the lift drops a book. What is the acceleration of the book with respect to the lift floor if: (1) the lift is going up, (2) the lift is going down? 36. A railway carriage moves over a straight level track with an acceleration a. A passenger in the carriage drops a stone. What is the acceleration of the stone with respect to the carriage and the Earth? 37. A man in a lift moving with an acceleration a drops a ball from a height H above the floor. In t seconds the acceleration of the lift is reversed, and in 2t seconds becomes equal to zero. Directly after this the ball touches the floor. What height from the floor of the lift will the ball jump to after the impact? Consider the impact to be absolutely elastic. 38. An overhead travelling crane lifts a load from the ground with an upward acceleration of al • At the same time the hook of the crane carrying the load moves in a horizontal direction with an acceleration a, relative to the crane. Besides, the crane runs on its rails with a constant speed Vo (Fig. 11). The initial speed of the hook relative to the crane is zero. Find the speed of the load with respect to the ground when it reaches the height h. 39•. Body A is placed on a wedge forming an angle a with the horizon (Fig. 12). What acceleFig. 10 Fig. 1J MECHANICS Fig. 12 Fig. 13 15 ration should be imparted to the wedge in a horizontal direction for body A to freely fall vertically? 1-3. Dynamics of Rectilinear Motion 40. A force F is applied to the centre of a homogeneous sphere (Fig. 13). In what direction will the sphere move? 41. Six forces lying in one plane and forming angles of 600 relative to one another are applied to the centre of a homogeneous sphere with a mass of m = 4 kg. These forces are consecutively 1, 2, 3, 4, 5 and 6 kgf (Fig. 14). In what direction and with what acceleration will the sphere move? 42. How much does a body with a mass of one kilogram weigh? 43. The resistance of the air causes a body thrown at a certain angle to the horizon to fly along a ballistic curve. At what angle to the horizon is the acceleration of the body directed at Bkgr Flg. 14 A B Fig. 15 16 PROBLEMS the highest point of the trajectory A, if the mass of the body is m and the resistance of the air at this point is F? 44. A disk arranged in a vertical plane has several grooves directed along chords drawn from point A (Fig. 15). Several bodies begin to slide down the respective grooves from point A simultaneously. In what time will each body reach the edge of the disk? Disregard friction and the resistance of the air (Galileo's problem). 45. What is the minimum force of air resistance acting on a parachutist and his parachute when the latter is completely opened? The two weigh 75 kgf. 46. What is the pressure force N exerted by a load weighing a kgf on the floor of a lift if the acceleration of the lift is a? What is this force equal to upon free falling of the lift? 47. A puck with an initial velocity of 5 mls travels over a distance of 10 metres and strikes the boards. After the impact which may be considered as absolutely elastic, the puck travelled another 2.5 metres and stopped. Find the coefficient of friction between the puck and the ice. Consider the force of sliding friction in this and subsequent problems to be equal to the maximum force of friction of rest. 48. The path travelled by a motor vehicle from the moment the brakes are applied until it stops is known as the braking distance. For the types of tyres in common use in the USSR and a normal air pressure in the tubes, the dependence of the braking distance on the speed of the vehicle at the beginning of braking and on the state and type of the pavement can be tabulated (see Table 1). Find the coefficient of friction for the various kinds of pavement surface to an accuracy of the first digit after the decimal point, using the data in this table. 49. Determine the difference in the pressure of petrol on the opposite walls of a fuel tank when a motor vehicle travels over a horizontal road if its speed increases uniformly from Vo= 0 to v during t seconds. The distance between the tank walls is I. The tank has the form of a parallelepiped and its side walls are vertical. It is completely filled with petrol. The density of pet'rol is p. 50. A homogeneous rod with a length L: is acted upon by two forces F1 and F" applied to its ends and directed oppositely (Fig. 16). With what force F will the rod be stretched in the cross section at a distance I from one of the ends? MECHANICS Table 1. Braking Distance In Metres 17 I Speed. krn/h Pavement surface 10 I 20 I 30 I 40 I 50 I 60 I 70 I 80 I 90 I 100 Ice 13.9115.6135.3162.9198.11141.41192.61251.61318.2' 393.0 Dry snow II.91 7·81 17·6131.41 49. 0 I70·71 96. 31 125.81 159. I 1196.6 Wet wood block 1.3 5.2 11.7 20.9 32.7 47.1 64.2 83.8 106.0 131.0 Dry wood block 0.78 3.1 7.0 12.5 19.6 28.2 38.5 50.3 63.6 78.6 Wet asphalt 0:97 3.9 8.8 15.7 24.5 35.3 48.1 62.9 79.5 98.2 Dry asphalt 10.6512.61 5.81 10.4116. 3 123.51 32.1\ 4~.91 53.0 I 65.5 D~ concrete10.561 2.21 5.0 I9.01 14. 0 I 20.21 27. 5 1 35.91 45.41 56.1 51. A light ball is dropped in air and photographed after it covers a distance of 20 metres. A camera with a focal length of 10 em is placed 15 metres away from the plane which the ball is falling in. A disk with eight equally spaced holes arranged along its circumference rotates with a speed of 3 test» in front of the open lens of the camera. As a result a number of images of the ball spaced 3 mm apart are produced on the film. Describe the motion of the ball. What is the final velocity of another ball of the same radius, but with a mass four times greater than that of the first one? Determine the coefficient of I~ l ~I F; I i I ~ I -t: .. ..I t • Fig. 16... ~I 2-2042 18 PROBLEM~ friction if the weight of the first ball is 4.5 gf. At high velocities of falling, the resistance of the air is proportional to the square of the velocity. 52. Two weights with masses m1 = 100 g and m2 = 200 g are suspended from the ends of a string passed over a stationary pulley at a height of H = 2 metres from the floor (Fig. 17). At the initial moment the weights are at rest. Find the tension of the string when the weights move and the time during which the weight m, reaches the floor. Disregard the mass of the pulley and· the string. 53. A weight G is attached' to the axis of a moving pulley (Fig. 18). What force F should be applied to the end of the rope passed around the second pulley for the weight G to move upwards with an acceleration of a? For the weight G to be at rest? Disregard the mass of the pulleys and the rope. 54. Two weights are suspended from a string thrown over a stationary pulley. The mass of one weight is 200 g. The string will not break if a very heavy weight is attached to its other end. What tension is the string designed for? Disregard the mass of the pulley and the string. 55. Two pans with weights each equal to a= 3 kgf are suspended from the ends of a string passed over two stationary pulleys. The string between the pulleys is cut and the free ends are connected to a dynamometer (Fig. 19). What will the dynamometer show? What weight a1 should be added to one of the pans for the reading of the dynamometer not to change after a weight O2 = 1 kgf is taken off the other pan? Disregard the 'masses of the pans, pulleys, thread and dynamometer. Fig. 17 Fig. 18 F Fig. 19 MECHANICS 19 Fig. 20 56. A heavy sphere with a mass m is suspended on a thin rope. Another rope as strong as the first one is attached to the bottom of the sphere. When the lower rope is sharply pulled it breaks. What acceleration will be imparted to the sphere? 57. Two weights with masses m1 and m, are connected by a string passing over a pulley. The surfaces on which they rest form angles a and ~ with the horizontal (Fig. 20). The righthand weight is h metres below the left-hand one. Both weights will be at the same height in 't seconds after motion begins. The coefficients of friction between the weights and the surfaces are k. Determine the relation between the masses of the weights. 58. A slide forms an angle of a = 30° with the horizon. A stone is thrown upward along it and covers a di-t.iuco of 1= 16 metres in t, == 2 seconds, after which it slides down. What time t2 is required for the return motion? What is the coefficient of friction between the slide and the stone? 59. A cart with a mass of M = 500 g is connected by a string to a weight having a mass m = 200 g. At the initial moment the cart moves to the left along a horizontal plane at a speed of Vo = 7 m/s (Fig. 21). Find the magni tude and direction of I{, ~ Fig. 21 .) .aFig. 22 20 Fig. 28 11/ r H ~--I----'" F_ PROBLEMS m, Fig. 24 the speed of the cart, the place it will be at and the distance it covers in t = 5 seconds. 60. Can an ice-boat travel over a level surface faster than the wind which it is propelled by? 61. The fuel supply of a rocket is m=8 tons and its mass (including the fuel) is M = 15 tons. The fuel burns in 40 seconds. The consumption of fuel and the thrust F=20,OOO kgf are constant. (1) The rocket is placed horizontally on a trolley. Find its acceleration at the moment of launching. Find how the acceleration of the rocket depends on the duration of its motion and show the relation graphically. Use the graph to find the velocity acquired by the rocket in 20 seconds after it begins to move. Disregard friction. (2) The rocket is launched vertically upward. Measurements show that in 20 seconds the acceleration of the rocket is 0.8 g. Calculate the force of air resistance which acts on the rocket at this moment. Consider the acceleration g to be constant. (3) The acceleration of the rocket is measured by an instrument having the form of a spring secured in a vertical tube. When at rest, the spring is stretched a distance of 10 = 1 em by a weight secured to its end. Determine the relation between the stretching of the spring and the acceleration of the rocket. Draw the scale of the instrument. 62. A bead of rnass In is fitted onto a rod with a length of 21, and can move on it without friction. At the initial mo- . M.ECHANICS 21 ment the bead is in the middle of the rod. The rod moves translationally in a horizontal plane with an acceleration a in a direction forming an angle ex with the rod (Fig. 22). Find the acceleration of the bead relative to the rod, the reaction force exerted by the rod on the bead, and the time when the bead will leave the rod. 63. Solve the previous problem, assuming that the moving bead is acted upon by a friction force. The coefficient of friction between the bead and the rod is k. Disregard the force of gravity. 64. A block with the mass M rests on a smooth horizontal surface over which it can move without friction. A body with the mass m lies on the block (Fig. 23). The coefficient of friction between the body and the block is k. At what force F applied to the block in a horizontal direction will the body begin to slide over the block? In what time will the body fall from the block if the length of the latter is I? 65. A wagon with the mass M moves without friction over horizontal rails at a speed of VO• A body with the mass m is placed on the fran t edge of the wagon. The initial speed of the body is zero. At what length of the wagon will the body not slip off it? Disregard the dimensions of the body as compared with the length 1 of the wagon. The coefficient of friction between the body and the wagon is k. 66. A weightless string thrown over a stationary pulley is passed through a slit (Fig. 24). As the string moves it is acted upon by a constant friction force F on the side of the slit. The ends of the string carry two weights with masses m1 and m., respectively. Find the acceleration a of the weights. 67. A stationary pulley is secured- to the end of a light bar. The bar is placed onto a balance pan and secured in a vertical direction. Different weights are attached to the ends of a string passed over the pulley. One of the weights slides over the bar with friction and therefore both weights move uniformly (Fig. 25). Determine the force 'which the pulley acts on the bar with and .the readings of the balance when the weights move. Disregard the masses of the pulley, bar, string and the friction in the pulley axis. Consider two cases: (1) m1 = 1 kg, m2 = 3 kg, and (2) m1 = 3 kg, m2 = 1 kg. 68. A system consists of two stationary and one movable . pulleys (Fig. 26). A string thrown over them carries at its ends weights with masses m1 and ms, while a weight with a mass 22 PROBLEMS 15kg, Fig. 25 Fig. Z7 Fig. 26 m; is attached to the axis of the movable pulley. The parts of the string that are not on the pulleys are vertical. Find the acceleration of each weight, neglecting B the masses of the pulleys and the string, and also friction. 69. Two monkeys of the same weight are hanging at the ends of a rope thrown over a stationary pulley. One monkey begins to climb the rope and the other stays where it is. Where will the second Fig. 28 MECHANICS 23 monkey be when the first one reaches the pulley? At the initial moment both monkeys were at the same height from the floor. J)isregard the mass of the pulley and rope, and also friction. 70. Determine the accelerations of the weights in the pulley system depicted in Fig. 27. Disregard the masses of the pulleys and string, and also friction. In what direction will the pulleys rotate when the weights move? 71. A table with a weight of at = 15 kgf can move without friction over a level floor. A weight of O2 = 10 kgf is placed on the table, and a rope passed over two pulleys fastened to the table is attached to it (Fig. 28). The coefficient of friction between the weight and the table k = 0.6. What acceleration will the table move with if a constant force of 8 kgf is applied to the free end of the rope? Consider two cases: (1) the force is directed horizontally, (2) the force is directed vertically upward. 72. An old cannon without a counter-recoil device rests on a horizontal platform. A ball with a mass m and an initial velocity Vo is fired at an angle of a to the horizon. What velocity V1 will be imparted to the cannon directly after the shot if the mass of the cannon is M and the acceleration of the ball in the barrel is much greater than that of free fall? The coefficient of friction between the cannon and the platform is k. 1..4. The Law of Conservation of Momentum 73. A meteorite burns in the atmosphere before it reaches the Earth's surface. What happens to its momentum? 74. Does a homogeneous disk revolving about its axis have any momentum? The axis of the disk is stationary. 75. The horizontal propeller of a helicopter can be driven by an engine mounted inside its fuselage or by the reactive forces of the gases ejected from special nozzles at the ends of the propeller blades. Why does a propeller-engine helicopter need a tail. rotor while a jet helicopter does not need it? 76. A hunter discharges his gun from a light inflated boat. What velocity will be imparted to the boat when the gun is fired if the mass of the hunter and the boat is M = 70 kg, the mass of the shot m = 35 g and the mean initial velocity of the shot Vo = 320 m/s? The barrel of the gun is directed at an angle of '(x = 600 to the horizon. 77. A rocket launched vertically upward explodes at the highest point it reaches. The explosion produces three fr~ents. Prove 24 PROBLEMS ~ 1 2 • ~ Fig. 29 that the vectors of the initial velocities of all three fragments are in one plane. . 78. A man in a boat facing the bank with its stern walks to the bow. How will the distance between the man and the bank change? 79. A boat on a lake is perpendicular to the shore and faces it with its bow. The distance between the bow and the shore is 0.75 metre. At the initial moment the boat was stationary. A man in the boat steps from its bow to its stern. Will the boat reach the shore if it is 2 metres long? The mass of the boat M = 140 kg and that of the man m = 60 kg. 80. Two identical weights are connected by a spring. At the initial moment the spring is so compressed that the first weight is tightly pressed against a wall (Fig. 29) and the second weight is retained by a stop. How will the weights move if the stop is removed? 81. A massive homogeneous cylinder that can revolve without friction around a horizontal axis is secured on a cart standing on a smooth level surface (Fig. 30). A bullet flying horizontally with a velocity v strikes the cylinder and drops onto the cart. Does the speed of the cart that it acquires after the impact depend on the point where the bullet strikes the cylinder? 82. At the initial moment a rocket with a mass M had a velocity VO' At the end of each second the rocket ejects a portion Fig. 30 Fig. 31 MECHANICS 25 of gas with a mass m. The velocity of a portion of gas differs from that of the rocket before the given portion of gas burns by a constant value u, i. e., the velocity of gas outflow is constant. Determine the velocity of the rocket in n seconds, disregarding the force of gravity. 83. Will the velocity of a rocket increase if the outflow velocity of the gases with respect to the rocket is smaller than the velocity of the rocket itself, so that the gases ejected from the nozzle follow the rocket? 84. Two boats move towards each other along parallel paths with the same velocities. When they meet, a sack is thrown from one boat onto the other and then an identical sack is thrown from the second onto the first. The next time this is done simultaneously. When will the velocity of the boats be greater after the . sacks are thrown? 85. A hoop is placed. on an absolutely smooth level surface. A beetle alights on the hoop. What trajectory will be described. by the beetle and the centre of the hoop if the beetle begins to move along the hoop? The radius of the hoop is R, its mass -Is M and the mass of the beetle m. 86. A wedge with an angle a at the base can move without friction over a smooth leveI surface (Fig. 31). At what ratio between the masses m1 and m, of the weights, that are connected by a string passed over a pulley, will the wedge remain stationary, and at what ratio will it begin to move to the right or left? The coefficient of friction between the weight of mass ms and the wedge is k, 1-5. Statics 87. A homogeneous chain with a length 1 lies ·on a table. What is the maximum length l1 of the part of the chain hanging over the table if the coefficient of friction between the chain and the table is k? 88. Two identical weights are suspended from the ends of a string thrown over two pulleys (Fig. 32). Over what distance will a third weight of the same mass lower if it is attached to the middle of the string? The distance between the axes of the . pulleys is 21. The friction in the axes of the pulleys is negligible. 89. An isosceles wedge with an acute angle a, is driven into a slit. At what angle a will the wedge not be forced out of the slit if the coefficient of friction betweenthe wedge and the slit is k? 2(; m Fig. 32 m Fig. 33 PROBLEMS 90. What is the ratio between the weights 01 and Gz if the system shown in Fig. 33 is in equilibrium. Bars AD, BC, CH, DI and arm 00. of the lever are twice as long as bars AE, EB, IJ, JH and arm FO, respectively. Disregard the weight of the bars and the arm. 91. A horizontal force F is applied perpendicularly to an upper edge of a rectangular box with a length 1 and a height h to move it. What should the coefficient of friction k between the box and the floor be so that the box moves without overturning? 92. A homogeneous beam whose weight is G lies on a floor. The coefficient of friction between the beam and the floor is k. What is easier for two men to do-turn the beam about its centre or move it translationally? 93. An overhead travelling crane (see Fig. 11) weighing a= 2 tonf has a span of L = 26 metres. The wire rope carrying a load is at a distance of 1= 10 metres from one of the rails. Determine the force of pressure of the crane on the rails if it lifts a load of Go = 1 tonf with an acceleration of a = 9.8 m/s". 94. A lever is so bent that its sides AR, BC and CD are equal and form right angles with one another (Fig. 34). The axis of the lever is at point B. A force of 0= 1 kgf is applied at point A at right angles to arm AB. Find the minimum force that should be applied at D for the lever to be in equilibrium. Disregard the weight of the lever. 95. A rod not reaching the tloor is inserted between two identical boxes (Fig. 35). A horizontal force F is applied to the upper end of the rod. Which of the boxes will move first? MECHANICS 27 A B C '- D Fig. 34 ~ -----,m m ~ 96. A heavy homogeneous sphere is suspended from a string whose end is attached to a vertical wall. The point at which the string is fastened to the sphere lies on the same vertical as the centre of the sphere. What should the coefficient of friction between the sphere and the wall be for the sphere to remain in equilibrium? 97. A homogeneous rectangular brick lies on an inclined plane (Fig. 36). What half of the brick (left or right) exerts a greater pressure on it? 98. A horizontally directed force equal to the weight of a heavy cylindrical roller with a radius R is applied to its axis to lift it onto a rectangular step. Find the maximum height of the step. 99. A sphere weighing a= 3 kgf lies on two inclined planes forming angles at = 30° and a, = 60° with the horizon. Determine the pressure exerted by the sphere on each plane if there is no friction between the sphere and one of the planes. 100. The front wall of a drawer in a cabinet is provided with two symmetrical handles. The distance between the handles is 1 and the length of the drawer a. The coefficient of friction between the drawer and the cabinet is k. Can the drawer always be pulled out by applying a force perpendicular to the wall of the drawer only to one handle? 101. A homogeneous board is balanced on a rough horizontal log (Fig. 37). After a weight has been added to one of the ends F FI,. 86 Fig. 36 28 Fig. 37 PROBLEMS o ;...-. ..... 8 D Fig. 38 of the boasd, equilibrium can be obtained when the board forms an angle ex with the horizon. What is the coefficient of friction between the board and the log? 102. The upper end of a ladder rests against a smooth" vertical wall and its bottom end stands on a rough floor. The coefficient of friction between the ladder and the floor is k. Determine the angle ex between the ladder and the wall at which the former will be in equilibrium. 103. Solve the previous problem, assuming that the wall is rough and the coefficient of friction between the ladder and the wall is also equal to k. 104. A homogeneous thin rod AB with a length 1 is placed onto the horizontal surface of a table. A string with a length of 21 is attached to end B of the rod (Fig. 38). How will the rod move if the other end C of the string is slowly lifted up a stationary vertical straight line DO passing through end A of the rod. Disregard the weight of the string. ·105. At what coefficient of friction will a man not slip when he runs along a straight hard path? The maximum angle between a vertical line and the line connecting the man's centre of gravity wtth the point of support is cx. 106. A ladder leans against a smooth vertical wall of a house. The angle between the ladder and the horizontal surface of the Earth is a = 60°. The length of the ladder is I, and its centre of gravity is at its middle. How is the force acting on the ladder from the Earth directed? 107. A ladder with its centre of gravity at the middle stands on an absolutely smooth floor and leans against a smooth wall MECHANICS 29 (Fig. 39). What should the tension of a rope tied to the middle of the ladder be to prevent its falling down? 108. A man climbs up a ladder leaning against 8 smooth vertical wall. The ladder begins to slip only when the man reaches a certain height. Why? . 109. A picture is attached to a vertical wall by means of string AC with a length I forming an angle a with the wall. The height of the picture BC = d (Fig. 40). The bottom of the picture is not fastened. At what coefficient of friction between the picture and the wall will the picture be in equilibrium? 110. Four homogeneous rods are pin-connected to one another at points B, C and D (Fig. 41). The two extreme rods AB and DE can freely revolve with respect to stationary points A and E on a horizontal straight line. The lengths of the rod AB = ED and BC= CD. The masses of the rods are the same. Show that the angles a and ~ are related by the ratio tan ex, = 3 tan ~ when in equilibrium. 111. What is the coefficient of friction between a floor and a box weighing one ton-force if a minimum force of 600 kgf is required to start the box moving? . 112. A weightless unstretchable string is wound around a cylinder with a mass m (Fig. 42). With what minimum force Fmin and at what angle T.) and are kept constant. Find the temperature T" on the plane between the panels. . 310. Assuming in Problem 309 that the panels have the same thickness d, determine the coefficient of thermal conductivity of the wall. HEAT. MOLECULAR PHYSICS 71 A A A d~ - A, AZ A, }.,8 d Z d Z - d AI A2 A, AS A, AZ Fig. 124 Fig. 125 Fig. 126 31t. A wall consists of alternating blocks with a length d and coefficients of thermal conductivity x, and A,2 (Fig. 124). The cross-sectional areas of the blocks are the same. Determine the coefficient of thermal conductivity of the wall. 312. Two walls I and II of the same thickness are made of heterogeneous metals, as shown in Figs. 125 and 126. In what case will the coefficient of thermal conductivity be greater? 313. During one second m grammes of water boils away in a pan. Assuming that heat is received by the water only through the bottom of the pan and neglecting the transfer of heat from the pan walls and the water surface to the ambient air, determine the temperature T of the pan bottom in contact with the heater. The area of the pan bottom is A and its thickness d. The coefficient of thermal conductivity is A. 2-3. Properties of Gases 314. The cap of a fountain-pen is usually provided with a small orifice. If it is clogged, the ink begins to leak out of the pen. What is the cause of this phenomenon? 72 PROBLEMS 315. A barometer gives wrong readings because some air IS present above the mercury column. At a pressure of POI = 755 mm Hg the barometer shows PI = 748 mrn, and at POt = 740 mm it shows P2 = 736 mm. Find the length 1 of the barometer tube (Fig. 127). 316. A glass tube with a length of 1= 50 ern and a cross section of A = 0.5 ern- is soldered at one end and is submerged into water as shown in Fig. 128. What force F should be applied to hold the tube under the water if the distance from the surface of the water to the soldered end ish = IOcm and the atmospheric pressurePo = 760mm Hg? The weight of the tube G= 15 gf. 317. A narrow tube open at both ends is passed through the cork of a vessel filled with water. The tube does not reach the bottom of the vessel (Mariotte's vessel shown in Fig. 129). Draw a diagram showing how the pressure p of the air in the vessel depends on the quantity of water Q that has flowed out. 318. Upon each double stroke a piston pump sucks in a volume Vo of air. When this pump is used to evacuate the air from a vessel with a volume V it performs n double strokes. -,- ------ - -- --_11 -_-_- • I Fig. 127 Fig. 128 h Fig. 129 HEAT. MOLECULAR PHYSICS 73 The initial pressure inside the vessel Po is equal to atmospheric pressure. After that. another pump with the same active volume Vo begins to suck in the atmospheric air, also making n double strokes. What will the pressure in the vessel be? 319. A mercury column with a length 1 is in the middle of a horizontal tube with a length L closed at both ends. If the tube is placed vertically, the mercury column will shift through the distance ~l from its initial position. At what distance will the centre of the column be from the middle of the tube if one end of the tube placed horizontally is opened? if the upper end of the tube placed vertically is opened? if the lower end of the tube placed vertically is opened? The atmospheric pressure is H ernHg. The temperature remains the same. 320. Bearing in mind that, according to Avogadro's law, the volume of one gramme-molecule of any gas under standard conditions (temperature ooe and pressure 1 atm) is 22.4 litres, find the constant in the equation of state of an ideal gas (the Clapeyron-Mendeleyev equation) for a quantity of a gas equal to one mole and prove that this constant is the same for all gases. 321. Write the equation of state for an arbitrary mass of an ideal gas whose molecular weight J.1 is known. 322. How would the pressure inside a fluid change if the forces of attraction between the molecules suddenly disappeared? 323. A vessel contains one litre of water at a temperature of 27° C. What would the pressure in the vessel be if the forces of interaction between the water molecules suddenly disappeared? 324. Is the pressure the same inside a gas and at the wall of the vessel containing this gas? 325. Is the concentration of gas molecules inside a vessel and at its wall the same? 326. Find the temperature of a gas contained in a closed vessel if its pressure increases by 0.4 per cent of the initial pressure when it is heated by 1° C. 327. A thin-walled rubber sphere weighing G= 50 gf is filled with nitrogen and submerged into a lake to a depth of h = 100 metres. Find the mass of the nitrogen m if the sphere is in a position of equilibrium. Will equilibrium be stable? The atmospheric 74 PROBLEMS pressure Po = 760 mm Hg. The temperature in the lake t = +4° C. Disregard the tension of the rubber. 328. Two hollow glass balls are connected by a tube with a drop of mercury in the middle. Can the temperature of the ambient air be appraised from the position of the drop? 329. A cylinder closed at both ends is separated into two equal (42 em each) parts by a piston impermeable to heat. Both the parts contain the same masses of gas at a temperature of 27°C and a pressure ofl atm. How much should the gas be heated in one part of the cylinder to shift the piston by 2 cm? Find the pressure p of the gas after shifting of the piston. 330. Dry atmospheric air consists of nitrogen (78.09 per cent by volume), oxygen (20.95 per cent), argon (0.93 per cent) and carbon dioxide (0.03 per cent). Disregarding the negligible admixtures of other gases (helium, neon, krypton, xenon), determine the composition of the air (in per cent) by weight. 331. Find the mean (effective) molecular weight of dry atmospheric air if the composition of the air in per cent is known (see Problem 330). 332. The density of the vapour of a compound of carbon and hydrogen is 3 g/lit at 43°C and 820 mm Hg. What is the molecular formula of this compound? 333. When will the change in the pressure of a gas be greater-if it is compressed to a certain extent in a heat-impermeable envelope or upon isothermal compression? 334. A gas that occupies a volume of VI = 1 lit at a pressure of PI = I atm expands isothermally to a volume of VI = 2 lit. Then, the pressure of the gas iii halved at the same volume. Next the gas expands at a constant pressure to V4 = 4 litres. Draw a diagram of p versus V and use it to determine the process in which the gas performs the greatest work. How does the temperature change? 335. A cyclic process 1-2-3-1 depicted on a V-t diagram (Fig. 130) is performed with a certain amount of an ideal gas, . Show the same process on a p-V diagram and indicate the stages when the gas received and when it rejected heat. 336. A gas heated geyser consumes Vo = 1.8 m" of methane (CHc) an hour. Find the temperature tt of the water heated by the geyser if the water flows out at a rate of v = 50 crn/s. The diameter of the stream D = I ern, the initial temperature of the water and the gas t1 = 110 C and the calorific value of HEAT. MOLECULAR PHYSICS 75 v ....---...2 r ,I I I I I I I I If ----:------------ I I I I I Fig. 130 Fig. 131 Fig. 132 ------ -~--~---~-----.. --n- ----M- -- methane qo = 13,000 calrg. The gas in the tube is under a pressure of p = 1.2 atm. The efficiency of the heater 11 =:: 60 per cent. 337. A closed vessel impermeable to heat contains ozone (03 ) at a temperature of t 1 = 5270 C. After some time the ozone is completely converted into oxygen (02 ) , Find the increase of the pressure in the vessel if q = 34,000 cal have to be spent to form one gramme-molecule of ozone from oxygen. Assume that the heat capacity of one gramme-molecule of oxygen at a constant volume is equal to Cv ::= 5 cal/deg- mole. 338. Twenty grammes of helium in a cylinder under a piston are transferred infinitely slowly from a state with a volume of V1 == 32 lit and a pressure of PI == 4.1 atm to a state with V2 =-= 9 lit and P2 == 15.5 atm. What maximum temperature will the gas reach in this process if it is depicted on the p-l! diagram as a straight line (Fig. 131)? 339. Will the energy of the air in a room increase if a stove is heated in it? Note. Assume the energy of a unit of mass of the-air u to be proportional to the absolute temperature, Le., u=cT. 340. The temperature in a room with a volume of 30 m" rose from 15°C to 25°C. How much will the mass 76 PROBLEMS of the air in the room change if the atmosphere pressure p = 1 atm? The molecular (mean) weight of the air is J..t = 28.9 g/mole. 341. A small tube filled with air and open at the bottom is placed into a water-filled open vessel with a screen on the top (Fig. 132). The tube cannot be turned over. Draw a diagram showing how the depth of submergence of the tube depends on the temperature of the water if the temperature first slowly increases and then gradually decreases. 342. A cylinder contains m = 20 g of carbon dioxide under a heavy piston. The gas is heated from the temperature t, = 20° C to t2 = 108°C. What work is performed by the gas? 343. What quantity of heat should be imparted to carbon dioxide (see Problem 342) expanding at a constant pressure as a result of heating? The molar heat of the carbon dioxide (heat capacity of one gramme-molecule) at a constant volume is Cv=6.864 cal/mole-deg, 2-4. Properties of Liquids 344. Is it more difficult to compress a litre of air to three atmospheres, or a litre of water? 345. How can a minimum and a maximum thermometers be made using the phenomena of wettability and unwettability? 346. The surface layer of a liquid is frequently likened to a stretched rubber film. In what respect does this analogy disagree with reality? 347. To remove a grease spot from a fabric it is good to apply some petrol to the edges of the spot, while the spot itself should never be wetted with petrol immediately. Why not? 348. Why is moisture retained longer in soil if it is harrowed? 349. Skiing boots are usually warmed up to ensure that they absorb grease better. Should the boots be warmed up outside or inside? 350. Why can an iron be used to remove greasy spots from clothing? 351. Why do drops of water appear at the end of a piece of firewood in the shade when it is being dried in the sun? 352. A vessel whose bottom has round holes with a diameter of d = 0.1 mm is filled with water. Find the maximum height of the water level h at which the water does not flow out. The water does not wet the bottom of the vessel. HEAT. MOLECULAR PHYSICS 77 353. A soapy film is stretched over a rectangular vertical wire frame (Fig. 133). What forces hold section abed in equilibrium? 354. A cube with a mass m = 20 g wettable by water floats on the surface of water. Each face of the cube a is 3 em long. What is the distance between the lower face of the cube and the surface of the water? 355. The end of a capillary tube with a radius r is immersed into water. What amount of heat will be evolved when the water rises in the tube? 356. A capillary tube is lowered into a vessel with a liquid whose vapour pressure may be neglected. The density of the liquid is p. The vessel and the tube are in a. vacuum under the bell of an air pump (Fig. 134). Find the pressure inside the liquid in the capillary tube at a height h above the level of the liquid in the vessel. 357. The course of reasoning- given below is usually followed to prove that the molecules of the surface layer of a liquid have surplus potential energy. A molecule inside the liquid is acted upon by the forces of attraction from the other molecules which compensate each other on the average. If a molecule is singled out on the surface, the resulting force of attraction from the other molecule is directed into the liquid. For this reason the molecule tends to move into the liquid. and definite work should a ~--------------- 6 C 1-0----____________ tl Fig. /33 .: h Fig. 134 78 -- II PROBLEMS -- --- - - - - - - -.- - - - - - Fig. 135 Fig. 136 be done to bring it to the surface. Therefore, each molecule of the surface layer has excess potential energy equal to this work. The average force that acts on any molecule from the side of all the others, however, is always equal to zero if the liquid is in equilibrium. This is why the work done to move the liquid from a depth to the surface should also be zero. What is the origin, in this case, of the surface energy? 358. One end of a glass capillary tube with a radius r = 0.05 cm is immersed into water to a depth of h:=. 2 ern. What pressure is required to blow an air bubble out of the lower end of the tube? 359. A glass capillary tube with an internal diameter of 0.5 mm is immersed into water. The top end of the tube projects by 2 em above the surface of the water. What is the shape of the meniscus? 360. Water rises to a height h in a capillary tube lowered vertically into water to a depth I (Fig. 135). The lower end of the tube is closed, the tube is then taken out of the water and opened again. Determine the length of the water column remaining in the tube. 361. Two capillary tubes of the same cross section are lowered into a vessel with water (Fig. 136). The water in the straight HEAT. MOLECULAR PHYSICS 79 Fig. 137 Fig. 138 tube rises to. a height h. What will the level of the water be in the bent tube and what form will the meniscus take? The lower end of the bent tube is at a depth H· below the level of the water in the vessel. Consider the following five cases: (1) H > h (2) H =h (3) 0 < H , from the centre of the sphere? A 8 C • •• Fig. 148 Fig. 149 -. Fig. 150 90 PROBLEMS 417. A metal leaf is attached to the internal wall of an electrometer insulated from the earth (Fig. 149). The rod and the housing of the electrometer are connected by a conductor, and then a certain charge is imparted to the housing. Will the leaves of the electrometer deflect? What will happen to the leaves if the conductor is removed and the rod is then earthed? 418. The housing of the electrometer described in Problem 417 is given a charge (the conductor is absent). Will the leaves of the electrometer deflect in this case? Will the angle. of deflection of the leaves change if the rod is earthed? 419. By touching different points on a metal bucket having a narrow bottom with a test ball connected by a wire to an earthed electrometer (Fig. 150), we can observe an identical deflection of the leaves of the electrometer at any position of the ball. If the wire is removed, the deflection of the leaves of the electrometer, whose rod the ball is made to contact, will depend on what point of the bucket surface (external or internal) we touched previously. Why? . 420. Why does an electrometer connected by a wire to the metal body shown in Fig. 151 make it possible to measure the potential of the body? Why do the leaves deflect in proportion to the density of the charge on separate portions of the body when the charge is transferred from the body to the electrometer with the aid of an insulated current-conducting ball? 421. An uncharged current-conducting sphere with a radius R is at a distance d from a point charge Q. What is the potential of the sphere? 422. An isolated current-conducting sphere with a radius R carries a charge +Q. What is the energy of the sphere? 423. A metal sphere two metres in diameter is in the centre of a large room and charged. to a potential of 100,000 V. What quantity of heat will be liberated if the sphere is connected to the earth with a conductor? 424. Two metal balls with radii of '1= 1 ern and '2= 2 em at a distance of R= 100 em from eaeh other are connected to a battery with an electromotive force of 1) than the previous one. The second output terminal and the lower ends of the resistances are earthed.. Find the ratio between the resistances R1 : R.: R, with any number of cells in the attenuator. 472. What devices are needed to verify Ohm's law experimentally, i.e. t to show that the current intensity is directly proportional to the potential difference? 473. A charge Q is imparted to two identical plane capacitors connected in parallel. lIa ~ /!, Un 1* 100 PROBLEMS It +r· Fig. 172 At the moment of time t = 0 the distance between the plates of the first capacitor begins to increase uniformly according to the law d1 = do+ ot, and the distance between the plates of the second capacitor to decrease uniformly according to the law d;=do-vt. Neglecting the resistance of the feeding wires. find the intensity of the current in the circuit when the plates of the capacitors move. . 474. Find the work performed by an electrostatic field .(see Problem 473) when the distance between. the plates of the first capacitor increases and that between the plates of the second capacitor simultaneously decreases by a. 475. A curious phenomenon was observed by an experimenter working with a very sensitive galvanometer while sitting on a chair at a table. (The galvanometer was secured on a wall and the ends of its winding connected to an open key on the table.) Upon rising from the chair and touching the table with his hand, the experimenter observed an appreciable deflection of the galvanometer pointer." If he touched the table while sitting on the chair there was no deflection. Also, the galvanometer showed no deflection when he touched the table without first sitting down. Explain this phenomenon. 476. The following effect was observed in a very sensitive galvanometer when the circuit was opened. If a charged body is brought up to one end of the winding of the galvanometer. its pointer deflects. If the body is brought up to the other end of the winding, the deflection is in the same direction. Explain this phenomenon. 477. How is the potential distributed in a Daniell cell when the external circuit is opened? cl~~ S, 8~ s, &z &, A s, ~~~ ~:;~ ~~~r, l'z Ii Ii Ii IZ Ii J! B 8 8 &1>Sz G,a::88 &;>&z 8r=&z 'i 0 where k = 0.12 mA/V, are connected to a circuit as shown in Fig. 190. Draw a diagram showing how the current I in the circuit depends on the voltage V if eft1 = 2 V, itI = 5 V, G8 = 7 V, and V can change from -10 V to + 10V. 543. Calculate the sensitivity of a cathode-ray tube to voltage, i. e., the deflection of the light spot on the screen caused by a potential difference of 1V on the control grids. The length of. the control grids is l, the distance between them is d, the distance from the end of the grids to the screen is L, and the accelerating potential difference is UeFig. 189 Fig. 190 112 PROBLEMS 3-4. Magnetic Field of a Current. Action of a Magnetic Field on a Current and Moving Charges 544. Determine the dimension and magnitude of the coefficient k in the expression for the intensity of the magnetic fie-ld of a solenoid H = k 4nJ ~. if H is measured in oersteds and I in egs electrostatic units. The dimension of the oersted coincides with that of the electric field intensity in cgs units. 545. Two windings connected as shown in Fig. 191 are wound around a thin iron ring with a radius R = 10ern. The first winding has 2,000 turns and the second 1,000 turns. Find the intensityof the magnetic field inside the ring if a current of I = IDA flows through the windings. . 546. A current I flows through an infinitely long conductor ABC bent to form a right angle (Fig. 192). How many times will the intensity of the magnetic field change at point M if an infinitely long straight conductor BD is so connected to point B that the current I branches at point B into two equal parts and the current in the conductor AB remains the same? Note. Take into account the fact thai the intensity of a magnetic field induced at a certain point by a small element of current is perpendicular to the plane containing this element and a radius-vector drawn from this current element to the given point. 547. A current flows through a conductor arranged in one plane as shown in Fig. 193. Find the intensity of the magnetic field TN J J I C_ _- ...-~I ~_--- P 0 "A Fig. 191 Fig. 192 Fig. 193 ELECTRICITY AND MAGNETISM 113 I I I J J I J j I J I J I ....---+-t-..J... J ~"'::J ~ I Fig. 194 Fig. 195 Fig. 196 at an arbitrary point on line AB, which is the axis of symmetry 0-1 the conductor. 548. How will a magnetic pointer be positioned if it is placed in the centre of a single-layer toroidal solenoid through which a direct current flows? 549. A current I flows along an infinite straight thin-walled pipe. Bearing in mind that the intensity of the magnetic field of an infinite straight conductor at a distance , from it is proportional to I [r, find the intensity of the magnetic field at an arbitrary point inside the pipe. 550. Remembering that the intensity of a magnetic field inside a long cylindrical conductor H = k 2njr, where j is the current density and r is the distance from the conductor axis, find the intensity of the field at an arbitrary point on a long cylindrical space inside the conductor (Fig. 194) through which a current with a density j flows. The axis of, the space is parallel to the axis of the conductor and is ata distance d from it. 551. Draw the distribution of the force lines of a magnetic field in the space of the cylindrical conductor descri bed In Problem 550. 552. Determine the dimension and the magnitude of the coefficient k in the expression for the force F = kHI1sin cp acting from a magnetic field on a current if H is in oersteds and I in egs electrostatic units. 8-2042 114 PROBLEMS 553. Will the density of a direct current flowing in a cylindrical conductor be constant across the entire cross section of the conductor? 554. A lightning arrester is connected to earth by a circular copper pipe. After lightning strikes, it is discovered that the pipe became a circular rod. Explain the cause of this phenomenon. 555. A very great current is made to flow for a short time through a thick winding of 8 solenoid. Describe the deformation of the winding from the viewpoint of quality. 556. The magnetic system of a galvanometer consists of a magnet, pole shoes A and B, and a cylinder made of soft iron (Fig. 195). The magnetic force lines in the gap between the shoes and the cylinder are perpendicular to the surface of the cylinder. The intensity of the magnetic field is H. A rectangular coil with n turns is placed in the gap on axis O. The sides of the coil are parallel to the diameter and the generatrix of the cylinder. The area of each turn is A. One end of a spiral spring is so attached to the axis of the coil that when the latter rotates through an angle a, the deformation of the spring creates a rotational moment ka that tends to turn the coil to a position of equilibrium. Determine the angle through which the coil will turn if a current I passes through it. 557. A current of I = 1 A flows through a wire ring with a radius R = 5 em suspended on two flexible conductors. The ring is placed in a homogeneous magnetic field with an intensity of H = 10Oe whose force lines are horizontal. What force will the ring be tensioned with? 558.· A wire ring with a radius R= 4 ern is placed into a heterogeneous magnetic field whose force lines at the points of intersection with the ring form an angle of a = 100 with a normal to the plane of the ring (Fig. 196). The intensity of the magnetic field acting on the ring H = 100 Oe. A current of I = 5 A flows through the ring. What force does the magnetic field act on the ring with? 559. A rectangular circuit ABeD with sides a and b placed into a homogeneous magnetic field with an intensity H can revolve around axis 00' (Fig. 197). A direct current I constantly flows through the circuit. Determine the work performed by the magnetic field when the circuit is turned through 1800 if initially the plane of the circuit is perpendicular to the magnetic field and arranged as shown in Fiz. 197. ELECTRICITY AND MAGNETISM o lJ 115 a Fig. 197 Fig /98 8 * H o 0' A B H Fig. /99 116 PROBLEMS tic field in the direction of side a with a velocity e. The intensity of the magnetic field H is perpendicular to the base of the block with the sides a and c (Fig. 199). . Determine the intensity of the electric field in the block and the density of the electric charges on the surfaces of the parallelepiped formed by sides a and b. 567. An uncharged metal cylinder with a radius r revolves about its axis in a magnetic field with an angular velocity ro. The intensity of the magnetic field is directed along the axis of the cylinder. What should the intensity of the magnetic field be for no electrostatic field to appear in the cylinder? 3-5. Electromagnetic Induction. Alternating Current 568. Determine the direction of the intensity of an electric field in a turn placed in a magnetic field (Fig. 200) directed away from us in a direction perpendicular to the plane of the turn. The intensity of the magnetic field grows with time. 569. A rectangular circuit ABeD moves translationally in the magnetic field of a current flowing along straight long conductor 00' (Fig. 201). Find the direction of the current induced in the circuit if the turn moves away from the conductor. 570. A non-magnetized iron rod flies through a coil connected to a battery and an ammeter (Fig. 202). Draw an approximate diagram of the change of current in the coil with time as the rod flies through it. 571. A current in a coil grows directly with time. What is the nature of the relation between the current and time in another coil inductively connected to the first one? Fig. 200 0' B..---.O A 0 a Fig. 201 Fig. 202 o+ a B 2 E ELECTRICITY AND MAGNETISM Fig. 20,1 A D Fig. 204 tZ a a F 117 572. Will the result of Problem 571 change if an iron core is inserted into the second coil? 573. A wire ring with a radius r is placed into a homogeneous magnetic field whose intensity is perpendicular to the plane of the ring and changes with time according to the law H = kt. Find the intensity of the electric field in the turn 574. A ring of a rectangular cross section (Fig. 203) is made of a material whose resistivity is p, The ring is placed in a homogeneous magnetic field. The intensity of the magnetic field is directed along the axis of the ring and increases directly with time, H" = kt. Find the intensity of the current induced in the ring. 575. A coil having n turns, each with an area of A, is connected to a ballistic galvanometer. (The latter measures the quantity of electricity passing through it.) The resistance of the entire circuit is R. First the coil is between the poles of a magnet in a region where the magnetic field H is homogeneous and its intensity is perpendicular to the area of the turns. Then the coil is placed into a space with no .magnetic field. What quanti ty of the electricity passes through ~ the galvanometer? (Express the answer in cou- lombs.) 576. Determine the current in the conductors h of the circuit shown in Fig. 204 if the intensi- j ty of a homogeneous magnetic field is perpendicular to the plane of the drawing and changes in time according to the law H = kt, The resistance of a unit- of length of the conductors is r. 577. The winding of a laboratory regulating Fig. 205 autotransformer is wound around an iron core ha- 118 Fig. 206 Fig. 207 PROBLEMS ving the form of a rectangular toroid (Fig. 205). For protection against eddy (Foucault) currents the core is assembled of thin iron laminas insulated from one another by a layer of varnish. This can be done in various ways: (1) by assembling the core of thin rings piled on one another; (2) by rolling up a long band with a width h; (3) by assembling the core of rectangular laminas 1X h in size, arranging them along the radii of the cylinder.Which is the best way? 578. A direct induced current I is generated in a homogeneous circular wire ring. The variable magnetic field producing this current is perpendicular to the plane of the ring, concentrated near its axis and has an axis of symmetry passing through the centre of the ring (Fig. 206). What is the potential difference between points A and B? What is the reading of an electrometer connected to these points? 579. A variable magnetic field creates a constant e.rn.f. C in a circular conductor ADBKA (see Problem 578). The resistances of the conductors ADB, AKB and ACB (Fig. 207) are equal to R1 , R2 and Ra, respectively. What current will be shown by ammeter C? The magnetic field is concentrated near the axis of the circular conductor. 580. The resistance of conductor ACB (see Problem 579) is Ra= O. Find the currents 11' I, and /a and the potential difference V A-VB" 581. A medical instrument used to extract alien particles from an eye has the form of a strong permanent magnet or an electromagnet. When brought close to the eye (without touching it) it extracts iron and steel particles (filings, chips, etc.), What current should flow through the electromagnet to extract, without touching the eye, metal objects made of non-ferromagnetic materials (aluminium, copper, etc.)? ELECTRICITY AND MAGNETISM 119 582. A wire ring secured on the axis passing through its centre and perpendicular to the force lines is placed in a homogeneous magnetic field (Fig. 208). The intensity of the field begins to grow. Find the possible positions of equilibrium of the ring and show the position of stable equilibrium. What will happen if the intensity of the field decreases? 583. A conductor with a length l and mass m can slide without friction along two vertical racks AB and CD connected by a resistor R. The system is in a homogeneous magnetic field whose intensity H is perpendicular to the plane of the drawing (Fig. 209). • How will the movable conductor travel in the field of gravity if the resistance of the conductor itself and the racks is negle- cted? 584. A conductor with a mass m and length I can move without friction along two metallic parallel racks in a horizontal plane and connected across capacitor C. The entire system is in a homogeneous magnetic field whose intensity H is directed upward. A force F is applied to the middle of the conductor perpendicular to it and parallel to the racks (Fig. 210). Determine the acceleration of the conductor if the resistance of the racks, feeding wires and conductor is zero. What kinds of energy will the work of the force F be converted into? Assume that the velocity of the conductor is zero at the initial moment. 585. Considering the motion of a straight magnet in a plane perpendicular to a wire and using the law of conservation of energy, prove that the field of a long forward current diminishes with the distance from the wire as I/R. 586. A cylinder made of a non-magnetic material has N turns of a wire (solenoid) wound around it. The radius of the cylinder R A .. t-C .m. 0N I m H ct E0N B 0 Fig. 209 Fig. 210 120 PROBLEMS is , and its length l (r ~ I). The resistance of the wire is R. What should the voltage at the ends of the solenoid be for the current flowing in it to increase directly with time, i.e., I =kt? 587. A solenoid (see Problem 586) is connected to a battery whose e.m.I. is cC. The key is closed at the moment t = O. What is the intensity of the current flowing through the circuit of the solenoid if the resistance R of the solenoid, battery and feeding wires is neglected? 588. Calculate the work of the battery (see Problem 587) during the time 1". What kind of energy is this work converted into? 589. A ring made of a superconductor is placed· into a homogeneous magnetic field whose intensity grows from zero to Ho• The plane of the ring is perpendicular to the force lines of the field. Find the intensity of the induction current appearing in the ring. The radius of the ring is , and its inductance L. 590. A superconductive ring with a radius r is in. a homogeneous magnetic field with an intensity H~ The force lines of the field are perpendicular to the plane of the ring. There is no current in the ring. Find the magnetic flux piercing the ring after the magnetic field is swi tched off. 591. Find the inductance of a coil wound onto the iron core shown in Fig. 211. The number of turns of the coil N, the cross-sectional area A, the perimeter of the core (medium line) I and the permeability of the core l-t, are known. Note. Take into account the fact that .the intensity of the magnetic field inside the core is practically constant and can be approximately expressed by the formula H = OAn ~ I. 592. Estimate approximately the coefficient of mutual inductance of the windings of a transformer. Consider the windings Fig. 211 Fig. 212 ELECTRICITY AND MAGNETISM 121 as coils of identical cross section. Disregard the dispersion of the force lines of the magnetic field. Note. The coefficient of mutual inductance of two circuits is the ratio between the magnetic flux . 620. Two identical coils perpendicular to each other are divided in half and connected to a circuit as shown in Fig. 221. The inductance of the choke Ch and the ohmic resistance R are so selected that the intensities of the currents in the coils are the same. The ohmic resistance and inductive reactance of the coils are much less than the inductive reactance of the choke. What will occur if an aluminium cylinder A secured on an axis is introduced into the space between the poles of the coils? CHAPTER 4 OSCILLATIONS AND WAVES 4-1. Mechanical Oscillations 621. A weight suspended from a long string oscillates in a vertical plane and is deflected through an angle a, from the vertical (a mathematical' pendulum). The same weight can rotate over a circumference so that the string describes a cone (a conical pendulum). When will the tension of the string deflected through an angle a, from the vertical be greater? 622. A clock with an oscillation period of the pendulum of 1 second keeps accurate time on the surface of the Earth. When will the clock go slower in a day: if it is raised to an altitude of 200 metres or lowered into a mine to a depth of 200 metres? 623. Two small spheres each with a mass of In = I g are secured to the ends of a weightless rod with a length d = 1 metre. The rod is so suspended from a hinge that it can rotate without friction around a vertical axis passing through its middle. Two large spheres with masses M = 20 kg are fastened on one line with the rod. The distance between the centres of a large sphere and a small one L = 16 em (Fig. 222). Find the period of small oscillations of this torsion pendulum. 624. What is the period of oscillations of a mathematical pendulum in a railway carriage moving horizontally with an acceleration a? ~~---- rl----~tFIg. 222 PROBLEMS 625. Find the period of oscillations of a pendulum in a lift moving vertically with an acceleration a. 626. A block performs small oscillations in a vertical plane while moving without friction over the internal surface of a spherical cup. Determine the period of oscillations of the block if the internal radius of the cup is R and the face of the block is much smaller than R. 627. How will the period of oscillations of the block in the cup change (see Problem 626) lf, besides the force of gravity, the cup is acted upon by a force F directed vertically upward? The mass of the cup M is much greater than that of the block m. 628. How will the period of oscillations of the block in the cup change (see Problem 626) if the cup is placed onto a smooth horizontal surface over which it can move without friction? 629. A hoop with a mass m and a radius r can roll without slipping over the internal surface of a cylinder with a radius R (Fig. 223). Determine the period of motion of the hoop centre if the angle

= -; were n IS an In e-n • B ger. A point source of light S is placed between the mirrors at the same distance from both of them. Find the num~~~~~~~~~~~~ her of images of the source in the mirrors. Fig. 241 690. Two flat mirrors AOand OB form an arbitrary dihedral angle cp = 2n, where a is any number greater than 2. A pointa source of light S is between the mirrors at equal distances from them. Find the number of images of the source in the mirrors. 691. In what direction should a beam of light be sent from point A (Fig. 241) contained in a mirror box for it to fall onto point B after being reflected once from all four walls? Points A and B are in one plane perpendicular to the walls of the box (i.e., in the plane of the drawing). 692. Why does the water seem much darker directly below an airplane flying over a sea than at the horizon? 693. Over what distance will a beam passing through a planeparallel plate be displaced if the thickness of the plate is d, the refraction Index is n and the angle of incidence is i? Can the beam be displaced by more than the thickness of the plate? ----C Fig. 243 A c........_ .....---......a Fig. 242 A _--6--___. GEOMETRICAL OPTICS 141 a ~, 8 Fig. 244 694. At what values of the refraction index of a rectangular prism can a ray travel as shown in Fig. 242? The section of the prism is an isosceles triangle and the ray is normally incident onto AB. 695. A rectangular glass wedge is lowered into water. The refraction index of glass is n1 = 1.5. At what angle a (Fig. 243) will the beam of light normally incident on AB reach AC en- tirely? 696. On bright sunny days drivers frequently see puddles on some parts of asphalt country highways at a distance of 80-]00 metres ahead of the car. As the driver approaches such places, the puddles disappear and reappear again in other places approximately at the same distance away. Explain this phenome- non. 697. A thick plate is made of a transparent material whose refraction index changes from n1 on its upper edge to n" on its lower edge. A beam enters the plate at the angle a. At what angle will the beam leave the plate? 698. A cubical vessel with non-transparent walls is so located that the eye of an observer does not see its bottom, but sees all of the wall CD (Fig. 244). What amount of water should be poured into the vessel for -the observer to see an object F arranged at a distance of b= 10 em from corner D? The face of the vessel is a = 40 em. 699. A man in a boat is looking at the bottom of a lake. How does the seeming depth of the lake h depend on the angle l formed by the line of vision with the vertical? The actual depth of the lake is everywhere the same and equal to H. 700. The cross section of a glass prism has the form of an equilateral triangle. A ray is incident onto one of the faces perpendicular to it. Find the angle tp between the incident ray and the ray that leaves the prism. The refraction index of glass is n = 1.5. 701. The cross section of a glass prism has the form of an isosceles triangle. One of the equal faces is coated with silver. A ray is normally incident on another unsilvered face and, being reflected twice, emerges through the base of the prism perpendicular to it. Find the angles of the prism. 142 PROBLEMS 702. A ray incident on the face of a prism is refracted and escapes through an adjacent face. What is the maximum permissible angle of refraction of the prism ex if it is made of glass with a refraction index of n = 1.5? 703. A beam of light enters a glass prism at an angle a and emerges into the air at an angle p. Having passed through the prism, the beam is reflected from the original direction by an angle y. Find the angle of refraction of the prism cp and the refraction index of the material which it is made of. 704. The faces of prism ABeD made of glass with a refraction index n form dihedral angles: L A = 90°, L B = 75°, L C = 1350 and L D = 60° (the Abbe prism). A beam of light falls on face AB and after complete internal reflection from face Be escapes through face AD. Find the angle of incidence a of the beam onto face AB if a beam that has passed through the prism is perpendicular to the incident beam. 705. If a sheet of paper is covered with glue or water the text typed on the other side of the sheet can be read. Explain why? 5·3. Lenses and Spherical Mirrors 706. Find the refraction index of the glass which a symmetrical convergent lens is made of if its focal length is equal to the radius of curvature of its surface. 707. A plano-convex convergent lens is made of glass with a refraction index of n = 1.5. Determine the relation between the focal length of this lens f and the radius of curvature of its convex surface R. 708. Find the radii of curvature of a convexo-concave convergent lens made of glass with a refraction index of n = 1.5 having a focal length of f = 24 em. One of the radii of curvature is double the other. 709. A convexo-convex lens made of glass with a refraction index of n = 1.6 has a focal length of f = 10 em. What will the focal length of this lens be if it is placed into a transparent medium with a refraction index of n1 = 1.5? Also find the focal length of this lens in a medium with a refraction index of nt = 1.7. 710. A thin glass lens has an optical power of D = 5 diopters. When this lens is immersed into a liquid with a refraction index nIt the lens acts as a divergent one with a focal length GEOMETRICAL OPTICS 143 B of f = lOa em. Find the refraction index ns of the liquid if that of the lens glass ls n1 = 1.5. F 0 F 711. The distance between an object and a divergent lens is m times greater than the focal length of the lens. How many Fig. 245 times will the image be smaller. than the object? 712. The hot filament of a lamp and its image obtained with the aid of a lens having an optical power of four diopters are equal in size. Over what distance should the lamp be moved away from the lens to decrease its image five times? 713. The distance between twopoint sources of light is 1= 24 cm. Where should a convergent lens with a focal length of f = 9 ern be placed between them to obtain the images of both sources at the same point? 714. The height of a candle flame is 5 ern. A lens produces an imageof this flame 15 cm high on a screen. Without touching the lens, the candle is moved over a distance of 1= 1.5 ern away from the lens, and a sharp image of the flame 10 cm high is obtained again after shifting the screen. Determine the main focal length of the lens. 715. A converging beam of rays is incident onto a divergent lens so that the continuations of all the rays intersect at a point lying on the optical axis of the lens at a distance of b= 15 em from it. Find the focal length of the lens in two cases: (1) after being refracted in the lens, the rays are assembled at a point at a distance of a1 = 60 em from the lens; (2) the continuations of the refracted rays intersect at a point at a distance of a2 = 60 em in front of the lens. 716. The distance between an electric lamp and a screen is d = 1 metre. In what positions of a convergent lens with a focal .B Fig. 246 144 .9' Fig. 247 PROBLEMS length of f = 21 em will the image of the lamp filament be sharp? Can an image be obtained if the focal length is I' = 26 em? 717. A thin convergent lens produces the image of a certain object on a screen. The height of the image is hi. Without changing the distance between the object and the screen, the lens is shifted, and it is found that the height of the second sharp image is h2 • Determine the height of the object H. 718. What is the radius R of a concave spherical mirror at a distance of a = 2 metres from the face of a man if he sees in it his image that is one and a half times greater than on a fiat mirror placed at the same distance from the face? 719. Figure 245 shows ray AB that has passed through a divergent lens. Construct the path of the ray up to the lens if the position of its foci F is known. 720. Figure 246 shows 8 luminescent point and its image produced by a lens with an optical axis NIN2. Find the position of the lens and its foci. 721. Find by construction the optical centre of a lens and its main foci on the given optical axis N IN 2 if the positions of the source S and the image S' are known (Fig. 247). 722. The position of the optical axis NIN2' the path of ray AB incident upon a lens and the refracted ray Be are known (Fig. 248). Find by construction the position of the main foci of the lens. 723. A convergent lens produces the image of a source at point S' on the main optical axis. The positions of the centre of the lens 0 and its foci F are known, and OF = 45°. Determine the angle e between the extreme rays of the spectrum when they emerge from the lens if the refraction indices of the prism for the extreme rays of the visible spectrum are n,= 1.62 and nv= 1.67. 805. White light is incident from a point source on the optical axis of a convexo-convex lens at a distance of a = 50 cm from it. The radii of curvature of the lens are R1 = R2 = 40 ern. A diaphragm with a diameter of D = 1 cm that restricts the cross section of the light beam is placed tightly in front of the lens. The refraction indices for the extreme rays of the visible spectrum are n, = 1.74 and nv = 1.8, respectively. What pattern can be observed on a screen arranged at a distance of b = 50 em from the lens and perpendicular to its optical axis? 806. Using the results of Problem 748, construct the elementary theory of the rainbow, i. e., show that the centre of a rainbow lies on a straight line drawn from the Sun through the eye of an observer and that the arc of the rainbow is a part of a circle all of whose points can be seen at an angle of 42° PHYSICAL OPTICS 159 (for red light) with reference to the straight line connecting the eye of the observer and the centre of the rainbow. 807. Explain from a qualitative point of view the appearance of a double rainbow. How do the colours alternate in the primary and the reflection rainbows? 808. Can a rainbow be observed at midday in Moscow during the summer solstice (on June 22)? Note. At this time the Sun is the highest above the horizon in the northern hemisphere. 809. The length of a wave in water diminishes n times, n being the refraction index. Does this mean that a diver cannot see surrounding objects in their natural colours? 810. The word "excellent" is written on a sheet of white paper with a red pencil and the word "good" with a green pencil. A green and a red pieces of glass are available. Through which glass can the word "excellentn be seen? 811. Why do coated lenses (see Problem 782) have a purpleviolet (lilac) tint? 812. Why do the colours of thin films (for example, oil films on water) and the colours of a rainbow have different tints? 813. A thin soapy film is stretched over a vertical frame. When the film is illuminated by white light it shows three bands coloured purple (crimson), yellow and light-blue (bluegreen). Find the arrangement and the order of the bands 814. Why does the Moon, purely white in the daytime, have a yellowish hue after sunset? 815. Why does a column of smoke rising above the roofs of houses seem blue against the dark background of the surrounding objects, and yellow or even reddish against the background of a bright sky? 816. Why do the colours of moist objects seem deeper and richer than those of dry ones? ANSWERS AND SOLUTIONS CHAPTER 1 MECHANICS 1-1. Kinematics of Uniform Rectilinear Motion I. During the first hour after the meeting, the boat travelled away from the rafts. During the next 30 minutes, when the engine was being repaired, the distance between the boat and the rafts did not increase. The boat overtook the rafts in one hour because its speed with respect to the water and hence to the rafts was constant. Thus the velocity of the current fI= ~ l+~·.~+l km/h=3 km/h 2. Ascan be seen from Fig. 262, S= H H h s. Since the man moves uniformly, H s=vt. Hence S=H-h ot. The shadow moves with a constant velocity HHfIh • greater than the speed of the man. For this reason the velocity diagram has the form of a straight line parallel to the axis of abscissas. 3. The time of the meeting was 8 a. m. The man's speed was 4 km/h. The other questions can easily be answered with the aid of the chart in Fig. 263-the man met the twelfth bus at a distance of 10.7 km from the mill; the cyclist was overtaken by four buses. I,.., Fig. 262 8:30 7 7:30 8 8:30 9 9:30 10~halr8 Fig. 263 MECHANICS s,km 90 7:80 8 B:JO 9 9:30 10 161 4. The distance between the trains is s=vt; on the other hand, s=vr+u't. Hence u=v(t-'t')=45 km/h. l' 5. In Fig. 264, AA1N shows the usual trip of the car. SC-the engineer's walk until he met the car at point C, CB- the motion of the car after it met the engineer. According to the conditions, BN =KM = 10 minutes. The time during which the engineer had walked before he met the car is SD=SM -DM =SM -~=55minutes 6. The chart in Fig. 265 shows the movement of the launches between the landing-stages M and K. It follows from the chart that the landing-stages are served by eleven launches. A launch travelling from M to K meets eight launches, as does a launch travelling. from K to M. Fig. 264 11-2042 162 ANSWERS AND SOLUTIONS 10 I 2 9 5 5 G B 7 ist; hlJlJrS Fig. 265 7. Each tourist will walk half the distance and ride the bicycle the other half. The entire distance will be covered -in t=-2 5 +2~=5 hours 20 mtnVI Vt utes. Hence, the mean speed is v=T=7.S krn/h, The bicycle remains unused - during half the time of motion, i.e., during 2 hours 40 minutes. 8. Assume that the first candle burns down by the amount ~hl and the other by ~h2 during the time At (Fig. 266). The shadow on the left wall (from the first candle) will then lower over the distance tlx= ~hl +(l1hl -l1h2) = 2l1hl -l1hs and that on the right wall over the distance· l1y= aha - (~hl - Ah2) =2Aha-l1h! h h Remembering that I1hl = t; flt and ~hs=-r; 4t, we get ax 2h h h Vl= I1t=7;-7;= tlt a (2t,-t1) Ay 2h h h Va= At=t;-t;= t 1ta (2t1-t,.> If t2 > tt' then VI > 0 and Va may be negative, l.e., the shadow on the right wal may move upward. 0. The bus is at point A and the man at point B (Fig. 267). Point C is the spot where the man meets the bus, a is the angle between the direction towards the bus and the direction in which the man should run, AC=v1tlt BC=V2t2, where tt and t2 are the times required for the bus and the man, respectively, to reach point C. MECHANICS Fig. 266 163 b sin a a A glance at triangle ABC shows that AC= sin p , where sin ~= Be. Consequently. sin (1,= a b ~t • According to the condition, t1 ~ t., and Va 2 therefore sin a.~ :~ = 0.6. Hence 36°45' C;;;; a. OS;;;; 143°15'. The directions in which the man can move are within the limits of the angle DBE. Upon running in the directions BD or BE, the man will reach the highway at the same time as the bus. He will reach any point on the highway between D and E before the bus. 10. The minimum speed can be determined from the conditions . aVl 1 t1 = t2, 81n a = oo l = a Hence VS=TVl=2.4 mte. Here a=900. Therefore, the man should run in a direction perpendicular to the initial line (AB, Fig. 267) between him and the bus. II. Since the man's speed in water is lower than that along the shore, the route AB will not necessarily take the shortest time. Assume that the man follows the route ADB (Fig. 268). Let us determine the distance x at which the time will be minimum. The time of motion t is y d2 +Xi +_s-_% = V-=2_V_t1_2.....;.+_X_2__~Ul;;:..%_+'"'-Vl__S VI . VI VIVS This time will be minimum if Y=VI Yd2 + X2 - VIX has the smallest value. obvi ously, the value of x that corresponds to the minimum time t does not ~~ 8 Fig. 267 II * 164 ANSWERS AND SOLUTIONS depend on the distance s. To find the value of x corresponding to the minimum value of y, let us express x through Y and obtain the quadratic equation 2YVt V~d2 - y2 xt - I 2 x+ 2 2 ~O V2- VJ V2- Vl Its solution leads to the following expression VtY ± V2 Vy2 +d2V~ - v:d 2 x v:-v~ Since % cannot be 8 complex number, y2+d2v~ ~ v:d2 • Th~ minimum value of y. is equal to Ymin=d V v:-v:, and X= V.dVl corresponds to it. vi-vl If s~ dVt ,the man should Immediately swim to point B along AB. V v:-v~ Otherwise, the man should run along the shore over the distance AD=sdVI , and then swim to B. V (J~-VI2 Let us note that sin a= ~ for the route corresponding to the shortest °2time. 12. (I) Graphically, it is easier to solve the problem in a coordinate system related to the water. The speed of a raft equal to the velocity of the river current is zero in this system, and the speed of the ship upstream and downstream will be the same in magnitude. For this reason tan ex.!: tan ('X2=Vl on the chart showing the motion of the motor-ship (Fig. 269). When the ship stops at the landing-stage, its speed with respect to the water will be equal to the river current velocity Va. Hence tan a=v2. B " a s Fig. 269 B a I MECHANICS s 165 J 2 B j t; hours Fig. 270 It can be seen from Fig. 269 that t BF tan ~·t3-tan(Xl·tl V2= an a.=n= t. 2.5 km/h (2) From the moment the ship meets the rafts to the moment It overtakes them, the rafts will cover a distance equal to s=v. (t1 +t.+t8) On the other hand, this distance is equal to the difference between the distances travelled by the ship upstream and downstream: s=t3 (Vi +V2)- t1 (VI-VI) Hence, and 13. The motion of the launches leaving their landing-stages at the same time is shown by lines MEB and KEA, where E is their meeting point (Fig. 210). Since the speeds of the launches relative to the water are the same, MA and KB are straight lines. Both launches will travel the same time if they meet at the middle between the landing-stages. Point 0 where they meet Iles on the intersection of line KB with a perpendicular erected from the middle of distance KM. The motion of the launches is shown by lines KOD and COB. It can be seen from Fig. 270 that AM AF is similar to I1COF, and, therefore, the sought time MC=45 minutes.. 14. The speed of the launches with respect to the water VI and the velocity of the river current v2 can be found from the equations s= t1 (VI +VI) and s= t2 (Vl - V 2) , where it and t2 are the times of motion of the launches downstream and upstream. It follows from the condition that 11= 1.5 hours and t,=3 hours. 166 Hence ANSWERS AND SOLUTIONS and therefore 5 (t1 +12) 15 k h VI = 2/ 1 t 2 m/ 5 (t -t )V2 = 2 1 =5 km/h 2t 1t2 The point of the meeting is at a distance of 20 km from landing-stage M. 15. Let us assume that the river flows from C to T with a velocity of Vo• Since the duration of motion of the boat and the launch is the same, we can write the equation _ 5 =2(_5 +_5) VI +Vo V2 +Vo v 2 - Vo where s is the distance between the landing-stages. Hence, v~ +4v2Vo +4V2V1 - v: = 0 Vo= -202 ± V 5v:-4vIV2= -20 ± 19.5 The solution Vo = - 39.5 krn/h should he discarded, since with such a current velocity neither the boat nor the launch could go upstream. For this reason. Vo=- 0.5 krn/h, i. e.• the river flows from T to C. 16. The speed of the boat v with respect to the bank is directed along AB (Fig. 271). Obviously, v=vo+u. We know the direction of vector v and the magnitude and direction of vector Yo. A glance at the drawing shows that vector u will be minimum when u1-v. Consequently. b Umin =Vo cos a. where cos a Ya2 +b2 17. Let the speed u be directed at an angle a. to the bank (Fig. 272). Hence t (u cos a-v)=BC=a. and tu sin a=AC=b where t is the time the boat is in motion. By excluding a from these equations. we get t2 (u2_v2)-2vat-(a2+b2)=O whence t= 15/21 hour. It is therefore impossible to cover the distance AB in 30 minutes. c A Fig. 271 a B 6 Fig. 272 c MECHANICS 167 18. Let Uo be the velocity of the wind relative to the launch. Hence the flag on the mast will be directed along Uo' If v is the speed of the launch with respect to the bank, then u=uo+v (Fig. 273). In triangle FCDthe angleDCF=~+Ct-~ and the angle FDC=n - p. According to the sine theorem, A Fig. 273 B v sin (~+~- ~ ) u sin(n- P) Sin(~+~-~) and therefore v=u . ( A) •sIn It-.., It is impossible to determine the velocity of the current from the known speed of the launch with respect to the bank, since we do not know the direction of the moving launch with respect to the water. 19. (1) If the speed of the plane relative to the air is constant and equal to u, then its speed with respect to the Earth with a tail wind (along side Be) is VBC=V+U, with a head wind vDA=v-u and with a side wind VAB=vCD= = YV2 _ U2 (Fig. 274a and b). Hence, the time required to fly around the square is a a 2a v+ YV2 _ U2 t =--+--+ 2a---......-- 1 V+U v-u Vv2 u2 V 2_U2 (2) If the wind blows along the diagonal of the square from A to C, then (see Fig. 274c) V2=v~B+U2_2uVAB cos 450 B C V V U ~ A 11 u l/ (d) (b) (c) (d) Fig. 274 168 ANSWERS AND SOLUTIONS \ \ \ \ \ \ \ \ \ \ \ \ U.'-______ ~ Fig. 275 The speed along sides AB and Be Is V2 r: u2 VAR=VRC=-2- u+ VV 2_ 2 The speed along sides CD and AD (Fig. 274d) vCD=vDA=-V/ u+ YV2 - u; Let us leave only the plus sign before the root in both solutions to preserve a clockwise direction of the flight. The time required to fly around the square is 4a -(V2 - ~2 t2 V2_U2 Fig. 276 MECHANICS 169 20. Let us use the following notation; u12=speed of the second vehicle with respect to the first one; U21 =speed of the first vehicle with respect to the second one. Obviously, U12=U21 and u~2=v~+v~+2vIV2 cos a (Fig. 275). The time sought is t = 3- .U12 21. During the time ~t the straight line AB will travel a distance ol~t and the straight line CD a distance u2~t. The point of intersection of the lines will travel to position 0' (Fig. 276). The distance 00' can be found from triangle OFO' or OEO', where OF= v~ At =£0' and OE= v~ At =FO', i. e., sin a sin ~ 00' = YOF2 +OE2+20F· OE cos ~=V &t whence 1-2. Kinematics of Non-Uniform and Uniformly Variable Rectilinear Motion 22. The mean speed over the entire dislance Om = t1+:2 +t3' where tl' 12 and is are the times during which the vehicle runs at the speeds UI • o: and Os respectively. Obviously, Consequently, 18 krn/h 23. The path s travelled by the point in five seconds is equal numerically to the area enclosed between Oabcd and the time axis (see Fig. 6): 51 = 10.5em. The mean velocity of the point in five seconds is VI = SI/t1 = 2.1 cm/~ and the mean acceleration of the point during the same time is Av at =7; =0.8 cm/s" The path travelled in 10 seconds is 52 = 25 em. Therefore, the mean velocity and the mean acceleration are u2 = ~2 =2.5 cm/s, a2 = O.2 em/52 2 24. During a small time interval ~t the bow of the boat will move from point A to point B (Fig. 277). The distance AB=vl ~t, where VI is the speed of the boat. A rope length of OA-OB=CA=v At will be taken up during this time. The triangle ABC may be considered as a right one, since AC ~ OA. v Therefore VI =--. COSeL o Fig. 277 B' B A.Ds Fig. 278 25. Assume that at the initial moment t = 0 the object was at point S (Fig. 278), and at the moment t occupied the position CD. The similarity of the triangles SeD and SBA allows us to write the equation AB =~=!!!.-3D Vii The velocity of point B at a given moment of time is v2 = ~~' if the time ~t during which the edge of the shadow is shifted by the distance BB' tends to zero. • I , hl ( 1 '1) hi ~t hl SlnceBB=AB-AB=v;- T-I+~t t(t+At),thenv2 Vlt(t+~t)' hl or, remembering that ~t~t, we have U2 = - t2' VI 26. For uniformly accelerated motion x=xo+Vol +a~2 • Therefore Vo= = 35 crn/s, a=82 cm/s2 , and xo= 11 em is the initial coordinate of the point. 27. It follows from the velocity chart (see Fig. 8) that the initial velocity vo=4 crn/s (OA=4 cm/s). The acceleration a=g~= I crn/s''. First the velocity of the body decreases. At the moment t1 =4 s it is zero and then grows in magnitude. The second chart (see Fig. 9) also shows uniformly variable motion. Before the body stops, it travels a distance of h= 10 em. According to the first chart, the distance to the stop equal to the area of triangle DAB is 8 em. Therefore, the charts show different motions. A different initial velocity v' = 2h =5 cm/s and a differ- 11 ent acceleration a' = 2h2 = 1.25 t1 em/s2 correspond to the second :r: chart. 28. The mean speeds of both the motor vehicles are the same MECHANICS and equal to 171 CJ m= 2VIV2 =36 krn/h 01+ v2 Therefore, the distance between points A and B is 72 km. The ftrst vehicle travelled half of this distance in t' = 6/5 h, and the other half in t"= 4/5 h. The second vehicle travelled all the time with the acceleration 25 a=-=36 krn/h'' t~ and reached a speed of v/;:=ato= 72 krn/h at the end of its trip. It acquired a speed of 30 krn/h in t1=V~o=~ h and a speed of 45 km/h in t2 = V 45 = 5/4 h after the moment of departure. a At these moments the first vehicle moved at the same speed as the second. At the moment when one vehicle overtook the other, both of them travelled the same distance, and therefore the following equalities should be true at2 6 vl t =7 for t~5 hand at2 6 V1t'+V2 (t - t')= 2 for "5 h<;t~2h In the first case t =0 (the vehicles run side by side at the initial moment) or t = 5/3 h, which disagrees with the condition that t <6/5 h. In the second case t = 2 h (the vehicles arrive simultaneously at point B), and t = 1/2 h. This does not satisfy the condition that t» 6/5 h. Hence, neither vehicle overtakes the other. 29. The maximum velocity of the ball when it touches the support is vmax = Y 2gH. During the impact the velocity of the ball is reversed, remaining the same in absolute magnitude. The velocity chart has the form shown in Fig. 279a. Figure 27gb shows how the coordinate changes with time. 30. The time during which the first ball falls is /1= V2;1 =0.3 s. The VI yh2 I ratio of the maximum velocities of the balls is - - - - -VI - hI - 2 · {a} o (bJ t Fig. 219 172 '0; U2 OIr----~---1-~-+--...lIII~t-~-+---'IIIr-----'~ ANSWERS AND SOLUTIONS Fig. 280 It follows from the velocity chart (Fig. 280) that the mmimum time 't'=0.3 s. Besides. the second ball may begin to fall in 0.6, 0.9. 1.2 s, etc., after the first ball begins to drop. The time t during which the velocities of the' two balls are -the same is 0.3 s. The process periodically repeats every 0.6 second. 31. The initial equations are gtl g (t-'f)2 n-l T=n, 2 where 't' is the duration of motion of the body over the n-th centimetre of its path. Hence, t= Y2;. t-t:= y2(n;1) ~= y; CVn-Yn-I} 32. Upon denoting the coordinate and the velocity of the first body with respect to the tower by Xl and VI and those of the second by X2 and V2_ we can write the following equations Xa= -Vo (/-1') VI=Vo-gt; (The upward direction is considered to be positive here.) The velocity of the first body relative to the second is U=VI -V2= 2vO-g-r and it does not change with time. The distance between the bodies is g-rl S=XI-X2=(2vO- g't') l-vo't'+ 2 The bodies move uniformly with respect to each other and therefore the distance between them changes in proportion to the time. MECHANICS Fig. 281 173 33. According to the condition, AA'=vt and cc= a~2 (Fig. 281). From the similarity of the triangles AA'O, BB'O and GG'D we have AA' BB' cc AD = BO = CO A glance at Fig. 281 shows that AO=AB+BD and CO=BC-BO. These ratios allow us to determine AA' -Ge' vt at2 BB' 2 2-4 Hence, point B moves with the initial velocity ~ directed upward and a constant acceleration ~ directed downward. After reaching the height v2 h =4a ' the point will move downward. 34. Let us denote the speed of the lett-hand truck at a certain moment of time by VI' of the right-hand one by V2' and of the towed one by Va. Then, 'during the time t the left-hand truck will cover the distance al t2 51=V11+ - 2the right-hand truck the distance a2t'~ s2=vat+-2and the towed truck the distance Q st2 5s=03t +- 2At the same time it is easy to see that S3 = 51 tS2 Since this equality must be true at any value of t , then 01 +U2 and al +a2 va =--2"- aa= -2- 174 ANSWERS AND SOLUTIONS 35. The acceleration of the book with respect to the floor of the lift depends on the direction of acceleration of the lift and not on the direction of its motion (the direction of its velocity). If the acceleration of the lift is directed upward, the acceleration of the book will be g+a. If it is directed downward. the acceleration of the book will be g-a. 36. The acceleration of the stone with respect to the Earth is g and with respect to the railway carriage Ya2 +g2• 37. If the speed of the 11ft did not change, the ball would jump up to a height H. In a system of reading having a constant speed equal to that of the lift at the moment the ball begins to fall, the lift will rise during the time t to the height hl =a~~ , and during the next time interval t to the height h2= t2 . =at2 - a 2 . Its total height of rising is h=h1 +h2 = at2• The sought height which the ball will jump up to above the floor of the lift is x=H-h=H-at2• 38. The time during which the load is lifted to the height h is t= V:~ . The speed of the load relative to the crane in a vertical direction at this moment is VI =att and in a horizontal one v2=a.t. The total speed of the load with respect to the ground is v=y v~+v:+v~ 39. In free falling, body A will travel a vertical distance 51 =g~2 during the time t. During the same ti me the wedge should move over a distance 52 =a 2 t 2 • If the body is constantly in contact with the wedge, then ~ = . 51 =cot a. as can be seen from Fig. 282. Therefore, the acceleration sought a=gcot ct. If the acceleration of the wedge in a horizontal direction is greater than g cot a. the body will move away from the wedge. 1-3. Dynamics of Rectilinear Motion 40. The force F applied to the sphere determines, according to Newton's second law, the magnitude and direction of acceleration of the sphere, but does not determine its velocity. For this reason the sphere can move in any direction under the force F, and may also have a velocity equal to zero. 41. The resultant of all the forces is 6 kgf and coincides in direction with the force 5 kgf. Therefore, the acceleration of the sphere is a=~= 14.7 m/s2 m and is directed towards the force 5 kgf. Nothing can be said about the direction of motion (see the solution to Problem 40). 42. In the MKS (or 51) system the weight of the body G=mg=9.8 N. The unit of force in the technical system is 1 kgf, I, e.t the force wi th which , Fig. 283 the Earth attracts a body with a mass of 1 kg. In the cas system the weight of the body is 980,000 dynes. When Newton's second law is used to find the force in the technical system of units, it should be borne in mind that the mass shoul d also be expressed in technical units. 43.. The sought angle Cl can be determined from the ratio (F ig. 283): tan Cl=~g 44. The body sliding along the vertical diameter AB will cover the entire distance in the time tAB= {2~B.For an arbitrary groove arranged at { 2AC an angle a. to the diameter AB, the time of motion is tAc- - -- gcos cx· Since AC= AB cos a, then tAC= tAB. All the bodies will reach the edge of the disk simultaneously. 45. The force of air resistance F will reach its minimum after the parachutist's speed becomes constant, and we have F=mg=75 kgf. 46. According tv Newton's second law, N -mg= ± rna. Therefore, N =mg+ ma if the acceleration of the lift is directed upward, and N =mg-ma if downward, irrespective of the direction of the speed. . When a=g, then N=0. (Here and below N denotes the force of normal pressure. or the force of normal reaction.) 47. According to Newton's second law, ma=kmg. Hence, the coefficient of friction k=!!:.... Since in an elastic impact only the direction of the velocity g changes, irrespective of the angle, then a= ~ , where s= 12.5 metres is the total path travelled by the puck before it stops. v2 Therefore, k=-2 =0.102. gs 48. Assuming the acceleration of a motor vehicle to be constant, we can write a= ~ . Since the maximum force of friction in braking is kmg, then, v2 according to Newton's second law, m 2i=kmg, where m is the mass of the motor vehicle. Hence, k= 2 V2 • Upon inserting the values of u and s from the table in gs this formula, we can find the coefficients of friction for various pavements. 176 ANSWERS AND SOLUTIONS ice dry snow wet wood block dry wood block wet asphalt dry asphalt dry concrete k=O.l k=0.2 k=O.3 k=O.5 k=O.4 k=O.6 k=O.7 The coefficient of friction does not depend on the speed if the accuracy to the first digit after the decimal point is wanted. 49. When the motor vehicle accelerates, the rear wall of the fuel tank imparts an acceleration vlt to the petrol. According to Newton's second law, the force F required for this acceleration is Alp ~ J where A is the area of the rear wall of the fuel tank. In conformity with Newton's third law, the petrol will act with the same force on the wall. The hydrostatic pressure of the petrol on both walls is the same. Hence, the difference of pressures exerted on the walls is Ap= ~ =Ip -7-. M 50. The mass of the left..hand part of the rod mt=T I and of the right-hand part m2= ~ (L -I), where M is the mass ()f the entire rod. Under the action of the appl ied forces each part of the rod moves with the same acceleration a. Therefore, Ft-F=mta F-F1 = m2a Hence the force F is F 52. Figure 285 shows the forces that act on the weights. The equations of motion for B Fig. 284 51. The motion of the ball will be uniform. The images of the ball on the film appear at intervals of t= 1/24 s. The distance between the positions A and B of the ball in space that correspond to the positions C and D of the images on the film is AB=CD ~; • as shown in Fig. 284. The focal length of the lens OF= 10 em, OE= 15 metres, and CD=3 rnrn, The velocity of the ball A AB vt = - t- = 10.8 m/s. When the ball is in uniform motion. mg= =Iro~. In the second case 4 mg=kv:. Hence, E v:=4v~ and v2=21.6 tnls. G k=- ~ 3.9XIO-o kgf·s2/m2 v~ MECHANICS 177 the weights can be written as follows: mta=T-mtR and m2a=mte-T where T is the tension of the string and a is the acceleration. (The accelerations of the weights are the same since the string is considered to be unstretchable. The weightlessness of the string and the pulley determine the constancy of T.) Therefore, m -m1 a 2 g=327 cm/s2 mJ+m 2 T=mt (a+g)=130,700 dynes=133 gf The time of motion t= y2:::::1 s. 53. If the mass of the pulleys and the rope is negligibly small, then (Fig. 286) 2F-T=O, and T-G=ma. Hence. F= ~ (I +;). When a=O, we have F= ~ . 54. If. the mass of the second weight is much greater than 200 g, both weights will move with an acceleration somewhat below g, the acceleration of the lighter weight being directed upwards. To make a weight of mass m move upward with an acceleration g, a force of 2mg should be applied to it. For this reason the string should withstand a tension of about 400 gf. 55. The dynamometer first shows F =3 kgf. If the read ing of the dynamometer does not change, the weight 2 kgf is acted upon by the upward force of the string tension equal to 3 kgf. Therefore, this weight rises with an acceleration of a = ~. The other weight lowers with the same acceleration. 12-2042 Fig. 285 F Fig. 286 178 ANSWERS AND SOLUTIONS The additional weight on the second pan 'can be found from the equation Hence 01 =3 kgf. 56. The sphere is acted upon by three forces: the force of gravlty, the force of tension of the upper rope and the force applied to the lower one when it is pulled (Fig. 287). The acceleration imparted to the sphere by the pull can be found from the equation ma=F1 +mg-F2• For the lower rope to break, the force applied to it should be greater than the tension of the upper one, Lett F1 > F2• For this condition, the acceleration imparted to the sphere is greater than that of gravity, i.e, t a > g. Fig. 287 57. According to Newton's second law, (ml +m2) a=l1hg sin a- m28 sin ~-kmlgcos a-km~cos ~ The weights will be at the same height after travelling the distance s, which conforms to the following equations: s sin a.=h-s sin ~ and s=a;s. Upon eliminating s and a from the three equations we obtain ml g-c2 (sin a+sin ~) (k cos ~ +sin ~) +2h m2 g-r:2(sin a+ sin ~) (sin a-k cos a)-2h 58. The equations of motion give the following formulas for the acceleration of the stone: al=g(sina+kcosa) in upward motion a2 = g (sin a-k cos a) in downward motion The kinematic equations can be written as follows: alt~ a2t:l=vot1- 2 ; l=2; vo-at t1=O We find from these five equations that 2l-gt2 1 sin a k ~ 0.37 gt~ cos a t2 =t1Vgt~si~a.-l 4.2s 59. For this case the equations of dynamics can be written as mg-T=ma, and T=Ma where T is the tension of the string. Hence. m 2 a=--g=-g M+m 7 MECHANICS 179 The equations of kinematics give x=vot-a~2. and vt=vo-at. Upon solving this system of equations, we find that in 5 seconds the cart will stay at the same place (x=O) and will have a speed vt=7 mls directed to the right. The car t will cover the distance 5=2 {vo ~ a(tr}= 17.5 metres 60. The ice-boat can move only in the direction of its runners. When the speed of the boat exceeds that of the wind, the velocity of the latter with respect to the boat has a component directed backward. If the velocity of the wind with respect to the boat also has a component perpendicular to the direction of motion, the sail can be so set that the force F acting on it will push the boat forward (Fig. 288). Therefore, the speed cf the boat can exceed that of the wind. In practice It can be two or three times greater. 61. (1) At the initial moment the acceleration is ao= ~ ~ 13.1 mjs2. it changes with time according to the law a= M F f1t' where f1=200 kgjs is the mass of the fuel consumed by the rocket in a unit of time. A diagram of the acceleration is shown in Fig. 289. In 20 seconds the velocity is numerically equal to the hatched area, v=== 300 m/s. (2) Newton's second law can be written as (M-Ilt) a=F-(M -Ilt) g-f According to the initial conditions, t=20 sand a=O.8 g. Hence, the force 0/111/33 Veloclt!l 0/wind relative toice-boat Fig. 288 12 * Fig. 289 180 --+- 0 9 2g 3g 4g 5g Fig. 290 Fig. 29/ of air resistance is ANSWERS AND SOLUTIONS f=F-(M -Jit) g-(M -~li) 0.8 g= 12,800 kgf (3) Newton's equation for the weight gives mla=kx-m1g, where m1 is the mass of the weight at the end of the spring, a is the acceleration of the rocket, k is the coefficient of elasticity of the spring, x is the elongation of the spring. According to the condition, m1g=klo• Therefore, x=!JL (a+g). The scale of the device should be graduated uniformly g (Fig. 290). An acceleration of g corresponds to a division of one centimetre. 62. The only force acting on the bead is the reaction force of the rod N directed at right angles to the rod. The absolute acceleration W a of the bead (relative to a stationary observer) will be directed along the line of action of the reaction force N. The relative acceleration w,. is directed along the rod (Fig. 291) wa=a+wr It follows from the triangle of accelerations that Wr =a cos Cl and wa=asin Cl. From Newton's second law, the reaction force is N = masin Cl. The time 't' during which the bead moves along the rod can be found from . a cos Cl·",2 (-2-[-the equation [= 2 . Hence, '1'= - - • a cos a. 63. When the bead moves it is acted upon by the friction force kN and the reaction force N. The absolute acceleration will be directed along the resultant force F. It follows from Fig. 292 that" N . d kN ( k .=rna sin a, an w,.=a cos ex--m=a cos (X- SIn ex) MECHANICS Hence (see Problem 62). .r 2l 't= V a (cos a.-k sin a) 181 If k ~ cot a, the bead will not move with respect to the rod, and the force of friction is rna cos a.. 64. The equations of motion of the block and the body have the form: ma=f (1) Mb::xF-f (2) where f Is the force of friction; a and b are accelerations. Let us assume that there is no sliding, then a=b. The acceleration and the force of friction can be found from the equations of motion. The force of friction is f=m M :m' For the body not to slide, the force of friction should satisfy the inequality f<.kmg, I.e., M~m";;;;'kg.lf P > k(M+m)g, the body will begin to slide. Here equations (1) and (2) will take the form: ma=kmg, and Mb=F-kmg These equations can be used to find a and b: F-kmg a=kg, and b M Obviously, b > a. The acceleration of the body relative to the block will be directed oppositely to the motion and will be equal in magnitude to F-kmg k M g. The time during which the body moves over the block is t;::: .. /' 21M = V F-kg(M+m)· Fig. 292 r II Fig. 293 182 ANSWERS AND SOLUTIONS and the body a distance 65. Initially the wagon is in uniformly retarded motion at a speed of v= vo- kt, where f is the force of friction equaI to kmg. The body is in uniformly accelerated motion at a speed of u=L t. m If the wagon is long, the speeds of the body and the wagon may become equal. This will occur at the moment of time 1'= f V o f . After this, both m+JW the body and the wagon will begin to move at a constant speed equal to M MVo • By this time the wagon will have covered a distance +m s=vo1'- 2~ 1'1 s=L't2 2m The distance covered by the body relative to the cart is 8-5. It should be shorter than 1. Thus, the body will not slip off the wagon if S-5 ~ I, Le., Mv~ 2gk (M +m) -ct 66. Let us consider the string element in the slit. Assume that the string moves downward. Then, the string element will be acted upon by the force of string tension on both sides and the force of friction (Fig. 293). Since the mass of this string element is neglected, T1- F - T2 =0. The equations of dynamics can be written as follows: ml g- T1=m Ia m~-T2=-m2a Hence (ml-m2) g'-F a mt+ m2 67. Since the weights move uniformly, the tension of the string is equal to the weight nu. Therefore, the force with which the pulley acts on the bar is 2m.g, i.e., in the first case it is 2 kgf and in the second 6 kgf. In both cases the balance will show the sum of the first and second weights, Le., 4 kgf. The force of friction equal to 2 kgf is applied to the bar at the side of the second weight. In the first case it is added to the force of pressure exerted by the pulley on the bar and in the second it is subtracted from it. 68. Since the masses of the pulleys and the string may be neglected, the tension of the string will be the same everywhere (Fig. 294). Therefore, mlg-T=m1al mJI- 2T = m2a'J m~-T=m3a3 a1+aa a2=--2- MECHANICS Fig. 294 (see Problem 34). Hence, 183 Fig. 295 4mlm3- 3m2m3+ mlm2 at g 4mlm3 +m2m3 +mtmt mlm2-4mlm3+m2m3 a2 g 4m1m 3 + m2mS+ m1m2 4mlmS-3mlm2+m2m3 as= 4mlmS+m2mS+mlm2 g 69. The second monkey will be at the same height as the first. If the mass of the pulley and the weight of the rope are disregarded, the force T tensioning both ends of the rope will be the same and, therefore, the forces acting on each monkey will equal F=T-mg. Both monkeys have the same accelerations in magnitude and direction and will reach the pul.. ley at the same time. 70. Since the mass of the pulleys and the string is negligibly small, the tension of the thread is the same everywhere. Therefore, mlg-T=m1a1 mgg- 2T = m2a2 2T-T=O Hence, T=O and a1=a2= g. Both weights fall freely with an acceleration g. Pulleys Band C rotate counterclockwise and pulley A clockwise. 71. (1) The forces acting on the table and the weight are shown in Fig. 295. The equations of horizontal motion have the following form: for the table with the pulley 01 F-F+F/r=-atg and for' {h'c weight 184 ANSWERS AND SOLUTIONS Let us assume that the force F is so small that the weight does not slide over the table. Hence a1=a2 and F1r=F Gl~G2 · By gradually increasing the force F we shall thereby increase the force of friction F1" If the table and the weight are immovable relative to each other, however, the forceof friction between them cannot exceed the value Ffr.max=kG2• For this reason the weight will begin to slide over the table when F F 01+01 G2 0 > 'f.max-a--=k-a ( I+G2)=10 kgf . 1 1 In our case F=8 kgf, consequently, the weight will not slide, and al=~=Gl~G2 g=:5g~ 314 em/5 2 (2) In this case the equations of motion fo the table with the pulleys and the weight will have the form: -F+F,,=G1 al g F-F,,=G2 a 2 g The accelerations of the table and the weight are directed oppositely, and the weight will be sure to slide. Hence, F1,=k02• . The acceleration of the table is -F+kG2 2 01= g=-- g=- 131 em/s2 01 15 and it will move to the left. 72. According to Newton's second law, the change in the momentum of the system "cannon-ballIt during the duration of the shot -r should be equal to the impulse of the forces acting on the system. Along a horizontal line mvocos a-MvI =FIf'" where FIf" is the impulse of the forces of friction. Along a vertical line mvo sin a=N-r-(Mg+mg) '" where NT. is the impulse of forces of normal pressure (reactions of a horizontal area), (Mg+mg) -r is the impulse of the forces of gravity, Remembering that Ftr =kN. we obtain m m. M+m [)t= M Vocos a-k ¥vosln a-k~g-r or since g'f ~ Vo VI. ZVo (cos a,-k sin a,) This solution is suitable for k <; cot cx. When k > cot a the cannon will remain in place. MECHANICS 1-4. The Law of Conservation of Momentum 185 73. The momentum of the meteorite is transmitted to the air molecules and, in the final run, to the Earth. 74. Let us divide the mass of the disk into pairs of identical elements lying on one straight line at equal distances from the centre. The momentum of each pair is zero, since the momenta of both masses are equal but oppositely directed. Therefore, the momentum of the entire disk is zero. 75. The propeller of 8 conventional helicopter is rotated by an engine mounted inside the fuselage. According to Newton's third law, oppositely directed forces are applied from the propeller to the engine. These forces create a torque that tends to rotate the fuselage in a direction opposite to rotation of the propeller. The tail rotor is used to compensate for this torque. In a jet helicopter the forces from the propeller are applied to the outflowing gases and for this reason do not create any torque. . 76. The velocity of the boat u can be found with the aid of the law of conservation of momentum. In a horizontal direction Mu=mv cos a. Hence, u=8 cm/s, 77. At the highest point the rocket reaches, its velocity is zero. The change in the total momentum of the rocket fragments under the action of external forces (the force of gravity) is negligible since the impulse of these forces is very small in view of the instantaneous nature of the explosion. For this reason the total momentum of the rocket fragments before and immediately after the explosion remains constant and equal to zero. At the same time, the sum of the three vectors (m1v1l m2v2, mava) may be zero only when they are in one plane. Hence it follows that the vectors Vi' v2 and Va also lie in one plane. 78. Let the mass of the man be m and that of the boat M. If the man moves with a speed v relative to the boat the latter will move at a speed - u with respect to the bank. and therefore the man moves with a speed VI =v-u with respect to the bank. According to the law of conservation of momentum, m (v-u)-Mu=O m Hence, u=m +M v and the speed of the man relative to the bank is M vt=m+M u. Since the signs of VI and v coincide, the distance between the man and the bank will increase whatever the ratio between the masses m and M. 79. The speed of the boat u with respect to the shore is related to the speed of the man v with respect to the boat by the equation u=m~M v (see Problem 78). The relation between the speeds remains constant during motion. For this reason the relation between the distances travelled will be equal to the relation between the speeds s m -Y=m+M 186 ANSWERS AND SOLUTIONS where s is the distance travelled by the boat and 1is its length (the distance covered by the man in the boat). Therefore, for the boat to reach the shore its length should be at least m+M1=--s=2.5 metres. m 80. When the spring extends, it will act on both weights. The weight at the wall will first remain stationary while the second weight will begin to move. When the spring extends completely (I.e., is no longer deformed), the .second weight will have a certain velocity. Therefore, the system will acquire momentum in a horizontal direction that will be retained in the following period, since the external forces will not act in this direction. Thus, the system as a whole will move away from the wall, the weights alternately converg ing and diverging. 81. The speed of the cart will not depend on the point of impact. The momentum of the revolving cylinder is zero, irrespective of its direction and velocity of rotation (see Problem 74). For this reason the bullet will impart to the "cylinder-cart" system the same momentum as it would to a cylinder rigidly secured on the cart. 82. Let us denote the velocity of the rocket at the end of the k-th second by Vk. Gas with a mass m is ejected from the rocket at the end of the (k+ l)-th second, and it carries away a momentum equal to m(-u+v,,) It follows from the law of conservation of momentum that (M -km) v,,=[M -(k+ 1) m] t'k+l +m (-u+v,,) The change in the velocity of the rocket per second is mu M-(k+l)m If we know how the velocity changes in one second, we can write the expression for the velocity at the end of the n-th second vn=vo+U(M m +M m 2 +...+-Mm )-m - m -nm 83. The velocity of the rocket will grow. This becomes obvious if we pass over to a reading system with respect to which the rocket is at rest at the given moment. The pressure of the ejected gases will push the rocket forward. 84. Let the mass of the boat be M, that of a sack m and the velocity of the boats Vo. When a sack is thrown out of a boat the latter is acted upon by a certain force in a direction perpendicular to vo. It should be noted, however, that the boat does not change its velocity, since the resistance of the water prevents lateral motion of the boats. The velocity of a boat will change only when a sack is dropped into it. . Applying the law of conservation of momentum to the "sack-boat" system, we can write in the first case: for one boat (M+m)vo- mvo= (M + 2m) VI for the other one -Mvo+mvt=(M+rn)v2 Here ~Vl and V2 are the final velocities of the boats. From these simultaneous ti Mequa Ions VI =- t'2= M+2m vo- MECHANICS 187 When the sacks are exchanged simultaneously, the final velocities of the boats v~ and 0; can be found from the equations: Mvo-mvo=(M +m) v~ ; -Mvo+mvo=(M·+m) v~ ..,-----........ »< "/' , / A '.I \ I \ \ I J J I o (71''...... ,,/J ...... -----------" , , M-m Hence. VI =-V2 = M +m Vo' Thus, the final velocity of the boats will be higher in the first case. 85. External forces do not act in a horizontal direction on the "hoop-beetle" systern. For this reason the centre of gravity of the system (point C in Fig. 296) will not move in a horizontal plane. The distance from the centre of gravi ty of the syst em .. to the centre of the hoop is CO=m;M R. Since this distance is constant, the centre of the hoop 0 will describe a circle with the radius CO about the stationary point C. It is easy to see t hat the tra .. jectory of the beetle is a circle with the radius AC=m~M R. The mutual positions and the direction of motion of the beetle and the hoop are shown in Fig. 296. 86. Since no external forces act on the system in a horizontal direction. the projection of the total momentum of the "wedge-weights" system onto the horizontal direction must. remain constant (equal to zero). It thus followsthat the wedge will begin to move only if the weights move. For the weight m2 to move to the right, the condition Fig. 296 should be observed. Therefore. ml E;; sin a-k cos a. Here the wedge will move to the left. For the weight m~to move to the leit.ithe following condition should be observed: mlg~ m~ sin ex+km~cos ex or Here the wedge will move to the right. Hence, for the wedge to be in equillbrtum, the ratio between the masses of the weights should satisfy the inequality sin a.-k cos a~ ml ~ sin a+k cos a. mj 188 1-5. Statics ANSWERS AND SOLUTIONS k 87. ll=lk+l. 88. In the position of equilibrium (Fig. 291)mg-2mg cos a=O. Therefore, a=600. The sought distance h=l cot a= ~)_ . Equilibrium will set in after J' 3 , the oscillations caused by the weight being lowered attenuate. 89. The equality of the projections of the forces onto the direction of the vertical (Fig. 298) gives the equation 2N sin i -2F/T cos ; =0 where N is the force of normal pressure and FIT~ kN is the force of friction. The weight of the wedge may usually be neglected. a Hence, tan 2"<; k and a ~ 2 arc tan k. 90. If the weight 01 lowers over the height h, then point F will lower by h/3. The weight G2 will rise by 2h/3. Applying the "golden" rule of mechanics, 2 we have G1h=G2 3"h. 2 Hence, 01 ="3 O2• 91. If the box is not overturned, the moment of the force F rotating the box in one direction, say counterclockwise, about a bottom edge is less than or equal to the moment of the force of gravity rotating the box cloc kwise. For the box to slide, the force should be greater than the maximum force of friction applied to it. Th erefore, l Fh~mg2 and F~kmg 1 whence k <; 2h . Fig. 297 Fig. 298 MECHANICS Fig. 299 189 92. To turn the beam, the moment of the forces applied to its ends should be greater than the moment of the forces of friction when they reach their maximum. The forces of friction are distributed uniformly along the beam (Fig. 299). The mean arm of the forces of friction acting on the left- or right-hand part of the beam is l/4, if the length of the entire beam is 1. The moment of all the forces of friction with respect to the beam centre is 2 k~ ~ • Consequently, to turn the beam around, the applied forces F should satisfy the inequality kQ whence F > 4' To move the beam translationally, 2F should be greater than kG. Therefore, it is easier to turn the beam. 93. The equation of motion of the load is Goa=F - Go (Fig. 300). The g sum of the forces acting on the crane vertically is zero. Therefore, G1 +G2 = = G+F. Since the sum of the moments of the forces relative to point A L is zero, we have Fl+G 2=LGt . Solution of these simultaneous equations gives G1 =::2.23 toni, and O2 === 1.77 tonf 94. For the lever to be in equilibrium. the force applied at point D should produce a moment equal to G· AB. The force will be minimum when the maximum arm is equal to BD. 0, ....-----l Ar=======t===r========IB B Fig. 300 Fig. 301 190 ANSWERS AND SOLUTIONS Hence, F=G ~g=;2' and it is directed at right angles to BD. 95. If there is no friction between the floor and the boxes, the latter will move simultaneously. If the coefficient of friction is not zero, the right..hand box will move first (see Fig. 35), since the" force applied to it by the rod will be greater than the force applied to the left-hand box. Indeed, the rod is acted upon from the side of the right-hand box by the force F1 directed opposite to F, and from the side of the left-hand box by the force F2 directed along FI The sum of the forces in equilibrium is zero. Therefore, F1 = F+F2' and the force F1 will reach the maximum force of friction of rest before F!I 96. The equality to zero of the sum of the moments of the forces acting on the sphere with respect to point A (Fig. 301) gives us the equation F'rR-NR=O Since FIr ~ kN, then k ~ 1. 97. For a body to be at rest, the total moment of the forces that tends to turn the body clockwise should be equal to the moment of the forces that tends to turn the body counterclockwise around a point (around the centre of gravi ty t for example). In our case the moment of the forces of friction that turns the brick clockwise should be equal to the moment of the forces of the pressure exerted by the plane on the brick. It follows that the force of pressure exerted by the plane on the right half of the brick should be greater than on the left one. According to Newton's third law, the force with which the right half of the brick presses against the plane should be greater than that of the left half. 98. To lift the roller onto the step, the moment of the forces turning the roller around point A (Fig. 302) counterclockwise should at least be equal to the moment of the forces turning it clockwise, i.e., G (R-h)=G VR2_(R-h)2 Hence, h 2 ±rrR. Since h < R, then h=(t_V 2 2 )R~O.29R 99. Since the force of friction is zero on one of the planes, it is also zero on the other one. Otherwise, the sphere would rotate around its centre, for Fig. 902 Fig. 303 MECHANICS 191 mg • lJ a the moment of all the other forces relative to this centre is zero (because the arm of each of these forces with reference to the centre of the sphere Is zero). The sums of the projections of the forces on the vertical and horizontal directions are equal to zero (Fig. 303). For this reason Nt cos CX2 - N2 cos at :z:: 0 G-Nt sin cx2 - N2 sin (Xl =0 where Nt and N2 are the sought pressure forces. Hence, N!. G S!i 2.6 kgf SIn a2 +cos <%2·tan al N2 • G ~ 1.5 kgf SIn at +cos at · tan a2 100. Let us denote the force applied to one handle by F. The force F will turn the drawer and induce elastic forces Nt and N 2 at points A and B (Fig. 304) which will act on the drawer from the side of the cabinet. These forces are equal: NI =N2=N. Since the moment of all the acting forces I relative to the centre of the drawer C is zero, N =F 2a . The drawer can be pulled out if the applied force F is greater than the maximum force of friction of rest: F > ft +f2 = 2kN. For the last inequality to be satisfied, k should be smaller than ~ . 101. A board on a rough log forming an angle cx with a horizontal plane Is similar to a body retained by the forces of friction on an inclined plane with an angle cx at its base. Therefore, in equilibrium, FIr==mg sin cx. Bea.. ring in mind that FIr ~ kmg cos a, we have tane ~ k. ~ ...-.....----...~... B B Fig. 804 Fig. 305 o ••• ~. ., N A Fig. 306 Fig. 907 102. The forces applied to the ladder are shown in Fig. 305. In equilibrium the sums of the projections of the forces along a vertical and a horlzontal lines are equal to zero. Therefore, N1 = F/' and N2 =mg. Since the sum of the moments of the forces relative to point B is zero, we can write another equation sin a, Nt cos a=mg- 2 Hence, Ffr=mg ta~ ~. Since the force of friction satisfies the inequality F/r ~ kN2' the ladder will be in equilibrium if tan C%<2k 103. The forces applied to the ladder are shown in Fig. 306. Since the sum of the forces and the sum of the moments of the forces are equal to zero: f+N,,=mg (I) Nt=F'f (2) 1 f sin a+ NJ cos a=mg '2sin ex (3) The forces of friction f and F/r satisfy the inequalities f<.kN t and F/r'-; kN2· By using the first inequality and equations (1) and (3), we get: N2 k . Nt l-kt cot a ~ 2N1 -2· SInce k ~ N2' then cot a ~ ~ · If we assume that k= tan B, the inequality can be written in' a more convenient form for cal- MECHANICS 193 culations: cot a ~ cot 2P or a.e::;; 2~ 104. If at the moment when end B of the rod begins to rise. the force of friction F/r ..;; kN proves sufficient for end A not to slip. the rod will begin to rotate around point A. Otherwise, end A will begin to slip until the force of friction FIT =kN can keep the rod in equilibrium (Fig. 307). After this the rod will begin to rotate around end A. Let us find the coefficient of friction k at which slipping stops with 8 definite angle a between the rod and the string. The equality of the forces at the moment when the rod is almost horizontal gives us the equations: F,,=Tcosa G=N+Tsln ex The equality of the moments of the forces with respect to point A can be written as ---/l~ , I I I I •/ II I I IJ , 0 1/ N --1---II II /1,I J, I G2=~lslna By using this system of equations, we find that FJr k=-;:r=cota For the rod not to slip at all it is necessary that k;;:;:: cot 60"= ,; .-, J' 3 105. The sum of the moments of the forces acting on the man relative to his centre of gravity is zero. For this reason the force F acting from the Earth is always directed towards the man's centre of gravity C (Fig. 308). The horizontal component of this force cannot be greater than the maximum force of friction of rest: F sin ex ~ kF cos a. Hence, k ~ tan a. M Fig. 308 Fig. 309 Fig. 310 13-2042 194 C~~7;7l'l;~~W 'lr 0 Fig. 31J A F,r B ANSWERS AND SOLUTIONS N Fig. 312 106. The ladder is acted upon by three forces: its weight Gt the force from the Earth F and the reaction of the support N (Fig. 309). Since the wall is smooth, the force N is perpendicular to it. It will be the easiest to determine the force F if we find the point with respect to which the moments of the forces G and N are equal to zero. This will be the point at which the straight lines ON and 00 intersect. The moment of the force F relative to this point should also be zero. Therefore, the force should be directed so that it continues past point O. Figure 309 shows that the direction of the force F forms with the ladder 1 the angle p=3 instead of k so that tan q> =k. Then, cos a+ksin a cos (a-q» cos cp or cos a+k sin a= YI +k" cos (~-q» F Hence, k= ..r,. al_FI 112. The forces acting on the cylinder are shown in Fig. 315. Since the cylinder does not move translationally, F1,-Fcosa=O F sin a-mg+N=O Since the maximum value of cos(a- q» is unity, then F . _ kG eun>: YI +ki 0.75. MECHANICS 197 The force of friction FIr = kN- Hence, F kmg cos a+k sin a The denominator of this expression can be written as A sin (a+ cp). where A=Yl+k2 (see Problem Ill). Therefore, the minimum force with which the string should be pulled Is kmg Fmin "Ir • f 1+k2 The angle al can be found from the equation cos a1 +k sin al = Y 1+kZ, and tan a1 =k. 113. The forces acting on the piston and the rear lid of the cylinder are F1=Fz=pA (Fig. 316a). Point C of the wheel is also acted upon in a horizontal direction by the force F2 transmitted from the piston through the crank gear. The sum of the moments of the forces acting on the wheel with respect to its axis is zero. (The mass of the wheel is neglected.) Therefore. FIrR =F2" where Ftr is the force of friction. Since the sum of the forces which act on the wheel is also zero, the force F3 applied to the axis by the bearings of the locomotive is F3 ~ FIr +F2- According to Newton's third law. the force F.=Fs acts on the locomotive from the side of the axis. Hence, the tractive r effort F=F.-FI=Fjr=pA R' In the second position of the piston and the crank gear, the forces we are interested in are illustrated in Fig. 316b. For the same reason as in the pre- r vious case, Flr=F. R' The tractive effort F=F.-F.=F/,=pA ~ , As could be expected. the tractive effort is equal to the force of friction, for the latter is the only external force that acts on the locomotive. 114. The maximum length of the extending part of the top brick is 1/2. The centre of gravity of the two upper bricks C, is at a distance of 1/4 from the edge of the second brick (Fig. 317). Therefore. the second brick may extend by this length relative to the third one. The centre of gravity of the three upper bricks C3 Is determined by the equality of the moments of the forces of gravity with respect to C3; namely. G ( ~ -x) =2Gx. Hence x= ~, i.e. the third brick may extend over the &- fa) Fig. 316 198 Fig. 317 99 .0 Fig. 318 ANSWERS AND SOLUTIONS ,%, ---------- J( fourth one by not more than 1/6. Similarly it can be found that the fourth brick extends over the fifth one by L/B, etc. The nature of the change in the length of the extending part with an increase in the number of bricks is obvious. The maximum distance over which the right-hand edge of the upper brick can extend over the right-hand edge of the lowermost brick can be written as the series L=f(l+ ;+~+++ ...) When the number of the bricks is increased infinitely this sum tends to infinity. Indeedt the sum of the series 1 1 1 1 1 1 1 1+2+3'+4 +5+6+7+8+ ... is greater than that of the series ~ Y2 ,..-"'-. ....---"'--~ 1 1 1 1 1 I 1 1+"2+4+4+8+8"+8+8+ ... and the latter sum will be infinitely great if the number of terms is infinite. The centre of gravity of all the bricks passes through the right-hand edge of the lowermost brick. Equilibrium will be unstable. The given example would be possible if the Earth were flat. 115. Let us inscribe a right polygon into a circle with a radius, (Fig. 318). Let us then find the moment of the forces of gravity (with respect to the MECHANICS 199 axis AK) applied to the middles of the sides of the polygon AB, BC, CD, DE, etc., assuming that the force of gravity acts at right angles to the drawing. This moment is equal to pg (ABx1 +8Cx2+CDxs+ +DEx.+ EFx"+FKxe), where p is the mass of 8 unit of the wire length. o From the similarity of the cor- F . 319 responding triangles it can be shown ~g. that the products ABx1' BCx,. CDxa. etc., are equal respectively to AB'h, B'C'h, C'D'h, etc., where h is the apothem of the polygon. Therefore, the moment is equal to pgh(AB' +B'C' +C'D' +D'E' +E'F' +F'K)=pgh2r If the number of sides increases infinitely, the value of h will tend to r and the moment to 2r2pg. On the other hand, the moment is equal to the product of the weight of the wire nrpg and the distance x from the centre of gravity to axis AK. Therefore, 2r2pg=itrpgx, whence x=.! r, n 116. Let us divide the semicircle into triangles and segments, as shown in Fig. 319. The centre of gravity of a triangle, as is known, is at the point of intersection of its medians. In our case the centre of gravity of each triangle is at a distance of ~ h from point 0 (h Is the apothem). When the sides are increased infinitely in number the centres of gravity of the triangles will lie on a circle with a radius of fr, while the areas of the segments wi11 tend to zero. Thus, the problem consists in determining the centre of gravity of a se.. micircle with a radius of ~ r, It follows from the solution of Problem 115 that x, which is the distance between the centre of gravity of the semicircle and point 0, is equal to 2 2 4 x="1t3 r= 3n ' · 117. By applying the method used to solve Problems 115 and 116, it can be shown that the centre of gravity is at point C at a distance 2 · ex 81n"2 CO= r from the centre of curvature of the arc (see Fig. 45). a lt8. From the solutions of Problems 115, 116 and 117, It can be shown • ex 4 810'2 that the centre of gravity is at point C at a distance of CO==3-a- r from point O. 119. When the centre of gravity is determined, the plate with a cut-out portion can formally be considered as a solid one if we consider that a se- 200 ANSWERS AND SOLUTIONS rnicircle with a negative mass equal in magnitude to the mass of the cut..out portion Is superposed on it. The moment of the gravity forces of the positive and negative masses with respect to axis AB is equal to ( r nr' 4) 1pg 2rB - - - - r =-r3pg 2 2 3n 3 if the force of gravity acts at right angles to the drawing (see Fig. 47), where p is the mass of a unit area of the plate (see the solution to Problem 116). On the other hand, this moment is equal to the product of the weight of the plate and the distance x=OC from its centre of gravity to axis AB. ( nrl) 1Hence, xPi 2r2 - 2 =3" r3 pg. 2 Therefore, %=3 (4-n) r. 1-6. Work and Energy 120. The work of a force does not depend on the mass of the body acted upon by the given force. A force of 3 kgf will perform the work W = Fh= 15 kgl-m. This work is used to increase the potential energy (5 kgf-m) and the kinetic energy (10 kgf-m) of the load. 121. k=O.098 J/kgf·cm. 122. First let us find the force with which the air presses on one of the hemispheres. Assume that its base is covered with a flat lid in the form of a disk with a radius R. If the air is pumped out from this vessel the force of pressure on the flat cover will be PI =pA=prtR2. Obviously, the same pressure will be exerted on the hemisphere by the air t otherwise the forces will not be mutually balanced and the vessel will perpetually move in the direction of the greater force. The number of horses should be FI/F, since the other hemisphere may simply be tied to a post. The tensioned rope will produce the same force as the team of horses pulling at the other side. 123. The change in the momentum of a body Is equal to the impulse of the force of gravity. Since the forces acting on the stone and the Earth are the same and act during the same time, the changes in the momenta of these bodies are also the same. The change in the kinetic energy of a body is equal to the work of the . forces of gravitational attraction. The forces are equal, but the paths traversed by the stone and the Earth are inversely proportional to their masses. This is why the law of conservation of energy may be written in a form which disregards the change in the kinetic energy of the Earth: mt+Ep=const. where m is the mass of the stone and Ep the potential energy of interaction. 124. According to the law of conservation of energy, mlv~ ml gh= - 2where ml is the mass of the pile driver, h the height from which it drops, and VI its velocity before the impact. MECHANICS 201 Since the Impact is Instantaneous, the force of resistance cannot appreciably change the total momentum of the system. Seeing that the Impact is inelastic mlvl =(ml +m.) v. where m2 is the mass of the pile, v~ the velocity of the driver and the pile at the first moment after the impact. The mechanical energy of the driver and the pile is spent on work done to overcome the resistance of the soil F: (m +m )vl 1 2 2 I +(ml+m2)gs=Fs where s is the depth which the pile is driven to into the soil. Hence, ml h F= + ·- mtg+mlR+m2g=32,500 kg! ml m2 s 125. As a result of the inelastic impact, the linear velocity of the box with the bullet at the first moment will be equal to u=M~m' where u is the velocity of the bullet. On the basis of the law of conservation of energy, the angle of deflection a is related to the velocity v by the expression. (M +m) u l m 2v2 (M+m) L (I-cos a)g 2 2(M+m) whence . aM+m ..rv=2stn2"-m y Lg 126. Since the explosion is instantaneous, the external horizontal forces (forces of friction) cannot appreciably change the total momentum of the system during the explosion. This momentum is zero both before and directly after the explosion. Therefore, mtVt +m2v2 =0. Hence, VI = _ m2 • V2 ml Since the carts finally stop, their initial kinetic energies are spent on work against the forces of friction: mlv~ m2v: -2-=kmt gsl' and -2-=km~s2 vr 51 Hence, .=-,and, therefore, s2=2 metres. VI 81 127. Let us denote the speed of the body and the wagon after they stop moving with respect to each other by u: According to the law of conservetion of momentum, (M +m) U= Mvo (1) 202 ANSWERS AND SOLUTIONS The wagon loses its kinetic energy in view of the fact that the force of friction f actlng on it performs negative work MV~_Mu'_fS 2 2where S is the distance travelled by the wagon. The body acquires kinetic energy because the force of friction acting on it performs positive work mul 2=ls Here s is the distance travelled by the body. It is easy to see that the change in the kinetic energy of the system Mv~ _ [MU2 + mU 2] =/ (S-s) (2) 2 2 2 is equal to the force of friction multiplied by the motion of the body relattve to the cart. It follows from equations (1) and (2) that mMv~ S-s=2f(M +m) mMv~ Since S-s ~ I. then 1~ 2/ (M +m) Bearing in mind that f=kmg. we have l;a. 2kg7~~+m) · 128. The combustion of the second portion increases the velocity of the rocket v by Av. Since combustion is instantaneous, then according to the law of conservation of momentum, (M +m) v=M (v+t\v)+m (v-u) where m is the mass of a fuel portion, M the mass of the rocket without fuel, u the outflow velocity of the gases relative to the rocket. The velocity increment of the rocket Av= Zu does not depend on the veloci ty v before the second portion of the fuel burns. On the contrary, the increment in the kinetic energy of the rocket (without fuel) se, M(vtliV)1 _M;I=mu(2~u+v) wi11 be the greater, the higher is v. The maximum altitude of the rocket is determined by the energy it receives. For this reason the second portion of the fuel can be burnt to the greatest advantage when the rocket attains its maximum velocity. i. e.• directly after the first portion is ejected. Here the greatest part of the mechanical energy produced by the combustion of the fuel will be imparted to the rocket, while the mechanical energy of the combustion products will be minimum. MECHANICS 203 129. It will be sufficient to consider the consecutive combustion of two portions of fuel. Let the mass of the rocket with the fuel first be equal to M +2m. dh After combustion of the first portion, the velocity of the rockmUl . my et v= M +m .where Ul IS the velocity of the gases with respect to the rocket. The initial velo,,11 city of the rocket is assumed to be zero. Fig. 320 The increment in the velocity of the rocket after the second portion burns is ~v = m;;2 t where U2 is the new velocity of the gases with respect to the rocket. Combustion of the first portion produces the mechanical energy dEl = (M +m) tJ2 mu 2 = 2 +T t and of the second portion the energy (M +m) v2 2 According to the initial condition, ~El = ~E2' Hence, 2 ! m2 m \ _ 2 (m2 m ) Ul \ 2 (M +m) +2 ) - U2 2M + "2 Therefore, Ut > u2 ; the velocity of the gases with respect to the rocket diminishes because the mass of the rocket decreases as the fuel burns. 130. Both. slopes may be broken into any arbitrary number of small inclined planes with various angles of inclination. Let us consider one of them (Fig. 320). The work done to lift the body up such an inclined plane is equal to the work against the forces of gravity mg~h plus the work against the forces of friction FIr~s. /,l But FIr = kmg cos ex and ~s =_0__ . Therefore, FJrt1s= kmg/1l. The total cos ex. work L\W = mg (L\h+k81). If we consider all the inclined surfaces and sum up the elementary works, the total work will be W = ~ ~w =mg (~L\h+k~ 8l)=mgh+kmgl The work is determined only by the height of the mountain h and the length l of its foot. 131. The force appl ied to the handle will be minimum if it forms a right angle with the handle. Denoting the force sought by F, we shall have from the golden rule of mechanics: 2nRF=Gh. Hence, F=2~~' 204 ANSWERS AND SOLUTIONS 132. According to the definition, the efficiency f) WI'.;W2 ,where WI = = GH is the work done to lift the load G to a height H, and W2 is the ~ork done against the forces of friction. Since the force of friction is capable of holding the load In equilibrium, the work of this force cannot be less than the work WI. The minimum work of the forces of friction is WI= WI- Therefore, rr ~ 5 0 % . ' 133. As the man climbs the ladder the balloon will descend by a height h. Therefore, the work done by the man will be spent to increase his potential energy by the amount mg (I-h) and that of the balloon by mgh (the balloon without the man will be acted upon by the lifting force mg directed upwards). Hence, W =mg(l-h)+mgh=mgl This result can be obtained at once in calculating the work done by the man in the system related to the ladder. If the man climbs with a velocity v with respect to the ladder, he moves at V-VI with respect to the Earth, where VI is the velocity of the ascending balloon. According to the law of conservation of momentum, (v-v.) m = Mv., Hence, m °1= M+m U 134. To deliver twice as much water in a unit of time, a velocity two times greater should be imparted to the double mass of the water. The work of the motor is spent to impart a kinetic energy m;2 to the water. Therefore, the power of the motor should be increased eight times. 135. (1) The work done to raise the water out of the pit is H 3 3 Wi= pg2 AX 4'lf=g pgAH2 where p is the density of water. The work 1 H W.=2"P 2 Av2 Is required to impart kinetic energy to the water. The velocity (I with which the water flows out of the pipe onto the ground can be found from the ratio ~ A = nR2ut. The tot al work is w-! 2 1 H8A3 - 8 pg AH + 16P n2R'T2 (2) In the second case the work required to raise the water is less than WI by &W;=pgAlh(H-~). The work required to impart kinetic energy MECHANICS to the water is 205 F Fig. 321 The total work W'=Wl-~W~+W~. 136. It is the simplest to solve the problem in a coordinate system rela.. ted to the escalator. The man will walk a distance l=_.h_+ ut relative sm a to the escalator, where ur is the distance covered by the escalator. He should perform the work W=(-.h_+v-r) mgsincx, since during the ascent the sin ex force mg is applied over the distance 1 and forms an angle of 900 - a with it. Part of the work mgh is spent to increase the potential energy of the man, and the remainder mgucsin a. together with the work of the motor that drives the escalator is spent to overcome the forces of friction. 137. The relation between the elastic force and the deformation is shown in Fig. 321. The work done to stretch (or compress) the spring by a small amount 8X is shown by the area of the hatched rectangle ~W =F~x. The total work in stretching or compressing the spring by the amount I, equal to its potential energy Ep• is shown by the area of triangle OBC: W=Ep=k~2. Let us recall that an expression for the distance covered in uniformly accelerated motion can usually be obtained by similar reasoning. 138. The man acting with force F on the spring does the work WI =- FL. At the same time, the floor of the railway carriage is acted upon from the side of the man by the force of friction F. The work of this force W2=FL. Therefore, the total work performed by the man in a coordinate system related to the Earth is zero, in the same way as in a system related to the train. 139. In the system of the train the work done is equal to the potential kl i energy of the stretched spring (see Problem 137) W= T' since the force of friction between the man and the floor of the railway carriage does not do work in this system. In the system related to the Earth, the work the man does to stretch the spring is equal to the product of the mean force k~ and the distance L-l, i. e., Wl=k~ (L-l). The man acts on the floor of the carriage with the same mean force ~ . The kl work of this force W2 = 2' L. The o ~-----~~~--f---!....IC total work in the given coordinate klt system W=W 1+W 2 = T is the same as in the system or the carriage. 206 ANSWERS AND SOLUTiONS 140. On the basis of the laws of conservation of momentum and energy we can write the following equations: mtvt +m2v2=mlv~ + m2V~ 2 2 '2 '2 mtVl + m 2V 2 = m1Vl + m2V2 2 2 2 2 where v~ and v; are the velocities of the spheres after the collision. Upon solving these simultaneous equations, we obtain , (mt-m2)vl+2m2v2. d ,_(m2-mt)v2+2mtVl Vt ,an V2 - - - - - - - -.. ml+ m2 mt+m2 (1) If the second sphere was at rest before the collision (v2 =0), then (m1 - m2) VI • ' 2m1V1 v. V2 m1 +m2 ' mt +m2 If ml > m2' the first sphere continues to move in the same direction as before the collision, but at a lower velocity. If m1 < ms. the first sphere will jump back after the collision. The se.. cond sphere will move in the same direction as the first sphere before the collision. , 2mv2 ' 2mvt (2) If m1=m2, then Vl= 2m =V2 and V2= 2m =Vl' Upon collision the spheres exchange their velocities. 141. The elastic impact imparts a velocity v to the left-hand block. At this moment the right-hand block is still at rest since the spring is not de- formed. Let us denote the velocities of the lett-and right-hand blocks at an arbit.. rary moment of time by U 1 and U2 , and the absolute elongation of the spring at this moment by x. According to the laws of conservation of momentum and energyt we have m (u 1 +u2)=mv mu~ +mu~ +kx2 _ mv2 2 2 2 - 2 or kx2=m [t12-(ui+u~)] Upon substituting ut +U 2 for v in the last equation, we obtain kx2=2mu 1u2 kx2 Therefore, U1U2 = 2m and U 1 +U2=V. It can be seen from the last two equations that u1 and U2 will have the same sign and both blocks will move in the same direction. The quantity x2 will be maximum when the product of the velocities u1 and U2 is also maximum. Hence, to find the answer to the second question, it is necessary to determine the maximum product U 1U2 assuming that the sum "1+"2 is constant and equal to v. MECHANICS .----------j; tL.. ...1 1 --------..,tZ2 a b Fig. 322 207 Let us consider the obvious inequality (UI-U2)2~O, or U~-2UIU2+U~;;a:O. Let us add 4U1U2 to the right and left sides of the inequality. Hence, U~+2UIU2+U~~4uIU2 or (Ul+U2)2~4uIU2' Since Ul+U2=0, then 4UI U2 ~ v2 • Therefore, the maximum value of "lUI is equal to 0 2 /4 and it is attained v when U 1 = u2 = 2 ' At this moment the distance between the blocks is I ± xmax=l ± CIV~ 142. The lower plate will rise if the force of elasticity acting on it is greater than its weight: kX2 > m2g. Here XI is the deformation of the spring stretched to the maximum (position c in Fig. 322). Position a shows an undeformed spring. . For the spring to expand over a distance of X2 it should be compressed by Xl (position b in Fig. 322), which can be found on the basis of the law of conservation of energy: kx~ kx: T=T+mJR(Xl +X2) Hence, 2mtg+ mzg xI>-k- T To compress the spring by Xl' the weight of the plate should be supplemented with a force which satisfies the equation F+mtg=kxl' Therefore, the sought force F > mlg+m2g. 143. In a reading system related to the wall, the velocity of the ball is v+u. After the impact the velocity of the ball will be-(v+u) in the same reading system. The velocity of the ball after the impact with respect to a stationary reading system will be -(o+u)-u=- (v+2u) 208 ANSWERS AND SOLUTION The kinetic energy after the impact is ~ (v+2U)2 and before the impact ~V2 2 · The change in the kinetic energy is equal to 2mu (u+v). Let us now calculate the work of the elastic forces acting on the ball during the impact. Let the collision continue for l' seconds. For the sake of simplicity we assume that during the impact the elastic force is constant (generally speaking the result does not depend on this assumption). Since the impact changes the momentum by 2m(v+ u), the elastic force is F 2m (v+u) 1: The work of th is force is W=F5=Ful:=2m (o+u) U1: 2m (v+u) u l' It is easy to see that this work is equal to the change in the kinetic energy. 144. (I) Up to the moment when the rope is tensioned the stones will fall freely g (t-'t')2 2 The moment of tensioning of the rope can be determined from the condition I =51-52. Hence, t =3 seconds, 51 =44.1 metres, s2=4.9 metres. The time is counted from the moment the first stone begins to fall. When the rope is tensioned, there occurs an elastic impact and the stones exchange their velocities (see Problem 140). At the moment of impact VI =gt = 29.4 tnt«, and v2=g(t-'t)=9.8 m/s. The duration of falling of the first stone t1 (after the rope is tensioned) can be found from the condition gt: h-S1 = V1t1 +-2and of the second stone t2 from the condition gt: h-52 = V1tl +-2Therefore t1 e:i 1.6 seconds, and t, ~ 1.8 seconds. The first stone falls during 4.6 seconds and the second during 2.8 seconds. (2) If the rope is inelastic, the veloci ties of the stones after it is tensioned are equalized (inelastic impact): v=VI t V2 = 19.6 mts. The duration of Ialling of the stones with the rope tensioned is determined from the equations; , gt~2 , ut~2 h- 51= vt1+- 2- and h-s2= ut2+- 2- 51 and S2 are the same as in the first case. MECHANICS 209 Hence, t;:=!: 1.2 seconds, and t~ ~ 3.3 seconds. The first stone falls during 4.2 seconds and the second during 4.3 seconds. 145. If only the right-hand ball is moved aside, the extreme left-hand ball will bounce off after the impact through an angle equal to that by which the right-hand ball was deflected. If two right-hand balls are simultaneously moved aside and released, then two extreme left-hand balls will bounce off, and so on. When the first ball strikes the second one, the first ball stops and transmits its momentum to the second one (see the solution to Problem 140), the second transmits the same momentum to the third, etc. Since the extreme left-hand ball has no "neighbour" at its left, it will bounce off (provided there is no friction or losses of energy) through the same angle by which the extreme right-hand ball was deflected. When the left-hand ball ~ after deviation through the maximum angle, strikes the ball at its right, the process of transmission of momentum will take the reverse course. When two right-hand balls are deflected at the same time, they wiJl transmit their momenta to the row in turn after a very small period of time. In this way· the other balls will receive two impulses that will be propagated along the row at a certain time interval. The extreme left-hand ball will bounce off after it receives the "first portion" of the momentum. Next, its "neighbour" will bounce off after receiving the next portion of the momentum from the extreme right-hand ball. If three right-hand balls are deflected, the row of balls will receive portions of momentum from the third, second and first ball following each other in very small intervals of time. If four balls are deflected and released at the same time, four balls will bounce off at the left while the other two will remain immobile. 146. The striking ball will jump back and the other balls up to the steel one will remain in place. The steel ball and all the others after it will begin to move to the left with different velocities. The fastest velocity will be imparted to the extreme left-hand ball. The next one will move slower, etc. The balls will move apart (see the solutions to Problems 140 and 145). 147. Assume that the weight 2m lowers through a distance of H. The weights m will accordingly rise to the height h (Fig. 323). . 2mv2 2mvll On the basis of the law of conservation of energy, 2mgh+-r+--r= =2mgH, or v~+v:=2g(H-h), where VI is the velocity of the weights m and V2 that of the weight 2m. When the weight 2m lowers, its velocity v2 approaches the velocity Vi' since the angles between the parts of the string thrown over the pulleys tend to zero. In the limit, v2 d VI- At the same time, H -h ~ I. Therefore, the maximum velocity of the weights is v=ygl 148. The velocities of the weights are the same If they cover identical distances ~s in equally small intervals of time. These distances will be the same at such an angle ANB at which lowering of the weight ml through ~s = NK (Fig. 324) is attended by an increase in part AN B of the string also by the amount Ss. Therefore, when the velocities are equal, HK ::::s 14-204~ 210 ANSWERS AND SOLUTIONS m I :hI o Zm m Fig. 323 As ~s. ~ BK-BN=2 and FK=AK-AN="2- The triangles NHK and NFK wi11 be the closer to right ones, the smaller is the distance 6.5. When 6.s -+ 0, the angles NHK and NFK tend to be right ones, while the angles K NHand /(NF tend to 30°. Therefore, the velocities will be equal when L. ANB= 120°. Let us use the law of conservation of energy to find these velocities mlgh=2 (2-V3) mzgh+ ml tm2 v2 Hence, VI = 2gh ml- 2 (2- ya) ms ~ 0 ml+m2 The weights will oscillate near the position of equilibrium, which corresponds to the angle ANB=2 arc cos 2 ml ~ 149°. The maximum deviation from m2 the position of equilibrium corresponds to the angle ANB= 1200. 149. Since there is no slipping of the board on the rollers and of the rollers on the horizontal surface, the distance between the axes of the rollers C A rna Fig. 824 MECHANICS Fig. 825 211 in motion remains constant. For this reason the board will move translationally in a horizontal direction and at the same time down along the rollers. When the rollers move through a certain distance it each point on the board (in particular, its centre of gravity A) will move in a horizontal direction through the same distance I and will also move the same distance along the rol1ers: AB=BC=l (Fig. 325). (This is particularly obvious if we consider the motion of the rollers in a coordinate system travelling together with the rollers.) As a result, the centre of gravity of the board will move along straight line AC incl ined to the horizon at an angle a/2, since ABC is an isosceles triangle. The motion will be uniformly accelerated. The board acquires kinetic energy owing to the reduction of its potential energy m;! =mgl sin a, or v2=2g1 sin a. On the other hand, in uniformly accelerated motion v2=2as, where s=AC=2l cos ~ . Hence, the acceleration v2 • a a=2i=gslnT· . 150. Let us calculate the difference between the potential energies for the two positions of the chain-when it lies entirely on the table and when a part of it x hangs from the table. This difference is equal to the weight ~ xg of the hanging part multiplied by x/2, since the chain is homogeneous and the centre of gravity of the hanging end is at a distance of x/2 from. the edge of the table. Ivfv2 _ MR 2 On the basis of the law of conservation of energy, -2- - 4f x or u= -Vg:z2 . At this moment of time the acceleration can be found from N ton' diM M Th r gxew on s secon aw: a=2/ gx. ere.ore, a=2l . To calculate the reaction of the table edge, let us first find the tension of the chain at the point of its contact with the table. It is equal to the change in the momentum of the part of the chain lying on the table M M Mg F =21 (2l-x) a-2f u2 = 2l3 (l-x) ~ 14 * 212 Fig. 326 ANSWERS AND SOLUTIONS Fig. 327 Let us now consider a very small element of the chain in contact with the table edge. This element is acted upon by the three forces (Fig. 326). Since this element is infinitely small, the sum of the three forces which act on it should be zero. Therefore, the force of reaction is N= F Y2= Y2 M (l;;Xl x x When x > I, the chain no longer touches the edge of the table. 151. Let us denote the .velocity of the wagon by v. The horizontal component of the velocity of the pendulum with respect to the wagon is u cos ~ (Fig. 327). Since the wagon moves, the velocity of the pendulum with respect to the rails is v+ u cos p. The external forces do not act on the system in a horizontal direction. Therefore, on the basis of the law of conservation of momentum, we have m(v+ucos~)+Mv=O (1) since the system was initially at rest. The vertical component of the pendulum velocity with respect to the wagon and the rails is u sin ~. According to the Pythagorean theorem, the square of the pendulum veloci ty relative to the rails is (v+u cos ~)2 +u2 sin2 p. With the aid of the law of conservation of energy, we can obtain a second equation interrelating the velocities v and u: ; [(ucosP+v)2+u2sin2pl+ ~ v2=mgl(cosp-cosa) (2) It can be found from equations (1) and (2) that v2 2m2g1 (cos ~ - cos a) cos2 ~ (M+m)· (M+msin2~) In a particular case, when p=O (assuming ~ ~ I) m2 v2 = 2 M2 gl (1- cos ex) or J\\ECHAN ICS Fig. 828 213 152. Let us denote the velocity of the wedge by u, and the horizontal and vertical components of the velocity u of the block with reference to a stationary reading system by Ux and uy (Fig. 328). On the basis of the laws of conservation of momentum and energy we can write Mv2 m (2 2)-Mv+mux=O, and -2-+2 ux+Uy =mgh It should be noted that the angle a with the horizontal surface is formed by the relative velocity urel' i.e., the velocity of the block with respect to the moving wedge, and not the absolute velocity of the block u, by which is meant the velocity relative to a stationary horizontal surface. u It follows from the velocity diagram (Fig. 329) that +Y =tan cx. Upon v Ux solving these equations with respect to v, we obtain , ;r 2mgh v= V M+m[(~r+(~+lrtan2a] == , ;r 2gh = V .~ +(~ r +(: +I r tan2a At the same moment of time the absolute .velocity of the block is u=Vu~+u;=V2gh" ;r1- M m ( M rV 1+ m+M I +m tan2a When the mass of the wedge is much greater than that of the block u tends, as should be expected, to V2gh. t 153. The velocity of the rod with respect to the moving wedge is directed at a~ angle ex. to the horizon. If the velocity of the wedge is added to this relative velocity, the result will be the absolute velocity u of the rod 214 Fig. 329 ANSWERS AND SOLUTIONS Fig. 330 (Fig. 330). The relation between the velocities is obviously equal to ~=tan a v Mv2 milt. It follows from the law of conservation of energy that -2- +T = mgh. Upon cancelling u {rom these t\VO equations, we get an expression for v: .. / 2mgh v= V M+mtan2 a We can "then wri te for the relative velocity of the rod: 1 .. / 2mgh Urel = cos fl. V M +m tan2 ex The velocity of the rod is u=tana .. / __2mgh__ = .. /2 mgta~h V m+Mtan2 a V m+Mtan2 a It can be seen from this formula that the velocity of the rod changes with the path h travelled according to the law of uniformly accelerated motion u= V~ah. Therefore. the acceleration of the rod is m tan2 ex a m+M tan2 ex 1-7. Kinematics of Curvilinear Motion 154. The driving pulley rotates with an angular velocity of WI = 2nn 1 and the driven one with a velocity of w2=2nn2• The velocity of the drive belt is v=w.'.=(J)'l.'2. Hence, ~=W2=n2. '2 00, nl The sought diameter is D1=D2 n2=100 mm. n1 155. (l) Let us denote lhe length of the crawler by L= na. Hence, L- 2nR is the distance between the axes of the wheels. 2 MECHANICS 215 The number of links taking part in translational motion is nl = ~ = L-2nR Th b f I· k · t t l ti t th E h= 2a • e same num er 0 In s IS a res re a ve 0 e art. In rotary motion there are n.= 2nR links. a (2) The time during which the tractor moves is to=!.. . During a full rea. v volution of the crawler the link will move translationally a distance 21 at a speed of 2t1. The duration of motion of one link during a revolution is :~. Altogether, the crawler will make N = ~ revolutions. Therefore, the duration of translational motion of the link is 11 = Nl . This is the time during which v the link is at rest. The link will participate in rotary motion during the time 2Nl s+2nRN-NL I'}. =/0 v v If s ~ L, the number of revolutions may be assumed as a whole number, neglecting the duration of an incomplete revolution. 156. The duration of flight of a molecule between the cylinders is t = R-r . v During this time the cylinders will turn through the angle wt and, therefore, I=R(J)t=wR R-r v 6>R (R-r) Hence, v= 1 • 157. Let us denote the sought radius by R and the angular speed of the tractor over the arc by (a). Hence (Fig. 331), til = CJ) ( R- ~ ) , and tlo = CJ) ( R+: ) Thus R-.!. VI =__2 and R=!!.vo+v1 = 6 m VO R+.!. 2 VO-Vl 2 158. First the observer is at the pole (point 0 in Fig. 332). The axis of the Earth passes through point 0 perpendicular to the drawing. Line 0 A (parallel to BC) is directed toward the star. The mountain is at the right of point A. The angle (%= ooat is the angle through which the Earth rotates during the time at with the angular velocity w. To see the start the observer OC should run a distance OC.OAcoat. The observer's speed v= dt =OAw=O.7 m/s. 216 Fig.3d1 ANSWERS AND SOLUTIONS Fig. 882 159. Let us take point A from which the boat departs as the origin of the coordinate system. The direction of the axes is shown in Fig. 333. The boat moves in a direction perpendicular to the current at a constant velocity u. For this reason the boat will be at a distance g from the bank in the time t=.1L after departure. Let us consider the motion of the boat up to the u middle of the river (y<; ~ ) • The current velocity is (1= ~o y at a distance y from the bank. Upon inserting y=ut into the expression for the current, we get 0= 2vout . c It follows from this relationship that the boat moves parallel to the banks with a constant acceleration a=200 u . The boat will reach the middle of the c Fig. 333 MECHANICS 1'-'~~---------""2 2 9 46------------..lIIII~-9 Fig. 334 217 river in the time T=;u and will be carried downstream during the same time over a distance of s=a~2 = ~~. When moving from the middle of the river (point D) to the opposite bank, the boat will again be carried away over the same dist~nce s. Thus, the sought distance is ;:. When the boat moves to the middle of the river X= a~2=(Itt2 and y=ut. Let us use these ratios to determine the trajectory of the boat from A to D. We get y2=:!f.. x (a parabola). The other half of the trajectory (DB) is of the Vo . same nature as the first one. 160. It is obvious from considerations of symmetry that at any moment of time the tortoises will be at the corners of a square whose side gradually diminishes (Fig. 334). The speed of each tortoise can be resolved into a radial (directed towards the centre) and a perpendicular components. The radial speed wiII be equal to (I,= ";2. Each tortoise has to walk a distance of a 1=V2 to the centre. Therefore, the tortoises will meet at the centre of the square after the time l a t=-=- elapses. v, v 161. Ship B moves toward ship A with the speed 0, At the same time ship A sails away from ship B with the speed v cos a (Fig. 335). Therefore, 218 ANSWERS AND SOLUTIONS a Fig. 335 the distance AB reduces with a speed of v (I-cos a). Point C (the projection of point B onto the trajectory of ship A) moves with a speed of v (I-cos a). For this reason distance AC increases with a speed of vcosa. Therefore, the sum of the distances s= AB +AC remains constant as the ships move. At the initial moment point C coincides with A, and therefore s= AB=a. After a sufficiently great interval of time point C will coincide with B, and AB= AC= ; =; ,and the ships will move at a distance of 1.5 km from each other. 162. With respect to the reading system shown in Fig. 336, the coordinates and the velocities of the body can be determined at any moment of time from the following formulas: x=voxt (1) gt2 y=voytT (2) ~=~ ~ vy=voy-gt (4) Here Vox=Vo cos'a and voy=vo sin a are the projections of the initial velocities on the axes x and g. Equations (1), (2), (3) and (4) provide an answer to all the questions stipulated in the problem. The duration of flight T can be found from equation (2). When y = 0, we . gTl 2v sin t% have Vo sin aT -2=0. Hence, T 0 g 6 :c Fig. 396 MECHANICS 219 V~ sin 2a The distance of the flight is L=vo cos ex T= . This distance g will be maximum when a=45°: VI Lmax=-o g The height at which the body will be after the time -r elapses is equal g'f2 to h=vo sin (X-r-T. The velocity of the body at the moment 'f is equal to v= V~,'where vx=vocosa and vy=vosina-g'f. Hence, v=Vv~+g2't'2_2voR'1'sinaand it forms an angle of ~ with the vertical that can be found from the equation tan ~= v.o cos ex . Vo sin a- g-r 163. The coordinates of the body x and y change with time according to the law . gt 2 y=Vo sin at-To x=vo cos at Upon excluding the time from these expressions, we obtain an equation for the trajectory y= - ~ g x2+tan a·x. This is an equation of a pa- 2vo cost a. rabola. By denoting the coordinates of the vertex of the parabola (point A in Fig. 336) by Xo and Yo, the equation of the trajectory can be written as y- Yo =k (X-XO)2, where and k= g . Y 2v~ cos> ex ' 0 v~ sin2 a. 2g v~ sin 2a Xo= 2g 164. The trajectory of the ball takes the form of a parabola passing through a point with the coordinates hand s. Therefore (see the solution to Pro-- blem 163), . h= g s2+tao as 2v~ cos 2 a Hence, gs2 V2-_~-~--~ n-2 cos> a (tan as -h) (s sin 2a.- h cos 2a)-h y 52 +h'sin (2a-q»-h where tan cp=h/s. The minimum velocity V gs2 Vtlo= V = g (h+ Vhl+SI ) s2+h'l.-h 220 " ...--~tI' " , I~ , ANSWERS AND SOLUTIONS Fig. 837 Fig. 338 Therefore, is attained when m 1t h+ Ys2+h2 s a=..I 2 +-4=arctan s arctan ¥h2 + s2-h (Fig. 337.) 165. The coordinates and the velocities of the body at any moment of time with respect to the reading system shown in Fig. 338 are determined by the same equations as in Problem 162. At the moment when the body falls into the water its coordinate y=- H. For this reason the duration of flight T can be found from the equation gT2 -H=vo sin aT-T Hence, Vo sin a ± V v~ sin2 a.+2gH T= ~--------- g Since T > Ot we shall retain the plus sign. The distance from the bank is sa • 2VoSID a.+vocosaV sa· I +2 H L=vo cos a.T 2g --g- Vo sin a g The body will be at a height h above the water after the time v sina ± V v2sin2a+2g(H-h) 't= -.,;;o~ _ g If Ih I < IH I, only the plus sign has a physical meaning. When h ~ n, both solutions have a meaning. During its motion the body will be twice at the same height above the water. It is the simplest to find the final velocity v with the aid of the law of conservation of energy 2 2 mvo + H mv 2 mg =T v=V v:+2gH MECHANICS O~----s------..t Fig. 339 221 166. In the reading system depicted" in Fig. 339, the coordinates of the stone are determined at any moment of time by the following equations: X=Vocos ai gt 2 y=ho+vosin aJ- T At the moment when the stone falls, y=O and X=S, where s is the distance covered by the stone. Upon solving these equations with respect to the angle a, we obtain tana= v~ (I ± V1+2;hO_g;:3)gs 0 0 This expression has a meaning when 1+2gho _g2s2 ~ 0 v~ v~ Vo V v2 + 2gho PoV v~+2gho Hence, 5 ~ 0 • Its maximum value issmax g g When s is smaller, two values of the angle a correspond to each of its values, the difference between which is the less, the nearer s is to its maximum value. Fig. 340 222 ANSWERS AND SOLUTIONS Therefore, for the maximum distance of Oight, v: 00 1 tan 1%= V ,r-; and 1%=30"gSm" v~+2gho f 3 187. The components of the velocities of the bodies along x and y at any moment of time are determined as 011/=VOsin al- gt ; (121/=00 sin ~ -gt; vtx=uo cos at; and Vlx=- Vo cos a2 Let u be the velocity of the second body with respect to the firstone. Hence u,=oo sin al-gt-vosin CXs+gt=vo (sin at-sin ( 2) uJ'=vo (cos at +COSCXs) Therefore, the velocity u is equal to u=V u~+u;=2cos (1%Itl%l) (/0 The bodies move with respect to each other at a constant velocity. After the time l' the distance between them will be s=2vocos (1%1tlXs) "C 168. The horizontal path of the bomb s= Yl2_h2 = vcos at, where t is the duration of falling of the bomb. The vertical path is h=v sin I%t +g~1 (Fig. 340). Upon excluding the time from these equations, we find tan a=- Vi ± .. I(VI)2+2hv t -1 gs JI gs gsl The solution with the plus sign has a meaning. The minus sign corresponds to ex < 0, i.e., the bomb is dropped when the dive-bomber is flying upward. 169. It will be convenient to solve this problem in a reading system rela· ted to the uniformly moving vehicles. In this system, the highway moves back with the speed of 0=50 km/h, the vehicles are at rest with respect to each other and their wheels rotate. The linear speed of points on the circumference of the wheel and tha t of the stuck stone IS also v. The stone will fly the maximum distance if it flies out when its speed forms an angle of 450 with the horizon. Let us find this distance. Neglecting the fact that the stone is somewhat above the level of 2 • 2 I the highway when it is thrown out, we obtain 1 v sin a v = 19.6 met. g g res. The minimum distance between the vehicles should be 19.6 metres. 170. It will be much easier to solve the problem if the axes of coordinates are directed along the inclined plane and perpendicular to it (Fig. 341). In this case the components of the acceleration of the ball on the axes x and y will be ~pectively equal to as=g.=g sin ex and a1l= g J/= - gcosa. ,~\ECHANICS Fig. 341 223 Fig. 342 Upon the first impact with the inclined plane, the velocity of the ball will be Vo= Y2gh. The initial velocity of the ball after the first impact is Vo and forms an angle a with the y-axis (Fig. 341). The distance between the points of the first and second impacts is . t2 11 =vo sin at) + g Sl~ a ,where it is the duration of flight and is determined by the equation gcos ai~ 2 o Hence, i1= 2vo and i1=8h sin a. The velocity of the ball at the second g impact can be found from the equations v)x=vOX+aXil =vo sin a+g sin at l =3vo sin a Vly=Voy + aytl =00 cos a,- g cos atl =- Vo cos a. After the impact these velocities are equal to v 2X=v t x , and V2y=- v 1Y The distance between the points of the second and third impacts is equal to . g sin at~ 12 = 300 sin at2 + 2 where 12 .is the time during which the ball is in flight. Since the initial velocity along the y' axis is the same as during the first impact, 12 = / }. Therefore, 12 = 16hsin a. Similarly, it can be shown that the distance between the next points la=24h sin a. Consequently. 11:l2 : [3 ••• =1:2:3, etc. 224 ANSWERS AND SOLUTIONS 171. The motion of the body can be considered as superrosftion of movement along a circumference With a radius R in a horizonta plane and vertical falling. Accordingly, the velocity of the body v at the given moment can be represented as the geometrical sum of two components: VI =v cos a directed horizontally and v2=v sin a. directed vertically (Fig. 342). Here a is the angle formed by the helical line of the groove with the horizon. In curvilinear motion the acceleration of a body is equal to the geometrical sum of the tangential and normal accelerations. The normal acceleration that corresponds to movement along the circumference is v~ v2 cost ex al n = 7[ = R The vertical motion is rectilinear, and therefore a2n = O. The sought acceleration a = V Q~"t +a:'t+a~n, where a1t and a2't are the tangential accelerations that correspond to motion along the circumference and along the vertical. The total tangential acceleration Q't Is obviously equal to a't=V a~1'+a:1'.. The value of a't can be found by mentally developIng the surface of the cylinder with the helical groove into a plane. In this case the groove will become an inclined plane with B height nh and a length of Its base 2nRn. h Apparently, a~=g sin a.=g y .h2+4n2R2 To determine DIn' let us find v from the law of conservation of energy: mv2 2 Bn"nhgR . T=mghn. Consequently, () =2ghn and a1n h2+4n2 R" Upon Inserting Fig. 343 MECHANICS 225 the found accelerations a'f and aln into the expression for the sought acceleration, we get ghYh?+4nI RI+ 64n4lnI Rs a= h2 + 4n2 RI 172. As usual, the motion of the ball can be considered as the' result of summation of vertical (uniformly accelerated) and horizontal (uniform) mo- tions. The simplest method of solution is to plot a diagram showing how the coordinates of the ball along the horizontal depend on the time for the limiting velocities 267 cmls and 200 cmls (Fig. 343). The lower broken line corresponds to the maximum velocity and the upper one to the minimum velocity. In the course of time, as can be seen from the diagram, the inde.. finiteness of the ball coordinate x shown by the section of the horizontal straight line between the lines of the diagram increases. The vertical hatching in .Fig. 343 shows the movement of the ball from M to N, and the horizontal hatching-from N to M. The cross..hatched areas correspond to indefiniteness in the direction of the horizontal velocity. (I) The diagram shows that after the ball bounces once from slab N the direction of its horizontal velocity will be indefinite when the duration of falling OK< t ~ OL or t > AB (where OK=0.15 s, OL=O.2 sand AB=0.225 s). gt 2 Hence, 10 em <; H < 20 em or h ~ "2 ;is -26 em. (2) The ball may strike any point on the base supporting the slabs if the duration of falling of the ball t'".;;:; AF-=0.3 s. Therefore, Hmin=44 em. 173. (1) During the time T of a complete revolution the disk will cover a distance equal to the length of its circumference, i.e., s=2nr, where r is the radius of the disk. Therefore, the translational velocity; of any point on the disk Vtr= 2;' =v. On the other hand, the linear vJlocity of rotation of points on the disk rim with respect to the centre 0 is Vlin=CJ)T, where (J) is the angular velocity of rotation. Since CJ)=2;,then Vlin=y=Vtr. (2) The veloci ty of points on the disk rim with respect to a standing observer will be the sum of the translational and rotational velocities. The total velocity for point A will be equal to 2v. For points Band D the velocities being added are equal in absolute magnitude and their sum is Y2v (Fig. 344a). For point C the total velocity with respect to 8 standing observer is zero, since the translational and rotational velocities are equal in absolute magnitude and oppositely directed. (3) The instantaneous velocities of points on diameter AC increase in direct proportion to the distance from point C. For this reason the motion of the disk may be considered at the given moment of time as rotation around the point where the disk touches the path. The axis passing through point C perpendicular to the plane of the disk is known as the instantaneous axis of rotation. When the disk moves, this axis constantly passes through the point of contact between the disk and the path. Therefore, all the points on the disk equidistant from point C at the given moment of time will have the same total velocity with respect to a standing 15-2042 226 ANS\VERS AND SOLUTIONS A 2u c (6) c Fig. 344 observer. The points which are at 8 distance equal to the disk radius from the instantaneous axis (point C) will have the same velocity (in absolute magnitude) as that of the axis, i.e., v (Fig. 344b). 174. The angle between adjacent spokes of the front wheel is cp= ~: . The wheel will seem stationary on the screen if it turns through the angle a=kq> during the time between the filming of two successive frames T= 1/24 s. Here k is a positive integer. On the other hand, the angle of rotation of the wheel during the time T is equal to a=co"t', where CI) is the angular speed of the wheel. Therefore. the front wheel will seem stationary if CJ)~2 N1tk, and IT the speed of the cart V= (J)T = 2 N MT • The minimum speed of the cart 0min = l't' 2n.r = N=8.8 rn/s. 1 The rear wheel will also seem stationary if NRHence, N2 = _ I _ = 9 when kl=k2=1. r 175. (1) The spokes will seem to rotate counterclockwise if during the time 'T (see Problem 174) the wheel turns through the angle ~1 which satisfies the condition ker > ~l > kff- ~ , where k= 1, 2, 3, ...• The consecutive positions of the wheel spokes are -I.own for this case· in Fig. 345a. It seems to the audience that each spoke t 'ns through the angle a < ~ counterclockwise. The possible angular velocit ies lie within the interval kq> (2k- 1) q> -;- > 001 > --2-:r- MECHANICS 227 (2) Since the front and rear wheels have the same number of spokes, the wheels will seem to revolve counterclockwise if the speed of the cart is kepr kfPr cpr -:r>v>-:r-2't (1) kcpR > v > kcpR _ fPR 't 't' 2-r Since R= 1.5r, the second inequality can be rewritten as foHows: I 5 kcpr 1.5kfPr I.SepT · T>v>-'t--~ Both Inequalities, which are congruent only when k= 1, give the permissible speeds of the cart in the form cpr > v > 0.75 q>r 't 't 2n . Or, since CP=6' we have 8.8 m/s > v > 6.6 m/s. (2) The spokes of the rear wheel will seem to revolve clockwise if during the time r the wheel turns through the angle ~2 which satisfies the condition (2k-l) : > Pi > (k- 1)cp (Fig. 345b). Hence, the following inequality is true for the speed of the cart: 1 5 (2k-l) cpt 1.5 (k-l) cpr • 2't >v> 't At the same time inequality (1) should be complied with. When k= 1, both inequalities are congruent if 0.75 cpr > v > 0.5 fPr . When k=2 they are can't 't' gruent if 2cpr > v > 1.5 cpr • If k > 2 the inequalities are incongruent. T " F'II 15 * D'O (h) Fig..845 228 ANSWERS AND SOLUTIONS 6.6 m/s > v > 4.4 mls 17.6 mls > t1 > 14.2 mls or 176. The instantaneous axis of rotation (see Probtern 173) passes through point C (Fig. 346). For this reason the velocity of point A relative to the block is R+,VA=V-,- . Point B has ~ Therefore, _.......~~--~ Fig. 346 R-r the velocity vB=v --. r Points on a circle with the radius r whose centre is point C have an instantaneous velocity equal to that of the spool core. 177. The trajectories of points A, Band c are shown in Fig. 347. Point B describes a curve usually called an ordinary cycloid. Points A and C describe an elongated and a shortened cycloids. 178. The linear velocity of points on the circumference of the shaft Ul=OO ~ and that of points on the race u2 = Q ~. Since the balIs do not slip, the same instantaneous velocities will be imparted to the points on the ball bearing that at this moment are in contact with the shaft and the race. The instantaneous velocity of any point on the ball can be regarded as the sum of two velocities: the velocity of motion of its centre Vo and the linear velocity of rotation around the centre. The ball will rotate with a certain angular velocity (00 (Fig. 348). Therefore, Hence, VI =vo-wo' V2=VO+OOO' 1 1 vO=T (vt +v2)=4 ( ~5 , the rod will not slip until N is zero, l.e., until tI';;;; 2 arc cos ~ , the equation m01R=mg cos tI- N will give us N 2mg (Fig. 356). Therefore, the system will not be in equilibrium. The right-hand weight will have a greater pull. 193. The direction of the acceleration coincides with that of the resultant force. The acceleration is directed downward when the ball is in its two extreme upper positions Band C (Fig. 357). The acceleration will be directed upward if the ball is in its extreme bottom position A and horizontally in positions D and L determined by the angle cx. Let us find ex. According to Newton's second law, the product of the mass and the centripetal acceleration is equal to the sum of the projections of the forces on the direction of the radius of rotation: mv2 -1- = T - mg cos ex On the other hand, as can be seen from Fig. 357, we have T= mg . On cos ex the basis of the law of conservation of energy: mol T=mgl cos a, 1 We can find from these equations that cosa= ya' and therefore ex ~ 54045'. 194. Let us denote the angular velocity of the rod by co at the moment when it passes through the vertical position. In conformity with the law of conservation of energy: co2 t 2 2 (miT!+m,T.)=g (I-cos a) (mITt +m2's) or MECHANICS 235 Fig. 357 whence Fig. 358 Vl=ror1= 2rt sln~'" /gmlr~+m2r: 2 V m1' 2 + m2' S . a. ml'l +m2'2vs=r2=2r2sln- g 2 2 2 mt'2+ m2'2 195. The resultant of the forces applied to the ball F=mg tan a. should build up a centripetal acceleration a=oo2r, where r=1 sin a. (Fig. 358). Hence, mgtan a-=moo2l sin (I This equation has two solutions: CXt=O cx! =arc cos ro~I Both solutions are valid in the second case: a1=0 (here the ball is in a state of unstable equilibrium) and ~ = 600 • In the first case the only solution is al =0. 198. Let us resolve the force F acting from the side of the rod on the weight m into mutually perpendicular components T and N (Fig. 359). Let us project the forces ontoa vertical and a horizontal lines and write Newton's equations for these directions mooll sin cp=T sin cp- N cos q> mg=T cos cp+N sin, Let us determine T and N from these equations T=m (CJ)11 sint cp+g cos cp) N=m 19-CJ)21cos cp) sin cp 236 H o 0' A ANSWERS AND SOLUTIONS a 0' Fig. 859 Therefore. Fig. 360 F =YT2+N2=m y g2+ro4l2sin 2 cp 197. The forces acting on the bead are shown in Fig. 360: f is the force of friction, mg the weight and N the force of the normal reaction. Newton's equations for the projection of the forces on a horizontal and a vertical directions will have the form f sin q> 1= N cos q>=mro2l sin q> f cos q> ± N sin q>-mg=O The upper sign refers to the case shown in Fig. 360 and the lower one to the case when the force N acts in the opposite direction. We find from these equations that f=mro2l sin2 cp+mg cos q> N= ± (mg sin q> - mro2l sin cp cos q» In equilibrium f ~ kN or k sin cp-cos cp g 1<.. (k +.) • 2 when k ~ cot cpsin q> cos cP sin q> (J) and k sin

g I ~. (k . ) • 2 when k ~ tan q>sin q> cos cp-sln q> co 198. Figure 361 shows the forces acting on the weights. Here T1 and T2 are the tensions of the string. Let us write Newton's equations for the proj ectlons onto a horizontal and a vertical directions. MECHANICS 237 For the first weight TI sin q>-T2 sin 'i'=mooll sin, T1 cos q> - Tt COS '\1'-mg=O (1) For the second weight mro21 (sin q>+ sin 'i')=T2 sin", (2) T2 cos"i'=mg Upon excluding T 1 and T2 from the system of equations (I) and (2), we obtain the equations a sin q>=2 tan cp- tan", a (sin q>+sin "p) =tan",· 002 l where a=-' g From these equations we get 2 tan q>- tan", < tan'P and, therefore, q> < W. 199. The forces acting on the weights are shown in Fig. 362. Here T I' NI and T 2' N 2 are the components of the forces acting from the side of the rod on the weights m and M. The forces NI and N2 act in opposite directions, since the sum of the moments of the forces acting on the rod with respect to point 0 is zero because the rod is weightless: N1b-N2 (b+a)=O. The equations of motion of the masses m and M for projections on a horizontal and a vertical directions have the form mro2b sin q>=T1 sin q>-NI cos +N1 sin q>=mg Mw2 (b+a) sin q>=T2 sin cp+N2 cos q>; T2 cos cp-N2 sin cp=Mg Upon excluding the unknown quantities T 1. T 2' Nt and N2 from the system, We find that g mb+M (a+b) (1) qJ=Ot and (2) cos q>= 002 mb2+M (a+b)2 o Fig. 361 Fig. 362 0' 238 ANSWERS AND SOLUTIONS The first solution is true for any angular velocities of rotation, and the second when 0) ~ .. / mb+M (a+b) (see the V g mb2+M (a+b)2 solution to Problem 195). 200. In the state of equilibrium mro2 x=kx, where x is the distance from the body to the axis. It is thus obvious that with any value of x the spring imparts the centripetal acceleration neFig. 963 cessary for rotation to the body. For this reason the latter will move after the impetus with a constant velocity up to stop A or as long as the law of proportionality between the force acting on the spring and its deformation is valid. 201. Let us write Newton's second law for a small portion of the chain having the mass ~ R Aa and shown in Fig. 363: 7R Aa(2nn)'R=2Tsin A; Since the angle Aa is small, sin A; ~ A; , whence T=mln'=9.2 kgf. 202. Let us take a small element of the tube with the length R Aa (Fig. 364). The stretched walls of the tube impart an acceleration a=~ to the water flowing along this element. According to Newton's third law, tne water will act on the element of the tube with the force ~~ Y <" i\ I " I \ I " '11, I " I , I " I, I " .dDt I, I '\7, I V a Fig. 364 nd2 Vi ~ F=p- Rl\a- 4 R where p is the density of the water. The force ~F is balanced by the tension forces of the ring T. From the condition of equilibrium. and remembering that ~a is small, we have AF=2Tsin A; ~ T Aa Therefore, the sought force is T - pnd2 2 - 4 v. 203. Let us divide the rod into n sections of equal length and con- MECHANICS i+T t-4----- 7j-- - - - -...... ?i+l------~ Fig. 865 239 sider an arbitrary section with the number i (Fig. 365). The acceleration of the various points in this section will be different, since the distances from the points to the axis of rotation are not the same. If the difference ';+1-'1 is small, however, the acceleration of the i-th section may be assumed as equal to ro2 'i+ 2+'1 , and this will be the more accurate, the smaller is the length of the section. The i-th section is acted upon by the elastic force T;+l from the side of the deformed section i +1 and the force Ti from the side of the section i-I. Since the mass of the i-th section is ~ ('1+1 -rd, on the basis of Newton's second law we can write that m '·+1 +,.T;-Ti+l=T (ri+l- f ; ) 00 2 I 2 t or mID? ( 2 2) Ti+1-T;=-y r'+l- r/ Let us write the equations of motion for the sections from k to n, inclusive, assuming that 'n+i=l and ',,=x: moo? a -Tn= - - (12_ r n) . 21 moo? I I Tn- Tn-l = -Y ~rn-fn-l) mID! t 2 Tk+I-Tk+1 =-21 (rk+I-'k+J mID'}, 2 Tk+l- T,,=-21(rk+l- X!) ;; The first equation in this system takes into account the fact that the elastic force does not act on the end of the rod, te., Tn+l =0. Upon summing up the equations of the system, we find that the sought tension • moo'/, IS equal to T"=2 (11_X2) . 240 ANSWERS AND SOLUTIONS The closer the sections of the rod are to the axis of rotation, the more wi11 they be stretched. 204. In the reading system that is stationary with respect to the axis, the force of tension of the rod does not perform any work, since this force constantly remains 'perpendicular to the velocity of the ball. In the moving system this force performs work other than zero, and this changes the kinetic energy of the ball. 205. Section AB of the hoop with a mass m at the highest point has an energy of mg 2R +m ~V)2. During motion the kinetic and potential energies of section AB begin to decrease owing to the work of the forces of the elastic deformation of the hoop whose resultant produces a centripetal force always directed towards the centre. The velocity of section AB forms an obtuse angle a with the force F (Fig. 366). For this reason the work of the force W1=F!1S cos a is negative, and, consequently, the energy of the section with the mass m diminishes. After section AB passes through the lowermost position, it is easy to see that the work of the force F becomes positive and the energy of section AB will grow. 206. Let us draw a tangent to the inner circumference of the spool (Fig. 367) from point A, which is the instantaneous axis of rotation (see Problem 173). If the directions of the thread and the tangent AC coincide, the moment of the forces that rotate the spool around the instantaneous axis will be zero. Therefore, if the spool is at rest, it will not rotate around the instantaneous axis and, consequently, it will not move translationally. The angle a at which the motion of the spool is reversed can be found from triangle AOB; namely, sin a = ; . If the thread is inclined more than a, the spool will roll to the right, If a Is smaller it will move to the left on condition that it does not slip. If the tension of the thread T satisfies the condition Tr ez.iR, where f is the force of friction, the spool will remain in place. Otherwise, when sin a= ~ it will begin to rotate counterclockwise around point o. 207. Break the hoop into equal small sections each with a mass of L\m. Consider two symmetrical sections (relative to the centre). All the particles of the hoop simultaneously participate in translational motion with the velocity Fig. 366 Fig. 367 MECHANICS Fig. 368 241 v and in rotational motion with the velocity VI =roR. The resultant velocity V2 of the upper section of the hoop can be found as the geometrical sum of the velocities v and VI (Fig. 368): v:=vr+v2 +2wI cos ex For a symmetrical section v:=vl+v~-2vvl cos ex The total kinetic energy of both sections is ~E,,= A I I = mvs +~= Amv2 + f1mO)' R2 2 2 Since this expression is valid for any two sections, we can write for the entire hoop Mvs MR20)2 EII=-2-+-2If the hoop rolls without slipping, v=roR and, therefore, Ek=Mvl • 2Gv2 208. Ek=-g- (",,+1). 209. The cylinder made of a denser material will obviously be hollow. At the same translational velocities without slipping the kinetic energy of rotation will be greater in the hollow cylinder, since the particles of its mass are further from the centre and, therefore, have higher velocities. For this reason the hollow cylinder will roll down an inclined plane without slipping slower than the solid one. At the end of the plane, the total kinetic energies of both cylinders are the same. This is possible only when the velocities are different, since when the velocities are the same, the energies of translational motion are identical, while the energy of rotational motion of a solid cylinder will always be smaller than that of a hollow one. 210. When the drum moves, the force of friction performs no work, since the cable and the drum do not slip. Hence, the energy of the system does not change: !!-v'+GR=G-px u2+o--:-rn; MECHANICS 243 If Vo < CiloT, the hoop will stop at the moment of time 't'= m;o when rotatVoing with the angular velocity 00=(00--. Then the hoop begins to move r with slipping in the reverse direction. In a certain time the hoop will stop slipping and will roll without slipping to the left with a translational velocity wor-vo v 2 (see Problem 213). If Vo > CiloT, then after the time 't'= m~Cilo elapses the hoop will stop rotating and will move to the right with a translational velocity v=vo-rroo. Next the hoop will rotate in the reverse direction, and in some time it will roll without slipping to the right. Its angular velocity will be vo-rwo 00= 2r Practice shows that the loop will also be braked when it does not sli p, We did not get such a result since the specific rolling friction was neglected. 215. Since the hoops do not slip, Vo (velocity of the centre of gravity of the hoops) and v (velocity of the weight) are related by the expression R vo=v R-r Assume that the weight lowers through the distance h. If the system was at rest at the initial moment. from the law of conservation of energy we have mv2 2 mgh=-2-+ Mvo· (see Problem 207). This expression can be used to find the velocit y of the weight: V 2mgh v= R 2 m+2M ( R-r ) Hence, the acceleration of the weight is a mg m+2M ( RR T r The weight lowers with the acceleration a under the action of the force of gravity mg and the tension T of the string. The sought tension T is 16 * T=m(g-a) 2mMg(h)' m+2M (RRr r 244 ANSWERS AND SOLUTIONS Since the centre of gravity of the hoop moves with an acceleration equal to a -RR under the action of the force T and the force of friction F, Newton's -r second law gives us the following equation for the force F R F=T-Ma-- R-r or F Mmg(h Y(2- ¥ ) Mmg('+i) m+2M ( RR r r m ( 1-; r+2M The friction force of rest cannot exceed the value kMg. For this reason slipping occurs when Mmg(hr(1+N--) ( R )2 > kMg m+2M -R--r or k<-~-~--~ l+!.. R 2~+(1-; r 216. The centre of gravity of the spool will not be displaced if the tension of the thread satisfies the equation T=Mgsina. Let us find the acceleration of the weight with a mass m to determine the tension of the thread T. Let the weight lower through a distance h. Since, according to the initial condition, the centre of gravity of the spool should remain at rest, the change in the potential energy is equal to mgh. If v is the velocity with which the weight having a mass m moves, the velocity of the points on the spool at the distance R from the axis of rotation is v R . r Hence, the kinetic energy of the system is mv" Mv2 RI E"=T+-2-fi It follows from the law of conservation of energy that (m+M ~:) v;==mgh or V 2mgh v= R2 m+M-rl MECHANICS 245 Therefore, the acceleration of the weight is a= mg RI m+M-;r If we know the acceleration <]f the weight, we can find the tension of the thread R2 M-,2 T=m (g-a)=mg R2 M-+m,1 Thus, for sin a we get the expression . 1 sin cx= M~-r--2 . m+R! The centre of gravity of the spool can be at rest only if M ,2 m+ R2~1 217. If the velocity of the board is [,It the velocity of the centre of gravity of each roll will be v/2 (see Problem (73). The kinetic energy of the system (the board and the two rolls) is Mv2 +2mvl =M+m tt2 2 4 2 By equating the kinetic energy to the work of the force Q over the distance s, we obtain M+m . V"2QS--v2 = Qs and v= - - 2 ' M+m (Fig. 369). (The forces of friction perform no work since there is no slipping.) It follows from the expression for the velocity of the board that its acce.. Ieration is a=-Q- M+m To determine the force of friction applied to the board from the side of a roll, let us write the equation of motion of the board Ma= Q-2F. Upon inserting the value of the acceleration a in this equation, we get mQ F 2(M+m) Since the velocity of the centre of gravity of the roll is half that of the board, the respective accelerations will be in the same ratio. Therefore, the 246 ANSWERS AND SOLUTIONS F F q equation of motion of the centre of gravity of a roll will have the form: m2..=F-/2 It follows from this equation that 1=0. 218. Let us assume, to introduce determinancy, that mtR > m2'. In this case the first weight will lower and the second one wil rise. If the first weight lowers through a distance of h, the other one will rise through hi. The decrease in the potential energy will be m1gh-magh; .gh ( ml-ma; ) If the absolute velocity of the first weight is v, that of the second weight will be v ~ . All the points of the first step of the pulley move with the velocity v, and those of the second step with the velocity v ~. The kinetic energy of the system will be ml+Mt 2+ m2+ M9 ,2 2 2 v 2 Ri v It follows from the law of conservation of energy that ml +M1 2+ m2+M s ,9 2- ( _ !.) h 2 v 2 RIV - ml ma R g or ..r 2 ( m1-mai )gh V= JI (ml+M1)+(ma+Ma) ;2 Therefore, the acceleration of the first weight is , ml-m2l[ MECHANICS Fig. 370 247 From the ratio Ul =!i, where a2 is the acceleration of the second weight, a2 ' we can find that On the basis of Newton's second law, the tensions of the strings T1 and Ta are equal to: , ,2 M1 +m2 R+R2(m2+ M2) ,a ml+M1 +(ma+ M2) R2 ml+M1 +-k(m1+Mz-k)1. ,2 .: ml+Ml+(ms+Ms) RZ t The force F which the system acts on the axis of the pulley with is F=T1+T.+(M1+Ms) g 219. Let the path travelled by the centre of gravity of the cylinder during he time t be equal to s, and the velocity of the centre of gravity be v at his moment (see Fig. 370). Hence, from the law of conservation of energy we have Mvl=Mgs sin ex, Thus. the velocity is v=Vgs sin a and the acceleration a=g s~n a . The velocity of the centre of gravity of the cylinder and the angular velocity of its rotation will be o=gs~nat and fJ)=g~~at 248 ANSWERS AND SOLUTIONS 1-9. The Law of Gravitation " "\ \ \,I I II /', B 220. According to Newton's second law, mig==F, where m, is the inertial mass-a quantity that characterizes the ability of bodies to acquire an acceleration under the influence of a definite force. mM On the other hand, according to the law of gravitation F='V ~2 g, where y is the gravity constant and mg and Mg are the gravitational masses of the interacting bodies. The gravitational mass determines the force of gravity and, in this sense, can be referred to as a gravitational charge. It is not obvious in advance that mi=mg • If this equation (proportionality is sufficient) is satisfied, however, the gravity acceleration is the same for all bodies since, when the gravity force is introduced into Newton's second law, the masses mj and mg can be cancelled, and g will be equal to y ; . Identical accelerations are imparted to all bodies irrespective of their masses, only by the force of gravity. 221. The gravitational acceleration g=y ~ (see Problem 220). Assuming that g=982 em/s, we find that y=6.68X 10-8 ems,-1 S-2. 222. The bodies inside the spaceship will cease to exert any pressure on the walls of the cabin if they have the same acceleration as the spaceship. Only the force of gravity can impart an identical acceleration in this space to all the bodies irrespective of their mass. Consequently, it is essential that the engine of the spaceship be shut off and there be no resistance of the external medium. The spaceship may move in any direction with respect to that of the force of gravity. 223. The force of gravity imparts the same acceleration to the pendulum and the block. Gravity does not cause any deformations in the system during free falling. For this reason the pendulum will so move with respect to the block as if there is no gravitation (see the solution to Problem 222). The pendulum will move with a constant angular velocity as long as the block falls. 224. On section Be A (Fig. 371) the force of gravity performs positive work (the angle 91 is acute) and the velocity of the planet will increase, reaching its maximum at point A. ~ C Fig. 311 Fig. 372 MECHANICS 249 On section ADB the force of gravity performs negative work (the angle a2 is obtuse) and the velocity of the planet will decrease and reach its minimum at point B. 225. For the satellite to move along a closed orbit (a circle with a radius R+h) it should be acted upon by a force directed toward the centre. In our case this is the force of the Earth's attraction. According to Newton's second law, mv2 mM R+h =1' (R+h)2 where M is the mass of the Earth, R=6,370 km is the radius of the Earth, and y is the gravitational constant. At the Earth's surface mM "W=mg Therefore, .. j--gti.2 v= V R+h ~7.5 krn/s 226. The resistance of the atmosphere will cause the satell ite to gradually approach the Earth and the .radius of its orbit diminishes. Since this resistance is small in the upper layers, the decrease of the ra.. dlus is insignificant during one revolution. Considering the orbit to be approximately circular, we can write mv2 mM y=yw where R is the radius of the orbit. Therefore, v= yY: • i.e., the velocity of the satellite increases with a reduction of R. This can be illustrated as follows. In view of the atmospheric resistance. a satellite placed, for example, mto a circular orbit (dotted line in Fig. 372) will actually move along a certain helix (solid line in Fig. 372). For this reason the projection of the force of gravity F onto the direction of the sa.. tellite velocity v differs from zero. It is the work of the force F (greater than the atmospheric resistance f) that increases the velocity. When the satellite moves in the atmosphere, its total mechanical energy diminishes but, as the Earth is approached, the potential energy drops faster than the total energy, causing the kinetic energy to grow. It should be stressed that the high force of resistance in the dense layers of the atmosphere does not allow us to consider, even approximately, the motion of the satellite as rotation along a circle, and our conclusion is not correct. . 227. If the container is thrown in the direction opposite to the motion of satellite A. it will begin to move along a certain ellipse 2 inside the orbit of the satellite (Fig. 373). The period of revolution of the container will be slightly less than that of satellite B. Therefore they can meet at the point of contact between the orbits only after a great number of revolutions. The container should be thrown in the direction of motion of satellite A. It will begin to move along ellipse 3. The velocity u should be such that • 250 Fig. 373 ",' ~ ,I I fI I I t,,\, ',2 ,' ...._-, Fig. 874 ANSWERS AND SOLUTIONS during one revolution of the container. satellite B also makes one revolution and in addition covers the distance AB This is quite possible. since the period of revolution along ellipse 3 is somewhat greater than along circular orbit 1. The container will meet the satellite at the point where orbits 3 and 1 coincide. 228. Assuming the Earth's orbit to be approximately circular, the force of gravity can be determined by the equation F=mCJ)2R. where m is the mass of the Earth, and 6)= 2.; is its angular velocity (T=365 days). On the other hand, according to the law of gravitation, F=V m;: ,where M is the mass of the Sun. Hence, mM "/f2=m0l2 R or Cl)2R3 M=--~2Xl()3s g y 229. Since both the Moon and the satellite move in the field of gravity of the Earth, let us use Kepler's third law T~ (h+H +2Ro)3 r: = BRa (Fig. 374). Therefore, h=2R (~:r/3-H-2Ro=220 km 230. Since the mass of the ball is greater than the mass of the water in the same volume, the field of gravity will be greater near the ball than away 251 from it. Therefore, the water near the ball will be compressed additionally. The pressure of the water acting on the bubble from the left will be smaller than the pressure that acts from the right. On the other hand, the forceof gravity between the air in the bubble and the ball is greater than the· force of volume of the water shown by dotted ---,'!...- -~- --! =~-=:-_t~l - ==- __--:t ).:= _.. _------~-~- - ~ -.-, - - - -- -- -- - ----R R R MECHANICS Fig. 375 attraction between the air and the circle a (Fig. 375). Since the mass of the air in the bubble is very small, the first factor becomes decisive. The bubble will be repulsed from the ball. Conversely, the motion of the iron baII will be determined by the fact t hat the force of attraction between the air in the bubble and the ball is much less than that between the ba11 and the volume of the water shown by dotted circle b. To calculate the force, let us reason as follows. A homogeneous medium (water) contains a sphere almost devoid of mass (the bubble) and a sphere with an excess mass (the ball). From a formal standpoint, this can be regarded as the presence of negative and positive masses. The force of interaction between the spheres in the water is equal to the interaction in vacuum of a negative mass equal to the mass of the water in the bubble and a positive mass equal to the mass of the iron ball that exceeds the mass of the water in the same volume. . Therefore, F ml (m2-ml) =-y R2 Here ml is the mass of the water in a sphere with a radius r, and m2 is the mass of the iron baII. 231. The field of gravity is smaller near the bubble than in a homogeneous liquid, and the liquid is compressed less. For this reason one bubble will move into the volume of liquid near the other one, and the bubbles will be mutually attracted. Two bubbles in a homogeneous liquid with negligibly small masses can be considered from a formal standpoint as negative masses superimposed upon the positive mass m of the medium in the volume of a bubble: (-m)(-m) m2 F=V R2 y R2 232. If the ball were solid, the· force of gravity F1 =1' ~:" where M = ~ :tR3p is the mass of the ball without a spherical space. The presence of this space is equivalent to the appearance of a force of repulsion F2=y~~m , where m'=: :trsp, and s is the distance between the centre of the .space and the material particle. The sought force F is the geometrical sum of the forces F1 and Fa (Fig. 376). (1 Fig. 976 1 A6000-"""'------------..li.....According to the cosine rule, F=V F~+F:-2FIF2 cos p= 4 .. /~R~8---r6---2""!!!'R.....8r-S-c-o-s-=p =3"nymp V zr+(12-d2) 2 11(lI-d2) ~5.8XIO-t gf 233. The sought force of attraction is the geometrical sum of the forces created by separate elements of the sphere. The small elements 01 and 0'2 (Fig. 377) are cut out of the sphere as cones with vertices at point A obtained when the generatrix Be revolves around axis 8182• The areas of the elements (AS )2ro (AS )1ro (AS )2rop are 1 and 2 respectively and their masses are _ .....1_~ cos (Xl cos (X2 " cos (Xl and (AS2)2 rop , where (J) is the solid angle at which both elements can be cos (Xl seen from point A; p is the surface density of the sphere (the mass per unit of surface); L. (Xl = L. (X2' since StOSI is an isosceles triangle. The forces of attraction created by the elements are equal, respectively, to: m (AS1)2 VI' then VI > V2. Therefore, the level of the water in the cup will lower. 241. In both cases the pumps perform identical work, since the same amount of pumped in water rises to the same level. 242. The inverted L will be stable on the bottom of the empty vessel, since a perpendicular dropped from the centre of gravity of the figure is within the limits of the supporting area. As water is poured into the vesseI, the expulsion force acting on the rectangle wiIJ grow in magnitude (it is assumed that the water can flow underneath the figure). When the depth of the water in the vessel is O.5a, the sum of the moments of the forces which tend to turn the body clockwise will be equal to the sum of those acting in the opposite direction. As the vessel is filled further, the figure will fall. 243. The length of the tube x can be found from the condition iX=Yo (x-h), which shows the equality of the pressures at the depth of the lower end of the tube. Here Vo is the specific weight of the water. Hence, x=-.1i!-= 50 ern Vo-y 244. Let us separate a column with a height of h inside the water (Fig. 378). The equation of motion of this column has the form: ma=mg-pA, where m= pAh is the mass of the water and p is the pressure at the depth h. Therefore, p=ph (g-a). 245. In accordance with the solution of Problem 244, the force of expulsion can be written as follows: F=pV (g-a), where V is the volume of the submerged portion of the body. The equation of motion of a floating body with a mass M has the form: Ma=Mg-pV (g-a). A, Flg.378 A-- 1/--- Fig. 979 ---8 ~~~;';~..i~~·----D MECHANICS 255 Hence, V= ~, as in a stationary vessel, and the body does not rise to p the surface. 246. If the tank were at rest or moved uniformly, the pressure at the depth h would be equal to PI =pgh. On the other hand, if the tank moved with an acceleration and the force of gravity were absent, the pressure at point A would be equal to P2 =pal. It is this pressure that would impart, in conformity with Newton's second law, the required acceleration a to the column of water with a length i. In accelerated motion of the tank, both the pressure PI and the pressure P2 appear in the field of gravity. According to Pascal's law, the pressure in the water is the same in all directions. For this reason the pressures PI and P2 are summated, and the result ing pressure at point A is P==P(gh+at). 247. Using the law of conservation of energy and Archimedes' principle. we obtain the following equation mgx= ( : nR3p-m) gh Where p is the density of the water and x is the sought height. Hence, ( ~ nR3p-m) h x= ~--------- m 248. If the level of the water in the vessel is the same, the level of the mercury will also be the same before the piece of wood is dropped. Dropping of the piece of wood gives the same result as adding of the amount of water that will be displaced by this piece, i. e., the amount of water equal to it by weight. Therefore, if the cross sections of the vessels are the same, the levels of the water and the mercury in both vessels wiII coincide. If the cross sections are different, the water will be higher and the mercury lower in the vessel with a smaller cross section. This will occur because the pressure on the surface of the mercury will increase differently when amounts of water equal in weight and volume are added to the vessels with various cross sections. 249. After the block is dropped into the broad vessel, the level of the mercury in both vessels will rise by the amount x and occupy position AB (Fig. 379). The required height of the water column in the broad vessel is determined by the equality of the pressures, for example, at level CD (y+x) PIg=hp2g where PI is the density of the mercury and P2 that of the water. The vaIue of y can be found by using the condition that the volume of the mercury is constant: . (x+ y) Al =V2 Where V2 is the volume of the mercury displaced by the block after the water is poured in. 256 ANSWERS AND SOLUTIONS Fig. 380 If the water covers the block entirely, then according to Archimedes' principle,Vopog=V2Plg+ (Vo- V2) P2i, where Po is the density of iron. Solving the equations, we obtain h- PI (PO-P2) Vo - P2 (Pt-P2) At If the water does not cover the block, Archimedes' principle can be written as VoPog=V2Plg+hAp~, where A=V:/3 is the area of a block face. In this case the sought height h PoVo P2 (At+V:/ 8 ) The first solution is true when A <; PI (PO-P2) V2/ 3 and the second one 1 P2(PI-PO) 0 when AI:;=:: PI (Po- Pi) V:/ 3 • P2(PI-PO) 250. It follows from the equality of the moments of the forces acting on the board with respect to point C (Fig. 380) that Ot (l-a- ~) cosa.=0 ( ~ -a) cosa. where 01 = Ax,\,o and G= Al'V~ Here A is the cross-sectional area of the board and Yo the specific weight of the water. Hence, ~=(l-a) ± .. 1(l-a)2 _.1. I (1-2a) V 1'0 Since x < I-a, only one solution is valid: x=(l-a)- .. I (l-a)2 --Y-I(I-2a) JI 1'0 251. The pressure on the "bottom" of the vessel is pgh and the force with which the hatched portion of the liquid (Fig. 381) presses on the table is pghn (2Rh tan a.-h2 tan2 a.). According to Newton's third law, an identical force acts on the liquid. The condition of equilibrium of the liquid at the moment when the vessel stops exerting pressure on the table has the form G+G1 =pghn (2Rh tan a-h2 tan2 e) where 01 is the weight of the hatched portion of the liquid (the truncated cone minus the cylinder volume) 01 =P~h { nR2+n (R-htana.)2+nR (R-htana.) } -pghn(R-htana.)2 Therefore. p G ( h tan a)ngh2 tan ex. R-- 3 - MECHANICS Fig. 381 Fig. 382 257 252. A liquid moves in a syphon by the action of the forces of cohesion between the elements of the liquid. The liquid in the long elbow outweighs the liquid in the short one, and pumps it over. It could be assumed on this basis that water can be pumped over a wall of any height with the ski of a syphon. This is not so, however. At a lifting height of 10 metres the pressure inside the liquid becomes zero. The air bubbles always present in water will begin to expand, and the water column will be broken, thus stopping the action of the syphon. 253. The device will first act as a syphon and the water will flow through the narrow pipe into the reservoir. Then an air bubble will slip through A and divide the liquid in the upper part into two portions. After this the liquid will no longer flow out. . 254. The pressure of the water directly under the piston of each pump is less than atmospheric pressure by pg(H +h), where p is the density of water. Therefore, to keep the piston in equilibrium, it should be pulled upward with the force F =pg (H +h) A, where A is the area of the piston. Hence, a greater force should be applied to the pistons with a greater area. 255. The pressure on the bottom is P=pgbH+h) (Fig. 382). On the other hand, since the vessel is a cylinder, p= :R~. The height h can be determined if we equate the forces acting on the piston pghn (RI-r2)= G Hence, 1 ( G r 2 )H=-- m----- ~ 10 cmnR2p g R2~r2 256. To prevent flowing out of the water, the vessel should be given such an acceleration at which the surface of the water takes the position shown in Fig. 383. The maximum volume of the water is ~: and the mass of the entire system is M+be'; p. The required acceleration can be found from the condition that the sum of the forces acting on a small element of the water with a mass 11m near the surface is directed horizontally (Fig. 383). 17-2042 258 ANSWERS AND SOLUTIONS N According to Newton's second law, ~ma==t1mg tan a. Therefore, the sought force is F = (M +b~~ P) g ~ 257. The lower part of a chamber is filled with denser air. The air leaves the chambers in the upper part. The pressure is gradually equa lized. The machine will continue to function only as long as the difference in the pressures between the two ha1ves of the vessel is sufficient to raise the water along the" tube to the upper portion. 258. Here the disk is not asymmetrical and the air pressure on the righthand side of the disk will be greater than on the left-hand one. The surplus force of the pressure acting on the right-hand side of the disk is F=(Pt- P2) A, where A is the cross-sectional area of the chamber. The weight of the chambers filled with water cannot exceed 0 = pgAh. Since h ~ 0 1 - O2 , then F ~ G. pg The disk will begin to rotate counterclockwise, and the chambers will rise from the bottom of the vessel to the top filled with air. The disk will rotate counterclockwise until the reduced difference of the pressure can no longer lift the water to the height h. 259. The bottom of the cylindrical vessel will fall off in all three cases, since the pressure exerted on the bottom from the top will always be equal to 1 kgf. In the vessel narrowing upward the bottom will fall off only when oil is poured in, since its level here will be higher than in the cylindrical vessel. In the vessel widening toward the top the bottom will fall off when the mercury is poured in, because its level will be higher than in the cylindrical vessel. This will also occur when the weight is put in. In this case the weight will be distributed over a smaller area than in the other two cases. 260. The reading of the balance will increase if the mean density of the body being weighed is less than the density of the weights. The reading will decrease if the mean density of the body is greater than that of the weights. The equilibrium of the balance will not be disturbed if the weights and the body have the same mean density. . 261. The man's aim will not be achieved, since, while increasing the expulsive force. he will also considerably increase the weight of the tube (the density of the compressed air in the tube is greater than that of the atrnospheric air). 262. The true weight of the body is G=Gt+yo (v-~:) d! 801.16 gf The error is equal to Fig. 383 G-Ot -0-100% ~O.14% 263. When the atmospheric pressure changes, the Archimedean force acting on the barometers from the side of the air will change owing to the change MECHANICS 259 '~ Fig. 385 I (b)Ia) Fig. 384 in the density of the air and in the volume of the barometers when the level of the mercury changes in their open parts. If all the conditions in the problem are taken into account, the barometers have not only the same weight, but also the same volume. For this reason the change in the expulsion force due to the first reason will be identical for each of them. The change in the volumes will be different. To change the difference in the levels by a certain amount in a U-shaped barometer, the level of the mercury in each elbow should change by only half of this amount. In a cup barometer the level of the mercury in the cup changes negligibly, and in the tube practically by the entire change in the difference of the levels. Here the change of the volume of the mercury in the tube should be the same as in the cup. Therefore, the change of the volume in the cup barometer will also become less, and it will therefore outweigh the U-shaped barometer. 264. The normal atmospheric pressure is approximately equal to 1 kgf/cm2• This means that an atmospheric column of air with an area of I em2 weighs 1 kgf. If we know the surface of the Earth, we can find the weight of its atmosphere. The Earth's surface A=4nR', where R=6,370 km is the mean radius of the Earth. ThE' weight of the atmosphere a2:: 4nRzx t kgf/cm2"E!!5 5X lOll ton f. 265. If the man stands on the mattress, his weight will be distributed over a smaller area (that of his feet) than if he lies down. Therefore, the state of equilibrium will set in in the first case at a higher air pressure in the mattress than in the second. 286. Let us first consider the tube inflated with air (Fig. 384a shows a cross section of the tube). For sections AB and CD of the tube to be in equilibrium it is obviously necessary that the tension of the expanded walls of the tube T balance the excess pressure inside the tube p, Let us now consider the forces that act on sections AB and CD when the tube is fitted onto a loaded wheel (Fig. 384b). The distribution of the forces '0 17 * 260 ANSWERS AND SOLUTIONS acting on AB does not. appreciably change in the top of the tube. There will be a difference in the bottom. Section CD will be acted upon by an elastic force from the side of the rim equal to the load applied to the wheel (the weight of the wheel and one-fourth of the weight of the motor vehicle). This additional force flattens the tube and the angle between the forces T tensioning the rubber increases. The total force of tension acting on CD diminishes and the excess pressure of the air in the tube equalizes- the force of tension, and also the weight of the wheel and of part of the motor vehic Ie. Thus, the rim does not lower because it is supported by the excess pressure of the air in the tube. In the top of the tube the excess pressure is equalized by the tension of the tube walls, and in the bottom it equalizes both the reduced tension of the walls and the force applied to the wheel. 267. The force per unit of length with which the cylindrical portion of the boiler is stretched in a direction perpendicular to axis 001 is 2Rl fl=21 P= pR where 2RI Is the cross-sectional area ABeD of the boiler and p is the pressure inside it (Fig. 385); 2Rlp Is the force acting on one half of the cylinder (see Problem 122). The maximum force per unit length of the hemispherical heads can be found from the formula nRt pR II "=2nR P=T =2' Consequently, the heads can withstand twice the pressure that the cylindrical portion of the boiler does if their walls are equally thick. For the strength of all the parts of the boiler to be identical, the thickness of the head walls may be half that of the cylindrical walls, l. e., 0.25 em. 268. The shape of the boiler should be such that the force applied per unit length of its cross section is minimum. This force is f=P~ , where A is the cross-sectional area of the boiler, 1 the peri meter of the section and p the pressure of the steam. The force f will be minimum with the smallest ratio between the crosssectional area and the perimeter. As is known, this ratio will be minimum for a circle. It is also known that a circle can be obtained by cutting a sphere with any plane. Therefore, a sphere is the most advantageous shape for the boiler. 269. The ceiling of a stratosphere balloon is determined not by the maximum altitude which it can ascend to, but by the altitude ensuring a safe velocity of landing. The envelope of a stratosphere balloon is filled with a light gas (hydrogen or helium) only partly, since when the balloon ascends, the gas in the envelope expands and forces out the air, making it possible to maintain the lifting force approximately constant. At a certain altitude, the gas will fill the entire envelope. Even after this the lifting force of the balloon continues to increase at the expense of the gas flowing out from the bottom hole in the envelope. The weight of the balloon decreases and it will reach its ceiling only after a certain amount of gas has leaked out. For the balloon to descend, some gas should additionally be expelled through the upper valve in the envelope, so that the lifting force is only slightly smaller than the weight of the balloon. At a small altitude the velocity of MECHANICS 261 descent will be too high, since the volume of the gas decreases and less of it remains in the balloon than during the ascent. The ballast is dropped to reduce the velocity of descent. 1·11. Hydro- and Aerodynamics 270. Let us denote the distance from the level of the water to the upper hole by h, the sought distance from the vessel to the point where the streams intersect by x, and the distance from the level of the water in the vessel to this point by y (Fig. 386). The point of intersection will remain at the same place if the level of the water in the vessel does not change. This will occur if Q= AVI +Av2 , where VI = Y2gh and V2 = Y2g (H +h) are the outflow velocities of the streams from the holes. On the basis of the laws of kinematics, where t1 and t2 are the times during which the water falls from the holes to the point of intersection. Hence, 1 ( Q2 2gA2) . %=2 2gA2- H2 ~ = 120 em I ( Q2 2gA2) Y='2 2gAI+ H2Q1 =130 em 271. The velocity of water outflow from a hole is v= Y2gh. The impulse of the force acting from the side of the vessel on the outftowing water FAt = Amv, where Am=pAvAt is the mass of the water ftowing out during the time At. Hence, F=pv2 A= 2pghA. The pressure at the bottom p=pgh and therefore F=2pA. The same force acts on the vessel from the side of the stream. • ..s..~- --- --.....=--- J.; -~_ .... - II ------.._-----1-- ~~ ~ ...... -------...:.....-- H --..- ............. _-.-.-....... ... ~ !I Fig. 886 262 ANSWERS AND SOLUTIONS Thus, the water acts on the wall with the hole with a force 2pA smaller than that acting on the opposite wall, and not with a force smaller by pA as might be expected. This is due to a reduction in the pressure acting on the wall with the hole, since the water flows faster at this wall. The vessel will begin to move if kG < 2pA or k < 2pghA G 272. According to Newton's second law, the equality pAo= 2pA should exist. Therefore, if the liquid flows out through the tube, the cross-sectional area of the stream should be halved A _ Ao -2 This compression of the stream can be explained as follows. The extreme streamlets of the Iiquid approaching the tube from above cannot, in view of inertia, flow around the edge of the tube directly adhering to its walls, and move towards the centre of the stream. Under the pressure of, the particles nearer to the centre of the stream, the lines of flow straighten out and a contracted stream of the liquid flows along the tube. 273. By neglecting splashing of the water, we thus assume the impact of the stream against the wall to be absolutely inelastic. According to Newton's second law. the change in the momentum of the water during the time ~t is limv=FM, where lim=p n:'vM is the mass of the water flowing during the time ~t through the cross section of the pipe. Hence, plld2 F =-- v2 === 8 gi 4 274. When the gas flows along the pipe (Fig. 387), its momentum changes in direction, but not in magnitude. A mass of pAv passes in a unit of time through cross section I of the vertical part. This mass brings in the momentum PI=pAUVl where VI is the J Fig. 387 p H-h A=Po-pgh Fig. 388 H MECHANICS 263 vector of the velocity with which the gas flows in the vertical part, numerically equal to the given velocity v. During the same time the momentum P2=pAvv2 is carried away through cross section II. Here v2 is the vector of .velocity in the horizontal part, also numerically equal to v. The change in the momentum is equal to the impulse of the force F that acts from the side of the pipe on the gas: F=pAv (Vi-VI)' In magnitude F=pAv2 Y 2. According to Newton's third law, the gas acts on the pipe with the same force. This force is directed oppositely to the pipe bend. 275. The initial velocity of the water with respect to the blade is v= Y2gh-roR. Therefore, a mass of water m=pA (Y2gh-roR) impingeson the blade in a unit of time. After the impact, the velocity of the water with reference to the blade is zero, and for this reason the change in the momentum of the water in a unit of time is mv. According to Newton's second law, the sought force is F=pA (y 2gh-roR)2 276. At the first moment the ship will begin to move to the right, since the pressure on the starboard side diminishes by 2pA, where p is the pressure at the depth h of the hole, and A is its area (see Problem 271). As soon as the stream of water reaches the opposite wall. this wall will be acted upon by the force F=pAv', where v is the velocity of the stream with respect to the ship (see Problem 273). The force F is slightly greater than 2pA, since tJ > Y2gh because the ship moves towards the stream. As a result, the motion will begin to retard. 277. The velocity of the liquid in the pipe is constant along the entire cross section because the liquid has a low compressibility and the stream is continuous. This velocity is v= y 2gH. The velocity of the liquld in the vessel is practically zero, since its area is much greater than the cross-sectional area of the pipe. Therefore, a pressure jump which we shall denote by PI - p" should exist on the vessel-pipe boundary. The work of the pressure forces causes the veloelt'i to change from zero to Y2gB. On the basis of the law of conservation of energy. Amvt -2-= (Pl - PI) AAh where A is the cross-sectional area of the pipe, Ah is the height of a small element of the liquid, and Am=pAAh is the mass of this element. Hence, . pv2 T=PI-pf,=pgH Since the flow velocity is constant, the pressure in the pipe changes according to the law P=Po-pg(h-x) as in a liquid at rest. Here Po is the atmospheric pressure and x is the distance from the upper end of the pipe. 264 ANSWERS AND SOLUTiONS The change of pressure in height is shown In Fig. 388. The pressure is laid off along the axis of ordinates, and the distance from the surface of the liquid in the vessel along the axis of abscissas. 278. The water flowing out of the pipe during a small interval of time At will carrywith it the momentum Ap=pAvIAt, where v= Y2gH is the velocity of the outflowlng stream (see Problem 277). According to Newton's law, FAt = 2pgHAAt. The same force will act on the vessel with the water from the side of the outflowing stream. Therefore, the reading of the balance will decrease by 2pgHA at the initial moment. 279. At the first mcment when the stream has not yet reached the pan, equilibrium will be violated. The pan will swing upwards since the water flowing out of the vessel no longer exerts pressure on its bottom. When the stream reaches the pan, equilibrium will be restored. Let us consider an element of the stream with the mass Am. This element, when falling onto the pan, imparts to it an impulse AmY2gh In a vertical direction, where h is the height of the cock above the pan. On the other hand, after leaving the vessel, this element will cease to exert pressure on its bottom and on the pan duri~g the time of falling t= Y2ii. This is equl. g valent to the appearance of an impulse of force acting on the vessel vertically upward when the element of the liquid is falling. The mean value of this impulse during the duration of the fall is .. /2h ,r~mg V g=Am f 2gh Thus, each element of the tiquid Am is accompanied during Its falling by the appearance of two equal and oppositely directed impulses of force. Since the stream is continuous, the balance will be in equilibrium. At the moment when the stream stops flowing, the pan will swing down, since the last elements of the liquid falling on the pan act on it with a force that exceeds the weight of the elements, and there will no longer be a reduction in-the pressure on the bottom of the vessel. 280. On the basis of the law of conservation of energy we can write Mv2 -2-=mgh where M is the mass of the water In the tube stopped by valve V. and m is the mass of the water raised to the height h. Therefore, pbtd 2 X ~=pVoRh 4 2 where Vo is the volume of mass m. The average volume raised in two seconds is lnd2v2 Vo= 8gh =1.7 X to-3 m3 One hour of ram operation wi11 raise V=1.7 X 10- 3 X 30 X 60=::= 3m3 MECHANICS F ::::::::a vt~ Fig. 989 265 281. The pressure of the air streaming over the roof is' Jess than of air at rest. It is this surplus pressure of the stationary air under the roof that causes the described phenomena. 282. Since the gas in the stream has a high velocity, the pressure inside the stream is below atmospheric. The ball will be supported from the bottom by the thrust of the stream, and on the sides by the static atmospheric pressure. 283. When air flows between the disks, its velocity diminishes as it approaches the edges of the disks, and is minimum at the edges. The pressure in a jet of air Is the lower, the higher its velocity. For this reason the pressure between the disks is lower than atmospheric. The atmospheric pressure presses the lower disk against the upper one, and the flow of the air is stopped. After this the static pressure of the air again moves the disk awayI and the process is repeated. 284. The pressure diminishes in a stream of a flowing liquid with an increase in its velocity. The velocity with which the water flows in the vessel is much smaller than in the tube and, therefore, the pressure of the water in the vessel is greater than in the tube. The velocity Increases at the boundary between the vessel and tube, and the pressure drops. For this reason the ball is pressed against the screen and does not rise. 285. The piston will cover the distance ut during the time t (Fig. 389). The force F will perform the work W;:::Fut. The mass of the liquid flowing out during the time t is pAut. The outflow velocity of the liquid v can be found from the equation Au=av. The change in the kinetic energy of the liquid during the time t is ( VI U' ) pAut 2-2 This change should be equal to the work performed by the force F: ( V' U 2) Fut=pAut 2-2 Upon eliminating u, we find that 2F I vZ=-x--Ap . as 1- AI .. /2F If a~ A, then.v= V Ap· 286. It was assumed in solving Problem 285 that the velocity of any element of a liquid in the pump is constant. The velocity changes from u to v when the liquid leaves the pump. This does not occur immediately after the force begins to act on the piston, however. The process requires some 266 ANSWERS AND SOLUTIONS time to become stable. i. e., after the particles of the liquid in the cylinder acquire a constant velocity. When a -+ A this time tends to infinity and. for this reason, the velocity acquired by the liquid under the action of the constant force becomes infinitely high. 287. Let us introduce the coordinate system depicted in Fig. 390. According to Torricelli's formula, the outflow velocity of a liquid is V = Y2gy, where y is the thickness of the water layer in the upper vessel. Since water is incompressible, aV= Av, where v is the velocity with which the upper layer of the water lowers, A is its area, and a is the area of the orifice. If we assume that the vessel is axially symmetrical, then A=nx2, where x is the horizontal coordinate of the vessel wall. Therefore, nx2 a --=-=-:- =const V2gy v since in confor mity with the initial condition, the water level should lower with a constant velocity. Hence, the shape of the vessel can be determined from the equation where :1;2V2 k=- 2ga2 288. The pressure changes in a horizontal cross section depending on the dis tance to the axis , according to the law p0)2 P=PO+T ,2 where Po is the pressure on the axis of the vessel and p the density of the liquid. The compressive deformation of the liquid will be maximum near the walls of the vessel, while the tensile deformation of the revolving rod (Problem 203) is maximum at the axis. - - 1 - -r- - - - 1 - - _I I Fig. 390 Fig. 391 MECHANICS 267 289. The excess pressure at a distance r from the axis of rotation is p=p;2,S (see the solution to Problem 288). On the other hand, this pressure is determined by the elevation of the liquid level in this section above the level on the axis: p=pgh (Fig. 391). Upon equating these expressions, we have (02 h=-r2 2g This is the equation of a parabola, and the surface of the liquid in the rotating vessel takes the form of . a paraboloid of revolution. Ftg. 392 290. The stirring imparts a certain angular velocity (0 to the particles of the water in the glass. The pressures in the liquid will be distributed in about the same way as in the soIution to Problem 288. The excess pressure inside the liquid balances the pressure due to a higher level of the liquid at the edges of the glass (see Problem 289). , When stirring is stopped, the velocity of rotation of the liquid near the bottom will begin to decrease owing to friction, the greater the farther the elements of the liquid are from the centre. Now the excess pressure caused by rotation wiII no .longer balance the weight of the liquid column near the edge of the vessel. This will cause the liquid to circulate as shown in Fig. 392. This is why the tea leaves will gather in the middle of the glass. CHAPTER 2 HEAT. MOLECULAR PHYSICS 2-1. Thermal Expansion of Solids and Liquids 291. l\t =4200 c. 292. Reinforced-concrete structures are very strong because the expansion coefficient of concrete is very close to that of iron and steel. 293. The quantity of heat transferred from one body to another in a unit of time is proportional to the difference between the temperatures of these bodies. When the temperatures of the thermometer and the surrounding objects differ appreciably, the volume of the mercury will change at a fast rateo If the temperature of the thermometer is nearly the same as that of the surrounding bodies, the volume of the mercury changes slowly. For this reason it takes longer for the thermometer to take the temperature of a human body. If the warm thermometer is brought in contact with relatively cool air in the room, the mercury column "drops" so fast due to the great temperature difference that it can be shaken down in a moment. 294. When the scale cools down from tl to to=O° C, the value of each graduation diminishes. Therefore, the height of the mercury column read off the scale with a temperature of to=00 C will be different and equal H =HIX X (1 +at1)' The heights of the mercury columns at different temperatures and identical pressures are inversely proportional to the densities: n, P I HI =~=l+ytl Hence, H =H1(1+at1) ::::: H (l+at _lUt) o 1+ yt l 1 1 r 1 295. The thermometer can be precooled in a refrigerator and shaken. If no refrigerator is available, put the thermometer into your mouth or in your arm-pit for a time sufficient for the entire thermometer to reach the body temperature, then take it out and shake it immediately. The thermometer will show the temperature of the body. 296. The difference in the lengths of the rulers at a temperature t1 is l~(l +a1tt)-l; (1+r.J.2t 1)=l At a temperature t2 this difference is equal to L~ (1 +alt2)~l; (1 +Cltt2)= ± 1 The plus sign corresponds to the case when the difference In lengths is constant (see Fig. 393a). The relation between the lengths and the temperature shown in Fig. 39311 corresponds to the minus sign. In the first case the system of equations gives 1;(u=~l=6.8 em; 1;(u=~l=4.8cm "s-~ a.-al HEAT. MOLECULAR PHYSICS 269 (0) . /l ~ (b) In the second case ~ t t Fig. 393 cm; When I=()O C, the iron ruler should be longer than the copper one. 297. A possible way of suspension is depicted in Fig. 394, where / and 2 are rods with a small coefficient of linear expansion (11 (e. g., steel rods), and 3 are rods with a high coefficient of expansion a, (e. g., zinc or brass rods). The lengths of the rods can be so selected that the length of the pendulum does not change with -the temperature.' With this aim in view it is essential that ClJ (11+11)= (1,', . 298. When the cylinder is heated, its volume increases according to the same law as that of the glass: (11=00 (1+yl/), where y is the coefficient of volume expansion of glass. If the densities o. mercury at the temperatures to and '1 are denoted by Po and PI' we can write that mo=vopo and ml=vIP1, where 270 ! Z 9 Fig,394 e ANSWERS AND SOLUTIONS This system of equations will give the following expression for 1': l' m1 (1+1'l t1)- mO~ 3X 10-~ deg-1 motl The coefficient of linear expansion a= ; ===10-& deg-1• 299. Let the pendulum of an accurate clock perform N oscillations a day. At the temperature tl the pendulum of our clock will perform N oscillations in n - 5 seconds (where n = 86.400 is the number of seconds in a day) and at the temperature t2 in n +10 seconds. The periods of oscillations will respectively be equal to n-5 n+10 Tl=~ and T2=-N-' T 1 n-5 15 Hence, T 2 =n+IO~ I-n° On the other hand, bearing in mind that the period of ren- rrr: dulum oscillations T=2n V ~. we obtain T fl+aLl ,r~~-- T: = l 1+aL2 ~ y 1+a(tl-t2)~1+ a +"2(t1-t:!) Upon equating the expressions for the ratio of the periods. we find that 30 a ~ ~ 2.3X 10-6 deg-I (t 2- t1) n 2-2. The Law of Conservation of Energy. Thermal Conductivity I 300. According to the Jaw of conservation of energy. the liberated heat is equal to the loss of kinetic energy , Q_Mv~ (M+m)v2 - 2 2 where v is the velocity of the cart after the brick has been lowered onto it. This velocity can be found from the law of conservation of momentum: Mvo v=M+m· I hani 1 it Q . Mmv~ d I th 1 u Q J Mmv~n mec anica urn s =2 (M +m) an In erma urn s = 2 (M +m)' where j is the thermal equivalent of work. HEAT. MOLECULAR PHYSICS 271 301. On the basis of the law of conservation of energy, mgl =mv 2 +k (1-/0 )2 +Q 2 2 where l is the length of the cord at the moment when the washer leaves it. • On the other hand, we can write that mv2 mgl=T+Wl + W2 where WI = flo is the work of the force of friction acting, on the washer (the washer travels a path of lo relative to the cord), and W s=f (1-10) is the work of the force of friction acting on the cord. Therefore, k (1-10)2 Q=W 1 +W2 2 Using Hooke's Law f=k (l-Io> we find that f2 Q=flo+ 2k The work W1 is used entirely to liberate heat. Only half of the work W2 = ~ , however, is converted into heat, the other half producing the potential energy k (/-10)2 2 . 302. The electric current performs the work W =PT. At the expense of this work the refrigerator will lose the beat QI=qH +qct; where c is the heat capacity of water and H is the heat of fusion of ice. According to the law of conservation of energy, the amount of heat liberated in the room will be Q. = W+Q2=Pt:+qct+qH since in the final run the energy of the electric current is converted into heat. 303. The temperature in the room will rise. The quantity of heat liberated in a unit of time will be equal to' the power consumed by the refrigerator, since in the final run the energy of the electric current is converted into heat, and the heat removed from the refrigerator is returned again into the room. 304. It is more advantageous to use a refrigerator that removes heat from the outside air .and liberates it in the room. The heat liberated in the room in a unit of time is P +Q2' where P is the power consumed by the refrigerator and Q, is the heat removed from the outside air in a unit of time (see Problem 302). It is only the high cost and complicated equipment that prevent the use of such thermal pumps for heating at present. 305. When salt is dissolved, its crystal lattice is destroyed. The process requires a certain amount of energy that can be obtained from the solvent. In the second case, part of the intermolecular bonds of the crystal lattice have already been destroyed in crushing the crystal. For this reason, less energy is required to dissolve the powder and the water will be higher in temperature in the second vessel. The effect, however, will be extremely negligible. 272 ANSWERS AND SOLUTIONS 306. The quantity of heat removed fromthe water beingcooledis m2c (tt-8), where 6 is the final temperature. The cold water receives the heat mlc (6-tl). The heat imparted to the calorimeter Is q(a-tl ) . On the basis of the law of conservation of energy, mlc (6-t1)+ q(6-tt )= mgc (t 2- 6) whence 8=(mItt +m"t,,) c+ qtl ~ 40 C (ml +m2) c+q - 307. The power spent to heat the water in the calorimeter is Pl=pVctJ l' where p is the density of the water, C is its specific heat, and J =4.18 Jzcal is the mechanical equivalent of heat. The sought value is P-Pt pVctJ Q=-p-=1--W~5 per cent A 308. Q=l[ (T1-To)At ~ 9,331 kcal 309. The quantity of heat Q passing through the first pane) a second is Q= Al TI ~T1 A, where A is the area of a panel. Since the process is statio- To-T nary, the sameamount of heat passes through the second panel: Q ="" T A. T,,-TI To-T" We find from the condition At - d-t - A=A2 ~ A that T _AtdlTo+Atd,Tt "- A2dl +Atd2 310. Upon inserting the temperature T" into the expression for Q (see Problem 309) when dl=d2=d, we find that Q= 2"'IA2 To-T1 A x,+A.:a 2d Therefore, the coefficient of thermal conductivity of the wall is A= 2AI At At+ A" 311. The quantity of heat passing a secon d through the cross-sectional areas of the blocks with coefficients of thermal conductivity Al and A2 is equal, respectively, to Q Al A2 T At=cr(TI-To) A and Q2={[ (T1- 0) The quantity of heat passing through two blocks whose entire cross-sectional area is 2A is Q=Ql + Q2 = Al+A2 Tt-T 0 2A 2 d HEAT. MOLECULAR PHYSICS Hence, the coefficient of thermal conductivity of the wall is equal to ~ _AI+A2 "'- 2 273 312. The coefficients of thermal conductivity of walls I and II are equal to '1 Al+A2 d ~ 2AIAI ""1=-2- an "'II = At+ A2 (see the solutions to Problems 310 and 311). It follows from the obvious inequality (A.I - At)! > 0 that (AI +A2)2 > 4"'1AI Hence, At+A2 2AIA2 • Il 'l -2- > Al +1 2 ,I.e., "'I > "'II 313. The quantity of heat supplied by the heater into the water through the pan bottom is A Q=7f(T-Tt ) A=mr where T1 is the boiling point of the water and, is the specific heat of va- porization. Therefore, 2·3. Prope'rties of Gases 314. The removable cap acts as a pump and under it a rarefied space is formed that sucks out the ink. The orifice serves to maintain a constant pressure under the cap. 315. Assuming that the temperature remains constant, let us apply Boyle's law to the volume of air above the mercury: (POI-PI) (1-748 mm)=(P02-P2) (1-736 mm) whence 1=764 mm. 316. In a position of eqnilibrium f-6--F=O, where f is the force of expulsion equal to yh1A (here y is the specific weight of the watert and hl the height of the air column in the tube after submergence). In our case the force of expulsion Is built up by the difference of pressure on the soldered end of the tube from below and lrom above: f=ptA-(lJo+'Vh) A, where Pi is the air pressure in the tube after submergence. According to Boyle's law, PolA=PlhiA. It follows from this system of equations that E= : [y (po+yh)2+4Poyl.:.....(po+yh»)-G=8.65 gf 18-2042 274 P A ~ --------------------- B ANSWERS AND SOLUTIONS 1/ Fig. 395 317. First. the pressure p of the air will decrease approximately isothermally owing to the drop of the level of the water in the vessel. This will continue until the total pressure at the level of the lower end of the tube becomes equal to the atmospheric pressure Po; i, e., p+pgh= Po. where h is the height of the water column in the vessel above the level of the lower end of the tube. From this moment on air bubbles will begin to pass into the vessel. The pressure at the level of the lower end of the tube will remain equal to the atmospheric pressure, while the air pressure P=Po-pgh will grow linearly with a drop in the water level. The water will flow out from the vessel at a constant velocity. The relation between p and Q is shown in Fig. 395. The negligible fluctuations of pressure when separate bubbles pass in are not shown in Fig. 395. 318. When the air is being pumped out of the vessel, the pressure in the vessel after one double stroke will become equal to Pt =vP+o~ • After the o V)2second double stroke PtV=pz(V +00) and, consequently, Pz=Po (v+00 • etc. After n double strokes the pressure in the vessel will be p'=Po (V:vJ" When air is being delivered into the vessel. after n double strokes the pressure will be _ ,+Ponvo_ {( V )n+nuo}p-p - - -Po - - V V+vo V Here p > Po at any n, since during delivery the pump during each double stroke sucks in air with a pressure (Jo, and during evacuation (10 of the air is pumped out at pressures below Po. 319. Applying Boyle's law to the two volumes of gas in the closed tube, we obtain L-I (L-l )P-2- A=Pl -2-- A1 A L-l (L-l)p -2- A=tJ2 -2-+~1 A Pl=P2+Vl HEAT. MOLECULAR PHYSICS 275 Here p is the pressure in the tube placed horizontally, Pl and P2 are the pressures in the lower and upper ends of the tube placed vertically when its ends are closed, y is the specific weight of the mercury, A is the cross-sectional area of the tube. Hence, the initial pressure in the tube is p=y.!- (l!._&1)2 III 10 Here. for the sake of brevity we have designated L 2 1 by 10, If one end of the horizontal tube is opened', the pressure of the gas Jn the tube will become equal to the atmospheric pressure. According to Boyle's law. ploA = yH11A (here H is the atmospheric pressure), and, therefore, l-~(~-~)1- 2H &l 10 The mercury column will shift through the distance L\11= lo- lt = l!JL [2H -(~-~)J2H I L\l 10 For the mercury not to flow out of the tube, the following condition is required &~ ~ V(~r+l+ ~ \Vhen the upper end of the vertical tube is opened ploA ="1 (H + 1)l2A Hence, tlI2 = 10-12 2(:+1) [27-(~~ - ~:) +2] The mercury will not flow out of the tube. if ~~ ~ j1{2(~+l)r+I+2(Hl+l) When the lower end is opened ploA =Y (H -I) 13A whence M3= 10- 13= 2(~~l) [27- (~~ -~:)-2] The following condition should be satisfied to prevent the mercury column from being forced out of the tube ~ 1//-4 (H -1)2 1+2 (H -I) t1l ~ r 12 + 1 18 *. 276 ANSWERS AND SOLUTIONS 320. Since for one gramme-molecule of any. gas at P= I atm and T =2370 K we have V,.= 22.4 lltres, then for one mole C=P~"=O.082lit.atm/mole.deg. This constant is usually denoted by R and called the universal gas constant. The values of R in the various systems of units are: R=0.848 kgl-rn/mole- deg=8.3X 107 erg/mole- deg= 1.986 cal/rnole-deg 321. At a fixed pressure and temperature the volume occupied by the gas is proportional to its mass. A volume of VfA- corresponds to one gramme-molecule and a volume of V to an arbitrary mass m. Obviously. VIi =V ~. where I..t is the molecular weight expressed in grammes. Upon inserting this expression into the equation of state for one grammemolecule. we have pV=!!!...RT J.I. ~ ~ • .. .. ~ ~ Jr Fig. 397 /1. "' ~ ~ -- .: .- -- .- ., I Fig. 396 322. If the attraction between the molecules suddenly disappeared. the pressure shou ld increase. To prove this, let us mentally single out two layers J and IJ inside a fluid (Fig. 396). The molecules penetrating from layer I into layer I J owing to thermal motion collide with the molecules in layer II, and as a result this layer is acted upon by the pressure forces Pt that depend on the temperature. The forces of attraction act on layer I I from the side of the molecules in layer I in the opposite direction. The resultant pressure of layer I on layer IJ P=Pt- Pi, where Pi is the pressure 'caused by the internal forces of attraction. When PI disappears, the pressure grows. 323. If the forces of attraction between the molecules disappeared the water would be converted into an ideal gas. The pressure can be found from the equation of state of an idea 1 gas: m RT p=~ V~ 1,370 atm HEAT. MOLECULAR PHYSICS 271 324. Let us separate a cylindrical volume of the gas in direct contact with the wall (Fig. 397). The forces acting on the side surface of the cylin.. der are mutually balanced. Since the volume is in equilibrium, the pressure on the gas from the side of the wall should always be equal to the pressure on the .other base of the cylinder from the side of the gas. We can conclude, on the basis of Newton's third law, that the pressure of the gas on the wall is equal to the pressure inside the vessel. 325. The pressure in the gas depends on the forces of interaction between the molecules (see Problem 322). The forces of mutual interaction of the molecules and of the molecules with the wa11 are different, however. Hence the pressures inside the gas and at the walls of the vessel (see Problem 324) can be identical only if the concentrations are different. 326. Since the volume is constant ~=.~, or P2-Pl=T2- T 1 = O.004 PI T 1 PI T 1 Hence, T =T2-T1=2500K 1 0.004 327. From Archimedes' law, mg+G=yV, where y is the specific weight of water and V is the volume of the sphere. The equation of state gives (Po+ yh) V=!!!:.... RT ~ Upon deleting V from these equations, we find that m GJ-t (Po +yh) ~ 0.666 g VRT - fJ-g (Po +yh) and equilibrium will be unstable. 328. When the tube is horizontal, the device cannot be used as a thermometer, since the pressures exerted on the drop from the right and from ·the left will be balanced at any temperature. If the tube is placed vertically, the pressure of the gas in the lower ball will be higher than in the ufper one by a constant magnitude. If the volume is the same, the pressure wil grow with a rise in the temperature the faster, the higher is the initial pressure. To maintain a constant difference of the pressures in the balls, the drop will begin to move upward, and in this case the device can be employed as a thermometer. 329. Since the masses of the gas are the same in both ends and the piston is in equilibrium, Hence, T2 =!.!. T1 =3300 K VI Applying Boyle's law to the volume of the gas whose temperature does not change, we obtain p= PoVo= l.05 atm VI 278 ANSWERS AND SOLUTIONS 330. When the external conditions are the same, equal volumes of various gases contain an equal number of molecules (Avogadro's law). Therefore, Vl:V,:Vs:V.=Nt:N2:Ns:Nc, where Vi is the volume of a gas and N,. the number of molecules of this gas. The mass of a certain amount of a gas is proportional to the nurnber of its molecules and the molecular weight of the gas: ml:m2: ma:m.= N1""'1: N2""2: N8""3: N.J.t. On the other hand, denoting the relative volume of this gas in per cent V· by nj=V 100%, we have . . . VI. V2 • Vs • Vc N1.N2.N3 • N. nt·n2 .na·n,=V·V·V· V=7j·N'N·!i If the composition of air in per cent is described by ni= mi 100% (com- m position by weight), we can obtain from the previous ratios that n ' 'n' ·n' 'n' _m1.m2.m3, m4 _ N1Jll . Na""2 . Naf.ls. N.Jl4_n II. ·n II. • n II. in u 1 • 2· 3· • - m · m · m · m - N · N . N · N - lr1' 2r2' 3r3· 4r4 Hence, , n~+n;+n;+n;n, n.,.,.. n1Jll+nl J.L2 + naJla+n.J1. I I Remembering that n;+n;+n;+n;= 100 per cent, we obtain , nilJi1OO% n1 nl""'l +n2J.t2 -l- nal-'a +n.J.t4 Therefore, n;==75.52%; n;=23.15%; n;= 1.28%; n~=O.05% 331. For each gas, the equation of state can be written as follows: PIV=.!!!.L RT J.ll P2V=~RT fJ2 P3V=~RT ""3 P4V=~ RT f.14 Hence, (PI +P2+P3+P4)V= (~+~+~+.!!!!.)RT f.tl ""2 f.La fJ4 On the other hpnd, for a mixture of gases pV=~ RT t where m=ml +f.1 + m2+m3+tn4 and Il. is the sought molecular weight. HEAT. MOLECULAR PHYSICS 279 I/otm 1 ---- ! 0.5 Z /lZ6 ------ 4.3 I I I J Z J 4 .J{lit Fig. 398 According to Dalton's law, p=p,+p!+Ps+P,. Therefore, 11= ml+mS+ma+m• n~+n;+n~+n~ =28.966 .!!!!..+~+~+~ !!i+ n; + n~ + n; ""'1 fi! J,Ls J,L. ftl fl.' fJ.af.t. where ni=~ 100% is the composition of the air in per cent by weight. m The result obtained in the previous problem allows us to find .... from the known composition of the air. by volume f..L ....l nt +""2 n2+""sna+J.t.n. = 28.966 nl +n2+nS+n. 332. On the basis of Clapeyron's equation, J1 = mRT =pRT =72 g/mole pV p The sought formula is C&H12 (one of the pentane isomers). 333. When the gas is compressed in a heat..impermeable envelope, the work performed by the external force is spent to increase the internal energy of the gas. and its temperature increases. The pressure in the gas will increase both owing to a reduction in its volume and an increase in its tempe.. rature. In isothermal compression the pressure rises only owing to a reduction in the volume. Therefore. the pressure will increase more in' the first case than in the second. 334. A diagram of p versus V is shown in Fig. 398. The greatest work equal to the hatched area in Fig. 398 is performed during the isothermal process (1..2). . The temperature does not change on section 1-2, and is halved on section 2-3. After this the temperature rises. and, T,=T1 when V4=4 lit. 335. Line 1-2 is an isobaric line (Fig. 399). The gas is heated at a constant pressure, absorbing heat. 280 p 1.....---+---.... v ANS,WERS AND SOLUTIONS Line 2-3 is an isochoric line. The gas is cooled at a constant volume, the pressure drops and heat is Iibera... ted. Line 3..l is an isothermal line. The volume of the gas diminishes at a constant temperature. The pressure ri... sese The gas is not heated, although it is subjected to the work of external forces. Hence, the gas rejects heat on this section. 338. The amount of heat liberated per hour upon combustion of the methane is Fig. 399 Q _90PVof.' 1- RT where 11=16 g/mole is the mass of one mole of the gas and T=t+273° = = 2840 K is its temperature. The amount of heat received by the water in one hour is nDI Q'=Tvpc (tt-tl ) 3,600 where p= 1 g/cms, is the density of the water and c= 1 cal/deg- g is the specific heat. According to the condition, Q Q: ='1=0.6 Upon solving these simultaneous equations, we find that t t + qopVoJ.LTJ - 930r 2 = 1 900nD2vpcRT = -\J 337. In the initial state plV =~. RT l' where J.tl is the molecular weight ~1 of the ozone. In the final state, P2V =m RT 2' where ....2 is the molecular 1-12 weight of the oxygen. The heat balance equation gives ~ q =Cvm (T2 - T I) J.Ll J.L2 Upon solving these simultaneous equations, we find that P'I.=-q...-+J.LI=lO PI CvTt ""2 338. In view of the linear dependence of pressure on volume we can write: p==aV+b. HEAT. MOLECULAR PHYSICS r I \ I \ I \ I \ \ I \ I \ I \ , \ I \ I \ \ I \ va ~O:G If II Fig. 400 281 The constants a and b can be found from the condition of the problem: a=P V l - P v 2 :!!:-0.5atm/lit 1- 2 b plV1- plV2 S!!E 20 atm VI-V' Upon inserting the expression for p into the equation of state of an ideal m gas pV=-RT=const T, we find that J.L aVI+bV=const T (1) The relation between T and V (see Fig. 400) is a parabola. The curve reaches its maximum at Vmax= - :a2E 20 lit when the roots of quadratic equation (1) coincide. Here b Pmax=aVmax+b=2 e: 10 atm Therefore, T _PmaxVmaxJ.L ~ 4900 KmQX- mR - 339. The energyof 8 unit volumeofgas"1 =CTp, where p Isthe density ofthe air. According to the equation of state ofan Idealgas, Pi=mB (B is a constant). Since p=~ , then pT= ; . Therefore, Ul=~ P is determined only by the pressure. The energy of all the air in the room is also determi ned only by the pressure. The pressure in the room is equal to the atmospheric pressure 282 ANSWERS AND SOLUTIONS Fig. 401 and does not change when the air is heated. For this reason the energy of the air in the room also does not change. As the air is heated, some of it flows outside through cracks, and this ensures a constancy of energy despite the heating. The energy would increase with heating only in a hermetically closed room. 340. On the basis of the equation of state. the sought mass of the air will be A _ f.LpV T2 - T1 ~ 13kum------ g R T 1T2 • 341. Let the tube first be near the bottom in a state of stable equilibrium. Upon heating, the air pressure in the tube and, correspondingly. the force of expulsion increase. At a certain temperature TIthe tube begins to rise to the surface. Since the pressure of the water gradually decreases upwards from the bottom, the volume of the air in the tube and, therefore. the force of expulsion continue to increase. The tube will quickly reach the surface of the water. Upon a further increase in the temperature, the tube will be at the surface. If the temperature lowers, the tube will not sink at T1. because it has a great reserve of buoyancy caused by an appreciable increase in the force of expulsion as the tube rises. It is only when T2 < T1 that the tube begins to sink. Here the force of expulsion will drop because, as submergence continues, the air in the tube will occupy a smaller volume. The tube will reach the bottom very quickly. The relation between the depth of submergence h of the tube and the temperature T is shown in Fig. 401. The tube will always be at the bottom when T < T2 and at the surface when T > T1. If T2 < T < Tit the tube will be either at the bottom or at the surface, depending on the previous temperatures. 342. The gas expands at a certain constant pressure p built up by the piston. The work W=p (V2- VI). where Viand V2 are the Initial and final volumes of the gas. By using the equation of state. let us express the product pV through the temperature T. Then. W=~ R (T2 - Td ~ 33.9 kgf-m LL HEAT. MOLECULAR PHYSICS 283 343. The heat imparted to the gas is used to heat it and perform mecha.. nical work. According to the law of conservation of energy, 2-4. Properties of Liquids 344. It is more difficult to compress a litre of air, since more work has to be done in this case. Water has a small compressibility, and a small reduction in volume is required to increase the pressure inside it to three atmospheres. 345. A maximum thermometer can be made as follows. A small unweUable freely moving body is placed inside the tube of a horizontal thermometer (Fig. 402). The position of the body will show the maximum temperature, since the body will move along the tube when the liquid expands and will remain in place when the liquid in the tube is compressed. To make a minimum thermometer, a wettable body should be placed inside the liquid in the tube. 346. When an elastic rubber film is stretched, the force of tension depends on the amount of deformation of the film. The force of surface tension is determined only by the properties of the liquid and does not change with an increase of its surface. 347. The surface tension of pure petrol is less than that of petrol in which grease is dissolved. For this reason the petrol applied to the edges will contract the spot towards the centre. If the spot itself is wetted, it will spread over the fabric. 348. Capillaries of the type shown in Fig. 403 form in a compact surface layer of soil. They converge towards the top, and the water in them rises to the surface, from which it is intensively evaporated. Harrowing destroys this structure of the capillaries and the moisture is retained in the soil longer. 349. Leather contains a great number of capillaries. A drop of a wetting liquid inside a capillary having a constant cross section will be in equllibrium. When the liquid is heated, the surface tension diminishes and the liquid is drawn towards the cold part of the capillary. The grease will be drawn into the leather if it is heated outside. 350. The grease melts, and capillary forces carry it to the surface of the cold fabric placed under the clothing (see Problem 349). 351. The end of the piece of wood in the shade is colder, and the caplltary forces move the water in th is direction. 352. The hydrostatic pressure should be balanced by the capillary pres- 4~ sure: pgh={f' Hence, h=30 em. 353. The following forces act vertically on section abed of the film: weight, surface tension Fab applied to line ab and surface tension Fcd applied to cd. Fig. 402 284 ANSWERS AND SOLUTIONS Equilibrium is possible only if Fall is greater than Fcd by an amount equal to the weight of the section of the film being considered. The difference between the forces of surface tension can be explained by the difference in the concentration of the soap in the surface layers of the film. 354. The force of expulsion balances the weight of the cube mg and the force of surface tension 4aa. Fig. 403 i.e., a2xpg= mg+ 4aa, where x is the sought distance. Therefore, mg+4aa. ~ 2 3 x a2pg - . em The forces of surface tension introduce a correction of about 0.1 em. 355. The water rises to a height h= 2a . The potential energy of the wa- pgr ter colurnn is E p = mgh = 2na;2 2 pg 4n(£2 The forces of surface tension perform the work W=2nrah=--. One half Pi of this work goes to increase the potential energy, and the other half to evolve heat. Hence, 2na2 Q=-pg 356. The pressure inside the liquid at a point that is at a height h above a certain level is less than the pressure at this level by pgh. The pressure is zero at the level of the liquid in the vessel. Therefore, the pressure at the height h is negative (the liquid is stretched) and is equal to p= -pgh. 357. The forces of attraction acting on a molecule in the surface layer from all the other molecules produce a resultant directed downward. The closest neighbours, however, exert a force of repulsion on the molecule which is therefore in equi librium. Owing to the forces of attraction and repulsion, the density of the liquid is smaller in the surface layer than inside. Indeed, molecule / (Fig. 404) is acted upon by the force of repulsion from molecule 2 and the forces of attract ion from all the other molecules (3, 4, ...). Molecule 2 is acted upon by the forces of repulsion from 3 and / and the forces of attraction from the molecules in the deep layers. As a result, distance /-2 should be greater than 2-3, etc. This course of reasoning is quite approximate (thermal motion, etc., is disregarded), but nevertheless it gives a qualitatively correct result. An increase in the surface of the liquid causes new sections of the rarefied surface layer to appear. Here work should be performed against the forces of attraction between the molecules. It is this work that constitutes the surface energy. HEAT. MOLECULAR PHYSICS 285 358. The required pressure should exceed the atmospheric pressure by an amount that can balance the hydrostatic pressure of the water column and the capillary pressure in the air bubble with a radius r, The excess pressure is p+pgh+ 2a =4,840 dyne/ems, r 359. Since in this case p gh < 2a, , the water rises to the top end of the r tube. The meniscus will be a part of a spherical segment (Fig. 405): The radius of curvature of the segment is determined from the condition that the forces of surface tension balance the weight of the water column: 2Jtra cos q:> = rtr 2hpg. rhpa Hence. cos p= 20.'. It is obvious from Fig. 405 that the radius of curr 2a vature of the segment R=--=h-=O.74 mm. cos q> gp 360. When the tube is opened. a convex meniscus of the same shape as on the top is formed at its lower end. For this reason the length of the water column remaining in the tube will be 2h if I ~ h. and 1+h if t «; h. 361. (1) The forces of surface tension can retain a water column with a height not over h in this capillary tube. Therefore. the water will flow out of the tube. (2) The water does not flow out. The meniscus is convex. and will be a hemisphere for an absolutely wetting liquid. (3) The water does not flow out. The meniscus is convex and is less curved than in the second case. (4) The water does not flow out. The meniscus is Oat. ----------------@ -------@------- -------@------- ------@------- Fig. 404 Fig. 405 ------ 286 JI!!!I!IIII!!!!~~~""------- --_.....---_... ANSWERS AND SOLUT IONS -------~~~~I!!!!!!! -..------ - - - (aj -------. -#-- -~ ~ .r.__~ ... ---.- -- (b) Fig. 406 (5) The water does not flow out. The meniscus is concave. 362. The pressure p inside the soap-bubble with a radius R exceeds the atmospheric pressure by the amount of the double capillary pressure, since the bubble film is double: P=Po+ ~ • The pressure inside the bubble with a radius R together with the pressure of the section of the film between the bubbles should balance the pressure inside the smaller bubble. Therefore, ~a. +::= ~, where u, is the radius of curvature of section AB. Hence, Rx = RRr • -r At any point of contact the forces of surface tension ba lance each other and are mutually equal. This is possible only when the angles between them are equal to 120°. 363. According to the law of conservation of energy, the cross will not rotate. The components of the forces of surface tension are balanced by the forces of hydrostatic 'pressure, since the hydrostatic pressure of the water. higher than the level in the vessel is negative (see Problem 356). 364. If the bodies are wetted by water, its surface will take the form shown in Fig. 4000. Between the matches, above level MN, the water is tensioned by the capillary forces, and the pressure inside the water is less than the atmospheric pressure. The matches will be attracted toward each other, since they are subjected to the atmospheric pressure on their sides. For unwetted matches, the form of the surface is shown in Fig. 40Sb. The pressure betweE'n the matches is equal to the atmospheric pressure and is greater than the latter on the sides below level M N. .;~ HEAT. MOLECULAR PHYSICS 287 =M~ ~ ---- - -~-_-~---== -::-.!L- -_-=~L -= (a) N~ --- ------~~~~------~~~~~~ w- ----- (h) Fig. 407 In the last case two various forms of the surface correspond to the wetting angles when the matches approach each other (Fig. 407). One of them, however, cannot be obtained in practice (Fig. 407a). The pressure at level KL should be the same everywhere. In particular, the pressure of columns AB and CD of different height should be the same. But this is impossible, since the position of the column can be so selected that their surfaces are identicaJ in form. In this case the additional pressure of the surface forces will be the same, and' the hydrostatic pressure different. As a result, when the matches approach each other, the surface of the water between them will tend to assurne a horizontal form (Fig. 407b). In this case, as can be seen from the figure, the pressure between the matches at level MN is equal to the atmospheric pressure. The pressure exerted from the left on the first match is also equal to the atmospheric pressure below level MN. The pressure acting on the second match from the right is less than the atmospheric pressure above level JWN. As a result, the matches will be repulsed. 2-5. Mutual Conversion of Liquids and Solids 385. Water will freeze at zero only in the presence of centres of crystallization. Any insoluble particles can serve as such centres. When the mass of the water is great, it will always contain at least one such centre. This will be enough for all the water to freeze. If the mass of the water is divided into very fine drops, centres of crystallization will be present only in a comparatively small number of the drops, and only they will freeze. 288 ANSWERS AND SOLUTIONS c 386. The water and the ice receive about the same amount of heat in a unit of time, since the difference between the temperatures of the water and the air in the room is approximately the same as that for the ice and the air. In 15 minutes the water receives 200 calories. Therefore, the ice receives 8,000 calories in ten hours. Hence, H =80 cal/g. 367. 0=2,464 m/s. 368. The heat balance equations have the form Q1 =mtcIL\#+CAt ~t At Q2=mlclT+mtH+mlc2T+C&t where ml and Cl are the mass and heat capacity of the ice, C Is the heat capacity of the calorimeter, c2 is the heat capacity of the water, and ~t=2°C. Hence, Q ( C2 HI) Q1 -+--+- - 2 2c1 cl~t 2 150 lid CI ~t At H :z: ca eg "C;T-T+C;- 369. The amount of heat that can be liberated by the water when it is cooled to O°C is 4,000 cal. Heating of the ice to O°C requires 12,000 cal. Therefore, the ice can be heated only by the heat liberated when the water freezes. .One hundred grammes of water should be frozen to produce the lacking 8,000 calories. As a result, the calorimeter will contain a mixture of 500 g of water and 500 g of ice at a temperature of O°C. 370. The final temperature of the contents in the vessel is O=O°C. The heat balance equation has the form m1c1(tt-8)=mtcl (8-t2)+ (m2 - ma) H where m1 Is the sought mass of the vessel and c, is the heat capacity of the ice. Therefore, m,Ct (9-tt )+ (mt - ms) H 200 ml = CI (tl - 6) g 371. (1) The sought mass of the ice m can be found from the equation mH =Mc (-t). Hence, m= 100 g. (2) The heat balance equation can bewritten in this case as MH = Me(- t). Hence, t=-80°C. 372. The melting point of the ice compressed to 1,200 atm will drop by At =8.8° C. The ice will melt until it cools to -8.8° C. The amount of heat Q=mtH is absorbed, where m\ is the mass of the melted ice and H is the specific heat of fusion. From the heat balance equation m1H=mc~t, where c is the heat capacity of the ice. Hence, cmtu oc: m1 =n - 5.6 g HEAT. MOLECULAR PHYSICS 289 2 -6. Elasticity and Strength 373. F AE (R-r) 60 kgf. r • 374. When the rod with fastened ends is heated by t degrees, it develops an elastic force F equal, according to Hooke's law, to F = AEAI = AEaJ I where E is the modulus of elasticity of steel and (X is its coefficient of therrnaI expansion. If one of the rod ends is gradually released, the length of the rod. will increase by Al= lat. The force will decrease linearly from F to zero and its average magnitude will be F/2. F I The sought work W=2 Al = 2 AEla2 t2 • 375. The tension of the wire T=2 ~g • It follows from Hooke's law sina Ai that T =2f EA. Since Al=2 (_/__ 1) , then T I-cosa AE=2~g . At small ancos ex cos ex. Sin a gles, sin a ~ a, and cos a= 1-2 sinS ~ e; 1-~s • Bearing this in mind, we obtain VMg cx= AE 376. The rod heated by At would extend by Al=loaAt in a free state, where Lo is the initial length of the rod. To fit the heated rod between the walls, it should be compressed by 111. In conformity with Hooke's law, IF Al=EA Therefore, F = E AaAt = 110 kgf. 377. When the rods are heated in a free state, their total length will increase by Al=All+Al,=(al'l+a21z)t. . Compression by the same amount ~l will reduce the lengths of the rods by ~l; and AI;, where AI; +~l;=L11. This requires the force F - ElA AI' EzAA1; - '1 t 12 Upon solving this system of equations, we find that F allt +asll At .!L+~ E) £2 The rods will act upon each other with this force. 19-2042 290 ANSWERS AND SOLUTIONS 378. It is obvious from cons iderations of symmetry that the wires will elongate equally. Let us denote this elongation by ~I. On the basis of Hooke's law, the tension of a steel wire Fs=~' AEs and of a copper one Al Fe=-l- AEc· It follows that the ratio between the tensions is equal to the ratio between he respective Young's moduli . Fe Ee 1 p;= E$ ="2 In equilibrium 2Fc +Fs=mg. Therefore, Fc ="'4g = 25 kgf and Fs = 2Fc= 50 kgf. 379. On the basis of Hooke's law, ~l ~l Fe=-l- AcEc and Fi=-l- AiE; F It follows that F~ = 2. Thus, two-third~ of the load are resisted by the concrete and one-third by the iron. FI 380. The compressive force F shortens the tube by A E and the tensile FI c C force F extends the bolt by A E · s s The sum A~~s+A~. is equal to the motion of the nut along the bolt: ..f!.-+~=hAsEs AcEc Hence, F -!!:.. AsEsAcEc - I AsEs+AcEe 381. Since the coefficient of linear thermal expansion of copper ac Is greater than that of steel as,. the increase in temperature will lead to compression of the copper plate and tension of the steel ones. In view of symmetry, the relative elongations of all the three plates are the same. Denoting the compressive force acting on the copper plate from the sides of the steel plates by F, we shall have for the relative elongation of the copper plate: ~'= F = act- AE . c Either steel plate is subjected to the tensile force F/2 from the side of the copper one. Upon equating the relative elongation of the plates, we obtain: HEAT. MOLECULAR PHYSICS 291 8 Fig. 408 On upper nut On lower nut O=Fgo F Fpllf----.-, Hence F _2AEcEs «(Xc-as) t - 2Es+Ec mvl 382. When the ring rotates, the tension T = 2- appears In It (see Prob- nr tern 201). For a thin ring m=2nrAp, where A is the cross section of the ring. Therefore. ~ =pv·. Hence. the maximum velocity v= V~Q!;~l m/s. :$83. Initially, an elastic force Po acts on each nut from the side of the extended bolt. . The load G~ F0 cannot increase the length of the part of the bolt between the nuts and change its tension. For this reason the force acting on the upper nut from the side of the block will not change as long as G~ FeThe lower nut is acted upon by the force F0 from the side of the top part of the bolt and by the force G from the bottom part. Since the nut is in equilibrium, the force exerted on it from the block is F=Fo-G. Thus the action of the load G~ F0 consists only in reducing the pressure of the lower nut on the block. When G > FOI the length of the bolt will increase and the force acting on the lower nut from the side of the block will disappear. The upper nut will be acted on by the force G. The relation between the forces acting on the nuts and the weight of the load G is shown in Fig. 408. . 2·7. Properties of Vapours 384. The calorimeter will contain 142 g of water and 108 g of vapour at a temperature 1000 C. 385. By itself, water vapour cr steam is invisible. We can observe only a small cloud of the finest drops appearing after condensation. When the gas burner is switched off, the streams of heated air that previously enveloped 19 ': 292 ANSWERS AND SOLUTIONS the kettle disappear, and the steam coming out of the kettle is cooled and condenses. 386. On the basis of the equation of state of an ideal gas p= ~ = ;';. . If the pressure is expressed in mm Hg and the volume in m3, then R:::;:: 760X 0.0224 mm Hg·ms = 273 deg.mole· 273 Therefore, p= 1.06 PT' At temperatures near room temperature, p~p g/ms. 387. It seems at first sight that the equation of state of an ideal gas cannot give values of the density or specific volume of saturated vapours close to the actual ones. But this is not so. If we calculate the density of a vapour by the formula p= ~ = k~ and compare the values obtained with those in Table 2 (p. 85), we shall observe good agreement. This is explained as. follows. The pressure of an ideal gas grows in direct proportion to the temperature at a constant volume of the gas and, therefore, at a constant density. The relation between the pressure of saturated vapours and the temperature depicted in Fig. 146 corresponds to a constant volume of a saturated vapour and the liquid which it is in equilibrium with. As the temperature increases, the density of the vapeur grows, since the liquid partially transforms into a vapour. An appreciable increase in the mass of the vapour corresponds to a small change in the volume it occupies. The pressure-density ratio becomes approximately proportional to the temperature, as with an ideal gas. The Clapeyron-Mendeleyev equation mainly gives a correct relationship between p, V and T for water vapour up to the values of these parameters that correspond to the beginning of condensation. This equation, however, cannot describe the process of transition of 8 vapour into a liquid and indicate the values of p, 1> and T at which this transition begins. 388. At 300C the pressure of saturated vapours p=31.82 mm Hg. According to the equation of state of an ideal gas, V=.!!!. RT e: 296 lit 1-1 p 389. When the temperature gradually increases, the pressure of the water vapours in the room may be considered constant. The vapour pressure p= ~;;o corresponds to a humidity of Wo=10 per cent, where Po= 12.79 mm Hg is the pressure of the saturated vapours -at 15° C. At a temperature of 25°C the pressure of the saturated vapours is PI =23.76 mm Hg. For this reason the sought relative humidity is w=L l00o/o=~=5.40/o PI PJ' 390. According to the conditions of the problem, the relative humidity outside and in the room is close to 100 per cent. The pressure of saturated water vapours outside, however, is much smaller than in the room, because the temperature of the air in the room is higher and much time is required to equalize the pressures owing to penetration of the vapours outside through HEAT. MOLECULAR PHYSICS 293 slits. Therefore, if the window is opened, the vapours will quickly flow out from the room and the washing will be dried faster. 391. (1) The water levels will become the same as in communicating vessels. The water vapours in the left-hand vessel will partly condense. and some water will evaporate in the right-hand vessel. (2) The levels will become the same because the vapours will flow from one vessel into the other. At a given temperature the pressure of saturated vapours is identical in both vessels at the surface of the water and will decrease at the same rate with height. For this reason the pressure of the vapours in the vessels at the same level is different. This causes the vapour to flow over and condense in the vessel with the lower water level. 392. When t2 = 30° C, the pressure of the vapours is equal to the pressure P20 of saturated vapours (P20=31.8 mm Hg) only if the air pressure is 10 at. Upon isothermal reduction of the air pressure to one-tenth, the volume of the air will increase ten times. Hence, at atmospheric pressure and a temperature of 30° C, the pressure of the water vapour is p= 3.18 mm Hg. It follows from the Clapeyron equation that at a temperature of 11 = 10°C, the T vapour pressure PI = PT:' where TI = 283° K and T1= 303° K. The sought relative humidity is w=Pt 1000/0 = P.. TTl 1000/0 ~ 32.60/0 Po Po 2 where Po=9.2 mm Hg is the pressure of saturated vapours at t1 = 10°C· 393. The pressure p=6.5 mm Hg is the pressure of saturated water vapours at t=5° C. A sharp drop of the pressure shows that all the water has been converted into vapour. The volume of the vapour pumped out until the water is evaporated completely is V=3,600 litres. On the basis of the Clapeyron-Mendeleyev equation of state, the sought mass of the water is v . m=P R / !!!!S. 23.4 g 394. An amount of heat Q1 =mc At =3,000 cal is required to heat the water to 100°C. Therefore, Q2= Q- QI =2,760 cal will be spent for vapour formation. The amount of the water converted into vapour is ml = Q =J=5.1 g. r In conformity with the equation of state of an ideal gas, this vapour will occupy a volume of V =m 1 RT. Upon neglecting the reduction of the YOJ.L P . lume occupied by the water, we can find the height which the piston is rai- V sed to: h=;r= 17 ern. CHAPTER 3 ELECTRICITY AND MAGNETISM 3-1. Electrostatics Q2 395. F=7=918 kgf. The force is very great. It is impossible to impart a charge of one coulomb to a small body since the electrostatic forces of repulsion are so high that the charge cannot be retained on the body. 396. The balls will be arranged at the corners of an equilateral triangle with a side ~31. The force acting from any two balls on the third is 4Q2 £=12 Y3' The ball will be in equilibrium if tan a,= f- (where a=300). Hence, mg I Q=2 Ymg~ 100CGSQ_ 397. Since the threads do not deflect from the vertical, the coulombian force of repulsion is balanced by the force of attraction between the balls in conformity with the law of gravitation. Therefore, in a vacuum Q2 p2V. -;:2=Y7 and in kerosene (taking into account the results of Problem 230) Q2 (p-PO)2 Vi E r r2 = 'V r 2 where V is the volume of the balls. Hence, P= Po yer ~ 2.74 g/cm8 Ye,.-l 398. The conditions of equilibrium of the suspended ball give the following equations for the two cases being considered: T - QQs Y2 0 1 sin (Xt - 2a2 X -2-= T +QQs Y2 QQs 0 1 cos al 2a2 X -2--7- mg= T · QQs V2 0 2 srn (X,2- 2a2 X -2-= T2cosa2+QQs-QQsx V2 -mg=O a2 2a2 2 ELECTRICITY AND MAGNETISM Fig. 409 Fig. 410 295 where T1 and T2 are the tensions of the thread, al and ~ the angles of deflection of the thread, +Q and ~ Q the charges of the fixed balls, + Qs the charge of the suspended ball, and mg is the weight of the suspended ball (Fig. 409). Upon excluding the unknowns from the above simultaneous equations, we get cot a1 -cot a 2=cot ai-cot 2al =2 (V2-1) whence, cot al =2 (2 Y2 - 1) ± V35-16 Y 2 Thus, C&J. =7°56' and a,=15°52' when mg > ~~s ( 1- -':2), and al=82°04' and a 2=164°OB' when mg < ~~s (1- -':2). 399. In uniform motion the drop is acted upon by the force of gravity G. the expulsive force of the air (Archimedean force) F, the force of the electrostatic field eE and the force of friction against the air kv=k ~ . All the forces are balanced. Therefore, G-F-eEO+k..!..=O t1 G-F+eE-k~=O t; G-F-k!..=O t where e is the charge of the drop, E the intensity of the electric field, and s the distance covered by the drop. 296 ANSWERS AND SOLUTIONS Upon solving the equations, we get t= 2tlt2 tl-t9 Fig. 411 400. It can, if we use the phenomenon of electrostatic induction. Bring a conductor on an insulated support up to the charged body and connect the conductor to the earth for a short time. The conductor will retain a charge opposite in sign to the given one, while the like charge will pass into the earth. The charge can be removed from the conductor by. introducing the latter into a metallic space. The operation may be repeated many times with a charge of any magnitude. Electrostatic machines operate on a similar principle. 401. The energy is produced by the mechanical work that has to be performed in moving the conductor from the oppositely charged body to the body .that accumulates the charge. 402. They can, if the charge of one ball is much greater than that of the other. The forces of attraction caused by the induced charges may exceed the forces of repulsion. 403. Since Q ~ q, the interaction between the separate elements of the ring can be neglected. Let us take a small element of the ring with a length RA.a, (Fig. 410). From the side of the charge Q it is acted upon by the force IJ.F =~~q • where t1q=q:na.. The tension forces of the ring T balance IJ.F. From the condition of equilibrium, and remembering that AeJ is small. we have t1F=2T sin (IJ.;) • rs« The sought force is the tension T =::.~.· 404. Let us consider the case of opposite charges Q1 > 0 and Q, < O. The intensities created by the charges Q1 and Qz are equal, respectively. to £1 = ~l and £.= Q.- . A glance at Fig. 411 shows that ,] '1 EI =Er+£:-2E1E, coscp From triangle ABC Hence, cosq> ,I+,:-dl 2'1" ELECTRICITY AND MAGNETISM 297 A 8 a Fig. 412 Fig. 419 If the charges are like E-VQ~ + Q: +QIQI( I 2 - 4 • 8 a rt+'t- dl) '1 '1 '1'. 405. Each charge creates at point D a field intensity of E1 = ~ . The a total intensity will be the sum of three vectors (Fig. 412). The SUIO of. the horizontal components of these vectors will be zero, since they are equal in magnitude and form angles ~of 120° with each other. The vectors form angles of 90°-a with the vertical, where a is the angle between the edge of the tetrahedron and the altitude h of triangle ABC. The vertical components are identical, each being equal to ~ sin ---ar- or Q EO= R2 be V= ~ =EoR=30,OOORV if the radius of the s~hereThe potential wilt is in cen timetres. 416. If the charged body is placed into the centre of the sphere, an additional charge -q uniformly distributed over the surface will obviously appear on the external surface of the sphere, and a charge +q on its internal surface. The potential VR at a distance of R from the centre of the sphere will be 415. The maximum charge that can be imparted to the sphere Is determined by the equation When the body moves inside the sphere, the outside field will not change. Therefore, the potential will be VR at any position of the charged body in the sphere. 417. The housing and the rod connected by the conductor will have equal potentials. For this reason the leaves will not deflect. After the conductor is removed and the rod is earthed, both leaves will deflect. This is the result of a potential difference appearing between the rod and the housing, since the work of the electrostatic field is zero when the charge moves along a closed path ABCDEFA shown by the dotted line in Fig. 415. The work on section AC is zero, and the work on section AB is equal to that on BC taken with the reverse sign. The potential difference between the earth and the housing is equal to that between the housing and the rod. 418. When the housing of the electrometer is given, for instance, a positive charge, electrostatic induction will charge the ball of the electrometer rod positively and the end of the rod negatively. A potential difference will appear between the housing and the rod, and for this reason both leaves will deflect. The potentials of the housing and the rod are positive with respect to the earth,. the potential of the housing being higher (that of the earth may be considered as zero). When the rod is connected to the earth, the potential difference between the rod and the housing will increase, as can be proved by the method used in Problem 417. Therefore, the angle of deflection of the leaves will be greater. 419. The electrometer measures the potential difference between the given body and the earth. Since the surface of the wire is equipotential, the leaves will deflect in the first case through the same angle wi th the ball in any position (if the capacitance of the wire is negligibly small). 302 ANSWERS AND SOLUTIONS Q! We=-Q

n =0, since the potentia of an uncharged sphere is zero. The average potential of the sphere is cP=~. Upon multiplying it by the initial charge, we obtain the energy of a charged sphere: ELECTRICITY AND MAGNETISM 303 The same result can be obtained if we use a diagram showing the change in the potential of the sphere as the charge decreases. The diagram will have the form of a straight line passing at an angle to the axis of the charges, and the work will be numerically equal to the area limited by the diagram and the axes. Q2 RU" 423. The energy of the charged sphere We=2R =-2-' where R is the radius of the sphere and U its potential (see Problem 422). Upon a discharge this energy will be liberated as heat. Expressing the energy in calories, we obtain We=0.13 cal. 424. The potential difference between the balls should be WI' Besides, the heat Q is liberated in the conductor. According to the law of conservation of energy, however, the total amount of energy in the balls should be the same in both cases. Since the work WI and, correspondingly, W 2' is the potential energy of the second ball in the field of the first one in the first and second cases, then W1 +Wet = W2 +Q+We2 2 2 where Wcl = ~: + ~; is the intrinsic energy of the balls before connection 2 q2 and We2 = ;, + 2, is the intrinsic energy of the balls after the charges are redistributed (see Problem 422). The energy liberated as heat is (qt -q2)2 (J.._-!-)Q = Wel - We2 +W1 - W2 4 r I 426. Assume that the radius of the envelope increases by 6, which may be an infinitely small quantity. The expanding force will perform the work W=4nR2f6, where f is the force per unit of area. This work is done at the expense of a reduction in the electrostatic energy. First the electrostatic 304 ANSWERS AND SOLUTIONS Fig. 416 energy is 2 QR 2 , and after expansion Q2 2 (R+6) The change in the energy Q2 QI QI 6 2R 2(R+6) 2 R(R+~) is equal to the work W, I. e., Q2~ 4nR'f~ 2R(R+6) Taking into account the fact that 6 can be infinitely small, we obtain the following expression for the force: f- Q2 -2'"'1'1'2 -8nRt- ".v Here, (1= 4~~ is the density of the electricity, Le., the charge per unit of area. The sought force can also be found directly. Let us consider a small area a on the sphere (Fig. 416). Let us find the intensity £1 of the electric field created on the area being considered by all the charges except the ones on the area itself. To introduce definiteness, let us consider the case when the sphere carries a positive charge. Let us denote by £2 the intensity of the electric field created by the charges on the area itself. Since the resulting intensity is zero inside the sphere, then £1 =£2· Q The resulting intensity on the sphere E1 +EJ =W' and, therefore, Q2£1 =W =4no. Hence, £1=2na. To find the force that acts from all the charges outside t he area oft the charges on the area, the intensity E1 should be multiplied by the magnitude of the electric charge of the area aa: F=E1oa=2no sa The force per unit of envelope area will be f=21tCJ1 • 427. For the charge q to be in equilibrium, the charges -Q should be at equal distances a from it (Fig. 417). The. sum of the forces acting on the charge - Q is also zero: QS Qq 4a2 -Qi'=O Hence, q= ~ . The distance a may have any value. Equilibrium is unstable since when the charge -Q is shifted along 001 to the left over a distance x, the force of attraction Q2 Fg= 4 (a+x)1 ELECTR ICITY AND MAGNETISM 305 acting from the side of the charge q is less than the force of repulsion Q2 F Q=(2a+x)2 and the charge -Q moves still farther from the position of equilibrium. When the charge -Q is shifted along 001 over a distance x towards the charge q, then Fq > F Q for x <; a, and the system does not return to the position of equilibrium. As can easily be seen, equilibrium is also violated by arbitrary motion of the charge q, The potential energy of the charge -Q in the field of the other two charges is ( q Q) QI 3y-a We1=-Q y-a+y =Ty(a+y) where y is the distance between the charge q and one of the charges - Q. When 0 es;; y < 00, the relation between Wel and y is shown by curve ABC for one charge and curve DEF for the other (Fig. 417). When the charges -Q are stationary, the energy of the charge q is ( -Q Q) Q2 a We2 = q a=z-a+z = -2 a2+z2 where z is the displacement of the charge q from the position of equilibrium. When z changes from 0 to a, the energy changes according to curve MNP (Fig. 417). It is interesting to note that the maxima of all the three potential curves correspond to the position of the charges in equilibrium. It is for this reason that equilibrium is not stable. 428. The work performed by the field of induced negative charges when the charge +q moves is equal to the work done by the field of the charge -q (see Problem 413). The work performed during the motion of both +q and -q is ;~. Hence, the sought kinetic energy of the charge, equal to the work of motion of only one charge, will be :~. . FiB. 417 20-2042 306 ANSWERS AND SOLUTIONS + + + 0 c + + + =- A 9 Fig. 418 0' Fig. 419 0' c' 429. Let us first prove that the intensity of the electric field at all points in the plane of section 00' is. directed perpendicular to this plane. Let us take an arbitrary point in the plane of the section and two small areas arranged -arbitrarily but symmetrically on the cylinder with respect to section 00'. It is easy to see that the resulting intensity of the field induced by the charges on these areas will be directed along the axis of the cylinder (Fig. 418). Since another element arranged symmetrically with respect to the plane of the section can be found for each element. it follows that the intensity produced by all the elements will be parallel to the axis of the cylinder. Let us now prove that the intensity will be the same at all points equidistant from the axis of the cylinder. Let A and B be two such points (Fig. 419). The intensity of the field inside the cylinder will not change if, apart from the available charge, each square centimetre of the cylinder surface receives the same additional negative charge so that the density of the charges at point C is zero. This is obvious from the fact that the field inside an infinite uniformly charged cylinder is zero. In this case the densities of the charges will be distributed on the cylinder surface as shown in Fig. 152. Therefore, the intensities at points A and B are the same. It now remains to prove that the intensities of the fields at points at different distances from the axis of the cylinder are identical. For this purpose let us consider circuit BKLD (Fig. 420). With an electrostatic field the work in a closed circuit is known to be zero. The work is zero on sections KL and DB because the intensity of the field is perpendicular to the path. The work on section BK is -EBI and on section LD it is EDl (as proved above, EB=EI(. and ED=EL ). Hence, -EBl+EDl=O, i. e., EB =ED It has thus been proved that the intensity of the electric field at all points inside the cylinder will be the same everywhere and directed along the axis of the cylinder. It should be noted that such an arrangement of the charge on the surface of a conductor appears when direct current passes through it. 430. The concept of capacitance can be used because the ratio between the charge imparted to a conductor and the increment of the potentia1 indu- ELECTRICITY AND MAGNETISM o = Fig. 420 307 Hence, bearing in mind that the capacitance of the sphere is equal to its radius, we find that c= V3X 6 X 10 4 e: 25 cm 4n Measurements give a close value: C a: 30 em. 432. The electrometer will show the e. m. f. of the galvanic cell irrespective of the capacitance. 433. (I) U=4Ead=8.4 X lOC V. (2) A voltage of U1 = Ead= 2.l X 10· V can be supplied to each "air" ca- pacitor. The charge on the capacitor will be Q= C1U1 = ~~. Upon series connection, the charge of all the capacitors is the same. Therefore, the voltage on the capacitor with glass will be U2 =cQ =~. The entire battery may 2 e,. thus receive not more than U=3U1+U2=6.6Xl04V. If the voltage exceeds 6.6X 10. V, all four capaci tors will be punctured. The capacitor with the glass dielectric will be punctured last. 434. When a charge moves in a closed circuit, the work of the forces of an electrostatic field is zero. Therefore The charges on the capacitors are the same, since the sum of the charges present on the upper and the lower conductors is zero. Hence, Q=C1U1 = = C2U2• 20 * 308 ANSWERS AND SOLUTIONS + - +0-~a.------....------. Fig. 421 Therefore, u1=e C2 c (cCl+cC2)=17.5 kV 1+ 2 u2 = C C t e <111 + We. The difference in the energies is Weo-We=c~f~s (U~+U:-2U1U2) > 0 When U1=U2, we have Weo- We=O, and when C1=C2 and U2=O, then Weo=2We· The electrostatic energy diminishes because when the capacitors are connected by conductors, the charges flow from one capacitor to the other. Heat is liberated in the connecting conductors. The quantity of heat evolved will not depend on the resistance of the conductors. When the resistance is low, the conductors will allow greater currents to flow through them, and vice versa. ELECTRICITY AND MAGNETISM 311 442. Since the dielectric is polarized, the intensity will increase at points A and C and decrease at point B. 2nQ 443. E=-A-=50.2 CGSE. t, 444. The capacitances and, therefore, the charges of the balls immersed in kerosene increase e, times: q~=e,ql' and q;=s,q. The force of interaction of the charges in the dielectric, on the contrary, diminishes e, times. Hence, q q e q q e,~2rllr! F =_1_1_=.L...!....!.. ~. 0 0088 d 8,R2 R2 R2(rl +r2)2 . yne The force of interaction increases e, times, while if the balls were disconnected from the battery it would decrease e, times. 445. As a result of motion of the plates, the charge on the capacitor will be increased by AQ=QI- Q,=!A (..!..._..!...)431 d. d1 tlA ( 1 1 )The battery will perform the work W = '~Q = 4n d. - d 1 The electrostatic energy of the capacitor will be increased by AW,=Wes-Wet =6~1_'~1=';~ (~I - ~J The mechanical work WI was performed when the plates were moved closer .to each other. On the basis of the law of conservation of energy, W =W1 +AWe. Therefore, W1=W-AW =,IA (..!..._..!...)e 8n d, d1 At the expense of the work of the battery, the electrostatic energy of the capacitor increased and the mechanical work W1 was done. 448. Let us consider for the sake of simplicity a dielectric in the form of a homogeneous very elongated parallelepiped (Fig. 426). Let us resolve the field Eo in which the piece of dielectric (for example, 8 rod) is placed into components directed along the rod and perpendicular to it. These components will cause bound charges to appear on surfaces AB, CD, Be and AD. The field of the bound charges between surfaces AD, BC, and AB. DC weakens the components' of the field Eo inside the dielectric, the component perpendicular to the rod being weakened more since the bound .charges on surfaces AD and Be are close to each other and their field is similar to the homogeneous field of a plane capacitor, while the charges on the surfaces of the small area are moved far apart. For this reason the full field inside the dielectric will not coincide in direction with the field Eo. The dipoles appearing will therefore be oriented not along Eo, but along a certain direction OP forming the angle p with Eo. (This refers to both ordi~ary and dipole molecules.) From an electrical standpoint, a polarized dielectnc can be regarded as a large dipole forming the angle ~ with the field Ea. 312 ANSWERS AND SOLUTIONS A JJ a Fig. 426 The dielectric will rotate in this field until it occupies a position along the field. The field of the bound charges is an internal force and cannot cause rotation of the dielectric. 447. (a) The capacitance of the capacitor will be equal to that of capacitors connected in parallel, one of which is filled with a dielectric, and the other not, i. e., C=ErAl1 + A (1-11) A {1+< -1)2}4n.dl 4ndl 4nd 8, I (b) The electric field between the plates of the capacitor will not change, and, consequently, the capacitance will not change if the upper surface of the dielectric is coated with an infinitely thin layer of a conductor. Therefore, the sought capacitance will be equal to the capacitance of two capacitors connected in series: c Hence, COCl where Co A and C _ erA CO+C1 ' 4n (d-d1) 1- 4M1 C erA 4", {d1 +e,. (d-d1) } 448. To simplify reasoning, assume that two parallel metallic plates carrying the charges +Q and -f2 are placed into a liquid dielectric. The intensity of the electric field between the plates is E=41t A Q • The intensity of e, the field induced by each plate will be. E1=E2=21t A Q 8r Let us find the force acting, for example, from the side of the first plate on the second. For this purpose the intensity of the field induced by the first plate should be multiplied by the charge on the second one. Thus, F= 2nQ2 erA Let us assume that the first plate is fixed and the second can slowly move (we disregard the change in the mechanical energy of the dielectric). The work that the electric field can perform in moving the plates up to direct contact Is equal to the product of the force F (constant) by the distance d, ELECTRICITY AND MAGNETISM 313 i. e., + + + + W=Pd=21tQ A 2 d 8r This work is done at the expense of a reduction in the electric energy of the capacitor. Thus, the electrostatic energy will be equal to 2:nQ2d Q2 QU we=e;r=2C o~ We=T where U is the potential difference. The above formula is true for any dielectric. 449. The intensity of the field . inside the space consists of E and Flg. 427 the intensity induced by the charges that appear on the surfaces of the space owing to polarization of the dielectric (Fig. 421). In the first case the influence of the polariziof charges is negligibly small, and E1 =E. In the second case the action 0 the charges exerted on the surfaces of the space is fully compensated oy the action of the charges on the surfaces of the dielectric adjoining the metal plates of the capacitor. For this reason the sought intensity is equal to that which would be induced if the dielectric were removed altogether, i. e., E2 = 8,E 450. If the dielectric is drawn into the capacitor over the distance x, the energy of the capacitor will be QI AU2 1 Wel =2C = 8nd x 1+(e,-I)T since C= 4~ {1+(8,-1) : } and Q= 4~ U (see Problem 447). If x is increased by 6, the energy will be reduced and become equal to AU' 1 W'et=81td 1+(8r-l)'~+6 . I The difference between the energies 6 AUI (8,-1)7 11' e1 - 11' 61 = 8nd {1+(8,-I)xt6}{I+(8,-I)~} will be equal to the work of the sought force F over the path 6. Generally speaking, the magnitude of the force will change over this path, but if a 314 ANSWERS AND SOLUTIONS sufficiently small value of ~ is taken we can write We1 - Wel=F6 It follows therefore that F_AU2 8,.-1 -SMi {l+O '1 '2 '1 '2 (b) /=<81; VBA=O '1 The potentials of the conductors connecting the elements are the same, but the current is not zero. (c) J Il- O. For the other circuit "2 = V~- V~ =(<82 +12' 2) + 'Z 501. In this case the potential of point A, when the key is closed, is higher than that of point B, since when the key is openVB > VA. For this reason 1/,,-/1' 2= Ul ' The other equations have the same form as in the s0lution of Problem 500. Therefore, "2 2 O. Therefore the following inequalities should exist or 504. There are two methods of connecting the storage battery cells. Either the batteries are connected in series in the separate groups, and the groups themselves in parallel, or vice versa. Denoting the total number of cells by N and the number of cells in a separate group by n, we shall have in the first case: J1 nCo Co R+,n 2 .B..+rn N n N since the e. m. f. of one group is nCo, the resistance of the group is rn and the number of groups connected in parallel is NIn. The current /. reaches its maximum if ~ + ~ n is minimum. The minimum of an expression of the type ax+!- can be found as follows The relationship x b y=ax+X- (I) Is shown graphically by the curve in Fig. 446 which has its minimum at point xo, at which the roots of quadratic equation (1) coincide. For this reason, Therefore, n=V~=4 and 11max=~O V:,=20 A 332 9 !/min ANSWERS AND SOLUTIONS IIOv Fig. 446 In the second case N - 0, I, e., when V > cfJlt the second at V > li". and the third at V > <830 For this reason, the diagram showing the full current versus the voltage will have the form of a broken line (Fig. 4 5 3 ) : ' . I=OatV<.G l=k(V-. ted in these units. The dimension of the coefficient can be found directly 342 ANS'WERS AND SOLUTIONS .Irorn the formula for the intensity H: [k] _ [H] [l] - [I] Bearing in mind that (H]=(E) and (E)= ~~~ =g". ocm-'I'os-l. (ll=cm. and [/]=gl/·.cm3/ 1.s- 2• we obtain [k]=s/cm. If a newconstant c is introduced instead of k. so that k =-.!.. . then c will be c equaI to the velocity of Hght in a vacuum. 545. The intensity of the field induced by the first winding is HI o.~::::1 400Oe The second winding induces the intensity H O.4n1N2 -200 0 2 2nR - e Since the fields HI and H'}, are directed oppositely. the sought field will be H=H 1-H2=200 Oe. 546. The conductor Be does not induce a field at point M lying on the continuation of Be. According to the rule giveil in the note, the magnetic field produced by any elements of conductor Be should be perpendicular to line 8M. For this reason the presence of a field other than zero at point M would disagree with the symmetry of the problem. because all the directions perpendicular to 8M have equal rights. Since the intensity of the field is proportional to the current. then H1 =kl without conductor BD. The fields from the conductors AB and BD are summated. Therefore after conductor BD is connected kJ H2=kl+7j H2 3 whence HI ="2. 547. At an arbitrary point on line AB any small element 0 f current of conductor ACB induces a magnetic field perpendicular to the plane of the drawing (see Problem 546). The element of conductor ADB symmetrical to it induces the same field, but directed oppositely. For this reason the field from any two elements arranged symmetrically will be zero. Hence, the field at the arbitrary point on AB induced by the entire conductor is zero. since the straight sections of the conductor also do not induce a field on AB. 548. In the main, the field of the solenoid will be concentrated inside the toroidal winding and will not act on the magnetic pointer. A singlelayer winding, however. can simultaneously be regarded as one turn of a large radius that induces a magnetic field perpendicular to the plane of the torus. The magnetic pointer will be positioned along the axis of the torus. The direction of its poles can be determined by the right-hand screw rule. 549. The current flowing along the pipe can be regarded as the sum of a great number of identical straight currents uniformly distributed over the ELECTRICITY AND MAGNETISM Fig. 455 343 Fig. 456 surface of the pipe. The intensity of the magnetic field at any point of space can be represented as the sum of the intensities of the fields induced by these currents. Figure 455 shows a cross section of the pipe along which the current flows. Let us compare the" intensities of the magnetic fields HI and H2 created at point A by the straight currents /1 and /1 passing through the small aRt a,R arcs 8 1 and 82. The lengths of the arcs are 81 =-- and 82 = - _ 2_ , cos' <1'1 cos The current in the pipe is distribu. / 8 / I ted uniformly, and therefore /1 =81 , whence R1 = R2 • • 2 2 I 2 The intensities of the magnetic fields created at point A by the currents /1 and /2 are equal, namely, J / HI=k RI =k R1 =H, and directed oppositely. Since a corresponding element that compensates completely ior the magnetic field of the first element at point A can be selected for each element of the pipe cross section, the resulting magnetic field of the current flowing in the pipe will be zero at any point inside the pipe. " __550. The conductor with the space is equivalent to a solid conductor ~ich carries 3 current with a density i, while through the volume that corresponds to the space there also flows a current of the same density in the opposite direction. The total current in this volume will be equal to zero, and this shows that a space exists in the solid conductor. 344 . Fig. 457 ANSWERS AND SOLUTIONS Fig. 458 The field created by the current with a density I at an arbitrary point of space A is equal to HI =k·2njR (Fig. 456). Here R is the distance from the axis of the conductor to point A. (It is assumed that the current flows toward us.) At the same point, the current flowing through the volume. corresponding to the space, but in the reverse direction, induces a field of H2 =k· 2njr. Figure 456 shows that the total intensity of the field is H=VH:+H:-2H1Ht cos ex Obviously, Therefore, the intensity H =k·2njd is the same for all the points of the space. . 551. Triangle AOC is similar to triangle BAD (Fig. 456) since they have one equal angle, and the sides confining these angles are proportional.' Therefore, L AOC=L BAD. But R l..H1, and therefore H l..d. The intensity of the magnetic field at any roint of the space is perpendicular to the line that connects the centres 0 the conductor and the space. The distribution of the lines is shown in Fig. 457. 552. k=..!.., where c is the velocity of light in a vacuum. c 553. No, it will not. Forces of attraction exist between the separate elements of the current. As a result, the density of the current increases somewhat toward the axis of the conductor. The effect is negligible. 554. When lightning strikes. a very high current flows for an instant through the pipe and the separate elements of the current are mutually attracted with a high force. It is this force that converts the pipe into a rod. ELECTRICITY AND MAGNETISM 345 555. The currents in the adjacent turns are parallel and flow in the same direction. For this reason the turns will be mutually attracted. At the same time the" currents in opposite sections of the turns flow in different directions. Therefore, opposite sections. are repulsed. . The turns of the winding wi11 tend to increase in diamet-ert and the distance between them along the axis of the solenoid will be reduced. HnAJ sse. a=-k-' 567. The action of the magnetic field will cause the ring to so turn that the force lines of the field are perpendicular to the plane of the ring and form a right-hand screw with the direction of the current. The tension of the ring will be maximum. Upon employing the method used to solve Problem 403, we obtain F=kJRH=5 dynes 558. The element of the ring 111 is acted upon by the force I1F=kJHI1l (Fig. 458). Let us resolve it into the components AF1 and AI. The component AF1 lies in the plane of the ring and At = !iF sin a is normal to the plane of the ring. The resultant of the forces !iF1 that act on the separate elements of the ring is zero. These forces only stretch the ring. The full force fact ing on the ring is equal to the sum of the forces ~f: f=~klH sin a.·lilj=klH.2nR sin a.~ 273 dynes &59. The forces acting on BC and AD are perpendicular to the motion of these sides and, therefore, perform no work. The forces acting on AB and CD are constant, form a right angle with the direction of the field and are numerically equal to f=kHla (Fig. 459). The sought work will be equal to the double product of the force and the .motion of side AB or CD in the direction of the force. Wlren the circui t is turned through 1800 , this motion is b. Therefore, W=2kHJab: &80. Assuming that all the electrons move with a velocit.y v, the intensity of the current can be expressed as follows (see Problem 521): J=nAev . Upon inserting the value of J Into the formula for F. we get F:::kHnAlev sin ex Since a piece of the conductor contains N = Aln electrons, the force acting on one electron is f=kHev sin a. The force f is known as the Lorentz force. The direction of the Lorentz force is determined by the left-hand rule (the magnetic field intersects the palm, four fingers are directed opposite to the motion of the electrons, or along the motion of a positively charged particle, and the thumb shows the direction of the Lorentz force). &81. The Lorentz force is always perpendicular to the velocity of a particle and therefore performs no work. The kinetic energy and, hence, the absolute velocity of the particle remain constant. &82. The electron is acted upon by the force f=kevH. If H is measured In oersteds and the charge in CIlS electrostatic units, then k =: (see Prob- 346 ANSWERS AND SOLUTIONS II /"----....., / " ~ I " I \. I ~ \ I \ '.0\ / , / " fill' '---_.--" H H Fig. 459 Fig. 460 lem 560). This force is constant in magnitude and perpendicular to the velocity v. For this reason the acceleration of the electron is also constant in magnitude and constantly remains perpendicular to the velocity. The velocity changes only in direction. With a constant acceleration perpendicular to the velocity, the motion at a velocity constant in magnitude is uniform motion along a circle. mv2 e On the basis of Newton's second law, R=c·vH. Therefore, the electron will move along a circle with a radius R= n;::· 563. Let us resolve the velocity of the electron into the components vII parallel to Hand V...L perpendicular to H (Fig. 460). The component vu does not change in magnitude or direction since the Lorentz force does not act on a particle whose velocity is directed along the field. The component V.1. changes in direction in the same way as in Problem 562. Thus, rotation along a circle in a plane perpendicular to H is superposed on the uniform translational motion along H. This produces motion along a helical line with a constant pitch h=vu1:, where 1: is the duration of one revolution of the electron along a circle with a radius R = mc:~n IX • . 2nR 2nmc 2nmc Since l' = --= -H ,then h= --;:;- v cos ex v~ e e'4 564. The action of the Lorentz force (see Problem 560) will cause the electrons to move towards the edge of the band. For this reason one edge of the band will receive a negative charge and the other a positive one. An additional electric field will be generated inside the band with an intensity E directed perpendicular to the current. The electrons will continue to move until the Lorentz force is equalized by the force acting on the electron from the side of the electric field E: eE=kevH. Hence, E=kvH. The potential difference CPA-CPB=Ea=kvHa or, since I = nevA, then I A-CPB= kHa-A·ne 565. q> A - CPB =:: 23JLV. 566. The Lorentz force (see Problem 560) acts on both the free electrons and the positive ions at the points of a crystal lattice, since both move in a -, ELECTRICITY AND- MAGNETISM 347 a cl It 0H I~+ + + + + + + + Fig. 46/ magnetic field. In accordance with the left-hand rule, the force f that acts on the free electrons will be directed as shown in Fig. 461. The electrons are displaced wit h respect to the lattice, and one side of the -parallelepiped is charged negatively and the other positively. An electric field is produced in the block, and when the intensity of this field satisfies the ratio eE=kevH, the electrons will no longer move with respect to the lattice. The sought intensity E=kvH. The density of the charges (J can be found from the equation 41(<1=£. 1 Therefore, (J = 4:rt kv H. 567. For no electrostatic field to appear, the electrons should not move with respect to the crystal lattice when the cylinder revolves. This motion will be absent if the Lorentz force acting on the electrons is equal to mro2r, i.e., mro"r=kevH. Since V=CJ>" then H=~:. The field should be arranged in the direction of the forward motion of a screw rotating in the same direction as the cylinder. 3-5. Electromagnetic Induction. Alternating Current 568. The direction of the intensity of the electric field isshownin Fig. 462. 569. When the circuit moves,the magnetic fluxpassing through area ABeD diminishes. Therefore, in accordance with Lenz's law. the induced current will flow clockwise. 570. As the iron rod flies through the coilt the magnetic flux passing through it changes. This induces an e.m.I, of induction in the circuit. Ac- I ~--------""'-~f Fig. 462 . Fig. 463 348 ANSWERS AND SOLUTIONS cording to Lens's law, the total current in the coil decreases when the rod enters it, and increases when the rod leaves it. A diagram of the change in the current is shown in Fig. 463. 571. The magnetic flux changes at a constant rate, and therefore the e. m. f. of induction in the second coil will also be constant. If the coil is connected to a closed circuit, it will carry a direct current, which will set In not at once, but depending on the coefficient of self..induction of the second coil and its resistance. 572. Yes, it will. The e.m.f. of induction is proportional to the rate of change of the magnetic flux, while the magnitude of the magnetic flux in the iron core does not change directly with the current. The relationship will be more complicated. 573. According to Faraday's law, A«D cCi=10-S AT=10-8 kA The e.m.f, of induction is numerically equal to the work performed by the electric field when a single positive charge moves in the turn, i.e., cCi=2nrE. Hence E= 28; .nr Thus, we finally obtain: E= 10_skn,r 2 =lO-skr 2nf 2 It should be noted that this electric field is induced not by the electric charges, but by 8 magnetic field varying with time. Let us recall that when an electric charge moves in a closed circuit in an electrostatic field the work is always equal to zero. By an electrostatic field is meant an electric field induced by electric charges. 574. Let us divide the ring into n=b 6 a small rings each with a width 6. Let us consider a ring with a height h whose internal radius is x and external radius is x+6. If 6 is small as compared with x, the resistance of such a ring can be expressed by the formula 2nx R=p?;ji The e. m. f. of induction acting in this ring (if 6~ x) is equal to L\(!) 8=10-8 AT=10-snx'k The intensity of the current flowing in such a ring is Al = +(a+26) +.··+ la+ (n-l) 6)l B.LECTRICITY AND MAGNETISM 349 The expression in the braces is an arithmetical progression. Therefore, 1= 10-8 kh (b- a) 2a+b-a-6 2p 2 This result will be the more accurate, the smaller is 6. Assuming ~ as tending to zero, we obtain kh 1= 10-8 - (b2-a~) 4p 575. On the basis of the law of electromagnetic induction and Ohm's law J we have for the quantity of electricity that passed through the galvanometer: AcD AQ=/At=lO-slf or 10- 8 Q=-r(cD-cDo) Since the initial magnetic flux «1>o=HAn and the final ftux «%>=0, the quantity of electricity in coulombs will be Q= 10;8 HAn if R is measured in ohms and H in oersteds. 576. The e. m. f. of induction tOt= 10- 8 kat acts in circuit ABeD, and 8,= 1O-8k a; in circuit BEFC. The simplest equivalent circuit with galvanic cells used as the e. m. 1. of induction wi II for our circuit have the form shown in Fig. 464. On the basis of Ohm's law, J3ar = tCI - / l 3ar= / 22ar- B 2 Since the charge is retained, '1=1,+1a- All three currents can easily be found from the given system of equations: 11 6Itl+ 2i12 / _2=U-2cp' where U is the initial potential difference between the plates of the charged capacitor and q> is the value by which this potential difference drops at the 590. The magnetic flux through the ring cannot change (see Problem 589). Therefore, = c. Hence, AWe=uCAq> A decrease In the potential difference between the plates of one capacitor by q> is attended by a similar increase on the other. Therefore, 23 * 356 u A u o IJ Fig. 468 UB 2 ANSWERS AND SOLUTIONS given moment. Since xp changes from zero to U/2, the diagram showing u versus

0 and does not flow through it when V a ~ O. In a quarter of a period the current will not flow during the time interval 0 ~ t ~ t1 (Fig. 471). where t1 is determined by the equation I/o ~ sin rot. -IR=0. Hence, t. = ~ arc sin 2~R . This is also the time during which the current does not flow in the following quarters of the period. Altogether in a period the current does not flow during 2T . 2/R """it arc sin ---0-=0.465 T 3-6. Electrical Machines 605. If the frequency of the alternating current remains the same, this means that the revolutions of the motor and the generator also remain as before. and the e.rn.I. of the generator wHI not change. When the external resistance in the circuit is high, the circuit will carry a smaller current and a lower power will be supplied. For this reason the power of the motor that revolves the generator should be reduced. 606. The work performed by a field in moving conductors carrying a current (armature windings) is not equal to the total work of the field. Apart from the work spent to move the conductors, the magnetic field performs work to 360 ANSWERS AND SOLUTIONS retard the electrons in the conductor, which produces an e.m.I, of induction in the armature winding. The first part of the work is positive and the second negative. The total work of the magnetic field is zero. The electromotive force of the source that generates a current in the motor armature performs positive work, and the latter compensates for the negative work of the magnetic field in retarding the electrons. In essence, the motor does its work at the expense of the energy of the source feeding it. 607. The power consumed by the motor is P= JV; here V = tCi +f R , where =2%A X lOS. Here tIli= ~ and 1=2~' In a unit of time the U2 · battery performs the work 2R • From this amount, one half is converted into mechanical power and the other half is liberated as heat (see Problem 607). The relation between P and 00 is shown in Fig. 473. 610 M _ kHAU _ k·lO~8H2A2 · - R R CIl, U.108 The moment will be equal to zero when ro=---n;r (see Fig. 474). Here 1=0, because tRi=U, 611. The naTure of the relation between P and H is shown in Fig. 475 (see the solution to Problem 609). The power reaches its maximum when U U U2 H=2ACI) X 108 • Here l1i=T and Pmax =4R J. p U 8 ZHA'O Fig. 473 N KHAU f { Fig. 474 362 p ANSWERS AND SOLUTIONS 612. The torque will reach its VI maximum Mmax =107 X 4Rro dyne- U -crn when H = 108 X 2AID. 613. As with a series motor, the power of a shunt-wound motor is p Fig. 475 where R is the resistance of the armature (see Problem 607). Two values of H > Ho, the speed increases when the current grows in the windings of the stator. and decreases when H>Hm: If the motor operates without load (M =0). the speed will be n= k~ , i.e., it always decreases with an increase of H. 616. The intensity of the current flowing in the winding of the motor will be determined by the e.m.I, of the mains t-1200). The potential difference U12 is equal to that on the winoings WI and W2' Le., UI 2 = Vh= Vg (R-r) (cos [2 R2 where J is the luminous intensity of the lamp, R the radius of the table, and cp the angle of incidence of the rays (Fig. 488). The maximum value of E is attained if the angle q> satisfies the equation 1- sin2 q>=J..- sin2 q> 2 I. e.• when q> = arc sin y ~ (see Problem 407). The lamp should be hung above the table at a height of h= ~2 R~0.71 R. 670. Tissue paper diffuses the incident light rays in all directions. If the paper is at some distance from the page, the diverging beams of Il ght reflected from the white portion of the page (between the letters) overlap on the side of the paper facing the text (Fig. 489). As a result the paper will be illuminated more or less uniformly, and the diffusion of the light will make it impossible to read the text. If the paper is placed on the text, the illumination of the paper side adjoining the text will not be uniform. Accordingly the intensity of the diffused Iight will be different at various portions of the paper. and the text can be read. 5· 2. Fundamental Laws of Optics 671. The shadow will be equally distinct everywhere only if a point source of light is used. The separate sections of an extended source throw shadows that are superimposed on one another. The boundary of the shadow will be the sharper, the smaller is the distance from the object to the surface on which the shadow is formed, since the distances between the boundaries of the shadows produced by the various sections of the source will be minimum. It is for this reason that a man's legs give a sharper shadow than his head. 672. The pencil should be held parallel to the lamp and as close to the table as possible. As a result the shadows sent by separate portions of the lamp will be almost accurately superimposed. If the pencil is perpendicular to the lamp, the shadows from it will be so mutually shifted that practically no shadow will appear. 673. This phenomenon can be observed only if the angular distance between the branches is less than the angular diameter of the Sun. Let us assume, to introduce definite conditions, that the lower branch is thicker than the upper one. GEOMETRICAL OPTICS 383 (a) (b) (e) (d) (e) Fig. 490 676. The height of the mirror should be equal to half the height of the man. The distance from the .lower edge of the mirror to the floor should be equal to half the distance from the man's eyes to his feet (Fig. 493). 677. Let h be the height of the object and ex. the angle of incidence of the rays on the mirror (Fig. 494). If the screen is at a distance of I ~ h tan (J. from the object, a direct and an inverted shadows withFig. 49/ To understand why the illumination inside the shadow changes as stated in the problem, assume that we look at the Sun alternately from different sections of the shadow. The Sun's disk can be seen entirely outside of the shadow. In section A of the shadow (Fig. 235) the eye is in the half shadow cast by the lower branch, and only this branch is visible in front of the Sun's disk (Fig. 490a). Since the branch covers a part of the Sun's disk, the illumination of this point will be weaker. Moving the eye farther to position B (Fig. 235), we shall see that the other branch also partly covers the Sun's disk (Fig. 490b), and for this reason the illumination will be still less. Moving farther, the eye will occupy position C (Fig. 235) in which both branches will be superimposed (Fig. 490c). Now, the part of the Sun's disk covered by the branches is smaller and the illumination greater. The disk as viewed from D and E is shown in Fig. 490d and e. This explains why the central stripe of the shadow is brighter than the .adjacent parts. 674. As can be seen from. Fig. 491, we have H =L sin a, while sin a= : ' since DE= b is the cross-sectional diameter of the light cone on the ground. With the angular dimensions of the Sun's disk ~, we obtain L= ~ . 1 b2 Therefore, H T a = 9 metres. 675. If the rays are turned in the periscope as shown in Fig. 492, the sought ratio of the widths of the prisms alb can be found from the similarity of the triangles: a L+l -r;=-l- 384 ANSWERS AND SOLUTIONS D A B =~-=-·-+-·--3a-·-----·_·-E~- - - - - - - - I I I 14; l .I~ l~ Fig. 492' H I Fig. 493 their bases fitted against each other will be seen on the screen. The total length of the shadow is 2h. The shadow is illuminated by the Sun and contrasts with the other portions of the screen illuminated by both direct and reflected rays. If the screen is nearer. the length of the shadow is smaller than 2h and will have portions that are illuminated neither by direct nor reflected rays. 678. A point source of light always produces a reflection that depends on the shape of the mirror. The dimensions of the Sun are finite. Each small section of the luminescent surface produces a bright spot that gives the shape of the mirror. These spots from various portions of the Sun are superimposed and produce a more or less diffused pattern. If the surface on which the reflection is observed is far from the mirror, the shape of the bright spot will not depend on the shape of the mirror. It is only at a small distance from the mirror that the spot will reproduce the shape of the mirror, since the angles at which the rays from the various portions of the Sun tall onto the mirror differ very slightly from one another. Zh Fig. 494 GEOMETRICAL OPTICS 385 B I I I I I I I -------,--~----- " ...... --~_ ,~................ I I <, ---_~-.......................... I· : <," ';--~J7{~'8' ',,', I ...........,', I ..... :\. I ..... , A Fig. 495 Bs I I I I ----t- I I / ;........ ) / ...... I I c-:--S ~--N---..."""" ZI ,I l/ I I I I I II / I / / I / / J / / J I / s :p/ 1 Fig. 496 25-2042 679. The reflected landscape is seen as if it were viewed from a point below the water surface at a distance equal to that from the camera lens to the water. 680. The whole of the image of straight line AB can be seen only if the eye is inside the hatched area in Fig. 495. 681. When mirror M N moves toward the wall, the position of light spot AB on the wall will be invariable as can be seen in Fig. 496 (81 and 8 2 are the images of source S with the mirror in two positions: M Nand M'N '). The dimensions of the light spot will not change either, constantly remaining equal to the double dimensions of the mirror. 682. If the losses in reflection are neglected, the illumination of the light spot will always be one-fourth of the illumination of the mirror. At the same time the illumination of the mirror changes in view of the change in the distance from the lamp to the mirror and the change in the angle of incidence of the rays. With a small mirror the maximum illumination will be observed when the distance from the mirror to the wall is I = ~2 d, where d is the distance from the source of light to the point on the wall which the mirror is brought up to. 683. When the mirror turns through an angle ex, the reflected ray will turn through 2a since the angles of incidence and reflection increase by a Hence, the angular velocity of rotation of the reflected ray is ro=2nnX2. The linear vee locity with which the light spot moves along the screen is v=4:nnR ~ 62.8 m/s. 386 ANSWERS AND SOLUTIONS 684. (a) The beam reflected from the first mirror forms an angle 2cx with the incident beam (a is the angle of incidence). During the time t the mirror will turn through an angle oot and the new angle of incidence will become equal to a+ cut, as will the angle of reflection. Therefore, the angle between the incident and reflected beams will increase by 2cut, Le., the reflected beam will turn through an angle 2mt. In view of this, the angle of incidence on the second mirror, provided it does not rotate, would be p+2m!t where p is the angle of incidence with immobile disks. But the mirror also revolves through the angle rot during the time t, and therefore the angle of incidence becomes P+300t. The angle of reflection will be the same. Thus, after two reflections the beam will turn through the angle 3mt from its direction with immobile mirrors. After three reflections the beam will turn through 5cot and after n reflections through (2n-l) 2mt. In this way its angular velocity will be Q =(2n-l) 200. (b) When the mirror moves from the source with a velocity e, the image will move away from the source with a velocity 2v and from the second mirror with a velocity 3v. Therefore, the second image moves with a velocity 3tt with respect to the second mirror and with a velocity 4v with respect to the source. The velocity of the third image with respect to the source will be 6v and the velocity of the n-th image 2nv. 685. (a) When the first mirror turns through an angle rot the reflected beam will turn through an angle 2mt (see the solution to Problem 684). 'Hence, the angle of incidence on the second mirror will also increase by 2m!. and. if the mirror did not revolve, the angle of reflection would also increase by 2rot. After two reflections the beam would turn through 2mt as compared with the case of immobile mirrors. Since the second mirror does rotate, however, the angle of the beam incident on it decreases by rot during the time t, The angle of reflection decreases by the same amount and for this reason the reflected beam will travel in the same direction as with immobile disks. Since this line of reasoning may be adopted for any two consecutive reflections, the angular velocity of rotation of the beam subjected to n reflections will be Q=O if n is even. and Q=2m if n is odd. (b) The first image moves away from the source with a velocity 2v and from the second mirror with a velocity v. Therefore. the second image moves with respect to the second mirror with a velocity - e, Le., it is immobile wi th respect to the source. /1' ~ ~ S ~~~~~---~'--t--- ~ ~a ~. o Fig. 497 GEOMETRICAL OPTICS Fig. 498 381 Fig. 499 Reasoning similarly, we find that the sought linear velocity of the n-th image is zero if n is even, and 2v if n is odd. 686. The beam reflected from mirror ON forms with the incident beam an angle

. The virtual source S; is reflected from mirror AO and produces image S; lyin g cn the same circle at a distance of 2cp arc degrees from source S. Image S; of virtual source S~ is formed in mirror DB in the same way. Continuing our construction, we obtain the third images S; and S; removed from the source by 3cp degrees and the fourth S; and S; (removed by 4q> degrees), etc. If n is even (n =2k), then image Sk coincides with Sk and will be on one diameter with the source. Altogether there will be 2k-l =n -1 images. 25 * 388 ANSWERS AND SOLUTIONS s F' D J I J I I I \ I \ I \ 0 a If n is odd (n=2i + 1) it can easily be seen that the i-th images lie on the continuations of the mirrors and th us coinci de with the (i +1) th and all the subsequent images. For this reason there will be 2i images, Le., n-l as before. 690. Using the solution of Problem 689, let us plot consecutively the first, second, third, etc., images of source S in the mirrors (Fig. 500). All of them will lie on a circle with a radius OS and the centre at pointo. If a is an integer, the last i-th images will either get onto poin ts C and D at which the circle intersects the continuations of the mirrors, or will coincide with point P diametrally opposite to the source. In both cases the . number of images will be a-I. If a is not an integer, for example a=2i ± ~, where ~ < 1, and i is an integer, the last i-th images will lie on arc CPD that is behind both the first and the second mirrors and there wiII be no more reflections. Thus, the total number of images will be 2i. 691. Let us plot the image of point B in mirror bd (Fig. 501). Let us then construct image B1 in mirror cd. Also, B3 is the image of B2 In mirror ac and 8 4 is the image of 8 3 in mirror abo Let us connect points A and B4 • Point C is the point of intersection of ab with line AB4 • Let us now draw line B3C from B3 , and connect point D at which this line intersects ac with 8 2• E with B1 , and F with B. It can be stated that broken line ACDEFB is the sought path of the beam. Indeed, since B3CB4 is an isosceles triangle, CD is the reflection of beam AC. Fig. 500 GEOMETRICAL OPTICS i Fig. 502 389 Similarly, it is easy to show that DE is the reflection of CD, etc. This solution of the problem is not unique, since the beam shouId not necessari1y be sent initiaIJy to mirror abo 692. The coefficient of reflection of light from the surface of water diminishes with a reduction in the angle of incidence. If the observer looks down, rays reflected at small angles reach his eyes. The rays reflected from the sea water at the , .D horizon reach the eyes at greater , angles. " 693. According to the law of refraction s~n i =n (Fig. 502). sin r Upon exit from the plate s~n ~ =- . Upon multiplying these expressions, we sin r, n get sin i=sini1, i.e., beam CD leaving the plate is parallel to incident beam AB. A glance at the drawing shows that a,=i-f. The sought displacement of the beam is x=EC=BC sin (i-f). Since BC=_d_. then cosr x= d sin (i - f) d sin i (1_ cos i ) cos, y n2-sin2 i And a maximum displacement of d can be obtained when i --.. 90°. 694. The angle of incidence of the ray onto AC and BC is 45°. For total internal reflection it is necessary that sin i > J. .n Hence, n > y2'~ 1.4. 695. The angle of incidence of the rayon face Be is equal to the sought angle cx. For the ray to be completely reflected from face Be, the angle ex should exceed the limiting one. . Therefore, sin a > ::' where nz is the refraction index of water. Hence, a > 62030'. 696. This phenomenon is nothing but a mirage frequently observed in de- serts. The hot layer of air in direct contact with the asphalt has a smaller refraction index than the layers above. Total internal reflection occurs and the asphalt seems to reflect the light just as well as the surface of water. 697. Let us divide the plate into many plates so thin that their refraction index can be assumed constant within the limits of each plate (Fig. 503). Assume that the beam enters the plate from a medium with a refraction index of no and leaves it for a medium with a refraction index of na· 390 ANSWERS AND SOLUTIONS Fig. 503 Then. according to the law of refraction, sin a nt sin ~ = no sin p n' --=-sin y nl sin 'V n" sin ~=n' sin q> n2 sin, = n(n) sin ~ na sinx=n2 Upon multiplying these equations we" get sin a ns SiiiX=no Hence, the angle at which the beam leaves the plate x=arcsin (:: ~in a) depends only on the angle of incidence of the beam on the plate and on the refraction indices of the media on both sides of the plate. In particular, if ng=no. then x=a.. GEOMETRICAL OPTICS Fig. 504 n" N Fig. 505 F 391 Generally speaking, the angle O. at which the beam is inclined to the vertical is related to the refraction index n at any point on the plate by the ratio n sin 8=const=no sin a.. If the refraction index reaches a value of n = nosin ex anywhere inside the plate, full internal reflection will take place. In this case the beam will leave the plate for the medium at the same angle ex at which it entered the plate (Fig. 504). 698. The minimum amount of water determined by the level x (Fig. 505) can be found from the triangle MNF. We have NF=x-b=xtanr. From the law of refraction . sin i 51nr=-- n Therefore, b x= -t--t-a-n-, since i=45° and 11= ~ . The amount of water required is V=xa2 =::: 43.2 litres, 699. The man's eyes are reached by rays coming in a narrow beam from an arbitrary point C on the bottom. They seem to the eye as issu ing from point C' (Fig. 506). Since di and dr are very small, we can write: AD=AC'~'=~arcos r AD'=AC'.~i=~~icos ~ By equating the values of AB from triangles ABD and ABD', we have -!!- ~r= .s..f1.t cos2r cos2 l 392 ANSWERS AND SOLUTIONS a Fig. 506 Using the law of refraction, we can find the ratio ~~. Indeed, s~n i =n and sin (i +ai) =n sin, sin (, +ar) Remembering that ~i and ~r are small, we have sin ~i ~ 8i, sin~, === ~" and cos L\i === cos ~, =:: t Therefore, the last equation can be rewritten as sin i +cos i· Ai=n sin ,+n cos r-br Ai cos r Hence, T= n --., Uoon inserting this expression into the formula relating a' cos l GEOMETRICAL OPTICS 393 Hand h, we find h=H coss i =!!.... . coslj i n cos 3 , n ( sin2 i ) 3/2 1--- n2 When i=O, we have h=!!..., i.e., the n depth seems reduced by n times. As i increases, h diminishes. The approxi.. mate dependence of the seeming depth on the angle i is shown in Fig. 507. The man's eye is above point A of the lake bottom. 700. q>= 120°. 701. The path of the ray in the prism is shown in Fig. 508. There is an obvious relationship between the angles cx and ~, namely. 2ex.+~= 180°, and a.=2~. Hence, cx=72°, ~=36°. Fig. 508 702. The path of the ray in the prism is shown in Fig. 509. To avoid full internal reflection on face BN, it is necessary that sin ~ ~ ..!... As can n be seen from the drawing, ~ = ex. -,. Hence, the greater the value of " the higher is the permissible value of cx. The maximum value of r is determined from the condition: sin r=...!- (angle of incidence gOO). n Therefore, CImax = 2 arc sin ~ ~ 83040'. B B Fig. 509 Fig. 5/0 394 ANSWERS AND SOLUTIONS B ------------t~----aD Fig. 51/ 703. When considering triangles ABC, AMC and ADC (Fig. 510), it is easy to see that r+rl=q> and y=ex+P-cp. According to the law of ·re- fraction, sin ex sin rt I sinr=n, and sin~=n Upon solving this system of equations, we find that cp=ex+~-y and .. /{ sin ex 1 } 2 n=sin~ V sin~sin(ex+~-y) +tan(a,+~-'V) +1 704. According to the initial condition, the incident beam and the beam that has passed through the 'prism are mutually perpendicular. Therefore, L q>= L. a, and also L. y= L. Ii (Fig. 511). The sum of the angles of the quadrangle AKMN is 360°. Therefore, L. KMN = 90° and beam KM is incident onto face Be at an angle of 45°. If we know the angles of triangle KBM it is easy to find that ~=30°. In conformity with the law of . sin ex • retraction ~=n. Hence, sm ... sin a=O.5n and a= arc sin O.5n Since the full internal reflection at an angle of 450 is observed only when n~V~ the angle ex is within 45°~a.EZ;;;90°. 705. Paper partially lets through light. But owing to its fibrous structure and the great number of pores, the dissipation of light is very high in all directions. For this reason it is impossible to read the text. When they fill the pores, glue or water reduce the dissipation of light, since their refraction index is close to that of the paper. The light begins to pass through the paper without any appreciable deviation, and the text can easily be read. GEOMETRICAL OPTICS 395 6-3. Lenses and Spherical Mirrors 706. n= 1.5. 707. f=2R. 708. The convex surface has a radius of curvature of R1 =6 cm and the concave one R2 = 12 em. 709. In the first case the focal length is determined from the formula f~ =(:1-1 ) (~1+~J Since in a vacuum the focal length of the lens is f, then 1 1 1 (n-l)f R1+R2 (n-l)f· Hence, ft= n_ 1 =90cm nt In the second case the sought focal length is I, (n-l)f -102cm ~-l n2 The lens will be divergent. 710. As shown in the solution of Problem 709 -f n 2(n 1 - l ) D(n1-n.J Therefore. n2 fDn l ==1.67 fD+ I-n, 711. The image will be m+ 1 times smaller than the object. 712. The lamp should be moved two metres away from the lens. 713. Obviously, one of the image will be virtual. Therefore, denoting the distances from the sources of light to the lens by al and a2 and from the lens to the images by bI and b2 , we obtain: 1 1 1 1 1 1 ---=- and -+-=-al bl t : a 2 b2 t According to the initial condition, at+a2 = l and b1=b2• Upon solving this system of equations, we get 1(1± V~) 2 The lens should be placed at a distance of 6 em from one source and 18em from the other. 714. Applying the formula of a lens to both cases, we obtain 1 1 1 1 1 1 -+-=- and -+-=-al bl t ' ala b2 f 396 Fig. 512 According to the initial condition, ANSWERS AND SOLUTIONS ~-"":;:'-----I--~--- (magnification in the first case); (magni fication in the second case). Hence, t=kk1k 2k [=9 em 1- 2 715. (1) For this case the path of the rays is shown in Fig. 512a. From the standpoint of reversibility of the light beams, point B can be regarded as a source of light and point A as its image. Then, according to the formula of a lens, I 1 I at -7i=-T f alb Hence, =--b= 20 em. al(2) The path of the rays is illustrated in Fig. 512b. In this case, both the image (point A) and the source (point B) are virtual. According to the formula of a lens f a2b 2 lienee, =a 2 + b= 1 em. 716. On the basis of the formula of a lens, 1 1 I a-+d-a =T where a is the distance between the lens and the lamp. Therefore, a2-ad+df=O GEOMETRICAL OPTICS Fig. 513 Upon solving this equation, we obtain d yd-2 a=- ± --df2 4 397 Two positions of the lens are therefore possible: at a distance of al = 70 em and at a distance of a2 = 30 ern from the lamp. When I' =26 em, there will be no sharp image on the screen, whatever the position of the lens, since to obtain the image it is necessary that d ~ 4'. 717. In the first case hHl = bl , where at and b1 are the distances from the al object and the image to the lens. In the second case Hh 2 = b2 • a2 It follows from the solution of Problem 716 that al =b2 and b1 =a2• Therefore, H=Vhlh2 718. On the basis of the formula of a mirror 1 1 1 -a--;;=T The linear magnification of the mirror. is H b 71=(i According to the initial condition, the angular dimensions of the image on the concave mirror are 1.5 times greater than those on the fiat mirror: p= 1.5C& (Fig. 513). It is obvious that h H tancx=2a and tan p=a+b 398 ANSWERS AND SOLUTIONS Fig. 514 Fig. 516 A'l M F F Fig. 515 B M ~ F F S' Fig. 517 GEOMETRICAL OPTICS 399 When h ~ 2a, ex. and ~ are small. For small angles H 01:: 15 h a+b - · 2a Upon excluding the unknown quantities ~ and b from the equations, we find 3 that 1=2: a. Hence, R=2f=3a=6 metres. 719. The path of the ray is shown in Fig. 514. Let us continue AB up to its intersection with the focal plane of lens NN. The beam of parallel rays after refraction in the lens so travels that the continuations of the rays should intersect at P'. Ray F'O is not refracted. Thus, ray C A passing to point A is parallel to F'O up to the lens. 720. If A is the source and B is the image, then the lens will be convergent. The position of the optical centre of the lens 0 and its foci F can be found by construction as shown in Fig. 515. If B is the source and A is the image, the lens is divergent. The respective construction is illustrated in Fig. 516. 721. The centre of the lens 0 is the point of intersection of straight lines SS' and NIN2. The foci can easily be found by constructing the rays parallel to the major optical axis (Fig. 517). 722. Point 0, which is the optical centre of the lens, can be found by dropping perpendicular 80 onto straight line N iN2 (Fig. 518). Let us draw an auxiliary optical axis DO parallel to ray AB and extend straight line BC until it intersects DO at point E lying in the focal plane. Let us drop a perpendicular from E onto N IN2 to find point F, one of the main foci of the lens. By using the property of reversibility of the ray. we can find the other main focus Ft. 723. The image S' may be real or virtual. In both cases let us draw an arbitrary ray ADS' and auxiliary optical axis BOC parallel to it to find the position of the source lFig. 519). By connecting the points of intersection B and C (of the auxiliary axis with the focal planes) to point D by straight lines, we can find the position of the source SI (if the image S' is real) and S2 (if the image is virtual). 724. Since the ray incident on the mirror at its pole is reflected symmetrically with respect to the major optical axis, let us plot point SI symmetrical to S' and draw ray SSt until it ina tersects the axis at point P (Fig. 520). This point wilJ be the pole of the mirA ror. The optical centre C of the mirror can obviously be found as the point of JJ intersection of ray SS' with ax is NN' . The focus can be found by the usual ____..o-.--;;:a~_o--_--_-construction of ray SM parallel to Ai ~ the axis. The reflected ray must pass through focus F (lying on the optical ax is of the mirror) and through 8'. 725.. (a) Let us construct, as in the solution of Problem 724, the ray BAC Fig. 518 and find point C (optical centre of the s*I I I I I I I I N P H' 8' Fig. 519 Fig. 520 B Fig. 521 GEOMET~ICAL OPTICS 401 mirror) (Fig. 521a). Pole P can be found by constructing the path of the ray AP A' reflected in the pole with the aid of symmetrical point A'. The position of the mirror focus F is determined by means of the usual construction of ray AMF parallel to the axis. (b) This construction can also be used to find centre C of the mirror and pole P (Fig. 52Ib). The reflected ray 8M will pass parallel to the optical axis of the mirror. For this reason, to find the focus, let us first determine point M at which straight line AM, parallel to the optical axis, intersects the mirror, and then extend 8M to the point of intersection with the axis at the focus F. 726. (a) The rays reflected from the flat mirror increase the illumination at the centre of the screen. The presence of the mirror is equivalent to the appearance of a new source (with the same luminous intensity) arranged at a distance from the screen three times greater than that of the first source. For this reason the illumination should increase by one-ninth of the previous illumination, i.e., Ea=2.51x (b) The concave mirror is so arranged that the source is in its focus. The rays reflected from the mirror travel in a parallel beam. The illumination along the axis of the beam of parallel rays is everywhere the same and equal to the illumination created by the point source at the point of the mirror closest to it. The total illumination at the centre of the screen is equal to the sum of the illuminations produced by the source at the centre of the screen and reflected by the rays: Eb=2X2.25lx=4.5lx (c) The virtual image of the point source in the convex mirror is at a distance of 2.5r from the screen (r is the distance from the screen to the source). The luminous flux (Fig. 533) determined by the equality ~in2(j) =n. Since the angles are small cp ~ 2an. sin ex, The second image of the source will be obtained at a distance d from the first image, namely d=f(fJ=f2an. d Hence, n = 2a.f . (2) The optical axis of the lens is perpendicular to the rear face of the wedge. The rays reflected from the front face will be deflected by an angle 2a (Fig. 534) and produce an image at a distance of d1 =2af from the source. The ray s reflected from the rear face will be deflected by an angle edetermined from the equations: sin a d sin (a+8) ---n an -n sin ~ - , sin (2a-~) When the angles are small, a= 2a (n - 1). For this reason the second image will be at a distance d7,=2a.(n-l) f from the source. The total distance between the images is d=d1 +d7, =2anf· d Hence, n=2af as in the first case. 740. Since the image that coincides with the source is formed owing to reflection from the part of the mirror not covered by the liquid, the source is obviously arranged at centre 0 of the hemisphere. Let. us find the position of the other image (point A in Fig. 535). According to the law of refraction sin ex .- a sin q>

2, the path of the ray is 8S shown in Fig. 539. The sought distance is f'=OF~CF-R As can be seen from Fig. 539, CF=ACcot (i-f) ~ .AC :2!!: ~ l-T &-T Hence, f'=-.!Ln-l 744. Let us extend ray BF until it intersects the continuation of the ray Incident on the sphere parallel to the optical axis (Fig. 538). It can easily be seen that section DO that connects the point of intersection with the centre of the sphere forms a right angle with the direction of the incident ray. Triangle ODF is a right one, since OFpe!! R _n_ 21 (n-I) Rt (see Problem 743) 2 n-l n For this reason the main planes of sphere M N coincide and pass through its centre. GEOMETRICAL OPTICS Fig. 539 411 745. The focal length of the sphere is R n 1=2"" n-l =15 em (see Problems 743 and 744). By using the formula of a lens. and this may be done because the main planes coincide, let us find the distance from the centre of the lens to the image b= at r:: 15 ern aThe image is virtual and is in front of the sphere. 746. The thin wall of the spherical flask can be regarded as a divergent lens fi th a focal length of ___~1_~~ ~ R' 11 ( 1 1 ) (n-l) ~R(n-I) - - R. R'J, After passing through two such lenses at a distance of 2R from each other (Fig. 540), the rays parallel to the major optical axis (the flask diameter) will be so refracted that their continuations will intersect at the focus F of 1'104----41- I I I I I I I lo+-ZH~ Fig. 540 412 ANS\\'ERS AND SOLUTIONS the system at a distance b from the second lens. According to the formula of a lens, b Hence, I I 1 'l+ 2R -1j=-t; 'I ('I +2R) 2 (/1+R) Point D of the intersection of AB (continuation of the incident ray) and CF (continuation of the refracted ray) lies on the main plane of the system at a distance x from the second lens. It follows from the similarity of triangles ACB and FICO, and also of triangles DC8 and FCO that x 2R 1J=2R+f. The main plane lies at the following distance from the second lens 2Rb t.« x 2R+tl 'I+R Therefore, the focal length of the system is f-b - ,~ oc'l _ R2 - -x-2(fl+R)-T- 2(n-l)&R In view of symmetry of this optical system, the positions of the second focus and of the other main plane are obvious. 747. A glance at Fig. 541 shows that the angle of refraction is r=L. OAB==L. ABO=L. OBC=L. OCB and L. BAD=L. BCD=i-r At point A the beam turns through an angle i -', at point B through an angle 1t - 2" and at point C through i -,. Therefore, the total angle through Fig. 541 QEOMETRICAl OPTICS 413 Fig. 542 which the beam is deflected from the initial direction is e=i-f +.n--2f +i -r=n+2i -4, . sin i The angle r can be found from the ratio -.-=n. sm r 748. When a parallel beam of rays is incident on the drop, the ray passing along the diameter has an angle of incidence of i =0°, while the angles of incidence of the rays above and below it may range from 0 to 90°. (1) Using the results of the previous problem and the law of refraction, we can find the values of 8 for various values of i: Table 3 e e 0° 180° 55° 138°20' 20° 160°24' 60° 137°56' 40° 144°40' 65° 138°40' 50° 139°40' 70° 140°44' (2) A diagram of e versus i is shown in Fig. 542. (3) The minimum value of the angle of deflection is approximately equal to Smin = 1380 • The rays leaving the drop are nearly parallel when e=8mi,.. since in this case, as can be seen from Table 3 and the diagram, echanges the slowest when i changes. An approximate path of the rays in the drop Is illustrated in Fig 543. 414 J- ------....,.c 8- ----~--- ANSWERS AND SOLUTIONS Fig. 543 749. In accordance with the formula of a lens l+-l=~ a b f The magnification is b b-f . k=(i=-f-=24 times 750. The condenser should produce a real image of the source on the lens having the size of the lens. Therefore, using the formula of a lens and the expression that determines its magnification, we can write two equations: 1 l i d x -X+y=,' and Do =y Here x is the distance from the source of light to the condenser and y is the distance from the condenser to the lens. According to the initial condition, x+y=l. By cancelling x and y from the expressions obtained, we can find the focal length: f diDo 7.1 cm The diameter of the condenser will be minimum if the slide is behind it. The minimum permissible diameter is D e:: It cm. 751. Ground glass is needed to fix the plane in which the image is obtained, and to i ncrease the angle of vision. • Transparen t glass is used to examine the image produced by a lens in a microscope. For this purpose a line is drawn on a transparent glass to fix the focussing plane, and this line and the adjacent portion of the image produced by the lens are focussed in the microscope. In this case ground glass cannot be used, since the microscope will show all the distortions due to the structure of the ground surface. 752. (1) The lanterns will appear Equally bright, since the illumination of the retina of the eye E= ~~ is the same for both lanterns. (Here L is the GEOMETRICAL OPTICS 415 luminance of a lantern, A the area of the pupil, and b the distance from the crysta Iline lens to the retina.) (2) The image of a far object is closer to the lens than that of a near object. For this reason a remote lantern produces a higher illumination of the film and its image will be brighter on the photograph. 753. The illumination of the photographic film is A (a- 1)9 E""",,-I""WS,--- b2 at where S,= :: is the lens speed, f is the focal length and a is the distance from the lens to the object being photographed (see Problem '152). It is obvious therefore that the exposure in a camera with a short focal length should be smaller. 754. The distances between the Sun and the Earth and between the Sun and the Moon are practically the same. For this reason if the Moon and the wall had equal coefficients of reflection, their luminance would seem identical. It can be assumed, therefore, that the surface of the Moon consists of dark rock. 755. In air, the external convex cornea of the eye collects the rays and produces an image on the retina. The crystalline lens only helps in this. The refraction index of the liquid inside the eye is very close to that of water. For this reason the cornea refracts almost no light and the eye becomes very far-sighted. The refractive properties of the cornea are completely retained when the swimmer is wearing a mask. 756. When the man looks at remote objects through his spectacles, he sees them as he would see objects at a distance of al=60 em without any spec- tacles. Therefore, when the man is wearing spectacles (see the solution to problem 735) 1 I 1 1 (i+7J=7+t; where a=oo. When the man is without spectacles -.!-+-!.=-!.a2 b f Here b is the depth of the eye, +the minimum optical power of the eye and *the optical power of the spectacles. It is assumed that the spectacles are fitted tightly against the eye. Hence, '0=-at· Let us now determine the position of the nearest point of accommodation of the eye with spectacles 1 I 1 I I 1 1 -+-=-. and -+-=-+-a1 bl fI a3 b1 f1 f0 416 Therefore. ANSWERS AND SOLUTIONS 1 1 1 1 1 -=-+-=---03 al to at at and aa=15 em 757. The long-sighted man when wearing his friend's spectacles can see only very far objects. Therefore, the distance a2 of best vision of the eye of the long-sighted man can be determined from the equation 1 1 ---=D1 at at where al is a very great distance (a1 -+ 00) and D} is the optical power of the spectacles of the short-sighted man. The optical power D2 of the spectacles that correct the defect of vision of the long-sighted man can be found from the formula 1 1 ---=D2 ao a2 where ao= O.25 metre is the distance of best vision of the normal eye. The distance a3 of best vision of a short-sighted eye can be found from the equation 1 1 ---=D1 ao aa If the short-sighted man wears the spectacles of his long-sighted friend, the distance of best vision, Le., the minimum distance a at which the shortsighted man can easily read a small type, can be determined from the formula 1 1 ---=D2 a aa Upon solving these four equations, we get a= 12.5 ern. 758. When an object with a height of 1 is examined from a distance of D, the angle of vision q>l is determined by the formula I l=T[=k T I' b f+b wherek=T=1J=-f- is the linear magnification determined from the formula of a lens (f is the focal length). QEOMETRICAL OPTICS 417 Fig. 544 Hence, N- .!!.-b+f_E.- L-r+f -, L-, L D (1) When L=oo, we hhveN=T. D r (2) When L=D, we have N=T+1-T. 71S9. The magnification of a telescope is N= ~: . where II is the focal length of the objective and /s that of the eyepiece. Since in a telescope adjusted to infinity the distance between the objective and the eyepiece is 11 +/1' . D 11 + I, 7=-bHere b is the distance from the eyepiece to the image of the diaphragm. According to the formula of a lens, _1_+.!.=J.. '.+/s b '2 Upon cancelling b from these equations, we find that D II 7=-,;=N 760. Sharp images of remote objects will be obtained with the convergent lens in three different positions. The lens can be placed in front of the d 1vergent lens or behind it. 27-2042 418 ANSWERS AND SOLUTIONS N Fig. 545 For the first position the distance d between the lenses can be found con.. sidering point K as the virtual image of point A in the divergent lens (Fig. 545). 1 1 1 -f,-d+ T=- t; Ray MN is parallel to the optical axis of the system. Hence, f fll d= ,-'l+l = 3.5 cm For the second position (the convergent lens is behind the divergent one) the path of the rays is shown in Fig. 546. Regarding point A as the image of K in the convergent lens, let us use the formula of a lens . 1 1 1 11+d+ l-d=t; Hence. d=l-fl ± l+/1 ../ l-~ 2 2 V t+'1 The distance between the lenses may be d2 = 35 em or da=5 em. 761. Let the rays coming from one end of the diameter of the visible disk of the Moon be directed along the optical axis of the system. The rays will produce an image on the optical axis at point A (Fig. 547) removed by a distance of l=45 cm from the divergent lens. IV ----.......... K Fig. 546 GEOMETRICAL OPTICS ....---l---~ A. B Fig. 547 419 The rays coming from the other end of the diameter form, according to the condition, an angle cp with the first rays. After passing through the system, these rays will produce an image (point B) lying in a plane perpendicular to the optical axis and removed by the same distance I from the divergent lens. To find the diameter of the image D1 = AB, let us consider the path of the ray passing through the optical centre of the first lens. In the first arrangement. the convergent lens is placed in front of the divergent one at a distance of d1 =3.5 ern. If we consider point E as the virtual image of point 0 we can write Using the similarity of triangles ABE and 0lPE. and remembering that 0IP = d1 tan cp, we obtain Upon canceJJing Xl from these equations, we find that D1 ' 0.72 ern, In the second arrangement of the lenses (d2=35 em), the path of the rays is shown in Fig. 548. The diameter of the Moon's image D2 can be found ~---+----J-----~ E A Fig. 548 420 ANSWERS AND SOLUTIONS Fig. 549 from the equations: d2 tan cp d2q> ---~(x2 + d2)- 1 X2 - Xz (considering triangles EOP, E AB ~nd OPOI ) and 1 1 1 -+-=-d2 .x?, /2 (considering E as the image of °1) . ' Hence DC}. ~ 0.011 em. In the third arrangement (d3=5 em) the path of the rays will be somewhat different (Fig. 549) than that ShO\\'D in Fig. 548. The equations for 'D8 can be written similar to the preceding cases as follows: Ds ds tan q> e: dscp and ..!.._..!..=..!.. (l- ds) +Xs Xs Xs ' da Xa fI Hence, Ds=O.18 em. 762. It follows from the formula of a lens 1 1 1 (i+1J=Fob that the magnification of the objective is b' Fob k1=(i=;-="P ob where b is the distance from the image to the objective. The real inverse magnification of the object produced by the objective can be viewed through the eyepiece as through a magnifying glass, the virtual image produced by this magnlfying glass being arranged from the eye at the distance of best vision D=25 em. According to the formula ~f a magnifying glass 1 1 1 al -D =Fey e where al is the distance from the image produced by the objective to the eyepiece. 1 he magnification of the eyepiece is D D+Feye k,,=-=---- at Fey e The total magnification of the microscope is Fob (D+Feye) k=ktk" 180 times (a-Fob) Feye CHAPTER 6 PHYSICAL OPTICS 6-1. Interference of Light 763. NOt it does not. The presence of Illumination minima on the interference pattern means that no quantity of light enters the given section of space. / 764. The maximum illumination will be observed at an arbitrary point of the screen C (Fig. 550) if the difference of the paths dl-d1 =kA, where k=O, 1, 2, ... is an integer. According to the Pythagorean theorem, d: = DI +(hk +;r d~=D'+( hk - ; rwhence d:-d~=(d,+dl) (d,-dt)=2hlc' In accordance with ~he initial conditlon d, +dt ~ 2D. Therefore, d, -d1 == =k).~ 2:~' .The distance tothe k-th light band fromthe centre of the screen is h,,-= k~D • The distance between the bands is Ah=hk+l-hk=).f . 785. The distance between Interference bands is Ah=~ (see Problem 764). In our case D==AB es:a+b, and 1=818, is the distance between the images Sl and 52 of the source in the ftat mirrors (Fig. 551). The value of I can cr----------------------------I I I I I 4 1 ~~~-----D -------:l~ Fig. 550 422 S, IT ...........In --.. I ~-----I U' .,---------- ~ ANSWERS AND SOLUTIONS N f7-=":_-=-1i:;§L~ ----- /I .---- ()(, r IV Fig. 551 be found from triangle SlSB: ; =2b ~ or 1=2ba Hence, 768. The second coherent source is obtained in Lloyd's experiment by reflection of the rays from mirror AD. In reflection the phase changes by n (loss of a half-wave) and the oscillations will be damped at point 0 where a bright band should be observed (a' minimum of illumination). As compared with Problem 764, the entire pattern will be shifted by the width of a light (or dark) band. 767. The illumination on the screen will be intensified when d2 - d1 =kA. The locus of points on the screen reached. by rays from both sources with this difference is a circle with its centre at point A (Fig. 552). For this reason the interference bands wi11 have the form of concentric circles. When l=n'A the illumination will be intensified at point A (an interference' maximum of the n-th order). The nearest bright interference band (circle) of k-l -~..------p----~/I Fig. 552 PHYSICAL OPTICS Fig. 653 423 the (n-I)th order is at a distance from point A determined from the equation d2-d1 = V(nA+Dr~ +h~-l-VD2+h~_1.=(n-l) A. Bearing in mind the conditions of the problem that A~ D and A. ~ I, we get hn-t~ V2D if an even number of zones can be accommodated in slit AB (Fig. 561). This figure shows four Fresnel zones. ·We have b=2kx, where x is the width of the Fresnel zone, k= I, 2, 3, .•.. Here AK is the difference in the path between the extreme rays sent by one zone: . ). AK=x sin q>=2 Hence, A. %=2 sin cp Therefore, the minimum will be observed in the direction

tha gives the direction to the first minimum (dark ring). According to the note, 2,sin

s=3A Hence, A=d (sin cps-sin cpt)=2d cos (CP'tCPa) sin (CPs; CPt) $!5 .d(CP8-CP2)=da~1.7 X IO-6cm 798. Sin q>= 1 corresponds to the maximum value of k. Hence, d k=r=4 197. For a spectrum of the first order to appear, d should be greater than or equal to A.. Therefore, the sought period of the grating must not be less than 0.02 ern. 798. The direction to the first maximum is determined by the expression d sin q>=A. The screen is arranged in the focal plane of the lens. Assuming dl the angle q> to be small, we have l =f are small and sin q> ~ ~ . 800. The spectra of different orders will be in contact if-k).,2=(k+ I) "'1· Therefore, _ Al _ k-r---;r-5 ""-/\'1 For this reason only the spectra of the sixth and seventh orders can be partially superposed. But this grating (see Problem 796) can give only the spectrum of the fourth order for this range of wavelengths. Therefore the spectra will not be superposed in our case. 801. When the rays are incident on the grating at an angle 8 (Fig. 563), the difference in the path between the waves issuing from the edges of adja- BI\ I \ I \ I \ \ \ \ \,\ ,\ To -first \mazlmum Fig. 563 434 cent slits Is ANSWERS AND SOLUTIONS 6=BD-AC=dsln tp-dsln a These waves add up and Intensify each other when d (sin cp-sin 8)=kl where k= I, 2, 3, ... for the maxima lying at the right of the central one (Jl=O) and k=~I, -2, -3, ••• for those lying at its left. The maximum order of the spectrum will be observed when q>=-9()O. Thus d (-1- ~ ) =kA. Hence, k=-6. The spectrum of the sixth order may be observed. The minus sign shows that the spectrum lies to the left of the central one. 802. As follows from the formula d (sin q>-sin 8)=kA (see the solution to Problem 801), the period of the grating will be minimum with tangential Incidence of the rays: 9=90°. In this case d d ~ . Therefore, the period of the grating should satisfy the inequality d~ ~ • 803. In the general case, as shown in the solution of Problem 801, the sought condition will be d (sin cp-sin 9)=kA It may be rewritten as 2dcos CPt 9 sin cP 2 9=kA q>+9 T 9 an cp S n cp PHYSICAL OPTICS S~-------+--+------~IF----';;;~~Fig. 564 or Upon solving this equation, we find that -1±Y4nl - l sin p 2 Y2 435 The solution with the plus sign has a physical meaning. For red. rays sin ~r e!! 0.26. Therefore, PI' S!: 15°6'. For violet rays sin Poee0.31 and Pv:::= ~ 18°6'. The sought angle is e=~o-p,. ~ 3°. Fig. 665 436 ANSWERS AND SOLUTIONS Fig. 566 805. For the red rays the focal length of the lens is t,.= 2 (n:-l) e: 27 em and for the violet rays If} = 25 em. According to the formula of a lens, the image produced by the red rays will be at a distance o"f b; = af'f = 58.7 em and that from the violet rays a- ,. at bfJ = 50 em. The image of the source on the screen (Fig. 564) will have the form of a spot whose edges are coloured red. The diameter of the spot d can be found from the similarity of triangles ABE and CDE: d=Db,-bf/ :=::: 0.15 cm b, 808. The sunrays falling on rain drops may be assumed to be parallel. As they emerge from a drop after being reflected once on its internal surface, the rays diverge in all directions. Only the rays subjected to minimum defle.. ction are about parallel. When these rays get into the eye, they produce the maximum visual impression. These rays travel, as can be said, with the maximum density. The other rays are diffused in all directions. As shown in Problem 748, the angle of deflection for parallel rays is 138°. Therefore, the angle between the sunrays and the direction to the rainbow is 42° (for red light) (Fig. 565). The eye will receive light from the drops that are in the direction forming an angle of 420 with the line passing through the eye and the Sun. For violet rays this angle is about 40°. 807 The first (primary) rainbow is observed owing to the rays that were reflected once inside the water drops. Upon refraction, the violet rays undergo the greatest deflection from the initial direction (see Problem 747) (L a grows with n, since r decreases). For this reason the external arc will be red and the internal one violet. The reflection rainbow is caused by rays that were reflected twice inside the drops. The approximate path of the ray is shown in Fig. 566. As can be shown, the direction to the rainbow is 51° with the line that connects the eye and the Sun. With two refractions and two reflections the colours alternate in the reverse order: the external arc will be violet and the internal one red. PHYSICAL OPTICS Fig. 567 437 The luminous intensity is much weaker after two reflections, for which reason a reflection rainbow is much less intensive than the primary one. 808. The geographical latitude of Moscow, i.e., the angle between the plane of the equator and a normal to the surface of the Earth, is q>= 56°. At this moment the Sun is in the zenith above the northern tropic (latitude a=23.5°). Hence, the angle between the direction to the Sun and the horizon (Fig. 567) is ~ = 90°- «p +(X = 57°30' A .rainbow can be observed only when the altitude of the Sun above the horizon does not exceed 42° (see Fig. 565). Therefore, no rainbow can be observed. 809. Our eye perceives a colour when its sensitive elements are irritated by a light wave of a definite frequency. The frequency of light waves, however, does not change during transition from one medium into another. 810. The green glass should be used. In this case the word will appear black against the green background of the paper, since the red colour of the word "excellent" does not pass through green glass. If the red glass is used, the word written by the red pencil will not be seen against the red background of the paper. 811. Camera lenses predominantly reflect the extreme parts of the visible spectrum: red and violet (see Problem 782). A mixture of these colours produces a lilac tint. 812. The colours of a rainbow are pure spectral colours (see Problem 806) since only a ray of a definite wavelength is seen in a given direction. Conversely, the colours of thin films are produced by extinguishing (totally or partially) of the rays of a certain spectral interval due to interference. The colour of the film will supplement the colour of this spectral interval. 813. Under the force of gravity. the soapy water drains down onto the lower portion of the film, which is always thicker than the upper one. Hence. the bands that show the locus of points of equal thickness should be arranged horizontally. The light-blue (blue-green) tint is obtained when the longwave (red-orange) part is excluded from the full spectrum (see Problem 812). When the midd Ie (green) part of the spectrum is extinguished, the remaining rays impart to the film a purple (crimscn) hue, and when the short-wave 438 ANSWERS AND SOLUTIONS (blue-violet) faTt is excluded from the solid spectrum. the film will appear as yellow. I the difference in the path of the mutually extinguishing rays forms the same number of half-waves in all three cases, the yellow band should be on top, followed by the purple band and by the light..blue band at the bottom. . 814. In the daytime the light-blue light diffused by the sky is added to the yellowish light of the Moon itself. This mixture of colours is yerceived by the eye as a white colour. After sunset the Hght-blue colour 0 the sky is attenuated and the Moon acquires a yellowish hue. 815. The smoke is seen against a dark background because it diffuses the sunrays incident from\ above. The particles of the smoke diffuse blue llght much more intensively than red or yellow light. Therefore, the smoke seems blue in colour. The smoke is seen in transmitted light against the background of a bright sky. The smoke seems yellowish since the blue light is diffused in all directions and only the long-wave part of spectrum of white light reaches the eye. 816. A thin film of water covering a moist object reflects the incident white light in cne definite direction. The surface of the object no longer diffuses white light in all directions, and its own colour becomes predominant. The diffused light is not superposed on the light reflected from the object, and for this reason the colour seems richer. TO THE READER Mir Publishers would be grateful for your comments Oil the content, translation and design of this' book. We would also be pleased to receive any other suggestions 1J0u may wish to make. Our address is: MIT Publishers, 2. Pervv Rizhsky Pereulok, Moscow. USSR. Printed in the Union of Soviet Socialist Republics