Tricky Physics Problems 6415i

Olutcly not sure that the high-jumpers can jump off at identical initial sptlC
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H . 25. The stone moves away from us until the component of its velocity parallcl tO its position vector decreases to zero. lf this never occurs then the condition of the problem is fulfilled. H. 26. The shape or the common envelope of t he jets of water st~g from the same place, at the same initial speed and following paraboric paths has to be detennined. Examine the condition of finding a jet of water cr~ing a given point in spac,•. H. 27. The process should be started from the top! The right strategy is to slide t he topmost body as far as possible, the n do the same thing with the one nn
4- k

H . 31 . From the mechanics point of view, a car is not a closed system, it is in with its surroundings, the Earth. H . 32. The focal length can be obtained using the relation between the lens law and the magnification. T he rat io of the brightness values does not only depend on the si?.c of the images but also on t.he amount of light. reaching t.he lens. H . 33. The magnitude of t.he vir tual image is not determined by the size of the image hut by its angle subtended at tile eye. H . 34. No light patcb can be seen either right by the quarter-cylinder or very far from it. T he closer light patch is excluded by total reflection, while the most distanL part of the light patch can be detennined by considering t.he part of the quarter of a cylinder dose to the t.able as a piano-convex lens. H . 35. The angular velocity of a suitably chosen simple pendulum can be the same in any position as that of the rod pendulum mentioned in this problem. Compare Lhc periods of swing of these two pendulums. The ratio of the periods of swing of two thread pendulums of different lengths deviated at the same angle can be 'guessed' using dimensional analysis.

123 TRICKY PHYSICS PROBLEMS

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H. 36. No unambiguous answer can b given to the qu tion : th direct.ion of h • motion d p uds on the magnitude of th coeffici nt of fricti n. How? H. 37. Write down th formula for th conservation of mat r. It is also to be n t.Jce
ur to the surface t nsion (pressure of curvatur ) of the water, th wate1 sc:on is low r than the atm sph •ri pr urc.

11. 38. Jll'L'!-iS Ht'<• i11 t hr

H. 39. Cal 11late th speed of the points of th thread in a given m m nt. H. 40. Sit into th (' elasti ') fram

fr feren

fixed to th thread.

H. 41. Find a . imple, math matically easy LO treat - motion in th rnurse of whi ·h tlw ball rca · 1es a speed as high as possibl , and th tim gained cover. thr. defidt cau. d by th long path. H. 42.

' l>rt>:i.'> Lile condition for equilibrium using

H. 43 . Sit

1111 11

tlw frame of reference fixed to the centre f mass of th beads.

H. 44 . 111 1he· 1"a1>I' of inelastic colJisions, th cluster of growin mass consisting m1 ru d b1•11
wlw t would happ

11

if the external force only acted until the first collision occurs.

H. 45. Th int rnal friction of the water can be as.c;umed not to b large, anno change the horizontal momentum of th jet of water. therefore thr gutt The diang m t h potential energy of th liquid should be n I cted compared to tb kincti(' nergy. Bernoulli's equation can be applied (indeed, several tim ). H. 46. Writ d wn th chaog in th pot ntfal and kine i ncrgy of the iuitia.lly standin liquid that starts to move at. (initial) accel rati n a, in a very . h 1 l tim int al D.t.

shows that the speed at which th sand runs dO\ n does not cl J><'J )(I on th amount sand in the upper part of the sand-glass. (Th explanation averag speed f t r this is that du to U1 friction between the grains of sand, tht> fflu ut and only d pends on its close surroundings, primarily on the diam ter f th orifi ., and n t n the effect of the distant paru . This is n t true for liquids, 30.) Thus, the i 'transmitted' to large dist.an ; see P bl 11 wh re th p tim in whi h th sand runs down has to b proportional to th ube of th initial h ·g in H of th sand. Find the other quantiti on which this time may depend. Apply t h m thod of dimensional analysis.

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H. 48. In tJt ' given circumstances, both the horfaontal and th vertical motion of th body can b approximated by harmonic osciUati ns. H. 49. Describe the motion in the (accelerating) frame of referen th train .

fixed Lo

H. 50. Examin th motion from the frame of refi r nee fixed to th wedge. H. 51. Find th condition for a long, thin thread to move uniformly on a 'syn uonous orbit', i.f!. at the same anglllar velocity as the Earth, above the Equator.

H. 52. The ater of the canal always tak the boat downwards by the sam distan as b which it approaches the given point at th opposite bank. H. 53. Calcula . how much the magnitude of the velocity f th thrust child and the component of his velocity pointing downwards on the slope chang in unit time. Find a relation between the rate of change of these two quantities.

H. 54. Compare the rate of decrease of the dist.a.nee between the ship of the smugglers and th r nu cutter to th sp d at which this latter moves away from th shor . H. 55. Due to the s mmetry of th problem, the bodies ar in th corners of a regular n-angle at ev ry moment and each of them mov as if only the gravitational attracti n of a single body (of a suitably hosen mass Mn) fixed at the centre acted on it. Th tim to fall into the centre of attraction can be calculated using Kepler's hird Jaw. H. 56. The force n.r.t.ing n th charge can be determined using th m U1od of what are called image charges. This for ·e - similarly to the force of gravitational attraction - is inversely proportional to the square of the distan , therefore th motion is similar to th motion (on a degmerat.e
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123 TRlCKY PHYSICS PROBLE MS

H. 58. A slow- motion picture of the phenomen on would show that right after the collision the first ball stops and rotates in place, while the second ball moves on but- with~t rotating. Thus, in the course of t.he extremely quick collision, t he first ball only transmits momentu m but no angular momentu m to the second one. After the collision, friction mO\-es the first ball forward, while it. slows its rota· tion. On the other ha nd. friction slmvs the translatio n of the second ball, while it makes it rotate rac;ter and faster. Hencefort h, the angular momentu m of (each of) the balls about some point of the t able remains constant.

H. 59. Examine the anguiar momentu m or tJ1e ball about some point of the table. H. 6(). From the e nergy and moment.um point of ,·iew, the system is not. closed, therefore these quantitie s are not conscrvt.'
H. 63. Calculate the force stretching the chain when it leaves the pulley. Use the principle of conservation of energy. H . 64. Gravitatio n can be ignored in the frame of reference moving together with the centre of mass of the chain, which shows the chain to be weightles s (but not massless!) . £xamine in what direction the forces acting on a small piece of the chain - with radius of curvature R and moving at a uniform spee
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H. 66. Examine the change of the horizonta l momentu m of the sand falling onto the band in some time interval (say 1 second). The considera tion of the energies is also useful.

H. 67. Apply the law of conservation of energy, then find out the force using the duUJge of the momentu m calculated from the speed of the 1011. H. 68. The gravitatio nal field of a thin spherical shell of uniform mass distribution is zero inside the shell, while outside the shell, it is tht• same as if thf' tota l mass of the shell were cooc.entra ted in its centre. H. 69. The gravitatio nal field inside a homogeneous sphere is directly proportional to the radius of the sphere (see the previous problem). The gravitatio nal field of the hollow sphere can be simulated by the superposi tion of the fields of .t homogeneous sphere and a smaller sphere of 'negative mass density'.

H. 70. Divide tJ1e hemisphere into equally Uuck hemisphere shells. Prove that these shells all cause tbe same gravitatio nal field in the point in question. H. 71. CaJculat-e the force a 'fairy giant' should exert to pull apart the two halves of the asteroid (cut in two) to 1 met.re. H. 72. The electric field exerts a force having a magnitud e proportio nal to the surface and a direction perpendic ular to the surface on each piece of the surface. This force is similar to that caused by the pressure cf liquids os gases.

H. 73. Find the maximum amount of hydrogen that tl1e container could initially bear without bursting. The material of the container can be cl1<)Sen freely but only from real materials. H . 74. The character s of the laws of gravitatio nal ancl electrosta tic fielcls aro very similar. Make use of this analogy and apply Ga uss' Jaw. ~o

H. 75. Examine t.he electric field caused by a small piece of the rod , at aogl<:> subtended at point C . H. 76. Two very long rods can be ed at their ends.

H. 77. Imagine that the solenoid is completed by another coil symmetrically to point P , in which the same current flows as in the original solenoid. Apply the principle of superposi tion. H. 78. The total charge ae<.-umula ted on each plate docs not chaugt: if the charge Q is 'spread' - b-eeping its distance from the plates.

123 TRlCKY PHYSICS PROBLEMS

H. 79. Imagin . that the medium is sliced into thin layers perpendicular to x. Then the individual layers can be considered as piano-parallel plates with difli rent raractiv • indic:cs aIICI th relation between the refrac.1.ive index of th layer and the angle of in ·iden of th light ray can be d tennined.

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HINTS

H. 87. The 'trap' hidden in the problem· related to the heat of vaporization. Th heat of vaporization of water at 100 °C and at a pressur of I atmosphere (th usual value of 2256 kJ/kg found in tables) is not only caused by the higher internal en gy of th vapour but also by th expansion work done against the atmosph i · pressure.

H . 80. Find th horizontal forces acting n the lay r of water thaL has ri · n . H. 81. During evaporation, the surfa e of th drop shrinks, th o its surfa · n rgy cl reasf!S. C mpare this ergy d ease to th energy necessary for 1~vaporat.im1 .

H. 82. Th p ure f saturated vapour 1s a littl • higher at th surfa e of th !'mall drops than at that of the large ones. Since the pressur of th vapour is uniform at the bottom of t he container, this value is higher than the equilibrium value for large dr ps, whil it is lower fi r small cl ps, therefore vapour evaporates from th small drops, making them ever smaller, and condensates onto th large n , making them ever Larger. T h relation hetween th pressure of saturated vapour and the curvature of th dr p can b deduced b examining the equilibrium of pressures in the case of a v ·set containing som liquid and a capillary tube hanging into it.

H. 88. A liquid st.arts boiling when the pressure of saturated vapour reacbe-; or sures that of th gas above the liquid.

H. 89. The cylind r keeps accel rating until the algebraic sum of the quantities of momentum receiv d per unit time from the parti I colliding from th right and from tlae left becomes zero. However, the cylinder can11ot move after a very long tim , because that would violate the second law of thermodynamics. H. 90. Th • entropy of the system cann t increase during the pro H. 91. Calculate the change of entropy of the air that gets into th contain r.

H. 92. Th electric field strength is zero inside the space ship, as in a Faraday Exam.in whether the electric potential of the space ship changes during the 111111n y.

r:a~ .

H. 83. Th in rease of the internal energy of the air enclosed in the container equals th > decrease of th potential energy of the body hung onto the piston. The friction l> tween the piston and th wall of th cylinder transforms these different types of nergy into each other. H. 84. If am untain is very high then its base m Its becaus~ of high presswe. ompare th n rgy needed to melt the bottom layer of a certain thickness of the mou11tain to the gravitational energy that would be released if the mountain could 'sink' . H . 85. Plotting th stat of th enclosed air on ap - V diagram, a straight tin is obtained. The beauty hidden in this problem is revealed when w realise what it means that th straight characterising the process is tangent to an isothermal or adiabatic line in certain points. H. 6. Th hydrostatic pressur of the water in the flue increas th pressure \ ater in th avity, therefore the wat boils at a temperature higher than the usual 100 °C. he relation betw the pressure and temperatur of saturated water vapour can b obtain :!
r th

p=

A

-L/(RT)

can b applied (where Lis th molar heat of vaporization of wat r, Tis its boiling point, p is the pressure of saturated water vapour at this temperatur , R is th gas constant and Ai· a constant). When the geyser bursts, the overheated wat in th cavity read1 its equilibrium stat by boiling until it cools down to 100 ° .

H. 93. Examine th change of energy of the spherical capacit-0r. H. 94. Compare the energies of the electrostatic fields of the two assemblies.

H. 95. The opposite charges of the capacitor plates attract each other, therefore work has to be done when pulling the plates apart. Meanwhile, the capacitanc C of the capacitor decreases, and as the electrostati energy of a capacitor is proportional t.o C (at a given voltage U0 ), th energy of U1e capacitor decreases! Thr solution of the paradox is that the capacitor connected to a battery cannot b considered as a closed system.

H. 96. Due to the current flowing in th same direction in each t.um, the sprin contra ts. Th force of traction caused by the current would be the sam if th · spiral spring were cooled to a very low temperature to mak~ it a superconductor. A urrent can How in such a supercondu ting coil ven when its ends are sh rt.cir ·uited. Examine the

l = 2E E2 •

Th dielectric slid between the plates of the capacitor clecrea.-;es th

elect1ic field (via its polarisation), therefore it decreases the energy of the system as well. The for e acting 1 the dielectric can be calculated from the change of energy (using th w rk theorem) .


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H. 98. Apply I flowing through the consecutive resistors and' th elements of the Fibonacci series. H. 99. onsid r two different cases: in th • first case, a current T flows into a grid point. In th second case, a urrent T flows out of a grid point neighbouring the previous on . ln both cases, make use of the symmetry f th system aud 'superimp ' thf! t.wo curreut and voltage distributions.

H. 100. Apply .t h method of sup position of the previous problem. B carefu l, h re th current has to flow out. f th circuit som wher because of th conservation of barges. Solve this dil mrna without spoiling th ymmctry of th problem. ff. 101. As in th previous twu prob! n, sup rposition can again b of great

help. H. 102. 111d equipotential points on the cube, while current I flow in at one end of the diagonal and current I flows ut at the other end. The circuit can be simplified by connecting th · equipot ' ntial points. H. 103. J\pply Gauss' law.

H. 104. Du to gravitation, th d starts moving downwards on th slope. lectromagn ic induction causes a current to flow in it, which brakes its motion according Lenz's law. Th equation or 111 tion of the rod (written iu of the ·urrent} is th sam in all three cases. Th difference is caused by the relation b tween th current flowing in th rod and the induced electromotive force.

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H. 109. A Moebius band is a surface that has no dir duction law can only be applied with great caution. lmagin the edge of th band is unfolded in sucl1 a way that it can wiU1out crossing itself. Exam in th surfa · , bound by th equival nt of this surface for the M bius banrl.

tion, therefore th inthat the wir following be laid out on a plau band here and fincl the

H. 110. An el tric field is inrlu ed in th rotating dis and this induces 1' current in th coil. In addition to th magn ti fl d of th E<:u-th, th magneti field of th cnil a.I n contribu · U> u1 • indu ·t.ion, increasing or rlccrea.sing it accordin to th dir ·tion of th r tatiou. H. 111. Th net magn tic flux of th super ondu tor ring ( 11"isting of thP flux f th . xt •rnal field and its own Hux) must not chang"C dm'ing rh • moti~n . The flux of th xterna.l ficlcl changP.S during the motion, but th magnetic flux of thP ·urrent i11du
H. 113. Describe the phCJ1011tl'1ton in a fram of referen moving togeth "';th th liquid. 'onsider firs ho' Lh cl tri aud magneti fields transform when one changes from some fram of refor 11ce to another one moving at. speed Vo wit.Ii respect to th first on . (Th s d 0 is much lower than the spee
two.

H. LOS. The d1ang of th speed of th rod is directly proportional to th chang of th chargf' of Ut capacitor. The rod accelerates until th induced electromotiv for e allows the capacitor di,charg ..

H. 115.

the relativistic formul.

ror energy and mom ntum cou rvation .

H . 106. A cording to the induction law, th current flowing in a coil cannot rhange suddenly.

H. 117. Examin the moti n of th

I trons in th wall of th Faraday cagt-.

ff. 107. xamin the forces I..-eeping the electrons in th m ta1 on circ:ular orbits. Knowing the el ctric field strength , Gauss' law helps determine the resulting charge distributi n .

H. 108. The magnetic field of the Earth induces a current in the rotating ring, which changes the average magnetic field at Uie centre of th ring, therefore the magnetic need! is deviated .

H .116.

7

B , cl ·ays via electro11 c:apt.ur .

IJ. 118. Th mass of Lh • pro n is much (nearly two thousand times} larg r than that of the positron . For this reason, t.he positrons mo • with a ruuch huger acceleration than th protons and ther will be a pe1iod in th ir motion when t,h positrons ar already far away but th protons have hardly moved at all . ff. 119. In t.h course of this pr (Comp n scattering}, the total en r and mom 1tum of the olliiling parti · es (photons + electrons} are constant. (It is useful to choos th r ·t energy Eo mc2 ::::: 510 keV of the el tron as the unit of energy for th cal ·ulations.}

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H. 120. Consider the electron as a spherical capacitor with radius r anrl wit.h a uniform surface charge distribution. The moment of inertia of a sphere with mass m and radius r is = K · mr2 , where K is a dimensionless constant depending on 2/5.) the mass dist.rihntion. (E.g. for a homogeneous sphere K

e

PART Ill. SOLUTIONS

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H. 121. Apply the uncertainty rcJation and the principle of energy minimum.

II. 122. When an electron is enclosed in a sphere of radius r, the unccrtamty relation prescribes a momentum p ~ h/r to it. Calculate the total energy (the sum of the electrostatic and kinetic energies) of the electron (using relativistic formu lae) as a func:lion of the radius ,. and find the minltnum of this funcliou.

H. 123. In tl1c case of small wavelengths, the speed of surface water waves i:. delea mined by surface tension. Examine the dependence of the speed of the (capillary) water waves on their wa\'elength and find out whether there is a lower limit for the wavelength of water waves.

S. 1. In the course of thermal expansion, the size of both spheres increases. The centre of mass of the sphere lying on the plate rises, while that of the sphere hanging on the thread sinks. Thus, the potential eJJergy of the first sphere increases, while that of the second one decreases. According to the rigorous application of the first law of thermodynamics, the heal transferred to the spheres does not only r.oven; the increase of th!! internal energy and the little expansion work done against the atm~pheric pressurl! (these are identical for the two spheres), but also the change of gravit.ational potential energy. The potential energy of the sphere lying on the insuJating plate increases a little, therefore the transferred heat cannot increase its internal energy so much, while the decreMe of the potential energy of the hanging sphere contributes to the increase of the internal energy. In summary, the temperature of t h1· sphcrP. banging on a thread will be higher. It is worth giving a numerical estimate. E.g. if the temperature of the iron baJJs with a radius of 10 cm is increased by 100 °C, a diffenmcc of 6.T ~ 5 - 10-s °C occurs in their temperatures, which is practically unnoticeable.

I

on

S. 2. The work done against gravity can be calculated from the increase of the height. of the centre of mass. The centre of m~ of the worm 'folded in two' can be found at the middle of the two halves, i.e. at one quarter of the total length of the worm, measured from one end. Thus, the centre of mass of the thin worm goes up by 5 cm, while that of the thick one by 7.S cm, the ratio of the amounts of work done is therefOl"e 2 : 3. Note. The centre of mass of the worm is not always at the same position with resp«:t t.o the worm; indeed. it may not be any point ol the worm at all. The centre or mass of the straight worm is obviously in its middle, that of the worm folded in two is at its quartes-. The centre of mass of a flexible body is therefore not 'stuck' to a given point or the body,

JLJl

its relative position may change. This is wbaL a high-jumper use when bis body arcs above the lath but bis centre of mass remaln.s below it (see also Problem 22).

S . 3. 1st solution. H the spatial (planar) and time coordinates are determined in a right-angled frame of reference with axes x, y and t (a spaoe-time diagram), the 'world-lines' of the snails in a rectilinear uniform motion will obviously be straight


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lines. Encounters occur when two snails are at tl1e same place at the same time, i.e. when two world-lines intersect. ' According to the statements of the problem, the four world-lines (a, b, c and d) certainly cross each other in 5 'points'. (Intersections mean events in i.- pace-timc.) Let us denote these events by A, B, C, P and Q. Points A, B and C determine a plane (the plane of world lines a, band c), and since P and Q also lie in this plane, world line d has to be in this plane a<; well. This means that world lines c and d also cross each other, therefore the sixth e.ncounter will by all means take place, too. 2nd solution. As five encounters have already occurred, there has to be a snail who has alrearly encountere
SOLUTIONS

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also go round this point at a tangential speed of v/2. It is easy to show that t he snails are init.i ally at a distance of (./3/3) · 60 cm from the centre of the triangle, therefore they meet in

( VJ/3} · 60 cm r.; ( v ->/2) · 5 cm/rrun

. = 8 mm.

It is worth generalising the problem for the calculation of the trajectory_ Examine the motion of a borly moving at a velocity of constant magnitude v around a fixed point, while the angle of velocity with the posotion vector of iuitial length r 0 is constant, 0 < er < 90°- If the radius turns by a little angle 6 , while its lengUi changes by -Ar (Figure 9) then the condition of constant angle between the velocity vector and the radius implies that Llr() A

Figure 3.

= - r() · cot a .

61')

This equation is very much similar to the equation 6 = -m(t) · >.of radioactive decay, the known solution of which is m(t) = mo · e-.>.t. Using this analogy, th.at the equation of the path of the motion is question (in polar coordinates) is

3

=

=

u

vVJ/2 = 5 cm/rrun · ../3/2 Figure 2.

(it is obvious that this is where they meet), while they

This is the equation of the so-called logarithmic spiral showing that the radius r only tends to zero after ma.king a turn of infinite angle, i.e. the point-lib.-e body reaches the centre (in finite time and covering a finite distance} after infinite turns around the origin. Note_ Flying night insects try to follow straight paths by keeping a constant anglP with respect to a distant light source (e.g. the Moon). If a nearby light source misleads them then, according to the problem, they follow a spiral path and fall into a trap. As neither the insects, nor the tight sources are point.-like, sooner o.r later the insects hit thP lamp.

t

S. 5. The avei·age speed of the litUe something is m/s. Since the motion slows down uniformly, Vgve•age Htl;,,itial + Vfu.aJ), its initial speed could not be greater than 1 m/s (as tlfinftl ~ 0). Thus, the speed of the body deaeased by a maximum of 1 m/s in two seconds, the absolute value of its acceleration was maximum !al = 2 0.5 m/s , i.e. 1/20 times the gravitational acceleration. The coefficient of kinetic friction could therefore not be greater than 1/20 between the little something and the surface of the table. This is much smaller than coefficients os friction between ordinary materials, therefore it is very likely that the little something does not slicie but rolls.

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S. 6. a) The short.est. path is in the direction perpendicular to the bank. The boat. goes in this direction if the boatman rows in the direction shown in Figure 1. Thus, the speed of the boat is ./5 m/s :::::: 2.24 m/s with

the bank. The boatman has to row against Lhe stream aL

angle o with respect. to the bank. Since coso = 2/3, ti.Us angle will be o:::::: 48.2°. river b) ln this case, the current of t.he river is so strong that the boat goes downwards even if the boatman rows F19ttrc: I . at. full speed against the stream, tl1erefore he cannot get ac1oss perpendicularly to the bank. This means that, in contrast to the previous caw., lw cannot go in an arbitrary direction with respect to the bank. Tht> possible directions he can take can be determined by adding alJ the possible <;t;,nding \WllP.r vrJocitie:s of the boat to the velocity of the river. Draw the velocity \"l'("tor of the river and, from the endpoint of this vector, draw velocity vectors i~ a.II directions, with a magnitude equal to the standing water speed of the boat . The endpoints of these vectors will form a circle as shown in (Figure!').

4 mis

SOLUTIONS

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torque about. point P in oontact wit h the inclined plane (see Figure J), therefore the sphere will surely roll down. It is to be shown that t his is the situation for a sphere made of t.wo homogeneous hemispheres - whatever the densities of the two halves.

Figure 1.

Consider t he part marked in grey in Figure 2. The centre of mass of this part is obviously in point A (for symmetry reasons). i.e. at distance r/2 from the centre 0 . The rest of the sphere 'pulls' the centre of mass even closer to point 0, i.e. OS < r/2. And this, according to the above consideratioos, implies that the sphere cannot remain in equilibrium on the 30° inclined plane. (In the course of the solution, rolJing resistance is assumed to be small, i.e. no torque can act at point P . In the case of a surface covered with velcro, this is obviously not true, the sphere may even stick to a vertical surface.)

nver

S. 8. Viewed from a coordinate system fitted to the inclination of the plane, the ball is seen to bounce on a horizontal plane in a field of gravitational acceleration g' g oos o, with an additional constant 'horizontal' force (of magnitude mg sin a) acting on it. The 'vertical' motion consists of bounces of identical heights, i.e. of identical periods. Meanwhile, the ball accelerates uniformly 'horizontally', its average speed between the bounces increases evenly, therefore the distances between two consecutive bounces follow an arithmetical progression.

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Figure 2.

Figure !J.

The possible resultant velocities of the boat can be obtained by ing the starling point of the velocity vector of the river to the points of this circle. The shortest path will be the steepest one with respect to the current of the river, i.e. when the line of action of the resultant velocity vector is tangent to the circle, see

Figure 3. Thus, the velocity of the boat with respect to the s hore is .,/7 m/s ~ 2.65 m/s. Again, t.hc boat.man has to row against the stream, but. this time at angle /3 witb the bank. Since 006(3 = 3/4, thus (3:::::: 41.4°. The figure also shows that in this C
S. 9. Determine the force of attraction between the Sun and t.he Earth ing for the centripetal force that keeps the Earth on its approximately circular orbit, and express the angular velocity w in of the periOd T of the revolution : mM

47r2

G-;:;- = mrw = mr Tl , 2

where m and M are the respective masses of the Earth and the Sun and r is the average distance between them. Simplify by m and express M in of the average density p and average radius R of the Sun: ~R37rp 4tr2 G ~= T2r.

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This yields

for the period of revolution. It can be seen that the period of the Earth only depends on the universal gravitational constant C, the average density of the Sun and the ratio r/ R, i.e. if the density of matter is left constant then any reduction of the solar system leaves the length of a )1'.aT w1rhanged. It can aJ50 he :;ecn t hat only t he density and the size of the Sun count are significant, the data of the Earth do not: any body that is small wit.h respect to the Sun would have the same period on the same orbit. Note. This result ca11 al'IO he obtained using Kepler's third law:

~:

= '/;,..: (a is the

major sentiaxis of the elliptkaJ orbit of the Earth) When the mass of the Su11 is expressed in of its average c!cnsity them a proportional reduction does clearly not changP. I.he period of planets in elliptical orbits either.

S. 10. The gravitational acceleration caused by the Sun at the position of the Earth is the same as tha t causrd h~· one of the stars at the position of the other one. According to Newton 's J;1w of µ.1 a\ it.ion, 9 (the gravitational force acting on a uuit mass) only depends on tlw ni . . ...,., nf the centre of attraction and on the distance from it, and these quantities at'f' itlt•nt k<1l in the two systems. Thus, the of lht' bina1y !olar move with the same acceleration as the Earth but the radius of their orbit is only half of that of the orbit of t he Earth, which means that the square of their angular velocity has to b e twice as large (n = rw2 ). The period of the biuary s tar in question equals therefore t hat of the Earth divided by ../2, i.e. 8 months and a half.

=

S. 11. n) The acceleration of a satellite moving at speed v in a circular orbit of radius R is 9 = V- / R. If R is the radius of the Earth (more exactly: a slightly larger value) theu g has to be the gravitational acceJeration on the Earth, which yields the 'first cosmic speed': v 1 = .J1f9 = 7.9 km/s for the speed of the satellite. Such an initial spec
SOLUTIONS

47

t.o be taken into if the satellite were launched from one of the poles or its neighbourhood; there are obviously tedmical obstacles w this.) The initial speed necessary to put a satellite on a polar orbit is therefore at least ~:: ::::: 1.06 limes greater , i.e. the kinetic energy necessary for this is at least 1.13 times great.er. This does not seem to be a big difference, but in rr.ality the slightest increase of the initial speed requires an enormous effort. The reason for this is that the carrier rocket has to be arcelerated together with the satellite, thus the mass to be launcl1ed increases exponentially with the aimed final speed. c) In order to leave the attraction of the Earth (i.e. to move very far from t be Earth), the escape speed or 'second cosmic speed', vi = ,/2iR = J2v 1 ::::: ::::: ll.2 km/s is the r~uired limit. The rotation of the Earth can again be used, therefore (law1ching eastward from the Equator) a speed of 10.7 km/s is sufficient.

d) The Earth revolves around the Sun at a speed of approximat.cly 30 km/s. In order to reach the Sun, a space proce has lo be lawiched at an initial speed of 30 km/s (ruore exactly: this value can be 0.5 km/s smaller due to tht> rotation of the Earth). If the aim is to leave the solar system, the space probe has to reach ../2 times the speed of 30 km/s, but the ' initial speed' caused by the rotation of the Earth can again be subtracted from this value if the launching is well timed. i.e. a speed of 12 km/s is :rufficient. Thus, it is easier to leave the solar system than to send a space probe of identical mass into the Sw1. (The previous atten1pt has been successful , the latter is still to be waited fo1·.) The situation is even more favourabl e if the possible effects of the outer planets (Mars, Jupiter , Saturn, ...) are taken into consideration . A space probe launched at the right time and in the right direction can significantly be ' pushed' by (reccive energy from) these planets: this is t he phenomenon called gravitational catapult.. This way, the space probe does not even have to go 'very far', it only has to reach Mars or Jupiter, the rest goes 'automatically'. S . 12. The rocket has to reach the hig).lest. possible total energy. Jf the zero le vel o f gravitational pot.P.ntial energy is 'infinitely' far away then the energy of the rocket standing on the surt-ice of the Earth is negative. The energy released during the operation of the engines increases the total energy of the rocket, and the rocket. can leave the attraction of the Earth if the sum of its potential and ki netic r.nergy becomes positive. ThP. P.nergy relea.cmcl in the course of the operation of thP. principal and the auxiliary engines increases the total energy of the rocket and of the ejected combustion products by a certain value. This increase is independent of the moment when the engines are switched on, but the speed at which the combustion products fall back onto the Earth does depend on the timing of the operations of the rocket. Indeed, if t he auxiliary engine starts working at a greater height, the combustion products start falling higher, therefore their speed and total energy are higher when they hit the ground. This means that the sooner the auxiliary engine switched on, t he higher the energy remaining with the rocket. (The same is valid for the principal


48

engine: if only the energies are taken into consideration then it is worth operating the engines for the shortest time and at the highest intensity.)

side oft.he blades warms up more, theref
S. 13. The mass distribution, and thus the position of the centre of mass changes from instant to instant due to the sinking of the ball. In some time t , the centre of mass is displaced by o,,V - 01.V M

s=vt · - - - (whP.re M is the total mass of the system, f!h and er. are the respective densities of t he honey and the ball), since the steel ball displaced by a distance of vt can be considered as having changed places with a 'honey ball' of identical volume. Thus, the t-0tal momentum of the system is

Ms

lcou.1 = -

t

= ver, V -

V{!h V.

The first term of the right- hand-side is the momentum of the steel ball, therefow t he second i.i; that of the honey: /honey

= -v{!hV = -2 g cm/s.

The negative sign shows that the momentum of the honey points vertically upwards. and its magnitude is identical to that of a honey ball moving upwards - with identical but opp osite speed to that of the ball.

S. 14. The (average) kinetic energy of the gas molecules is proportional to t he square of their velocity, the internal energy of the gas is proportional to the temperature, t herefore v 2 ""' T. If the wall is warmer than the gas (T1 > T) then the average speed of the reboWlding gas molecules will be higher than before the collision (the wall warms the gas). If the wall i.s colder than the gas (T1 < T) then t he situation is just the opposite: the molecules rebound with a lower speed (the gas cools down). l n the language of molecules, gases press the walls of their container because the momentum of molecules hitting the wall and rebounding from it. changes. The change in the momentum of molecules rebounding from a warm wall is greater than the change in the momentum of molecules rebounding from a cold wall. Thus, the gas exerts a higher pressure onto a warm wall than onto a cold one. Note. This phenomenon gives the explanation for the strange rotation of the racliometer ('light wheel'). If on.e side of each blade of a wheel mounted with a fine bearing is black ancl t.he other one shiny then the wheel starts turning when it is lit. At first sight, one could think that the wheel is rotated by the pressure d light. ThiS, however, is not true, sinc.e experience shows the shiny side (that reftects light, causing a greater change of momentum) of the blades to move forward! The acceptable explanation is that the black

49

SOLUTIONS

·---- ------.!

S. 15. T he simple circuit consisting of two identical resistors shown in the figure fulfills the given requirements.

S. 16. When the bucket is filled with water, the centre of gravity of the water is above that of the bucket, therefore the common centre of gravity is the highest and the period t he shortest. AB the water starts leaking out, the common centre of gravity moves away downwards from the point of suspension and the period becomes longer. When there is a medium quantity of water in the bucket, the centre of gravity of the water is below that of the bucket and the common centre of gravity moves even lower, significantly lower than the centre of gravity of thP. bucket, thus the period increases further. When there is no water left in the bucket, the centre of gravity is higher than in the previous cases (at the position of the centre of gravity of the empty bucket), i.e. the centre of gravity stops moving downwards at some point and starts moving upwards again. The reason for this is that there is so little water in the bucket at the end that. despite its low position, its small mas.c; cannot significantly affect the position of the common cent.re of gravity. In summary: the longest period can be experienced when the common centre of gravity is at its lowest position, then as the centre of gravity moves upwards, the period shortens. Nott!.. A similar explanation can be given to the question: when do bottles of drinks tnrnble over the least easily (e.g. on t.he table of the restaurant in a moving train)? The most stable state is when the common centre of gravity of the system (drink+ bottle) is in its lowest position, i.e. when there is some drink left in the bottle.

S. l 7. Assume for a moment that the cent.re of t he pot remains at the same height as originally. + Then the pot only turns about its centre (maybe moves sideways) and does not sink any deeper into t he water than previously. In such a case, the rim of the pot sinks to the surface of the water a t the side opposite to the chain, and water flows into the pot. It will be proved that this cannot occur. The upthrust acting on the body in the assumed situation remains the same as before, i.e. it equals the toLal weight of the pot and the soup. Thus, the chain would not exert any force onto the pot. On the other hand, the centre of mass of the pot is not on the line of action of the upthrust when the pot is tm;.ned away, therefore the equilibrium of the torques can only be fulfilled if the chain pulls the pot upwards.

I


50

The two contradictory conditions show that our assumption was wrong. The gl'<>m<>tiical centre of the pot cannot stay in place but it has to rise (since a smaller 11pthnist is sufficient when the chain is tight). And this implies that even the lowest point of the rim of the pot has to remain above the surface of t he water. T he p<>S.'>ibility of the soup flowing out into the water has also to be taken into consideration. This could occur if the level of the soup were higher in the pot than the water level in the lake, i.e. if the density of the soup were lower than that of wat ca. In 1ealistic circumstances, this is not the c:ase. S. 18. The soap buhble floats, therefore tlw mass of its waJI and of the helium inside it equals that of the displaced a ir. Since the density of helium is lower t.han half of t he dern;ily uf air, the mass of the helium is smaller than half of the mass of the displaced air. Thus , the waJl of the bubble has to be heavier t han the gas rnclosed in it. S. 19. I.u case 11}, the ,vat.er can obviously not flow out of the tube, or else a perpetuum mobile could be built: a paddle operated with the water Aowing out could move perpetually. Cases b} and c) are not this simple The en H); on t.he other hand, the water does not Aow out of tube b). S. 20. The system described in the problem is spherically symmetrical, therefore l he mngnetic field that is built up has to be spherically symmetrical as well. A spherically symmetrical vector field has to be of radtaJ direction everywhere and its magnitude can only depend on the distance from the origin: B{r) B(r) · r/lrl. On the other hand, u magnetic field is sourceless: the magnetic flux crossing any closed surface has to be zero at. any moment. E.g. choose a spherical surface of radius r around the origin. The condition of sourcelessn ess can only be fulfilled if B(r) O for any r . This means that the current described in the problem builds up no magnetic field, either outside the spherical capacitor OI" insidP. it. Note. It is worth examining how the basic laws of electrodynamics are fulfilled between the plat.es of I.he spherical capacitor. Is it true that a magnetic field builds up

.U .U

,LJ

=

=

51

SOLUTIONS

around a current Bowing in a conduct«, and the rotation of this field is proportional to the current? S. 21. T he radiation has to be spherically i.-ynunetrical, since both t he distribution and the motion of the charges are spherically symmetrical. The magnetic field is radial in every moment and should be of identical magnitude at identical distances from the sphere. This, however, is impossible, since such a magnetic field {unless of zero magnitude) would imply a magnetic charge (magnetic monopole) that does not exist in nature, according LO experience. A similar reasoning shows that the electric field is also spherically symm etrical and its magnitudP. only depends on the total charge of the surface of the sphere and not on t he pulsation! Therefore, only the st atic Coulomb field can be observed outside the sphere, t he sphere emits 110 radiation at all! If particular pieces of the sphere are examined, these do behave Like dipole antennae and emit radiation. But the radiation of different. pieces has to be summed up taking into . the phases as we ll, uot only the magnitudt:l!. The radiations of the many dipole antennae cancel each other out, which can be proved by a direct {but rather long) calculation without referring to spherical symmetry. S. 22. No high~jump competition bas been held on the Moon so far; however, we shall try to give a numerical estjmate for the result to be expected. The record of men's high-jump on Earth is a \'alt1f' ;i hove 240 cm nowadays. A good highjwnper is taller than 190 cm with a ma~" .1rnund 80 kg, and his centre of mass is approximately 110 an high. In the c~ of a successful jump, all the points of his body above t he la th, but not his centre of mass. This achievement requires a special jumping t echnique that can be observed when watching video records of jumps slowed down. The centre of mass of the jumper remains by approximately 20 cm under the lath when the jumper is at the top of his jump. The most difficult part of the estimation is the comparison oC the circumstances of the movement when the jumper leaves the ground on the &rth aud on the Moon. Assume that the centre of mass of the jumper rises by 40 cm from the lowest point (right before the jump) and the highest point {when leaving the ground) in the stadium both on the Earth and on the Moon. Then his muscles do a work sufficient to lift his centre of mass from 110 cm to 220 cm , i.e. by h = 110 cm on the Earth. (The effect. of running up is not taken into a ccount or it is assumed to be identical on the Earth and on the Moon.) The basi~ of our estimation is the assumption that. the some wm* and the same jumping technique is a pplied by the jumper at both places. The work done increases t he kinetic energy of bis body and the potentiaJ energy of his centre of mass: W = mv 2 /2 + mgs. When leaving the ground, hi'> speed can be calculated using the relation v 2 - 2gh. Thus, the total work done is

W

= mg(h + s) = 80 kg· 10 m/s2 · 1.5 m = 1200 J .

We assume that the work done by the jumper is the same on the Moon, while the rise s of his centre of mass before leaving t.be ground is also the same. The

52

123 TRJCKY PHYSICS PROBLEMS

gravitational acceleration on I.he Moon is only one sixth of that on the Earth, i.e. (mg/G) . (s + h'). which the ('nergy C'Qtmlion of the jump on the Moon is 1200 J yields h' ;:::: 8.G m for the rising on the Moon. The height of 1.1 m of the centre of mass and the extra 0.2 m due to the special technique have to be ;idded lo this ' yielding a record of 9.9 m ::::: 10 m .

=

Note This problem as usually settled only mentiorung the difference in the gravitational accelcrntious hy saying that the world record on the Moon wo uld be six times that ou the Earth. approximately 14 - 15 metres, which is au exaggerat.ed expectation ac.cording lo thP. nbow reasoning Our estimate is probably still P.xaggcrated, since in th1~ ra.~ the J UnlJWf laa.o; to achiP.ve the same wo rk m a short.er tune, 1.r . a greater power o u1 put is nf'e ult o f 8 Ill can be obtained, which i,ci too low n value. Coru.-idcring all thi5, t l1r m1,.;t r>roh;lhlc~ valuP. : .ru.o; t.o be n height of about 9 m .

S. 23. a) The vertical acceleratiun of ball B falling from the table is g all lhe time, whid1 makes it possible to determine the lime (apprmumately half of a S<'Cond) it takes lO fall I metre. The motion of the bob of the simple pendulum is rather complicated (swinging with large deviations), therefore the time necessary for it is uol !'limple to determine. What can be stated with certitude is that its V<'Ilical 11rceleiation is smaller than g all the time (since the thread partly pulls it upwa rds), therefore a longer time is necessary fo r lhe vertical motion of ball A than for the free fall of ball l:J. b) The bob of the pendulum describes a quarter of a circle (a path of approximately 1.5 m). The other ball (ball l:J) follows a parabolic path, the length of which cannot Ix- determined by elementary methods. However, it can be seen that it hits ~he g,-oun
~.·.

/ Figure I .

p

CD Figure 2.

53

SOLUTIONS

One can expect to find the shortest time bet.ween these two directions. First, let us prove the following au.xiliary theorem: bodies starting at the same time, from the same point and following frictionless slopes in different directions are always situated along a circle. According to Figure ~ the topmost point of the circle is the starting point. When time t has elapsed after the st.art, the diameter of the circle has to be d = gt2 /2 (free fall). A body moving on a slope at angle o with the vertical plane has an acceleratio n or gcoso, iLs path to the circumference of the circle is dcoso, i.e. it takes this body the same time t to cover this distance as to the vertically moving one. Thus, the auxiliary theorem is proved. The original problem is easily solved using the auxiliary theorem. The bodies starting at the same time from point P in different directions are always on a cirde that grows with time and its topmost point is always point P . After some time, this growing circle reaches the inclined plane: the slope is tangent to the circle in point P'. In order to obtam the straight line along which the time to reach the inclinoo plane is the shortest, points P and P' have to be ed. (In fact, the problem is three-dimensional. The bodies starting from point P are on a sphere in each time instant and the direction with the shortest. t.ime is given by ing point P and the point ci the sphere tangent to the inclined plane. However, it is sufficient to examine the vertical plane perpendicular to the inclined plane, as we have tacitly done so far.) It is clear from Figure 3. that in the case o( a plane inclined at <1nglt> ''- the line PP' corresponding t.o the shortest time is at angle a/2 with respect to the vertical plane, Le. the most favourable direction is exactly the bisectrix of the two directions mentioned at the beginning of the solution.

=

S. 25. The motion of the stone can b e described by the following relations in

t.he coordinate system shown in the figure:. z

= t.totcoso,

y V.,

y

= tJoCOSa,

. = Vo t Slll

v 11

g 2 ' 2t

Q -

= t.tosino -

gt.

The stone is at the greatest distance from the origin when its velocity is at right angles to iLs position vector. The condition for this is II

v,.

x

v.,'

- = --

Figure 3.

which yields a second order equation for the time: = 0. t2 - 3t1o ;in at +

2'j

54


. ..

.

. .

. .

When, the d1scnmmant of this equation as negative, 1.e. when

< 4( at

;j)

2

(3Vo sin o)

2

9

<

then the stone permanently moves away from the origin, it is never

th~test

c_listance'. The

con~ition

< .,/8/9 = 0.94, 1.e. when o < 70.5 .

is fulfilled at the angle for which sin o

<

S. 26. The water bell is cylindricalJy symmetrical with respect to thP vertical straight line cross in.g the sprayer ,

therefore it is sufficient to solve the problem in one plane, iudeed i11 oue halfplane. Let the point-like sprayer be at the origin of a coordinate systt>m :r: - y , thus the jets of water follow parabolic paths starting from the origin. Mathematica!Jy, our task is to find the envelope of this multitude of parabolas. It is well known that the equation of the path of a body thrown with initial speed v and at angle o with thP. horizontal plane is x

y=:rtano -

?

9

2trCOS2

0

x

1

,

which can be written as

gx2 . u 2

-

2.,2

-

gx2) x .u + ( y + . 2t12

=0

.using the notation u = tano. If (x, y) is fixed then the above relation is a second-order equation for " that has a real solution if its discriminant is non-negative:

gr ( 2v

x 2 - 4 -·-2 i.e.

vz

2 gx ) > O y+2 2v

g

-

'

t

Y ~ 29 - 2v2 x · This inequality divides the plane x - y into two regions separated by a parabola. Water can reach the points under the parabola (paraboloid of rotation in three ilimensions) but not the points above it, the limit corresponding to I.he aimed en-

velope. The water bell is therefore a paraboloid ofrotation. It is clear from the equation of the limiting curve that the height of the water bell is ~ /(2g}. as one can expect it according to vtlrtical throwing. The water bell draws a circle into the water of the

55

SOLUTIONS

=

basin the radius of which can be determined from the condition Y = 0: r ti / g. This :neans that the diameter of the basin should be at least four times the height of the water bell to receive the fa!Jing water. S. 27. Start the process from the top. (Choose the length of the bricks to be unity.) The topmost brick can be displaced until half of it sticks out, then the top two ones have to be moved on top of the third one as shown in the figure. The common centre of gravity of the top two bricks caunot be further than the edge of the third one, therefore the second one can be displaced by ~ . The strategy is to displace the bricks until their common centre of gravity is above the edge of the next brick downwards. Before the t.hird displacement, the common centre of gravity of the third and the fin,-t two bricks bas to be found. The common centre of gravity of the first two bricks is above the edge of the third one and counts with a double weight, i.e. the distance of ~ has to be divided in the proportion 2 : l ; the third brick has to be displaced by In the case of the folJowing brick, the fourth one, the three placed above its edge count with a triple weight, the distance of between the centre of gr:11vity of the fourth brick and its edge has to be divided in the proportion 1 : 3, 1.e. the fourth brick can be displaced by ~- Adding the displacements up, the result is ! + ! + ! + ! = ~ > 1, i.e. the topmost brick can hang 01Jtside the table. 2 4 6 s equilibrium 24 • · ·in rea1·t Unstable positions have been coDSJdered al I th e time; 1 y, the displacements should be chosen to be a little smaller than the above calculated values but it can still be realised that the topmost brick hangs outside the table. The solution can be extended for an unlimited number of bricks. The distance of ! between the centre of gravity and the edge of the kth brick has to be divided in ihe proportion as the common centre of gravity of k - 1 bricks is situated above its edge. The kth brick can therefore be displaced by a maximum of units. If a total of fl bricks is available then the displacement of the topmost one can be calculated as

i·

.

t

6,

A

!2 (1 + 2~ + !3 +!+ .. . +_kl + ... + .!.) . 4 fl

Since the sum of the reciprocals of natural numbers tends to infinity

1+

111

a.5

11111

-2 + -3 + -4 + +-5 + -6 + -7 + -8 + -9 + ... >

(J

1 1 1 1) 1 I l >21 +2+ 4+41) + (18+8+8+8 + ... =2+2+2+ ... , an arbitrary displacement can be realised using a suitable number of bricks.


56

. 28. Air resistance i neglected and the balls are considered as perfectly elastic. If the balls are dropped from height h, they reach the ground at speed v = ./29fi. First, the bottom ball then the top ball hits th ground. The top ball re ives the greatest part of the available energy if the bottom ball is at rest after the two collisions. The bottom ball first bounces back from the ground at speed v, then collides with U1 • top ball moving downwards at speed - v. Since the speed of the ball of mass m 1 is zero afte1· the collision, th equations expressing the conservation of mom ntum and nergy ru· and The speed 11 of the f.op ball after the collision and the ratio of the masses can be calculated from these equations: u 2v and mi/m2 l /3. Due to the double speed when bouncing back, the top ball rises at a height of 4h . It is urprising that the top ball can bounce even higher than this. If »1'1 :;$> m1 then the top ball only takes a very little share of the total energy after the coUisions, but it spe rl is 3v and the height of the bounce is (in an ideal case) 9h. This sounds rather incredi ble>. hut it is in agreement with the principle of conservation of energy. Readers interested in theoretical problems may generalise the problem for n balls, while those interested in experiments may try to drop different balls. The balls bounce in very amusing ways!

=

Figure 1.

=

S. 29. Since the dumb-bells approach each other at identical speeds, the sum of their momentums is zero in the frame of reference fixed to the aircu hioned table (to their common centre of mass). Thus, the conservation of momentum means that the centres of mass of the two dumb-bells always move at identical speeds and in opposite directions. When the dumb-beJJs collicle, both their energy and their angular momentum are conserved, since the collision is perfectly elastic and no external torque acts on them. Write down the equations for energy and angular momentum conservation, the latter with respect to the point of P of the dumb-bells. The states before and after the colJision are shown in Figure J. Before the colJision, the dumb-bells only have translational kinetic energy, while a rotational term appears after the collision:

57

SOLUTIONS

2(

~ 2mv2) = 2 ( ~ 2mV

2

+

~ 2ml w 2

2 )

•

Before the collision, the dumb-bells only have orbital angular momentums, but a tenn describing their spin appears after the collision, due to their rotation about their centres of mass. 2(l · 2mv) = 2(l · 2ml'

+ 2ml 2 w) .

v The non-trivial solution {t• fc V, w 'I O) of the above set of equations is round to be \I -:: o. i.e. the centres of mass of the dumb-bell stop after the collision. The angular velocity of the rotation is then w = v/l, which implies that the :rc/(J) colliding point mas.5e5 change velocities, while the non-colliding ones keep their old velocities. This can be interpreted the following way: the point masses ed by a rigid but weightless rod are . £ not aware of each other's presence in the course Figure · of the collision occurring in a very short time, the rod only starts playing a role right after the collision, when the dumb-bells rotate in place. · . . The hidden point of interest of the problem is that the dumb-bells collide ag:UO after one turn (a time t fr/w). Using the previous results, the consequent motion ~an be foretold without writing equations. The rotation of the dumb-bells stops, they move again with the same speed as before the first collision, their path is th same straight line but their motion is 'upside down'. In other words the dumb-bells spend the time b etween the two collisions by 'turning upside down' . The speed of the dumb-belJs in function of the time i · h wn in Figure ~-

=

s. 30. Let x denote the ratio of the actual wat.er level to the maximum level in the basin. (This ame number shows the ratio of the actual volumes of water to the ma.ximum volume.) During filling, x increases uniformly with time, and sinc.e it reaches the value x l in time Ti,

=

1

When water flows out, the speed of efflux - therefore the rate of decrease of x - is proportional to the height of the column of water, i.e. to the square root of x:

(Ax) At

= -K . .,JX. out


58

T he coefficient of proportionality has to be chosen so that x decreases from 1 to 0 in ex~ly time T2. Since the equation for the efflux is of the same form as the relation v .J2aX between the speed and the displacement for an uniformly changing motion, it can be concluded that the liquid level decreases to zero at an uniformly changing speed. The initial rate of decrease is K, the final is zero, therefore the average rate of decrease of x equals K /2. This rate of decrease can be expressed in of the total time: !(/2 = 1/T2. When both the tap and the plug-hole are open then tht> rates of change caused by the water flowing in and out add up:

=

t.x _ (t.x) + tl.t -

tl.t

in

(t.:x) _Ti2_ _ T2~ Vx. tl.t out -

ln an equilibrium (stationary) state, the water level {and x) do not change. The condition for this is that

E .g. if Uae basin is filled up iu the same amount of time as in w l1id1 it becomes en1pty (T1 = T 3 ) then the stationary state realised when opening tli<' u1p and the plug-hole together corresponds to x 1/4 (regardless of the initial condit.ions), while with the data given in the problem, this ratio is 1/9. It can also be reen that the O\'Crflow of the basin is only to be feared if water flows out of the basin in more than twice as much time as it takes to fiU the basin (T2 > 2Ti). Note. One condition for the validity of Torricelli's law of efflux is that the size

=

· (diameter) of the orifice be much smaller than the dept.h of t.hc water. This condition is Certainly not fulfilled at the end of the efflux, therefore our results are only approximate. II the orifk-e is very small tlten the internal fri ction of the liquid (neglected so far) a lw plays an imp<>rtant role.

S. 31 . /st solution. During the acceleration, a car of mass m 'pushes' the Earth back a little and changes its rotational angular velocity. This very small effect has to be considered in order to resolve Ute paradox. For the sake of simplicity, consider the car as running on a body of mass M (M » m) t hat can move freely in the direction of the motion of the car. (Jn t he real situation, the Earth can rotate freely under the car.) The 'standing' observer says: if the car accelerates to some speed Vo and then to 2Vo then t.he body of mass M reaches a speed 111 -mVo/M and then tt2 = -2mt1o/M, while its kinetic energy increases from the initial zero to Mtt~/2 and then to Mtt:li/2. Since M ~ m, the kinetic energy of the body of larger mass and the chang-e in this energy can be neglected, thus t he ratio of the fuel consumption values has to be 1 : 3.

=

SOLUTIONS

5!)

The situation is different with the observer 'moving' at speed v 0 . He can sec the speed of the car increasing from Vo to 2Vo and then to 3tl(>. while, due to the law of conservation of momentum, the speed of the other body changes from the init.ial -Vo to ( 1 -

: ) tti and then to ( 1 -

the total system (car

~) '-\). The change in the kinetic energy of

+ Earth) is therefore first

and then 2

1 {(3t~) 2 t.&- = -m 2

- 2vo2 )

1 ( 1 - -2m) v2 + -M 0 2 M

-

1- M ( 1 - -111 )1v , ::::: -mv 3 ? 0 0. 2 M 2

It. can Le seen that the ratio of the energies (fuel consumption values) is I : 3 fur both observers. 2nd solution. Friction pushes the ground backwards, i.e. accelerates the car forward througl1 the wheels. In the frame of reference fixed to the ground (Peter's frame), work is only done on the car but not on the ground. On the other hand. in the frame of reference moving together with the train (Paul's frame). static friction acting on the ground also does work. Viewed from this frame, the ratio of tht· works done on the cars is 3 : 5, since the displac.ements of the cars are 3v0 t/2 and 5Vot/2, respectively, in time t. The work done on the ground is -2 units, as Uw displacement of the ground is VQt. The total work done is 3 - 2 = 1 units in the first stage of the acceleration and 5 - 2 = 3 in the second, their ratio is thereforr again 1 : 3, viewed from the train. S. 32. Since the object distance u and the image distance v can be exchanged iu the lens law and their ratio shows the magnification to be (t1/t1)2 9 (or I /9) , therefore v/t1 = 3 (or 1/3). Thus, the object distance is 30 cm (or 90 cm) and the image distance is 90 cm (or 30 cm). The focal length can be calculat.ed from I.hr lens law: / 22.5 cm. ff the same amount of light ed through the lenses in both cases theu tlH· 9 times smaller image would be 81 times brighter, as tlae smaller image occupies a surface 81 times smaller on the screen than t he larger one. However, the lens pl;wf'd at a greater distance only receives one ninth of the light reaching the nearby lens. t herefore the small image is only nine times brighter than the large one. [t can be shown iu general that in such cases the small image is as many Lime.<: brighter than the large one as the large ooe is larger.

=

=

S. 33. The leru;e; of the glasses of :short-sited people are divergent. Let f denote the (negative) focal length of a divergent lens, d the distance between the


60

object and the eye and 0 the size of the object (see the figure) . According to the lt"ns law, the distance bet.ween the (virtual) image and the lens is 1

1

1

-_ ,, + f' u =- -

i.e.

u/

v ~ u+f '

while the size of the image is

J v I =-0=0--. u+/

u

The seeming s ize of the image is determined by the angle rp subtended a t the eye, which (assuming an object of small size) is

d-u+u

0

OJ _ I . tt(d - u) + df d- u+v

=

S. 34.. Consider t he light beam as consisting of parallel light rays. Tbe light rays cross the vertical plane face of the quarter~cylinder without changing their direction , the n reach the curved surface of the cylinder at cliffes-ent angles c:J incidence. T he normals belonging to the light rays are always perpendicular to the tangents to lhC' cylinder, i.e. they are radial. The higher the position where a light ray enteo; the quarle<-
I

2

= n= J'

and

=

1.71 an, i.e. this is the minimum distance behind the quart.erwhich yields x cylinder at which light reaches the table. As the angle of incidence of light rays striking lower is smaller, these rays are from their original directions by refraction, therefore they reach the deviated less surface of the table further away. One is inclined to think that. in principle, the light patch can reach any distance on the table, since the direction o f the light ray arriving in parallel with the surla.ce of the table is not altered. This, however, is false! The path of each light ray can be followed (e.g. as a function of the angle of incidence), and it can be shown that they do not get that far on the table. Instead of the tedious calculations, the furthest point of the light patch can be found out by m<'.ans of a simple 'trick'. Consider the part of the quarter-cylinder close to the table as a piano-convex lens. (Then the glass body 'preceding' the lens behaves like a piano-parallel lens and can be ignored.) The focal length of the piano-convex le~ can he calculated using the formula for thin lenc;f'-'1: I

-R

R

+x

= coso111

n- 1

7 =[ l·

_

This angle is greatest as a function of u when the denominator in the right-hand-side of the above formula is the smallest, which is d/2. In other fulfilled when u words, the seeming size of the object is the greatest when the lens is equally far from th<' eyE' and the objoct. It is interesting that this condition is independent of the focal length of the le11s.

.

61

SOLUTIONS

This yields f = IO cm for the distance of the furthest point oC the light. patch from the quartcr-cylinrler S. 35. Consid"a <1 simple m __1:_ ___ -<>jm... ~-a pendulum ofle ngLh L ;met a ho_ mogeneous physical pendulum consisting of uniform rod oflength l suspended fromm one end. Both are released from a horizontal position. What will their speeds be after a tum of angle a (see the figure)? The principle of conservation of energy yields 1 2

- mL2 w 2

= mgL sin o '

i.e.

w=

~ o VTsm

Le.

w=

~ o VT'Slll

for the simple pendulum and

=

21/3 then the angular velocities of the two motions for the rod pendulum. H L are equal for all values of o. In this case, the two motions are identical in all time instants and their periods of swing are the same. The result so far is that the period of swing of the rod pendulum of length 'U/3. But what. is this tim~? l equals that of a simple pendulum of length L The formula T = 2-rr../L79 valid for small deviations cannot be applied, as the deviation is large. The exact calculatiOD8 would require a complicated mathematical

=


62

apparatus, but these are not. necessary if instead of calculating the period T, one only wants to determine the rlendence of T on L . The period of swing of the simple pendulum may depend on the length L, the 11ui.ss m of the hob, on the gravitational aeeeleration g and on the maximum angle of deviation Om,.• · If the dimensions of the quantities are taken into , this function.'ll dependence can only be or the form

T(L, m , 9, 0

111" " )

= /% · /(am"").

(Thi' 1limcnsion of mass is the kilogram, and sinr.e the kilogram occurs in the di111e11sion of nonr. of the o ther quantities, the period cannot depend ou the mass of tht! boh. 011 tire other hand, seconds only occur in g, therefore the required dimension of sf'1·m11lc; 01n ouly be obi ainc.' (' proportio nal to the square root of L.) The fom1 of the function /(am"") cannot he determi ned via dimensional analysis, since the angle is dimensio nless; the only available mfonnation is that for small a.ugles, it is /(am'AX) s:::: 271". Ac.:conling to th<' ;ibove reasoninr,, it can be concluded that (with identical iniliaJ deviations} LhP peliod of swiug of a lllrea.d !)f'lnrlulum of length 2l/3 is times that of " thread pendulum of length l . Thus , the periocl of swing of a rod pendulum of leng th l is only approximately 82 3 of that of the thread pendulum of the ~<1mc length. (This conclusion is not only valid for a horh.ontal, but for an arbitrary but ident.ic-,1 initial positiou.)

../213

S. 36. Imagine that the surface of the table is very smooth (the coefficient of friction is very small). Fil-st, lllC centre of mass of the pencil increasingly accelerates in t.hP direction of the fall. This acceleration is caused by friction, and since the surface of thr. table is ~mooth . the point of the pencil slips sooner or later, in the direction oppo.~tle to that of tJ1e fall. If friction is '·ery large, the pencil does not slip for a long time. The horizontal velocity of the centre of mass moving on a circular path increases in the direction of thr fall) in the beginning, then starts to decrease and - if the pencil continued moviug this way until its horizontal position - it would tend to zero. The sign of the horizontal acceler ation of the centre of mass and of the friction changes during the motion. ff the peucil does not slip in the first period then it, can only slip 'forward' , i11 the direction of lh.e fall. (This bas to occur at some point, as the point of the pencil presses the table with a decreasing force that becomes zero when a = arc:cos{l/3) ~ 700, thus slipping occurs however large the coefficient of friction is.) A detailed calculation can show that the two cases are separated by a (static} coefficient of friction of µo = 0.37. lf µ < µo then the point of the pencil considered as a thin, homogeneous rod slips 'backward' and ifµ> µo then it slips 'forward'.

SOLUTIONS

63

S. 3 7 . The pressure ill a soap bubble of radius R is greater than the atmospheric pr~ure Po by 6p 4o/ R. (The pressure caused by surface tension is taken with the coefficient. 4 because both directions of the curvature of both sides of the

=

membrane have to be taken into aecount.) Clearly, the pressure is higher in the bubble of smaller radius, therefore the air flows into the larger bubble, so to say inflating it, an
For bubbles of ordinary size, the pressure of curvature is many o rders smal ler Lha11 the external atmospheric pressure. H the pressure of curvaturP. i.<> neglected then the radius of the resulting bubble is R 3 :::::: + R~ . If the radii arc: measun•d 'accurately' in order to find out the surface t.ensiou theu the for111ula

VRt

has to be applied. In practice, however, this metbocl 1·a1111111 he applied, as the denominator is found to be zero with the available :in·111 :in u f the measurements. Thus, the measurement data only provide us with a ro11gli l':>t.imate of the upper limit of the surface tension. S . 38. The cross-section of the I' water scx>ne is limited by semicircle:; of d radius r =
=·'-= Po

=

»

F= D"J11" . 2a 4 d 'pulling' the glass plates together.

Note.. If d is much smaller than D, this force cnn be quite considerable. IL is 1n fac.t very difficult (if not impossible) to puU a part two glass plates iu the direction perpendiculM Lo their plane when there is water between llit!ru. In order to be separat('(), thP.y havP. to be slid in the direction parallel to their plane.


64

-

S. 39. The speed of the threa d at a distance of x metres from the wall is o bviously proportionally smaller than the I mc:tcr x mc1cr speed of the end of the thread, i.e. X · tJo . If this value is greater than the speed of the caterpillar then the caterpillar moves away from the wall and its situation becomes more and more hopeless: it will never reach the wall. If, on the other hand, v..,.l~ri>•llar > Vo then the resultant speed of the caterpillar points to the wall and this speed increa.c;f!S as time es, so the caterpillar certainly reaches the wall. Tht' limiting case corresponds to x Vcalerpill11ro/Vo = 0.1 metres: starting fro m this point. the caterpillar does not move in any dirt'Ction.

=

S. 40. In Your imagination, draw signs onto the thread and write beside them tbc ratio of the distanc-" of the giv~l poiut from the wall to the total length of the thread. These figures are exactly the coordinates x mentioned in the solut ion of the previous problem (x = 0 corresponds to the position of the wall, while x = 1 to that of the spider), but now these 'stretch' together wit!J the thread of the spirier- web. Calculate how long it takes to the caterpillar to get from a point x to a nearby point x - C:,.x if the spider has been moving for time t . Since the distance between the points in question is (1 + t1ot).O.x and t he caterpillar moves at speed c,

lix

=

cfit =~ L - ~ 1 + t1oL

1T

C

(The s um turns into an integral as the size of the step8 is infinitely decreased.} Considering that the above integral c-.an be made arbitrarily large by a suitable choice of T , the surprising result. is that however fast the spider is in pulling the end of the thread (with an arbitrarily large Vo, an arbitrarily small c and an arbitrary xo) , the caterpillar surely reaches the wall within a finite time interval T .

S. 41. a ) Suppose that the ball falls freely for 1 m, then reaches point B bouncing prac tically horizonta!Jy on the row of closely placed nails at the bott om edge of the drawing-board. The duration of the vertical fall is t 1 0.45 s , by the end of which the ball reaches a speed of v 1 = 4.4 m/s and covers lhe remaining distance of 2 m in time t 2 0.45 s . As the ball would have reached point B in time t3 = 1 s > t1 +ti sliding on the straight line AB, the answer to the first questio n is: The quickest way to ge t to point B is no t the shortest one for the ball.

=

=

S. 42. Clearly, the centre of mass S of the triangle has to be below the point of suspension. Denote the vectors pointing from the centre of mass S to the vertices of the triangle by r1, r 2 and r :\ as shown in the figure, and the vectOr pointing to the point of suspension by m . Then the respective forces F 1 , F 2 and F 3 exerted by the threads onto the plate can b e determined i11 of the vectors pointing to the threads: Fi = >.i(m - r ;).

(i = 1, 2, 3) G

c - dt = - In (I + t1oT) . o l +Vol Vo

-

Note. It can be proved by means of complicated mathematics that the curve along which the duration of motion is the shortest bas to be a cycloid. b) The vertical velocity of the ball at any point of the board is maximised by the velocity of a body in free fall at the same height dropped from point A with no initial speed . (The kinetic energy and therefore the speed of both bodies are determined by the magnitude of the vertical displacement .} Thus, the bouncing ball cannot reach the bottom of the drawing-board faster than the body in free fall , 1.e. iu less than t 1 = 0.45 s . The answer is: no!

c..O.t 1 +vot

holds. Let us sum up both sides of the above equation for the whole motion of the catc>.rpillar. If x0 denotes the original position of the caterpillar and T denotes the time interval in which it reaches the wall then Xo

65

SOLUTIONS

The plate is in equilibrium, the vectorial sum of the forces acting on it is zero:

Make use of the fact that the vector pointing to the centre of mass of the triangle is the arithmetic mean of the vectors pointing to the vertices, i.e.

(The origin of o ur frame of reference is in the centre of mass of the triangle.) No tice that G and m are parallel, therefore G = - km. Eliminating e.g. r3 from the a bove equa tions,

is obtained . Since ri, r 3 and mare not in the same plane, their linear combination

(the sum of the vectors multiplied by scalars) can only be zero if the coefficient of each of them is zero. Thus, >., = >.2 = >.3, which means that the forces stretching the threads are proportional to the lengths of t he threads. (This statement becomes invalid if one of the threads is loosened, since the planed the plate then becomes vertk.al.)

123 TRICKY PHYSICS PRODLEN!S

66

v.,

·--:.-

==<9l::;::;====:::============ m.q • ('M

S. 43. Viewed from the frame of rnference fixed to the centre of mass, the initial speed of the beads (considering the rlirection to the right as positive) is m

VM

= ----Vo

67

SOLUTIONS

Jn the limiting case when the initial kinetic energy is just sufficient to allow the two beads to approach each other at a distance d then the two beads stop with respect. to each other, i.e viewed from the lab frame, they move on at a common speed mvo . Vm = VM = M (Figure 3).

CM : Ill,

q : Slopped

=========M=.=lQ~F~=========

m+M

m+

Figure 3.

and their ratio is constant , since their cen-

Flgu1'<: I .

tre of mass is at rest in this frame of .reference (Figure 1). As the beads approach each other, their speeds decrease if the signs of q and Qare identical and increase if the charg('.S are opposite. In t his la tter case, the beads each other by then at a sufficieutly large distance, their speeds reach their initial values, since their energy is conserved. Returning to the original frame of reference, the speed of the body of mass m decreases to the original value Vo after a temporary acceleration, while the body of mass M is displaced by a certain distance (to the left), but finally stops. The case of repelJing beads requires a - ll., more detailed discussion. If t heir initial en~ ergy is sufficient, the beads each other ~========;:::============= by, then moving away from each other m, q ! CM (viewed t>.ither from the centre of mass frame or from the 'lab' frame), their speeds ~ vM get. back to the original values. If, on the . other hand, their initial kinetic energy is Figure 2. too low to allow them to approach each other to a distance rl then they 'turn back', and in the centre of ma.s..c; system, the body of mass m moves to the left at speed -vm, while the body of mass M moves to the right at speed -vM (figure 2). T he condition for this is tha t l

2

2

m

-mv

i.e .

l

2 1 qQ + -21 M vM <- -. 47rco d •

mM +M

1

Note. The three cases discussed above model the on<' dimensional ruechanical collisions. The limiting case, when the two bodies move on together models au inelastic collision. In this case only mechanical momentum remai11s constant , mech1u1ical energy decreases. The velocities of the bodies approad1ing eacl1 other a11cl then mo\'ing away from each other are identical to those calculated from the laws of el astir. collisions (conserv~tion of energy and momentum). The motion of the bodies ing eacli other by (these do essentially not collide and keep their original velocities) is obviously 111 agreement with the conservation laws. The solution corresponding Lo t h.is case is usually not used for mechanical collisions {since the bodies cannot through each other).

S . 44. a) Let Vo denote the speed (when it is already constant), d the original dist.a.nee between the beads and m their masses. Jn a certain time interval 6.t, the cluster colJides with v 0 6.t / d further beads, which increases its mass by 6.m = mvo6.t/d and its momentum by 6.p = t1o6.m = mv56.t/d. According to Newton's law of motion,

=

F

6.p

= -Llt = mv~/d,

which yields iu = JFd/m for the spec
qQ

- - - - v02 < - - - -. 47Tco d

2m

mM

{The quantity - - - is called the retluc.etl mass of the system composed of 1.wo

m+M

masses.) The velocities of the bodies in the laboratory frame of reference can be obtained by adding the velocities in the centre of mass frame to the rela tive velocity of the two frames, which is lvMI:

v~

m-M

= m+MVo'

and

(accelerating uniformly) until the first {elastic) collision. Then it 'transmits' its speed to the second bead and stops, then starts acceler ating again due to the external force. What happens to the bead it has pushed? It mov~ at a constant Spet'.d v 1 , collides with the third bead and stops. The third, fourth, etc. beads behave similarly, the 'shock wave' propagates forward at speed v1 • Meanwhile, the bead on the left is again accelerated to speed v 1 , colJides with t he second bead so far at rest, and the process is repeated by starting a new shock

123 TRICKY PHYSICS PROBLEMS w<WC'. The speed of Lhe accelerated bead varies uniformly from zero to v 1 , its average va.lue is " • /2 Vo/ ../2 = J F ado; pu.slung each other condense into a single cluster that behaves like a perfectly Fd/m. The lime necessary for the duster to in<'last1c body: its final .speed is Vo = (t.hr collL.;ion number-). The more elastic inelasticity of dcgret< the on co11Jt:111't' dc11ds Ll1t> t>lem.,ut.n1 y w llisiom,, t he longer t he time Deces.5al)' for an inelastic cl11stec to condense.

=

J

S. iJ5. If the cross-section of the incoming watR.r jet is A audits speed is t1 then t he mass o f the water of den!>ily ll flowing into the gutter in unit time is of uAv. Tl1is qua11lity of watc1 has a kinetic ene rgy of uAv3 / 2 and a horiwntal moment um Clf o:lc/ sit1a These quantities cannot change if internal friction is neglected: :s

l

I

l 2uAv = 2eA1v1 + 2uA2u,,

(J )

3

69

SOLUTIONS S. 46. The level of t.he liquid starting with acceleration a decreases by t::.h = a(6t) 2 /2 in a time interval 6t, thus the mass of the liquid flowing out (D1 1r /4)l:iliu. This means a decrease of is 6m 6mgh in the potential energy of the liquid. Meanwhile, the whole o f the liquid is accelerated to speed a6t, its kine tic energy increases therefore by 6t1 2 2 (D 1r/4)/tu · (6v) /2. (The speed of the effluent liq. uid is higher than this, hut this can be neglected as its quantity is very small compared to the total amount

=

=

a'

D

•..

J

: _ _:r 1Jiu=a61

of water.) According to the law of conservatiou of energy, the absolute values of tbe changes in the potential and kinetic energies are identi· cal:

3

(2 )

m11l 1111· ia" · ·! • oil:-<'rvati<m o f matter has to be fulfilled as well: (3)

which yields a = g. This means that a part of the liquid starts a free fall at the beginning of the efflux. According to t he law of conservation of matter, the speed of the effluent 2 liquid is (D/d) times higher than that of the surface of the water, therefore its acceleration is greater than g by the same fact.a. (E.g. if the diameter of the orifice is one tenth of that of the pipe then the initial acceleration of the effluent liquid is lOOg!)

(.4 1 1rnd A :i art· tlie respective cross-sections of the water flowing out to the right aud to t he lc:ft, while i 11 and Vi are the respective speeds.) Equatious (1), (2) an .s (the cross-sections and the speeds), a further relation has to be found . 2 Acoordi11g t.o OeruouJh 's law, the quantity {!IJ /2 + p + e.9h i.-; constant a long the Inside the liquid and far from friction. internal str earr1li11es of a liquid with no the atmospheric pressure. equals and constant is pressure the ><>int., 1 tlw hit.ting If t he differci1r.c be Lweeu the heights, or more exactly: the changes in the energy c-orrespo nding to these differences are neglected (this is right for a rapidly s treaming liquid), tht' coaseque11ce o f Bernoulli's equation is t1 t11 = 1>2· This means that the Hq11id l c>av~ t hP gutt.er at the same speed at both ends! This is rather surprising b11l n g ht, within the accuracy of the above approximation. T lw equal ious for Lhe conservation of matter and momentum are therefore of A1 + A'l and A sin o = A1t11 - A2t>1, which yields LhC' rorm A

=

=

A1 A1

l+sina sino

= 1-

T his r-ntio can be examined experimentally and a surprisingly good agreement can be found with Lhe calculated value;, which proves that the applied approximation was right .

For how much time is it true that the liquid is practically in 'free fall'? Ac· cording to Torricelli's law of efflux, the speed of the efflux is of ./'IgTi and the rate 2 ofdecrease of the Hquid level is {d/D) ../29Ti.. The time interval r elapsed between of the (nearly constant) speed of efflux can reaching the and efflux the of the start roughly be estimated using the relation

rP

gr ;::: 0 2

.Jiih.

U, for example, h == 20 cm and the ratio of the diameters is 1 : 10 then r ~ 0.002 s , which is quite a short time. S. 47 · The sand flows down roughly evenly, therefore the total operating time

T of t.J1e sand-glass is proportional to the volume H3 of the sand. (As our aim is

onJy to give a rough estimate, the differenoo between the volumes o r the cone and the cube is not taken into ac.count.) The time in which the sand runs down may also depend on the gravitational acceleration g , the diameter d of the orifice and the density e of the sand: T:::::: H3 · f (g, d, u). The function I has to be proportional to the reciprocal of the square root of g to yield a result with a dimension of seconds (as time does not occur in the dimensions of the other quantities). A similar reasoning shows tha t T cannot depend on u.


70

and it is proportional to the power of - ~ of the diameter d: T == H 3 / The coefficient of proportionality is a dimensionless number, and sinee it does not depend on anything, it has to be of the order 1. Consider actual realistic
.JiFY.

the parameters relevant to the phenomenon

in

hand.

s. 48. First the body practically falls freely - due lo the looseness of the spriug. The le.ngth of the spring soon re.aches a value seve1;1) times larger than the original length that can be neglected during the .remainini:, Set:ti~n of.the. uiot1on . In this approximation, the motion of the body will be a harmonic oscilla tion both verticalJy and horizontally. As the body is released with no initial speed, it arrives under the Point of suspension after a quart.er of the penod of the horizontal motion. Meanwhile, the vertical motion also makes a quarter of a pE'riod, therefore the body sinks to its equilibrium position at a depth of mg/ D . (TI11:- i~ tnuch larger than L .) The motion can be follrnwil "11 h calculations. In the frame of reference s hown in t lw jigm-e, the equations of motion o f the body in point (.1:, 11) a re L Y----~--_.

~ = - D(J x 2 + y 2 mav

L) Jx x+11. • 2

= - D (Jx 2 +y2 - i) Jx-i.y+y2 +mg.

then it is the original length of the sp1ing that cau be neglected and t.he equations of motion take the followiug simple Conn: ma,.,

= - Dz

and

1nay

= - Dy+ my.

These equations describe harmonic oscillations of identical periods (about the origiu in directior. x and about the equilibrium position tA> mg/Din direction 11). Taking the initial conditions into , the solution is

=

x{t) =Leos (

~t),

y(t)

=

71

The body is under the point oC suspension wheu z(t) = 0, and y this time inst~t. in agreement with the above reasoning.

= !A> =

mg/ D at

Note. In the first section of the motion (when the equations of motion are in reality not valid yet), the functions describing the harmonic oscillations can l>e npproximatro hy x(t) == L and y (t) :::: gt 2 /2, which is in agreement with tbe formulae describing free fall (valid for that section). S. 49. If the car is braked with de<-elerat.ion a then observing the motion of th<· weight of mac;s m from a frame of referi>na- fixed to thP. car, ;1 virtual 'inertial fon·1" of magnitude mn aud Pointing in the dirL'Ct.iou of the motion of the car ;1ppe;1rs. If this inertial force acted permanently, the pendulum could ce1 tainly nol rPad1 its top standstill, since the resultant work o f the iuertial force done- up lo rhal. i>oiut would be z-ero ancl the gravitational force would bf> negative, i.e. the kine t.ir cnPr~y of the pendulum would be nllgative, which is impos.o;ible. Consicler now the fact that the car is only hrakecl for ;i c~• 1.ai11 timr. (until 11 stops). If this moment comes when the thread of the pendulum is hori7.ontal, LhP work done by the inertial force is W moR (R is the length of the thread) . lf tire pendulum rear.h~ its top standstHI pooition with speed v then, according lo tlw work-energy theorem

=

mnR - 2mgR

= -mv:z . 2

~e thread remains s tretched even at the top Point if mt?/ R > mg, which, together Wlth the above relation for the speed, yields the constraint a > 2,5 9 for the slowin~ of the railway carriage.

2

ln the first section of the motion (when the expansion of the spring is not much larger than L). the force exerted by the spring can be neglected. On the other hand, wbe u

y

SOLUTIONS

m; [1 -cos (~')] .

S. 50. The forces acting on the wedge are the gravitational force and a force/\" perpendicular to the inclined plane (Possibly of a magnitude changing with time) Due to these two forces, the component of the acceleration a of the wedge parn llcl to the inclined plane is gsin o {from an inertia.I frame). Newton's equations of motion ~nly ~emain ~id }n. au accelerating frame of reference fixed to the wedge if an mertial force - m a is added to the forres really present in the motion of a body. where m' is the mass of the examiJ1ed body (e.g. a tiny part of the liciuicl). The resultant of the gravitational and inertial forc:es acting on the body of mass m' is pe.rpendicular 111 '1< ~'" n Lo the inclined plane as the component.<; parallel to the inclined p lane cancel each other. TI1e bodies 011 the wedge (the glass and the water in it) 'feel' as if they were in a gravitational field perpendicular to the inclined plane, therefore the surface of the water is perpendiculru- to this gravitation, i.e. parallel to the inclined p lane. This statement does not depend on the motion of the inclined plane: it can btfixed or .n~ve freely or ma~be . due to a small force - move to and fro. As long as the fnct1on between the mclmed plane and the wedge is negligible and it is su1e


72

that the wedge does not rise off the inclined plane, the shape of the surface of t.he . water cannot be other than a plane parallel to the inclined P_lane. The limiting case m > M deserves a separate discuss10n. In th~ case, the wedge ' thrusts' the inclined plane out and falls nearly freely. The :"'e ights of the bodies on the wedge (including the water} are nearly ~mpletely 'lost, but even the small force keeping the water inside the glass is sufficient to set the surface of the wa tCJ parallel to the inclined plane. Note. If the system is initially at rest, the water is spilled from the ~Jass. The w~ter surface parallel to the inclined plane would only be ~isecl af~er a sufficiently long time, a sufficiently long inclined plane. This is why this mterestmg phenomenon cannot be 0 11

expenmentally observed in o rdiuary circumstances.

s. 51. Assume that the string of uniform cross-section and mass di~t~buti~n, free a.t both ends, orbits the Earth in such a way that i~s r~ati~e pos1t10n with respect to the Earth is always the s~e. O~~iously, this ~tuat1on can only be realised above the Equator, in the vertical poo1t1on of the strmg. From a frame of reference moving together with the Earth, a body_of mass ·b"ting the Earth above the Equator at distance r and angular velocity w exm 01 1 GMm . . · nces a gravitational force of and a centrifugal force of mrw2 · (M is pene . r2 .. the mass of the Earth and G is the gravitational constant} T_he condition for ~he equilibrium of the string is that the resultant of the grav1tat1onal force ~angmg from point to point equals the resultant of the centrifugal forces also changin~ from point to point. This coodition can easily be given ~g integral calculus but 1t can also be found without such sophisticated mathematics. Imagine that the st.ring is pulled down a little by som~ ~tern~ for~: ~ince (as seen from the rotating frame of reference} the string "'.'as 1~t1ally ID equih?num, it can be moved from its equilibrium position by an arbitra~y small f~rce, 1.e. an arbitrarily small work. The work done in the course of this ~all displacement ca.n be calculated without a need for integral calculus. ~he d~placement of the the slow whole string _ from the point of view of the work done - IS equivalent displacement of a litUe piece (of mass 6m) of the string from its top. to its botto~ The work done is the sum of two : the change of the gravitat•on:11 potalt1al ener gy and the work done by the average of ~he (linear!y changing) centrifu.gal force. If the bottom end of the string of length L is at the distance R of the radius of the Earth from the centre of the Earth, the work in question is

!'°

1 1 ) - A R W-GM6m -'""m (R -L+R

+ (R + L) tJl. L = o. 2

This is a second order equation for L, the solution of which is L =

~ ( - 3+ J1+~~)~140000 km

SOLUTIONS

73

for the length L of the string after substituting the known data. This length is several times the distance r.

= ( GM)1/3 w2 ~ 42000 km of the telecommunications

satellites (measured from the centre of the Earth)! What is the maximum stress in the string? It is easy to find out that the stress is the highest at the position r = r,, namely Uma:x/P = 4.8 · 107 Nm/kg, where 11 is the mass density of the string. This figure is much greater than the ratio tensile strength/density of known materials (this figure is 2.6 · 1<>5 for steel 1.7 · 106 for carl>on). Therefore in principle, the 'hook to the sky' could be realised according to Newton's laws of motion, but up to now this is impossible (at least with a string of constant cross-section) because of the structure and characteristics of the available materials.

S. 52. Denote the width of the canal by d and d raw a straight line perpendicular to the bank of the c:anal at distance d from the initial place of the boat downstream (see the figure}. T he boat is initially at distance d both from the pomt F on t he opposite side and from this straight lill('. As both the speed of the water and that of the boat with respect to the water are v, the water of the canal takes the boat downwards by the same distance as that covered by the boat in the direction of point F. This means that the boat is always equally far from J>oint F and from the straight line. The path of the boat is therefore a parabola with F as its focus and the straight line as its principal straight. After a very long time, the boat approaches the opposite bank to a distance of d/2. It cannot escape from this situation as the speed of the current equals that of the boat. S. 53. If the children slid down at an even speed after a slight push then the component F of their weight parallel to the inclined plane would be of the sain<' magnitude as the force of friction S: F + S = 0. The force of kinetic friction - the direction of which is always opposite to t.he instantaneous velocity - decreases the magnitude of the velocity, while the force F aims a t increasing the component of the velocity parallel to the inclined plane. These two effects a.re naturally present together, finally causing a rather complicated motion (on a curved path and with a changing acceleration}. Despite this, the final velocity can be determined without the detailed description of the motion. Establish the frame of reference shown in the figure in the inclined plane iUld denote the magnitude of the instantaneous velocity of the sliding child by v, and its component y by Vy . Calculate the change in these two quantities in a short time

123 TRICKY PHYSICS PROBLEM S

74

={- S + Fcoso)M,

mD.v11

= (F -

Add these two equations up and make use of the fact that F = S: D.t1

75

S . 55. Due to the symmetry of the layout and of the initial conditions all the towards the centre of then-angle with the same acceleration (changing with lime). The format.ion keeps its original shape but the distance x from the centre decreases at an accelerating (but not uniformly acceJerating) rate. The resultant force acting on one (say the nth) body when it is at distance r from the centre is

interval D.t. According to Newton's 2nd Jaw,

mD.u

SOLUTION S

lx_>di~ £all

S cosa)LH.

tr~· · ·.\~·

+ .6.t111 :::: D.(t1 + tiy) = o.

m2 ~ 1 F(x)=G 7 · L..,,4 . 1rk . b

a

i.e.

v

+ v11 = constant.

v,· -.. v F

Due to the initial conditions, the magnitude of t.his constant is v0 = 1 m/s. The final speed t'mnr of the sliding child has to point in the direction of the slope, and its magnitude is uniquely determined by the above 'conservatio n law': Vmoz

\

n

This force - consisting of the gravitation al force of all the other bodie3, more exactly of the component s of these forces pointing to the centre - is identical to the gravitation al attraction of a fixed body of mass

'°'

n- 1

M,.

m L..,, -.-1T-. 1 =4 l=t

M2 revenue cutter, i.e. let k denote the ratio oft.he speeds of the two ships. In an (arbitrary) situation shown in the figure, the distance between the ships (initially a) decreases by

= 0.25;

n

in time D.t. Meanwhile, the distance of the cutter from the shore decreases by

= ktJ sin

Q

M3

=0.58;

M4

= 0.96;

. .. M 10

= 3.86 .. .

The time T of the fall from distance R towards the centre of att.ra.ction of mass M can be consid_ered as ha.If of the time of revolution Te oo a strongly flattened (degenerate d) elbpse of major semiaxis R/2: T = Te/2. On the other hand the time of revolution Tc on a circular orbit of radius R can be calculated fron: the dynamics equation of circular motions:

6a = kv6t - v sin a 6t

6y

sm-k

The values of the masses Mn (in units of m) can be calculated numerically for all values of n:

S. 54. Let k · v denote the speed of the

(2)

Sin -

y

= ~ = 0.5 m/s.

(1)

l

(2n)

G Mm 1(2 = mR Tc

2

and, according to Kepler's third law,

.6.t.

(o is the angle of the instantaneo us velocity of the cutter with the shore.)

. Sum up the small displaceme nts occurring in equations (1) and (2). Knowing that both sums should eriual a, the angle o causing the difficulties can be eliminated aud a surprisingly simple second order equation is obtained for k: k2

-

k- 1

= o,

with

l +J5

k= - - ~ 1.618 2

as its positive ro<>t. This figure is the famous 'golden ratio'. This is the ratio of the spc.'Cd of the revenue cutter to that of the smugglers if it reaches them in the situation described in t.he problem.

which finally yields T

= 1TJ ~:n

for the time.

Note. The limiting case n > 1 is interesting. As I.he number n 0£ the bodies iacreascs M., increases even if the total mass of the system tends to a fixed value Mo, i.e. m = Mo/n'. !he more finely a given amount of matter is spread along a circle, the shorter time it takes 1t to 'col~ilfl'5t'' under its ~~ gra~itational attraction Tb1111, there is no point in examining the continuous matter llli.1.r1but1on spread along a line; the transversal dimensions cannot be neglecte.d.

123 TRJCKY PHYSICS PROBLEMS :harge distribution at the surface of the metal plane, caused by ;es an electrostatic field identical to that of a charge - q situated of the plane, symmetrically to q. This is the prinaple of image ~ fori;e of attraction acting on the body (not moving too fast) can g Coulomb's law: '12 F(x} :;::: k - 1 , 4x :<1ntaneous <listaur.e between the plane and the hotly (1.c. the body rom the image charge). This expression for t.he for(-e could also be 2 kq /(4Gm) ! gravitational force exerted by a body of mass M 2 GMm/x . ly of ml\SS m situated at 11istance x from it: F analogy, Kepler·:; third law cau be applied, which gives the relatiou Kl of revolution T of the planeL'i anil the major semiaxis a oft.heir

=

=

ri

47rz a 3 =GM.

d body is released at distance d from the metal plane then its orbit I as a \'er}' flat (degenerated) ellipse with major semiaxis a= d/2. T,, is half of the period on the degenerated orbit, and it can be Luting the respective quantities:

SOLUTIONS The charge distribution inside the spherical metal shell is in potential of the shell If the shell were not earthed then charge +Q its outeT surface. The charge distribution on the outer surface is w of the inner charge distribution. The electric field stceogth is zero 1 the material of the shell, therefore the charge on the outer surfa< the very presence of the charge on the inner surface. The exteJ looks as if the charge were in the centre of the sphere. The force acting on the charge Q inside the shell equals the Coulomb force between the charge and the corresponding image cho.rge. Applying the notations of Figure J!, the charge +Q is at distance d from the centre of the spherical shell of inner radius R, while the image charge - nQ is at distance x from the shell The ratio of I.he charges is therefore n, which Figun can easily be formulated in two cases (for the points in which the straight line connecting the charges intersects with the spherical shell): n

=

= _x_ R- c

Rfd and This yields the following solution for n and x: n The force acting on the charge inside the spherical metal shell is t F

=-k

Rd - - kQ2 nQ2 (.R2 _ d2)2 · (x + R - d)2 -

It is ~ear ~hat this fotce is zero when d =0, while il tends lo infinitJ S. 57. The method of spherical image charges cau be applied. Let the two point charges of opposite sibrns be +Q 1 alld - Q 2 • The points of zero potential have to be found in the field of the t wo point. charges:

where r 1 and r 2 are the distances of tht: g1ve11 poitlt from U1e t\\-"O charges. After a straightforward transformation, QifQ2 = r ifr2 is found. According to Apollonios's tbeun:m, the geometrical position of such Points is a sphere sphere}, therefore the zero potential su rface is a sphere. -ical metal shell mentioned in the problem is earthed then it is a t T he point charge +Q inside it causes an inhomogeneous charge he inner surface of the shell as shown in Figure J. The electaic field is the same as if it were caused by a nt>gative point charge outside >here shell. This is what is called a spherical image charge.

negative sign shows that it points in the direction of the image cl sign.) No~. The electric charge distribution on the inner surface ol ti shell can be calculated using Gauss's law. The quantity of charge on < surface charge clensjty) is proportional to the electric field strength tru by superimposing the fields'of the real charge and of the image charge.

-Fl'nc Ftgure I.

S. 58. Friction betw can be neglected, therefo exert forces perpe.ndiculai in the course of the coll' first ball stops after the a: second takes its speed Vo tion of the balls, oo the ' not change, so the first ba and the second slides wit speed ti() .


78

Th Cri tion between th balls and th table is indeed important and changes the motion of th balls. The first ball is accel rated forw-Md by the force of kinetic fri tion Fr"'c = 1nng, while the second is slow d by th same £ re as shown in Figure I . The rotation of the first ball is decreased by friction, that of the second one is increased. The role of friction las witil both balls reach th stat of r lling without slipping. Then th ir further motion is unchanged . It will b shown that the final motion of the balls d pends n "tlier on th friction c , fficient, nor on th pos ible chaug of this latter w:ith th position. Aft r the llision, the initially mo1•ing ball rotates at. angular 1• lo ity w = . I b . . . th _, o .. 2 v :-, its augu a.r momentum a out its ax!!; u; .r.,.ore ow = 52 m1•2 vo ;5nwo1-.

=

0

=

I

Tb anbrular momentum about any point P of th ball or of th table i · th same,

since th ntr of mass of th ball i at rest, i.e. th angular momentum cau · d by translation is zero. The angular mom ntum f th ball about P cannot b hanged b ' fri ction any more, as the line faction f this fore crosses P . (The sum of Lh gravitational fore and of th ing force of the tabl is zer , t11 refo1e tlwse hav no torqu either.) Th angular momentum of th ball W= 11/I" rolling without slipping at speed v is the sum f its own angular momentum ~mvr w : and the angular momentum mvr caus d by ·. .· > th motion of its centre of mass. Figure 2 P N Ffr.-: sho'\\'S on of the balls (th one on the I ft) and the fore actiug on it right after th Figu 2. collision and in th stat r Hing without slippiug. A ording to the law of conserv-atio11 f in mentum,

(.~oi::~;. t

~O'· u

which yi lds A similar reasoning shows that th final speed of the oth r ball has lo b<.' ~Vo (regardlesi; of the magnitude of the £ri tion) .

S. 59. Cb a point P n th surface of the tabl and examiu · be ;u1g11lar mom ntum of U1 body about this point. The Jin of action of th f. IT . of fri tiou ·1 Of;Ses this point, th r fore friction has no tor<Ju in t11is case. Th ravit11tionaJ for e and the ing force of th table eliminate ach other. No oth r for ce acts on th body, ther fore the angular mom tum is i;-on tant about this hos •n point, ru1d as th ly is initially at rest, the valu of th angular momeutu m iR zero. As the table-cloth is suddenly pulll'd nut, th body starts sliding and rolling. The angular momentum J of th body of mom nt of inei·titl n1" I and mass m an be written as th sum of two E (foll wing the notations of the figm·e) :

J

= Iw + r x (mv) ,

SOL TIONS

79

where th first term is what is the self angular momentum (spin) and the second term i the orbital anguJar momentum due to the motion of the centr of mass. ( onsidering th direction f the vectors, the magnitude of the angular momentum can be written in th form J = 0w mvR. where R is the radius f the cylindl"ical body.) Th sum f the two t rms can only be zero if th sign of the s If nngular momentum is opposit to that of th orbital angular momentum, i.e. if th body starts rolling ' backwards' when the table-cl th is pulled out. lt is easy to se that the sigu of I.he orbital anguJar momentum has to he identical to that of th elf angular momentum in the case of rolling without slipping. two On th other hand, th sum of the two has to remain zero all th time. Th conditious ca11 only be fuJfiU d at th same tim if the body finally st ps. ThP Reader can ch ck this experimentally. The final st.at depends n ith r on th value of the friction, nor on th way thp tabi -cloth is pulled out.. (The taLle-cloth can be pulled out ev nly, with a u11iform acceleration or even with sevei·al sudden movern nts.) How ver, it is important that n ither air r ·stanc , n r rolJing resistance be significant, sine t,h se effects an ·hange the angular momentum about point P . S. 60. Observe the motion of the baU exactly from the original directi n of tnwslati n. Th ball can be seen to move away but with no 'sideways' motion. This is why it can b stated that its anguJar momentum about any point P of th table ausecl by a sideways motion and rotation - is zero. When the ball reaches the rotating dis , th occurring for e of friction causes it to move ideways and to rotate. A , however, friction has no torque about point P, the anguJar momentum about this point has to r ma.in zero all the time. (If th ba!J moves to th right, its ntre of mas.-; has to rotat anti- oekwise at a suitable angular velocity and vie versa when it moves to th left, always in such a way that its angular momentum is z ro!) Th ame is true wh n the. ball Jeav the dis and rolJs on the standing table. H r , howev r, the hall moving to th right should roll clockwise - du to th ondition of l'olling with ut slipping-, which is impossible as the anguJar momentums of these m tions houlcl hav th sam sign and their resultant could not be z ro. the only way ut is to exclud any moti n of the ball perpendicular to the original dir tion of its motion, i. . it rolls n without changing its direction. A similar r ·15 ning shows that the magnitude of its velocity cannot b ditfi rent after leaving the di · thau originally. This is a consequence of the con · rvarjou of th angular rnomentum perpendicular to the direction of the motion. Thus, th ball leaves the table at th same velocity, ther fore wit.h the sam ~omentum ve ·to_c as it originally had, and, in addition, its total kineti nergy is onserved . This latter fact is especially strange, as both when th balJ arrives at th di and when it I av it, friction does work on the baJI and changes its kin ti energy. The algebraic: sum f the two works is zero which is n t due to t.h conservation of mom ntum or of energy but it is a co~uence of th tb r m of angular momentums.


80

Nole. A little more complicated calculations (ui.;ng the equations ol translational and rotatio nal motions ) show that (viewed from the frame of reference of the table) the ball follows a uniform circular motion on the e venly rot;\l.ing disc. The cent.re of this circle is, however, uot iclentkal to the axis of the disc, and its angular velocity is also different from that of the dii;c, l>eing ~ times smaller. Uthe disc rot.ates evenly, the ball continues 1ts wny on the prolongation of its original track, which is not true if the disc does no t rot.ate uniformly Jn this latter cao;e, the momentum of the balJ is unchanged, but its track deviates sideways dut' to the disc.

S. 61. Hold one end of the threa
Fo and pull its other end with force Fm"" such that the thread does ju. t not slip on the cylinder. Let the section of the thread lying on the cylinder cover a length oorresponding to anglP n on the cylinder. The thread winding onto the cylinder tightens itself, providing the pressing force necessary for friction. The stretching force in the thread increases with o, which is balanced by the force of friction acting on each piece Fig11n• I of the thread. If the ~11 ..11111111!. force is Fat one end of the little piece ~l1F of thread :.ho"·11 111 F19Ttre l , while it is F + t:lF aft.er an increase or t:ln Ill I be angle then the increase t::.F balanced by friction cau h<> c alculated as

~ N

(l}

6F

= 1iN,

Figure 2. whe re N is the pressing force perpendicular to the surface and 11 is the require
(2)

N

Clo = 2Fsin T ;: : Fllo,

and substitutfog this into equation (1), the equation describing the relation between F and the angle a is found to be t:lF(a)

= µF(o) · Clo.

This relation is formally similar to the equation t:lm(t)

= - >. m(t) · t::.t

of radioactive decay (where m(t) is the mass of the decaying matter, tis the elapsed time and >. is t he decay constant). As it is wcJJ known, the mass or the decaying matter decreases exponentially:

m(t}

= mo e- "1 ,

SOLUTIONS

81

thus (using the correspondence - >. lated: (3)

= 11) the law of 'thread friction' can be formu-

F(o}

= Foe''" .

In the circumstances of the problem , Fml\ll

l

= Foe''" = 2Fo,

whichyiclds /' = - In 2

R:

7r

0 .22.

Note. The force of the thread increa.."es very rapidly, exponentially with the angle o. The ratio of the forces a t tlie two ends of the thread can reach an order of magnitude afteT only a few turns . Alpinists make use of this interesting fact when they protect themselves

from falling, and bare-handed sailors 'stop' large boats with the same technique.

S . 62. Denote the radius of the ring by R , t he mass of a unit-long piece of it by p, and its angular velocity with respect to its centre by w . The centripetal acceleration of the points of the ring is R · w2, a unit-long (small) piece of the ring therefore experiences a centripetal force of p pRw2 • \ .: R If a 1-metre-high cylindrical container were surrounded by a gas of pressure p then the force exerted by the gas onto the wall of the cylindei· would exactly be the force required . for the rotation (Figure J). Figure 1. In reality, the small pieces of matter forming the ring are not kept on their circular orbit by some imaginary external pressure but by the internal tangeutial pulling fon;e, whooe magnitude has to be

=

F

.. ..

= pR2 w 2 •

This can be proved e.g. by examining a I-metre-high container with a semicircular base surrounded by a medium of pressure p . A force of p · 2R is exerted onto a rectangle with area 2R, which has to be equilibrated by a force or magnitude 2F acting in the wall (Figure 2).

''\\ 11 1

p

-Lii--~ I,!, 'l '-.....

/

-

-

,(I I

Figure 2.

I

'

in

-· . . s~ iG;\ ·.·+... . . ..

Figure 3.

.. -·

F

123 TRIC KY PHYSICS PROBLEMS

82

Examine now a piece of a ring with central angle 2a. rotating with angular velocity w, and sec how the Newtonian motion equation is fulfilled, continues Johnny. The resultant of the forces acting on a body of mass m 2Rpo is 2Fsin o , and the 2 acce leratio n of the centre of mass is a ~ s w , where s denotes the distance between the centre of mass and tbe centre (Figure 3). According to Newton's la w,

SOLUTIONS

83

The displacement and the speed can be calculated from the above equations:

=

2F - sino = mo, which, using thP pn>vious formulae, yields s = R · ~i:n . For a semicircle, for eimmple 7r /2, therefore /1 = 2R/7r . {this was Charlie's orig111al problem), o If U1e centre of mass of a sector has w be determined , this cau be divi
=

J·

•i:

S. 63. ACCO£ding to the solutiou of the previous problem, a flc.xihlt· 1O pl' (chain) of mass e per unit length moving at speed v is stretched by a fo1 cc·

F

x

vr;g

1 rr;L ~ 0.35£, = 2v2

v= -2- .

Thus, the chain becomes detached of the pulley when 15 percent of its length is still moving upwards and its speed is smaller than what one would obtain by naively substituting z = L/2 into the principle of conservation of energy. (This fal~ value would be JLg/2~0.7 1 ./£9.) Note. The further motion of the chain is also interesting. The loop dccil("hed of the pulley starts moving upwards and with increa..<>ing speed; the speed of this piece of chain of decreasing ma.."'5 tends (in principle) to infinity. (In realit.y, the finite size of the liuks of tliP chain and of the pulley, and air resistance impose a limit on the increase of the speed.) TI1P kinetic energy of the piece of chain moving upwards remains fmit.e - despite tbc rapidly increasing Speed because the decrease of its mass is more rapid than the incrca"" of iL'< speed. The same phenomenon can be observed when a whip crads. a t:~taiu p iece (of decreasing length) of the whip moves a1 increasing speed, and when it readies the speed of sound, it causes a sharp supersouic bang.

S. 64. It is to be shown that a chain (a flexible rope) in the state of weightles.c;ness, moving at a uniform speed along a closed space curve of arbitrary shape. keeps this motion even if no constraint (e.g. chain conducting elements in the cun·es) is applied.

= (!tl2

when tbe rope is in cont.act with no other body. In the case of the pulley, this condition is fulfilled when the chain just leaves the pulley (becomes detached of it) because of the a ccelerating mot.ion. Denote the displacement of the chain by x and the acceleration of the right hand side of the chain by u. The equations of motion of the two pieces are

F

U2

F

Figure I .

~r. . . . ···1··· .(

F

The speed of the chain can be detennined using the principle of conservation of energy. The decrease in the potential energy with respect to the original situation (of negligible speed) is the same as if a piece of chain of length x had moved lower by x.

Figure 2.

Consider the chain shown in Figure 1 - moving evenly along some winding space curve a t speed v. The force stretching the chain has to be the same (of magnitude F) everywhere, as the tangential acceleration of the links is zero. lf the radius of curvature of the chain (of mass o per unit length) is R in some point (this can vary from place to place) then the equation of motion of a piece of length Rt::.o (of mass t::.m oRll.o and acceleration v 2 /R, shown in Figure£) is

=

v2

oRllci · -

R

~

Ft::.ci,


84

whi 1 ields the relati n F = (!11 2 • Noti e tba R d not oo ur in t~ equation, i.l-'. r gardl of th radius of curvature, the resultant of the tang ntiaJ forces ·1 · ust th for e needed to make the chain curve a the given place. If the cham • ti · more is Jstraight th n th r ultant force acting on a ht piece o f ·'"• ·IS . zero. The . . urvcd th chain, the greater the resul~t fore 8:cting on e:1ch (littl~) piece of :t, ancl th di rection of the resultant force is1ust what IS needed (1t can point ou~ard .) . This means that the chain falls keeping its original shap and speed, JUSt as

!'

(In this latter formula, the wo signs correspond to loops moving 'above' or 'under' the rope, i.e. to the two possible 'polarisati rlS'.) It can be seen tha pro\.ided that only circular excitati ns of a single frequency are allowed - the energy, momentum and angular momentum can only take up di rete values that an be written as the produ t of an elementary quantum and an integer, and the following relations are valid between these {frequency-depending) elementary quanta: E(w) P(w) w · J(w) .

= ·

Frank v,u ssed .

. 65. a) ·xamine the motion of the loop from a frame of reference mo~ng · a t sp e<:d r: . Io this sy:;tem, the pieces of a loop o f rad'ms R are .m um·rorm rotation · Sine th a ·celerntion of a piece of rope of central angl 60., 1 ngth R 60. and_mass p R tJ.cr; . r? / R, and the force caused by the streu:rung_is F 60. (s th solution of hr. pr vio us pr blem) , th Newtonian law of IJ'IOUOn yields

=

VFfP, which is identical to the speed at whidi small transversal waves propagate in a stretched rope. . b) Th angular frequency of the rolling loop of radius R is w c~ R , and its 2n R p. The energy carried by the loop an be expressed an of 111 < s is m I h s q ua ntiti s:

i. . th ' loop-wav • spreads at speed c

=

=

E

1 E(w) = 27TFC· -

w

l

En(w) = n · K · -

w

1 = /( -. w

(n

= 0, 1, 2,3, . . .).

1 = n . (21TR p) · c = n -K · c;:;

(n

= h · w;

Ppho lon

= h · w/c:,

Jphoton

= ±h

of the photon! Naturally, this formal analogy mu t no be taken very seriously thinking that the photon is 'like this'. Still, this similarity can be used to show that, e ven in classical physics, there are objects more complicated than a point mass but easy to understand, the excitations of which have discrete nergy, momentum and angular momentwn values.

=

S. 66. Au a mount of sand of mass 6m = 50 kg reaches a speed of v 1 m/s in tim 6t l , I he change in its horizontal momentum is therefore 6p = llm · v = 50 kgm/ . T his means that a force

=

=

F = 6p flt

= 6m · v = 50 N 6t

sow.

m mcntu m carried by th loop(s) can be calculated similarly: P.,(w)

Epholon

has to accelerate the sand, which corresponds to a torque of 10 Nm for a driving wheel with a radius of 0.2 m . The work done by the engine - if only the a.cceleration of the sand is taken into - is of 50 J per second, i.e. its power output is of

T he quanti ty K - 21T Fe is a constant characteristi of t~e rope and of its tightn~. Naturall y, s era! (n ) loops can simultaneously be exoted on the rope and their total n gy is

=

It i not diffi ult to recognise that the same relations are valid for the quanta

= E,..,.,,,. 1 + Erot = 21 · (27r Rp) · (c2 + R2w 2) ,

which an be writ.t n in the form

Tl

85

SOLUTIONS

= 0, 1, 2, 3, . . .),

The sand loses its vertical mom ntum when it arrives onto the conveyor belt. (It pr the belt vertically with a fo greater than its weight.) The sand fallen onto the belt and slowed down is accelerated horizontally, the belt in.c reases its kinetic energy. The kinetic energy of the sand increases by ~m v2 /2 = 25 J in 6t I s. This means that half of the power o( the engine (25 watts) is converted into the kineti energy of tlie sand, the rest is I.he work done against friction.

=

Note. Tl1 average speed of the sand during the acce1eration is v/2, therefore the

=

=

power of the force of friction of magnitude F 50 N is F · v/2 25 W . The helt experi n a force - F, the power of which is - Fv -50 W. Thus, exactly half of th.e

=

power of tl1e engine is used to accelerate the sand.

just like th angular moment.um: J.,(w)

= ± n· (21TR p) . Re = ±n · ]( · w12

(n

= 0, 1,2,3, .. .).

S. 67. a) First, determine the speed of the roll as a function of the distanee it has covered . As the mass of the moving part of the hose if is m(z) = M ( 1 -

i')


86

afte r a distance x , the speed v (x) can be determined using the Jaw of conservat.ion of energy. -I M v02 2

+ -1(1- MR2 22

)(Vo) R

2

2 l = -mv 2

2

2 + -1(1 - mr )(v) 22 T

0

(The change in the potential energy and t.hn little vertir.al speed caused by the ciccrP.ase of the ra
t he mao;.'I,

t111:1

speed ·is finally found to fulfil t he relation v(x) -

-

T

= 1 =1TT

o

dx

dt

o v (x )

= - 1L J t - x/ 1

VO

o

L dx = -L

VO

1' .Jl""=U o

du = -2 -L

3 Vo

.

As the hose rolls faster and faster , t his time is obviously shorter than the time corresponding to a uniform motion, namely it is two-thirds of tha t . b) The system consisting of the roll of decreasing mass and increasing speed a nd of the motionless horizontal part of increasing length can obviously not be cousiclcred as n poiut mass, the basic Ja w of dynamics cannot be applied to it in the simpl<' form F mn uut in t he more general form

=

where p is t he momentum of t he system. T he tota l womeutum of t.he moving roll (and of the whole system) is 7J(:r.) == m (x) · v(:c)

= M ( 1 - x/ L) J 1Vo = M Vo Jl - x/ L . - x/L

Cfoarly, as :r. iucreascs, p(x ) decreases (as the mass of the piece in motion decreases fa.st.er t.hau the ra te a t which its speed increases), the direction or the external force K (x) a cting on the system is therefore opposite to the direction of the motion, it ha." to point ' backwards': dp

J((x )

tlp(x)

= dt = ~v(x) = -

shell were concentrated in its centre. HlO km-deep un
=

Vo --:====== .,/1 - x/L

As :r inn ei\SCS, the speed increase> as well, i.e. the roll a t.celerates ' forward'. T he total time the hose takes to unroll can be obtained by integrating t he func tiou v(x) d:r/dJ.:

=

87

SOLUTIONS

Mv~

l

L 2(1 - x /L)

S . 68. The gravitational potential of a thin spherical shell of uniform mas.5 distribution is zero inside the shell and the same outside as if the total mass of t he

S. 69. The narrow uial boring (of a volume negligible compared to that of the whole asteroid) is considered first. Let the density of the asteroid be denoted by (! and its radius by R. Examine the gravitational a cceleration at distance r from the centre. The gravitational acceleration in this point is I ht> same as if only the sphere of radius r ' under' it were present:

g(r)

=Gm(r} = G e (4/3 )r 3 11 = ·l11G£1 r . 2 r

1.2

3

The gravitational acceleration is directly proportional to the distance from the centre and it always points in the direction oC the centre of the asteroid. This means that the first unfortunate little green man approached his fate making a harmonic oscillation oC amplitude R, and reached it after the first quarter of the period of the oscillation. The coefficient of r in the above expression corresponds to w 2 , i.e. to the square of the angular frequency, therefore the dura tion of the fall was

T1

= ~ = ~/fe-

The speed of the collision can be obtained as the product of the amplitude and the angular frequency :

{;Ge

v , = Rw =2RVT· By the time the second a ccident occurred, the little green people h a d already exploited one eighth of the material of the asteroid.The gravitational field has to be determined in the spherical cavity reaching from the surface oC the planet to its centre. This can be done by a 'tricky' application of the principle of superposition: the cavity is imagined as if it were filled by a mixture of 'n<>rnlal' titanium and titanium of ' nega tive density'.

123 TRJCI
88

The gravitationaJ fields of the compact asteroid and of t.he sphere of ' negativl! titanium' have to be added. The vector r pointing to an arbitrary point P of I.be cavity, the vector c pointing from the centre of the cavity to the centre of the a.steroid and the vector r + c pointing from the centre of tbe cavity to point P are shown in Figure J. The gravitational accelerations (that of tbe homogeneous asteroid, of the cavity and their result.ant) are proportional to the position vectors , the coefficient of proportionality being the constant previously denoted by - w2. (The 'lack of matter' in the cavity is represented by a negative density.) A

Figure 2.

Figure 1.

The resultant acceleration is (Figure 2)

g

= -Jr+ w2 (r + c) = w2c.

This gravitational acr.eleration is a constant vector magnitude inside the cavity, regardless of the position of point P. This means that there is a homogeneous gravitational field in tile cavity, with a constant magnitude identical to the magnitude of -w2 R/2 of the gravitational acceleration in the middle of the trial boring (in the centre of the cavity). Note. It can similarly be shown that. there is a homogeneous field m any spherical cavity, U1e magnitude of which equals the gravilational acceleration of the solid i.-phere at the centre of the future cavity.

SOLUTIONS

89

It can be seen that. the second unfortunate little green man fell during a little longer time than the first one, but they both hit the centre of the asteroid with the same speed. The identity of the speeds - as the comparison of the energies shows it - is not a coincidence! In the first case, the gravitational potential is lower at. the centre of tbe asteroid than at its surface and the first unfortunate man is accelerated and gains his kinetic energy according to this. The second case is different from the first one because the gravitational potential of the cavity with a 'negative' density has also t.o be taken into . The men fell· from one edge of the cavity to the other, thus the part of his gravitational potential caused by the cavity did not change. Applying the principle of superposition to the gravitational potentials, the final result is that the total change in the potential is the same in the two cases, t,berefore the change in the speed of the two victims is identical.

S. 70. Imagine that the hemisphere is divided into many concentric hemisphere shells of identical thickness. What is the force exerted by these shells onto the probe of unit mass in the centre of the sphere? Since the mass of a shell is directly proportional to the square of its radius and the force exerted by each piece is inveraely pre>R Rln portional to the square of the distance, the gravFigure 1. itational accelerations caused by the hemisphere shells of identical thickness and different. radii are identical in the centre (Figure I). If there are n shells then t he volume of the outermost shell is of 27rR2 (R/n)e, where R is the radius of the asteroid and e is its density. The tot..il gravitational field of the hemisphere is n times the field of the outermost shell, Le it is the same as if there were a mass M = 27rR3g on the surface of the hemisphere. (This mass is 3 times the actual mass of the hemisphere.)

Tbe duration T 2 of the fall and the speed of collision t'2 of the second unfortunate little green man can be calculated applying the well-known relations for evenly accelerating motions: and

t'2= 2R~.

Tbe ratios to be reported by the expert are therefore and

p

Figure 9.

What is the force exerted by the hemisphere shell of mass M onto the body of unit mass placed at the centre? From the magnitude point of view, this force is the same as that exerted by the probe onto the shell, Le. p = GM/(27rR2 ) · (l/R2 ) ::;: == GoR per unit surface. A liquid of pressure p would exert the same force onto the shell (Figure 2). Furthermore, the resultant of the forceB coming from the pressure


90

of the liquid (acting on the surface of the hemisphere shell) is the same as the force of the pressure acting on the plane surface of the hemisphere (Since the resultant force of a liquid acting on a solid sphere is z.ero), i.e. g = p · R 2 tr = G{!RTr (Figure 9). If it. also taken into consideration t..hat the gravitational a.cceleration was Yo G(41TR3 /3)e · (1/ R 2 ) at the surface of the asteroid, the above calculated value 2 is found to beg 3go/4 7.36 crn/s . Knowing the d ensity of titanium, the (original) radius of the asteroid can be determined: R ~ 78 km.

SOLUTIONS

91

T he above sum tends to an integral - as the thickness of the layers goes to zero:

=

=

=

=

S. 71. If one could calculate the work W necessary tO pull a part the two halves = 1 m from eacl1 other then the pulling force would also be known using the formula W = F · d. But how can W easily be determined? Certainly not directly, as the product of the force and the displacement, since that would lead us back to the original problem, thf' calculation of the force. Some other method has to be found. The gravitational p otential energy of the system increases when the two hemispheres are pulled apart and W equals this increase. This increase can however be calculated by determining the work done by the little green people when carrying the titanium (originally in the disc of radius R and thickness cl) to the surface. (The extracted metal can be considered as carried to the surface by means of easily rolling devices and then spread evenly on the surface.) Calculated the work done in the course of thP extraction using the increase in the potential energy of the metal brought up from different depths. If g denotes the gravitational acceleration at the surface then the gravitational acceleration at a distance x from the centre is g(x) xg/R. (l! 3M/(41TR3 ) denotes the density oC the asteroid.) Consider the amount of metal of volume 6 V d · 2trx · l:l.x and mass 6m {!6 V originally situated between the values x and x + l:l.x -l 1- d of the radius. This matter has to be brought up from the same depth, it is therefore suitable to consider it as one piece from the point of view of the work done. A force of Cl.m · gx/ R acts on this piece at its original pl.a ce, while this force is of l:l.m 9 at the surface. Since gravitation increases evenly when moving from inside the asteroid to its surface, the aritlunetic mean of the initial and fina) forces can be used . The total displacement is R - x, the work done is therefore of an asteroid of mass M and radius R to a distance of d

= =

l:l.W

1 + (x/ R) = l:l.mg -- - (R 2

x ) =Cl.mg

R2 - x2 R 2

=

and the work done is W 3M gd/16. Notr.. The same result can be obtained witl1out using such sophisticated u1athematie>< (int.tgral calculus}. lnlrocluu the new variable u = x 2 / R2 instead of x (O :S u :S I) , and replaa> the sum mat ion O\W the layers at differeut depths by t.he summation over the t.erros correspomling to different u's. Using t.hr relations u x 2 / R 2 and u+Cl.u = (:r: + tu:) 2 / W (anct neglectinA the containing the square of Uie small Cl.:r:). tJ1e work done can he

=

expressed a:;

This sum ran easily b" calcuJat.ecl, since I - 11 changes evenly Crom I t.o 0, tl1crefore iu average it ca11 be taken t.o be 1/2, and the s um of t.he D.u is 1. Finally, the result presented above is found for the work. The fairy giant does a wotk W 3Mgd/l6 while pulling the two hemispheres apart to dist.a.nee d, and the force necessary for this is F = 3Mg/16. This is the force of attraction between the two hemispheres and this is the force the props !
=

F

=

=

gd ~ -2 =43 M y ( R -x )xl:l.x.

(It has been used that Cl.m can be expressed in of the volume and the density and this Jat.ter can be given using the mass and the radius of the asteroid.) The total work is the sum of the works done when bringing up the matt.er from different depths:

IV = ~ Mgrl L(l - u) Cl.u.

=~ M . GM = ~ (47rR3e)2 !!.__ - GR:'t11T2 . r3 16 R2 16 3 R23 -4.5 10 N.

In order to give a feeling of tlre orders, calculate the average load on a unit surface: 71

= RF21T = 1.4 · Hr

Pa, which is one and a half times the atmospheric pressure on

Earth. This is tlre weight (on Earth) o f 14 tons of matter; sufficienUy strong pr-ops should bear s uch a weight.

S. 72. At the surface of the charged sphere (whether it consists of a single part or of two parts close to each other), t.he electric field strength is 1

Q

E--- 4Treo w · while the electric charge per unit surface is

Q CT=

41f[{J-.


92

=

=

This ctric field exerts a force b.F tE6Q onto the charge 6Q u6A o upying a surface 6A . (The reason for the factor of is that the cl tri 6eld strength is at the ou r urfac of the phere and zer in id , its average value is therefore /2.) The force ex onto th charges on th ph rical sections is th refore

l:lF 6A

=

4

Q1 32-;r 2 c~ R 4

:::::

P

S. 73. Some hydrogen of pressur

(2)

(7

3Po

(}

2Qo,,.

->--.

(3)

It is interesting that neither the radius, nor the thickness of the walls of the container o cur in (3)! (A container with thicker walls bears a higher pressure but its weight i · greater, too.) Paging the tables of physical constants, no material can be found that would fulfil condition (3)! None of t.h materials known today is strong enough (compared to its density) lo lift itself up with t.he help of the upthrust of the gas it contains. S. 74. Gravitational acceleration al the surface of the Earth of radius R, mass M and density p can be given in the following form :

p is enclosed in a contai11rr nf radius R with walls f thirJcnes ti. L1·1 fJ cl note

the density of th mat ·al of t lie walls aud u its tensil str ngth . Calculate th maximum pressure th c ntainer can bear without bursting. lf the container w re cut in two, the gas would push the pieces apart with a force of pR2 7r (the external pressur is negligihl compared top) . This force mu t not b greater than th product of t he surface of 2RTrd of th cut and the tensil str ngth: p Fi?Tr

The weight of the container is of 4R2 Trdeg, and according lo the above reasoning, this cannot b greater than the maximum lifting force:

Inequalities (1) 8lld (2) yi Id a relation betwe n the material constants of the contain r and the pressure and density of the external air:

per unit area, orr p ndi11g Lo the for with which a liquid of pressure P would pu.-;h th two parts of th sphere apart.. On the oth r hand, this force is the product of p an
(1)

93

SOLUTIONS

I'

••••• ••. Po

In order lo solve the problem, the magnitude of gravitational acceleration has to be determined at the surface of a very large disc of thickness H and density {! (far from the edge of the disc). The result is relatively easy to obtain using the analogy of the laws of force of electrostatic and gravitational interactions. A mass m ('gravitational charge'} corresponds to an electri charge q, the gravitational constant f to 1/(4'11'Eo), the gravitational acceleration g F/m to an electric field strength E F/q. Thus, the new task is to determine the ele tric field strength of an infinitely large disc (out.bide t.11~ disc) with a homogeneous charge distribution, then lo substitute the respective quantities to obtain he gravitational acceleration for a mass distribution of similar geometry. The electric analogy can be determined using Gauss' theorem: l

=

=

< 2R7r u d.

If the gas ii; I t into a large balloon , its pressure decreases to the value Po of th external pressur , its volume is therefore

v = 4.R37r . .!?... 3

Po

An amount of gas of such a volwne can c rtainly not lift a load h avi r t han tlair9 balloon and of th hydrogen have to ·ubtracted from th upthrust.

(f!,.;r is the d nsity of air), as the weight of th

~i;; =- L:q. £0

=

where ~B 2AE is th electric flux (see t.he figure) . If the electric charge density is f!q then the total charge surrounded by a closed surface is E q = u9 AH, therefore 1

2EA::::: -eqAH, Co


94

SOL TIONS

95

whicl! yield

= _!__ /lqH

E

2 for the electri · 6 Id strength. Substituting the resp ive quantities, the gravitational acceleration is, from (2) : g 21r0oH. This has to b equal to th gravita tional acceleration mea.5ured at the !.urface of the Earth: £0

£~ '

£

=

¥5P"

re = 9<1isc

Ir :::::: =======~ :: ::: :::: "

oH = G-4R'3-ITQ = 41TGT,

A

Figurt J.

whid1 yields

~3 R = ~3 ·6370 km = 4250 km

H =

The magnitude E of the ele tri field strength can be det.f'.rmincd using the following ' trick'. Imagine two very long, venJy charged rods d at th ·r encls. The resultant flP.ld str ngth will be the vectorial sum of th field strengths of th<' two infirute half-rods (Figttrr. I) . The clir tion of the resuJtan wilJ obviou ly b perpendi ular the rod - considering the symmetry - , and its magnitude ( .J2 times U1e field strength E of the individual rods} can be determined using Gauss· law of electric field lines. Surround a section of a certain length l of the ' infinitely long' rod by a cylinder of raclius h (Figure 2) . There is a charg Q = la insid th cylinder (a is the amount of charge per unit length), and th number of field lin t'~ leaving the ylinder is

for the thickucss of the ' flat Earth'. S. 75 . Due to the similarity of triangles FHC and GI
r r 6x = CH · h = r6a · h

2

r = -;:;6a.

4>e

=

There i a charge 6Q u6x on ucb a piece (wh re a i the charge per unit length}, and U1e magnitude of the electric field strength corresponcling to this charge is 6E

I

D.Q

= -47r£o- -r -2

K

Accorcliug to Gauss' law, Figure

1.

a == - - - 6a. 47r£o h I

8

S. 76. Using the result of the previous problem , it can be stated that th direction of tl1e electric field in a point (in the perp nclicular plane) at distanc h from the nd of a very ('infinjteJy') long rod is at an angl of 45° with respect to the rod. Figure 2.

= ./2E · 21rhl.

tu

This quantity is independent of the angJ o, it only depends on the angl at which the piec of rod is seen. Thus the electric field strength vectors of the charges situated symmetrically with respect to the bisectrix of angle C of the triangle ABC are of the sam magnitude and their resultant points in the direction of th bisectrix (Figure 2) . A

Figun: 2.

which yields

E

= 41T1S ../2 ~ . o Ji

S. 77. n) H the magnetic induction in point B is denoted by B1 th n the symmetrical coil causes the same induction (Figure 1). However, since the resultant of the superimpooed fields ha.5 to be Bo, B 1 == Bo/2 is ilircctly obt.ained. b) A similar reasonfag shows that the axC I i 0 I j I !!) I I ial component of the magnetic induction vector in th plan p rpcndicular to the axis of the Fi!)'Ure I . sol noid and crossing point P has to be 8 0 /2 as long a.5 the dist.an from the axis i smaller than Rand zero otherwise (Figure 2). A total mag:n tic flux of R 2 7r • (Bo/2) crosst~ therefore th 'cl ing surface' of the solenoid, which if exactly half the flu x insidr the solenoid. Wher docs the other half of the flux 'clisappear'?


96

97

SOLUTIONS

~L

x

~

BI ,~P ,

~r

R- x )(

Figure 2.

Figure 3.

<") A qirnlit.at.ive sket.ch of the field lines can be seen in Figure ~- The fi~ld liuP LW:;Sing the outermost tum (point Q) of the solenoid h~ to contmue outside perpendicularly to the solenoid. Half of the flux goes left of this, the rest leaves t~e coil between the turns. The field lines that form the limit go inside the solenoid (far enough from its end) at a distance of R/J2 from its axis, since there, the total flux has to be divided into two as weU. S. 78. The charge placed between the capaciwr plates could be divided into two parts that could be moved further away from each other in the plane parallel to t hf! plates. In this case, the charges on the capacitor plates w~~d also mo~e ~ut their sum would be the same on both plates (principle of superpoS1Uon). Contmumg tJ1is process, the charge Q can be chopped further to be finally 'spread' evenly along

plane (parallel to the capacitor plates and of the same size as these). This has lead us to two plane capacitors connected in parallel (the distances betweell the plat.es are x and d- x). The voltage of the capacitor plates is identical, the field strength between them is therefore inversely proportional ~o ~e~r dis~ance: £ 1 / E1 (d - x)/x . The electric flux starting from the charge Q 1s d1V1ded m tbe same proportion (Gauss' law), and the ratio of the charges on the earthed pla~s is the san1P., t.oo. Since a t.otal charge of -Q accumulates on the plates, respective charges of

<1

=

d -x

Q1 = -Q-cl- ,

and

:1cc-1Jmulatc on each plate.

S. 79. Fi!]11.re J shows a light ray ing through piano-parallel plates of different refractive indices placed one after the other. According to Snell':; law, sina2 sina1

= n1 n2,

sin a3 sina2

n2

- - = -, ... n3

It can be seen that t he product of the sine of the angle ofincidence and the absolute refractive index is constant: n sin a = constant.

Figure 2.

Figure I .

This relation is also valid for a 1111:dium of coutiuually dlangiug refractive index, since the medium can be c:onsiclerecl as consisting of thin piano-parallel plates. Place the origin of the coordinate system to the point where the light ray enters the medium. Here, the angle of incidence is 90° and the refractive index is no, which gives the above constant: n(x) · sin a = no. Assume that the light prop;i~ates along an arc of radius R and examine the place of coordinate x shmn1 in Fi•J•irt· 2. Apply the above relation: 110

=

R -x

11

s in n = n(x) · ~·

which yields the space-dependence- of t he refractive index to be

.

n(x)

R =- ·no. R -x

The known material with the greatest refractive index is diamond, but even the refractive index of this material does not reach the value nmax 2.5. It is practically this limit that ::sets tJ1e maximum arc the light ray can cover. U the refractive index changes from 11<> = 1 to n.nax = 2.5 then the maximum value of x can be 3R/5, which corresponds to an arc with a central angle of 66.4°. In practice, it is difficult to 'direct ' light along an arc. However, solutions can be made in which the concentration, therefore the refractive index shows a continuous vertical change. ln such a solution, the light ray does not propagate along an a rc but. along some continuous curve.

=

Note. Why does the ligbt ray entering along a,..qs z st.art to ' bend' ? The reason for this is that there is no infinitely thin light ray; a 'ray' always means a beam of firute thickness. The refractive index (i.e. the speed of propagation) changes along the cross-section of the beam, therefore the wavefront bends and the beam is deviated.

S. 80. The pressure of the water risen to height h changes linearly with the height: it is equal to the external atmospheric pressure Po at the bottom and to

!IK

I~•

123 TRICKY PHYSICS PROBLEM S

The average pressur ex rted onto the wall is Pnt1ern9e = corr sponding to this value ( oonsidering a water layer of d• •1it Ii ti) ii< 1"1 = lPatternge · h . <'111111id1·r I.he? horizontal fore ;u·11111'. 1111 llw volum limited by d1u. l11·•I h111~1 iu t.ltc figure. Th wall • 1•11:.111·11 it 111 lit • right with force ,-- F I ? Vi . 1111· •·s t 1·111al air to he left with 1 1" 11 •· /.'_. 1,~,1, and th res of = ......--~=:;:_-~~ ·:::· -····... : ·· :::.-. ·:.··. ::·-~ : ..-.i 1, .. w1111·1 pulls it. t.o th right by "" " " "' 11f IJ11· s11rfa · tension with forces has to be zero, since Lb examined 1. '" •· I··, 1,. The imltan of th ''"1111111· 1: •• 1t 1e•sl.. This m ~ thal

/~,

l'!tlt 111. t lu· top. l'!tlt /'J., I.he for

SOLUTION S formation anyway), but the above r asoning is suitabl for estimating th size of molecules using macroscopi c characteristic.

P,. •

·--

2 . 0.073

1000 . 10 '' '

1

t J... '"'tr.ht . \\'1111•1 I L'< •'S hy approxi mately 4 millim tr

~ -~--~..... _

at. lhe wall or the aquarium.

tension o (with energy o per unit surface} decreases by a mall 6 R, the energy of the surface d creases by

M anwhil , the volume of t11e drop decreases by 41f R2 6R , the evaporation f this quantity of water uses an energy of pomlion

= 4Ltnr R 6R, 2

I '11·

density of water and L is its heat of vaporizatio n. Tiu· il1·n1•nse of th surface energy ould cover the evaporation of the drop if I~ I·.'_,,,,.,,.,., 6Eevap0
R<

20

pL

~1 -

10- 11 m

111'111. llt1w1'\·1•1· such a drop of water does not exist, since this size is smaller than thnt 11r 11111• , a ter molecule (it is of the same order of magnitude) . So th re is no w11t... r tlr11p Lh, t c.;: n vaporate in it.self wit.houl cooling (it could not be a stabl 1

for the equilibrium of th h. drostat.i · µi sure and the preAAure caused by ! rel="nofollow">'Urf • ill rht' curved surface of the liquid than at the plane surface, the ctiffererice b 111~ rh pressure oft.he column of vapour of height h and density Uv: 6JJ.,

s. 81. When the radius R of a drop of water with surfac

6Ee wlll'11 · 1•

= 0.0038 m

h

S. 82. Imagine a cl sed vessel with a liquid up to a cert.aln level and th saturated vapour of ltis liquid filling the rest of the v el. A capillary tub of radius r i immersed into the liquid. The liquid does not adhere to the wall of the tube. In such circum ances, the level of the liquid in th capillary tub fall to some depth h. The magnitud of t his fall can bf> det.erruin d from the condition

= g.,gh = -2o- -e . 1" l!I

As Q.,
S. 83. Let A denote the ross-section of the piston and y its vertical displa ement bet n its initial and new equilibrium positions (s the figure) . he decrease of the potential energy of th body of v.'t!.ight G hung onto the pist.on increases the internal energy of the air inside th yHnder and the kineti energy of the bodies starting to move. The equation cxpressinp; th r:onservation of en rgy after the dumping of t.b oscillations is (1)

Cy =

5

2 (p 1A{l1 -

v)+112A(h

+ y)

h

,,

A

.J

- 2PoAh],

where P1 is th pressur in the bottom part, p 2 is that. in the top part an
c

123 TRl KY PHYSICS PROBLEMS

100

d rcase 1n 1 potential energy (therefore th increase of th interual energies of th g< ) i very large and the initial internal n rgy of th air can be n glected:

( l') Th load hunp, onto the piston is at rest, the quilibrium condition for this is

(2) Furth r, th temp rature and the mass f the gases are identical in th t.wo parts, thcr f re: their internal energies have also t.o b id ntical :

(3)

5

- 711:\(h - y) 2

= -251>iA(h + y) .

=

Equati ns ( l ') , (2) and (3) yield v ,/577h for th displa n n of the piston, i.e. th p,as in th bottom part is compressed to 1 ~ 15 % of its original olum . Not e. The s1wprising result is that th volume of th gas in th bottom part does 11ot t('ud w zero, however large the weight is (though are supposed be easy to rompn~s)! The larg loa.cl iucreases th internal energy, tbcr fOl"C th temperature of the encl cd gll~ , whi h com;iderably increases not only th pr · ure but also the difference bl'twcen th prCSl''t1res. Iu !lr:U-lir<' the increase o( the load 1. limited by tJ1e mechanical weight-bearing capaoty' of the ~tructurc and the me.lting point. of tl1e applied materials. It is also 1.o be consich'rcn whether the given temperature is below or above th temperature up to which air ' :au lw ·nusiclerrd all an id<'al gas of two-atom molecules. If th initial internal energy of the air is not n gl ted then tbe result is y 11

=

./517

J35G2 + 25j)gA1 7G

-

SOLUTIONS

101

Approximating the melting heat by the melting heat of 200 - 300 kJ/kg of metals, the maximum possible heigh of mountains on Earth is found to be 20 - 30 km, which is in the right order of magnitude. H it is al taken into that the base of the mountain does not have to melt but the yield point is the real limit of the growth of the mountains, the actual height of the high t mountains oo Earth is surprisingly well obtained. Gravitational acceleration is significantly smaller on Mars than on Earth 2 (9Mnr• ~ 4 m/s ), therefore mountains (consisting of similar rocks) can be ltigher on Mars than on Earth. Indeed, the highest mountain on Mars, Mons O lympus, is 26 km (!) high. S. 85. In order to simplify the calculation, the system of unit.s should be chosen in such a way that the initial volume and the external pressure ar unity. The units of the mole number and the gas constant should b cho.c; n in urh a Wc:\Y that their product is also unity, which reduces th equation pV nRT t.o p\ • = T . Then th unit of temperature cannot be chosen arbitrarily, it is determined by the product of the pressure and the volume. In this system, the molar hea is a number that has to be multiplied by the gas constant to yield the usual molar h at. The pressure of the gas is initially given by the sum of the pressure of the 76cent.imetre-higb column of mercury and the atmospheric pressur , ;i 1Ot(l l of 2 units. The increase of the gas in centimetres equals the decrease of height of tl1e column of mercury, therefore as the gas expands from 1 unit t.o 2 uni , it pr<'l>sure linearly decreases from 2 to 1 as shown in the Figure 1. The temperature of the enclosed air is initially of 2 units and still of 2 units at the end of the straight section shown in the figure, since these points are on the isothermal pV 2. In this representat.ion, the isothermals are symmetrical to the bisectri:x of the axes p - V, therefore the highest temperature is obtained when the corresponding isotl1ermal does not cross the straight but is tangent to it, as shown in Figure 1. This occurs at the middle of the process, in the point p ~. V ~. the maximum temperature is therefore ~ ·

=

=

=

=

c

5Po~ 6

.... -. ':

. 84. ' nsider a mountain of average density e. ba.sc ar a A a.nd height h. Au en gy of AdgJ, would be necessary t.o melt the bott m lay r of thickness d (L is Lh meltin , ll at of the mateJ·ial of th mountain) . Th tal mass of tJ1 mountain (assu111c1J tu h11 the shape of a rectangular box) is appro ·imat.ely Ahq, he energ r I , · d in th cours of its sinking by d v.'Ould therefore be Ahggd. The condition thaL th ha'" of a mountain does not melt under th weight of th mountain is:

AdeL > Ahogd,

i.e.

[, h < -. g

)! :

-- -- - - ~ --- ~- ---------,• ,'

2

Figure 1.

-3 Figure 2.

123 TRJCJ
102

SOLUTIONS

103

H.

The equation of the straight representing the process is p = 3 - V. The first principle of thermodynamics for an arbitrary elementary section of the process reads as

up to this point (in this peculiar syste m of units) is while the increase of the inlernal energy is t.he required heat transfor is therefore Q ll. In the usual 6 units this means that if the column of mer-cury originally encloses ~ moles of air in the tube then the heat Lransfer necessary to remove it is of nR.

where the l.h.s. is the change of the internal energy of air (since the molecules of air consist of 2 atoms, Cv = ~ R), and on the r.h.s., C is the molar heat in question. Express the pressure in of the volume (p = 3 - V), and transfonn the change of volume into a change of temperature. TI1is can be done by substituting the T , the result is (3 - \l)V T , tl1e u take the pressure into the equation 7N' change of both s ides: (3 - 2lf}6 V t:.T . Performing all thf! operations. t.he first principle takes the fonn

S. 86. Assume that the cavity can be found at depth h . When the flue is filJed with water, the pressure in the cavity is p = Po + ugh, where Po is the atmospheric pressure and {} is t.he density of water. Boiling starts at the depth and tempera ture where the pressure p = A e-L/(RTJ of saturated water vapour equal:; this pressure. T he value of the specific heat of vaporization occurring in the relauo11 is approximate.ly L = 40 kJ/rnol and the molar gas constant is R 8.3 J /molK. Having obtained these data, the constant A can be determined usjug the boiling temperature of 373 K of water at. aunospheric pressure: A = 4.1 . 10 10 Pa. Fbr the calculations, the temperature of the ground at the surface can lw t akc,i1 to be To = 290 I( , and as it increases by one degree with each metre, tl1~ tempera•'" " at depth his given by T = To + h if the depth is measured in metres. The followiur, equation can be formulated using the equality of the pressures:

5 7.D.T = C 6T - p6V,

=

=

=

3-V

5 2

- D.T =C ;);.1' - -

3- 2

\.. 6T.

The P.quation can be simplified by D.T, then the molar heat C of the be expressed: C 2 1 - 12V _ 6 - 4V

p~ocess

cau

M,

=

N

=

=

Figure 2 s hows the molar heat versus the volwne. It can be seen lo take an infinite value at V 3/2 because then the straight approximates weU the isothermal corresponding to the maximum temperature. In the course of an isothermal process, heat is transferred but tbe temperature does not rise, this is why the molar heat is infinite. After this, the mola1 heat becomes negative, which means that despite the positive he.at transfer, the inte rnal energy of the gas decreases because the work done by the gas is greater than the transferred heat. The most interesting section of the process occurs when V = ~ (only a quarter of the mercury is left in the tube by then): the molar heat becomes zero. This corresponds to an adiabatic process, siuce there is a change of temperature without any heat transfer. In this point, t he straight describing the process is tangent to the adiabatic line. After this, the temperature should decrease furthe r but the molar heat becomes positive, which could only be done if the syst em were cooled. This is how the process characterised by a straight could be termiuatcd through equilibrium states. If the system is heated further, it does not follow the straight any more hut sta.J·ts a life on its own and follows the adia ba tic line. The tempera ture of the air does decrease, but not as much as t.he straight would prescribe; the decrease of the internal energy of the air covers not onJy the work done against the atmospheric pressure (hy lifting the mercury) but also the acceleration of U1e mercury: the mercury i<> P.jected from the tube. Re turning to the original question , the heat transfer necessary to remove the mercury from the t.ube, it can be seen that heat only has to be transferred from \I = 1 to \I f. Simple arithmetic operations show that the work dooe by the gas

=

=

which can be solved by means of numerical methods, i.e. by carefulJy m11~t
=

=

=

=

=

S. 87. First, examine the state of the ice and the pressure in the container a t. 100 °C. Since the density of saturated water vapour is of 0.5977 kg/m3 ~ 0.6 kg/ 1113 and its pressure is of 1 atmosphere, the 0.6 kg of ice are completely transfor med into vapour at 100 °C, indeed into saturated vapour of a pressure of 1 atmosphere. How does the -10 °C ice reach the state of 100 saturated vapour ? H temperature is increased very slowly, the system always es through equilibrium states. First, the ice sublimates and the ice phase is in equilibrium with the vapour phase, which lasts until the temperature and pressure at the triple point (0.01 °C, 610 Pa).

cc


104

At the triple point, liquid state appears beside the ice and the water vapour. Further heating makes the solid phase disappear, there remains water and saturated water vapour in the container. It is interesting to notice that afterwards, t he wate1 is constantly at its boiling point, until the whole quantity of water is transformed into vapour. From the point of view of the heat absorption, only the initial and the final states play a role, the heat taken up by the system ice-water- vapour (increasing its internal energy} is independent of the intermediary st ates. In the course of t he calculations, it is s11itable to decompose the heating p rocess of the 0.6 kg of it:c into fom· sections (wanning U1e ice, melting the ice, warming the water and boiling the wal<'r): Q

=<;mD.T1 + L1m+c 11U::.T2 + l,,m, where c, = 2. J

heat of ice, 6.T1

10

= 10 ° C,

L1

= 334 ~~ =

kJ C is the :
i:; I.he heat of fusion ice, r-.,.

= 4.2 k~~C

is tlw specific heat of water , D.T2 100 ° C and L., is the heat of v
= 2256

kJ given in tables is not only due to the higher kg internal enc>rgy of the vapour but it also contains the work done against the atmospheric' pressure of 1 atmosphere. ln the problem, this work was done previously when the container wac; evacuated, i.e. the heat to be transferred to t he system is smaller by this amount: - pl:!. V - 101.3 kJ. As this value corresponds to 0.6 kg of water ,

=

L•

= 2256 -

lOl,3 0,6

= 2087 kJ ::1:: 2090 kJ kg

kg

is the more accuratR value to be used for t he calculation. After substituting the data, Q = l720 kJ is fow1d for the heat transfer. Neglecting the work done against the atmospheric pressure would have caused an inaccuracy of nearly 6 %, while the slight dependence of the other coefficients on temperat.ure and pressure cause errors of approximately l %. Tue main reason for this is t hat the change of t he specific volume is very significant (of a factor of 16{X)!) in the course of t he water- vapour transition. The volume of water or ice is negligible compared to that of vapour , aml they have indec
SOLUTIONS

105

of the saturated vapour. This means that the water cannot start boiling at any

temperature. If the container is strong enough, it.

p

can bear a significant rise of the tempera.. ture. We can ask the quest.ion: until what temperature can liquid water be found in the container? Data of tables concerning wruer saturated water vapour tell us that the critical temperature of water is 374.2 °C, its critical density is 326.2 kg/m3 , '".,hich ice means that the wholP. amount of water cannot evaporate before reaching the critical T vapour point C (as in that case the density would 3 be around 500 kg/m in the container) . In365 °C 372.S °C deed, a strange change of state occurs at loll.·er temperatures! The density of water is only 500 kg/m3 at approximately 365 °C, and the density of saturated vapour 3 is 160 kg/m at this temperature. This means that at 365 °C, the liquid water fills the whole container. The explanation for this is that as the temperature rises, tlw decrease of the density of water is more rapid than the increase of the densi1.v of water vapour in this temperature range. At temperatures above 365 ° C, water remains in liquid state in the (constant) volume of the container, but its pressure increases. Then the system leaves the liquid-vapour boiling curve of the phase diagram as shown in the figure, the process goes on above the curve, in the liquid phase- The liquid and vapour state lose their meaning above the critical temperature. Rigorously applying the terminology, water is then in gaseous state. In what state can the air be in the container? In the temperature range of 360 - 380 °C in question, the pressure is around 200- 300 atmospheres. At such a pressure, some part of the air is dissolved in the water, while the rest is compressed into very small air bubbles only filling I - 2 per millc of the container. Thus, th1> air has no noticeable effect on the previously described behaviour of the water.

c

S. 89. a) The particles colliding from the nght (rebounding elastically) transmit a greater momentum to the body at the beginning than those colliding from the left (inelastically). For this reason, a resultant force pointing to the left acts on the body, accelerating it. The faster the body moves to the left, the fewer particles collide "'ith it from the right (and "';th a lower relative velocity than initially), while the number of particles (and their relative velocity) colJiding from the left increases, therefore the force acting on the body decreases. After a sufficiently long time, the body moves at some uniform speed v 1 •• The condition for this is that the tiny particles transmit the same momentum from the

123 TRICKY PHYSI S PROBLEMS

106

1igJ1t as from th left in unit time. They reach t.h bocly at a relative speed Vo - v 1 from th right and rebound at th same speed . In a h rt tim interval 6t, the particles coming from a distance of (Vo - v 1 )6t from the right reach th c finder, the.r efore the numb of bits is proportional to (Vo - v 1 ) and the momentum transmitted b a singl pa.rticl is proPQrtional to 2(vo - vi ) 2 . Similarly, th particl colliding from th lefl at a reJative speed of {v1 Vo) transmit a momentum of {Vo + v, )2 to th body, as their colli ions are inf'Jastic. The condition for the constant spt"C
"'2 - 1

Vo = --r,:;-Vo:::::: 0.17vo. v2 +I b) Th oUiding particles cau th mot ul · of a gas at" r.crtain mperatu . If th body moved uniformly af a long tim , this would ma.k the pr paration of a s«t>nd order perpctuum mobile possibl , i . . a heat gi11 coulci b built that c:au permanentJ take out nergy from a single h at r r oir. According to experi ce, such an engine cannot be mad . Where is the mistake in th above solution? The heat produced by th in lastic ollisions has not been taken into yet! The particles bombarding (in lastically) th left of th body heat th left half of th hody up. If the h at produced her is perman ntly taken away (th body is led) th n the motion descri~ in part a) i su tainabl . In this case, th result is a r gular heat ngin working v.ith two heat reservoirs (th gas onsi ing of th bombarding particl and the cooling medium) . However, if the body is u t cooled tiien soon or later it wami up. Th molecul forming th sid of th warm body move to and fro at the average speed 01T ponding to the temperature, and th particles of the gas rebound s m times at a greater, som limes at a lower peed tl1an in th of th Id bod . Th rns ult of this process is that finally tlie ocfy does not take up any m r It at from the gas on any sid (or it would warm further) . But then tb collisions are elasti on oth sid · - considering th a ra~ in time - , therefor th momentum transfer {pushjng force) is th sam n both ·d Thus, th body has to stop after a ry loug tim (whf'n thermal equilibrium i reached).

S. 90. When a body of absolute temperature T takes up beat 6Q th n be chang of the 1tropy is l:l ~ l:l0 /T. (Th equality is \alid when the process is r versibl .) Let 6T1 arid l:lTi dcno th littl chan~ of teinperatur of th bodies of temp rature T 1 and T2 , resp tively. Th change of the entropy of th wh le system is l:lS, + l:l 6Q1 + l:lQ2 > 0 2 1i T-i. - '

=

107

The ratio of _th hea~ transfer to I.he change of the temperatur i identical for th two bodi , smce their masses are equal, t11 above equation can th refi re be written as

lll'1T2

+ 6T2T1

= 6(T1T2) ~ O.

This means that tb geometric mean of the temperatures of the two bodi cannot cl during the pro (but it may increase). . T11i relati n is valid for ach step, ther fore it is also valid for th initial an
(T•; Tz _y'T;T;) me,

whf!re m is the total mass of the water and

is its specific heat..

S. 91 . During the operation of the pump, /> each conipr ion cycl can be divided into two tions. Initially, both valves are closed. The piston compr isotl1erma1Jy the gas in th 1 litre part. Wh n th pr ur in th pump reaches that in th ntain r, th inner valve opens up _and th total amount of gas is compressed isotherma!Jy. Th moment wh n the 1nne.i· valve opens is more and mor delayed, which makes th calcuJatiou of th total w rk d n rather coruplicatecl. Fortunately, I.her is a much simpJ m thod!

~

Consid r th am unt of gas that ts into I.he container at th nd of th ~r cess. Th. is th 10 litres of air initially present and furth r 9 times this amount 1.e. 90 litres of air (initially at atmospheri pressure) in all. This is amount of ga~ compr to JO litres. According to the firs rin ipl of t.he.iniodynamics, the sum of th w rk ~V clone on the gas and th heat Q transferred it incr ase the internal en rgy of tlie ~as: In th present situ~i n, the temperatur of the gas does not chang , th refor its m tcrnal en rgy remains th sam as it initially was. Thus, w Q o, i. . lhe rk don n the gas equals th heat - Q given out by the gas. ~rther, the ~ange of the entropy of th gas at constant temperature T, i 11 a uld b rever bl pro ess, 1 l:lS 6Q/T. If the cl1a.ng of the entropy of the g cal ulate :-states is 1/10N times their original numbE'r. A rcliug to the st~ust1cal m_terpr tat1on f entropy, th change of the entr py f th g. i.c; the I gantltm of thJS number multiplied by th Boltzmann constant:

=

=

which yields

6S = kin

( 10~) = -Nk In IO.


108 This yit>lds - t:::.Q

= -Tt:::.S = NkT In 10 ::::: nR.Tln 10 =10poV0 ln 10

for thl' heat. given out by the gas. and the work done is W

= 10 · 105

Pa -10- 2 m 3 ·In10::::: 23 kJ.

S. 92. Thet~ has to be an electric field between the Earth and the distant planC:!t, as there is a potential difference between them. This electric field causes a charge separation in the wall of the spaceship, therefore the electric fielrl is zero iuside (Faraday cage). The p.d. between the spaceship and the planet changes in I.he course of t.ln: motion of the spaceship: it is very large at the beginning, slowly clecrea.<;eS and becomes zero at the end. (The p.d. is the product of the average electric field strength a.ud the distance. The former does practically not change, the latter slowly decreases due to the motion of the spaceship.) Thus, when the spaceship lands on the planet, its electric potential is exactly equal to that of the planet a nd 'the astronauts can get out.

S. 93. The energy of a capacitor of charge Q and capacitance C is Q2 /(2C). If the change of the energy of the capacitor could be found, the change of its capaci~ce could be c:alculated, too. The capacitance of the capacitor is higher when it is indented, since the sw-face r.hru·gf'1; 111ovP in the direction o opposite to the force acting on them. Jt can also l>I' said t.hat an electrostatic field of field strength E has an energy EEY /2 per unit volum(' and when the capacitor is indented, the electric field penM.mt es a volume where it. was not. present previously. If the surface of the capacitor is only changed a little, the electric field cau be considered as identical to the original one near the surface. Thus, the change of the energy only depends on the change of volume and not on the actual shape of the indent.. Imagine that. the capacitor is hammered so that its volume decreases by 3 %, but its shape remains spherical. Its radius has to be reduced by I % for this (as the volume of the sphere is proportional to the cube of its radius). The ratio of the cuergy of such a capacitor to the energy of the original one is the same as the ratio of the energy of the indented capacitor of the problem to that of the original one. Thus, the relative change of their capacitance is identical as well. Further, the capacitance of a spherical capacitor is proportional to its radius. The capacitance of the new capacitor (of reduced size) is therefore l % smaller, and the capacitance of the capacitor of the original question decreases by t.he same amount.

SOLUTIONS

109

S. 94. It will be proved that if the same amount of charge is carried onto F and p· then the electrostatic energy of the configuration belonging to Fis lower. Let us start from the new surface (F•). Imagine that the charges on it are 'fastened' to the s urface, then the surface is hammered in such a way that the dis1->lacement of the charges is pt>rpendicular to the original surface. The charges have then moved in the direction of the force acting on them, as the surface was originally equipoteJ1tiaJ, the field strength v.·as therefore perpendicular to the surface. Tims, tbe energy of the system decreases in the course of the deformation. (A force acts on_ th~ surface charges that pulls them outward, regardless of their sign, as a field pomtmg outward starts from the positive charges, v.·hile a field pointing inward starts from the negative ones.) The new surface will not be equipot.ential at all, but if the fuce :> can be repeated until surface F" is transformed into surfru:e F. T he electrostatic energy decreases all the time, while the total charge Q on the metal does not change. Since the electrostatic energy depends on the capacitance Q2 . C as C, the capaatance of the capacitor with surface F has to be greater than 2 that of the metal plate with surface F" .

S. 95. The energy of the plane capacitor can be expressed in of its capacitance C and of the voltage U of the battery connected to it: W ::::: ~ClP :::::

= Eo A21;, where the capacitance of the plane capacitor of surface A is C = Eo~-

When the distance between the plates is increased from d to 2d then the capacitance and the energy of the capacitor decrease to half of their original values. T~ _is a surprising result, as pulling apart the plates of the capacitor requires a po51t1ve work, and the energy of the capacitor still decreases. It can be noticed, however, U1at. the charge leaving the plates of the capacitor increases the energy of the battery (charges the battery) at a rate 6.Q U. Since 6.Q 6.CU, the increase of the energy of the battery can be seen to be exactly twice the decrease of the energy of the capac.itor. The increase of the energy of the battery is due half to the increase of the energy of the capacitor and half to the work done when pulling t.he plates apart. Note. The above statement. can be verified by direct calculations the work done when pulling the plat~ apart is determined. The force of attraction acting between the plates of a plane_c~tor can be calculated using the relation F = Q2 /(2coA}. Substituting the charge wntten m of the voltage and the capacity:

=

F

2 1 = C2coA lfl = e.oAU 2cf'l .

If the distance bet.ween the plates is dMoted by :z:, the work done can be calculated as 211

W

={ } .,

F(x}dx = EoA U2 {"" dx

2

}.,

x2

= EoAU2 _!_ = EoAU2 2

2d

4d

.


110

So the p05itive work done when pulling Ute plateS apart is equal to the decrease of the eu ergy of the capacitor, and these two quantities increase the energy of the battery together.

S. 96. Assume that current / 0 flows iu a superconducting, short-circuited coil of N turns, with cross-section A = R"J.tr and length ;to. Magnetic induct.ion is then B :::: 1J<JIN/x0 inside the coil and the total magnetic flux is 4> = BAN = = µ 0 J0 N 2 .4/xn. If the length of the coil changes for some reason, the magnetic flu.-x does not change, as that would induce a voltage, which wo uld induce au 'infinit.PJy' large current iu the coil of rero I"esistance.. Thus the current depends on the length x as J(x) = J0 x/:to Further, the self-inductance of a coil of length x i.'> L(x) = = 1J<JN'l A/x, and the magnetic energy of a coil with current I flowing through it. is 1 N2f5A \.V,,. =- L I 2 1 i o2 - • x. 2Xo 2

=

The energy of the coil is proportional to x: W,,.(x) - x. The r.oefficknt of proportionality F0 is the magnetic force of contraction (as the coil can be stretched to length x against a force F0 by a work W = Fo x ). The coil is in equilibrium when the magnetic force of contraction balances the elastic force F(x) k(xo - x) , i.e. when the change of the length of the spring is

=

SOLUTIONS

11 l

The magnitude of the force can be calculated from the equality of the e.lemcntary work and the change of energy (FD.x D.Wr,.p). and with some differential calculus, F = 6.W = _ (E:.r - l)lclQ2 6.x 2E:oA [(er - l)x + l]2 .

=

The magnitude of the force is the greatest at x = 0, as the denominator increases with x in the expression of the force. It is easy to show that when the situation x = l is reached, the magnitude of the fo1·ce decreases by a proportion E~ . Note. How does the electnc field exert a force parallel to the plates onto the ins ulatrng sheet? If the electric fielrl betw~n the plates were homogeneous and peTpendicular to the plat.cs and zero o utside (this is our usual picture of capacitors) then no force would act on the insula ting sheet. The pbe:nornenon can be explained by the bendiug of the electric field at the edge of the plates.

(b) The force acting on the insulating sheet cannot depend on the connect.ion of t he capacitor to its surroundings, therefore the previous result has to be valid if the voltage of the capacitor is kept constant. In this case, however, the charge of the capacitor changes a cco1·ding to the formula Q = CU as soon as the insulating sheet is placed between the plates. Substituting the charge of the capacitor depending on x into the above expression,

F = _ coA(cr - 1)2 U2 2ld

S. 97. (a) Let the distance between the plates of the plane cap acitor be d, the charge of the plates be Q, their area A, and examine tbe sitI- x uation when a piece of length x of thP insulating sheet is already between the plates of length l as shown in the Ji!f!tre. ThP capacitance of the pla ne capacitor can be calculated as the sum of two cr1pacil01~ c:onuect ed in parallel (one filled with a dielectric and the other one empty):

A

c

C

l - x e-oA = t;oE,.dA · 1x +i;o dA · -,=W [(c,. -

l)x+l) .

The capacitance of the plane capacitor can be seen to increase proportionally to x from E:oA/d w E r times the initial value. As the charge is constar1t in case (ti) , the energy ·w..a p of the capacitor can be given as Qi

Wr,,p

Q2ld l)x + lr

= 2c = 2E:0 A[(i;, -

It is clear that the energy of the capacitor decreases as x increases! This means that the work done while s.liding the insulator in is negative, i.e. the plate::; llua..<;i suck the insulator in, the insulating sheet experiences a force pulling ii. inwa rd.

is found, i.e. in this case the force acting on the insu.lating sheet is constant.

An interesting conclusion can be made if the energy of the capacitor is expressP.<1 at constant volt.age:

W

_ Clfl _ E:oA ((er - l)x + l] lP 2 2/d

....I' -

which shows that in such a case, the energy of the capacitor increases linearly with Thus , the change of energy D.WCJ>p corresponding to a small displacement D.x of the insulator can be calculated by elementary methods:

.i:.

6.WCAp E:oA(E:r - l)U2 ~= 2ld Apart from the sign, this formu.la is exactly the same as that obtained when calculating the force acting o n the insulating sheet. This result can be summarised the following way: when the sheet is slid in, the system does some work on it (it pulls the sheet in), while the energy of the capacitor increases (!) by the samt> amount. This is pos..cable because the energy of the battery decreases by a double amount in th.is process, as the capacity (therefore the charge) of the capacitor im:reases, therefore the uattery emits charges. The calculation of the work done Ly the battery is left to the Reader.


112

-

SOLUTIONS

113

~I··

S. 98. Number tJ1e resist.ors st.art.ing by the last element. of the chain backwards. As a current of 1 A flows lhrough the first resistor, a current of l A has lo flow through the second one as well, thus the p.d. is of I V on each rePrgun; I . sistor. As a consequence, the p.d. 2 V, i.e. the current flowiug through it is of of t.hr thircl re;istor is (I + 1} V 2 A. T hr. current ftowiug through the next resist.or is of (1 + 2) A = 3 A, then the c-111 rent of the fifth re>istor can be determined using the p.d. of (2 -1 :5} V = 5 V of the resistors with 1espective currents of 2 A and 3 A flowing through them. 11nd so 0 11 a .-; shown in Figure J. If I hl' rhain of resist.ors is considered a.'> buill uv 1>1.i:u tiug ft om the last. element I h1•11 the <:011srcutivc elements are connected in series, then in parallel, then agaiu in ser1c-s aud again in parallel lo the chain. The sum of the current. flowing through the J>rt'vious two resistors Hows through each resistor in series (according t.o Kirchhoff's l;,w for meeting-points}. The element connect.eel in parallel creates a new loop in 1lw d1c1m, therefore the p.d. on this resistor equals the sum of the p.d. 's of lhe .. ., p1 1•\·iolL'; 0 11es (according to Kirchhoff's law for loops}. Since the numerical •. 1liw 11f I.Ill' J>.d . a.nd the current is identical fOt" a 1 ohm resistor, the sum of the • " " ' 11i-. of tli~ two previous resistors is again identical to the current of the new t 1•!-t~ I OI ("l)llUC'Ctecl in parallel. Thus, Kirchhoff's laws or meeting points and of loops a11• fulfilled i11 such a way in this scrcalled ladder-circuit. that the currents flowing 1.hrough ea.ch resistor (and the p.d.'s) are identical to the sum or the mrrents (and p.cl. 's) of the two previous elements. Notice that. the numerical values of the currents (or p.d .'s) are just the elements of tl11• Fi honacci series: 1, ·1, 2, 3, 5, 8, 13, 21. The p.d. of the two resistors built iu la.stir-; (21 -I 13) v = 34 v. which is exactly the p.d. connected l(.I the input or thf' drr.uit. As a total of 21 A flows due to a p.d. of 34 V, the equivalent resistance of tl1t~ circuit. is or* n = 1.61905 n. If one more clement is connected to the chrun (in parallel) then the p.d. on it is 11lso 34 V with 110 change in the input p.d., U1erefore the total current increases to 0.618 18 0 . (21 + 34) A 55 A . In tbis case the equivalent resistance is of ~ 0 If the11 a further clement is connected w Lht: circuit, a current of 55 A Hows through 11 and the input p.d . increases to {34 + 55) V = 89 V. The resultant rPSistance of th u1ain is then ~ n 1.61818 n. If the ladder-circuit is built further and further, an 'infinite' chain is obtained. The equivalent 1esistance of this chain can be calculated using the fact that. adding two mo1e elements does not change the resistance any more. Thus, the wholP. chain cau be replaced by a single resistor of resistance R, to which t'-1.-0 resist.ors of l ohru each arc connected, in parallel and in series, and it. is expected that the ec1uivalent resistance of this circuit of three resistors is also R (Figure £). The condition for this is: 2 1A

8A

----

3A

IA

Figure 2.

=

=

=

=

R

= I fl + .....,.1- -1-, !fl+n

which yields the following second order equation:

R2

-

R - L =0.

The only positive solution of this equation is the resultant resistance of the 'infinite' chain:

R

= l + ../5 i::::: 1.61803 0 . 2

It. can be seen ho"· well the equivalent. resistance of a chain of eight to ten elements approximates the resistance of the 'infinite' chain, therefore the ladder cin:i1it can be considered as infinite after relatively few elements. Now. I . In practice, the zero wire o( electric networks in villages can be considered as ladder-circuits: the rero wires are fastened to poles and earthed at, say, every tenth pole Such a lat.Ider circuit consists of two types of resiswrs and the result.ant resistance of the 'infinite' chain can be calculated using the above method

2. It is unportant to notice that the above second order equation 15 the famous golden ratio equation whose solution yields the golden ratio: (1 + ../6)/2 1.618 03 .. ., which can be seen to be identical to the numeric value of the equivalent resistance of the infinite chain made of 1 0 resistors. F\irthennore, the ratio of the consecutive elements of the Fibonacci series can be seen to tend to the golden ratio, which it approaches surprisingly quickly. It is also easy to prove that dividing the even elements by the previous odd ones, the golden ratio is approximated Crom below, while dividing the odd elements by the previous t>Ven ones, the ratio is apprmcimatP.d from ahove.

=

Finally, for the sake of the aesthetic delight caused by multiple-level fractions, it is worth expressing with l 's the equivalent. resistance of the infinite chain composed by l ohm resistors. This time, the consideration of the elements in series and in parallel is not started at the end but at the beginning, i.e. at the left hand side or the chain:


114

R = l

=

1

1+ - - -- - - - 1 l+ - -- - - 1

l

I +-1 + ... If the . . . at the end of the formula is replaced by a 1 then the equivalent resistance of the eight element d1a in of t.J1e original problem is obt ained again.

S. 99. The grid of ohmic resistances is discussed first. The following reasoning can be applied. Ch1lSc the grid point into which the current I l\rill flow. For symmetry re.a.c.ons, currents of id1m tical maguitude I /4 will s tart from this point into the four possible directions as shown in Figure J.

-+---iFigure 2.

=

= R/2.

=

R.- = 2(c- I ) . This figure is ~ ohms for the dodecahedron and ! ohms for the infinite square grid. and it can similarly be calculated for any reguia} shape.

V4

Now consider a neighbour of t he previous point and take out current / from it independently of the previous reasoning. Again, identical currents of magnitude 1 /4 flow through the fou r identical resistors (Figure 2). Now take the s uperposition of these two ca.c;es and consider them together. Scalar quantities (current, potential} simply add up due to the linearity of the circuit equations. A volt.age (p.d.) U appears between the two neighbouring points, current / goes iuto one of the points and current I flows out. of the other one. The currents flowing in the edges of the grid cannot. be easily determined , but the current flowing between the two neighbouring points can. The two currents discussed above Figure S. add up in this edge, i.e. a total current / /2 flows through it (Figure 3). But if cu rrent I /2 flows through a resistance R then a p.d. of U R · I /2 appears. The equivalent resistance between the two neighbouring points is therefore

R r = U/I

S. l~O. D_e note the number of gTid points by c and the number of edges starting from a gnd point by u . (E.g. in the case of a dodecahedron, c = 20 and u = 3). If a current of 1 amp is let in at one grid point and l/(c - 1) ampers are taken out at all other grid points then the current flowing in the edges sLarting from the edge where the current is let in is of I = 1/11 ampers due to the symmetry. Superimpose a current dist1ibution caused by a current of I amp flowing out at a point neighbouring the previous one and currents of I/(c - 1) ampers flowing into ~J the other points with the pre vious current distribution. A p.d. of U = 2/u volts is ca used by a cun-ent of I = 2/1i amps flowing through the 'common' resis tor and since the current flowing through the whole circuit is of / I + - '- == c ' amps, the resultant resistance is c- i c:=r

flC

~--1-11-4 -1-------------1-

Figure I.

115

The resultant values for the infinite grid of capacitors and self-i11ducting coils can be calculated similarly: C.- == 2C and c,.. L/2.

+ - - - - - -- - -- 1 1 + ---------

I+

SOLUTIONS

S. 101. The ' trick' of superposition is again applied by separately discussing the cases of the currents flowing in and (I out, and then the mea.o;urement described in the problem A B can be considered as the superposition of these tv;o cases. a Let A denote the corner of the square dra"t11.T1 onto the plane a where the current. Io flows in and B the neighbouring comer a o-- --- c wliere this current / 0 flows out, while the p.d . U is measured between the two other comers (C and D} as shown in the figure. lf first the current l o is let in through point .4 (imagining t he points infinite lv far away _to be_ at zero potential) then the current is distributed (hemi)sphcricall~ ~yrnmetncally 111 the ~alf-plane, i.e. a t distance r from point A, the current density J(r-} (the current Howmg through a unit s urface) is j(r) = Io/2Trr2 . This relation i:; not valid for very small distauces r (the current density does not tend to infinity}, as r approaches the size of the electrode in point A. The local version of Ohm's law is the clifferential law that expresses the curre11t density j in tenns of the electric field strength E and the local resistivity l!: j(r) = E(r)/f!. The magnitude of the e lectric tield sti-ength in the half-plane can be determined using this relation: .

E(r)

= -2loo

?.

7TT-

The potential function (and its changing, the p.d.) can usually be de termined from the field strength by means of integration, in this case, however, it can be


116

obtained using a simple analogy. It is useful t.o that the electric field of a point cl1arge Q is inversely proportional to r 2 , its potential is inversely proportional to r and the proportionality coefficient is t he same in both cases (E kQ/r2 and U = kQ/r, respectively). This means that the following potential field corresponds to the e lect ric field strength determined above: U = loe/(2rrr). The nearer a point to the electrode where the current flows in, the higher its potential. The potential Uv /oQ/{2rra) of point D is therefore higher than Uc 0 p/(2./2rra) . The p.d. between the tv.-o points is l oe(2 - ./2)/(4rrn). In the second step, the situation when current / 0 flows out through point B is discussed . Everything is identical t.o the previous case, except that the signs of the c111antities (current, current density, field strength, potential} is opposite. T bp poteutial in the examined half-space is described by the function U - lnl!/('J.rrr'), whern 1·1 is the distance fro m point B. The potential of point C is lower than that of point D, i.e. point D is again more posit.ive that D. Th~ p.d . betw1'<'11 Uw t wo points is the same as previously. U the above two cases are superimposed then the original situatio n of t he problem is realised and the p.d. between points C and D is exactly twice the above two identical p.d .'s, i.e. U0 0 e(2 - ./2)/{27ra). Apart from the given data . this C'.xprcssion only depends on the resistivity e in question, therefore thf• solution of the prob lem is (2 + ./2)1TaUo e= . lo

=

=

=/

=

=/

117

SOLUTIONS

points denoted by 0 and X in the figure are equipotential, i.e. they can be ed. When the equipotential points are ed, the circuit can be redrawn as shown in Fi!JUre 2, then the equivalent resistance can even be calculated mentally. If all the resistors are of 1 n then the equivalent resistance is of n. A o ne-dimensional 'cube' is simply a straight section, its resul tant resistance is itself, i.e. 1 n. The two-dimensional 'cube' is the square. T'\\-o resistors start from one of the e.nds of a diagonal and two resistors nm into the opposite end of the diagonal. If current flows through the square, the f>lldpoints of the other diagonal can be seen to be equipotential, t herefore two times two resistors connected in parallel are connected in series. In the case of the square, the equiv-cl.lent resist.1'nce is again 1 n. In the case of the three-dimensional cu be, three resistors st.art from each endpoint of the diagonal, the remaining six resistors equipotential surfaces. A four-dimensional 'cube' can be obtained by a parallel displacem ent of a cuhe in the direction of the fourth dimension and by ing the corresponding points. T he four-dimensional 'cube' has therefore 12+ 12 + 8 32 edges (12 for each normal cube plus eight to them). A distorted projection of such a 'cube' can be made for experimental purposes as shown in Figure S, replacing the disp lacement by a magnification. Figure S also shows one diagonal from the eight possible ones.

i

=

8

Note. T his method is widely applied in practice, e.g. to determine Lh.- average resb-iivity nf rocks. T he measurements are naturally not canied out on infimt<' half-spaces but on rnlumes and planes with all linear sizes much larger t han the side a of tlie square, so U1at n is negligible compared to them.

8

A

Figure 9.

A

Figurel.

Figure 2.

S. 102. Let current I flow into the cube at one point, Rowing out at the poim. diagonally opposite to this one, as shov.-n in Figure 1. Symmetry prescribes tha t the C\lrrent flowing through the three resistors that belong to each endpoint of the diagonal is I / 3 , therefore the voltages are identical on these resistors. Thus , the

Fi!}ure. 4.

In the case of the four-dimensional 'cub e', four edges start from a point., t.he endpoints of U1ese are equipotential, and the same is true at the other end of the diago nal. T he shortest way from one of these equipotential surfaces to the other goes through two resistors of the remaining 24. This means that the 'inner' points of the 24 resistors are also equipotential, i.e. 2 times 12 r esistors are connected i11 parallel, then these are connected in series. The redrawn circuit is shown in Figure ./, the equivalent resistance is ~ !l. Note.. The problem can be generalised for n-dimensional 'cubes' if the circuit can be decomposed into identical resistors connected in parallel and in series after connecting the


118

cq1lipctential points. In general, the equivalent resistance of an n-dimcn.sional 'cube' made of l resistors across a diagonal is

n

R., =

!

I 19

rod makes a uniform motion after reaching a maximum speed. The maximum speed can be calc\llated from the equation of motion with the condition a = O:

in in in in 1(~) + 2{;) + 3{;) + ... + n{:)"

VITllU

S. J 03. Let I denote the current flowing in the wire, A the cross-section of 1.he wire and p 1 and fl'~ the resistivites of the occurring materials. Ohm's law for a wire of length l reads as U I el/A , which yields E U/ l ul /A for the electric field strength in the wire. The resistivity of copper is lower than that of iron, therefore the electric field strength Co :_ =: Fe has to be smaller in the copper than in the iron. According to Gauss' law, the difference in the electric field strengths implies an accumulation of the charges at. the ho11ndary of the two metals (see the figure). The sum of the accumulated charges is

I;

SOLUTIONS

=

Q

=

= eoEA = Eo l (tJPe -

It is interesting that this quantity only depends on the current and on material constan1 s, but not on the cross-section of the wire. Substituting the kno~-n data, the charge is found to be Q i:::i 5 · 10- 21 C, which is only 1/3-0 of the elementary charge! Though a practically realisable, measurable macroscopic current flows through the wire, the aa:umulated charge is only a small portion of the microscopic elementary charge.. This strange result shows that da'>Sical electrodynamics (imagining charge carriers as small balls) cannot always correctly describe electric phenomena. Only the application of the more sophisticat.P.
mgRsino:

82 (l

b) lf the rails are closed through a capacitor of capacitance C then the equation of motion is the same bl•t the relation between the induced voltage and the c urrent is different. The charge of the capacitor is determined by the induced volt.age:

=

i?Cu)-

=

Q =CU= CBlv.

Not.ice that the current flowing through the rod equals the derivative of the charge of the i:apacitor with respect to time:

6.Q 6.t

I =-

l::i.v =Ct::i.U - = CBl - = CBla, 6.t 6.t

i.t'. t l1e nu rent Aowing in the rod is directly proportional to the accelention of the> rncl. If the above expression for the current is substituted into the equation of motiuu, t hf' rod is found to move with a constant acceleration on the slanting rails.

a;

mg sin a . m+B2 l 2 C

Induction decreases the acceJeration caused by gravitation. The speed of the rod and the d1arge of the capacitor are directly proportional to the elapsed time. c) If the rails are closed through a coil of self-inductance L, the relation between the induced volt.age and the current is

6./

L 6.t

tu

= Blv = Bl 6.t .

(It has been ~ t.hal the speed of the rod is the derivative of the displacement with respect to time.) Notice that I = 0 and x 0 at the beginning of the motion, therefore the above formula implies that the current itself is proportional to the coordinate x.

=

LI = Blx. ma= mgsino: - Bl/.

Sub!;tituting the c urrent from this relation into the equation of motion,

T his equation is identical in the three cases, the difference comes from the different relation between the induced voltage and the current flowing in the rod. a) H the circuit is clOOt!tl through an ohmic resistor R then the currP.nt. I and the induced voltage U Blv are connected by the relation I U/ R Blv/ R. This shows that the braking force increases proportionally with the speed, i.e. the

=

=

=

1na

=mg sin o -

B2l2 --x L

is found . The force acting on the rod is the sum of a constant term and a term proportional to the displacement and opposite to it. This is the equation of motion

123 TRJCKY PHYSICS PROBLEMS into a spring and released from the unstretched position of the rod makes a harmonic oscillation about its equilibrium position

SOLUTIONS coefficient of proportionality can be regarded as the efficiency '1 o: (considered as an electromagnetic gun), which can be transforme

sino

xo :: mgL 8 112 • the c&illation is A x(t)

( rm

= xn t.he dependence of the displacement of

= A(l -

coswt) ,

B l../C

+

mvc) ../m

2 •

The product. of the two t.erms of the sum in the denominator is 1, the inequality between the a1;tbmetic aud geometric means) the This means that the efficiency of the elec.1romagnetic gun carm

25%. U the instant when the capacitor is switched over, a current I = in the rod, which causes a force F = B l/, i.e. the initial 1P ro
Bl

_ Qmin

tlmlUC -

c .

motion of the rod is 6.v ma = 1n6.t

6.Q = Bil= Bl ""A""", ut

ration and the cWTent have been expressed as the change of the r:harge, respectively. The rate of change of the speed_ is pr~p~rtion~ uige of the charge of the capacitor and this proport1onnhty is valid llges as well. The speed of the rod increases from rero to vm_"", the pacitor decreases from Q 0 = CUo to Qmin• therefore equation (2) ~a.<\

mvmax

= Bl(Qo -

=

2 2

Nou. If the condition m CB l of the maximum efficiency expression of the fiual charge of the capacitor , Qmin = UoC/2 is fow initial charge of the capacitor ill left. Thus, only one quarter of t be enei is left, one quarter or it is transformed into the ldnetic energy of the r or the initial energy warms t he rod or resistance R in the fonn of J ou)I

S. 106. When the sv.;tch is close cmTents shown in Fi circuit. (The value of the current I is d1•t('r111ined by the voltage c the resistances of the lamps.) A ver}· short time after turning tl cunent flowing in the coil is practically unchanged. (If this were 1 would correspond to a very rapid change of the magnetic flux, wl a very high voltage in the coil.) Currents of 21 and I flow therefo these determine the currents flowing through the lamps (Figure J

- -21

s

I

21

(~t# JOO

Figure 1.

Qmin) ·

and the remaining charge of the capacitor can be calculated using d (2):

Figur

This means that the lamp closest to the swi tch suddenly flash of the two other lamps does not change. (This is only true fo~ a on all three lamps fade and go out.)

BlCUo

Vm..,.

= m + B2Z2C' B 2 l 2 C2Uo

Qmin

= m+B2 l2C"

e relation shows that the maximum speed of the rod is proportional tage Uo of the capacitor . Thus, the final kinetic ener gy of the rod is JJ (for given C and m), i.e. to the initial energy of the system. The

S. 107. Write dO"WD the forces keeping the free electron (inside the coil): eE ± erwB mrw2 ,

01

=

where e is the charge of the electron, m is its mass, r is its dista of rotation and E is the electrostatic field strength in the metal c: the charge distribution inside the cylinder. The sign ± shows t:


122

Loretitz force can point either inward or outward, depending on t.he direction of the rotation of tbe cylinder. Express the electric field strength from the equations of motion of the electron: 2

E = ( " : ' ± wB) · r

SOLUTIONS

123

and the induced voltage is U = -

~

U I= -

R

= K · r.

= r 2 7r8w sin wt . The curTent

r21' 8W sin wt = --R

flowing in the ring causes a magnetic induction of magnit ude The fie ld su eugth can be seen to be dirE'CUy proportional to the radius. (I< is a constant. depending oa the anguJar velocity, the magn etic field strength and the da ta of the electron.} K nowing t he electric field s trength and using Gauss' . . 61 . R :----= ----: ~ law, the quantity or electric charge in unit volume, i.e. the electric charge d~msity of the cylinder can be determined. ~Consider the thin cylindrical ring shown in the figure and denote the electric charge density at distance r from the E: axis by p{r}. An electric flux with a magnitudt: of E(r) · ~ 27rr L enters the ring and a flux of E(r+llr}· 27r (r+llr) L leaves it.. According to Ga~· law, :...... . ..... . ..... ... .. . K(r+ llr ) ·27T(r +llr) L - Kr ·21rr L = (l /£o}P2T1' llr L,

81

P

= 2£(JW7n (w± eB) e

sin wt

in the centre of the ring. The direction of the magnetic induc tion B 1 is perpendicular to the plane of the ring and rotates together v.ith it. Decompose the vector B 1 into a component. parallel to B and a component perpendicuJar to B . The paraJJel componPnt is (l/2}sin2wt Lhe time average of which i:l zero Oo v1 oportional to coswt ·sin wt t.he other hand, the perpendicular component can be v.ntten as

=

,

which yields

nrBw I -= µo = µo2R 2r

BJ..

nrBw

r7rBw .

= "'°2R sm2 wt= µu~ (1 -

cos2wt} .

This expression consists of a term varying (relativeJy rapidJy) wit h Lime, th•· 11111r average of which is zero, and a oonstant term that deviates the magnl'tic n....•Jlr 2° from its original (ncrth-south} direction. Since the needle takcs 11µ the by a direction of the (average} result~t field,

=

m

for the charge deni:;ity of the cylinder, whid1 is a constant value independent of r. The charge density inside the cylinder can be positive, negative or zero - depending on t he direction and magnitude of t he magnetic field and on the angular velocity. This la tter case can occur if w = lel 8 /m. Then positive and negative · charges are not separated inside the cylinder, as the Lorentz force can provide a sufficient centripetal force in itself for the circular motion. Nol e. 111 order to get an idea of the or ders involved, l\SSUDle that the induction of the 3 · 10-s T . magnetic field is like the magnetic field on Earth , near the Equator, i.e. 8 1 6 l.ero rJtarge density corre.-ponds theu i.o angular velocity w = eB/m = 53 · 10 11- , ,vhich means morf' than 50 million (!) revolutions per minute. Such a q niclc rotation cannot. br realised m practice, !'iuce no material could beM it T he concll1Sl011 of this problem is Lhal the mllSS o r the e lcctrou is very small compared to iLc; r.harge. T hus, the resuJtant of the electric and magnetic forC£S has to be z.cro {to a. good approximation) in macroscopic cases even when the electron is accelerating_

=

S. 108. Derompose the magnetic field of the Earth into horizontal and vertical components. The vertical component induces no current in the ring, since its Hux is zero all the time. Let B denote the horizonta l component of the magnetic field of the Earth and w the angular velocity of the ring. The magnetic Oux of the ring is

T1'W B..t = 1Jo-4R B

tan a= -

holds. (The magnetic needle makes small oscillations about the above position, the a m plitude of which is determined by the mechanical and magnetic characteristics of the magnetic needle and by the dumping force.) It is interesting that the angle of deviation of t he magnetic needJe does not depend on the magnitude of I.he magnetic field of the Earth. T he onJy impo1ta11t thing is that the horiwntal component of B is non-zero. The resistance of the ring can be calculated from the above formula and is found to be of 1.78 . 10- 4 n.

S. 109. Consider the two discs (of radius R

= 2~

=

each} shown in Figure 1, slightly touching each other. H this assembly is placed into a homogeneous magnetic field changing u11iformly with time and perpendi01Jar to the plane of the paper, the voltage induced in the piece of wire conducted a long its edge is U

=

D.B 2R w • flt . 2

Figun: / .


124

Twist the disc on the right by 180° about the symmetry axis e (Figure 2a) , then its top (say dark) side becomes the bottom side (Figure 2b). Turn now the same disc by (almost exactly) 180° about axis t (Figure 2c ). Thus, the dark sides of both discs are at the top and the limiting curve is exactly the limiting line of the Moebius band mentioned in the pmhlem

SOLUTIONS

125

B, B' and I tan easily he determined from the above three equations. The two signs occurring in equation (1) have been taken into by allowing both the p
B' -

2RBo

- 2R - µonr2w'

1

Fsgure 2.

2

L = 2R 2 7r • -t:..B = k -27f t:..t

= i:.f

in a umfonn ly changing magnetic field. This value is much highei- t.hau what one woold naively expect by reason of the area of the paper band. The area of the (one-sided) surface covering t he Moebius band is not iclentical to the area of tlte paper ba nd , it is much larger for narrow bands! T he induced voltage can also be calculated after cuUing the wire into two 'coils' of one turn each at the 'twist' , and adding up the algebraic values of the voltages {according to the direction of the turns ) induced in each turn. In the present case, the direction of the t~'O turns is identical, therefore the voltage 1/0 kR2 7r of onP. .turn is doubled : U 2U0 •

=

=

B' =Bo±B .

T he current flowing through the coil is determined by the induced voltage U, and the 1 esi:;tam;e R.: (2)

1

= U; = B' r2 w ..!._ R

2 R"

(The ind uced voltage has been calculated using the average Lorentz force qB'rw/2.) T he magnetic induct ion of the coil itself is (3)

B = µon/ .

the current flowing in the coil tends to B 0 µon {this is the current that can eliminate the magnetic field

w ···· · -~- -~· · ··--··· 1· ·7

S. 110. The resultant magnetic induction B' is the sum of the magnetic inductions Bo and B of the Earth and of the coil, respectively: ( 1)

Bor2w . 2R- µonr2 w

When the disc is at rest, the current is obviously zero and the resultaut magnetic induction inside the coil equals the magnetic induction Bo of the Earth. B• When the direction of the rotation is such that the field of the coil decreases the external magnetic field (w < 0) then the resultant magnetic induction decreases monotonically as the (negative) angular frequency of the disc increases, and tends to zero as =-----~-_..:w_. .ru;o ;_·-=-w w becomes very large. At such a very fast rotation,

Thus, the voltmeter reads

U

=

µon

Figure 1.

j

of the Earth). In the case of the opposite direction of the rotation (w > 0), the current flowing in the coil increases the resultant magnetic induction, thus a higher voltage is induced and a larger current flows, which leads to a further increase in the magnetic induction {positive ). At a certain 'criti2 cal' angular frequency, i.e. when Wcrll = R2, µonr

both the magnetic induction and the CWTent want to rea.c'h infinitely large values (Figure 1). Such a state can obviously not be realised. H the angu-

lar frequency is increased, the etµrent and the heat given out by the coil can be seen to increase until the wires bum away! The strange behaviour of the syst.em can more easily be understood if the relation between the current of the coil and the resultant magnetic induction is graphically tepresent.ed (Figure 2). According to equation (2), I is proportional to 1 8 in such a way that the proportionality coefficient depends on w. This corresponds graphically to a straight line aossing the origin, whose slope is proportional to w . Jn the meantime, (1) and (3) prescribe that B' =Bo+ µon · I, which is also a linear relation whose graph is a straight line that does not cross the origin; the slope of this line is of ~n (independent of w). The intersection of these two straight

123 TRICKY P HYSICS PROBLEMS

126

lines determines the actually realised current and the resultant magnetic induction. Hw Wen• then the slopes of the two straight lines are identical and the set of equations has no sol ution: the state corresponding to the critical angular frequency cannot occur. (Considering real data, the critical angular frequency is such a high value that it cannot even be approached.) The Joule heat given out by the coil has to be covered by the mechanical work clone while rotating the disc. ll> < O The electric power is P.,1 = PR, while the mechanical work done per second is the product of the torque and the angular freq uency: P,.,.,rn Mw = B' Jr2 w/2. (The torque is given by the product of the force B' Ir and the .,,·erage perpendicular distance of the line of action of this force from the axis (r /2).) Using the relation bet-ween B ' Figure :?. and l , Pei = Pmech can be directly obtained. The strange device described in this problem is called a unipolar dynamo.

=

=

=

S. 111. Tbe resultant magnetic flux of the ring is composed by the effects of

thP external magnetic field and of self-induction:

~

= B: · ~7T +LI.

The change of the magnetic Hux induces a current in the ring:

SOLUTIONS

127

The Lorentz force can be seen to be approximately direct1y proportional to the vertical displacement of the riug, the coefficient of proportionality can be calculated from the given data. (The approximation is only valid for small displacements. since the magnetic induc tion does not change accord ing to t he given formulae in the case of large displacements.) The equation of motion of the ring is TIU!:

= F~ -

mg

= - kz -

mg.

This me.ans that the ring makes hannonicoscillations about the equilibrium position Zo = - mg/k: z (t) - Zo = A coswt, where w

= JkFii. Considering the initial conditions, A = -z

0,

z (t)

y

= w2 (cos wt -

i.e.

1) .

The coordinate of the vertical position z never takes up positive values, thercfo1e lhe Lorentz force always points upwards and it is zero a t the topmost point of the oscillation. The current flowing in the ring always follows the same direction. Substituting the numerical data: w 31.2 s- 1 and A 1 cm are found. The time dependence of the current flowing in the ring can easily be expressed in of z(t): 1 2 1 2 l = zBo cu·0 7T · z (t) = zBo or0 ?TA(coswt -1).

=

=

The maximum value of the current can be obtained at the bottom standstiJJ: l m:u = 39 A.

6~

=

R1 =Lit' which has to be zero, since the ohmic resistance of a superconducting ring is zero. Acc:ordingly, the magnetic flux of the ring is constant:

~

= 8 0 (1 -

oz) r~7T + Ll

=

constant.

=

Taking I.he initial conditions (z 0, I = O) into considerat ion, the value of the c.-onstant is 1> = Bo~7T. T he currt!.llt in the ring can be determined using the above two equations:

l

= -L1 · Boa~?T · z.

The magnetic Lorentz force acting on t he ring (which can onl y be vertical, ~ue to the symmetry of the assembly) can be expressed as F=

= - B .. · / (z) · 2ro7T = - B 0 f3ro · L1 · Boor02

7r •

2r o7T · z

= - k · z.

S. 112. Imagine that you sit in a frame of reference moving horizontally at cons tant speed t1 at right-angles with the induction lines, and you take a charge q with you . From a frame of reference fixed to the ground, the charge moving at speed v can be seen to experience a force of magnitude qvB (p0il1ting upwards or downwards, depending on the direction of the motion). In the moving coordinate system, the charge is at rest, the Lorentz force does no t act on it., but it still has to ' feel' the force. (A force caused by some interaction cannot depend on the coordinate system in which it is described.) The seeming contradiction can be resolved by presenting the transformation of electric and magnetic fields when chang1ng from a standing coordinate system to a moving one. In the present case, there is only a magnetic field in the standing frame of refe1·en and no electric field. In a system that moves {at a speed much lower than t he speed of light), this same magrietic field can be observed, but an electric field of field strength E v B appears besides it to pr ovide the force F = qE = qvB that would otherwise be missing.

=

123 TRICKY PHYSI S PROBLEMS

128 Let th speed of th moving frame of r ~ rence be such that the electric force described abov compensates for the gravitational force mentioned in th problem:

F

q

.

·-

... - .

.··

..

~

.

. ...

u

= qE =qvB =mg.

Oescril> th motion of th observ d body in this coordinat system. Sine the effects of the Jp.ctric and gravitational fields cancel each other and the body moves at sp d -v in this system, a for of --qvB ac on the body (th magnetic Lorentz force), making it move ou a circular track I.urning downwards: qvB m1i2 fr, wh re r is the radius of tht> circl : r mv/qB. sing the values of the speed and of th radius, th tim in which th body mak on turn is found to T = 21rfTl/qD, regardless of l11t! speed . ln the system fixed to the ground , a uniform rect.ilinear motion is add d to this uniform circular motion, th refore the particl follows a cycloid path as shown il1 the figure, goes down to 2r = 2mv /qB and then rises again and reaches its initial height in tim T, during which its horiwntal displacement is Tv 2nr. Here, the particle stops for a mom nt, tb n SLarts a n w cycloid curv .

=

= Q(E + v x B}

Q moving at v locity v in a field of electric fi d rength E and magn ti indu tion B (th Lorentz force). U on auges to anoth r frame of r i rence moving at I city vo with respect to th original one, th locity of the particle is seen to be v' = v - v 0 . lf all physical quantities are denoted by a prime in the n w frame of reference then th expression for the Lorentz force i

a.c;-ts on a small body of charge

F'

= Q'(E' + v' x B').

mpare th two descriptions. The force acting on th particle (whi h can be measur d .g. via its nccelP.ration) does not. change: F' F . The a.me is tru for the electric ·barge: Q = Q' . (If the charge of particles d pended on their velocity then an initially neutral body could show a n t electric harge when heated, no trace of whi h c:an be in nature!) Using th transformation formula for the velocities yield

=

Q(E + v x B)

= Q(E' -

v0 x B

+v

x B).

This relation has to b fulfilled for any v , sp iflcally for standing particles, there ore

E'

=E -

Si~ into the fram of reference moving together with the liquid. An tri field ~1th . a fi~d strength - vo x B appears in this system. The standing liquid is polarised m thlS field, therefore the magnitude of the electri field strength decreases the original frame of reference (at by l/~ r : E' = - vo x B · (l/Er)- Returning rest with respect to th plane capacitor), the electric field is

E

= E' + v 0 x B = v 0 x B · (I, - ..!..) , Er

while a p.d . of

=

=

S. 113. F1 rce

129

SOLUTIONS

vo x B

and

B'=B

found . It can be seen that ctric and magn tic fields are not separated physical quantities, their values depend on the hoice of the (velocity of the) frame of ref rence from which th phenomenon is d cribed.

appears between th pla . I~ the case of noi;polarisable materials (Er= 1), this voltage is obviously zero, while 1n th case of very w II' pola.-isabl 111.at~rial (metals}, U ....:. VoBd. This latter, called the ~all ~ffect, can b interpreted the following way. The charges of the conductor moving in th magnetic field are displaced by the Lorentz for . The migration of charges lasts as long as an eJectri field trong enough to balan the Lorentz force proportional to the magnetic induction is built up. From the condition QE = QvaB, the above determined relation U = Ed= VoBd is inck"'ed obtained.

S. 114. The volume of the uranium nucleus is the same as th sum of the volumes of the fission products, but th total surface of the fission products is greater than that of the uranium nucleus. Thi means a temporary loss of energy (this is the en~rgy that_has to b 'credited' in th course of the activation of th nuclei), but this energy ts well r funded by the d as in Coulomb energy when th nuclei move a~ay_ from each other. H uranium split into three parts, the energy needed at the begmmng of the process would be much higher than for two fission products therefore in practice this does not occur (or only with negligible probability) . ' S. 115. When atoms are heated to a fi w thousand degrees, thermal energy is naturally much smaller than th binding energy per nucleon. This is why th speed of nucl ar reactions · u ually inde.pen
i

and

mv p

= ,/I -

v2 / c2

t) mass

LUTIONS

130

(c cl notes the sp<..00 of light). It is useful to express th energy in of th mom ntum : E = Jpici + m2c4. urse of th collision of high energy protons, n w particl , e.g. protonln th a.ntiproton pairs can com izito being. (The rest mass of the proton and of the antiproton is the sam .) Th total nergy of the produced partid is th lowest possible if they stay together, moving at a 11 gligibl speed. ln this ca c, the fi ur particles (three pr tons aud an antiproton) can be considered as on par ti cl of m s 4m, whos mom ntum equals th initial mom ntum of th acc.-elerat.ed prot n. The law of con vation of energy prescribes that

After squaring the quation, p = mc"'48 and Ep = 7mc2 can be found . Th proton 2 6mc2 ~ 6 G V in th our . of tbP hav to gain a kin tic energy of Ep - mc 9 accel rating vultag . ac I ration, which is possibl by means of a 6 · 10

=

S. 117. EI rons can move freely in th wall of the Faraday cage to rea ·h th state wh n the resultant of the el ctric aud gravitational forces adi11g 11 t h m is zero in ach point. This is satisfied by a ·light condensation of th I trous a1 th bottom of the m tal, cau ·ng a surplus of positive cbarg (a lack of l'kn rim') at th top. 111 charg displacement goes on until th magnitude f Lb<.' 11'Sull "''' electri fi d strength is E:o = mg/lel (m is the mass of an electron and 1· '" ii:. charge). Th resultant field strength points vertically downwards, since thi i.o; h \\" tric field can exert a for of magnitude mg, pointing vertically upwards, the on the el ron having an gative charge of - lel. has to As a homog neous electric field is form d in the wall f the cage, th be th sam homogeneous eJectri field in id the cage as w U. (H this wer not th . cru , the motion of a charged particle vertically upwards or downwai ds in the wall of th cag and then back to its initial positi n through th inner part of th ca P would cau a non-?.ero resultant work of th electri field, and a perp tuum mohil could be built .) Thus, du to th charge clisplacement in the wall, ther i an el tr~tatic field the force f gravitation fi r a fr elec:tr n cau.ing an I ctri fi rce that equali placed insid the cage. ln other words, the cage shield the gravitational field as well! This shielding, how ver, only works with electrons, the condition of uali ation of the for es i not fulfilled for particles with a different pecific charg (e/m) . Cn th 2mg, i. . th P cas of positrons, fi r example, the resultant force is E:olel +mg 2g. of acceleration an wiLh downward vertically pa.rticl start moving Num rically, Eo = 5.6 · 10- 11 V /m, which is a very small valu , th efore th n in rdinary circwnstauces. Electrons abo described phenomeii n cannot be cannot be placed into the Faraday cag with no initial peed, ince ven in th of the photo I tri · efft:ct, the electrons pushed out of the m ta! with an energy f only 0.1 eV travel at a speed of 200 km/s. Th relative change in th speed after 1 metre of fr faJI is only about 10- 10 , which is imm asurably small.

=

131

Note. So far, we have not con "dered the ffect of the displacement of the charges iu t.h wi\11 due lo lhe J 1-esen of the )XJ6itron iu the ~e. It is known that this effi t. can b handled by means of the method of image charges. A charge near th wall at dist.an rl from ii , exp 'en es a force aused by the p larised charges in the metal th~t is th Mnw aw d by an imag charge - at. distance 2d from the first charge: as th fo -

F1

(-e)e

"""6• - k (2d):.I .

their w 1ght 1; lo the case 0 £ lectrons or positrons, thi
using measur mcnt data, b houJd build such a larg Faraday cage that th parttcl~ , ..in further away from the wall than a few metres, or else it is no the shielding rff<'c-t r.r the cag that he could examine but the attra tion of U1e image charges.

of th positron and that of the prot n, · _118. L t m and M denote them mpare th 2000m). and I t denote the I mentary charge. respectively (~ nergy of th ~t1al state to the situation when the positrons have moved 'far away ' and a.re trav lhng at speed v1, while th protons have practi ally not moved a t of the motion du w thr all. (Such a situation can really be found u1 th cou difference or order in the masses of the particl .) The prin "pl or co I rvatiou or energy prescribes that

::i

1 )

2 . rr; v2a

2

1 1 e + 2 -mv2 -= -2 47T£o ../2a 1

'

=

350 m/s. In the remaining time. thus, substituting th_e num rical data yields v 1 the proton , 11 th ot.h r significantly; chan~ l n does positrons the of the speed rated to speed vi. Th th orem f r.nuscrh~d, tart slowly to move and are a vat1on of ner can again b applied:

c2

I

I

- - - - =2 - Mv22 ' 2 47T£o ,/'ia whi h yields th valu v.i

R::

2. 7 m/ .

. 119. The en rgy of a photon with frequ n Y f is hf, while its mom ntum is hf /c. (h denotes Planck's constant and c is th peed f light.) Th en rgy of th pho n decreases to hf' after th collision, and its momentum is of magnitud hf' /c and p rpendicular to th original momentum. Acrding to th notations of the figure, th

dcetmo

;/o

~. _1___..._IOl_,_OO_

"' - - -

0

l

P.

~•>too

/'


132

=

hf /c and P11 <.'Omponents of the momentum oC the electron are Pz The principle or conservation of energy then prescribes that hf+ Eo = hr +

= h/' /c).

SOLUTIONS

13.1

~ (h is l/(27r) times the Planck constant). In such circumstance;, the t::.p electron has a potential energy of

=6

J~ + w+~>Cl.

(The l'elativislic formula for the energy, W = Eg + -p2c1 has been used.) The abcwe relations yidcl the change in the wave-length of the photon is therefore

and a kinetic energy of EkineLK-

= -h ~ 2.4 · io- l:l m - -jC = -he Eo m e /'

Cl>. = >. - >. = -C I

Note. Thas iulcrest.iJag quantity with a dimension of lcnglb - only depending o n the nrn.ss " ' ~hr. Plrctron, lbc speed of lir;ht and Planck's constant, i.e. on natural constants 1s called Lhe Compton wavelength of the electron.

S. 120. Applying the energy formula for spherical capacitors, 1 e2 ---<me 47r£o 2r

e2 1 is found for the ' classical electron' of charge e, which yields r > 4 71'£o 2m.c2 :: 1.4 . 10-rs m. (This is the quantity called classical electron radius.) If the electron is considered as a rotating sphere of the size of the classical C'lectron radius, with a homogeneous mass distribution and with angular frequency w , aud if it!. angular momentum is assumed to ec1ual the mechanical angular m<>2 rnt>nl111n of this rotating rigid body then ~ · mr w 4':, can be written. This gives this speed is many times higher Since speed'. 'equatorial the rw:::::: 350c for than the limiting speed of relativistic physics, the speed of light, the original version uf the classical electron model - worked out by Lorentz and Abraham before the birth of quantum theory, in the early IOOO's - had to be abandoned. A modified version suggested that the electron could be a sphere with radius r, i:harged on iLs surface, which would not rotate but would contain a magnetic dipol<' inside (Experiments confirm that electrons do have a magnetic momentum i\S well.) 011 tside the sphere, the vector Ex B produced by the electric and magnetic fi<'lds together (multiplied by a suitable proportionality coefficient) would describe a current of electromagnetic momentum. This current or momentum has an angular momentum as well, and a suitable choice of the parameters can make this angular momentum equal the measured one.

''"'·"' =

=

S . 121. Assume that the electron occupies a horizontal layer of thickness H above the bottom side of the box. Its vertical coordinate is then known with accuracy H, therefore the w1certainty in its vertical momentum has to be at least

6-p2 = -2m

Ti2 i:::: - 2

2mH

'

thus its total energy is

E(H)

I

= Ep,,,_tiAl(H} + Elci...,.uc(H} = A · H + B · H2 .

where A and B are constants that can be read from the above equations. If H is small the potential energy is low but the kinetic energy is high. H, on the other hand, H is large the low kinetic energy goes together with a high potential energy. Total energy will be minimum if Es--tW and Eirinetic are of the same oroer of magnitude. It can be seen with the help of differential calculus that the optimum L'< 2 : 1 in favour of the potential energy. Considering that we only want to gh-e a rough estimate, the ratio of the energies can be taken to be I.} This gives

for the thickness in question, which is quite a large value compared to the usual microscopic sizes. The reason for this is that gravitation is very weak compared to the electromagnetic interaction determining the binding energy and size of at.oms (molecules). S. 122. When a nucleus of atomic number Z keeps an electron within distance r {in a sphere of radius r) then the electrostatic energy of the electron is

- kZe2 /r and its momentum can be estimated to be h/r. H Z

»

Eei

~

-

1 (i.e. the nucleus in hand is a heavy one), the kinetic energy of the strongly bound electron can be calculated using the relativlstic formula:

(c is the speed of light, his the Planck coostant divided by 21f.) The total energy of the electron is E(r}

he 1 zt? - r + -. = Ee_1(r} + Esnn(r) = -4treo r


134

Sinr.e the fine-structure constant is for small values of

r

is

.-!!__ h = -1-, the toLal energy of 471't:o c

137

2

E(r)

e = -4n-c

.

the electron

1

(137 - Z) · - . r

0

(The above expression is not valid for large values of r , since a small, non-relativistic momentum wo uld correspond to those.) According to the a bove formula, t he e lectron falls into the nucleus if Z > 137, or raU1er: it is restr<~ued to a small volume of a size comparable to that of the nucleus. The figure J 37 is only an estimate of the critical at.omic number: more accurate calculations tl1at also take the finite size of the nucleus into a ccount yield val ues around 150 - IGO for Z<>rit · These calculatfons show that the electron could find room in the nucleus above t he at.omit number 150, if transuranic elements of such high atomic numbers could IJe macle at all. S. 123 . The speed of propagation of surface water waves can depend on the surface tension o and density{! of water, and on t he wavelength~- The dimensions of these quantities are: N m

kg s

(n) = - =-, 2

kg

[u] = ms•

[~J = m.

Speed c:u1 l>e ' mixed' from these quantities if the ratio of o to f! occurs in the expression (or else ' kg' would not be eliminated!). However , as

this expression has further to be divided by the wavelength, then it is the squareroot. of the result tha t has the dimension of speed. In summary: according to dimensional analysis, the speed of propagation of capillary waves is proportional to the reciprocal of the square-root of the wavelength:

From this functio11al dependence (and the given datum), the conclusion is that the speed of propagation of surface waves would reach Lltat of sound a t wavelengths in the range of 10- R cm. Since, however, the speed of propagation of s urface waves cannot be higher than that of sound (the molecules cannot 'push' each other faster at the surface than inside the m atter), wavelengths under approximately 10- e cm have no meaning. And in fact , this is I.he order of magnitude of the size of water molecules!

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