Problem 1. Answers: 1. 216v i j =+; 8a j = ; 7.13︒.(cos a v av θ⋅= ) 2. 1/3(3/)f t v k = 3. a-e, b-d, c-f. 4. [d]: 222x y L +=, 0dx dy xy dt dt+= dx v dt =, B dy v dt =, 0B xv yv +=, cot B xv v v yθ== 5. (a)32(102)3t r i t t j =+-, (Answer)(b) 912r i j =+, (3)(0)343avg r r v i j -==+ , (Answer) (3)(0)343avg v v a i j -==- (Answer)(c) 92v i j =-2tan 9y x v v θ==-, 12.5θ=- (Answer)6. Solution: From the definition of acceleration for a straight line motiondva dt =,and the given condition a =-dv dt-=. Apply chain rule to d v /d t , the equation can be rewritten asd v d x d vv d x d t d x-== Separating the variables givesv k dx =- Take definite integration for both sides of the equation with initial conditions,we havexv d v k dx =-⎰⎰, or 3/2023x v k = (Answer)Problem 2. Answers:1. 13.0 m/s 2, 5.7m/s, 7.5 m/s 2,2.1)R .Solution: At initial momentwhen the ball is just kicked out:2ir va g ρ==, 2iv gρ=. In order the ball not to hit the rock,2i v R g ρ=≥,For the vertical motion:212gt R =, t i v t R -=1)R (Answer)3. d.4. d.5. Solution: at any moment the speed of the projectile isv =Tangential acceleration:2t dv a dt ==,Radial acceleration:r a ==,From 2r v a ρ=,we have the radius of curvature is given by2223/220()r v g t v a gv ρ+== (Answer)6. 86.7θ=Solution: sd sg gd v v v =+ sg dg v v =-3510/3600t a n17.360.8dg sgv v θ⨯=== 86.78642θ︒'==Problem 3. Answers : 1. Solution: 212cos cos v T T m Rθθ+=12sin sin T T mg θθ=+(a)21108N 2cos 2sin mv mgT R θθ=+=, (b) 2156N sin mgT T θ=-= 2. (a) Mg ,(b)3. d.4. e5. Solution: from the given condition ct i v v e -= and with 10.0m/s i v =, at 20 s,5.0m/s v =, we have (a)10.0347s -,With 40.0 s, we can get (b) 2.50 m/s,Solution: (c) From ct i v v e -=, we havect i dvcv e dt-=- Therefore a cv =-, (Answer)which means the acceleration of the boat is proportional to the speed at any time.6.Solution: From Newton ’s second law, we have2k m vm a -= By the definition of acceleration, this equation can be rewritten as2dvkmv m dt-=Separating the variables obtains2dvkdt v-=Take the definition integral with the initial conditions, we hve020v t v dvkdt v -=⎰⎰11kt v v -= Then 00/(1)v v ktv =+.(Answer)Fig. 3-1Problem 4.Answers: 1.(a) 21J, (b) cos θ=, θ =19.4︒ 2. 23(79)3F x y i x j =-- . 3. c. 4. d5. (a) 002()x x U F dr Ax Bx dx ==-+⎰⎰ =2323A Bx x -,(b) 51923A B U ∆=- (c) 0U K ∆+∆=, 19532B AK U ∆=-∆=-6. Solution: (a) In the process of the pendulum swinging down, only thegravity does work , mechanic energy is conserved:21(1cos ),2mgL mv θ-=When the sphere is released from a certain height, in order to make the ball will return to this height after the string strikes the peg:21()2mv mg L d ≤-, (1cos )()mgL mg L d θ-≤-cos L d θ≥. (Answer)Solution: (b) If the pendulum is to swing in a complete circle centered on the peg, at the top of the path, the tension T on the cord must not be zero.2v T mg m L d+=-,Because 0T ≥ So 2()v g L d ≥-From the conservation of mechanical energy, we have212()2m g L m v m g L d =+- Inserting 2()v g L d ≥-into above equation, obtains1()2()2mgL mg L d mg L d ≥-+- 35L d ≥ (Answer)Problem 5. Answers: 1.(a) 52104i j - 32()p Fdt ∆=⎰2.(a) 0.284, (b) 114.6 fJ, 45.4 fJ.Solution: (a) Momentum and kinetic energy are conserved: 0n c p p p =+ (1)2220222n c p p p m m M=+, 12M m = So, 22201212n cp p p =+ (2) From Eq.1, 0n c p p p =-, insert Eq.2222001212()c c p p p p =-+ 22200122412c c cp p p p p =-++ 202413c c p p p =,02413c p p =222001242()0.28412132c c p K M p K m===. (Answer) 0000.284n c K K K K K =-=-15114.610J 114.6fJ -=⨯= (Answer)1500.28445.410J 45.4fJ c K K -==⨯= (Answer)3. [a]. 12Fd K K ==,22121222p p m m =,12p p =4. [a]Solution: For m : in the elastic collision process,00()F t mv mv m v v ∆=--=-+,0()m v v F t+=-∆ in the inelastic collision: 00()F t mv mv m v v ''''∆=-=--,0()m v v F t '-'=-'∆ 00,()()t t v v v v ''∆>∆-<+ F F '∴>5. v ≥ Solution: /2mv MV mv =+, 2m V v M=2122M V M g l ≥, v ∴≥6. proof problemSolution: ()mv M m V =+, mV v m M=+212h g t =,d Vt ==v ∴=Problem 6. Answers: 1. 144 rad. Solution: 106t α=+, d dtωα=， 20(106)103t t dt t t ω=+=+⎰2103d t t dtθω==+, 4223400(103)(5)144rad t t dt t t θ∆=+=+=⎰(Answer)2. (a)22853I Md md =+,Solution:2221(2)2(2)3I md m d M d =++(b)212K I ω=22245()32Md md ω=+ 3. c. r F τ=⨯(4i 5j)=+⨯ (2i 3j)+ 2k =4. a.Solution: The initial angularacceleration of the rod and theinitial translational acceleration of the right end of the rod:I τα=, 2123L Mg ML α=, 32g L α=32t ga L α==5. 21.5 NSolution: For the flywheel:()u R R T T r I α-=212R I M R =, 21.67rad/s R α=, 135N u T = 21.5N T = (Answer)6.Solution: from the conservation of mechanical energy:222111sin 222mgd kd mv I θω+=+v R ω=,ω⇒=Fig. 6-2Fig. 6-4Fig.6-5Fig. 6-6Problem 7. Answers: 1.(a) 230L t k =- , 60dL tk dtτ==-Solution: (a)L r p =⨯ , 2(3i 5j)m r t t =+ , 2.0kg m =, (b) dLdtτ=.2. (a) Mvd , (b) 2v v '=, (c) 23Mv . Solution: (a) Angular momentum is conserved:22f i d dL L Mv Mv Mvd ==+=;(b) ,f i L L =24dMv Mvd '=2v v '= (c)2222211222243W K Mv Mv Mv Mv Mv '=∆=⨯-⨯=-= 3. d. 4. a5.(a) 222233ML ml v v ML ml -=-+,022603mlv ML ml ω=>+, (b)212I Mgh ω= Solution: (a) In the process of collision, angularmomentum and mechanical energy are conserved:201()3mv l mvl ML ω=+ (1)222201111()2223mv mv ML ω=+ Or 222201()3mv mv ML ω=+ 222201()()3m v v M L ω-= 22001()()()3m v v v v ML ω+-= (2)Form Eq.1, we have201()3v v ml ML ω-= (3)E q.(2)E q.(3): 0v v l ω+=(4)Fig. 7-2Fig. 7-5From Eq.(3), we have 2013v v ML mlω-= (5)Eq.(4) + Eq.(5) 20123v l ML mlωω=+22603mlv ML ml ω=>+ (Answer)Inserting ω into Eq.(4), obtains2202233ML ml v v ML ml -=-+ (Answer) Solution: (b) 212I Mgh ω=, 213I ML =22I h Mg ω==220226()3mlv L g ML ml=+6. Solution: (a) angular momentum isconserved221()2i mv d mR MR ω=+22202i mv dMR mRω=>+ (Answer)(b) Mechanical energy is not conservedbecause the clay and the solid cylinder undergo an inelastic collision. In this process, some kinetic energy must be lost.:2222212(2)i f m v d E I M m Rω==+ 22220112111/22f iI E d E R M mmv ω==<+， 0f E E <Problem 9. Answers : 1. (a) 3.07 MeV （Or: 6mc 2, 4.92⨯10-13 J ）, (b) 0.986c . Solution: (a)22(1)5mc mc γ-=, 6γ=,For an electron , 20.511MeV mc = Total energy: 2 3.07MeV mc γ=Fig. 7-6Solution: (b)6=, 0.986v c =2. 2mc 2 = 2⨯0.511=1.02 MeV . (1.64⨯10-13 J )3. c.Solution:21/u v u u v c -'='-2220.750.7510.75/c c c c --=+0.96c =-4. dSolution: 2K mc =,,v c K →→∞.5. 1.63⨯103 MeV/c.Solution: 222mc mc γ=, 2γ=, v =2/p mv c γ===,For a proton, 27162191.6710910939MeV 1.610mc --⨯⨯⨯==⨯23/ 1.6310MeV/p c c =⨯ 6. (a) 5.37⨯10-11 J = 335 MeV (5.36⨯10-11 J) (b) 1.33⨯10-9 J = 8.31 GeV (1.33⨯10-9 J) Solution: W K =∆2222121(1)(1)()K mc mc mc γγγγ∆=---=-(a) 221()335MeV W mc γγ=-==;(b) 221()8.31GeV W mc γγ=-==7. Solution: 22222(),E p c mc =+ 2E mc γ= 224222m c p c m c γ∴=+, 2422242211/m c p c m c v c∴=+-,222424221()/1/p c m c m c v c∴=+-,224222241v m c c p c m c -=+,2242222242221v m c p c p c m c p m c=-=++v =Problem 10. Answers: 1. 1.2⨯1022 for 37︒C. Solution: pV RT ν=,56(761/760)1.0110500100.02mol 8.31(27337)pV RT ν-⨯⨯⨯===⨯+23220.02 6.0210 1.210A N N ν==⨯⨯=⨯2. 385 K, 7.97⨯10-21J, molar mass of the gas.Solution: pV RT ν=, 631.6107.010385K 3.58.31pV T R ν-⨯⨯⨯===⨯ 2321331.38103857.9710J 22K kT --==⨯⨯⨯=⨯v =M is the molar mass of the gas. 3. b.λ== 4. a.p v =5. (a) 2/3v 0, (b) N /3, (c) 1.22v 0, (d) 1.31v 0. Solution: (a)()1f v dv ∞=⎰,00112v a v a +=, 023a v =; (b) 02.001001.502()(2 1.5)233v v v NN Nf v dv N v v a Nv ==-=⋅=⎰ (c) 00022000011() 1.29v v v v av v vf v dv v dv vadv v v v ==+==⎰⎰⎰(d) 02222220()()()v v v v v v f v dv v f v dv v f v dv ==+⎰⎰⎰002220v v v avv dv v adv v =+⎰⎰4330000(8)43v a v v a v -=+20114()69v =+01.31rms v v = (Answer)6. Solution:(a) 5252310, 1.8101.3810400p p nkT n kT -====⨯⨯⨯.(b) 23262222 1.3810400 5.3110kg 456p p kT v m v --⨯⨯⨯====⨯,Molar mass: 23266.0210 5.311032.0g A N m -=⨯⨯⨯=, oxygen gas. (c) 30.96kg/m nm ρ==. (d)232553551.3810400 1.8102.510J/m .22kTn -=⨯⨯⨯⨯⨯=⨯Problem 8. Answers: 1. 0.92c L L =0.92v c == 2. (a) 2.2⨯10-6 s, (b) 653 m.Solution: Lt v∆=, 0t ∆=∆60 2.210s t -∆==⨯,00653m x v t ∆=∆=3. b. (the constancy of the speed of light)4. b. As the figure shows0010.25V AH V AH ==== 5. (a) L 0 = 17.4m , (b) θ = 3.3︒Solution:(a)0/cos30/x x l l l == 0sin30,y y l l l ==017.4m l == Solution: (b)000tan tan 30y xl l θ==0 3.3θ=6. (a) 2.50⨯108 m/s, (b) 4.97 m, (c) - 1.33⨯10-8 s. Solution: (a)From the given condition, 0x '∆=From the Lorentz transformation, ()0x x v t γ'∆=∆-∆=Fig. 8-58953 2.5010m/s (91)10x v t -∆-===⨯∆-⨯ Solution: (b) From the Lorentz transformation(),x x v t γ'=-89() 2.51010) 4.97m,RB R R x x x vt γ-''==-=-⨯⨯=Solution (c) From the Lorentz transformation2()vxt t c γ'=-, 82() 1.3310s R RR vx t t c γ-'=-=-⨯ 92() 4.9810s B BB vxt t cγ-'=-=-⨯. In S ', red flash occurs first. Problem 11. Answers: 1. (a) 3.5⨯103 J, (b) 2.5⨯103 J , (c) -1000 J.Solution: 3,778.31120 3.510J 22p m Q C T R T =∆=∆=⨯⨯=⨯3,558.31120 2.510J 22V m E C T R T ∆=∆=∆=⨯⨯=⨯Work done by the gas:1000J W Q E =-∆=Work done on the gas: 1000J W =- 2. 227K,Solutio: 111122TV T V γγ--=, 12112(),V T V T γ-= 121120.40.412300(),227K 22V T T T V T γ-==== 3. b . As the figure shows.4. e.Solution: For a free expansion:0,0,0E W Q ∆===, 1122pV p V =2112,V V p p >>5. Solution : (a) ,a a a c c c p V RT p V RT ==355(4252)105kJ 22c c a a V p V p V R E C T R -∆=∆==⨯-⨯⨯=-(b)Q E W =∆+, 31(25)2107kJ 2W =+⨯⨯= 572k J Q E W ∴=∆+=-+=. (c) 5kJ E ∆=-, 3521010kJ W =⨯⨯=5105k JQ E W ∴=∆+=-+=6. Solution: From the definition for molar specific heat at constant pressure, wehave(),p p dQC dT= From the first Law of thermal dynamics, we have()p V dQ dE dW C dT pdV =+=+, Here, V dE C dT = Combining theses equations, obtains()V p p C dT pdV dQC dT dT+== p V pdVC C dT=+,From the ideal gas law, for the constant pressure process with one mole of ideal gas, we havep d V R d T =Then p V RdTC C dT=+Therefore p V C C R =+Fig. 11-5Problem 12. Answer: 1. 3.28 J/KSolution: The entropy change of the Universe due to the energytransfer by radiation from the Sun to the Earth is331010 3.28J/K 5800290universe sun earthS S S -∆=∆+∆=+=. 2. 57.2 J/K.Solution: Suppose the process can be replaced by a reversible isothermalprocess, then 57.2J /K a i r l a k em g hS T +∆==. 3. d. 0gas enr S S S ∆=∆+∆≥4. c.Solution: free expansion is an irreversible process which occurs in an isolatedsystem. 0S ∆>5. (a) -0.390νR , (b) -0.545νR ,Solution: (a) 27325273185298ln 0.3902255VC dT S R R T ννν+-∆=-=-=-⎰ (b) 27325273187298ln 0.5452255p C dTS R R Tννν+-∆=-=-=-⎰6. (a) 4500 J, (b) - 4986 J, (c) 9486 JSolution: (a) dQdS T=dQ TdS =, Q TdS ==⎰The area under the T -S curve(b) (c)Problem 13. Answer: 1. 1.862. 1/3, 2/3. ,， .3. b.4. d.5. (a) 4.10⨯103 J, (b) 1.42⨯104 J, (c) 1.01⨯104 J, (d) 28.9% Solution: (a)(b) A → B:C → A:.(c) B → C: (d)(e) , the efficiency of the cycle is much lower than that of a Carnotengine operating between the same temperature extremes.6. Solution: (a) The work done by the gas from state a to c along abc is given byIn the constant volume process along b to c , W bc = 0.Therefore, (Answer)Solution: (b) the change in internal energy form state b to c is ,Using the ideal gas law to this constant volume process, obtainingFig. 12-6Fig. 13-5Then(Answer)From the definition of change in entropy,we haveUsing the first law of thermodynamics to this constant volume process with dW = 0, obtainingSubstituting this into the expression of entropy change, we have = (Answer)Solution: (c) Because both internal energy and entropy are state properties, for one complete cycle,(Answer)Problem 14. Answers: 1. 40.9 N/m. , 2. .Solution: For block B : ,For block P :If block B is no to slip , , , ,3. c. ,4. c. , ,, , .5. (a) x m = 2.00 cm, (b) T = 4.00 s,(c) , (d) , (e) , (f) .6. proof problem.Fig. 13-6Fig. 14-2Fig. 14-5Solution: (a) the net force acting onthe ball is given by(b) Comparing the force acting on the ball to the Hooke ’s law ,obtains, the ball moves in SHM, with angular frequency (Answer)Problem 15. Answers: 1. 3.5 s.Solution: For a torsion pendulum, ,, , 2. 0.944 kg m 2.Solution: For a physical pendulum, , here his the distancebetween the com and the rotation axis.3. d. , *4. a.Solution: when or resonance occurs, here f is the naturefrequency of the system. The oscillation of the system is due tothe spring property of the diving board, so it can be treated as a torsion pendulum: , , 5. . Solution: Method one:Method two:,The mechanical energy is conserved: , , 6.Solution: Method one , ,Fig. 14-6Fig. 15-1For small angle, such as , ,Method twoFor small angle, such as , ,,Problem 16. Answers: 1. 0.319 m.Solution: , the speed of the wave:2. (a) 0.25m,(b) 40.0 rad/s,(c) 0.300 rad/m, (d) 20.9 m,(e) 133m/s, (f) positive x direction. Solution: Standard SH wave function:, ,3. d. Standard traveling wave function:4. c. ,5. (a) 31.4 rad/s, (b) 314 rad/m, (c) , (d) 3.77 m/s, (e)118 m/s 2. Solution: (a) ; (b)(c)(d)(e) 6.Solution: wave speed for a transverse wave in a string is given by ,The time interval required for a transversewave to travel from one end of thestring to the other isFig. 15-6Fig. 16-5Fig. 16-6(Answer)Problem 17. Answers: 1.(a) 2.0 Hz, (b) 3.38 m/s.Solution:(a)Hz(b) , v d = 3.38 m/s*2.1029 m/s, 58.3 s.Solution: (a) From the definition of March number,,(b) In the time interval t, the sound travels adistance h, ,3. d.4. c.5. (a) 308 m/s, (a)163 N, (b) 660Hz.Solution: (a) For the fundamental mode,(b) the wave speed:163 N.(c) ,.6.(a) As the figure shows(b) 2 cm /s,(c) , with y and x in cm and t in s.(d) 2.5 cm/sSolution:From the curve:, ,,(a)Transverse velocity, .At t = 0,(b)(c)(d)Fig. 17-2Problem 18. Answers: 1. 1500 nm, 5Solution:, ,;2. Because the half-wave loss in the reflection,at point O is dark, interference pattern onlylocates above O, and reverse the positions ofdark and bright fringes.3. c. ,4. d.,5. 632 nmSolution:For m = 1,(Answer)6.Solution: ,,Problem 19.Answers: 1. 0.50 cm，2. 0.221 mm.Fig. 18-1Fig. 18-23. d.Solution:4. d.Solution: For , optical path length differencecorresponds to the second minimum. 5. 8.7 μm6. 1.32Solution : For m = 10, ,When a liquid is introduced into theapparatus,(Answer)Problem 20. Answers: 1. 5λ/2, 5Solution: In single slit diffraction, bright fringes correspond to the path lengthdifference for the two rays emitted from the top and the bottom of the slit , for the second order bright fringe, , , , 5 half-wave zones.2. 650 (red) , 430 (blue),In a diffraction of single slit, half angular width for the bright fringe , at A, smaller wavelength iszero intensity, 650 nm is non-zero intensity. At B, 430 nm is non-zero intensity. 3. b.“No diffraction minima are observed ” means the first minimum appears at infinity Or ,If , , the first minimum will appear ontheFig. 19-4Fig. 19-5Fig. 20-5viewingscreen.4. b.,5. 25 cmSolution:For the third-order minimum: ,The position of the third-order minimum:6. 0.284 mSolution:Problem 21.Answers:1. 7Solution:For the grating diffraction,brightlines satisfy the relation:For a single-slit diffraction,the minima satisfy therelation:Therefore, the missing orders occur for,Within the central envelop of each single-slit diffraction pattern, m = 4 is a missing order. There are 7 orders can be observed within the central envelop: .2. (a)0.109nm, (b) four.Solution: for x-ray diffraction from a crystal, maxima satisfy the relation:(a) ,(b)Letm can only take integer number: 4 (Answer)3. c. , , , for the same m and , ,the spread of each spectral order increases.4. a.;(out of the wavelength range of the light source); (within the wavelength range of the light source)(out of the wavelength range of the light source)5. (a) 3.53⨯103, (b)11= 5⨯2+1Solution: (a) ,(b)m can only take integer number: 5There are 11 maxima for the diffraction pattern:6.(a) Let m and m+1 be the orders for the two adjacent maxima. Thus we have.Subtraction of the first equation from the second leads to.Solving for d yields. (Answer)(b) The fourth-order maxima are missing. So the ratio of grating spacing to slitwidth satisfies.When the ratio is 4, the slit width is smallest. Thus. (Answer)(c) The diffraction maxima of gratings are determined by.When , the order m reaches its maximum that is given by.Since the diffracted light with diffracted angle of 900 cannot reach the screenand the fourth-order and eighth-order maxima are missing. The orders ofmaxima produced are. (Answer)Problem 22.Answers: 1. ,Solution: during the rotation of the polarizing filter,the maximum transmitted intensity,the minimum transmitted intensityfor the unpolarized lightfor the polarized light2. 60.5︒,3. b.4. a.,5.0.375Solution: ,6.54.7︒, 63.4︒, 71.6︒.Solution:(a) ;(b)(c)。
