物理經(jīng)典力學(xué)

1、mh1. A sa lite orbits the Earth in a circular orbit. An astronaut on board perturbs the orbit slightly by briefly firing a control jet aimed toward the Earthscenter. Afterward, which of the following is true ofthe salitesth?(A)(B)(C)(D)(E)It is a ellipse.It is a hyperbola.It is a circle with larger

2、radius.It is a spiral with increasing radius.Itexhibitsmanyradialoscillations revolution.3. As shown above, a block of mass m is releasedfrom resa distance h above a vertical masslessspring with spring constant k, what is thekinetic energy of the block?umper(A)mgh1 m2 g 2解：圓軌道受到小擾動后仍是閉合軌道，因此是橢圓。選（A）

3、。mgh (B)4k1 m2 g 22.rson standing on the surface of the Earthmgh (C)2kthrows a ball. The ball leaves the throwers hand withinitial velocity vi and has final velocity vf just beforem2 g 2it is caught. If airis negligible, which ofmgh (D)kthe following diagrams correctly represents asible sequence of

4、velocity vectors for the ball?(E)2mghdvVerticalVertical解：物體速度最大時 0 ，即加速度為零，此dt時物體受力平衡。設(shè)速度最大時彈簧壓縮量為 x，則( A )Horizontal( B )Horizontalmg kx ，VerticalVerticalmgkx 。( C )Horizontal( D )Horizontal由機(jī)械能守恒11mgh x mv kx22。Vertical22將 x 的表達(dá)式代入，( E )Horizontal2 2m g 1h x 1Ek2 mv mg2kx mgh222k解：因?yàn)榍蛟谒椒较虿皇芰?，所以其?/p>

5、平動量不變，即速度的水平分量不變，只有（E）符合。選(E)。選（C）。1vB = 0BeforeAfter4. A bullet of mass m traveling at speed v strikes a block of mass M, initially at rest, and is embedded in it as shown above. How far will the block with the bullet embedded in it slide on a rough horizontal surface of coefficient of kinetic fric

6、tion k before itcomes to rest?F = 10 N2 kg5. A 2-kilogrom box hangs by a massless rope from aceiling. a forlowly pulls the box horizontally tothe side until the horizontal force is 10 newtons. The box is then equilibrium as shown above. The anglet the rope makes with the vertical is closest tom Mv2(

7、A) ()()m2k g(A)(B)(C)(D)(E)arctan 0.5arcsin 0.5arctan 2.0arcsin 2.0 45m Mv 2(B) ()()M2k gm Mv 22k g)2 (C) ()解：物體重 20N，角度10M arctan arctan 0.5 。20v 2m(m M2k g)()(D)選（A）。6. A 5-kilogram stone is dropped on a nail and drivesv 2m)2 (E) ()the nail 0.025 metero a piece of wood. If the stonem M2k gis movin

8、g at 10 meters per second when it hits the解：射入過程很短，時間及此過程中所走過nail, the average force exertednearlyo the wood is most距離可忽略不計(jì)。由動量守恒，mv m M V 。(A)(B)(C)(D)(E)10 N100 N1000 N10,000 N100,000 N之后由于摩擦做勻摩擦熱，則運(yùn)械能全部轉(zhuǎn)換為1 m M V 2m M gs ， k2解：力對石頭做功等于其動能減少，F(xiàn) 1 mv 2 / h 1 5 102 / 0.025 10000N 。V 2v 2ms ()()2 。2選（

9、D）。22k g2k gm M選（E）。7. A machine gun fires bullets of mass 20 grams eachat a rate of 1200 bullets per minute. The bullets hit athick woodenat a speed of 600 meters persecond and are stoppedhe. The averageforce exerted on the144N240Nby the bullets striking it is2mvM(C) 14.4103 N(D) 24.0103 N(E) 14.41

10、03 N解：由沖量定理由平行軸定理，圓盤繞桌子中心的轉(zhuǎn)動慣量為1I I 0 MR 2 Mr MR ，所以它相對于洞的角動量為222 Fdt P ，11L I M (R 2 r 2 ) 2 2 。220 103 1200 600mVtF 240 N 。選（E）。60選(B)。9. If the string is pulled slowly downward sot thecenter of the puck moves in a circle of smaller radius,Questions 8-9tiesfollowing?t are conserved include which o

11、f theI.II. III. (A)(B)(C)(D)(E)Angular momentum Linear momentum Kinetic energyI only III onlyI and II only II and III onlyI, II, and IIIRMr解：由于是有心力，所以冰球相對洞的角動量守恒，I 正確。在圓運(yùn)動中動量本身就不是一個守恒量，變化中冰球又顯然受力，所以動量守恒無從談起，II 不對。變化中冰球受到了指向洞的力，沿此方向又有位移，所以繩子對其做功，動能不守A uniform cylindrical puck of radius r and mass Mis

12、 attached toof a cordtsses through ahole in a fixed horizontal frictionless table as shown above. The center of the puck moves in a circle of radius R wi ngular speed 8. The magnitude of the angular momentum of thepuck about the hole is1恒。或者由動能為I 2 ，而角動量守恒，2I constant ，動能顯然不能守恒，III 不對。選（A）。3(A)MR 22

13、(B) M (R 2 r 2 )1(C) M (R 2 r 2 ) 2 210. An expresfor the potential energy of twoions iskq2bP.E. .r9r(D) M (R 2 1 r 2 )What is the constant b as a function of the equilibiums cing r0?2kq 2r011(E) M (R 2 r 2 ) 2 2(A)102解：首先，圓盤繞自身圓心的轉(zhuǎn)動慣量為I 1 Mr 2 ，2kq 2r(B)03(B)(C)(D)(E)800 J4000 J9600 J19,200 J8kq 2(

14、C)9r0kq 2r0 10(D)解：剛體定軸轉(zhuǎn)動的動能為1T I ，22kq 2r 80 9(E)動能變化為T 1 I 2 2 1 4 802 402 9600J0解：因?yàn)?r0 處勢能最小，由22。dUkq 29bP E 0選(D)。r 2r10dr解出13. If the cylinder takes 10 seconds to reach 40 radius per second, the magnitude of the app d torque is80 Nm40 Nm32 Nm16 Nm8 Nm解：由剛體定軸轉(zhuǎn)動的動力學(xué)方程M I ，其中 M 為外力矩，為剛體的角加速度。對于本題

15、9b8kq 2r0，kq 2r 8b 0 。9選（E）。11. A rigid cylinder rolls at constant speed withoutslipon top of a horizontal plane surface. Theacceleration of a poon the circumference of thecylinder at moment when the poistouches the plane 80 40M I I 0 4 16N m 。t10(A)(B)(C)(D)(E)directed forward directed backward dir

16、ected up directed downzero選（D）。Questions 14-15A nonrelativisticrticle of mass m moves in aplane. Itsitionis described bythe polarr& and & .解：因?yàn)橛|地點(diǎn)相對于圓心做勻速圓周運(yùn)動，所以相對于圓心的加速度為豎直向上方向，而圓心做勻速直線運(yùn)動，加速度為零，所以觸地點(diǎn)相對于圓心的加速度就是其對地的加速度。選(C)。coordinates r and , with time derivativesThere exists a potential energy U =

17、 kr2, where k is a constant.14. Which of the following is the Lagrangian of therticle?Questions 12-13A cylinder with moment of inertia 4 kgm2 about a fixed axis initially ro es at 80 radians per second about this axis. A constant torque as app d to slow it down to 40 radius per second.12. The kineti

18、c energy lost by the cylinder is(A) 80 JL 1 mr& 2 r 2& 2 kr 2(A)2(B) L 1 mr& 2 2 kr 224(C) L 1 m 2r& 2 r 2& 2 kr 2(B)(C)(D)(E)0.80 h0.75 h0.64 h0.50 h2(D) L 1 mr& 2 rr& r 2& 2 kr 22L 1 mr& 2 2rr& r 2& 2 kr 2解：由機(jī)械能守恒(E)1 mv 2 mgh ，2解：對于有系，Lagrange 函數(shù)定義為L T V 。2h v 2 0.82 0.64 。本題中，由于極坐標(biāo)系下的速度表達(dá)式為v r&e

19、r r& e 。而兩分量之間垂直，所以h v 選(D)。L T V 1 mv 2 v 2 Vxr2m1m2mr& r kr1&22 22217. Two masses, m1 and m2 , are joined by a masslessspring of force constant k and placed on a horizontal選(A)。frictionlesrface as shown above. The system is15. Whichconstant?of thefollowingtiesremainsreleased from rest when the se

20、ration betn themasses is x. If the unstretched length of the spring isx0, the speed of mass m1 when the two masses are a distance x0 a rt is(A) mr& 2 r 2& 2 (B) mr 2& 2km(x x0 )2(A)1(C) kr 2km(x x)2(B)(D) mr&02(E) mr 2&k(x x0 )2(C)m mdUdr12解：F 2krr ，或者由于U kr 與無關(guān)，2km2m(x x0 )2(D)為有心力場，角動量守恒。利用上題中速度的表

21、達(dá)式，徑向速度為 r& ，所以角動量表達(dá)式為mm112km1(x x)2(E)mr 2& 。0mm m選(E)。212解：設(shè)恢復(fù)原長時兩物體速度為 v1、v2，由動量守恒16. A ball dropped from a height h. As it bounoffthe floor, its speed is 80 percent of what it was just before it hit the floor. The ball will then rise to a height of most nearly(A) 0.94 hm v m v 。1 12 2由機(jī)械能守恒51 kx

22、 xm vm11correctly concluded aboutobjects?the force betn the222 。v01 12 2222(A)(B)(C)(D)(E)It is central.It is inverse-square. It is conservative. It is graviional. None of these聯(lián)立以上二式，解得kmv 2 (x x。)210mm m112選(D)。s is justified.解：角動量17. What is the number of degrees of freedom for 6rticles moving fr

23、eely in one plane?ddrdtr mv m r 2m，dtL(A)(B)(C)(D)(E)68101218其中d 1 r dr2為矢徑 r 在 dt 時間內(nèi)掃過的面積。由本題所給條d度為 2，而且相互解：二維運(yùn)動，每個粒子的件，為常量，故角動量守恒。有心力可保證dt角動量守恒。選（A）。之間無關(guān)聯(lián)，沒有限制條件，所以總數(shù)為 26=12。選（D）。18. A mass m1 atharmonic motionof a spring executes simpleperiod T1. The period ofQuestions 20-21wioscillation of a di

24、fferent mass m2 on the same springism2 m(A) Tl11lm(B) T 1 1m2wo experiments, two cylinders, X and Y, are(C) T1released from resthe top of the same inclinedrand roll down without slip. Let tX and tY bem(D) 2 1 m2the respective times taken for the cylinders to reach articular line ll on the rshown abo

25、ve.20.heexperiment, both cylinders are solid,m(E) 2 2 m1uniform, and of identical dimenmade of different materials sos; but they aret the mass of X istwicet of Y, MX = MY. Which of the followingm解：一維諧振子公式，T 2。relationships betn tX and tY is correct?選（A）。ktX 2tY2tY tX tY tX = tY2tX tY tXtY 2tX(A)(B)(

26、C)(D)(E)19. Under the influence of a mutualeraction, anobject orbits another objectt is fixed. The orbits in a plane and the areas st out by the radius爲(wèi)vector in equal times are equal. What can be爲(wèi) f6mg sin f ma 爲(wèi) f fr I r a vcos2 0g(C)2vtan20g(D)2vsin cos20g(E)解：飛行時間為mg sina 2v sinIr 2t 0 ，gm 1水平位移

27、為 I mr 2222v 2 sin coscos t 0 。S va g sin 0g3 mr C選（E）。21.he second experiment, both cylinders are solid,23. A spherical neutron star has a uniform mass density . What is the period of ro ion below which material will fly off the equator? (Use nonrelativistic mechanics and let G be the universal gra

28、vi ionalconstant.)uniform, of identical length, and of the same density;but their radii are different sot the mass of X istwicet of Y, MX = 2MY. Which of the followingrelationships betn tX and tY is correct?34GtX 2tY2tY tX tY tX = tY2tX tY tXtY 2tX(A)(B)(C)(D)(E)(A)43G(B)13 2(C)8G(C)22. A golf ball

29、is hit from ground level win initial1 2velocity v at an angle with respect to the ground. G (D)0If airis negligible and the magnitude ofthe graviional acceleration is g, the ball hits the1 3 2ground at what distance from the powas hit?at which it(E) G 解：臨界狀態(tài)質(zhì)點(diǎn)所受萬有引力全部用來提供所需向心力：2v 0 g(A)GMm m r ，2r 2

30、vsin2 0g(B)4GGM ，r 337所以I and II onlyII and III onlyI, II, and III解：力場為保守場，能量守恒，I 正確。又為有心力場，且 F 指向原點(diǎn)，相對于原點(diǎn)的力矩為零，II正確。顯然相對于原點(diǎn)的角動量守恒，III 正確。選（E）。12 3 2T G 。選（E）。24. Article of mass M moves alongthe x-axisunder the influence of a conservative field with thepotential energy V b x 2 .vIf therticlestarts0

31、2km326.rticlem2m1 andm1from resx = 1, itsum velocity is1 of massrticle 2 of massMb2m 1 m(A)are coupled by a massless spring of212force constant k andat rest on a horizontalMb(B)frictionlesrface as shown above. A thirdrticle1of mass m3 m2 2 m1 and speed v0 strikes rticle 2 along the axis of the sprin

32、g and sticks torticle 2. The speed of the center of mass of the2Mb(C)b2M(D)system after the colliisv04v03v02v0bM(A)(E)(B)解：因?yàn)樗芰楸Ｊ亓Γ瑱C(jī)械能守恒。勢能最小時速度最大：(C)b221mvmax ，22xmax(D)解得8k3mv0 (E)bMv1。max解：質(zhì)心動量等于各部分動量之和。碰撞前后動量守恒，選(E)。m3v0 (m1 m2 m3 )VMC ，25. Article moves in a force field given by F = r2 r,wher

33、e r is theition vector. If there are no othert remain constant include whichmv 3 v 0 V。for,tiesMC0m m m4123of the following?Total energyTorque about the originAngular momentum about the originI onlyIII only選(A)。27. A planet of mass m moves about the Sun of massM. G is Newtons constant, r is the plan

34、ets distance from the Sun, and v is the planets speed. Except for8an additive constant the planets potential energy is1 mv2 GMm1 m3(A)2r1 mv2 GMm(B)2rGMm(C) r GMm2 m(D) r 2(E) mgr解：取無窮遠(yuǎn)為勢能零點(diǎn)，則萬有引力勢為1m3GMGMU dr 。r 2r2 mr選（C）。29. The weight of a door is entirely supported by the upper hinge U which is

35、 2 meters from the lower hinge L as shown above. A me the mass per unit area of the door is constant and the hinges have negligible size. If the door weighs 200 newtons and is 2 meters wide, of what magnitude is the horizontalforce exerted by the lower hinge L?(A)(B)(C)(D)(E)80 N100 N120 N140 N200 N

36、Q28. The diagram above show a top phonograph turntable mounted on abearing. Initially, the turntable is atviewofafrictionlessrest and a解：因?yàn)殚T的重量完全由 U 點(diǎn)支撐，所以 L 點(diǎn)無豎直方向的力，L 處受力沿水平方向。系統(tǒng)對 U點(diǎn)力矩平衡，massive bug is asleet poQ. The bug wakes up,takes a walk such as the one indicated by the dotted line, returns

37、to Q, and goes back to sleep. Afterward,the turntable is1 m ，L(A)(B)again at restat rest, buthe samehe sameitionition only if the walkF 100N 。Ldid not encircle that rest, but not nevot posarily選（B）。(C)he sameitionwhether or not the walk encircled thvot po30. A thin uniform rod of mass m and length l

38、 is hinged at one end to a level floor and stands vertically. If allowed to fall, the rod will strike the floor wi n angular speed . If the same rod werecut in half to length l/2 and the initial conditions(D)at rest only if the walk did not encircle thpovot(E)not nesarily at rest, whether or not thv

39、otpowas encircledremained unchanged, it would strike the floor wi angular speed most nearly equal to2 2n解：因?yàn)橄到y(tǒng)沒受到相對圓心的外力矩，所以相對于圓心角動量守恒，最后轉(zhuǎn)桌必處于狀態(tài)。小蟲和轉(zhuǎn)桌間有相對運(yùn)動，轉(zhuǎn)桌可有任意的角位移。選（C）。9UL(C) gr9.8 5 105ve 1565 。(D) /(E) /2解：由能量守恒，勢能轉(zhuǎn)換為轉(zhuǎn)動能量222選(C)。mg l 1 I 2 。32. A solid ball weighs 5.0 newtons in air and 3.022

40、對于質(zhì)量為 m 長度為 l 的均勻桿，相對一端的轉(zhuǎn)動慣量為newtonnewtonbmerged in water. If the ball weighs 2.0bmerged in an unknown liquid, thespecific gravity of the unknown liquid is most nearlyI x m dx 1 ml 2 。(A)(B)(C)(D)(E)0.661.001.251.501.75l2l30解得3gl ，解：由重力與浮力和拉力平衡可見只與桿長有關(guān)。第二次桿長變?yōu)橐话耄?G F 3.0N ，水 2 。選（B）。G F液 2.0N 。31. A

41、n asteroid has a radius of 5105 meters and an1定律 F浮 gV ，得由acceleration due to gravity oft on Earth. The4F液液，velocity of escfor an objectstartingontheF水水surface of the asteroid is most nearly(A)(B)(C)(D)(E)150 m/s720 m/s1560 m/s5550 m/s11,200 m/sFG 0.25.1。液液水G 3水F水選（D）。解：所謂逃逸速度，是恰好使物體的機(jī)械能為 0。所以GMa m1

42、2mve 0 ，2r2GM av 。erv2vv11而由重力加速度33. Water flows through a Venturi tube as shown inthe diagram above. The radius of the large crossGM14a g ，r 2section of thpe is 2 centimeters and the raduis ofGM 1 gr 2 ，the constricted portion of thpe is 1 centimeter. Ifa4the speed of the waterhe large cross sectio

43、n is 1er second, the prere difference (p1 p2) is所以metmost nearly10p2p1(A) 0.6102 N/m (B) 3102 N/m (C) 1.5103 N/m (D) 7.5103 N/m (E) 37.5103 N/m解：由流量守恒35. The force on the body is12mv2(A)4kx3kx4(B)(C)kx55(D)v S v S ，1 12 2(E)mgv 4v 。21解：直接對勢能求導(dǎo)得 dU由 Bernoulli 方程， 4kx3 。F12v P gh 1 v 2 Pdx2 gh ，111222

44、2選(B)。從量綱上看，（A）、（C）為能量，（D）為1能量乘以距離，均不是力的。P1 P2 2 (v2 v1 ) g(h2 h1 ) 。22代入數(shù)字計(jì)算，選（B）?，F(xiàn)場有可能不記得36. The Hamiltonian function for this system isBernoulli 方程，其實(shí)它是機(jī)械能守恒在流體力學(xué)中的一種體現(xiàn)。再結(jié)合量綱分析，很容易推出這個形式。p 22mkx 4(A)p 22mkx434. A rock is thrown vertically upward with initial(B)speed v0. Ame a friction force propo

45、rtional to v,kx41 mv 2 kx 421 mv 2where v is the velocity of the rock, and neglect the buoyant force exerted by air. Which of the followingis correct?(C)(D)(A)(B)The acceleration of the rock is always equal to g.The acceleration of the rock is equal to g only at the top of the flight.(E)2解：對于變換方程不顯含

46、 t 的有系（這一條件(C)The acceleration of the rock is always lessg.n在 SUB中幾乎必然滿足），H=T+V。選（A）。(D)The speed of the rock upon return to its starting37. The body moves from x1 at time t1 to x2 at time t2.pois v0.Which of the followingties is an extremum for(E)The rock can attain a terminal speed greaterv0 befor

47、e it returns to its starting po.nthe xt curve corresponding to this motion, if endpos are fixed?解：在頂點(diǎn)，石塊的速度為零，此時空氣摩擦力為零，故加速度為 g，選（B）。對于選項(xiàng)（E），石塊有可能獲得收尾速度，但由于有阻力，必小于 v0。 14 t2mv kx dt2(A) 2t1 12 t2mv dt(B) 2t1Question 35-37The potential energy of a body constrained to move on a straight line is kx , w

48、here k is a constant. The ition of the body is x, its speed v, its linearmomentum p, and its mass m.t2 mxvdtt(C)41 1x2mv kxdx24(D) 2x111從另一個角度看 F 等于重力和離心力的合力(E)xmv2dxx1F mg 。1 r4 222解：本題涉及分析力學(xué)的變分原理部分。由Hamilton 原理，定義 Hamilton 作用量t2S Ldt ，t選（E）。BAkkC1則真實(shí)運(yùn)動使 S 取極值，其中 L 為L T V 。選（A）。日量，mm2m39. Three mas

49、ses are connected by two springs asshown above.A longitudinal normal mode with12kmfrequencyis exhibited byl(A)A, B, C all movinghe same direction withequal A and equal A andequallitudeC moving in op(B)ite directionswithrmlitude, and B at rest(C)C movinghe same directionwithite38. The figure above re

50、presentslitude, and B movinghe opapomass mdirection with the sameA and C movinglitudehe same directionattached to the ceiling b a cord of fixed length l. Ifthe pomass moves in a horizontal circle of radius(D)withiteequallitude, and B movinghe opr witiform angular velocity , the tencord ishedirection

51、 with twice thenone of the abovelitude(E)解：（B）所屬情況是可實(shí)現(xiàn)的，A、C 對 B 的作用 r (A) mg l 力始終大小相等方向相反，B 保持普通彈簧振子完全相等。選（B）。，A、C 與 (B) mg cos 2 mr(C) sin 2 L(D) m 2r 2 g122v(E) m 4r 2 g 122View from above40. A uniform stick of length L and mass M frictionless horizontal surface. A po解：豎直方向受力平衡F cos / 2 mg ，s on

52、article ofmass m approaches the stick with speed v an amgcos / 2mglF 。straight line pendicular to the sticktersectsr 2l 212Mthe stick at one end, as shown above. After thebe writtenm dv u dm 0colli, which is elastic, therticle is at rest. Thespeed V of the center of mass of the stick after thecollii

53、sdtdtwhere m is the rockets mass, v is its velocity, t is time, and u is a constant.42. The constant u represents the speed of themMv(A)m(A)(B)(C)(D)(E)rocket=0v(B)M mm v Mrocket after its fuel spentrocket in its instantaneous rest frame rockets exhaust in a s ionary frame rockets exhaust relative t

54、o the rocket(C)解：由動量守恒 mv(D) m v dv Vdm ，m dm vM m3mM其中 V 為被拋出后的速度。mdv V vdm 0 ，v(E)解：由動量守恒，碰撞后桿的動量為 mv。由于質(zhì)點(diǎn)組相對質(zhì)心動量為零，所以質(zhì)點(diǎn)組的動量等于質(zhì)心速度乘以總質(zhì)量：令u V v 為的相對拋出速度，則dvdmm u 0 。dtdtmv (M m)V ，c選(E)。mVc M m v 。43. The equation can be solved to give b as afunction of m. If the rockes m=m0 and v=0 when it starts,

55、 what is the solution?選（B）。41. Article of mass mt move s along the x-axisum / m(A)0has potential energy V(x)=a+bx2，where a and b are itive constants. Its initial velocity is v0 at x=0. Itu expm / m(B)0will execute simple harmonic motionfrequency determined by the value ofwiu sinm / m(C)0(A)(B)(C)(D)

56、(E)b aloneb and a alone b and m aloneb, a, and m aloneb, a, m, and v0u tanm / m(D)0(E)None of the above.解：由前題結(jié)論m dv u dm 02bm解：由一維諧振子公式， 。選（C）。dtdt得dmmdv顯然與 a 無關(guān)，因?yàn)閯菽艿牧泓c(diǎn)選取有任意性，或者說質(zhì)點(diǎn)受力 F 2bx 與 a 無關(guān)。dx 。udV兩邊積分得 vt v0 u m em t ，Questions 42-430The equation of motion of a rocket in free s ce can因此131MR

57、 gh2M2 ghm0m0 v u ln u lnv t。mt mt 0選（E）。(D) Mgh44. The period of a hypothetical Earth salite1orbiting at sea level would be 80 minutes.erms of(E)Mgh2解：圓環(huán)的轉(zhuǎn)動慣量為 I MR 2 。由能量守恒，the Earths radius Re, the radius of a synchronoussalite orbit (period 27 hours) is most nearly(A)(B)(C)(D)(E)3 Re7 Re18 Re320

58、 Re5800 Re1 Mv 2 1 I 2 Mgh ，2由無滑滾動，2R v ，解：由向心力公式聯(lián)立上面兩式，解得m 2r GMm ，gh r 2。R2其中 ，得T所以角動量4 2ghT23L I MR 2 MR gh 。r ，GMR選（A）。周期比為27 6081，80446. Article is constrained to move along the x-axisunder the influence of the net force F=kx with由此得半徑之比約為 7.4。選(B)。或者直接用 Kepler 第三定律，周期的平方與半徑的立方之比為常數(shù)。但是要切忌，這只是對繞

59、同一天體運(yùn)行的星體適用。litude A and frequency f, where k is aitiveconstant. When x=A/2, the(A) 2fArticles speed is(B)3fAR(C)2fAh(D) fA1(E)fA31解：質(zhì)點(diǎn)顯然做簡諧振動，勢能為 kx 2 。由能量45. A hoop of mass M and radius R is at resthetop of an inclined plane as shown above. The hoop2rolls down the plane without slip. When the守恒，1

60、kA2k()2 1 mv 2 ， 1Ahoop reaches the bottom, its angular momentumaround its center of mass is2222解得(A) MR gh14(E) 10解：功率全消耗在克服摩擦力上，所以3k4mA 3 A 。v 2UI120 9Pf vv摩擦系數(shù)為 108 N ，因?yàn)?01f T ，fN108 1.08 1.1。所以100選(D)。v 3fA 。選(B)。 0Rv047. A system consists of two chargedrticles ofIPequal mass. Initially thertic