物理经典力学

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：圆轨道受到小扰动后仍是闭合轨道，因此是椭圆。选（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：物体速度最大时 0 ，即加速度为零，此dt时物体受力平衡。设速度最大时弹簧压缩量为 x，则( A )Horizontal( B )Horizontalmg kx ，VerticalVerticalmgkx 。( C )Horizontal( D )Horizontal由机械能守恒11mgh x mv kx22。Vertical22将 x 的表达式代入，( E )Horizontal2 2m g 1h x 1Ek2 mv mg2kx mgh222k解：因为球在水平方向不受力，所以其水

5、平动量不变，即速度的水平分量不变，只有（E）符合。选(E)。选（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：物体重 20N，角度10M arctan arctan 0.5 。20v 2m(m M2k g)()(D)选（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：射入过程很短，时间及此过程中所走过nail, the average force exertednearlyo the wood is most距离可忽略不计。由动量守恒，mv m M V 。(A)(B)(C)(D)(E)10 N100 N1000 N10,000 N100,000 N之后由于摩擦做匀摩擦热，则运械能全部转换为1 m M V 2m M gs ， k2解：力对石头做功等于其动能减少，F 1 mv 2 / h 1 5 102 / 0.025 10000N 。V 2v 2ms ()()2 。2选（

9、D）。22k g2k gm M选（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：由冲量定理由平行轴定理，圆盘绕桌子中心的转动惯量为1I I 0 MR 2 Mr MR ，所以它相对于洞的角动量为222 Fdt P ，11L I M (R 2 r 2 ) 2 2 。220 103 1200 600mVtF 240 N 。选（E）。60选(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，所以冰球相对洞的角动量守恒，I 正确。在圆运动中动量本身就不是一个守恒量，变化中冰球又显然受力，所以动量守恒无从谈起，II 不对。变化中冰球受到了指向洞的力，沿此方向又有位移，所以绳子对其做功，动能不守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。或者由动能为I 2 ，而角动量守恒，2I constant ，动能显然不能守恒，III 不对。选（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，圆盘绕自身圆心的转动惯量为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：刚体定轴转动的动能为1T I ，22kq 2r 80 9(E)动能变化为T 1 I 2 2 1 4 802 402 9600J0解：因为 r0 处势能最小，由22。dUkq 29bP E 0选(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：由刚体定轴转动的动力学方程M I ，其中 M 为外力矩，为刚体的角加速度。对于本题

15、9b8kq 2r0，kq 2r 8b 0 。9选（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选（D）。Questions 14-15A nonrelativisticrticle of mass m moves in aplane. Itsitionis described bythe polarr& and & .解：因为触地点相对于圆心做匀速圆周运动，所以相对于圆心的加速度为竖直向上方向，而圆心做匀速直线运动，加速度为零，所以触地点相对于圆心的加速度就是其对地的加速度。选(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：由机械能守恒(E)1 mv 2 mgh ，2解：对于有系，Lagrange 函数定义为L T V 。2h v 2 0.82 0.64 。本题中，由于极坐标系下的速度表达式为v r&e

19、r r& e 。而两分量之间垂直，所以h v 选(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选(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 与无关，2km2m(x x0 )2(D)为有心力场，角动量守恒。利用上题中速度的表

21、达式，径向速度为 r& ，所以角动量表达式为mm112km1(x x)2(E)mr 2& 。0mm m选(E)。212解：设恢复原长时两物体速度为 v1、v2，由动量守恒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由机械能守恒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联立以上二式，解得kmv 2 (x x。)210mm m112选(D)。s is justified.解：角动量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为矢径 r 在 dt 时间内扫过的面积。由本题所给条d度为 2，而且相互解：二维运动，每个粒子的件，为常量，故角动量守恒。有心力可保证dt角动量守恒。选（A）。之间无关联，没有限制条件，所以总数为 26=12。选（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：一维谐振子公式，T 2。relationships betn tX and tY is correct?选（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vector in equal times are equal. What can be爲 f6mg sin f ma 爲 f fr I r a vcos2 0g(C)2vtan20g(D)2vsin cos20g(E)解：飞行时间为mg sina 2v sinIr 2t 0 ，gm 1水平位移

27、为 I mr 2222v 2 sin coscos t 0 。S va g sin 0g3 mr C选（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：临界状态质点所受万有引力全部用来提供所需向心力：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：力场为保守场，能量守恒，I 正确。又为有心力场，且 F 指向原点，相对于原点的力矩为零，II正确。显然相对于原点的角动量守恒，III 正确。选（E）。12 3 2T G 。选（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：因为所受力为保守力，机械能守恒。势能最小时速度最大：(C)b221mvmax ，22xmax(D)解得8k3mv0 (E)bMv1。max解：质心动量等于各部分动量之和。碰撞前后动量守恒，选(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选(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：取无穷远为势能零点，则万有引力势为1m3GMGMU dr 。r 2r2 mr选（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：因为门的重量完全由 U 点支撑，所以 L 点无竖直方向的力，L 处受力沿水平方向。系统对 U点力矩平衡，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选（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：因为系统没受到相对圆心的外力矩，所以相对于圆心角动量守恒，最后转桌必处于状态。小虫和转桌间有相对运动，转桌可有任意的角位移。选（C）。9UL(C) gr9.8 5 105ve 1565 。(D) /(E) /2解：由能量守恒，势能转换为转动能量222选(C)。mg l 1 I 2 。32. A solid ball weighs 5.0 newtons in air and 3.022

40、对于质量为 m 长度为 l 的均匀杆，相对一端的转动惯量为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：由重力与浮力和拉力平衡可见只与杆长有关。第二次杆长变为一半， G F 3.0N ，水 2 。选（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水选（D）。解：所谓逃逸速度，是恰好使物体的机械能为 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：直接对势能求导得 dU由 Bernoulli 方程， 4kx3 。F12v P gh 1 v 2 Pdx2 gh ，111222

44、2选(B)。从量纲上看，（A）、（C）为能量，（D）为1能量乘以距离，均不是力的。P1 P2 2 (v2 v1 ) g(h2 h1 ) 。22代入数字计算，选（B）。现场有可能不记得36. The Hamiltonian function for this system isBernoulli 方程，其实它是机械能守恒在流体力学中的一种体现。再结合量纲分析，很容易推出这个形式。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：对于变换方程不显含

46、 t 的有系（这一条件(C)The acceleration of the rock is always lessg.n在 SUB中几乎必然满足），H=T+V。选（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：在顶点，石块的速度为零，此时空气摩擦力为零，故加速度为 g，选（B）。对于选项（E），石块有可能获得收尾速度，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从另一个角度看 F 等于重力和离心力的合力(E)xmv2dxx1F mg 。1 r4 222解：本题涉及分析力学的变分原理部分。由Hamilton 原理，定义 Hamilton 作用量t2S Ldt ，t选（E）。BAkkC1则真实运动使 S 取极值，其中 L 为L T V 。选（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）所属情况是可实现的，A、C 对 B 的作用 r (A) mg l 力始终大小相等方向相反，B 保持普通弹簧振子完全相等。选（B）。，A、C 与 (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：竖直方向受力平衡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：由动量守恒 mv(D) m v dv Vdm ，m dm vM m3mM其中 V 为被抛出后的速度。mdv V vdm 0 ，v(E)解：由动量守恒，碰撞后杆的动量为 mv。由于质点组相对质心动量为零，所以质点组的动量等于质心速度乘以总质量：令u V v 为的相对抛出速度，则dvdmm u 0 。dtdtmv (M m)V ，c选(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?选（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：由前题结论m dv u dm 02bm解：由一维谐振子公式， 。选（C）。dtdt得dmmdv显然与 a 无关，因为势能的零点选取有任意性，或者说质点受力 F 2bx 与 a 无关。dx 。udV两边积分得 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选（E）。(D) Mgh44. The period of a hypothetical Earth salite1orbiting at sea level would be 80 minutes.erms of(E)Mgh2解：圆环的转动惯量为 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由无滑滚动，2R v ，解：由向心力公式联立上面两式，解得m 2r GMm ，gh r 2。R2其中 ，得T所以角动量4 2ghT23L I MR 2 MR gh 。r ，GMR选（A）。周期比为27 6081，80446. Article is constrained to move along the x-axisunder the influence of the net force F=kx with由此得半径之比约为 7.4。选(B)。或者直接用 Kepler 第三定律，周期的平方与半径的立方之比为常数。但是要切忌，这只是对绕

59、同一天体运行的星体适用。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：质点显然做简谐振动，势能为 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摩擦系数为 108 N ，因为101f T ，fN108 1.08 1.1。所以100选(D)。v 3fA 。选(B)。 0Rv047. A system consists of two chargedrticles ofIPequal mass. Initially thertic