2013泛珠三角及中华名校物理奥林匹克邀请赛答案.pdf

1 Part I Q1 9 points Rotational inertia of the beam 杆的转动惯量的一般表达式 2 2 1 1 23 12 3 l l l l m I l ll m Ldll L a 2 0 3 m IILL Conservation of angular momentum 角动量守恒 2 00 0 2 3 4 Lmvv ImLmLv IL mL Energy lost 动能损失 22222 00 111 8 222 EmvImLmv 3 points b 2 317 4448 IILLmL or 22 2 2 7 1241216 11 4 1 8 ImLmLmL L m Conservation of momentum and angular momentum 角动量 动量守恒 0 0 2 0 0 2 6 5 2 4 4 mvmv v L LL Immv vv Energy lost 动能损失 2 22222 00 111 10 2224 L EmvImmvmv 3 points c 2 111 2212 IILLmL Conservation of momentum angular momentum and energy 角动量 动量 能量守恒 0 012 0 201 2222 0 2102 12 5 2 225 11113 22225 v mvmvmv L vLL mvmvv v Imvmvmvv I 3 points Q2 10 points a 1212 1 121 33 121 2 212 2 mmmm mrrrm rrr rrrr GG 1 point b 21 12 1212 cc mm rrrrrr mmmm 2 12 121212 33 12 00 cc mm rr m mm mmm rGrrGr mmrr 2 points c 0 c r 1 point d The equation for ris the same as a uniform circular motion r满足的方程与匀速圆周运动满足 的方程一致 Therefore 因此 122 12 23 2 G mmG mm a aa 00 cos sin r tat xat y 3 points e 12 mmm 1 31 3 3 2281130 22 1 3 279 229 106 7 104 10 442 6 116 101 8 10 m aT Gm Ta Gm 3 points Q3 12 points a 2 qQR xxRx 2 2222 00 1 1 4 4 QqQ Rx f xxxR 22 22 0 0 1 8 L L Q Wf x dx R RL 3 points 不用积分扣 2 分 b Since the outside charge does not create any field inside the conductor sphere the work done here is no different from the one in a 外电荷对导体内部无影响 所以结果与 a 一样 22 22 0 1 8 L e W R RL 2 points c The static electric field on the charge 2 qLQ R outside the conducting sphere is equivalent to a field generated by three point charges 1 qQ at 0 0 image charge 22 qq R x at 2 0 lRl and 22 qq at 0 0 Therefore 球外 2 qLQ R 受的静电力来自以下三个点电荷 1 qQ 在 0 0 镜像电荷 22 qq R x 在 2 0 lRl 以及 22 qq 在 0 0 22212 22 0 1 4 q qq qq f x xxx 2 2212 2223 0 1 4 q xRqxqRq xRx 2 points And the work done by electric field is then 电场做功 3 2 22 2221 323 0 2 1 8 RL L qLq LqRq f x dx RL RR 422222 5325 0 1 2 8 L QLRL Q RR LR 242242 522 0 1 22 8 L LL RR Q RRL 3 points d in c WW 2 points Q4 9 points a cosrr t vc sinsin cos rcv v tcv 2 points 0 9 0 707 1 75 1 0 9 0 707 c vc 1 point b sin 0cos0 cos s s s v v vvv vv 2 points Thus cos s vv 1 point c Let the emission angle of the beads be Then to ensure the net velocity of the beads is along the Y direction we have 令小球的发射角为 为使球的合速度沿 Y 方向 须有 sinsin b vv 2 coscos co1os sc bbbb vvvvvvv v cos b rr t vv 2 2 sin cos 0 1 1 bb bb vvvv v vvv v 2 points So it will not happen 不会发生 1 point Q5 a The initial number of mole of air in the tire is 轮胎内原有气体摩尔量 0i i a PV n RT The final number of mole of air in the tire is 打完气轮胎内有气体摩尔量 0f f a P V n RT So number of mole of air pumped into the tire is 因此 打进轮胎的气体摩尔量为 r v b v b v 4 0fi fi a PP V nn RT Inside the compressor this amount of air has volume 在压缩机里 此摩尔量的气体的体积为 0 The work done by the compressor is hence 压缩机做功 0ccfi WPVPP V 1 point The internal energy of the gas now in the tire increases by 轮胎内气体内能增加为 0 1 fiVacfiafi R Enn C TWnn TPP V 1 point The maximum temperature is then 因此这时的温度 0 0 1 1 The maximum pressure is 气压 0 1 1 The minimum Pc required is 最小 Pc必须满足 1 cmaxfi PPPP 2 points b The total number of stokes is 打气总次数 0 fi fi apaap PP Vnn N PVRTPV 1 point During the j th stroke in the adiabatic compression inside the pump from Pa to Pj 第 j 个斯托克循环绝 热压缩使压强从 Pa 变为 Pj where V is the volume of the air inside the pump after the adiabatic compression V 压缩后气体体积 The internal energy of the air at this moment is 此时气体的内能 1 1 1 1 1 The amount of work done to inject this amount of air into the tire is 打气所做的总功 1 Hence the change in internal energy of the air inside the tire during the j th stroke is 内能的变化 1 1 111 a jjjjp j P UPVPVPVPV P 2 points 5 On the other hand 另一方面 1 1 1 1 0 1 1 0 1 1 1 So 因此 1 0 1 Replacing the finite difference equation by differential equation and integrate 改用微分形式表示 1 0 1 1 0 1 0 We have 我们得到 1 1 0 1 1 1 1 1 1 fi a maxi ia PP P PP PP 3 points 6 Part II Q1 20 points a 2 00 2 2rBJ rBJr Energy per length is 单位长度能量 2 2 0 0 0 1 2 8 R I WBrdr Compare with 相比较 2 0 1 24 WLIL H m 1 point b Newton s second law gives us the differential equation 牛顿第二定律引出的方程 1 point c 0 0 0 1 e E x im 2 points d The current density is given by 电流密度 2 0 1 enE Jnex im 2 points e 2 2 1 IenV rim D so 22 1 imD VI ren 2 points f The real part of the impedance represents the resistance 阻抗实部 22 Dm R r en 1 point The imaginary part represents 虚部 22 I D m L r e n so 22 I Dm L r e n 1 point g 9 2 0 10 I LHR 7613 0 1 0 10101 0 10 4 Faraday LDH 2 points h 1 1 1 1 1 1 7 11 2 20 m mmm V i C VVV LC 5 points i 2 1 2 0 1 1 sin 42 ka LC 2 points j Yes 是 1 point Q2 30 points a Largest最大 compressed压缩 Smallest最小 stretched伸展 or vice versa反之亦然 1 point b Joule s N m s kg m2s 3 Bbkg2m4s 6 5 1 point 5 2 4 6 1 point c 1 2 2 1 2 1 2 2 2 with 1 2 3 1 2 2 Kinetic energy 12 12 M M M MM 12 MMM 222 32 32 22 112 3 2 1 2 11 22 1 22 MG KEM GM M MRMTR R 2 points d Neutron binary 双中子星系统 2 32 35 3 1 2 3 k dE M GMT dT 1 point kk G dEdE dT L dtdT dt 1 point 5185 5 3333 1212 3 2 dT G c M MMMT dt 1 point 8 15 676046760814 33 1 4 104 10 2 3 10 1 4 104 100 79 3 4 101 5 10 dT dt 1 point e For Sun Earth system 日 地系统 with 1s 5 673024121325 3 1 4 102 106 10 365 24 86400 1 7 103 2 105 4 10 dT dt 1 point 2525 2518 6 1 8 101 8 10 1 8 105 7 10 365 24 864003 155 10 dt tsyy dT 1 point f 2 1 point g 2 s RGMc or 21 s R c GM 1 point Also 2vR 1 point 6 55 5246542226 5 401153 161616 3 1610105 9 10 Order of magnitude only 6 7 Gss cvc LGc M RGc c G R R v GcG W 1 point 数量级正确即可 h 0 2 1 2 points i 2 1 2 with 104light year 9 46 1019 2 1 49 103 32922 1 5 1 5 10 10 0 952 4 10 s RR 3 points j 1 2 6m 1 1 1067Hz 2 points k 1st resonance 第一个共振态 1 2 1 2 24 2217 233 1142 1 4 101 3 3 10 221 1 1021 07 10 B B k T mxk Txm m 4 point