1. Length of Daytime (7 points) 白昼长度 (7 分 )

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1 Pn Perl River Delt Physics Olympid 年泛珠三角及中华名校物理奥林匹克邀请赛 Sponsored by Institute for Advnced Study, HKUST 香港科技大学高等研究院赞助 Simplified Chinese Prt- (Totl 6 Problems, 45 Points) 简体版卷 -( 共 6 题,45 分 ) (9:00 m :45 m, 22 Februry, 208) Plese fill in your finl nswers to ll problems on the nswer sheet. 请在答题纸上填上各题的最后答案 At the end of the competition, plese submit the nswer sheet only. Question ppers nd working sheets will not be collected. 比赛结束时, 请只交回答题纸, 题目纸和草稿纸将不会收回. Length of Dytime (7 points) 白昼长度 (7 分 ) () In the figure, ABCD is rectngle lying on n inclined pe mking n ngle q with the horizontl pe. ABEF is the projection of the rectngle on the horizontl pe. If the mesure of the ngle DAC is f, derive n expression for the ngle. [2] 如图所示, 矩形 ABCD 位于斜面上, 斜面与水平面夹角为 q ABEF 为该矩形于水平面的投影 设角 DAC 为 f, 试推导角 的表达式 [2] D C f F q E A B Let h = AC. Then CE = h sin α. BC = AD = h cos f. CE = BC sin θ = h cos φ sin θ. Equting the expressions of CE, h sin α = h cos φ sin θ Þ α = rcsin(cos φ sin θ). (b) The ecliptic is the pe on which the Erth revolves round the Sun. The xis of rottion of the Erth is inclined t n ngle of 23.4 o with the norml to the ecliptic. The dy of the Winter Solstice (in the Northern Hemisphere) is 2 December. Using the result of () or otherwise, clculte the incident ngle of sunlight reltive to Erth s equtoril pe tody (22 Februry). [2] 黄道面是指地球围绕太阳公转的平面 地球的自转轴相对于黄道面的法线倾斜, 角度为 23.4 o 在北半球, 冬至的日期为 2 月 2 日 试用 () 部结果或其他方法, 计算今天 (2 月 22 日 ) 阳光相对于赤道面的角度 [2] In the figure bove, consider ABEF to be the equtoril pe of the Erth, nd ABCD the ecliptic. Then q = 23.4 o. When the Erth revolves round the Sun, sunlight is incident on the Erth from different directions lying on the pe ABCD. For exmple, on 2 December, sunlight is incident on the Erth in the direction AD, since this is the southernmost direction of sunlight. Similrly, during Spring Equinox nd Autumn Equinox, sunlight is incident on the Erth in the direction AB or BA.

2 Axis of rottion 自转轴 Equtoril Pe 赤道面 q Winter Solstice 冬至 Tody 今日 On 22 Februry, the time is 63 dys fter the Winter Solstice. Hence on the ecliptic, sunlight is incident from the ngle φ = : :9; = 62.48, nd reltive to the equtoril pe, it is incident from the ngle α = rcsin cos φ sin θ = rcsin[cos sin( )] = (c) The ltitude of Hong Kong is b = o. Clculte the length of dytime in Hong Kong tody (22 Februry). Give your nswer in hours to 3 significnt figures. [3] 香港位于北纬 b = o 试计算今天 (2 月 22 日 ) 香港白昼的长度 答案请以小时表达, 给三位有效数字 [3] P R Q P b Sunlight Q R S Ecliptic Tody Ltitudinl Cross Section t b b Consider the the ltitudinl cross section t t b. PR = R cos β. (R is the rdius of the Erth.) QR = R sin β tn α. Hence the ngle QRS is given by cos x = JK LK = M NOP Q RSP T M UVN Q = tn β tn α. x = rccos(tn β tn α) = rccos(tn tn ) = Length of dytime in Hong Kong = 24 Y;.;Z =.4 h [Y8 2. Rotting Bll (6 points) 滾動的球 (9 分 ) A bll of mss m nd rdius is t rest on the surfce of sphere with rdius R. The bll is initilly t the ngle θ 8 t t = 0. The bottom sphere is fixed nd cnnot move, but there is friction so the bll rolls without slipping until it leves the surfce of the sphere t ngle θ [. 2

3 The moment of inerti of the bll bout its xis is I = ; m. The grvity is directed s shown. 一个质量 m 和半径 的球放在一个半径 R 的球体表面上 球最初处于角度 θ 8 处 底部球体是固定的并且不能移动, 但是存在摩擦力, 因此球作纯滚动, 直至它以角度 θ [ 离 开球体表面 球围绕其轴线的惯性矩是 I = ; m 引力方向如图所示 () Wht is the frictionl force cting on the bll t ngle θ? [3] 在角度 θ 时作用在球上的摩擦力是多少?[3] (b) Wht is the norml force cting on the bll t ngle θ? [3] 在角度 θ 时作用在球上的支持力是多少?[3] (c) Find θ [ in terms of θ 8. [] 试推导 θ [ 的表达式, 用 θ 8 表示 [] (d) Wht is the speed of the bll t ngle θ [? [2] 在角度 θ [ 時球的速率是多少?[2] & $ % ' "! # Solution: () (b) $ " # Δ& " Δ" " + Δ" If the bll is rolling without slippering, RΔθ = Δφ θ + Δθ + Δφ θ ω = dθ Δt dt + dφ dt = dθ dt R + Δθ = vδt + R 3

4 v = R + Δθ Δt = ω By the conservtion of energy nd Newton s 2 nd lw, we hve mg R + + cos θ 8 mg R + + cos θ = 2 mv + 2 Iω mg R + cos θ 8 cos θ = 2 mv + 2 v m 2 5 = 7 0 mv v = 0 7 g R + cos θ 8 cos θ () From Newton s 2 nd lw on rottion f = I dω dt f = 2 dω m 5 dt Newton s 2 nd lw long the tngentil motion, mg sin θ f = m dv dω = m dt dt = 5 2 f f = 2 mg sin θ 7 Another wy to get the result is from eqution () without using the Newton s 2 nd lw. Since v = R + lm, Eq. () implies R + dθ = 0 dt 7 g R + cos θ 8 cos θ Differentite on both sides, 2 R + dθ dt And (b) Newton s 2 nd lw gives The norml force is d θ dt = 0 7 dθ g R + sin θ dt d θ dt = 5 7 f = 2 dω m 5 dt = 2 5 m + R d θ dt = 2 mg sin θ 7 mg cos θ N = mv R + g sin θ R + N = mg cos θ mv = mg cos θ m 0 R + R + 7 g R + cos θ 8 cos θ = mg( 7 7 cos θ 0 7 cos θ 8) (c) At θ [, the norml force vnishes, we hve (d) the velocity of the bll t the ngle θ is cos θ [ = 0 cos θ 7 v = g R + cos θ 8 cos θ [ = 7 7 g R + cos θ 8 4

5 3. Nerest Exopet Discovered (7 points) 发现最近的系外行星 (7 分 ) On 24 th August 206, stronomers discovered pet orbiting the closest str to the Sun, Proxim Centuri, situted 4.22 light yers wy, which fulfils long-stnding drem of science-fiction writers: world tht is close enough for humns to send their first interstellr spcecrft. 206 年 8 月 24 日, 天文学家发现在距离太阳最近的恒星 比邻星 (Proxim Centuri), 有一颗行星围绕着它运行 比邻星距离太阳 4.22 光年 这发现实现了科幻小说作家的长期梦想 : 一个足够接近的世界, 人类可以把第一艘星际航天器送达 Astronomers hve noted how the motion of Proxim Centuri chnged in the first months of 206, with the str moving towrds nd wy from the Erth. In the figure below, the rdil velocities of the str re mesured nd the direction of the rdil velocities chnged regulrly. This regulr pttern cused by n unseen pet, which they nmed Proxim Centuri B, repets nd results in tiny Doppler shifts in the str s light, mking the light pper slightly redder, then bluer. 天文学家注意到比邻星的运动在 206 年的头几个月的变化 恒星规律性地朝着和远离地球移动 在下图中, 透过测量恒星的径向速度, 發現径向速度的方向有规律地变化 径向速度的规律性变化是由一颗看不见的行星引起的, 该行星称为比邻星 B, 会重复导致恒星光线发生微小的多普勒频移, 从而使光线稍微变红, 然后变蓝 It it is given tht the str, Proxim Centuri, hs surfce temperture of 3000 K nd rdius of R = 0.4R qrs nd the orbit of the unseen pet, Proxim Centuri B, round the str is circulr. (Rdius of the Sun R qrs = Y m, the grvittionl constnt G = w[[ m : / kg s ) 恒星比邻星的表面温度为 3000 K, 半径 R = 0.4R qrs, 行星比邻星 B 围绕恒星的轨道是圆形 ( 太阳半径 R qrs = Y m, 引力常数 G = w[[ m : / kg s ) Figure: Mesurements of the rdil velocity of Proxim Centuri (denoted s RV in the figure) from Jnury 206 onwrds. The blue curve is the best fit curve of the dt. 图 : 从 206 年 月 日起, 比邻星的径向速度 ( 在图中以 RV 表示 ) 测量结果 蓝色曲线是数据的最佳拟合曲线 () Proxim Centuri is red dwrf str, unlike our Sun, with mss of only 0.2 M qrs. Estimte the rdius of the pet s orbit using the given informtion. (The mss of sun, M qrs = :8 kg) [2] 比邻星是一颗红矮星, 与我们的太阳不同, 质量仅为 0.2M qrs 试用所给资料估算行星轨道的半径 ( 太阳的质量 M qrs = :8 kg) [2] 5

6 (b) Estimte the mss of the pet in terms of Erth mss. (M ~ n = kg) [2] 试估算行星的质量, 以地球质量为单位表示 ( 地球质量 M ~ n = kg)) [2] (c) Estimte the equilibrium temperture of the pet by ssuming tht both the str nd pet re blck bodies. [3] 假设恒星和行星都是黑体, 试估算行星的稳态温度 [3] Solution: () The period is the time intervl between two consecutive peks of the curve. From the rdil velocity curve, the period is dys (Full mrks for the period within ±2 dys) Using Kepler s 3 rd lw: (Assuming M n M ˆs n ) T : = 4π = Œ m GM n (b) The orbitl velocity for circulr orbit is: v = 2π T = 47200m/s Here is the distnce between the str nd the pet, tht is, the sum of the orbitl rdii of the pet nd the str, nd we ssume tht the orbitl rdius of the str is negligible. In the center of mss frme, p Ž = 0. Mv mv = 0 From the figure, v = 6.5 km/h =.806 m/s. m = v M =.806 v ( :8 ) = 9. 0 kg.5m ~ n (c) By ssuming therml equilibrium, the het bsorbed by the pet (due to the rdition of str) is equl to the het rdited by the pet (s blckbody) 4πR σt 4π πr = 4πR σt T = R 4 T q T = T 4. L-Shped Conductor with Wire (8 points) L 形导体和导线 (8 分 ) R 248K = 25 C 2 A L-shped conductor consists of two semi-infinite conductors in the xz nd yz pes where the cross section is shown in the figure. The L-shped conductor is grounded nd centered t the origin. A line of chrge, with liner chrge density runs prllel to the z-xis is locted t (, b) where b > > 0. L 形导体由 xz 和 yz 平面中的两个半无限平面组成, 图中显示了导体的横截面 L 形导体接地, 中心点为原点 一条线性电荷密度为 与 z 轴平行的电荷线位于 (,b), 其中 b > > 0 () Compute the electric potentil V(x, y, z) for x > 0 nd y > 0. [3] 计算 x > 0 和 y > 0 时的电势 V(x, y, z) [3] (b) Compute the cpcitnce per unit length of thin wire of rdius r, plced t the point (, b). Assume tht the wire rdius is much smller thn nd b (i.e. r, b) so tht the solution of prt () is pproximtely correct in the region exclusive of the conductors. [3] 6

7 计算放置在点 (, b) 处 半径为 r 的细导线, 其每单位长度的电容 假定线半径 r 比 和 b 小得多 ( 即 r, b ), 使得在 () 部的解在除导体之外的区域中近似正确 [3] (c) Compute the force per unit length on the wire (s vector). [2] 计算导线上每单位长度的力 ( 作为矢量 ) [2] " % # line chrge extends out of the pge & $! Solution: () We pply imge method by dding 3 imge line chrge in the following wys:., b : Chrge density 2., b : Chrge density 3. (, b) : Chrge density Hence the totl electric field long the x-xis (t the point (x, 0)) is E = 2πε 8 x + b x ı bȷ x + + b x + ı bȷ + x + + b x + ı + bȷ x + b x ı + bȷ = 2b 2πε 8 x + b + 2b x + + b ȷ which is long the y-direction nd hence the electric potentil which is constnt long the x- xis. Similrly, we cn show the electric potentil is lso constnt long the y-xis. By setting the potentil V 0 = 0 t the origin, the electric potentil of n infinite line of chrge t (, b) is V 8 (x, y, z) = + b 4πε 8 (x ) + y b Similrly, we cn get the electric potentil for other imge wires V [ x, y, z = + b 4πε 8 (x + ) + y b V (x, y, z) = + b 4πε 8 (x + ) + y + b V : x, y, z = + b 4πε 8 (x ) + y + b The totl electric potentil becomes, V x, y, z = (x + ) + y b (x ) + y + b 4πε 8 (x ) + y b (x + ) + y + b 7

8 (b) C = Q ΔV = Q V r, b, 0 V(0,0,0) = = 4πε 8 4πε 8 L 2πε 8 L 2b = ( + b )r 2b r + b (2 r) ( r) L (r) + 2b (2 r) + 2b C L = 2πε 8 2b r + b (c) The force on the wire is the electric force on the wire from three imge wires. From Guss s lw, the electric field creted by three imge wires t the point (x, y, 0), E = 2πε 8 x + + y b x + ı + (y b)ȷ + x + + y + b x + ı + (y + b)ȷ Substitute x, y = (, b), we get E = 4πε 8 The force on the wire is F = qe = L 4πε 8 x + y + b x ı + (y + b)ȷ (ı + bȷ) + b ı b ȷ (ı + bȷ) + b ı b ȷ F L = 4πε 8 5. A Flying Squre Loop (8 points) 一個飛行的方形環 (8 分 ) (ı + bȷ) + b ı b ȷ A squre loop of side nd mss m is mde of resistive mteril with totl resistnce R. At t = 0, the loop is locted t x = 0 nd moves with velocity v 8 x. The loop lies in the x-y pe. There is mgnetic field B = B 8 z where B 8 > 0 is constnt. In the problem, we neglect the effect of grvity. 一边长为 质量为 m 的方形环由电阻材料制成, 总电阻为 R 在 t = 0 时, 环位于 x = 0 并以速度 v 8 x 移动 环位于 x-y 平面中 有一磁场 B = B 8 z, 其中 B 8 > 0 是一常 数 在这问题中, 我们忽略重力的影响 () Wht is the induced current on the loop when the center of the loop is t the point x with velocity vx? Wht is the direction of the current? Is it clockwise/nticlockwise from bove? [2] 当方形环中心位于 x 速度为 vx 时, 环上的感应电流是多少? 方向是什么? 从上方观看是顺时针 / 逆时针?[2] (b) Wht is the velocity of the squre loop v(t) t time t? [3] 方形环在时间 t 的速度 v(t) 是多少?[3] (c) How fr does the loop trvel before stopping? [3] 8

9 方形环在停止前的行进距离是多少?[3] #! ( Solution: () When the center of the squre loop is t the point x, the mgnetic flux psses through it is, «Φ = B 8 x w The induced emf is The induced current is dx = B 8 2 x + 2 E = dφ dt = B 8 x 2 dx dt = B 8 = B 8 2 v 2x = B 8 x I = E R = B 8 R v nd the current is clockwise from bove. (b) The dissiption power of the induced current compensted by the work done to decelerte the loop, P = IE = E R = B 8 v = mvv R v v = mr The velocity of the loop is (c) Totl distnce trvelled is $ = 0 x = B 8 ³ n 8 v t dt = γ m v t = v 8 e w ² n = v 8m γ = Rmv 8 where γ = R B 8 6. The Phenomenon of the Hlo (6 Points) 光暈現象 (6 分 ) B 8 Bright hlos round the sun cn be observed s in Figure. As shown in Fig. 2, this opticl phenomenon is cused by the refrction of the sun's rys on ice crystls in the cirrostrtus, cloud genus tht reches height of pproximtely 5.5 km. 我们有时候可以观察到太阳周围的明亮光晕圈, 如图 所示 如图 2 所示, 这种光学现象是由太阳光线在卷层云中的冰晶折射而产生的, 该云层高度约 5.5 km To understnd the phenomenon of the hlo, we simplify the problem in two dimensions. In the following, we denote the ngle of incidence on n ice crystl θ, the ngle of refrction t the first interfce θ, the ngle of refrction t the exit of the crystl θ µ, nd the ngle of deflection between the ingoing nd the outgoing sun ry θ. $' 9

10 为了理解光晕现象, 我们将问题简化为两维 在下文中, 我们以 θ 表示冰晶上的入射角,θ 表示为经过第一个界面的折射角,θ 8 表示为光线离开晶体的折射角, 以及 θ 表示为入射和出射太阳光线之间的偏转角 Sun! # Ice crystl Sun! $! %! #! " Observer Fig. : A hlo surrounding the sun Fig. 2: Formtion of hlo. 2 Fig. 3: Light refrcted by n ice crystl 3 () We consider n ice crystl in the form of regulr hexgon (Fig. 3). Derive n expression for θ s function of θ, n nd n Ž, where n denotes the refrctive index of the medium i. (n =, n Ž.3) [3] () 我们考虑一个正六边形的冰晶 ( 图 3) 试求出 θ 的表达式, 并以 θ, n 和 n Ž 作为其函数表示, 其中 n 表示介质 i 的折射率 (n =, n Ž.3) [3] (b) Estimte the ngulr rdius of the hlo s mesured by the observer on the ground. [2] The identity my be useful: 试估算观察者在地面上看到光晕的角半径大小 [2] 这个公式可能有用 : d dx rcsin x = x (c) A closer look t hlo revels the light spectrum long its entire circumference. Which of the colors, red or blue, is on the inner, which on the outer side of the hlo? [] 仔细观察一个光晕, 可以在圆周上观察到不同颜色 问光晕的内侧是哪一种颜色, 红色或蓝色, 在光晕的外侧又是哪一种颜色? [] Solution: By Snell s lw, n sin θ = n Ž sin θ sin θ = n Ž sin θ n Ž sin θ : = sin θ µ θ + θ : = π 3 θ = θ θ + θ µ θ : = θ π 3 + θ 8 θ = rcsin n Ž sin π 3 rcsin n Ž sin θ θ = rcsin n Ž sin θ π 3 + rcsin n Ž sin π 3 θ (b) The hlo is observed t the sttionry point of θ 0

11 dθ dθ = n Ž sin θ n Ž cos θ n Ž sin π 3 θ n Ž cos π 3 θ = 0 θ = π 3 θ i. e. θ = π 6 nd the corresponding θ is: θ,ºop = 2.8 The ngulr rdius of the hlo is θ» = θ,ºop θ q θ,ºop = 2.8 (c) The refrctive index of red color is smller thn tht of blue. The deflection ngle corresponding to the blue color should be bigger nd hence the outer of the hlo should be blue, nd inner of the hlo is red.