Multiple Chices: HKPhO 香港物理奧林匹克 04. Answer: A r D. The phne is min t cnstnt elcity. The nly frce ctin n the phne is the erth rity. Remrk: The nswer is D. Hweer, since the questin sks fr "instntneus sclr redins", it is nt unresnble t ls cnsider A s lid nswer.. Answer: D. Ste : h = (/) (.5 ) (5) = (.5 ) m Ste : = (.5 ) (5) =.5 ms ; h = (.5 ) / = (78.5 ) m Mximum Heiht = (.5 + 78.5) = (09.75 ) m. Answer: C. = / r West = 5 / 5 West = 5 ms West 4. Answer: B. F = (0.0 x 00 x 600) / 60 = 0 N. 5. Answer D. Apply Newtn's lw f uniersl rittin t the rth s ner side nd its fr side. 6. Answer B. m Bttm Wire: T ct cs r m Bttm Wire: T T 8. N sin 7. Answer. Wrk Dne = Are under the fllwin cure. 8.4 N 60 50 40 F(x) 0 0 0 0 0 4 6 8 0 x 8. Answer. m Apply T t perfrm clcultin. k
9. Answer D. Tw nrml mdes. HKPhO 香港物理奧林匹克 04 First Mde: f l Secnd Mde: Cnsider the center f mss f the msses bein unchned, the effectie sprin cnstnt m is K K K m K K f l m l m 0. Answer A. GMm m r r T G. Answer C. liquid Fliquid F wter liquid. Answer D. nery Cnsertin: ( m M ) ( m M ) h - 0. ms Mmentum Cnsertin: m ( m M ) b b 6.9 ms Initil Kinetic nery: K m b 9.8 J -
. Answer B r C. x 0 t cs y 0t sin t When the bll hits the inclined plne in, y x tn limintin x, 0t sin 0t sin t 4 sin t x cs t x 4 s cs 4 sin sin cs HKPhO 香港物理奧林匹克 04 Remrk: Students redin the nlish ersin my chse nswer B. Students redin the Chinese ersin my chse nswer C. Due t this discrepncy, bth nswers re ccepted. 4. Answer A. 5. Answer. Cnsider the system with spheres:. sphere f density nd rdius R with y = 0 (center f mss). b. sphere f density nd rdius R/ with y = R/. c. sphere f density 5 nd rdius R/ with y = R/. (b) nd (c) re equilent t sphere f density 4 nd rdius R/ with y 4 = R/. Center f mss f the new sphere: Y R 6 m y i m i i R 4 4 R (0) 4 R 4 4 R 4 R 6. Answer: C. Ptentil nery:
A z dz Ah HKPhO 香港物理奧林匹克 04 A h 5007.8h.9h [J] [Wh] 7. Answer D. T l TM T 6 M T 8. Answer B. GM rthm m Usin the equtin nd perid f the rth T = 86400 s t clculte the heiht f R R esttinry stellite. 9. Answer A. Fr lre nle scilltin, the equtin f mtin becmes nnliner, nd the pprximtin sin is nw n lner hld. The restrin frce is prprtinl t sin, rther thn ; therefre its mnitude is less thn in the cse f SHM. A weker restrin frce ls results in slwer scilltin, tht is, the perid becmes lner. 0. Answer C. It is plt f GM R, tht is, inst (M/R ). 4
Open ended Questins HKPhO 香港物理奧林匹克 04. Venus Trnsit P () Kepler Lw: Cnstnt, P = Orbit Perid, = Orbit Rdius rth Venus rth Venus 65 5.806 AB AV AV (b). 67 AB AV AA AV rth Venus AB.67800 km 479 km (c) (d) Dimeter f the Sun = 479 km x 90 =.7x0 6 km Let = elcity f rth reltie t Sun Since plnetry elcity is ien by. 750 V. GM Sun, elcity f Venus reltie t Sun = r As bsered frm rth, elcity f Venus = 0. 750, nd elcity f Sun =. V Prjected n t the surfce f Sun, elcity f the shdw f Venus =. 0. 46 V V Hence the elcity f the shdw f Venus sweepin n the surfce f Sun =. 46 V V r ( )(.5 0 ) 9886 m/s r T (65)(4)(60)(60) GM r Sun (6.67 0 )(.99 0.50 0 ) 9747 m/s Time difference = 479 (.46)(9886) 479 7 s. min r 8 s. min (.46)(9747) 5
. Terminl Velcity f Free Fllin Object () d m dt F Grity F d Dr Frce m HKPhO 香港物理奧林匹克 04 At lw elcity limits, F 0. d m F dt d d At hih elcity limits, m 0. dt L t F F d m H m (b) m Cd A m C A d 7. ms - 0 9.8 0.0 0.5. (c) t 7. 0.7 s 9.8 6
. Ttl nery in Surfce We HKPhO 香港物理奧林匹克 04 z z / x / x () Gin in ptentil enery V frm z t z f n elementl mss m = x z: m z zx z V Ttl Ptentil nery: V x x0 x x0 x x0 z A zasin x z0 sin z dz dx zasin x x dx dx x x sin 4x Gien tht sin dx 8, V x A A 4 / 0 Ttl P per welenth: V A 4 (b) Assume equiprtitin f enery (P = K), the ttl enery er whle welenth A 7
(c) Pwer f We Perid: HKPhO 香港物理奧林匹克 04 P w A A TA 8 TH T V T T = Welenth T = Perid We mplitude is hlf f the we heiht 8
4. A Slidin Blck up Slpe Pltfrm HKPhO 香港物理奧林匹克 04 () Mmentum: m cs m M Rise in Center f Mss, h : Ttl Ttl m cs m M h L sin45 h L h sin45 h h (b) h h h h h L sin L 45 sin45 Applyin the trinmetric identity: sinx y sinxcsy csxsiny h h L sin nery: m m m cs m m M K m cs m M mh h m M m M m M m cs m M m M m cs m M m cs m M Ttl h m cs m M h Lsin, m cs m M m M mcs h mh mh 9
5. HKPhO 香港物理奧林匹克 04 The center f mss is rised by: h b L L cs The center f mss is rised by: h L L cs Ptentil nery, L Lcs P m L Lcs ml Lcs ml cs ml x L x ml L m x 4L Kinetic nery, 7 K m 4 P + K = Cnstnt m x 4L 7 m 4 Cnstnt (b) The ttl enery is equilent t tht f mss sprin system with n effectie mss f 7 m meff m nd n effectie sprin cnstnt f keff. 4L Hence k m eff eff T m L 7m 7L 8 8 7L 0
(c) HKPhO 香港物理奧林匹克 04 Assume tht the initil elcity is 0. When the initil displcement is x 0, the simple hrmnic mtin is ien by x x 0 cs( t) When x =x 0 / fr the first time, x0 x cs( t) 0 cs( t) t 7L T t r t 8 6