Theoretical Competition: Solution Question 1 Page 1 of 7. I. Solution. r 2. r Let O be their centre of mass. Hence MR mr. GMm. GMm 2.

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1 Theoetical Competition: Solution Question Page of 7 I. Solution M R O m. Let O be thei cente of mass. Hence MR m () m M R GMm R GMm R () G M m Fom Eq. (), o using educed mass, R Hence, G( M m) GM Gm ( R ) ( R ) R( R ). ()

2 Theoetical Competition: Solution Question Page of 7. Since is infinitesimal, it has no gavitational influences on the motion of neithe M no m. Fo to emain stationay elative to both M and m we must have: GM Gm G M m cos cos R GM Substituting (4) GM Gm sin sin (5) fom Eq. (5) into Eq. (4), and using the identity sin cos cos sin sin( ), we get sin( ) M m m R sin (6) The distances and, the angles and ae elated by two Sine Rule equations Substitute (7) into (6) sin sin R sin sin R (7) R M m 4 R m () m R Since,Eq. () gives M m R R () Gm By substituting fom Eq. (5) into Eq. (4), and epeat a simila pocedue, we get R () Altenatively, sin 8 R and sin sin R m sin M sin sin Combining with Eq. (5) gives

3 Theoetical Competition: Solution Question Page of 7 Hence, it is an equilateal tiangle with 6 6 The distance is calculated fom the Cosine Rule. ( R ) ( R )cos6 R R Altenative Solution to. () (4) Since is infinitesimal, it has no gavitational influences on the motion of neithe M no m.fo to emain stationay elative to both M and m we must have: Note that GM Gm G M m cos cos R (4) GM Gm sin sin (5) sin 8 R sin (see figue) sin sin sin R m (6) sin M Equations (5) and (6): (7) sin m (8) sin M (9) The equation (4) then becomes: M m M cos mcos R Equations (8) and (): sin Note that fom figue, M m sin M R sin () () () sin

4 Theoetical Competition: Solution Question Page 4 of 7 Equations () and (): sin Also fom figue, M m M R sin R cos cos () (4) sin sin (5) cos 8 8 (see figue) cos, 6, 6 Equations () and (4): Hence M and m fom an equilateal tiangle of sides R Distance to M is R Distance to m is R R R R R R Distance to O is. The enegy of the mass is given by E GM Gm d (( ) )..(5) dt Since the petubation is in the adial diection, angula momentum is conseved ( and m M ), 4 GM d E ( ) dt Since the enegy is conseved, de dt 4 de GM d d d d (7) dt dt dt dt dt..(6) d d d d.(8) dt d dt dt 4 de GM d d d d.(9) dt dt dt dt dt R O 6 o R 4

5 Theoetical Competition: Solution Question Page 5 of 7 d Since, we have dt 4 GM d o dt d dt GM 4 The petubation fom and gives. () and. Then 4 d d GM ( ) dt dt () n Using binomial expansion ( ) n, d GM..() dt Using, d GM..() dt GM Since, d dt d 4 dt d 4 dt Fom the figue, coso, 4.(4).(5).(6) d (7) dt 4 4 5

6 Theoetical Competition: Solution Question Page 6 of 7 7 Angula fequency of oscillation is. Altenative solution: M m gives R and G( M M ) GM. The unpetubed adial distance of is ( R R) 4R R, so the petubed adial distance can be epesented by R whee R as shown in the following figue. Using Newton s nd GM d law, ( R ) ( R ) ( R ). / { R ( R ) } dt () The consevation of angula momentum gives ( R) ( R ). () Manipulate () and () algebaically, applying and binomial appoximation. GM d R { R ( R ) } dt ( / R) ( R ) / GM d R {4R R} dt ( / R) ( R ) / GM ( / R) d R 4 R ( / R) dt ( / R) R / d R R 4R R dt R d 7 dt 4.4 Relative velocity Let v = speed of each spacecaft as it moves in cicle aound the cente O. The elative velocities ae denoted by the subscipts A, B and C. Fo example, v BA is the velocity of B as obseved by A. The peiod of cicula motion is yea T s. (8) The angula fequency T L The speed v 575 m/s (9) cos 6

7 Theoetical Competition: Solution Question Page 7 of 7 The speed is much less than the speed light Galilean tansfomation. In Catesian coodinates, the velocities of B and C (as obseved by O) ae v CB v ĵ î C L v CA O L L B v BC v v BA v AB v A v AC Fo B, v vcos6ˆi vsin 6ˆj B Fo C, v vcos6ˆi vsin 6ˆj C Hence v ˆ ˆ BC v sin 6j vj The speed of B as obseved by C is v 996 m/s () Notice that the elative velocities fo each pai ae anti-paallel. Altenative solution fo.4 One can obtain v BC by consideing the otation about the axis at one of the spacecafts. v BC L s 6 (5 km) 996 m/s 7

8 Q_THEORY_MARKING SENT_TO_LEADER.DOCX Theoetical Competition:Making Scheme Question Page of 4 Geneal making guidelines. Mino mistakes in the calculations e.g. copying expessions incoectlyfom line to line. Missing units in the final numeical answes (fo each pat). Final answes (fo each pat) containing too few o too many significant figues (fom + o - positions, say) Deduct % of the final answe Deduct. point Deduct. point 4. Using wong physical concepts (despite coect answes) No points awaded 5. Eo popagated fom ealie pats: mino eos Full points (except fo the final answe of the same pat. No maks ae awaded fo the final answe.) 6. Eo popagated fom ealie pats: majo eos (such that the solution becomes tivial). Deduct - 5% fo a paticula pat

9 Q_THEORY_MARKING SENT_TO_LEADER.DOCX Theoetical Competition:Making Scheme Question Page of 4 Theoetical Question : A Thee-body Poblem and LISA Questions. (Total.5). (Total.5) Points Concepts/Details..a Use the centipetal acceleation (.5) and gavitational foce (.5). (.5 =. fo concept +. fo coect fom).5.banswe:any of the thee following answes: G M m GM Gm R R R R,,..a Newton s nd law (.) fo two components of adial foces (. +.4 coect expession) and ciucula motion (. +. coect expession) GM Gm GM m cos cos R.5.bNewton s st law (.) fo tangential foces (.4 coect expession) GM Gm sin sin..c -Using at least two sine ules o sensible geometic elations (.)e.g. sin sin R sin sin R -showing adequate undestanding of geomety and/o tigonomety in the poblem (.) -algebaic manipulation to find a coect expession fo (.) -algebaic manipulation to find a coect expession fo (.) -ealize that (.).4.duse the cosine ule o algebaic manipulation to find (.4).6.eAnswe: R (.), R (.) R R (.)

10 Q_THEORY_MARKING SENT_TO_LEADER.DOCX Theoetical Competition:Making Scheme Question Page of 4 Questions. (Total.). (Total.) Altenate Solution Making Scheme Points Concepts/Details.8.a Expess enegy in tems of potential (.) (. fo each tem) and kinetic enegy (.) (. fo each tem) and the consevation of angula momentum (.4) (coect fom of angula momentum. and coect substitution.) to obtain 4 GM Gm E ( ).5 de.b Use the consevation of enegy as dt.4.c Use the equilateal tiangle: R, (.) cos o equivalent (.).4.d Expess d d d d (.)to obtain dt d dt dt 4 d d GM GM ( ) o dt dt..e Expess petubation fo adial components and ( ) (.), (.) 4 (.).5.f -Substitute and to obtain expession fo a simple hamonic d motion (.) dt - Expess angula fequency of oscillation in tems of only (.). 7.g Answe: fequency of oscillation.4..a Initial Condition fo G( M M ) GM (.) ( R R) 4R R (.)...b Evidence of petubed adial distance e.g. R (.)...c Use of Newton s nd law.6..d Gavitational foce(.)= mass(acceleation due to change in (.) +centipetal foce acceleation(.)). Note: Fo each. point. will be awaded fo evidence fo using coect concept and. fo the coect expession)...e Coect distance between and M R ( R )...fcoectly poject foce into adial diection.

11 Q_THEORY_MARKING SENT_TO_LEADER.DOCX Theoetical Competition:Making Scheme Question Page 4 of 4 Questions Points Concepts/Details.4..gUsing consevation of angula momentum (.) to obtain elationship between and (.).7..hApplying appoximation (.)and using binomial expansion(.) and algebaic manipulation (.) to obtain simple hamonic equation of (.). 7..i Answe.4.4.4a Find the angula velocity using T s (Total.8) T (. fo coect elation between and T and and. fo knowing numeical value of peiod = y).5.4b Apply the ciculation motion (.) and find the coect expession fo adius (.) fo each spacecaft to obtain L v cos (.).6.4c Coect expession of elative velocity e.g v BC v B v (.) C using dawing o vectos fo each v B (.) and v C (.) v ˆ ˆ BC v sin 6j vj o vbc v (.)..4d Answe: vbc 996 m/s. m/s.4 note Note.4a and.4b: Total of.9 will be awaded fo any coect method fo finding v fom T. Points fo the altenate solution using vbc L (axis of otation is at one of the spacecafts) will be given equivalently to the fome solution.

12 Theoetical Competition: Solution Question Page of 7. SOLUTION.. The bubble is suounded by ai. P, T, i i i R P, T, a a a O s, t Cutting the sphee in half and using the pojected aea to balance the foces give P R P R R i a 4 Pi Pa R () The pessue and density ae elated by the ideal gas law: RT PV nrt o P, whee M = the mola mass of ai. () M Apply the ideal gas law to the ai inside and outside the bubble, we get M iti Pi R M ata Pa, R iti P i 4 ata Pa R Pa ()

13 Theoetical Competition: Solution Question Page of 7.. Using of the atio is.5 Nm,. cm 5 R and. Nm, the numeical value P a iti T 4. (4) R P a a a (The effect of the suface tension is vey small.).. Let W = total weight of the bubble, F = buoyant foce due to ai aound the bubble W mass of film+mass of ai 4 4 R st Rig 4 at a 4 4 R stg R g Ti RPa g (5) The buoyant foce due to ai aound the bubble is 4 B R ag (6) If the bubble floats in still ai, B W 4 4 at a 4 R ag 4 R stg R g Ti RPa (7) Reaanging to give R at a 4 Ti Ra st RP a 7. K (8) The ai inside must be about 7. C wame.

14 Theoetical Competition: Solution Question Page of 7.4. Ignoe the adius change Radius emains R. cm (The adius actually deceases by.8% when the tempeatue deceases fom 7. K to K. The film itself also becomes slightly thicke.) The dag foce fom Stokes Law is F 6Ru (9) If the bubble floats in the updaught, F W B Ru 4 R st R i g R ag () When the bubble is in themal equilibium Ti T Ru 4 R st R a g R ag RP a a Reaanging to give u 4 4 R ag 4 R s tg RP a () The numeical value is.6 m/s u. The nd tem is about odes of magnitude lowe than the st tem. Fom now on, ignoe the suface tension tems..6. When the bubble is electified, the electical epulsion will cause the bubble to expand in size and theeby aise the buoyant foce. The foce/aea is (e-field on the suface chage/aea) Thee ae two altenatives to calculate the electic field ON the suface of the soap film.

15 Theoetical Competition: Solution Question Page 4 of 7 A. Fom Gauss s Law Conside a vey thin pill box on the soap suface. P, T i a i O R Pa, Ta, a q E E = electic field on the film suface that esults fom all othe pats of the soap film, excluding the suface inside the pill box itself. E q = total field just outside the pill box = 4 R = E + electic field fom suface chage = E E q Using Gauss s Law on the pill box, we have as a esult of symmety. E pependicula to the film q Theefoe, E Eq E 4 R () B. Fom diect integation R A R o O chageq q q Rsin. R 4 R 4

16 Theoetical Competition: Solution Question Page 5 of 7 To find the magnitude of the electical epulsion we must fist find the electic field intensity E at a point on (not outside) the suface itself. Field at A in the diection OA is E E A A q 4 R R sin q 4 R sin cos 4 R sin q 4R 8 q 4R cos d () The epulsive foce pe unit aea of the suface of bubble is q 4 R q E 4R (4) Let P i and i be the new pessue and density when the bubble is electified. This electic epulsive foce will augment the gaseous pessue P i. Pis i elated to the oiginal P i though the gas law. 4 4 P i R Pi R R P P R P i i a R R (5) In the last equation, the suface tension tem has been ignoed. Fom balancing the foces on the half-sphee pojected aea, we have (again ignoing the suface tension tem) P i q 4 R R q 4 R a R P P a P a (6) 5

17 Theoetical Competition: Solution Question Page 6 of 7 Reaanging to get 4 4 R R q R R R Pa (7) R Note that (7) yields R when q, as expected..7. Appoximate solution fo R when q 4 RP a Wite R R R, R R Theefoe, R R R R, 4 R R R R 4 (8) Eq. (7) gives: q R (9) 96RP a q R R R 4 96 R Pa 96 R Pa q ().8. The bubble will float if B W 4 4 R ag 4 R stg R ig () R Initially, Ti Ta i a fo and R R R 6

18 Theoetical Competition: Solution Question Page 7 of 7 4 R 4 R ag 4 R stg R ag R 4 R ag 4 R stg 4 q ag 4 R stg 96 RP a q 96 R t Pa s a () q 9 56 C 56 nc Note that if the suface tension tem is etained, we get R q 96RPa 4 RP a 4 R 7

19 Q_THEORY_MARKING_.DOCX Theoetical Competition: Making Scheme Question Page of Theoetical Question : An Electified Soap Bubble Questions Points Concepts/Details. (Total.7)..a Know that the diffeence between pessue (o foce) inside and outside the bubble comes fom the suface pessue...b Suface tension with two sufaces..5.c use the concept of suface tension de = γda with coect. da = d(4π ) (.) de = Fd = ΔPAd (.) (othe methods ae also acceptable e.g.f = γl de dx = γ da dx ) If the sign of suface tension pessue is wong, no mak awads..d Coect usage of Ideal gas equation (.) (.. coect expession).e Answe: -If the sign of suface tension pessue is wong, no mak awads. -No double penalty fom pat.b - The tem t cannot be included in this pat since poblem specify so. (Total.4). (Total.).4.a Answe: Fo the answe :. majo eo 5% Fo the answe.5 :. majo eo 5%.6.a Total weight fom the mass of the bubble (.) and the inside ai pulling downwad (.), and substitute fo (.):.6 In case that the student doesn t include the suface tension tem, deduct. point if the answe in.a is geate than. (a majo eo) Othewise, full points..b Use (.) Use the coect volume tem (.). The tem R + t instead of R is acceptable.4.c Setting up..4.d Answe: The ange of answe within [5,9] is acceptable.

20 Q_THEORY_MARKING_.DOCX Theoetical Competition: Making Scheme Question Page of Questions Points Concepts/Details.4 (Total.6).5.4a Setting the foce balance ( equal sign also acceptable) (.5, but only give. fo incoect sign)...4b Coect expessions fo the weight of the bubble (.) plus the inside ai (.)..5.4c Themal equilibium means (.) and substitute fo (.).4.4d Answe:.5 (Total.4).6 (Total.) Method A If the tem due to suface tension is neglected in.a, the second tem above can also be neglected In.a, if the student uses R + t instead of R, thee will be an additional thid tem. That is acceptable..4.5a Answe: o u min =.6m / s -The numeical value in ange of [.5,.7] is acceptable..6a Gaussian Law leading to the electic field outside the soap bubble: *If no facto /, no mak fo the following pat b,c..6b Gaussian Law leading to the electic field on the pill box:..6c Symmety lead to the electic field fom all othe pats of the film excluding the pill box itself: O Method B..6a Chage on a small stipe of the bubble film:..6b Fom the integation with a coect stipe...6c Do the integation coectly:.6 cont...6d Repulsive foce pe unit aea of the bubble:

21 Q_THEORY_MARKING_.DOCX Theoetical Competition: Making Scheme Question Page of Questions Points Concepts/Details.4.6e Use Boyle s Law to find the new pessue...6f Balancing the pessuized foce pushing inwad and outwad..6g Answe:.7..7a Apply the appoximation: (Total.7).4.7b Answe:.8 (Total.).7.8a Newton s Law (.). The balance between the weight (.) and the buoyancy (.). Check the coect fomula fo weigh and buoyant foce fom () in the solution. No double penalty fo the wong fomula of W fom.4b. If the student wite down the weigh W in tem of the new adius, R, and new density, that solution is acceptable too as long as it is coect...8b Answe:..8c Answe: nc -The numeical value in ange of [5,6]nC is acceptable.

22 Theoetical Competition: Solution Question Page of QUESTION : SOLUTION. Using Coulomb s Law, we wite the electic field at a distance is given by q q Ep 4 ( a) 4 ( a) q Ep 4 a a.() Using binomial expansion fo small a, E p q a a 4 4qa qa = + =+ 4 p 4..(). The electic field seen by the atom fom the ion is Q E ˆ ion 4.. () The induced dipole moment is then simply Q p E ˆ ion.. (4) 4 Fom eq. () p E ˆ p 4 The electic field intensity The foce acting on the ion is E p at the position of an ion at that instant is, using eq. (4), Q Q E ˆ ˆ p Q f QE ˆ p.. (5) 8 5 The - sign implies that this foce is attactive and Q implies that the foce is attactive egadless of the sign of Q.

23 Theoetical Competition: Solution Question Page of. The potential enegy of the ion-atom is given by U f. d. (6) Using this, U Q f. d 4 [Remak: Students might use the tem (7) p E which changes only the facto in font.] 4. At the position min we have, accoding to the Pinciple of Consevation of Angula Momentum, mvmax min mvb b vmax v.. (8) min And accoding to the Pinciple of Consevation of Enegy: Q mvmax mv 4.. (9) Eqs.() & (): 4 Q mv b b 4 min b min 4 min 4 6 min Q b b mv b The oots of eq. (4) ae: b 4 Q min 4 mvb.. ().. () [Note that the equation (4) implies that min cannot be zeo, unless b is itself zeo.] Since the expession has to be valid at Q, which gives b min We have to choose + sign to make min b Hence, b Q 4...() min 4 mvb

24 Theoetical Competition: Solution Question Page of 5. A spial tajectoy occus when (6) is imaginay (because thee is no minimum distance of appoach). is eal unde the condition: min Q 4mvb 4 4 Q b b 4mv.. () 4 Q Fo b b the ion will collide with the atom. 4mv Hence the atom, as seen by the ion, has a coss-sectional aea A, A Q 4mv b.. (4)

25 Theoetical Competition: Making Scheme Question Page of Theoetical Question : To Commemoate the Centenay of Ruthefod s Atomic Nucleus: The scatteing of an ion by a neutal atom Questions Points Concepts/Details. (Total.)..a Use Coulomb s law - Wite down invese squae law (. pt) - Coect constant (. pt)..b Take electic field fom chages - Wite down supeposition of electic field (. pt) - Coect chage polaity/diection (. pt)..c Coect distances - If the student didn t use the figue povided (-.pt). 4qa qa p.d Answe: Ep + o + o 4 4. (Total.)..a Wite down that the foce is the poduct of electic field and chage. { f QE p }.4 4qa qa p.b Answe: f + Q ˆ o + Q ˆ o Q ˆ c Use the electic field seen by the atom fom the ion.4.d Use Coulomb s law to wite down Q E ˆ ion (magnitude. pt, sign. pt) 4..e Use the given expession fo polaisability and wite down Q p E ˆ ion 4.5.f Use the concept of induced dipole by substituting Q p ˆ in equation () of question (.) 4 Q { Ep 4 4 E Q 8 ˆ }.(. pt) Get p 5 Q Q f ˆ ˆ g Answe: ˆ (magnitude. pt, sign. pt)..h Point out that the negative sign implies attactive foce.

26 Theoetical Competition: Making Scheme Question Page of..i Point out Q implies that it is egadless of the sign of the ion.. (Total.9).4 (Total.4).5 (Total.5).5.a Use the definition of potential enegy to wite down U. (wong limit -. pt).4.b Answe: Q U 4 (magnitude. pt, sign. pt).6.4a State consevation of angula momentum (. pt) Wite down mvmax min mvb (. pt).6.4b State consevation of mechanical enegy(. pt) Wite down Q mvmax mv (. pt) 4.5.4c Substituting v max in tem of min (.pt) Aange in tem of quadatic equation (. pt) Answe: b 4 Q min 4 mvb (.pt)..4d Choose + sign and wite down f d min b Q 4 4 mvb.5.4e State the easoning of the sign with Q o.4.5a Recognize that a spial tajectoy happens when min is imaginay because b b. (.7 pt) Recognize that min Q (.7 pt) is imaginay when 4 4mvb Q.5b Wite down b b 4mv.5c Answe: Q A 4mv