2000AM1S - 93535150 @355 FOR TEACHERS’ USE ONLY E % % it...

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Unformatted text preview: 93535150 @355 FOR TEACHERS’ USE ONLY E % % it E HONG KONG EXAMINATIONS AUTHORITY zoomseemeeee HONG KONG ADVANCED LEVEL EXAMINATION 2000 EmQE eaters fiitfi— APPLIED MATHEMATICS A—LEVEL PAPER 1 $EF%%%E%EE$%%EZ§E%EWWEE ’ esseeeez % ° fifiéfifififififill’fifé ' fi%$?%%%ifilfifiifi¥fi§%fli E102H3lfi$5$5§ ’ 211E7Fi~§li¥i ’ lfiéEtUEE ‘ EEW‘I‘EETfi$T§= fifi‘ti‘lifi’jr‘fiifiiEfilgéiql ° $éE%f‘¥§E§ElZXHEUtt%3U¢ ’ Efifi E fifififiifiEfiEfilifitfi ’ $$lfifil§§ifififi$%fifi%%fi§$ ’ L‘Afi {ESEEEEE'EEEE I fi§l$fi ° Efifitfifi’flgfix‘fiélfi ' Efi$féfifilfi$fi§ LEEU ’ HF’EEgeéfifiififiifiEflEflifiiififififiZ E ° filth ’ 7S EJEEEE élfifiE/fififiififié‘l’fi ’ 5%?LB‘JEEEU ° This marking scheme has been prepared by the Hong Kong Examinations Authority for markers’ reference. The Examinations Authority has no objection to markers sharing it, after the completion of marking, with colleagues who are teaching the subject. However, under no circumstances should it be given to students because they are likely to regard it as a set of model answers. Markers/teachers should therefore firmly resist students’ requests for access to this document. Our examinations emphasise the testing of understanding, the practical application of knowledge and the use ofprocessing skills. Hence the use of model answers, or anything else which encourages rote memorisation, should be considered outmoded and pedagogically unsound. The Examinations Authority is counting on the co-Operation of markers/teachers in this regard. ’ §$4?F%é}%fi%fifiétfié‘$fizfifii¢lb ’ fitflzfilfizfifirfi “ After the examinations, marking schemes will be available for reference at the teachers’ centre. eeeeae Efi’fifité Hong Kong Examinations Authority All Rights Reserved 2000 ZOOO—AL—A MATH l—l Rflfiafilfiifigflfl FOR TEACHERS’ USE ONLY Rflfiaflflifiéfifi FOR TEACHERS’ USE ONLY General Instructions To Markers 1. It is very important that all markers should adhere as closely as possible to the marking scheme. In many cases, however, candidates will have obtained a correct answer by an alternative method not specified in the marking scheme. In general, a correct alternative solution merits all the marks allocated to that part, unless a particular method has been specified in the question. Markers should be patient in marking alternative solutions not specified in the marking scheme. 2. In the marking scheme, marks are classified into the following three categories : ‘M’ marks ~ awarded for knowing a correct method of solution and attempting to apply it ‘A’ marks — awarded for the accuracy of the answer Marks without - awarded for correctly completing a proof or arriving at an answer given in the ‘M’ or ‘A’ question. In a question consisting of several parts each depending on the previous parts, ‘M’ marks should be awarded to steps or methods correctly deduced from previous answers, even if these answers are erroneous (i.e. Markers should follow through candidates’ work in awarding ‘M’ marks). However. ‘A’ marks for the corresponding answers should NOT be awarded (unless otherwise specified). 3. In marking candidates’ work, the benefit of doubts should be given in the candidates’ favour. 4. For the convenience of markers, the marking scheme was written as detailed as possible. However, it is likely that candidates would not present their solution in the same explicit manner, e.g. some steps would either be omitted or stated implicitly. In such cases, markers should exercise their discretion in marking candidates’ work. In general, marks for a certain step should be awarded if candidates’ solution indicated that the relevant concept/technique had been used. 5. Unless the form of the answer is specified in the question, alternative simplified forms of answers different from those in the marking scheme should be accepted if they are correct. 6. Unless otherwise specified in the question, use of notations different from those in the marking scheme should not be penalised. _ . — . _ _ _ . _ _ _ _ _ -.. 7. In the marking scheme, steps which can be skipped are enclosed by Edotted rectangles : , whereas ZOOO—AL-A MATH 1—2 FIFE-#51 Eifigfifi FOR TEACHERS’ USE ONLY Rflfifllfiifiéfifly FOR TEACHERS’ USE ONLY Solution Remarks 1. (a) Consider the vertical motion : 2 Using v =u2+2as, 0=v2 —2gh v=1l2gh 1A After the first impact, vertical component of the speed of the ball = ev =e§l2gh As the ball rises to a height g after the first impact, 7 h 2 h 0=(e1/2gh)“ —2gZ 1M For (ev) =2g(Z) 92 = l 4 e = ~1— 1 2 (b) Let Ti be the time of flight of the ball between the (1' — l)th and ith impact. OR Using s=ut+éat21 Using vzut+atz 1 2 OszpEgT] Ozv—gt T 1: 3 1A t = l g g 2 .l. T h: —v‘ 8’ As the vertical component of the speed of the ball becomes ev after the first impact, Zev For T1, T2, T3, . . .forming a T2 : — tr" ‘th g 1M geome 10 sequence w1 ' common ratio e . . 2 ezv Similarly T3 = , g Total distance travelled “4271) 2v 2ev 292v =u(—~+——~——+ +---) g g g =2—u\:(1+e+eZ+---) g 2101 1 41w 1 :._.__ :-— (T e = —) g l—e g 2 2000—AL—A MATH 1— El ,‘fifliflflifigfifi FOR TEACHERS’ USE ONLY D E351: $5 3; FOR TEACHERS’ USE ONLY Solution Remarks Let v be the speed of the bead, R be the reaction of the wire on the bead when it has rotated through an angle 19. By conservation of energy, %m(1/3ga)2 =%mv2+mga(l—c056) lM+lA v2 = 3ga w 2ga(l 4:059) : ga (1+ 20036) Equate forces acting on the bead along the radial direction : 1M 2 R — mg cos 9 = 3”— 1A a R = mgc056+mg(l+20056) : mga +3 C056) {M For expressing R in terms of 6 Put R : 0 -' and putting R = 0 mg(l +3 cos 9) = 0 cos 6 : —l 3 the angle rotated by OP is cvos*‘(~~31—) . 1A (OR = 109°, 1.91 rad, Emu-Q) etc.) Altgmati'ye solution R 1 1 §m(,/3ga)2 =~2—mv2 + mg[a+acos(7r—a)] IMHA v2 = ga(1+2cos a) mvz . . mgcos(7r—a') — R = a 1M+1A Marking cr1ter1a same —mgcos a— R = mg(l + 2 cos a) as above Put R = O : } 1M —cosa=l+2cosa _1 1 1A a = cos *— ( 3) _§_ ZOOO-AL-A MATH 1—4 Rflfliflflifiéfi FOR TEACHERS’ USE ONLY Rflflfifiifigflfi FOR TEACHERS’ USE ONLY Solution Remarks (a) Let u, v be the speed of the bead and prism respective} when the bead reaches B. By conservation of momentum, m u = 8 mv , u = 8v By conservation of energy, Loss in P.E. of bead = Gain in KB. of bead and prism 1 2 1 2 m h = —mu +— 8m v g 2 2 ( ) mgh = %m(8v)2 +—12-(8m)v2 mgh=36 mv2 1;:ng (b) Time for the head to travel from B to A ZOOO-AL-A MATH 145 RBEflEfligfi-fi a FORTEACHERS’ USE ONLY RFEflZEifiéE-fl FOR TEACHERS’ USE ONLY Solution Remarks 4. Let W be the weight of the cube, T be the tension in the string, R be the reaction acting by the wall on the cube. For the 3 forces W, Tand R Resolve forces acting on the cube horizontally and vertically : TcosB = W ————(1) QR Using Lami’s Theorem, T R W sin 90° sin(l 80° ~ 9) : sin(90° + 9) TsinH = R —~——(2) Take moment about E .' J33 R(2Z cos 6) = W—-2-——cos(45° —0) ———- — (3) QB Take moment about A : 2 e R(€cos€) = W J; sin(4S°—t9) From(l) and (2): R =Wtan6 Substitute into (3) : For eliminating W, Tand R, and expanding cos(45°—€) etc. 2 E W tan 6(2Z cos 6’) = W lC—(cos 45°cos6l + sin 45° sin (9) 2fismflzl—cos6+-l—sin0 fi J5 4sir16 =cos€+sin6 3sin0 =cos€ QR Resolve forces vertically : T cos 6 = W Take moment about D : 2 W12:- 6 cos(45° — «9) = T(2£’ cos 6) sin 0 1A 213 (TcosH)‘/_T(45° — 6) = T(2Zcos€) sin 9 1M £(—1—cosfi+—l——sin6)=25in6 } 2 5 J5 c056+sin6=4sin6 3sin9=cos€ l tan6=-3— 1 2000-AL-A MATH 1—6 Rflfifllfiifiéflfi FOR TEACHERS’ USE ONLY Rflfiflflifigfifi FOR TEACHERS’ USE ONLY Remarks : I . l i Let W be the weight of the cube, T be the tension in the string, R be the reaction acting by the wall on the cube. Since the cube is in equilibrium, the 3 forces are concurrent. 1M Suppose their lines of action meet at point P. 1A E I R P D For the 3 forces W, T and R DE = 26 cos 6’ IA DPzghosMTmQ) 1A DP tan 6 : —— DE 2M 2/Z€cos(45°—6) tan 6 : .424....______ 25 0056 zfisin 6 = ~—l—(c059+sin 9) J5 3shi6=c086 _ 41 9—m (3) 2000-AL-A MATH l—7 Rflfiiflfilfiéflfi FOR TEACHERS’ USE ONLY Rflfifizfiifiéflfia FOR TEACHERS’ USE ONLY Solution Remarks i A i Let v be the resultant speed of the aeroplane, 8 be the angle between AB and the steering direction of the aeroplane. Resolve velocities perpendicular to AB : 1M 93 By Sine Law u sin 9 =£sin 60° 1A So the aeroplane has to steep along the direction N[60°+ sin"‘([3—)]E. ' 1A Acce t N68°E 12 p (b) Resolve velocities along AB : 1M V ELK By Sine Law v=uc056+iécos60° 1A u v _ sin 60° sin(120° — 6) Substitute sin 6 =£ : 12 J122—3 12 _(J141+1 12 )+ a? 1M For finding v ) u (m 1.073 it) By cosine law, 1M u2 = v2 + ($2 — 2(-6“—) (v)cos 60° 1A 2 6 36v2 —6uv—35u2 = o : 61‘/36+4(36) (35) 1M For finding v Time taken to travel from A to B _ E v =———1—2—€——- 1A Accept 0.93£,——e- (J141+1)u u 1-0711 zooo-AL-AMMfi-gfigggmggg FOR TEACHERS’ USE ONLY Rflfiifilfifigfifi FOR TEACHERS’ USE ONLY Solution Remarks 6. (a) Since the line of action of 1:“ lies on a plane perpendicular to the x axis, the x component of I? = 0 . lM Let [3 = b} +cl€ EZxfi:(2F—2]+3IE) 1M Nomakforfixafi:(zf—2j+3£) (?+])x(b]+c/€)=2?—2]+3k‘ c?—c]+bE=27-2]+3I€ Equate components : lM For equating components 6 = 2 , b : 3 17* = 3] + 212 1A (b) Since I? has zero moment about B( p, 0, q) , B lies on the line of action of I?“ . 1M Line of action of 17' is 2=(?+j)+,1(3j+212) (,1 isreal) 1M . ForE=20+zfi =f+(1+3,1)]+2/1i€ Substitute B( p, 0, q) into the line of action : 1 = p 1 + 3,1 = 0 Elixfizé E; =54—07é For finding E3 =(f+]‘)—(p?+ql€) =(1—p>?+J‘-ql€ [ti—p)F+]—q121x(3i+2l€) 6 (2+3q)?+(2p—2)]+(3—3p)/E=6 2+3q =0 2p—2=0 3—3p=0 p=l 2000-AL—A MATH 1—9 Rflfiiflfiifigfifi FOR TEACHERS’ USE ONLY Rflfiiflfiifigfifi FOR TEACHERS’ USE ONLY Solution Remarks 7. (a) Let p be the density of the disc. (can be omitted) Consider the disc as made up of a number of concentric rings‘ Let the width of the ring at distance r from the centre bedr. Moment of inertia of the disc 2 J‘rzdm 1M = J (27zrpdr)r2 1A 0 r4 a :2 _ 7rp 4 O 4 :zpa 2 71' a4 m =_'0_ ( 2) 1M 2 arpa =§ma2 —l—— __4__ (b) (i) By conservation of energy, 5c Loss in PE of block = Gain in KB of block and cylinder + Gain in elastic 1M PE of spring mgx=lmjc2+llw2+lkx2 1A 2 2 2 Since the cylinder is perfectly rough, is = rm . 1M For i = m; m =—mx +— 2mr - +—kx 8x 2 2 ( )(r) 2 3 .2 1 2 m = —mx —kx gx 2 2 3mx2+loc2—2mgx=o —-—-(1) 1 (ii) Differentiate (1) with respect to t : 1M 3m(2)'c5c')+2/odtv2mgi:0 1A 3ij + kx — mg 2 0 ' 56 + —k—x = E 1 3m 3 '. The motion is simple harmonic. 2000—AL—A MATH 1—10 Rflfliflfiifigfifi FOR TEACHERS’ USE ONLY Rfiflfllflifigfifi 1 FOR TEACHERS’ USE ONLY Solution Remarks AL . 1 . m£=mg_T2 (2mr2)é=(TZ—T,)r TI =kx 552M Solve the equations : ‘ IFor solving the equations 56+ 3m 3 the motion is single harmonic. .l k (in) x + ——x = 1g— 3m 3 Complimentary solution . 2 k xc = cl cos COI+62 sm wt where a) 2—, , 3m For findmg xc and xP Particular integral g 3m mg x :— — :....._. P 3( k ) k xzxc +xp mg . =~~k~+c1 cosmt+c2 smwt Alternative solution .. k x:—__(X_E 3m k mg . 2 k * x-———= cl coscot-i—c2 smcot where a) 2—, 1: 3m For using the boundary conditions to find cI , c2 5c =c‘cosinwt+cza) cos col 2 62 =0 '. x:%(l—cosa)t) Tension in the string = kx = mg(1— cos cut) [OR = mg(1— cos J2 1)] 3m ZOOO-AL-A MATH 1—! l Rfifiiflgfliéflfi FOR TEACHERS’ USE ONLY Rflfiiflflifigfifi FOR TEACHERS’ USE ONLY Solution Marks Remarks 8. (a) Let VI, v2 be the components of the velocity of P afier Q the collision along and perpendicular to the line of centres. ’0 123 be the velocity of Q afier the collision. By conservation of momentum perpendicular to the line u of centres, . I v2 = u sin 6 . lA I By conservation of momentum along the line ofcentre, mucosl9 = mvl + mv3 - — - — «(1) 1M for setting up either equations. By Newtons Law of restitution, l lMHA 1A if both were correct v3 —v1 = ~(0~ucos€) = ucosB - — - - - (2) Solve (l) and (2): v1 = 0 , v3 = u cos (9 I 1A P moves with a speed usin 6' perpendicular to line of centres, and Q moves with a speed ucos 19 along the line of centr So their directions of motion are perpendicular. QB. As vI = 0, so P and Q always move in perpendicular directions. 1 g (b) (i) After the collisions, R I will move along the line of centres of RI 1A and W, R2 will move along the direction CB (fl AB), 1A W will move in a direction perpendicular to CB. 1A Alternative solution The directions of motion of the balls are shown below : lA+1A+lA ZOOO-AL-A MATH 1—12 Rllflfllafligfifi FOR TEACHERS’ USE ONLY Rflfiiflflmgfifi , FORTEACHERS’USEONLY Solution Remarks (ii) Let D be the centre of Wat the moment it hits R I: E be the foot of perpendicular from D to AC. AC220—3 x2314 AD = 3 X 2 1 6 From (a), velocity of RI and Ware perpendicular after the collision, so ABC 2 Let AACD = 6. sine = i6; =—:~ For finding sin 6(or c056) AADE=AACD=B Using (a), speed of R1 afier lst collision, = u cos AADE = u c056 2f 10 u 7 Accept 0.9 u speed of W afier lst collision, v=usinAADE 3 =usin0=~li 7 After the second collision, speed ofR2 = vcos AACD - - - -(1) 311 6m = ~COSB = 7 49 speed of W = vsin AACD - - - - (2) 3u . 911 =— 9=— Accet 0.18 7 5m 49 ( D If) u (Accept 0.39 u) 1M for (1) and (2), 1A awarded if both answers were correct 2000-AL-A MATH 1—13 R Ffiflfilfigfifi FOR TEACHERS’ USE ONLY Rflfigflfiifigfifi FOR TEACHERS’ USE ONLY Solution Remarks 9. (21) Equation of motion of the particle along the radial direction : m (F — r492 ) = ~-%mg 1M For setting up the equation ’2 F—rw22—ig where a): i 6 0 Characteristic equation : ,12 —a)2 = 0 xi = in) Complimentary solution rc = Clea" +626 1A 5 _g 5 Particular integral rp = £2— : i (i) 1M 0) 6 28 -5_a 12 r = rc +rp = Clea" +c2e_‘”' +5—a 1A 12 At t = 01 r z E ’ ,2 : 0 1M For using the boundary conditions 2 to c, and c2 5a a C1 +C2 +— — — 12 2 r = clwew' _c2a)e_m C160 " C20) : 0 Cl : 02 l a 50 c : c =—— -n— l 2 2 (2 12) _ a 24 r=~2%(e"" + ‘a”+10) l __6_ (b) (i) When the particle reaches the rim of the disc, r = a. a = fie“ + e““’T +10) 1M For substituting r = a 60” +e_“’T =14 (ewT )2 ~14ewT +1: 0 1M+1A 1M for transforming into a quadratic equation M _14iJ196—4 2 =7+4J§ or 7—4J3 wT =1n(7 + 4J3) or ln(7 ~ 4J3) 1 For rejecting 7 —4J§ (rejected ln(7 —— 4J3) < H e r=iln(7+4J3) 1A (0 ZOOO—AL-A MATH 1—14 Rflfiflfiifigfifi FOR TEACHERS’ USE ONLY Rflfiflfiifigéfi .. FOR TEACHE-RS’ USE ONLY Solution Remarks (ii) Radial component of velocity of the particle at time T V] : ’2 WT) 1M for either one, 1A if both were correct a . 2 — (mew * we 24 Transverse component of velocity v2 :rfi :aa) ll 24 ll 1 7+4J§ :____Z~4____ 7+4J37—(7~4J§) =J§ 7T (9:— or 60° 3 ( ) H443— ZOOO-AL-A MATH 1—1 5 Rflflflflifigfifi FOR TEACHERS’ USE ONLY Rflfiiflfifigfifi FOR TEACHERS’ USE ONLY Solution Remarks (5c,- +y',2) 1M Let GR 2 r,- and GP, makes an angle 8i with the x axis. x,- = x0 +ricos6’i - - ~ - (l) X, =J'cG —r,a)sin«9, y, :ya +r,-sim9, —~—-(2) For(l)and(2) yi = yo. +riwcos 61. For finding is, and 5»; Ahemgtijisolmigm rim y 1 x K.E. = 23mm2 v, is the vector sum of v6 and rim . Equate horizontal and vertical component : . . 71‘ x,- = x0 + riwcoswi + —2—) : ch — r, to sin 6,- . . . 7r y,- : yG + r,-wsm(6, +3) = yG +r,-(0cos€,~ l . . . K.E.=§Zmi[(xG —r,wsm 6,)2 +(yG +r,a)cosB,-)2] (Note : All summations are from 1': l to N.) l . 2 . . . =52mi[xc —2xGr,-casm 6i +rl-2a)2 sm2 6,- . . 2 +sz +2yGr,co cosHi +r,. (02 cos2 0,] 1 . 2 . 2 . . :_Zmi(xG +J’G )_xGwzmirismgi 2 For collecting terms . l 2 +yGa)§jm,-ri cosfi, +§Zmin (02 2mm Sing: =ZmiU’1 _yG) =27"in ~Zmrya =(Zmi)yc -(Zmi)yc =0 Similarly, Z m,r,- cos 6,- : 0 K.E. =~3~Zmi0302 +j’52)+%602 Zmi’iz For showing 2 mir,» sin 6 = 0 1 2 1 2 =—MV +—I a) 2 G 2 G 2000-AL-A MATH 1—16 Rflfiiflfilfigfifi FOR TEACHERS’ USE ONLY wastes a. FORTEACHERS’ USE ONLY Solution Remarks (b) (i) As the forces acting on the disc (its own weight 1M for considering forces and the normal reaction) are all vertical, so G } lM+1A acting on the disc moves vertically downwards under the action of force acting on the disc, so G moves vertically a couple. 1M+1A same as above downwards under the action of a coule. By conservation of energy, Loss in PE. = Gain in K.E. 1M Loss in PE. = Mgasinfl ' 1A Gain in K.E. = 941/0.2 + £1002 Downward displacement of G = a sin 6 1A d . ' Velocity of G = a (a sm 6) 1M . = aw cos 9 Mgasin6=%M(awcosfi)2 +éMk2w2 ' 1M Mgasin 6’ = %M(a2 cos2 6 + k2)a)2 2 a sin 6 “)2 z 2 g 2 2 *L-— a cos (9 + k _3_____ 2000-AL-A MATH l—l7 Rflfiflfiifigfifi FOR TEACHERS’ USE ONLY I . l . Since the table is smooth, there is no frictional force acting on the disc. As there is no horizonta Rflfiifizfiifi FOR TEACHERS’ USE ONLY Solution Marks Remarks 11. (a) Let u be the speed of P immediately before the collision. By conservation of energy, Loss in elastic potential energy in the string = Gain in PE and KB of the bead 1M — -——€ : m — +—mu 1A 2 g )( 3 ) g( 3 ) 2 10 4 1 2 —m E =—m Z+—mu 3 g 3 g 2 u = ZJ—gj _1A __ _3__ (b) (i) Speed of P immediately after the collision = eu‘ lM J3 = (7) GM) -— l 5 _ _ (*) 2 3 Let v be the speed of P immediately before it collides with Q. By conservation of energy, ‘IS 8 l m ( g )2 + mg! = lmvz lM+1M lM for using conservation of energy, 2 2 2 following through answer obtained in (*) 13 m f — 1 mv2 8 g 2 1 v : —— 13 Z 2 8 By conservation of momentum, speed of the combined beads immediately after the collision mv = 1M m + 2m _ X 3 1 = — 13 5 l 6 g . (ii) (1) Equation of motion-of the combined beads : ' 3m55=3mg~60£mgx 1M+1A IMfor M56=Mg—T .. 20 g x + -—— x = e g 2 xz—a)2 x——— ,whereco =—. ( ) ( 20) e The motion is simple harmonic. When the combined head has the maximum speed, 56 = 0. 1M x = A 1A 20 2000-AL-A MATH 1—18 Rflfiflfiifiéflfi FOR TEACHERS’ USE ONLY Solution maximum speed at a distance ——'— below the point ofcoliision (QR at a distance —— below A.) (3) Time taken 1 = Z (period of the motion) 1 271' 4 ‘IZOg/E -351. 4 5g ZOOO-AL-A MATH 1—19 Rflfiiflflifi%fifi the combined bead attains the ) P 20 211? 20 FOR TEACHERS’ USE ONLY Rflfiiflfiifig‘fifi FOR TEACHERS’ USE ONLY Solution Marks Remarks 12. (a) Let p be the density of the lamina. (can be omitted) By symmetry, G lies on the line of symmetry as shown 1A Consider a small sector at an angle 6’ from 0G . Area of sector : 1 c1de 2 1A . ’ 2 Distance of centre of mass of the sector from 0 = : a Distance of centre of mass from O = Irdm + mass of lamina 37: erm=2f(%a2pd9)(§acosfi) 1M+1A 1M forfinding rdm /—ara p 1M For 0G=_ rdm+MaSS Let p be the density of the lamina (can be omitted) Consider the cut away quadrant. By symmetry, the centre of mass G’ lies on the line 1A of symmetry as shown. Consider a small sector at an angle 6 from OG’ . Area of sector = % c1de lA . 2 Distance of centre of mass of the sector from 0 = —- a 3 Distance of centre of mass from 0 = J‘rdm + mass of the quadrant Irdm=2f(%azpd6)(§-acost9)----(1) 1A ZOOO-AL-A MATH 1——20 R Bfiiflfiifiaififi FOR TEACHERS’ USE ONLY Rfifliflfiifiéfifi 7. FOR TEACHERS’ USE ONLY Solution Remarks 3 E zzpkmek fia3p 3 (373/0 E0210 3 / 4 1M For 06': J-rdm+Mass _4J§ 37: Consider the whole circular lamina. Take moment about 0. mass of sector (0G) = mass of quadrant (OG’) - - - - (2) 1M For (1) and (2) 3 1 4J§ 2 2 _ OG _ __ 47m p( ) Ira p( 3 a) 0G=fla 1 00' = a (b) (i) Let R’ be the reaction of B on A. All correct — 2A R' 2 or3 correct— 1A ‘ ‘r r i . . a. < .‘ ( N. ‘ a V3“). *0» «A a. x a six x‘x‘x 03-h“ x‘i‘k‘x 's‘§*~f‘€~.v«.‘«."«fxx 31'». a a. x. x (ii) Equate vertical forces acting on the three cylinders : 2R = W + 2W + W R=2W (iii) Take moment about the point of contact of A and B for cylinder A : 45a 971' Fa+W(a+ c0545°)=Ra Fu+nha+¥19=2wa 971' F = (1 —i)W 971' (iv) Let ,u be the coefficient of friction. F s ,uR (1 —i)W s #(ZW) 97r l 4 2—1~-—- 'u 2( 971') the least value of y is (-1—~i). l A Accept 0.43 2 97! ZOOO-AL-A MATH 1—21 Rflfliflflifiéflfi FOR TEACHERS’ USE ONLY ...
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