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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
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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 coOperation 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—ALA 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, EmuQ) 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) _§_ ZOOOALA 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 ZOOOALA MATH 145 RBEﬂEﬂigﬁﬁ a FORTEACHERS’ USE ONLY RFEﬂZEiﬁéEﬂ 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 2000ALA 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) 2000ALA 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 10711 zoooALAMMﬁgﬁgggmggg 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=272]+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 2000AL—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 coscoti—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 ZOOOALA 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 ZOOOALA 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 2000ALA 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—ALA MATH 1—14 Rﬂﬁﬂﬁiﬁgﬁﬁ FOR TEACHERS’ USE ONLY Rﬂﬁﬂﬁiﬁgéﬁ .. FOR TEACHERS’ 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— ZOOOALA 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 +rl2a)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 2000ALA 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_____ 2000ALA 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 motionof 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 2000ALA 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 ZOOOALA 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 ZOOOALA 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': Jrdm+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 03h“ 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! ZOOOALA MATH 1—21 Rﬂﬂiﬂﬂiﬁéﬂﬁ FOR TEACHERS’ USE ONLY ...
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 Spring '09
 DRwONG

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