DXLAX; DYLBC; DZLAB.
頁四第張六第 日四初月十年申庚歷夏
WAH KIU YAT PO
報日僑華
二期星
日一十月一十年〇八九一曆公年九十六團民華中
from the smooth
(1)(a)(g) (104)
(Tr3)(f) (6) (r)
wall of the
1981
3r
中學會考試題預習專欄
(b)
r = √3r
cylinder
weight of the ball reaction between the balls.
101 -
o (c) (2A)
(7)(d)(g) (10A)
Given: ABCD is
quad.
а
a cyclic
(0.5)(a)()(8A)
物理
Ans
i«e <DX€+ZDYC=180°
(Given)
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·PHYSICS (6)
W.E.IO.
(MILL & DALE PRESS LTD.)
Answers to Exercise 3.
1(a) Take the direction to the right as positive Initial momentum of the target
= (100) (-3).
-300kgms
Initial momentum of a
bullet
(0.01)(600)
6 kgms
Assume n bullets are
required to stop the motion of the target, the momentum of n bullet
6n kgms
Since the target stops finally, therefore,
the final momentum of
the system is zero,
6n-300 =
As shown in the figure,
the centre of mass of the combined body is G and is x cm from G2
Hence, moments of the weights of the cone and the hemisphere about G are equal in magnitude (− πr2n) († ) (6) ( 3h+ğr-x)
(rr3) (4) (5) (x)
-x)=2rx
h =
x)=2x
2
(ii) The system is in
stable equilibrium because the c.g. is
below 0.
(a) The direction of the
reaction from the smooth wall of the cylinder on the upper
ball is normal to
the wall and towards the centre of the ball.
(b) As shown in the
diagram,
BA
x (radius of the ball) 0.2m
BP (diameter of the
cylinder)-2x
(radius of the
ball) 2x0.15
0.1m
2x0.1
√BA?
數學 (六)
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Mathematics (6)
CP. Man
MILL & DALEEPRESS LTD.
Solution to exercise 2. Section B. 11.
Solution: (■) AB is the
diameter of
- LACB = 90° circle. .• ¿CAB+LCBA - 90°
£NCB+ZNBC2 = 90
- LCAB .NCB In AACN
EN
tan≤CAN
AN
In ACNB
tan NCB
NB
CN
CN NB
AN
CN
CN"
(c)
Area of square of side CN = a(12-4)
To prove; (a) <DYX = ZDBA
(b) X,Y,Z are collinear
Proof: (a) DXLAX, DY1BC
LDXC=ZDYC=90
D.,X,C,Y are concyclie
(opp a supp.
¿DYX=4DCX (48 in the same
ZDCX÷ZDBA
ZDYX=4DBA
segment)
(Ext. eyelic
(b) DZLAB (Given)
ZDZB-90
i.e #ZDZB÷<BYD=90***
‚D,Y,Z are concyclic
(Converse, Ls in
the same segment)
ZABD+ZZYD=180°
(opp. ́s cyclic
quad.)
(Subst.)
ZDYX+ZZYD=180* X,Y,Z are collinear (Adj.Ls supp
B
50
Ans-
(b) The energy lost
2(a)
during the impact
1(100) (3)2 + (0.01)
(600)2(50)
450 + 90000.
90450J
As shown in the figure,
and
are the tensions T Ang
2 in the string.
Let a be the acceleration of the 18kg-block, then the acceleration of the 8kg-block is 2a.
8(2a)
16a
T2 = 271
(1) (2)
sine
Since the balls care
in equilibrium,
Roine
(1.5)(10) 17.321
AN.NB
(b) Area of square of side
2
CN CN
AN NB 8q.cm
a(12-a) sq.cm
≈ 36~36+12a-a 8q.cm
36-(6-1)2 8q.cm
2
The max.area of square of side CN 36 sq.ca
12, Solution: Let $C be the
basic cost and V km/hr be the speed of the cable.
C+ky where k is a
68 +8)
constant.
770x+125k---. -(2) (2)-(1) 702 - 117k-
k = 6
Put ke6 into (1)
68-86 20
13. Solution: (a)
Given: ABGR, BCFG and CDEF
are squares
AGF (a) A.
EGA are similar
(b) 8+&+ß=90°
Proof: (a) Let x cm be the
length of the sides of the
squares.
+x cm=2xcm
(Pythagoras
Theorem)
No2
AGGE
GF
ZAGF÷ZEGA Common)
AGFA EGA
are similar (ratio
of 2 sides,
• Inc..).
(b) * =LGAE (Corr. ́s similar
60+LGAE
But B
(Ext.
ofA)
Let f density of
copper
Weight of the copper
cone
= (} #r2h)(Y) (E)
Weight of the copper
hemisphere
Since the centre of mass of the combined body is at
and the ments or
therefore, the wei
of the cone hemisphere
about C
are equal in
mani tude.
185-212 18a. ·(3)
Substitute (2) into (3),
186-212 = 18a
9ɛ-T2 = 9a
(4)
Substitute (1) into (4),
98-16a = 9a
25a = 96
(10)
Let d
3.6 ms. Ans
acceleration of the 18k-block is 3.6ms
and that of the block is 7.2 ms (b) T2 = 16(3.6)
(0) 12
= 57.61
115.2N
Ans
Ans
Hot N the ren Lon
#
density of water f = relative density of the wooden block
Upthrust on the block due to oil
(0.5) (the
where A is
cross-
coctional area of ube
Upthrust on the block due
to water
Weight of the wooden block
「
二
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意 人共有四子一女,她,月薪一千五百元,雙慶年
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人士能盡。
春青好華年樣花
程前誤毒吸必何
慶祝五週
躲就鏃
#I-
(b) Let
5b+2c
2642
k
x+y=Z+Y+2;
7+8
22
8+9
X+Y=Z+Z
7:49
*
#
而且支
(Subst.) (properties
square)
Solution: Selling price
for 4 of the estate at
a gain of 20%, 83×240000x(1+20%)
$72000.
Selling price for - of
the estate at a loss of 15%.
$=x240000×(1-15%)
$81600
Total sale for the estate at a gain of 10%.
$240000×(1+10%) $264000.
The selling price of the remainder estate, $264000-$72000-681600.
$110400.
Gain percent of the remainder.
170500-240000x(1-1-3)
240000×(1————)
31.4%
100%
小
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