其四第張六第 日一廿月九年中庚后夏
1981
中學會考試題預習專欄
數事
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Mathematics 4
C. P. Man
MILL & DALE PRESS LTD. Exercise 2.
Answer All questions in Section A and any Six questions in Section B
Section A.
1. If p and q are the roots of the equation 2x-3x-1.0. Find the value of p2+q2 and
(p+2}{q+2).
Find the simplast form of
Sin
x(tanx
tanx
D
30cm
In the figure, AB is a
diameter of a circle, CN is perpendicular to AB.
日九廿月十年〇八九一腊公年九十六徽民者中 ★教儒
WAH KIU YAT PO
郭日僑華
三期星
The initial velocity is Oms
Hence,
2(a) (2)
= 2(2)(2)
2.8284ms
Ans
(a) Show that AN. NB - CN2
(b) If AB=12cm, AN-a cm,
express the area of a
square of side CN in terms of a.
(c) Find the maximum area
of the square of side CN.
12. The basic cost of running
a cable car is $X per hour whether it is moving or not, but when it is moving the additional cost varies as the cube of its speed. If the total cost per hour is $68 when the speed is 2km/hr and $770 when the speed is 5km/hr, find the value of X.
13. (a) if
3b+2c 2c+aa+3b show that X must equal either or -1.
(b), if
x+y-z = y+z-x - z+x=x 8
find x y z
(d) Let the time required for the 15kg-block to reach the floor be t (seconds).
Apply the formula
s = v. t + ŝat2
initial velocity
= 0 ms
-1
Substitute s#2,
Work done against F loss in kinetic energy + loss in potential energy Work done against ?
F(0.5)
Loss in kinetic energy = 4(10+12) v2
3(22)(7.042)2
= 545.53
Loss in
enery
potential
(10+12)(10)(0.5)
F(0.5) = (545.5+110)
Bat2
*(2)+2 = √2
3(a)(i)
1.41428. Ang.
(e)
30*9
E = 312.5N.
Ans.
(d) Let the tension in the
string connected to
the 50k-block be
newtons
effort.
1.6
When effort = 60ON"
600 = 1.6
T960N
Hence, the net force acting on the 50k- block
# 960-508 -960-500
#460N
Apply Newton's 2nd law
1311. Ans
460
LOSA
Goo H
50(a)
9.2 s
upwards with
Ans-
cceler-
40cm
In the figure, the area of the parallelogram is equal to 600 sq.cm. Find ZA. 4. If the surface area of a
sphere increases by 440 what is the percentage increase in its radius?
A
B
In the figure, DC is a diameter of the circle, Prove that
Zabc
a2+b2+c2+
In the figure, AP is tangent and BD is a diameter of circle ABCD, I BAP = xo· ZBPA = find a relation between X and y. (Geometry
theorem need not be quoted when used.)
B
In the figure, ABCD is a square, AP - AC and ANIDP. find the ratio of PN to ND. (Geometry theorem need not be quoted when used.)
8. Find the value of
Section B.
5/5
9: A tripod is made of three
rods OA, OB, OC each 12cm long, hinged at 0. The ends
Crest on horizontal od so that each of the angles BOC, COA, AOB is 60°. Calculate
(a) the height of above
the ground.
(b) the angle which OA makes
with the ground,,
10. A man starts his career
with a salary of $450 per month and is promised an an increase of $20 per month at the end of each; year's service.
(a) Write down an expression
for the total amount earned in n years.
(b) How many years does it take him to earn a total: of $36000 ?
In the figure, ABCD is a cyclic quad. DXLAX, DYLÈC and DZLAB. Prove that (a) ¿DYX = LDBA (b) X, Y, Z are collinear. 15.
H
In the figure, ABGH, BCFG and CDEF are three
consecutive squares,
Prove that
(a) AGF
As
EGA
are similar.
(b) 0+x+6 = 90-
16. An estate was bought by a
000. He sold
man for $240 a gain of
of the estate
20% and of it at a loss of 15%. At what gain percent
be sell the remainder must
so as to make a profit of 10% on the whole.
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PHYSICS (4)
W.K.LO
(MILI & DALE PRESS LTD.)
Answers to Exercise |1. (a)
1514
[10 kg]
As shown in the figure
above
Let T tension in the
string,
= acceleration of
the system.
Apply fewton's
Is 2nd law to the blocks separately, 15(10)-T = 15á..........(1) T-10(10) 10a....(2) (~)+12)=
50254
(b) Substitute a = 2 into
(2)
~10(10) = 10(2)
2
T = 1201. Ang (c) apply the formula
+2as Let v be the velocity of the 15% -block when it strikes the floor.
As shown in the figure
let the velocity above,
of the 10kg-block Le
2 when it reaches the level
at A.
2(a)(2) 2(2)(2)
Hence, the string slackens,
assume the block moves h meters farther before drops down,
Ans.
2(g)(h)
2(10)
0.4 st
The maximum height reached by the 10kg-block 18
0.4+2 2.4 m above
the floor.
2.(a) Let the velocity of
the pile driver as it
strikes the pile be
me
By conservation of energy,
gh =
where m = mass of the
Sirith horizontal plane
As shown in the figure above, let the tensions
in the strings be T1 and T2, the acceleration of the system be a. Apply Newton's 2nd law. to the blocks separately,
600-T
(1)
10a........ - 20a.....(2) 30a. (3)
(1)+(2)(3):
= (10+20+30) a 600
-2 a = 10 ms. ans
(11) Substitue a - 10
and (2),
into (
we have
= 600-10(10)
500N
30(10)
3001.
2.
Ans
As shown in the figure
above, f, f2 and 13
The block will move
ation 9.2 ms
5(a)(1) Upthrust of water
on the cube.
weight of the
cube
900(0.1) 6 900(0.1)3(10)
ON
Ans
(11) Let the volume of the cube immersed
in water be V
Upthrust on the
cube
(1000)(g) (V*) 1000(10) V
- 9x10
Since the volume of
the cube
(0.1)3
1 x 10-3, 3
fraction of the
cube immersed
below.
9 x 10
1 x 10
10
are the frictional forces (b) as shown in the figure on the 10kg, 20k, and 30kg-blocks - ́spectively.
f2 = 0.2(10)(10) = 201
0.2(20)(10)
= 60N
= 0.2(30)(10); Let the acceleration of
the system be a teng ons in tires
(-)**
pile-driver
Apply ewton'b 2nd low
to.
bloc
10kg
and h
height faller
12 m
·40
= 10a
20
·(1)
• (2)
201
[(1) + (2) + (5) ;
GOC-120
(11) Substitue e!
2(10)(12)
240
15.492 ms Ang.
(b) Let the common velocity
required be V HC Initial momentum of the system
≈ (10)(v ̧) kyns
Final momentum of the
system.
= (10+1214
~y conservation of linear momentum,
10%
(10+12)v
글을
10
19(15.4927. 72042 20
(c) Let the average re-
Ang
carding force exerted by the ground on the'. pile be hertons.
into (1),
1- Substitute anto (3)
3000.
Tet the inájdum. equired La F.
= 1.6
Load=150,i
5001
Foil
Let h depth of oil
cross-sectional area of the
cube
(=0.01 sq.m.)
Uptlirust due to oil on
the cube.
(720) (6) (4)
7200hA
Upthrust due to water on the cube
1000(¿)(0.1–h) A
= 10,000(2.1−}}A
Total upth ist exerted on the cube.
= 7200+10,000(0.1-h)A
1000A-2800hA
= (1000-2800h) A weight of the cube
= 900(071)(A)(6)
= 900A DA
"Total upthrust
weight of the cube (1000-2800h) A; -900A
12
= 0.0357m
the cube rises
= 0.0357m - 1cm 0.0357m-0.Clm
0.0257m