真四第張六第日十三月九年午戊歷

WAH KIU YAT PO

報日僑華

二期星

1979

meters farther before drops

down

600 - 10(10) 500N (Ans.)

Since the volume of the cube

· (0.1)3

中學會考試題預習專欄

v122 - 2(g) (h)

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To 30(10) 300N {ADE,).

(b)(i)

10-3

日一卅月十年八七九一年七十六國民華中 青收堡饰

6. Find the values of a and

if 3x-2=a(x−1)+b(x+2)

is satisfied for all

values of x.

2

If 5sin -2co

Find

物理

PHYSICS (4)

8

fraction of the cube in-

mersed

the values of tarß.

8

~ 9 x 10"

Suggested solutions to

exercise two

·30kg

SOON

20kg

་ཚ་

The value of a cachine: depreciates from $2000 to 81280 in 2 years. the yearly decay factor, assuming it to be constant. Section B

Find

1(a)

115 kg

jokg.

As shown in the above,

lét ↑ - tension in the

string

acceleration of the

system.

Apply Newtond: Law

the blocks seperatěly,

15(10) T➡ 15a

(1)

(2)

-

10(10)

-10a

50-25a

· (1) **

शोण

= 0.4 (m)

Ans: The maximum height reached by the 10-kg block is 0,42 - 2.4m above

the floor.

2(a) Let the velocity of the

pile driver as it strikes

the pile be v

ma

By conservation of energy,

mgh - tmv-

where m - mass of the pile-

driver 10kg

height fallen.

As shown in the figure above, fje fg and f. are the frict- ional forces on the 10 kg, 20kg and 30-kg blocks res-

pectively.

20N

f1 = 0,2(10)(10)

0.2(20)(10) 4ON 60N

2

fz = 0,2(30)(10)

Let the acceleration of the

system be a the tensions in the strings be T and Tj. Apply Newton's 2nd law to

Ti - 20 - 10%. (1)

and h

12m

the blocks seperately.

600

2gh

2(10)(12)

TI

T

4020

240

60

30

-15.492

(Ani

(1).

(2) ) + (3)

600

(b) Let the common velocity

required be v ma

(2)

Initial momentum of the

system

-1

(10)(v,) kgma

Final momentum of the

system

120 ·60m2

。(2).

(3)

8 ma (Ann.)

(ii) Substitute a' - 8 ints (1)

500N (Aus)

Substitute a' - 8 inte (3)

T 300N (ang.)

V.R.

2

(b) Substitute a

(Ans into (2)

4.(a)

(Ang.)

(10 + 12)▼ kgos

(b)

MA

· 10(10)

10(2)

By conservation of linear.

x 100%

momentu

x 100 40:

(1012)

M.A. 1.6 (An■.)

(c) Let the minimum effort

required be E

Load

E

1.6

500N

500

1 x 10

9.

10 (An.)

(b) As shown in the figu

below.

Leth

Plastic

Cube

(720 km)

depth of oil

-water-

A cross-sectional

area of the cube

(~ 0.01 m2)

Upthrust due to oil

the cube

(720)(g)(ha)

7200hA

Upthrust due to water-

the cube

=(1900)(g)(0,1-h)A

10000(0.1 h)A.

Total upthrust exerted on

the cube

7200A + 10000(0.1–h)A

1000A -28001A

(1000 2800h)A

Weight of the cube 900(0.1)(A)(g)

900A

Total upthrust.

weight of the cube

· (1000 2800h)A = 900A.

h 0.0357m

the cube rises

0.0357m- 1cm

0.0357m - 0,01m

0.0257m (Ans.)

9. manufacture produces two.

different models A and B of a product. Each model. must be

marlines. and Mo. To

complete one unit of each model, the two machines must work the number of hours inlicated in the following table.

A B

1

Ma 1 3

No machine may operate more

than 12 hours per day," The profit is $3 on each unit of model of A and $6 on each unit of model B. How many of each should be produced daily in order to maximize his profit?

19. If an A‚P, and a G.P. are

added together, the sum of their first term is 6 of their second term is −1. and of their third term "i

"If the first term of the G.P. is twice the fir term of the A.P、 find the common difference and the "common ratio: 11.A right circular cone

divided into 3 portions A, B and C by planes parallel to the base as shown in the figure. heights of the portions- are 10, 2m and 3mm* respectively. –Calculate () The ratio of the

volume of A, B and C. (b) The ratio of the

„curved surface ares

of A, B and C.

B

2.

C

3m

T = 120N (Ann.)

(c) Apply the formula

+2as

Letv be the velocity of the 15-kg block when it strikes the floor.

The initial velocity is Ons

(d) 1

2(a)(2) 2(2)(2)

8

2.8284

(Ana.)

Let the time required for

the 15 kg block to reach the floor het (seconds) Apply the formula

initial velocity

-tat2

Substitute

2

-12

■ 1,4142 • (Ani.)

15kg

15 y 10kg As shown in the figure above let the velocity of the 10-

kg block be when it rea-

ches the level at AĢ

2

2

2(a)(2) 2(2)(2).

8

the string slackens, hence,

assume the block moves h

-39(15.492)

7.012 maTM (Ans.)

(c) Let the average retarding force exerted by the ground

on the pile be F newtona.

Hence,

work done against F loss in kinetic energy

+ less in potential energy

Work done against P.

F(0.5)

Loss in kinetic energy

· 4(10 + 12),2

--1(22)(7.042)2

545.5 J

Loss in potential energy

• (10 + 12)(10)(0,5) ➡ 1103

P(0.5) (545.5+110)

Load: ➡ 50g.

E- 312,5N (ADE.)

(d) Let the tension in the

string connected to the 50-kg block be T newtona

-1.6

effort.

When effort. 600N

Hence,

- 1,6

600

T➡ 960N

the net force acting on

the 50-kg block

P = 1311N (Anm..)

- 960 - 50g

➡ 960 - 500

3(a)(i)

'Q ms

- 460N

30kg

20kg 10kg

smooth horizontal floor

600N

As shown in the figure above

let the tensiona in

strings be T

Tand

the

T2 the

acceleration of the system

bé a.

Apply Newton's 2nd law to the blocks seperately,

600 1 - 10■ ...(1) T1 - T2 - 20

(2)

(1),

T2

30a ...(3)

+ (2) + (3)

600 m

(10 + 20 + 30)a 10 ms (Ans.)

(a substitute

Apply Newton's 2nd law,

460- 50(a)

录 = 9,2 mm

(Ans.)

Ana: The block will move

upwards with acceleration 9.2 s

-2

5.(a)(1) Upthrust of water

on the cube

of the cube

E

900(0.1)3(10)

(ADS.)

9N

(ii) Let the volume of the

cube immersed in water be y

Upthrust on the cube (1000)(g)(V*)

-2

a

= 10 into (1)

บา

and (2), we have

1000(10)vi

- 9 x

9

104

Exercise 2

12. A man borrows $8500 to

repaid with interest at 124 in 2 equal annual instalments, the first payment being due at the ́end of the first year.

Find the annual payment correct to the nearest $10.

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數部 #

Mathematics 4

Attempt All questions in

Section A and any Six questions in Section B.

Section A

13.

12Aix 10% more than B and

Bis 10% more than C. What is the percentage of A more than C?

2. It the roots of the

equation x2+kx+9=0 a.e real. Find the values of k. 3. In the figure, the area of

the parallelogram. ABCD. is equal to 600sq.cm. Find 24.

50cm

40cm

In the figure, AB is the diameter of the circle. BAP, RQP are straight lines. 4Q9R=36", Lapq=20°. Find Z RQB. (Geometry theorem need not he quoted when used.)

5. In the figure, AP÷PB=2;1;

AC/./PQ//BD_\\\\If &C=8em, BD=5¢1), find Pų. (Ge one try theorem need not be quote); when used.) c

In the figure, the squRTES and the circle are of equal area and have the same centre. Calculate ZP9Q.

18. A hill-side is a plane inclined at 10" to the horizontal, Astraight path across it slopes upwards at 4" to the horizontal. Find the. bearing of the path

if the line of greates slope bear dne North. 15. In the figure, ABC in

equilateral triangle and ZPAQ-120, PBCQ is a straight line. Prove that:

2 PB.CQ-BC2 PB: CQ-AP:AQ]

16. IP, TQ`are the tangente:

from a point Ț to a circle; N in the mid- point of the chord PQ+I/ is the mid-point of To If PH cuts the circle at , prove that q, N, R, I are concyclic,

T

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