1979-10-30 — Page 22

其二第張六第 日十月九年未己腦裏

1980

中學會考試題預習專欄

物理

(四)

明德出版社魯榮家提供資料

Physics (4)

Anever to Exercise 2

1(a)

A

WAH KIU YAT PO

報日僑華

二期星

日十三月十年九七九一縻公年八十六國民中青教儒鞌

2

Load

2gh

1.6

In the figure, AC -

AB - HD

(u) LAQB

E

123

hoad

=

240

50g

500N

and (BAD = 4/BAC. Find the values of x and y.

(b) PO = P8,

-1

-

i

500 E

1.6

(c) If BP

LAPR.

is produced to cut the circle with centre (at 1, then i AR // BO

2(10)(12)

15.492ms1, Ans.

(b) Let the common ve

velocity

required be v ma

·

Initial momentum of the system

- (10)(~;) kgmu ̄1.

Fian momentum of the system

(10412)v kgme

-1

By conservation of

linear momentum,

10v i

(10+12)v

E' = 312,5N. Ans.

(d) Let the tension in the

string connected to the 50kg-block be T newtons

T

1.6 effort When effort = 600N

B

T

600

=.1.6

* 960 N

Hence, the net force

3. A manufacturer's catalogued

prices are 25% above cust,

but buyers are allowed a discount. of 8% off these

"What is the

prices.

manufacturer's nét gaine per cent?

4. Divide a load of 120kg

between A, B, C so that

A bas half as much again as B, and B has twice as much as C?

R

B

14.

115 kg

a mg

10 kg

As shown in the figure above, let T. tension in the

string,

acceleration of the system.

Apply Newton's 2nd law to

the blocks seperately,

15(10)-T - 15a

T-10(10) 108

(1)+(2):

-

50 - 25a

-2

a

2 ms

+

Ans

(b) Substitute a - 2 into (2)

T-10(10)

10(2)

Tx 120N. ADS

(c) Apply the formula

2

Let Vi

2

* + 2as

be the velocity of

the 15kg-block when it

strikes the floor. The initial velocity is Oms-1

•

V

M

10 22i

10

•

(15.492)

- 7.042 ms-1.

5.

acting on the 50kg-block

- 960-50g

- 960-500

= 460N

ABB.

Apply Newton's 2nd law,

460 =

50(a)

#

- 9.2 me-2.

Ans,

(c) Let the average retarding

force exerted by the ground on the pile be

F newtons, Hence,

work done against F

- loss in kinetic energy

+ loss in potential

energy

Work done against P

·F(0.5)

=

+

Loss in kinetic energy

19

(10+12)v

$(22)(7.042)2

- 545.5J/

Loss in potential energy

(10.12)(10)(0,5)

110J

.*. P(0.5) (545.5+110)

F1311 N. Ans.

3(a)(i)

a ms2

Ans: The block will move

upwards with accelera- tion 9.2 ms-2.

5(a)(i) Upthrust of water.

on the cube

- weight of the cube

900(0.1)3g 900(0.1)3(10) -

- 9N

Ans.

(ii) Let the volume of the

cube immerged in water

be Vi

が、

Upthrust on the cube

(1000)(g)(V1)

1000(10) 9"

y = 9x10-4

Since the volume of

the cube

(0.1)3

*

6.

In the figure. The circle RPQ touches BC at P. AB produced at R and AC

produced at 0. If [ARC = 70° and B&C

50°.

Calculate

BRP and PRQ. (Geometry theorem need not be quoted when used.)

B

In the figure, XY// BC, BC-8 cm, XY5cm; the lines RC, XY are 2cm apart. Find the ares of AAXY.

(Geometry theorem need not be quoted when used.)

7. In AABC, if a:bre - k:5:6,

find cosa.

8. If A+B+C = 180°,

14. In the figure, AD is a

tangent to the circle ABC and AC is, tangent to the. circle AHD. Prove that (a) AANC and ADBA are

sigilar,

(b) AB = BC BD,

(c) If the tangent at C

and Ŋ meet at T and

[CTD = 120°, calculate ZCAD.

15. Find, in c.e. the volume of

a closed cylindrical tube, internal diameter 2mm, internal length 35cm (take X-2). is filled with

which weighs 13.6 g

mercury

per c.c., find the weight of the mercury, correct to

10 KD.

16. A's income is 60% of B's

income and A's expenditure is 70% of R's expenditure. If A's income in 75% of

B's expenditure, find the rate of A's,eaving to B's saving.

1980

中學會考試題預習專欄

simplify

2

K

-

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

- 8

= 2.8284 ms

Ans.

(d) Let the time required for

the 15kg-block to reach the floor be t (seconds). Apply the formula

(e)

2

+ jat2

initial velocity

*4at2

Substitute & 2

2.

t-

+(2)+2 √

- 1.4142.

T

130kg +2014+104600N

smooth horizontal plane

As shown in the figure above, let the tensions in the strings be T. and T. the acceleration of the system be a,

(1)+(2)+(3):

I

(10+20+30) a

a = 10 ms «*.*

(ii) Substitute a 10 into (1) and (2), we have

T1

==

T

2

(b)(i)

a ms2

600-10(10) 500N.

30(10)

300N.

Ans

Ans

1 x 10-3

CosAsinC – sin(A+B)cos(B+C) COSĄCONC2 + Cos(A+B)cos(B+C}

Section B

fraction of the cube intersed

9.

2008

9 x 10

Д

1 x 10TM

4cm

Sin

數字推理 (+)

9 10

600-T1 T1-2

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

- 10a

D

5cm

(b) As shown in the figure

below

(1)`

*

208

(2).

=

T2 300

(3)

600 -

-2

(7:20 kgm2>

10.

P

Water

(1000 m2)

A U

B

-

C

R

S

In the figure, AB // DC,

AB =2cm, BC • Gem, NC = 5cm,

AD 4cm, find

(a)

{8} ZADC of ABCD.

數字推理練習七

***B*****

選出下列有關的正確答案,並在它的下面畫

15kg

115 kg 10 kg

As shown in the figure above let the velocity of the 10kg-block be v2 when it reaches the level at A.

hence,

2 2

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

8

the string slackens, sume the block moves à meters farther before drops

down, 2

2

2(g)(h)

2

་

8

30kg 120kg 10 ke

- 6:00 N

fi

As shown in the figure above, f,, f, and f, are the frictional forces of the 10kg, 20kg and 30kg-blocks respectively.

13

-0.2(10)(10) - 20N

0.2(20)(10)

0.2(30)(10)

40N

*

GUN

Let the acceleration of the system be a', the' tensions in the strings be T' and

'2

1

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

600-T*,-20 =.10a '

1

ནཱ་དྡྷ – ཡ''ཊྛམནཾ༦ * 20'

Let his depth of oil

A cross-sectional

area of the cube (= 0.01 sq.m.) Upthrust due to oil on the cube

"(720)(g) (ha)

72001A

Upthrust due to water on the cube

1000(g)(0.1–h)À

- 10,000(0,1-4)A

Total upthrust exerted ou the cube

-

7200A+10,000(0.1-h)A 1000A-2800h A (1000-2800k) A Weight of the cube 900(0.1)(A)(g)

900A

Total upthrust

weight of the cube (1000-2800h) A - 900 A

h = 0.0357m

the cube rises,

0:0357m ». 1cm

≈ 0.0357m-0.01m

-0.0257m

In the figure, ABCDEF in a regular hexagon of side 7 cm and the triangles are all isosceles, each having its equal sides 8cm long. It is then folded along the edges of the hexagon to form a right pyramid with ABCDEF as base and. with the six points P, Q, R, S, T, U coming together at the vertex, say V, calculate

(a) height of the pyramid, (b) volume of the pyramid,. (c) angle between the face

VBC, and the base ARCDEF.

11. A particle P starts from

a point 0. and moves along a straight line so that its velocity, v m/s, after time t sec in given by

v 30 6t5. Calculate (a) tlle velocity with which

-the particle leave the

point 0),

(b) the time taked for the vulicity of P to reach 50 /

(c) the minimum velocity

of Pa

1.下列五姐線段中,那一組的線段互相垂直?

A

B

C

D

E

X X 11 = xx

2.下列五怨線段中,那一組的線段互相平行?

A

E

B

X

D

3.下列五個圖形中,那一個與其他四個不相同?

A

B

E

B

4.下列五個圖形中,那一個的周界最短?

A

-C

E

D.

MOUWB

5.下列五個圈形中,那一個的有斜蝶部份佔全

A

?

C

6.下列五個子

P

個真其他四個不相网?

S

T

Өшефа

7. 在下圖中,那一線段與AB平行?

PAC

An:

(1)

1980

(2)

T-60 - 30a1

.(3)

中學會考試題預習專欄

(1)+(2)+(3):

600-120 6Qa'

數學心

** If (x+y)∞ (*~y), show that

1(10)

= 0.4m

Ann: The maximum height

reached by the

10kg-block is

0.4+2 2.4m above the floor.

2(a) Let the velocity of the

pile driver as it strikes the pile be v, IE

F

V

2

By conservation of energy

mgh

where m

"mags of the pile-driver

- 10kg

and h-height fallen

- 12

4(a) V.R. ** 4.

.*. M.A. (c) Let the minimum effort

required be E

a' - 8 ms2.

-2

Ans.

(ii) Substitute a'

into (1),

500N.

Substitute a' into (3),

明德出版社交長连接 Mathematios 4

hence deduce that

(x2+xy+y2) ac (x2-xy+y2).

Exercise 2

13.

ANE

8

Anaver ALL questions in Section A and any Six

questions in Section B

T

- 300N.

2

Section A

1. If x

y

(b)

H.A.

· x 100% V.R.

MAL 100%

40%

40

22

2

3/2 .. 2/3 and 3/2+2/3, simplify

In the figure, two unequal circles intersect at A and

ху

·

Ans..

B. The centre 0 of une cirole Jies on the circumference of the other. The straight line APQ outa the circle at P and Q.

* OC OD

ZA OE OB

OP

Prove that

QCD

RDE

SFG

TEF

B.在下面中,那一線侵觀日日亙相叠道?

PBC

QCF

RAH

SAB

TFE