1980-10-14 — Page 24

Take the

the horizontal

component of R

frictional force

百四第張六第日六和月九年中庆层夏 WAH KIU YAT PO

31:019 #

R2 =

10N

(ans.)

1981

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物理

()

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W.K.LO

PHYSICS (2)

(HILL & DALE PRESS LTD.)

Suggested solutions

Exercise one

(Take g = 10ms

(a)(i)

4m

100g.

As shown in the force

diagram above, take moment

about A.

7(0B) = 100g(AB).

where F is the horizontal

force required.

OA

I 100m 5m

OB = 5 4 = 1m

upward

direction

as positive

-Let the tension

in the spring of the

spring balance be T, the acceleration of the

elevator be a and the

mass of the block be m weight of the block W = BE..

mg:

(a) when a = 2.5mg

二期星

between the ladder.

and the floor.

weight of the ladder.

(ii) Since the Ladder is

ilibriú, the N2 - £ = 0 ....(1)

· = 0....(2)

Taking moment about

811

No x 8 = 0

= 0 ....(3)

(1)

From (3)

3 x 160(10)

T100N

Substitute into (1)

Ans

m = 8kg

True weight of the block

CON (Ips.)

(b) T = 641

Substitute into (1) 64 - 8(10) 8a a-2fs

(ins.)

Ans. The elevator

moves dowwards

with acceleration 2ms (or the

elevator has an

SOON

Tue direction of the force acting on the upper end of the ladder is

Serpendicular to the vertical wall with magnitude 60CN.

(iii) The frictional force

f is horizontal. From (1)

- 6007

(iv) The resultant R

日四十月十年〇八九一届公年九十六國民華中宵教僑華

(1) As shown in the force diagram below

+6x-9

6x = 53

length of AB is 82

си

The bouyant force B will

apply

at tie mid-point part of the rod

merved in water.

(ii) W - vei hit of the

rod

1.26 12

weight at the end of the rod to be determined.

The volume of the cod

m:

X

2.4

(where d is the density of

water)

Volume of rod submer ea

in water

2 x 2.4

(Ans.)

4. Solution: Let the length

of a = 4x

The length of b- 5x The length of G

6x

2

2

2(5x)(6x)

5. Solution: Cost price

shoes.

← $270 ÷ (1-10%)

- $300. SN

Selling price of shoes if gain of 10% is intended

= $300 × (1+10%)

- $330.

6. Solution:

Vi+Sing

1+sinx/1-sinx

130

√1+Sin30.

-Sin30

√ √

AB JOA

OBS

OB

F(1)

=√24m = 4.899

100(10)(4.899)

(Ana.)

— 7 = 4899N

(ii) The minimum force F

should tangent to the

wheel as shown in the diagram below.

is tangent to the min wheel at point C which is

one of the extremes of diameter AOC of the

circulaz wheel

Taking moment about

min

(AC) = 100g (AB)

(10) = 1000(4.899)

F

min

= 489.9M

(Ans.)

(b)(1) If the force is

applied to the 10kg block,

5 kq

I 10. kg R1

where Ry is the force

between the blocks.

the reaction of 5-15 block on 10kg block Let the acceleration of the

-2

system be a lis

F - R1 = 10a.

R1 = 5a

also, F = 15%

(2)

solving R from (1) and (2) we have

Ry 5N

(Ans.)

(ii) Similarly, let the

force between

e blocks be R and the acceleration.

upward decelerat- -ion 2ms 2)

(c) If the elevator moves

upwards with uniform

speed, then a = 0 From (1)

Tmg

= 80%

(Ans.)

Fonce, the reading of the balance is 8011.

(a) If the elevator moves

downwards with unifora speed, hence, we have

al so

O

SON

(ins.)

Therefore the reading of the spring balance is 80

(e) If the cable wire

breaks, then the

élevator will fall

*

5002

+1600

1708.8N

(Ane.)

Also,

tane

0.375

(Ans.)

220°33

The resultant at the end of the ladder is 1708.01 making 20°33′ with the

vertical,

.(a)(i) Let v and v1 be

the velocities of the

block A after the

bullet emerges and

the bullet as it

energes from block

respectively.

conservation of

energy

1 (2) v (2)(g)(0.2)

1

2ms

(ans.)

down with acceleration (ii) By conservation of

equal to the gravitat- ional acceleration g. Hence

mg

Therefore, the reading of the spring balance will then be of. 5.(1) The force diagram is

shown below:

where,

N Normal reaction

acting on the upper end from the smooth wall.

R the resultant force

linear moi entum The initial momentum

(0.01)(2400) 24koms

The final momentrum

(2)v、 + (0.01)(vg) (2)(2) + 0.6lv1

4 + 0.011

+0.0171

2000m

The bouyant force B

weight of wate displaced by the rod (density of water)

x (volume of the rod

submerged) x g

a xả 10

=20N (Ans.)

(iii) Taking moment about

the hinge.

(Bsine)(2.5) = (Wsine)(3).

+(wsine)(6)

2.5 x 20 12 13 +6w

72-3331 (kas.)

1981

中學會考試題預習專欄

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Mathematics 2

C. P. Kan

MILL & DALE PRESS LTD.

Solution to Exercise 1. Section A

1. Sölution:

10810(x-3).

-9x+18

-10g10(x-6) 108,18 10610(x-6)

» 19810

10

~~9x+8 = 0.

10

x- 8 or 1 (rejected)

Since

2

pr3+872+qx−2 is divisible by (x+2)(x-1),

f(x) =

(Ans.)

f(1)

ive.

(b) Let v be the velocity of the block B and the bullet after the bullet hit the block B.

Since the bullet is ebedded in bloc. B, hence, by conservation of mo-

sentum (0.01)(2000)

* p + 8 q = 2

P + q • -6 -8p+32-2q-2

1.e. kp + q = 15

(2)-(1)

3p 21

P. 7

Put p-7 into (1)

(3+/5)

Solution. -H is the

orthocentre

AABC.

BFCBEC 901

B,F,E and C lie on circle with BC as

diameter

BM MC

Mis the centre"

circle BFEC

ZABE 180

40%

FME 2LABE

80

-50°

Solution: Simple intere

for 3 years

$10000 x 3% × 3 $900

Compound interest for 3

years compounded yearly

= $10000 x

-$10000

- $927.3:

+3%)3

The difference between simple and compound interest

= $927.3 - $900

- $27.3L

Section B.

9. Selution: (1) The equation

2

**-3kx+Jk+8 * 0 bas equal -(-3%)2=4(3k+8)

2

roots

0

9k12k-32 (3k+4)(3k~8) × 0

or:

8 3.

(b) The equation

x2-3kx+Jk+8 = 0 has ne

-(-3k)2-4(3k+8)

real roots

(1)

9k -12k-32 < 0

0

(2)

(3k+4) (3k-8) < 0

<<

8

Solution: Let OP - r em

“Aren of circle-Area of ABCD

ON CH

(2x)2

(0.01

0.99)v

= 20ms 1

Solution: Let AB-xcm;

BC=ycm.

1. 128x

402

(1)

Com PON

(3)2 (1)-(2)

AD

- 100

(2)

-(x-3) 2.

44

agting on the lower By conservation of

end of the ladder.

R2 R2

of the system be a'ma".

15

=5a

10a

N2 the vertical

From which

Component of R.

400.01 +

(0.01 +

h = 20m

30

28x 27 36:

4P0Q 24PON

55 12.