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
中學會考試題預習專欄
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.