買四第張八第日六十月九年午戊
1979
-2
(a) when a - 2.5 ms
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物理
Physics (2)
Suggested solutions to
Exercise one
(Take g➡ 10 ma
1(a)(i)
100g
As shown in the force diagram
ve, take moment about A.
r(03) = 100g(AD)
where F is the horizontal
force required.
-
AB -
1 x. 10m - 5m
5
OF
24m 4.899m F(1) 100(10)(4.899) F■ 4899N (Ans.) (ii) The minimum force F
min should tangent to the wheel
as shown in the diagram below
min 19
Joog
is tangent to the wheel at point C which is one of
the extremes of diameter AOC
of the circular wheel, Taking moment about A
Tên (AC) . 1002(AH)
min
Fain
(10) - 1000 (4,899)
F
min
489.9N · (Ans.)
(b)(i) If the force is applied
to the 10-kg block,
10 kg 5kg
10kg
By is the force between
the blocks
the reaction of 5-kg block on 10 kg block.
water
- 2 x 204
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二期星 日七十月十年八七九一靨公年七十六國民華中 南教僑華
➡g ma
the vertical wall with
magnitude 600N.
Volume of rod submerged in
AD-BC
DAG=L.FCB
ADG-4 FRC
AANG ACBF
m- 8kg
is horizontal.
From (1)
f
GOON (Ans.)
-AG=CF\
GE
AF
The bouyant force B
weight of water displaced
by the rod
T = 10kg 10(10) 100N (iii) The frictional force f Subatitute into (1)
True weight of the block
8kgt = 80N (Ans.)
"
(b) I£ T = 6,4kgf = 64N Substitute into (1)
64 8(10) Ва
2
-= -2ma (Ans.)
Ana The elevator moves
downwards with acceler
2
ation 2 me2 (or the elevator has an upward deceleration 2 ms
(c) If the elevator moves
uniform upwards with
speed, then a 0
From (1)
T
mg
mg
80N 8kg
(ans.)
Hence, the reading of the
balance is 8 kgf.
(d) If the elevator noves downwards with uniform speed, hence, we have also
q 8kgf (Ans.) Therefore the reading of
the balance is 8kgf.
(e) If the cable wire breaks,
then the elevator will fall
down with acceleration equal
to the gravitational acceler-
ation g. Hence
Therefore, the reading of
the spring balance will then
be 0 kgf.
3.(i) The force diagram is
shown below
where
Let the acceleration of the
system be a s
10a
5a
F 15N
R
Normal reaction acting on the upper end from the
resultant force act→
ing on the lower end, of
the ladder,
N2
the vertical component
of R.
(iv) From (2)
-W-16:00N
No
The coefficient of sliding.
friction between the ladder and the horizontal floor
1600
0.375 (Ans.
(v) The resultant R
+ N2
√600 + 1600
1708.8 N. (Ans.
Also,
tang-
=
0.375
20° 33° (Ans.) Ans: The resultant at the
of the ladder is 1708,8N making 20°33' with the vert
ical.
4(a)(i) Let
and
be the
velocities of the block à
after the bullet energes and
the bullet as it emerges
from block A respectively.
By conservation of energy
- (2)(g)(0,2)
(2)v
2 ms (Ans.) (ii) By conservation of
linear momentum
The initial momentum
(0,01)(2400)
-1
- 24 kg ma The final momentum
(2) ▼ + (0.01) (v; } (2) (2)
4
+ 0.01v2
24 = 4 + 0.01)
-2000 s (And,
Let v be the velocity of
the block B and the bullet
after the bullet hit the
block B..
Since the bullet is embeded
in block B, hence, by cons
servation of momentum. (0,01)(2000) (0.0140.99)▼
=
20 ms
By conservation of energy
±(0.01+0.99)(20)2
(0.0140.99)gh
20m (Ans.)
-(density of water) x (volume
of the rod submerged) x g
x · 10
- dx
- 20% (Ans)
(iii) Taking moment about
the hinge
(Bsino)(2,5)
(Wsino) (3)
+
(wsinė)(6)
5 x 20 - 12 x 3 4 6w
2.333N (Ams.)
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數學
Mathematics 2.
Solution to exercis
Section A
1. Solution:
Let p and q be the roots. the equation
•p+q=1__and⋅ pq» P2+q (p+q)-2pq
=
►1-2(-5)
p2-(-5)2-25
The required equation
X2-11x+25=0
Solution:
51ogX=1+4log2-10g5
15
5
log} =log10+1pg2*-log! -logx log10x2
X-2
3. Solution:
The ratio of AF to CF is 3:1
Solution: Let LCDE-X°
ZABC=¿ CDE=X" ZDCE 2 ABC+Z_F
=X°+40°
In A DCE
X+X+40+5)=180
Section
2X-110 X-55
CDE-55
9. Solutio, dami
(a) Since the graph_y=a~(x+b) passes through the points (1,0) and (0,-3).
a=(1+b)==0)
(1) ~(2)-(1+b)2
1-2b-b
-2b=4
b-2
(1)
(2)
Put b into (2) a-4=-3
Since the point (C,0) lies on the graph
(b).
0-1-(C-2)2
C-2=+1
C-3 or
C=3
The maximum value of
(a) AE= cm
HE = √({})2 + (£)2
10, Solution:
sinA-cosĂ. sinA COSA
sina-cosA
COSA sinA+coSA
COBA
tanA-1
Solution:
8sin 0-9cose-6sinecose
ain 0-6sinəcos0-9cos ̃A=0
8tan-0-6tan8-9=0
(4tan8+3)(2tanė-3)=0
tand or
0-1458 523.8*;
56°19' or236 19%
5(1) As shown in the force
diagram below
5. Solution:
The perimeter of the
second square
x4cm
em
The ratio of the perimeter
of the second square to the first square
導
+
The perimeter of the
fifth square is 1,5cm
solving R. frær (1) and (2)
we have
5N (Anm.)
(ii) Similarly, let the force
between the blocks be B
and the acceleration of
the system be aa ms
15 R2 - 5a' 10a
From which
Take
the upward
direction
as positive
Let the tension
in the spring of
1ON (Ann)
上回
the spring balance be T, the acceleration of the elevator
be a and the mass of the
bluck be.
.. weight of the block
f the horizontal component.
of R
friction force between
the ladder and the floor.
W - weight of the ladder,
(11) Since the ladder is
in equilibrium, therefore,
Ne
W = 0
Taking moment about B
f 0
(1) (2)
N18-0
The bouyant force B will apply at the mid-point of the part of the rid sub- merged in water.
(ii) wom
OW From (3)
BN1 - 0 (3)
weight of the rod 1.2g
12N
W x 3
3 x 160(10)
8
600μ (Ans)
Ans: The direction of the
force acting on the
upper end of the ladder
is perpendicular to.
weight at the end
the rod to be
determined.
of
The volume of the rod
1.2 0.5 x
2.4
(where d is the density of
water)
percentage of water in the glass
5x22+7x27%
... 1.5-p()
риб
(c) sum of perimeters of the
five .squares
5+7
-3-015x100%
-25%
Solution:
x100%
Hatio of the rent for A and B:
=23x27:21x59.
69:91
The rent paid by'
A $14500:
#6210
The rent paid hy
B-14400–$6210.
=88190
7. Solution:
Draw DG BF
CF GE 1
FG ED
CM
cm
WE-2
3(6.712)
си
CD
cm
CM