頁二第張八第
日七十月四年午戊磨豆
“WAH KIU YAT PO
報日僑華
二期星
物理科建議答案
魯榮家
(c) Let T and a be the tension
in the string and the
leration of the syst pectively,
4.(a)(i) According to the
lense formula
res
SUGGESTED SOLUTIONS TO
PHYSICS PAPER I 1978
SECTION A
MECHANICS
The centre of mass
of the lamina lies on the straight line EB (Ans.)
(ii)
As shown in the figure;
Gịp G2 and
centres of
u
are the
8 of the laminas
ABE, BCDE and ABCD respective-
Let
density of the
lamina
thickness of the
lamina
Weight of the triangular
lamina ABE
2
(F)(g){}h2)(t)
Weight of the rectangular
lamina BCDE
· (P)(g)(ha)(t)
Moments of the weights of the triangular lamina ABE and the rectangular lamina BCDE about EB are equal in magnit-
ude.
~~ (P) (z) ( +h2 ) (+)})
(P) (g)(ha)(t)(ža)
Ja
2
- 1.732a (Ans,
(b) From the result of part
(a), the c.g. of the lamina lies on EB when
h = 1.7328. When h> 1.732a
the cog. will lie within the triangle. (In other words, when h≤ 1.732a, the cag. Will lie within the rectangle,)
As the slab has h
and a 2m
h> 1.732(2) » 3,464
m
b 7 1.732a
Therefore, the cag. of the slab lies in the triangle ABE and the sláb is not able to stand with BC on the horizon- tal ground. (Ans.)
I hi
(a) Upthrust of water on Y
- weight of water displaced
(F)(g) (v) ̈
(1000)(10)(0.0015) (Ana.)
15N
(b) Y will move upwards
after the system is released (Ans.)
ISN
TOM
As shown in the force diagrams above, apply Newton's 2nd Law of motion to X and T seperate-
19%
From the graph
7(10):
T + 15. (1) + (2)
TA-7a 8(10)
(1)
Ba (2)
when
0.102
1
+ - 0.102 cm
-15a -2
ms (-Ans.)
when
(-0.3333 ms
(d) Y will finally stay at
the position of its volume immersed in water.
(Explanation : Initially, the system is not balanced due to the fact that the up- thrust on Y is large enough to push the block Y upwards. If it is reduced by 1/3 of its value (i,e, decreased by 5N) the system will be ba- lanced. Therefore, the system. is balanced if block Y is immersed 2/3 of its volume
in water such that the up thrust on Y is reduced from 15% to 10N :)
3(a)
Let the velocity of A after collision be
Takak
the direction to the
as positive.
Initial momentum of the
system -(1)(12).
12 kg ms
Final momentum of the systen
• (1) (v)
(15
(3)(5)
kg ma
By conservation of linear
momentum-
1215 +
Therefore,
*3m (Ans.)
Velocity of to the left..
A is 3 ms (b) The average force on A
rate of change in momentum
of A
(1)(-3) = (1)(12)
30N
(Ans.)
The negative sign indicates the direction of the force on
A is to the left.
(c) Initial kinetic energy-
(1)(12) 2
72J
Final kinetic energy
4(1)(3)2 + 4(3)(5)2
42J.
The difference in kinetic
energies
-
72J 42J
30J (Anm.)
The 30J of mechanical energy converts mainly into the energy of the spring, heat energy and sound energy,
HEAT, LIGHT, SECTION B
SOUND AND WAVE MOTION.
0.098 am
0.098 cm
The average value of
(0.102 0.098) +
0.1 cm
-0.1
10 cm
0.1m (Ans.)
(11)
sifying
As shown in the figure, the observer and the object are on the opposite sides of the magnifying Lena. The object should be placed between the focus F and the lens.
(iii) REAL-IS-POSITIVE
H
From the diagram, the imag
is virtual.
Apply the formula
日三十月五年八七九一腊公年七十六國民中 青激佳瑩
sine
sine,
1.65
0.6061
4 - 37°18′ 45′′
102
ine
1.35
= 8.7407
= 47°48' >
the pair with refractive
ndex 1.35 should be used,
5.6)(i)The wavelength
10ст
4cm (-0,04m) (Ans.) (ii) The frequency
2.5 0.25
■ 10 Hz (Ana.) (111) ▼ = fa
10 x 0.04
0.4 ms (Ans.)
· (iv) The part move
downwards at A and upwards.
at B at the "moment shown. (v) The frequency remained
unchanged
f = 10 Hx (Ans.)
The new velocity
= 0.4 x
0,2 s
The new wavelength
0082009
0,02m
(Ans.)
(b)(i) Sound wave is a long- itudinal wave while water
wave in a transverse wave..
(ii) Let the wavelength be λ
end correction be
†λ • (32 + e)
Assume end correctingen
negligible,
6.(a)
20 cm
cm
-0.2m (Ans.) The negative sign indicates the image is virtual.
The image so formed is erect, virtual and enlarged,
(b) A periscope formed by two (45°-90°-45°) triangular prisms should be arranged as
shown in the figure.
A
Total internal
Place
• reflection takes
The periscope works if
total internal reflection.
takes place at the inclined
surfaces AB and CD.*
•
object
Let 91 and 82 be the critical angles of perspex with refractive index 1.65 and the perspex with refract- ive index 1.35 respectively.
32cm
2 128 cm
∙12
1.28
-(256)(1.28)
327.68 ms
is
(Ans
From the graph, the length.
of nitrogen column is zero when the temperature
Absolute zero
-265°C (Ans.)
-2658
(b) The volume of a fixed Dans of gas is directly proportional to the absolute temperature at constant
pressure.
(c) A thin wallcapillary tube should be used because the
temperature may be more uniform between the liquid and the gas in the tube. (d) Apply the conclusion
drawn in (b). **
265
20
V 265
+360
where V is the volume of
nitrogen at 360°c
1.5 x 625
285
3.29 cm3 (Ans.)
(- 3.29 x
(e) Let the cross-sectional
area of the tube be A.
From the graph, the length
of the nitrogen column at
is 2,8cm.
Apply Boyle's Law
(760)(2.8A)
(750)(LA)
where Lis the length of
the nitrogen column when the atmospheric pressure is
750mmlig.
760 750
x 2.8 cm
2.837 cm
0.02837 m (Anṣ,
SECTION C ELECTRICITY,
MAGNETISM AND ATOMIC PHYSICS
(a) The readings of the an-
meters in ascending order.
magnitude is Aya Aŋo Azı Ay
(b)(i) Let
resistan
vire P
-resistance
wire
- () ()
(Ans.)
across P
(ii) Since Power
and the voltag
and R are the same,
Power C
rate of heat pro-
duced in P
rate of heat pro-
duced in R
Let Pp
8; 3 (Ans.)
Az
(c) If K is closed, the read
and ings of amasters A, will increase and the readings of ammeters A2 and A will drop to zero.
If 3 is absent, A¡Az
would be ruined.
8.(a)
The direction of the force (F) is upwards, perpendicular to both the motion of the
electron and the magnetic
field as shown in the figure
above,
(b)Since there are free elect-
rons
on the surface of the
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