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CCC HOH FUK TONG COLLEGE
Second Term Examination, 2001 - 2002
S.4 Physics Paper 1 Answer
Section A (30 marks)
1. (a) Raising the temperature of a fixed mass of gas at constant volume will increase the average random kinetic energy of air molecules. (1)
Consequently, the molecules will collide with the wall (a) more frequently and each time with (b) a larger speed. (1)
These two effects combined together result in an increase in the rate of transfer of momentum to the wall due to bombardment of gas molecules, so the air pressure in the flask is also increased.
(1)
(b) By Pressure Law:
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=
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(1)
(3+1C)
P = 31291.9 Pa (1) (2)
2. (a)(i) P: liquid
(ii) R: liquid and solid
(1)
(1)
(1)
Latent heat is evolved which compensates the heat lost to the surrounding.
(c) By m

L
0.2
 f
= P

t
L f
= 100

[(20 – 8)

60]
L f
= 360000 J kg
-1
(1+1C)
(2)
3.
(a)
T x
3 m
T y
0.5 m
0.2 m
P X Y
Q
P
1.5 m
0.6 m
20 N
10 N 40 N
(b) A change of state occurs.
Taking moments at X:
20(0.2) + T y
(3-0.5-0.2)
4 + 2.3 T y
4 + 2.3 T y
2.3 T y
T y
= 10(1.5-0.2) + 40(3-0.2-0.6)
= 10(1.3) + 40(2.2)
= 13 + 88
= 13 + 88 – 4 = 97
= 97/2.3 = 42.1 N
Since T x
+ T y
= 20 + 10 + 40 = 70
T x
= 70 – 42.1 = 27.9 N (3)

4.
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(b) Since T y

T x
, string Y will break first. (1)
Taking moments at X:
20(0.2) + 45(3-0.5-0.2) = 10(1.5-0.2) + A(3-0.2-0.6)
= 10(1.3) + A(2.2)
= 42.95 N (2)
4 + 2.3 (45)
A
(a)
(3)
2 labelled axes with units (1) correct speeds at t = 0 and t = 6 min (1) straight line (1)
5.
(b) Distance travelled = area under v-t graph
=
1
2


 (40

120)

1
60

6 
 km = 8 km
(c) The reading on the milometer = 2271 + 8 = 2279
(a) By 2as = v 2 – u 2
2(-10)s = 0
2
- 15
2 s = 11.25 m
(b) acceleration = gravity = -10 m s
-2
(accept 10 m s
(c) By s = ut + ½ a t
2
0 = 15(t) + ½ (-10) t
2
15t = 5 t 2
t = 3 s
-2
)
(3)
(2)
(1)
(2)
(1)
(2)

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Section B (60 marks)
6. (a)
Immersion heater thermometer
OUT IN
OUT IN
Power supply to IN of joulemeter
OUT of joulemeter to heater
Heater in middle hole
Thermometer in side hole
(1)
(1)
(1)
(1)
(4)
(b)(1) energy supplied = 153602 – 113451 = 40151 J (1)
By H = m

s
  t
40151 = 1.5

s

(86 – 25) (1) s = 438.8 J kg
-1
K
-1
(2) power =
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=
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= 334.6 W
(1)
(4)
(1)
(c) s = 438.8 J kg
-1
K
-1
(2)
(d) higher (1) because the actual energy absorbed by the block is less than 40151 due to
(3) energy loss. (2)
(e) Surround the block with cotton wool. (2)
7.
(a)(i) By s = u t + ½ a t 2
500 = 0 + ½ a (20)
2 a = 2.5 m s -2
(ii) Uplifting force - weight of the helicopter = ma
i.e. F – 1500(10) = 1500(2.5)
F = 18750N
(2)
(2)

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8.
(b)(i) acceleration / m s
-2
8
Correct shape
From t = 0 -2.5 s, a = 8 m s -2
From t = 2.5 s, a = 0 m s -2
0 2.5 time / s
(ii) (I) At t = 2 s, a = 8 m s
-2
F = 80 x 8 = 640 N
(II) At t = 4 s, a = 0 m s
-2
F = 0 N
(c) Total distance travelled = area under the graph i.e. 500 =
[ (t

2.5)
2
 t]

20 t = 26.25s
(1)
(1)
(1)
(3)
(2)
(2)
(4)
(a)(i) The block moves at a constant velocity.

frictional force = applied force = 5 N (1)
(ii) By Newton’s second law of motion, F = ma
10

5 = 5a a = 1 m s

2
(b)(i)
75
5
A
R 10
75 – A = 5 a --------------------- (1)
R = 10 a
Adding:
--------------------
75 = 15 a

a = 5 m s
-2
(1)
(2)
(1) Into (1) Net force = 75 – A = 5 (5)
= 25 N (to the right)
(1+1)
(2)
(2) Into (2) R = 10 (5) = 50 N (to the right) (1) (4)
(ii) The forces are a pair of action-and-reaction.
By Newton’s third law of motion, the magnitudes of these forces are the same(but they are in different directions.) (2)
(c)(i) Let T be the tension of the string.
For the 5-kg block: 50

T = 5a ------------- (1)
For the 10-kg block: T = 10a ------------------ (2)
Resolving (1) and (2), a = 3.33 m s

2
(3)
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9.
(ii) In (c)(i), the weight of the 5-kg block (50 N) is used to accelerate both blocks. (1)
Now the 50-N force is used to accelerate the 10-kg block only.
(1)
The acceleration of the 10-kg block is greater than that in (c)(i).
(1) (3)
(a) (i) By Conservation of Momentum
2(10) + 1(2) = (2 + 1)v v = 7.33 m s -1
(2)
(ii) By F = Ошибка!
= Ошибка!
= -267 N
(2)
(to the left)
(1)
(b) (i) By Conservation of Momentum
(5)
2(10) + 10(0) = 2(-2) + 10(v) v = 2.4 m s
-1
(iii) By F =
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(2)
=
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= -1200 N (2)
(to the left) (1)
(c) (i) By Conservation of Momentum
0 = 2(v) + 0.5(2) v = -0.5 m s
a = -0.5 m s
-2
-1
(- means recoil)
(ii) By 2as = v
2
– u
2
2(a)(0.25) = 0
2 – (0.5) 2
(2)
(2)
By F = ma
F = 2(0.5) = 1 N (1)
(5)
(5)
*** END OF SOLUTION OF PAPER I***
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