Answers for March Holiday Homework
(Adapted from Geylang Methodist Prelim 2010)
1D
2C
3A
4B
5C
6A
7C
8C
9A
10B
11C 12C 13A 14C 15B 16B 17B 18A 19D
1 The variation with time t of the speed of two cars A and B is shown in speed-time
graph below.
Both cars travel along a straight road in the same direction.
(a)
Car A travels at a uniform speed. Car B is at rest before it starts moving at
t = 4 s with a uniform acceleration.
(i)
State what is meant by uniform speed.
[1]
Uniform speed refers to the constant rate of changes of distance.
(ii)
State what is meant by uniform acceleration.
[1]
Uniform acceleration refers to the constant rate of change of velocity.
(b)
At time t = 0, car A passes car B. Determine the distance between car A
and car B at time t = 14 s.
Distance between car A and car B
= area of shaded trapezium
= ½ (4 + 14) x 20
= 180 m
GMS(S)/Sci(PhyChem)/P2/Prelim2010/4E/5NA
distance = 180 m [3]
2 The resistive forces that act backwards on a car are air resistance and friction, as
illustrated in the following figure.
The car has a mass of 850 kg. It is moving at a constant speed of 60 km/h and the
total resistive forces acting on the car is 6000 N.
(a)
Calculate the forward driving force acting on the car by the engine.
Constant speed => 0 acceleration => F = 0
F = forward driving force – total resistive forces = 0
Forward force = total resistive forces = 6000 N
forward driving force = 6000 N [1]
(b)
Calculate the acceleration of the car if the forward driving force has
increased to 8125 N.
Resultant F = forward driving force – total resistive forces = ma
8125N – 6000 N = 850kg x a
a = 2.5 m/s2
acceleration = 2.5 m/s2 [2]
3 (a)
Figure 3.1 shows a catapult used to project an object.
Force F pulls back the object, creating tension in the rubber cords.
Figure 3.1
Figure 3.2
The tension force in each rubber cord is 10.0 N and the two cords are at
60° to each other. Figure 3.2 shows the direction of the two tension forces
acting on the object.
2
GMS(S)/Sci(PhyChem)/P2/Prelim2010/4E/5NA
By making a scale drawing, find the resultant of these two tension forces
acting on the object.
State the scale used.
object
scale: 1 cm represents ………… [1]
resultant force = ………………… [2]
(b)
When the object is pulled back, the average value of the force F is 16 N
and the object moves a distance 0.20 m in the direction of F. Calculate the
work done.
Work done
= force x distance moved in the direction of the force
= 16 N x 0.20 m
= 3.20 J
3
GMS(S)/Sci(PhyChem)/P2/Prelim2010/4E/5NA
work = 3.20 J [2]
4 The figure shows the same vertical force of 200 N exerted by a cyclist on the
pedal of a bicycle in three different positions A, B and C.
(a)
State the position A, B or C, in which the force exerts the largest moment
about the pivot.
Give a reason for your answer.
Position: C [1]
Reason: This is because the perpendicular distance between the force and
the pivot C is the greatest among the three positions A, B, C. [1]
(b)
Figure 4.1 shows a support for a leg in plaster and Figure 4.2 shows a
simplified diagram of the forces acting on the leg.
Figure 4.1
Figure 4.2
Calculate the force F needed to keep the leg in a horizontal position.
By the principle of moment, about the pivot
Anti- clockwise moment = Clockwise moment
F x (34 cm + 46 cm) = 120 N x 34 cm
F = (120 x 34) / 80 N
F = 51 N
4
GMS(S)/Sci(PhyChem)/P2/Prelim2010/4E/5NA
force F = 51 N [3]
5 The figure shows a conveyor belt carrying suitcases into an aeroplane.
An electric motor is used to drive the conveyor belt.
(a)
A suitcase of mass 20 kg is lifted from the ground into the aeroplane,
through a vertical distance of 4.0 m.
(i)
State two differences between mass and weight.
[2]
Mass is a measure of the amount of substance in an object and is constant
whereas the weight of an object is the product of its mass and the
gravitational field strength 'g' acting on it.
Since 'g' varies on different planets or locations, the weight will also vary
accordingly but its mass remains the same regardless of location.
(ii)
Calculate the increase in gravitational potential energy of the
suitcase.
gain in gravitation potential energy = mgh
= 20 kg x 10 m/s2 x 40 m
= 800 J
potential energy = 800 J [2]
(b)
The conveyor belt delivers five suitcases each minute. The average mass
of each suitcase is 20 kg. Calculate the useful power output of the electric
motor.
Power = Total work done / time
= 800 x 5 / 60
= 66.7 W
5
GMS(S)/Sci(PhyChem)/P2/Prelim2010/4E/5NA
power = 66.7 W [2]
6 The figure shows a refrigerator.
A fluid pumped through the pipes takes thermal energy (heat) out of the ice box.
This energy passes into the air at the back of the refrigerator through the black
metal fins.
(a)
Explain how the ice box at the top of the refrigerator keeps the whole of the
food compartment cool.
[2]
The air in the ice box has a higher density and lower temperature than the
air elsewhere in the food compartment. Thus, the cold air from the ice box
will sink down to displace the warmer air below it. The warmer air that has
risen into the ice box will be cooled and the colder air that has sunk will be
warmed up. The repeated process will set up a convection current that will
keep the whole compartment cool.
(b)
Explain why the fins are black.
[2]
The fins are black because the colour black is a good radiator of heat as
compared to lighter colours. Thus the heat loss to the surrounding is faster.
6
GMS(S)/Sci(PhyChem)/P2/Prelim2010/4E/5NA
8 Figure 8.1 shows words seen through a lens. Figure 8.2 shows the same words
without the lens.
Figure 8.1
(a)
Figure 8.2
State two properties of the image formed by the lens.
[2]
Image is virtual and upright.
(b)
On Figure 8.3, draw two rays to show how the image in Figure 8.3 was
formed by the lens. Mark clearly the focal point F of the lens and the image
formed.
[3]
F
image
object
lens
Figure 8.3
7
GMS(S)/Sci(PhyChem)/P2/Prelim2010/4E/5NA
9
X-rays, microwaves, ultraviolet rays and infra-red waves are different types
of radiation in the electromagnetic spectrum,
(a)
Write the name of one of these types of radiation in each of the boxes,
placing them in order of increasing wavelength.
shortest wavelength
X-rays
(b)
[2]
longest wavelength
ultraviolet
rays
infra-red
waves
microwaves
State one use of ultraviolet radiation.
[1]
Astronomy to detect very hot objects in space from space observatories
orbiting around the Earth or
Induces the production of Vitamin D in the human skin or equilvalent.
(c)
State two properties that all types of radiation in the electromagnetic
spectrum have in common other than their speed in vacuum.
1. All are transverse waves
2. They obeys the wave equation, v = f λ
End of Section A
8
[2]
GMS(S)/Sci(PhyChem)/P2/Prelim2010/4E/5NA
SECTION B (20 marks)
Answer any two questions on the writing paper provided.
10
(a)
Figure 10.1 shows a ray of light being refracted at a glass surface.
Figure 10.2 shows the incident ray at a different angle.
Figure 10.1
Figure 10.2
Not to scale
Calculate
(i)
the angle of refraction in Figure 10.2;
(ii)
the critical angle of this type of glass.
(b)
[2]
[2]
The following figure shows part of a long, thin spring used to demonstrate a
transverse wave.
The wave shown in the figure has a frequency of 4.0 Hz.
(i)
What is the direction the hand must move to make a transverse
wave?
[1]
(ii)
Describe how the hand must move to make the transverse wave
with frequency of 4.0 Hz.
[2]
(iii) The speed of the wave is 0.80 m/s. Calculate its wavelength.
[2]
(iv)
State what must be done to double the wavelength of the wave on
the spring.
[1]
9
GMS(S)/Sci(PhyChem)/P2/Prelim2010/4E/5NA
10 (a) (i)
Using formula: Refractive index of glass,
n glass = sin i / sin r
= sin 600 / sin 360
= 1.47
Let angle of refraction be r.
sin 30° / sin r = 1.47
r = 19.8°
10 (a) (ii)
n glass = 1/ sin c
Critical angle, c = sin -1 (1/1.47) = 42.9 0
10 (b) (i)
(ii)
up and down
The up-and-down continuous movement of the hand must produce 4
complete cycles of waves in 1 second.
v=fλ
λ = v / f = 0.8 / 4 = 0.2 m/s
To double the wavelength, from formula: v = f λ ,
the spring must be oscillated at half the initial rate of 2.0 Hz.
(iii)
(iv)
10
GMS(S)/Sci(PhyChem)/P2/Prelim2010/4E/5NA
11
A battery, an ammeter, a 2 Ω resistor and a resistance wire is connected as
shown in Figure 11.1.
Figure 11.1
(a)
The p.d. across the 2 Ω resistor is 4 V.
Calculate
(i)
the current flowing through the 2 Ω resistor;
(ii)
the current flowing through the resistance wire.
[1]
[1]
(b)
A voltmeter connected across the resistance wire shows a reading of 16 V.
Calculate
(i)
the resistance of the resistance wire;
[1]
(ii)
the combined resistance of the wire and the resistor;
[1]
(iii) the potential difference across the batteries.
[1]
(c)
The wire and resistor are disconnected and then reconnected in parallel.
Calculate the combined resistance of the wire and resistor.
[2]
(d)
Walls in buildings sometimes develop cracks. The width of a crack can be
monitored by measuring the resistance of a thin wire stretched across the
crack and firmly fixed on either side of the crack, as illustrated in Figure
11.2.
Figure 11.2
The wall moves and the crack widens slightly.
State what happens to the
(i)
length of the wire.
(ii)
area of cross-section of the wire.
(iii) resistance of the wire.
11
[1]
[1]
[1]
GMS(S)/Sci(PhyChem)/P2/Prelim2010/4E/5NA
11 (a) (i)
(ii)
I=V/R=4V/2Ω=2A
I = 2 A since the arrangement is a series circuit.
11 (b) (i)
(ii)
(iii)
R = V / I = 16 V / 2 A = 8 Ω
Combined resistance = 8 Ω + 2 Ω = 10 Ω
p. d across the batteries = p.d across the resistance wire & resistor
= 16 V + 4 V = 20 V
1/R = 1/R1 + 1/R2 = 1/8 + ½ = 5/8
R = 8/5 = 1.6 Ω
11 (c)
11 (d) (i)
(ii)
(iii)
the length of the wire increases.
the area of cross-section of the wire decreases.
the resistance of the wire increases.
12