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NANYANG JUNIOR COLLEGE
JC 2 PRELIMINARY EXAMINATION
Higher 1
CANDIDATE
NAME
TUTOR’S
NAME
CLASS
PHYSICS
8866/02
15 September 2010
Paper 2 Structured questions
2 hours
Candidates answer on the Question Paper.
No Additional Materials are required
READ THESE INSTRUCTIONS FIRST
Write your name and class on all the work you hand in.
Write in dark blue or black pen on both sides of the paper.
You may use a soft pencil for any diagrams, graphs or rough working.
Do not use staples, paper clips, highlighters, glue or correction fluid.
Section A
Answer all questions.
Section B
Answer any two questions.
For Examiner’s Use
At the end of the examination, fasten all your work securely together.
The number of marks is given in brackets [ ] at the end of each question or part
question.
Section A
1
2
3
4
5
Section B
6
7
8
Total
This document consists of 21 printed pages.
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DATA AND FORMULAE
Data
8
speed of light in free space
c =
3.00 x 10 m s
elementary charge
e =
1.60 x 10
the Planck constant
h =
unified atomic mass constant
u =
me =
rest mass of electron
mp =
rest mass of proton
acceleration of free fall
6.63 x 10
1.66 x 10
9.11 x 10
1.67 x 10
–19
–34
–27
–31
–27
–1
C
Js
kg
kg
kg
–2
g =
9.81 m s
s =
ut + ½ at2
Formulae
uniformly accelerated motion
v2 =
u2 + 2as
W =
pΔV
hydrostatic pressure
p =
ρgh
resistors in series
R =
R1 + R2 + ....
work done on/by a gas
resistors in parallel
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1/R =
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1/R1 + 1/R2 + ....
3
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Section A
Answer all the questions in this section
1
A 450 g football is kicked off the top of a building as shown in Fig. 1 below.

30 m
Building
24 m
Ground
Fig. 1
The football leaves the building at an angle andtakes 3.0 s to hit the ground.
Determine, for the football as it is kicked off the building,
(a)
the horizontal component of the velocity of the football,
sx  u xt
24  ux (3.0)
ux  8.0 m s1
horizontal component of the velocity =
(b)
m s-1 [1]
the angle .
Taking upwards as positive,
C
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1
s y  u yt  a yt 2
2
1
30  u y (3.0)  (9.81)(3.0)2
2
u y  4.715 m s-1
u
tan   y
ux
  310
angle  =
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[3]
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below the initial launch point.
(c)
Find the speed at which the football lands on the ground.
v y  u y  a yt
 4.715  9.81(3.0)
 24.715 m s-1
Speed of football as it lands =
(8.0)2  (24.715)2  25.978
=
 26 m speed
s-1
[1]
The football lands on the ground and does not rebound.
(d)
Calculate the magnitude of the average force acting on the football as it lands if the
football stops in 0.12 s.
Final momentum of the football = 0 N s
Initial momentum of the football = 0.450 x 25.978 = 11.690 N s
Taking upwards as positive,
Average force =
p f  pi 0  (11.690)

 97 N
t
0.12
average force =
2
(a)
N [2]
State the conditions necessary for a body to be in equilibrium.
The vector sum of forces acting on the body must be equal to zero.
The resultant torque of a body must be zero
[2]
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(b)
Fig. 2.1 shows a 1000 N uniform thin rod being towed and moving at constant
horizontal velocity.
.
T

R = Total reaction force from the ground
W = Gravitational force on the rod

Fig. 2.1
R
30o W
A
floor
P
(i)
Draw and label the 2 other forces on Fig 2.1 and show that the forces acting on
the block meet at a point. Mark that point P.
[2]
(ii) Given that T = 1500 N, show that  = 40o.
Taking moment about point A,
l
: W ( ) cos 30o  T (l ) cos   0
2
o
  73.22

[2]
A
    30o  90o
  46.8o  49o
(iii) In practice, wheels are installed at point A to reduce wear and tear at A, where
the block is in contact with the floor. Given that that  is fixed, explain how the
motion of the rod will change.
When wheels are installed, the frictional force at A will reduce
significantly and since  is fixed, T must remain the same to maintain
vertical equilibrium of forces, therefore, there must be a net force in the
horizontal direction and the rod will accelerate to the right.
[1]
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6
3
Fig. 3.1 shows a light dependent resistor (LDR), a 200  resistor and a light bulb of
resistance 1.0 k when operated normally connected to form a potential divider. The
resistance of the LDR is 1000  and 100 in the dark and in bright light conditions
respectively.
P
power
supply
Q
200 
R
S
Fig. 3.1
The light bulb requires a potential difference of 12 V to operate normally and it is designed to
be turned on when the room is in the dark.
(a)
Calculate the effective resistance between P and Q when the LDR is placed in a dark
room and the bulb is operating normally.
1
1 1
R(

)  500 
1000 1000
effective resistance =
(b)
 [1]
Calculate the potential at P when the room is in the dark.
The potential between P and Q is 12 V
By potential divider,
500
 VPR  12V
(500  200)
VPR  16.8V
(-16.8 V is also accepted)
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potential at P =
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V [2]
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(c)
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Calculate the power dissipated by the bulb when operated normally
P
V 2 122

 0.144W  0.14W
R 1000
power dissipated =
(d)
W [1]
Sketch the -V characteristic graph of a filament bulb and thermistor below.
I
I
V
V
Filament bulb
(e)
4
(a)
[2]
Thermistor
A student decides to connect a semi-conductor diode between S and R such that the
bulb will not light up. Draw the semiconductor on Fig. 3.1.
[1]
Define the term tesla.
One tesla is defined as the strength of a magnetic field in which a force of one newton
must act on a wire of length one meter carrying one ampere of current in a direction
perpendicular to the field.
(b)
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Fig. 4.1 shows a piece of wire carrying a current of 2.0 A placed perpendicularly to a
uniform magnetic field.
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Region of uniform
magnetic field
B
2.0 cm
2.0 A
300
10 cm
Fig. 4.1
The uniform magnetic field is of dimensions 10 cm by 2.0 cm and its magnetic flux
density is 0.10 mT. The magnetic field is into the page.
(i) Calculate the length of the wire in the magnetic field
2.0  l sin 300
l  4.0 cm
 0.040 m
length of wire =
m [1]
(ii) Hence calculate the electromagnetic force exerted on the wire.
F  BIl sin 
 0.1103  2.0  4.0 102  sin900
 8.0 106 N
force on wire =
(c)
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N [3]
Fig. 4.2 shows a negatively charged particle of velocity v entering a region of uniform
magnetic field.
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Region of uniform
magnetic field
v
Direction of
magnetic
field
Fig. 4.2
The electromagnetic force acting on the particle is into the page.
Draw the direction of the magnetic field on Fig. 4.2.
5
[1]
In the design of structure, such as buildings, towers or bridges, an engineer may use a
cantilever beam to allow for overhanging structure without external bracing. A cantilever is a
beam supported only on one end.
The engineer will make calculations to ensure that the cantilever beam is strong enough to
withstand any forces applied on it and ensure that there is not too much vertical deflection.
An appropriate beam can then be chosen based on the maximum allowable load to be
applied.
Fig. 5.1 illustrates a cantilever of length L loaded with a point load P at its end. A vertical
deflection y of the free end of the cantilever will result from the loading.
L
Load
y
Fig. 5.1
A student was asked to investigate the behaviour of such an arrangement and found out
from a book that the expression relating the vertical deflection of a loader cantilever. This
was given as Equation 1:
y
where
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4PL3
kbh 3
(Equation 1)
k is a constant
b = breath of cantilever
h = height of cantilever
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(a)
The student first used two steel cantilevers to investigate the relationship between y, P
and L. He first kept L constant and varied P and obtained the results as shown in Fig.
5.2:
P (N)
10
20
40
50
70
90
y (mm) for Cantilever A
1.9
3.9
7.8
9.9
13.8
17.7
y (mm) for Cantilever B
10.8
16.0
26.9
34.8
46.2
57.9
For
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Fig. 5.2
By plotting this result on a graph, he was able to deduce a relationship between y and P
for both cantilevers. This appeared to be of the same form for each, although there
seemed to be some form of systematic error in the results for B. The graph showing the
relationship between y and P for Cantilever A has been sketched on Fig. 5.3.
(i)
On Fig. 5.3, draw a best fit line through the plots to show the relationship
between y and P for Cantilever B.
[1]
Graph of Deflection y (mm) against Point Load P (N)
70
60
Cantilever B
y / mm
50
40
30
20
Cantilever A
10
0
0
10
20
30
40
50
P/N
Fig. 5.3
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60
70
80
90
100
11
(ii)
For
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From Fig. 5.3, discuss whether the data in Fig. 5.2 support Equation 1.
The graph shows a linear relationship between y and P.
The data support Equation 1 as y is proportional to P.
[2]
(iii) Suggest a reason why the student thinks the measurements for Cantilever B
were subjected to systematic error.
When no point load is being applied to cantilever B (P = 0), there should
not be any deflection (y = 0). However, from the graph, there seems to
be a deflection when P = 0 for Cantilever B, suggesting the presence of a
systematic error.
(iv)
[2]
From the graphs plotted, estimate the amount of this error.
Error =
(b)
any answers between 5 – 6 mm [1]
The student continued his investigation by keeping P constant and varying L, obtaining
the results for Cantilever A as shown in Fig. 5.4.
L (m)
1.0
1.5
2.0
2.5
3.0
3.5
4.0
y (mm)
1.6
5.3
12.5
24.6
42.4
67.2
99.9
Fig. 5.4
These results did not suggest direct proportionality between y and L. The student
proceeded to sketch a graph showing the variation of lg y against lg L as shown in Fig.
5.5.
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Graph of lg (y/mm) against lg (L/m)
2.5
lg (y/mm)
2.0
1.5
1.0
0.5
0.0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
lg (L/m)
Fig. 5.5
(i)
Determine the gradient of the graph. Show your working.
Gradient 
2.000  0.500
 3.00
0.600  0.100
choose a gradient triangle more than half the area of the grids
Gradient =
(ii)
[1]
Hence discuss whether the data in Fig. 5.4 support Equation 1.
The data supports Eq 1 as the gradient of the graph lg y against lg L
gives the order of the relation between y and L. From the data, y is
proportional to L3 which support Equation 1.
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[2]
13
(c)
After researching further into the subject, the student found out that the constant k in
Equation 1 is known as the Young’s Modulus which is a property of the material which
the cantilever is made from.
(i)
Use Equation 1 to show that the S.I. unit of k is Pa.
[2]
4PL3
kbh 3
4PL3
k
ybh 3
y
[k ] 
[P ][L3 ]
N ( m3 )
N

 2  Pa
3
3
[ y ][b][h ] m(m)(m ) m
(ii)
A steel cantilever has width 0.12 m, height of 0.040 m and length of 2.0 m as
shown in Fig. 5.6. If a point load of 1.0 kN is applied at the end, a vertical
deflection of 2.0 cm was measured at the end. Calculate the Young’s Modulus of
steel.
2.0 m
Point
Load
0.040 m
Fig. 5.6
0.12 m
4PL3
kbh3
4PL3
4(1.0  103 )(2.0)3
k

 2.08  1011 Pa
ybh3 0.020(0.12)(0.040)3
y
Young’s Modulus of steel =
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Pa [1]
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14
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Section B
Answer two questions from this section.
6
A sports car of mass 1500 kg is speeding along a straight road at 110 km h-1. A lorry
suddenly appears in front of the car. In the time interval between the lorry being spotted and
the brakes on the car coming into operation, the car moves forward a distance of 10.6 m.
With the brakes applied, the front wheels of the car leave skid marks on the road that are
4.2 m long, as illustrated in Fig. 6.1.
110 km h-1
Skid mark
10.6 m
lorry
4.2 m
Fig. 6.1
It is estimated that, during the skid, the magnitude of the deceleration of the car is 0.85 g,
where g is the acceleration of free fall.
(a)
Determine
(i)
the speed v of the car before just before it collide with the stationary lorry.
110 km h-1 = 30.6 m s-1.
v2 = u2 +2as = (30.6)2 + 2(0.85)(-9.91)(4.2)
v = 29.1 m s-1.
speed v =
(ii)
m s-1 [2]
the time interval between the lorry appearing and the collision taking place.
t1 = 10.6/30.6 = 0.346 s.
t2 = v-u/a = (29.1 - 30.6)/(0.85)(-9.81) = 0.180 s
total time = t1 + t2 = (0.346 + 0.180) = 0.53 s
time interval =
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s [2]
15
(b)
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Fig. 6.2 shows the side view of the
passenger of mass 65 kg in the front seat.
The forces acting on him at the instant
when the brake is applied is also shown.
C.G.
x
Normal
reaction
force.
Frictional force
Weight of passenger
Fig. 6.2
(i)
With reference to the forces acting on him, explain why the driver will lean
forward during this instant.
From the diagram shown, the frictional force will provide a clockwise
moment about the C.G. resulting in the driver rotating clockwise. Hence
the driver will lean forward during this instant.
[2]
(ii) State the net force acting on the passenger.
F = ma = (65)(0.85)(9.81) = 542 N
\\
net force =
(c)
N [1]
Given that the sport car make a completely inelastic collision with the stationary lorry
of mass 2500 kg.
(i)
Explain what is meant by a completely inelastic collision.
A completely inelastic collision is one in which the 2 bodies move with a
common velocity after collision and the total kinetic energy of the system
before and after collision is not conserved.
[2]
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(ii) Calculate the velocity of the car right after collision, stating any assumption made.
pinitial = pfinal
(1500)(29.1) = (1500+2500)(v)
v = 10.9 m s-1
m s-1 [3]
velocity =
(iii) The airbag in the sport car is deployed almost immediately after the collision.
Explain how the airbag help to minimize the injuries to the front seat passenger.
The airbag helps to increase the time of impact and by Newton 2nd law,
F = p/t, this will greatly reduce the force of impact on the passenger.
[2]
(d)
After the accident, the sports car of mass 1500
kg had to be towed by a 2000 kg lorry up a slope
at a constant speed of 5.0 m s-1 as shown in Fig.
6.3. The power P delivered by the lorry is 55 kW.
5.0 ms-1
10o
Fig. 6.3
(i)
By considering the sports car and the lorry as a whole system, find the rate of
increase of the gravitational potential energy of the system.
Rate of increase of GPE = mgvy = (3500)(9.81)(5.0sin10o)
= 29 800 J
rate of increase of GPE =
J s-1 [2]
(ii) Using the principle of conservation of energy, show that the total resistive force,
fR acting on the whole system (sports car + lorry) is approximately 5000 N.
Power delivered = Rate of increase of GPE + rate of work done against fR
55 000 = 29800 + fR(5.0)
fR = 5000 N
[2]
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(iii) Hence determine the driving force, F acting on the lorry, explaining your answer.
F = 0
F - mgsin - fR = 0
F = (3500)(9.81)(sin10o)+ 5000 = 11 000 N
driving force =
7
N [2]
The diagram below shows visible light waves spreading through a single slit a before
spreading at slits b and c.
D
N
M
S1
S2
(a)
Y
screen
Image
seen
Error! Not
a valid
link.
on screen
State the name of this experiment.
Young’s Double Slit experiment
[1]
(b)
With reference to the waves spreading from the slits b and c, explain why interference
occurs only where waves from both sources overlap.
Interference is the result of superposition of two wavetrains, thus will
only occur where there are waves from both sources.
[2]
Double Slit experiment
(c)
Explain what diffraction is, and describe the part played by diffraction in this experiment.
Diffraction is the spreading of light waves past an opening. Diffraction
from b and c ensures that waves from two sources meet so that
[2]
interference can occur.
Interference is the result of superposition of two wavetrains, thus will
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only occur where there are waves
from both sources.
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Double Slit experiment
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(d)
(e)
Label one part of the screen view where destructive interference has occurred with the
letter D.
[1]
Explain in detail why constructive interference occurs at N.
Constructive interference occurs because the path difference between
aN and bN is 2λ. The waves will meet in phase, and thus the result of
superposition is reinforcement: the resultant amplitude is the sum of
[2]
the individual amplitudes
(f)
State a condition necessary for observable interference, and explain how this
experiment ensures that the condition is met.
Coherence of the sources is necessary for observable interference. Slit
a is used to spread light from the same source to b and c so that they
are coherent sources.
OR The superposing waves must be of the same amplitude. Making slit
[2]
a and b equidistant from a ensures that the amplitude of the waveforms
(g)
The
monochromatic
lightsame.
used in this experiment has a wavelength 589 nm. The
from
b and c are the
distance between S1 and S2 is 0.200 m while the distance between S2 and the screen is
2.5 m. The distance between slits b and c is 0.80 mm.
Calculate the distance between two consecutive bright lines on the screen.
x=
D
(589x10-9 ) (2.5)
=
= 2.9 x 10-3 m
-3
a
(0.8 x 10 )
distance =
(h)
m [2]
If a piece of plastic is placed in front of slit b, the light waves from b will be caused to
slow down, without any change in frequency. Some of the wave energy will be
absorbed by the plastic as well.
(i)
State how the wavelength of the wave from b will change.
The wavelength will be smaller (from v = f λ)
[1]
(ii)
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State how the amplitude of the wave from b will change.
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The amplitude will be smaller.
[1]
(iii)
Based on your answer in (h)(i), predict whether there will be more or fewer
waveforms in the plastic piece as compared to the number of waveforms in an
equal thickness of air.
There will be more waveforms in the plastic piece.
[1]
(iv) Using your answers in (h)(ii) and (iii), explain how the image on the screen will
change, in terms of position, separation and contrast, when a piece of plastic is
placed in front of slit b.
Because there are more waveforms in the plastic piece than in the
same thickness of air, the centre of the interference pattern (where
there are exactly the same number of waveforms between both the
sources and the centre) would be nearer the side X.
Because the amplitude of one wavetrain is smaller, the amplitude
of constructive interference would be smaller while the amplitude
of destructive interference would be bigger, so the contrast
[5]
between the fringes would be smaller.
The separation of the fringes remains unchanged.
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8
Electromagnetic radiation is incident normally on the surface of a metal. Electrons are
emitted from the surface and these are attracted to a positively charged electrode, as shown
in Fig. 8.
electromagnetic
radiation
electrode
evacuated
enclosure
metal
surface
μA
Fig. 8
(a) Name the effect which gives rise to the emission of the electrons.
Photoelectric effect.
(b) State a word equation, base on the principle of conservation of energy, which describes this
effect
Energy of a photon is equal to the sum of the minimum energy required to remove an
electron from the metal surface and the maximum kinetic energy of the electron emitted.
(c) The current recorded on the mircroammeter is 2.1 μA. Calculate the number of electrons
emitted per second from the surface.
I = (N/t) e
(N/t) = 2.1 x 10-6/ 1.60 x 10-19 = 1.31 x 1013 s-1
(d) The incident radiation has wavelength 240 nm. Show that the energy of a photon incident on
the surface is 8.28 x10-19 J.
hf = hc/ = 6.63 x 10-34 x 3.00 x 108 / 240 x 10-9 = 8.28 x10-19 J
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(e) The intensity of the incident radiation is 8.2 x 103 W m-2. The area of the surface is 2.0 cm2.
Calculate
(i) the power of the radiation incident on the surface,
power = I A = 8.2 x 103 x 2.0 x 10-4 = 1.64 W
(ii) the number of photons incident per second on the surface.
(N/t) hf = power
N/t = 1.64 / 8.28 x10-19 = 1.98 x 1018 s-1
(iii) Hence determine the ratio
number of electrons emitted per second
number of photons incident per second
= 1.31 x 1013/ 1.98 x 1018 = 6.62 x 10-6
ratio =…..…………….
[1]
(f) Comment on your answer to (e) (iii).
Very few photons successfully emit an electron.
(g) When the wavelength of the radiation is gradually increased to 310 nm, the reading in the
microammeter just drops to zero. Explain this phenomenon and calculate the maximum kinetic
energy of the electrons emitted from the metal surface when the wavelength of the radiation
remains at 240 nm.
The wavelength 310 nm is the threshold wavelength [1], above which the energy of the
photon is not enough to remove an electron to the surface of the metal. [1] Hence no
current is detected.
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hf = hfo + ½ mvmax2
½ mvmax2 = 8.28 x10-19 - hc/o
= 8.28 x10-19 - 6.63 x 10-34 x 3.00 x 108 / 310 x 10-9
= 8.28 x10-19 – 6.40 x10-19
= 1.87 x10-19 J
(h) (i) Calculate the momentum of the photon given in (d).
[1]
p = h /  = 6.63 x 10-34 / 240 x 10-9 = 2.76 x 10-27 N s
(ii) If there are altogether 1.5 x 1018 photons striking onto the metal surface per second and are
all absorbed, determine the force exerting on the metal surface. Show your working clearly.
By Newton’s 2nd law, force on photon
N p 1.5 x1018 x(0  2.76 x1027 )
F

 4.14 x109 N
t
1.0
By Newton’s 3rd law, force on metal surface = 4.14 x10-9 N
NYJC 2010
8866/02/PRELIM/10