Preliminary Examination
JC2 H2 Physics 2011
Answers to JC2 Preliminary Examination Paper 1 (H2 Physics)
Suggested Solutions:
1
2
3
4
5
6
7
8
9
10
A
C
B
B
C
D
A or B
B
B
C
11
12
13
14
15
16
17
18
19
20
C
C
B
A
A
C
D
C
B
B
21
22
23
24
25
26
27
28
29
30
A
C
C
D
A
D
B
A
D
A
31
32
33
34
35
36
37
38
39
40
D
D
C
B
A
C
A
B
B
D
MCQ 1: A
Reasoning:
SI unit of energy = kg m2 s-2
SI unit of term g  hV = (kg m s-2 kg-1)(kg m-3)(m)(m3) = kg m2 s-2
MCQ 2: C
Reasoning:
Let X denotes the range in temperature rise.
Let Y denotes the temperature rise between a temperature T and 40°C.
X = (100 – 40) °C = 60 °C
Y = (T – 40) °C
∆Y = 0.5 °C + 0.5 °C = 1.0 °C
 Y
 X
So, 
Y
1.0

 %  X  100%  60.0  100%  1.7%

{% uncertainty does not need to be 1 s.f.}
MCQ 3: B
Reasoning:
Option A is not possible, kinetic energy cannot be negative.
v 2  u 2  2as
1
1
mv 2  mu 2  mas
2
2
EK  constant  mgs
i.e. is of the form y  kx  c
 option D is not possible
Option C is not possible as displacement s is at most h.
Preliminary Examination
JC2 H2 Physics 2011
MCQ 4: B
Reasoning:
Vy  u y  gt
Vy  u sin   gt
u sin   u sin   gt
2u sin 
t
g
MCQ 5: C
Reasoning:
v
v
F
mgsin36o
mg sin 36o  f  F
4(9.81) sin 36o  f  31
f  7.94 N
f
f
F
mgsin36o
mg sin 36o  F  f
F  4(9.81) sin 36o  7.94
 15 N
MCQ 6: D
Reasoning:
By conservation of momentum and considering the astronaut and the hammer as one system,
initial momentum will be zero, hence final momentum has to be zero as well. So if the hammer
moves forward, the astronaut must move backwards (vice-versa) so that final momentum can
be equal to zero.
MCQ 7: A or B
Using conservation of momentum,
The third particle has to account for the initial forward momentum of the nucleus, hence options
C and D are incorrect.
Since α > β, the momentum of the third particle cannot be along the x-axis. Hence answer is B.
Answer could be A as well if the speeds of the particles are not the same.
MCQ 8: B
Reasoning:
Taking pivot at the pin,
75(5) = Tcos30o (8)
T = 54 N
MCQ 9: B
Reasoning:
FB = FA = mg and since both masses are equal, hence FB = FA
Preliminary Examination
JC2 H2 Physics 2011
P = ΔE / t = (mgh) / t since B is raised twice as fast as A in the same time interval, B will reach
twice the height of A, hence PB = 2PA
MCQ 10: C
Reasoning:
mgh
= mgv .
t
Rate of energy loss =
A: Kinetic energy is constant since speed is constant.
B: Mechanical energy is not constant. The ball is losing gravitational potential energy.
D: Total resistive force cannot be zero; it is constant.
MCQ 11: C
Reasoning:
If we take UO = 0
UA = - mgℓsin 30o
UB = - mgℓ
Hence,UA – UB = 0.5mgℓ
UO – UA = 0.5mgℓ
MCQ12: C
Reasoning:
The vertical component of the acceleration is the centripetal acceleration which is present since
the particle is performing circular motion.
The horizontal component of the acceleration causes the speed of the object to increase.
MCQ13: B
Reasoning:
g
GM
M
g 2
2
r
r
Let the quantities with subscript ‘J’ represent that of Jupiter while those with subscript ‘E’
represent that of Earth.
2
gJ
M r
 J  E2
g E M E rJ
2
 6400  10 3 

 314  
7 
 7.23  10 
-1
Since gE is 9.81 N kg-1, g J  24.1 N kg
Preliminary Examination
JC2 H2 Physics 2011
MCQ 14: A
Reasoning:
The antenna will have the same speed as the space-craft and is still bound in orbit. There is still
gravitational force acting on it which causes it to move in circular motion.
MCQ 15: A
Reasoning:
a-t graph for body in SHM is sinusoidal.
Maximum potential energy occurs at maximum displacement, where a is maximum.
MCQ 16: C
Reasoning:
x = xo sin ωt = 2.0 sin (4πt)
ω = 4π rads-1
vo = ωxo = 4π(2.0) = 25 m s-1
MCQ 17: D
Reasoning:
 

 
2

 360  = 180°
MCQ 18: C
Reasoning:
I
1
x2
and
1
r
AQ
r

AP 2r
r
6 = 3.0 μm
AQ 
2r
A
MCQ 19: B
reasoning:
For a closed pipe,
Fundamental mode:
¼=L
v = f
f1 = v/4L
1st Overtone:
I  A2
Preliminary Examination
JC2 H2 Physics 2011
¾ = L
v = f
f2 = 3v/4L = 3f1
Of all 4 options, only B satisfy f2 = 3f1
MCQ 20: B
reasoning:
∆𝑥 =
𝜆𝐷
𝑎
To increase fringe separation, , D should increase and a should decrease
MCQ 21: A
Reasoning:
Let the amount of added heat be Q.
So,
Q  mcw Tw ……………. (1) {heat added to water}
Q  mci Ti ……………. (2) {heat added to iron}
Eqn. (1) = (2),
cw Ti

ci Tw
Since cw > ci, ∆Ti > ∆Tw  the final temperature of iron is going to be higher than the
water’s final temperature.
MCQ 22: C
Reasoning:
The total volume of the cylinder is not changed. Neither is the total number of molecules.
Hence, the average space given to each molecule remains constant. This implies that
the average distance between molecules is unchanged.
MCQ 23: C
Reasoning:
Electric Force on the electron is in the Y-X (left) direction. Hence
the field is doing negative work.
This also means that work done by external agent is positive.
Hence the potential energy of the electron increases.
MCQ 24: D
Reasoning:
Y is at higher potential than Z
Preliminary Examination
JC2 H2 Physics 2011
MCQ 25: A
Reasoning:
V =  - Ir
5.3 =  - (1.0)r ---(1)
4.6 =  - (2.0)r ---(2)
Solving simultaneously,
r = 0.7 
 = 6.0 V
MCQ 26: D
Reasoning:
𝐼
From V1 – V2, the ratio 𝑉 decreases, thus 𝑅 =
(resistance is NOT given by
𝑉
𝐼
increases.
dV
)
dI
For a I vs V graph
Since R 
V
1

, the resistance at any point along a the black curve of a I vs V graph is
I
I
V
 
just the reciprocal of the gradient of the red straight line, drawn from the origin to the point of
interest.
MCQ 27:B
Reasoning:
Current through the bottom branch (2 x 5  resistors) I1 = 10.0V / (5.0+5.0)  = 1.00 A
Therefore, current through the middle branch, I2 = 1.75 – 1.00 = 0.75 A
Therefore, potential difference across 3.0  resistor, V = IR = 0.75 x 3 = 2.25 V
MCQ 28:A
Reasoning:
0 V across WX, YZ
220 V across WY, WZ, XY, XZ
Preliminary Examination
JC2 H2 Physics 2011
MCQ 29: D
Reasoning:
MCQ30: A
I
BZ  o z
2 rz

BY 

o  20 A 
, into the page
2  2r 
o I Y
2 rY
o 10 A 
, out of the page.
2  r 
Net B field at X is 0, thus EM force on X is 0
MCQ 31: D
reasoning:
𝜀=
𝑑(𝐵𝐴)
𝑑(𝐵)
=𝐴
𝑑𝑡
𝑑𝑡
At T/2, the gradient of the graph is the steepest, hence  is the largest.
MCQ 32: D
reasoning:
The act of opening and closing the switch does not change the amount of flux cutting the right
coil. No e.m.f. is induced therefore no deflection is observed.
MCQ 33:C
reasoning:
The d.c. equivalence of 12 A r.m.s. is 12 A.
MCQ 34: B
reasoning:
Energy dissipated per unit time, P = V2/(Rt)
𝑃 ∝ 𝑉2
Power = 1000/2402 × 3402 = 2000 W (to 2 s.f.)
Preliminary Examination
JC2 H2 Physics 2011
MCQ 35: A
Reasoning:
3  2  1
f3  f2  f1
Consider a simple 3 level scheme
f1  f2  f3
1
1
1
1


1
2
1
2


1
3
1
3
MCQ 36: C
Reasoning:
E = ½ m v2 ; (mv )2 = 2mE
de Broglie wavelength,
 = h/p = h /  (2mE) ;
  1/ E
The graph of  vs 1/ E is a positive gradient straight line through the origin .
MCQ 37: A
Reasoning:
Electrons in the metal can tunnel through the small gap (this represents the barrier width)
between the tip of the microscope and the surface and allows a current to flow. The magnitude
of the current detected is a reflection of the transmission coefficient for the tunnelling process.
MCQ 38:
Reasoning:
The picture illustrates the p-n junction at equilibrium. Mobile
electrons from the n-type zone of the depletion region combine
with and neutralize mobile holes in the p-type zone of the
depletion region.
This exposes fixed negative ions in the p-type zone and fixed
positive ions in the n-type zone of the depletion region.
The net charge of the depletion region is still zero.
The net charge of the portions outside the depletion regions is
also zero despite the presence of mobile electrons and holes.
MCQ 39: B
Reasoning:
α-particle: 42 He
β -particle: -01 e
Nucleon number = 217 – 4 – 4 = 209
Proton number = 85 – 2 – 2 –(-1) = 82
Preliminary Examination
MCQ 40: D
Reasoning:
A  Ao e
 ln 2 
t


 t1 / 2 
400  8.0  10 e
3
 ln 2 

t
 15 
 ln 2 
ln 0.05  
t
 15 
T = 64.8 hr
JC2 H2 Physics 2011
Preliminary Examination
JC2 H2 Physics 2011
Suggested Solutions JC 2 Prelim 2011 Paper 2
1
(a)
(b)
Perpendicular to surface:
1
sy  uy t  g cos  20  t 2  15.0t  4.61t 2
2
Parallel to the surface:
1
s x  u x t  g sin(20)t 2  26.0t  1.68t 2
2
sy  0
4.61t 2  15t
t  3.25 s
sx  26.0(3.25)  1.68(3.25)2
2
(a)
 102 m
When an object is submerged in a fluid, the top of the object experiences a lower
pressure compared to the bottom of the object.
This results in a net upward force acting on the object, which is upthrust.
(b)
3
(a)
Resultant force
F = W slab + W swimmer – Uinitial
(mslab + mswimmer) a
= (mslab + mswimmer) g – waterVslab g
a = g – waterVslab g / (mslab + mswimmer)
a = (9.81) - (1000)(0.857)(10 x 10-2)(9.81)/ [(300)(0.857) (10 x10-2) + 75]
= 1.5 m s-2
(i)
1. its wavelength = 8.0 cm = 0.080 m {read from graph}
2. the wave crest travelled 3.2 cm in 0.2 s. {read from graph}
 wave speed = 0.032/0.2 = 0.16 m s-1
(1 mark is awarded if correct answer is given but no working is
provided)
3. speed = frequency x wavelength
frequency = 0.16/0.080 = 2.0 Hz
period = 1/frequency = 1/2.0 = 0.5 s
 note that the period is longer than 0.2 s.
(1 mark is awarded if correct answer is given but no working is
provided)
(ii)
The stationary wave is formed by the superposition of two waves of the same kind
with equal frequency and amplitude, travelling in the same plane and moving in
opposite directions.
The product f  in a stationary wave refers to the speed of the two travelling
waves.
Preliminary Examination
JC2 H2 Physics 2011
Due to the two waves having the same speed f  but travelling in opposite
directions, their velocity vectors cancel out resulting in stationary wave not moving
in either direction.
(b)
(i)
h = L – Lcosθ = L(1-cosθ)
mgh = ½ mv2
…
v=
2gL(1  cos )
(ii)
(above diagram is not drawn to scale)
At A (maximum angular displacement), bob is momentarily at rest.
No circular motion:
Fresultant = mgsinθ = ma {because mgcosθ = T}
a = gsinθ
(iii)
At B (equilibrium position), bob is at the lowest position and its velocity is v.
Circular motion (centripetal acceleration):
a = v2/L or 2g(1-cosθ)
(iv)
a=0
Preliminary Examination
4
(a)
JC2 H2 Physics 2011
KEmax  eVS = 1.6 x 10-19 x 1.19 = 1.904 x 10-19 J
E    KE max
  hf  KE max
  (6.63  10 34 )7.4  1014   1.904  10 19
(b)
= 3.0 x 10-19 J
KEmax  hf  

KEmax  6.63  1034
 6.4  10   3.0  10
14
19
 1.24  10 19 J
(c)
Energy of the photons is independent of the intensity. Since the metal and the frequency
of the light is unchanged, the maximum kinetic energy of the photoelectrons remain
unchanged.
The photocurrent decreases since the rate of arrival of photons at the surface decreases.
Hence the rate of photoelectron emission decreases.
(d)
The maximum KE of the photoelectrons emitted due to the blue light is lower than the
maximum KE of the photoelectrons emitted due to the violet light (or equivalent).
Therefore, the maximum KE of the emitted electrons should be that due to the violet
light.
Therefore, both students are wrong.
5
a
A long life-time for spontaneous emission will enable the accumulation of excited atoms
there allowing population inversion to occur.
It will also ensure that the excited atoms stay sufficiently long at the upper lasing level for
stimulated emission to occur.
b
The valence band and the conduction band in an intrinsic semiconductor are separated
by a narrow band gap.
At room temperature, only a few electrons have sufficient energy to be excited across the
narrow band gap thus explaining the high resistance of the intrinsic semiconductor at
room temperature.
At higher temperatures however, the energy of the electrons increases, thermal
excitation of the valence electrons across the narrow band gap becomes more probable.
With more thermally excited electrons in the conduction band and holes in the valence
band to conduct electricity, resistance of the intrinsic semiconductor decreases
Preliminary Examination
JC2 H2 Physics 2011
131
0
6 (a)(i) 1. 131
53𝐼 → 54𝑋𝑒 + −1𝑒
129
129
∗
2. 52𝑇𝑒 → 52𝑇𝑒 + 𝛾
)
(* denotes excited state. Accept answer without *
(ii) 382120+1887300+894720+922680+35416 = 4,122,236
= 4.1222 x 106 Bq
(iii) 1. A = N
𝐴𝑡1⁄2
𝐴
 𝑁 = 𝜆 = 𝑙𝑛2
=
12
=1.89 x 10 .
1887300(8.02×24×60×60)
𝑙𝑛2
2. 1887300 x (7.5 x [103]2)
= 1.42 x 1013 Bq
(b) Ao  ¼ Ao in 2 half-lives assumes a pure radioactive sample that is not
replenished.
In the case of Fukushima, more radioactive fallout released from the
nuclear power plant would further contaminate area and add to the
radionuclide already present.
(c) (i) The half-life of 137Cs is 30.1 years whilst it is only 8.02 days for 131I.
Once the emergency at the Fukushima Daiichi nuclear power plant is
under control and no new radiation is released into the atmosphere,
the activity of 131I will rapidly decay to an insignificant amount after a few
weeks, whilst the activity of 137Cs will persist for many decades.
(answers simply stating “radioactivity due to 137Cs will last longer” with no
reference to the half-lives shall only be awarded 1 mark)
(ii) Iodine-131 emits beta rays/particles, which is ionizing radiation.
Beta particles emitted within the body will kills the surrounding cells and
tissue destroying the thyroid and/or possibly causing cancer in future (if
cellular DNA is damaged).
(answers stating “because it can cause cancer” shall only be awarded 1
mark)
(iii) Consumption of potassium iodide pills prior to exposure will saturate the
thyroid with iodine.
This prevents accumulation of radioactive iodine if it is ingested
subsequently, and will result in the radioactive iodine passing out of the
system.
(Also award marks for answers saying that the potassium will tend to bind
to the radioactive iodine and remove it from the body, as this is suggested
by some sources, but saturation of thyroid preventing further absorption of
Preliminary Examination
JC2 H2 Physics 2011
iodine is main reason)
7
(a)
(i)
==-
d
d
=(NBA cos )
dt
dt
d
(NBA cos t)
dt
= NBA  sint
(ii)
max = NBA
29 = 100 × B × 120 × 10-2 × 10-2 × 2


(ii)
f = 1/(60/3600) = 60 Hz
 = 2f = 376.99 rad s-1
max = NBA = 100 × ×120 × 10-2 × 10-2 × 376.99
= 17.41 V
OR
(b)
max =(3600/6000)× 29
= 17.4 V
max = 0.31 / 2 = 0.155 V
max = NBA =1 × 0.80 × ½  × 0.22 ×  = 0.155
 = 3.08 rad s-1
Question 8 (Planning)
Basic Procedure [2]

Determine pressure between the air panes and amplitude/quality of
sound.
Preliminary Examination
Diagram [1]
Method of
 Varying and
Measuring/
Determining the IV.
 Measuring/
determining the DV
 Analyzing the data
[3]
Control of other
Variables [max 2]
JC2 H2 Physics 2011













Other details (reliability)
[max 3]






Safety Precautions
[max 1]


Repeat for different pressure values.
Diagram of workable arrangement with acceptable source of soundusing signal generator, alarm bell and set-up to show how pressure
is varied and sound is detected etc.
Devices to measure pressure, amplitude of sound,
Use of suitable source of sound described.
Start with no air pumped out. Subsequently air pumped out of gap
between panes. (IV)
Measurement of air pressure using pressure gauge. (IV)
Measure the amplitude of sound with microphone attached to CRO.
or any other suitable method. (DV)
Plotting of graph of amplitude versus pressure (analysis)
Distance of sound source from window.
Alignment of sound source to window (e.g. source of sound
perpendicular to window).
Alignment and distance of sound detector.
Temperature of room.
Frequency and loudness of source of sound
Steps taken to reduce random error/uncertainty & represent data
Experiment repeated and results averaged under same conditions (to
ensure repeatability)
Other Good Design features/reliability measures
Method to reduce sound reflection and interference: E.g. surrounding
sound detector and detector side of glass pane with soundproof
materials (foam)/ placing detector very close to glass/ attaching glass
panes to sound proof box with sound detector within box.
Performing experiment in quiet room.
Allowing for temperature of room and pressure to stabilize before
taking measurement.
Details on how amplitude of sound is measured (e.g. usage of
oscilloscope-reading vertical-gain).
Ensuring sound source is loud enough on one side of window pane
but not too loud so that
Relevant safety precaution related to either the use of glass or
intensity of sound e.g. switch on sound source for short period of time
only, careful handling of glass panes to prevent breakage e.g.
making sure glass panes are held securely with backing.
Other protection (safety scree/goggles?).
Preliminary Examination
JC2 H2 Physics 2011
Suggested Solutions JC 2 Prelim 2011 Paper 3
1
(a)
(b)
2
(a)
(b)
1. size of an atom = 1.0 x 10-10 m to 5.0 x 10-10 m (or order ~ 10-30 m3)
2. speed of the fastest man on Earth =
9.3 m s-1 to 10.3 m s-1
3. power rating of a domestic light bulb =
10 W to 100 W
Zero error = 0.14 mm
Therefore, actual diameter of wire = (2.59 – 0.14) = 2.45 mm
By the principle of conservation of energy, the increase in internal energy of
a system is equal to the sum of heat transferred into the system and work
done on the system. Internal energy is a function of state.
density of water = 1000 kg m-3

Vwater = m/
= 1.00 / 1000
= 10-3 m3
Q = mLv
= 1.00 × 2.26 × 106
W = pV
= 1.01 × 105 (1.67 – 10-3)
U = Q + W
= mLv + pV
= 1.00 × 2.26 × 106 - 1.01 × 105 (1.67 – 10-3)
= 2.09 × 106 J
(c)


Molecules of a liquid are in continuous random motion and collide
frequently with one other. Upon collision, some molecules will gain
kinetic energy while others will lose kinetic energy.
OR


(d)
A very fast moving molecule (i.e. with large EK) near the surface of the
liquid may have enough energy to overcome the attractive forces of the
neighbouring molecules and leave the liquid.
As the more energetic molecules escape from the surface of the liquid,
the remaining molecules in the liquid will have smaller kinetic energy.
The overall K.E. of the remaining liquid molecules decreases and
cooling occurs.
Any of the following answers:
Increase the surface area of the liquid.
More molecules exposed at the surface, hence greater chance for a
molecule with more Ek to escape.
Preliminary Examination
JC2 H2 Physics 2011
Increase the temperature of the liquid.
Average Ek of molecules is increased, hence a greater probability of a
molecule having sufficient Ek to escape.
Presence of draught/wind.
Remove vapour molecules before they have a chance of returning to the
liquid.
Reduce the air pressure above the liquid.
Decreases the probability of a vapour molecule rebounding off an air
molecule and returning to the liquid.
3
(i)
Within the velocity selector:
FE  FB
qE  qvB
E
v
B
On leaving the velocity selector,
Fnet  qvB
(ii)
mv 2
 qvB
r
mv
r
qB
m E

qB B
mE

qB 2
On leaving the velocity selector,
Fnet  qvB
mv   qvB
qB

m
2 qB


T
m
2 m
T
qB
Each ion travels through a quarter circle before colliding with the detector plate.
T
Hence the time taken 
4
Preliminary Examination
JC2 H2 Physics 2011
2 m
4qB
m

2qB
m
Time taken 
2qB
Time taken 
(iii)
1.

2.
R 12 C 
 (1.99  1026 )
2(1.6  1019 )(0.100)
 1.95  106 s
m12 CE
qB 2
1.99  10  50 000
1.6  10  0.100 
26

2
-19
 0.622 m
R 13 C 
m13 CE
qB 2
 2.16  10  50 000

1.6  10  0.100 
26
2
-19
 0.675 m
3.
RC13
RC12
45o
RC12
RC13
Preliminary Examination
JC2 H2 Physics 2011
R 13 C  R 12 C  0.053 m
 distance bet. points of impact

 0.053 
2
  0.053 
2
 0.075 m
4
(a)
(i)
1. phase difference = 0
They meet constructively to give a bright fringe hence the waves meet
in phase. Phase difference is therefore 0.
2. path difference = n 
P is the nth order bright fringe and the sources start in phase.
(ii)
path difference = (n – ½) or
P is the nth order dark fringe.
1st order dark fringe  path difference = ½ = (1 - ½) 
2nd order dark fringe  path difference = 3/2 = (2 -½) 

nth order dark fringe  path difference = (n -½) 
(b)
(i)
When red and blue light passes through the diffraction grating, both
light will meet constructively at the center (path difference = 0) to give
a central bright red and blue fringe that overlaps. When the red and
blue wavelength overlaps, the resultant color is magenta.
(ii)
From the diagram, the next fringe can either be a 2nd order red or 3rd
order blue fringe.
Using nλ=dsinθ, we have:
2 λr = dsinθr = 2(600) = 1200 --------- (1)
3 λb = dsinθb = 3(400) = 1200 --------- (2)
Comparing (1) and (2) we see that since sinθr = sinθb, θr = θb
Since the 2nd order red fringe and 3rd order blue fringe overlaps each
other at the same angle, the color of fringe X is magenta.
(iii)
Method 1:
There is a maximum order of n = 2 for red fringes.
Using nλ=dsinθ,where sinθ 1 we have:
Preliminary Examination
JC2 H2 Physics 2011
2(600) d
dmin = 1200 nm = 1.2 × 10-6 m
Method 2:
Same method except using maximum order of n = 3 for blue fringes.
5
(a)
1.
Progressive wave: all points within a wavelength have different phase.
2.
Stationary wave: all points between two adjacent nodes are in phase,
all points at 2 sides of a node are anti-phase.
Progressive wave: energy is transferred.
3.
Stationary wave: no net transfer of energy
Progressive wave: Maximum KE is the same at all points
Stationary wave: Maximum KE varies with position
(b)
Observable diffraction occurs when the wavelength of the wave is
comparable to the dimension of the slit/obstacle it meets.
The wavelength of sound waves is comparable to the dimensions of a room
therefore it is able to spread more such that it bends around corners. The
wavelength of light is much smaller than the dimension of a room and
appears to travel in a straight line.
6
a
Resultant force acting on the object is zero.
Resultant torque acting about any axis is zero.
b
TA sin 30  TB sin 50
Considering vertical forces,
TA cos 30  TB cos 50  mg
sin 50
cos 30  TB cos 50  mg
sin 30
TB  4.98 N
TB
Taking moments about the left end,
mgx  (4.98 cos 50)(6.5)
x  2.12 m
ci
dm Ah 

dt
t
 Av
Mass of air pushed by bullet per unit time 
dm
v  A v 2
dt
By N3L, Force of air on bullet  Av 2
By N2L, Force of bullet on air 
Preliminary Examination
cii
JC2 H2 Physics 2011
It will not accurately determine the drag force of the air on the bullet.
In practical situations, the bullet is more aerodynamic in design and not just
shaped like a cylinder.
Or
There is also friction between the air and the surface of the bullet, which was
not considered in the above expression.
di
Mechanical energy of A and bullet
1
 mgh  mgx sin  30   kx 2
2
 (4.20)(9.81)(0.30)   4.20  (9.81)(0.20)(sin(30)) 
1
(10000)(0.20)2
2
 216 J  220 J
dii
1
mv 2  216
2
v  10 m s-1
diii
By COM,
mbullet ubullet   mbullet  mA  v
ubullet  213
 210 m s-1
div
The collision between A and the bullet was perfectly inelastic, hence kinetic
energy is not conserved.
Part of the kinetic energy of the bullet is transferred to A as heat energy,
causing A to experience a rise in temperature.
7a (i)
It is possible at a point such that the net gravitational field due to surrounding
masses is zero (gravitational field due to surrounding masses cancels out each
other since it is a vector) but the gravitational potential at that point is non zero as it
is due to scalar addition of gravitational potentials due to masses.
(ii)
Positive Source
Charge
Negative Source
Charge
Change in Electrical Potential Energy
with increasing distance from source
charge
Negative test
Positive test
charge
charge
Increase
Decrease
Decrease
Increase
Preliminary Examination
b
JC2 H2 Physics 2011
(i)
B- Q
ER
A
C
+Q
(ii)
B
GR
A
C
(c)
(i)
mu 2
= FG
R
mu 2 GMm
=
R
R2
u
(ii)
GM E
R
1. GPE at infinity = 0. For space vehicle to just reach infinity means its
KE = 0 once it reaches infinity. Hence total energy at infinity = 0
Energy of vehicle during orbit at R away from Earth = Energy at infinity

GME (2m) 1
 (2m)v 2  0
R
2
Velocity of space vehicle (escape speed) =
2. By COM:
(3m) (
GM E
)
R =
VB = (2 2  3)
2m
2GME
 mVB
R
GME
R
VB = (2 2  3 )u  0.172u
Speed = 0.172 u
2GME
R
Preliminary Examination
(iii)
JC2 H2 Physics 2011
Considering energy before explosion and immediately after;
Explosion energy + Initial KE + GPE = GPE + KE right after explosion
2GME 2
)
R
Explosion energy + ½ 3 m u2= ½ m (0.172)2 u2 + ½ (2m) ( 2 )2 u2
Explosion energy + ½ 3 m u2= ½ m (0.172)2 u2 + ½ (2m) (
Explosion energy = 2.0148 m u2 – 3/2 m u2
= 0.515 m u2
(d)
(i)
Potential at P = V
𝑞
V = 4𝜋𝜀
𝑜𝑅
q =4𝜋𝜀𝑜 𝑅𝑉.
(ii)
Force on sphere = q E = q (2V/x) =
Acceleration = F/m =
8 o RV 2
x
8 o RV 2
mx
s = ut + ½ at2
(iii)
s= x
8 o RV 2 2
x=0+½(
)t
mx
t=
x
V
m
4 o R
time taken = 2 t = 2 (
8
(a)
x
V
m
4 o R
)=
x
m
V  o R
(i)
The ohm is defined as being the resistance of a material through
which a current of one ampere is flowing when a potential difference
across it is one volt.
(ii)


V = IR is actually the defining equation for resistance, R = V/I.
Ohm’s law is when I  V, i.e. when R is a constant (not necessarily
true in general)
Preliminary Examination

(iii)
JC2 H2 Physics 2011
Student B is thus correct.
Across the battery, it (the 2.0 V) represents 2.0 J of energy converted
from other forms/chemical to electrical energy for every coulomb of
charge flowing through it.
Across the resistor, it (the 2.0 V) represents 2.0 J of energy converted
from electrical energy to other forms/thermal energy for every coulomb
of charge flowing through it.
Answers leaving out values (e.g. no mention of 2 J) but otherwise correct,
award 1 mark.
Answers such as “energy supplied by the battery” and “energy dissipated by
the resistor”, award 0 marks.
(b)
(i)
The total charge that flows through the ammeter can be found by the
𝑡
area under the graph (from 0 to t), 𝑄 = ∫0 𝐼 𝑑𝑡
(ii)
V
time
(c)
(i)
- The higher temperature results in promotes the breaking of more
covalent bonds between the atoms, generating more free electrons to
move inside the material.
- The effect of more free electrons outweighs the increase in
resistance due to larger lattice atoms’ vibrations. The flow of electrons
still increases and thus current increases. Hence, the resistance of the
semiconductor decreases with an increase in voltage.
(from COE lecture notes pg13 on semiconductor(diode) I-V
characteristics)
(ii)
At T= 20C, R = 12000 
10
𝐼=
12000 + 10000
= 4.5454 x 10-4 = 4.54 x 10-4 A
(iii)
Potential difference across the thermistor at 20C,
V = (4.5454 x 10-4)(12000) = 5.45 V
Preliminary Examination
JC2 H2 Physics 2011
12000
(Alt can use Vthermistor = 12000+10000 × 10= 5.45 V)
𝑉2
𝑅
102
22000
(iv)
P=
(v)
Reasoning that since resistance R decreases, P = I2R also decreases
is incorrect as it ignores changes current I.
=
= 4.54 x 10-3 W
𝑉2
Using P = 𝑅 , it is clear that for the same emf source, as R decreases,
power supplied by the battery increases.
[changes in current are inversely proportional to changes to R. i.e. if
R ½ R, I  2I, and since P = I2R, changes in I will dominate.]
(vi)
1. Reading off 2 correct set of values.
Correct working/manipulation
Answer (range of ~3300 to 3600 K accepted)
2.
𝑅 = 𝑅𝑜
1 1
𝛽( − )
𝑇
𝑇𝑜
𝑒
1 1
lnR = β  -  +lnRo
 T To 
1 1
Plot a graph of ln R against   
 T To  .
 can be obtained from the gradient ( = gradient)
(where To is any chosen reference temp.
The vertical intercept will be ln Ro, where Ro is the resistance at temp To.)