SOLUTIONS TO CAPE PHYSICS
2003 TO 2015
CARIBBEAN ADVANCED PROFICIENCY EXAMINATION
(CAPE)
JUNE
PAPERS 1 & 2 (2003–2007)
PAPER 2 (2008–2015)
UNIT 1
Samlal Mannie
B.Sc., Dip. Ed. (Admin), Dip. LCCI
Examiner for CXC since 1986
© Caribbean Educational Publishers (2003) Ltd.
All rights reserved. No part of this publication may be reproduced,
stored in a retrieval system, or transmitted in any form or by any means,
without the prior permission in writing from the Publishers.
CAPE 2003–2015, Physics Solutions Unit 1
ISBN: 000-000-000-000-0
First Published 2010
Second Published 2016
CARIBBEAN EDUCATIONAL PUBLISHERS (2003) LTD.
TEDDY’S SHOPPING CENTRE,
GULF VIEW LINK ROAD,
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E-mail: admin@mbsceptt.com
DEDICATION
This work is dedicated at the Divine Lotus Feet of the Lord
and children of the world
“Education is not mere accumulation of information or even
the a
­ cquisition of skills. It is the cleansing of the mind, the
­strengthening of the unselfish ­tendencies, and the discovery
of truth, goodness and beauty that lie dormant in every being.
It is culmination of integrity, tolerance and compassion. It is the
­revelation of the Divine, which is the very core of every created
­being and thing”.
“LOVE ALL, SERVE ALL
HELP EVER, HURT NEVER”.
Sri Sathya Sai Baba.
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Contents
2003 – 2015 June Exams CAPE Physics UNIT 1
Perface. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vi
Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ix
About the Author . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xi
2003 paper 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
2003 paper 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2004 paper 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
2004 paper 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
2005 paper 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
2005 paper 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
2006 paper 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
2006 paper 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
2007 paper 1 (Specimen paper – Multiple choice). . . . . . . . . . . . . . . . . . . . . . . . . . . 134
2007 paper 1 (Exam paper – Multiple choice) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139
2007 paper 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
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vi
contents
2008 paper 2 Trinidad & Tobago . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
2008 paper 2 Other Islands . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179
2009 paper 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190
2010 paper 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203
2011 paper 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217
2012 paper 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229
2013 paper 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242
2014 paper 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255
2015 paper 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268
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PREFACE
Solutions to CAPE Physics Unit 1 2003 to 2009 (first edition) covers the
­suggested solutions for June past papers. Alternative solution methods are given
where necessary.
This book contains the following features:
• Solutions to the questions from the past papers with suggested alterations
to questions that are “ambiguous”.
• Hints on answering questions and points to be careful about.
• Colour highlighting of important points.
• Proper, well-labeled diagrams accompanying answers where necessary.
• Solutions to 2007 Specimen Multiple choice paper and solutions to the
­actual exam multiple choice paper of 2007 with explanations for each
­answer.
The author wishes to advise that these solutions are not necessarily those
used by CAPE during their marking exercise, and is not meant to be interpreted
as such.
This book is intended to help both students and teachers.
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ACKNOWLEDGEMENTS
The author would like to thank his students and fellow teachers for their
­encouragement and support in writing these solutions to CAPE Physics.
Special thanks to the Caribbean Examinations Council for giving me the
­opportunity to be an assistant Examiner in Cape Physics.
A heartfelt thanks to members of my family for their support in this project.
All praise and thanks to God Almighty for using me as His instrument.
Samlal Mannie
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ABOUT THE AUTHOR
Samlal Mannie graduated from the University of The West Indies (UWI) with a
Bachelor of Science degree in Physics, Environmental Physics and Mathematics.
He holds a Diploma in Educational Administration (UWI), a diploma from the
London Chamber of Commerce (LCCI) in Marketing, Advertising and Public
­Relations and a Certificate in Radio Broadcasting from the Announcers and
Broadcasters Academy (ABA).
Samlal Mannie has been an examiner for CXC Physics since 1986. He is also an
assistant examiner for the Advanced Level (CAPE). At present he teaches both
levels at Carapichaima East Secondary in Trinidad. He also continues to produce
and present programmes on the local radio stations.
Samlal Mannie is an ardent bridge player, and has represented his country,
­Trinidad and Tobago on many occasions. He is a member of the Sri Sathya Sai
Baba Organization and integrates education in human values within his teaching
of Physics.
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LIST OF PHYSICAL CONSTANTS
Unit I.indd 1
Universal gravitational constant
G
=
Acceleration due to gravity
g
=
1 Atmosphere
=
Boltzmann’s constant
atm
k
=
Density of water
ρw
=
Specific heat capacity of water
Cw
=
Specific latent heat of fusion of ice
Lf
=
Specific latent heat of vaporization of water
Lv
=
Avogadro’s constant
NA
=
Molar gas constant
R
=
Stefan-Boltzmann’s constant
σ
=
Speed of light in free space
c
=
Planck’s constant
h
=
Triple point temperature
Ttt
=
6.626 × 10–34 J s
1 tonne
t
=
1000 kg
6.67 × 10–11 N m2 kg–2
9.81 m s–2
1.00 × 105 N m–2
1.38 × 10–23 J K–1
1.00 × 103 kg m–3
4200 J kg–1 K–1
3.34 × 105 J kg–1
2.26 × 106 J kg–1
6.02 × 1023 per mole
8.31 J K–1 mol–1
5.67 × 10–8 W m–2 K–4
3.00 × 108 m s–1
273.16 K
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(2003) PAPER 1
Question 1
(a) Equations must have the same base units on either side.
(b) O
ne limitation of using base quantities to check the b
­ alance of
­equations is that unitless constants are not able to be ­determined.
(c)
∆F ∆A ∆ρ 2∆V
=
+
+
F
A
ρ
V
 0.005  0.1 
 1
=
+
+ 2 


 0.1   1000 
 30 
(
)
= (0.05) + 1 × 10−4 + 2(0.0333)
= 0.12 (or 12%)
(d) The unit for V is ms–1.
Substituting into V = agλ,
Units ⇒ ms–2 × m = m2 s–2
Not possible.
Substitution into V = b gλ
Units ⇒ ms −2 × m = ms −1
V = b gλ
(e)
So
Unit I.indd 3
Possible.
16 = b 9.8 × 160
b = 0.404
V = 0.4 gλ
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Question 2
(a) A
cceleration is the rate of change of velocity whereas acceleration
due to gravity is acceleration a body experiences when in the earth’s
gravitational field.
(b) (i) For horizontal motion:
d = ut
So
t=
d
u
For vertical motion:
1
h = ut + at 2
2
1  d
= 0 + g 
2  u
2
1 d2
h= g 2
2 u
(ii) (a) F
or horizontal motion, there is no resultant force, hence no
acceleration (from second law).
(b) F
or vertical motion, there is now a resultant downward
force (mg), so there is acceleration.
(iii) T
he resultant velocity can be found by finding the vector sum of
the horizontal and vertical velocities.
u
V = gt
Unit I.indd 4
R
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Question 3
(a) (i) The submerged portion of the cork will remain the SAME.
(ii) The reasons can be given in two ways:
–The weight of the block becomes less, hence weight of
­displaced liquid becomes less.
OR
–Use the fact that Archimedes Principle still holds, so weight
of cork = weight of displaced water.
On Earth, Mc g = Mw g,
where g cancels.
Similarly on the moon.
(b) (i) (a) Volume of water displaced = 75 cm3
= 75 g
Mass of water
= 0.74 N
Weight of water
= 0.74 N
\ Weight of block
(b) Volume of L displaced = 120 cm3
So
So
120 × ρL × g = 0.74
0.74
120 × 10−6 × 9.8
= 625 kg m −3
ρL =
Alternative method:
ρL × 120 = 1000 × 75
1000 × 75
120
= 625 kg m −3
ρL =
(ii) Frictional forces only occur when there is relative motion.
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Question 4
(a) T
wo conditions necessary for a body to be said to describe simple
harmonic m
­ otion are:
1.
2.
(b) (i)
he acceleration of the body must be proportional to distance
T
moved from a fixed point.
hat acceleration must always be directed towards THAT fixed
T
point.
a = –w2x
V = ω A2 − x 2
Maximum V occurs when x = 0
So
So
Vo = wA = wxo
ω=
So using
Vo
0.2
=
= 5 rads −1
−2
x o 4 × 10
ω = 2π f =
T=
(ii) a = ω 2 x o
2π
T
2π
= 1.26 s
5
= 25 × 4 × 10–2
(iii)
= 1.0 ms–2
E
Total Energy
PE
KE
–X0
+X0
Displacement
(iv) Kinetic energy + Potential energy = Total energy.
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Question 5
(a) R
efractive index is defined as the ratio of the sine of angle of
­incidence to the sine of angle of refraction, where the angle of
­incidence is taken as the angle in the faster medium.
n=
Sin i
Speed in faster medium
or n =
Sin r
Speed in slower medium
Critical angle is the angle of incidence in the slower medium for
which the refracted ray comes out perpendicular to the normal
(or refracted ray comes out parallel to the surface).
(b) (i)
n=
1.5 × 103
330
= 4.55
=
(ii)
(c)
Speed in faster medium
Speed in slower medium
1
Sin c
1
Sin c = = 0.22
n
c = 12.71°
n=
Water
Air
q2
q1
r2
q1 < C
q2 > C
Note:
Unit I.indd 7
Make sure θ2 = r2 in the drawing.
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Question 6
(a) D
iffraction is the spreading of waves as they pass through small
openings or around small obstacles.
(b) T
he diffraction of sound can be noticed as a consequence of being
able to hear around a corner.
(c)
For light: If a beam of monochromatic light is made to pass through
a small ­opening (e.g., a pin line or a painted glass slide) a diffraction
pattern is observed (a series of bright and dark fringes).
d Sin θ = nλ
nλ
Sin θ =
≤1
d
1
d=
mm = 3.33 × 10−6 m
300
λ = 6.4 × 10−7 m
nλ
≤1
d
d 3.33 × 10−6
n≤ =
λ 6.4 × 10−7
≤ 5.2
So
So
Hence, maximum n is 5 (since n is integer).
The wavelength of blue light is less than the wavelength of red light.
So from the formula:
n≤
d
λ
n could be greater.
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Question 7
(a) T
he specific heat capacity of a substance is the amount of heat
­needed to raise the temperature of one kilogram of the substance by
one degree celsius (or 1 kelvin).
C=
H
m∆θ
The heat capacity of a substance is the amount of heat needed to
raise the temperature of the entire substance by 1 K (or 1°C).
(b) (i) Heat energy taken in by container = H.C × Dθ
Heat taken in by water
So total energy supplied
Total time taken = 9 × 60
E 106,920
=
T
540
= 198 W
∴ Power =
= 90(79 – 25)
= 4860 J
= MCDθ
= 0.45 × 4200 × 54
= 102,060 J
= 106,920 J
= 540 s
(c) T
he energy supplied goes to increase the internal energy of the
water and container; i.e., the kinetic energy of the molecules and the
­potential energy of the molecules (they move further apart).
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Question 8
(a) Tensile stress is force per unit cross-sectional area.
Stress =
F
N/m2
A
Tensile strain is the ratio of the extension to the original length.
Strain =
x
l
Young’s modulus is the ratio of stress to strain.
Y=
Stress
Strain
(b) Let the original length be x.
0.15
×x
100
0.15
×x
So strain = 100
= 1.5 × 10−3
x
So extension =
Stress = Young’s modulus × strain
= 1.2 × 1011 × 1.5 × 10–3
(c) (i)
= 1.8 × 108 Nm–2
F = Stress × Area.
= 1.8 × 108 × 2 × 10–6
= 360 N
(ii) Work done = Area under force / extension graph.
1
= F×x
2
1
= × 3.60 × 102 × .0015
2
= 0.27 J
Note: x = 0.15% of 1 m
= 0.0015 m
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Question 9
(a) (i) C
onduction of thermal energy along a metal bar takes place in
two ways:
1.
2.
By movement of free electrons, and
By lattice vibrations.
(ii) In a piece of wood there are no free electrons and so conduction
takes place only by lattice vibrations.
(b) F
or good conductors, the bar should be about five times the diameter.
This allows for a measurable temperature gradient. The bar must be
lagged properly.
For a poor conductor, the specimen must be thin with a large
­cross-­sectional area. This allows for a greater rate of energy flow.
(See Lee’s discs experiment).
(c) T
he materials could have different specific heat capacities. Also heat
loss from the sides due to radiation will depend on the colour and
texture of the material.
Unit I.indd 11
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(2003) PAPER 2
Question 1
0.361 + 0.383 + 0.374 + 0.365
4
= 0.371 ± 0.0005 s
(a) Average =
1 2
(b) Using s = ut + at
2
1
2
0.65 = 0 + (a)(0.371) s = 0.65 m
2
t = 0.371 s
a=
0.65 × 2
(0.371)2
= 9.44 ms −2
(c) S ystematic errors are normally due to faulty instruments. They lead
to values that are either always too big or too small compared to the
true value.
Random errors are normally due to the experimenter. They lead to
values that are on both sides of the true value.
(d) R
andom errors can be reduced by repetition and finding the average,
or plotting a graph of h – vs – t2 and using slope.
(e) 1.The timing device may have a zero error. That is, it may not start
at zero.
2.
3.
Unit I.indd 12
The ball may not have been released instantly.
he value of h may not have been taken from the correct height.
T
(Any one)
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Question 2
(a)
1/mm
T/S
log (1/mm)
log (T/S)
231
292
411
515
859
0.94
1.06
1.27
1.42
1.86
–0.027
0.025
0.104
0.152
0.270
2.36
2.47
2.61
2.71
2.93
Note: 1. W
hen completing table, make sure significant figures are
maintained in each quantity.
2. W
hen finding “logs”, it is the “log” of the quantity and its
units together written as shown. This will mean that there
will be no units for “log” values.
T = aln
Taking logs,
log T = n log l + log a
(graph attached).
(b) n is slope of graph
So
Unit I.indd 13
(3.0 × 10
Slope =
n = 0.51
−1
− 0.2 × 10−1
(3.0 − 2.45)
)
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Log10T ×10-l
3.6
3.4
3.2
3.0
2.8
2.6
2.4
2.2
2.0
1.8
1.6
1.4
1.2
1.0
0.8
0.6
0.4
0.2
0
2.3
-0.2
2.4
2.5
2.6
2.7
2.8
2.9
3.0
3.1
Log 10l
-0.4
-0.6
-0.8
-1
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(c) Take point (3, 0.3)
(Note: The y-axis is multiplied by 10–1)
Use y = mx + c
0.3 = 0.51 (3) + c
c = 0.3 − 1.53
So
= −1.23
log a = −1.23
a = antilog ( −1.23)
= 0.059
OR
a = 10 –1.23
= 0.059
(d) T
ime 20 or more oscillations (for a total time of more than 20 s) and
find average time. Repeat this procedure at least once more and find
average.
Unit I.indd 15
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Question 3
(a) A
s the temperature increases, the rate of heat loss from the side of
the kettle ­increases, so the temperature will not increase uniformly.
(b) Consider a temperature rise from 35°C to 80°C,
Heat supplied = Pt
So
= 1.6 × 150 × 103 = mc ∆θ
1.6 × 103 × 150
1.1 × 45
= 4.85 × 103 J kg −1k −1
C=
If the entire graph up to 275 s is taken, then
C=
(c) 1.
Unit I.indd 16
1.6 × 103 × 275
1.1 × 74
= 5400 J kg −1k −1
 ∆θ = 100 − 26


 = 74°C

2.
Insulate the kettle to minimize heat loss.
3.
(Recall: Rate of heat loss is proportional to excess temperature
above ­surroundings.)
et the temperature rise above room temperature be small,
L
about 5°C, so that heat loss will be negligible.
Start with a temperature below room temperature.
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(d)
Heater
Scale
At boiling point (100°C), note the “loss” in mass of water for a time t.
Energy supplied by heater = Power × Time
So
So
Unit I.indd 17
Pt = ∆ m L
Pt
Lv =
∆m
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Question 4
(a)
VB
r
∆q
S
B
A
VA
Consider a body moving in a circle as shown, moving from A to B in ∆t.
s = r ∆θ
s
∆θ
V=
=r
= rω
∆t
∆t
∆V = VB − VA
So
−VA
Q
∆q
∆V
(1)
P
VB
R
ˆ ≈ 90°
If ∆θ is small, then < PQR ≈ PRQ
and ∆v = v ∆ θ
So
a=
∆v
∆θ
=v
= vω
∆t
∆t
Substituting 2 in 1, a = (rω) ω = rω2.
(2)
Acceleration is directed towards the centre of the circle, since RQ ⊥ to
both VA and VB.
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(b) (i) N
ewton’s law of universal gravitation states that for two ­bodies
in space, a force of attraction will exist between them that
is proportional to the product of their masses and inversely
­proportional to the square of their distance apart.
F =G
m1m2
r2
(ii) m
8
0
4.
×
10
m
E
The force of attraction of the earth on the moon is the force that
acts on the moon.
The moon attracts the earth with an equal but opposite force, in
keeping with Newton’s third law of motion.
(iii)
Unit I.indd 19
GmE mM
4π 2
2
=
=
m
R
m
R
ω
2
2
R2
T2
4π 2R3
2
⇒ T = 2.53 × 106 s (29.3 days)
T =
GmE
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cape physics - Unit 1
(c) (i)
38.3
50 g
0.20 m
mg
T = 0.8 s
2π 2π
=
= 7.854 rad/s
ω=
T 0.8
⇒ 7.85 rad/s
Centripetal acceleration = rω 2
= 0.2 × 7.852
= 12.3 ms–2
(ii) Centripetal force = mrω 2
50
=
× 12.3
1000
= 0.617 N
T Sin 38.3 = mg
mg
50
9.81
=
×
T=
Sin 38.3 1000 0.620
= 0.791 N
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21
Question 5
(a) (i) (a) Kinetic energy is energy a body possesses by virtue of its
motion.
1
KE = mv 2
2
(b) P
otential energy is energy a body possesses by virtue of its
position or state (arrangement).
e.g., G.P.E = mgh
(ii)
Distance
S
t=0
V=0
t=t
V=V
∆v v − 0 v
=
=
t
t
∆t
v
So Force = m × acc = m = ma
t
2
2
V = u + 2as
acc =
v2
2a
v2
as =
2
s=
Since u = 0
Work done is force × distance moved in direction of force
= F × s = Mas
v2
2
But work done = KE gained
1
So
KE = E k = mv 2
2
=M
Unit I.indd 21
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cape physics - Unit 1
(iii) (a) A
t terminal velocity there is no change in speed, so no
gain in kinetic energy. So all loss in gravitational ­potential
­energy dissipates as heat due to friction (drag forces)
within the medium.
(b) L
oss in G.P.E is as a consequence of loss in height. The
medium through which the ball bearing is falling provides
resistance (frictional) forces.
(b) (i) (a) W
hen the force is +4N, it means that the force acts in one
direction and is constant.
(b) W
hen the force is –4N, it acts in the opposite direction and
reduces to zero in 3 m.
(ii) Work done = Force × Distance moved (area undergraph )
= 4× 5
= 20 J
(iii) W
ork done opposite to the motion of box = Area undergraph
from 5 m to 8 m
=
1
×3× 4 = 6 J
2
∴ Net gain in KE = 20 – 6 = 14 J
(c) (i) Using conservation of energy
1 2
mv = mgh
2
v = 2 gh
= 2 × 9.81 × 11000
= 465 ms −1
Unit I.indd 22
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1
1
(ii) Loss in KE = mv12 − mv22
2
2
1
= m 2502 − 502
2
(
23
)
1
= m (250 + 50)(250 − 50)
2
=
1
× 170 × 103 × 300 × 200
2
= 5.1 × 109 J
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cape physics - Unit 1
Question 6
(a) (i) A
longitudinal wave is one in which the direction of vibration of
the particles is the same as the direction of travel of the wave.
The frequency of a wave is the number of waves passing a point
in one second.
(ii) (a)
v= fλ
v 340
λ= =
= 0.68 m
f 500
d/m
.34
.68
x/m
Scale: 4 cm ≡ 0.34 m
(b) T =
d/m
1
1
=
= 2 × 10−3 s ≡ 2 ms
f 500
1
2
t/ms
Scale: 4 cm ≡ 1 ms
(b) (i) T
he intensity of sound is the energy per second incident on an
area of one metre square.
Intensity unit W m–2.
The threshold of hearing is the lowest intensity that can be
heard. This is taken as 10–12 W m–2.
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25
I
dB level = 10 log  
 I0 
(ii) (a)
 I 
80 = 10 log  −12 
 10 
So
I
= Antilog of 8 = 108
−12
10
I = 10−12 × 108 = 10−4 Wm−2
So
(b) Power = Intensity × Area
= 10–12 × 0.4 × 10–4
= 4.0 × 10–17 W
(c) For fundamental l =
For first overtone
λ
4
=
1v 1
= × 1.214 = 0.304 m
4f 4
For second overtone
l=
3v
= 0.911 m
4f
5v
= 1.52 m
4 f
At 3rd ⇒ l = 2.13 m
l=
So at length 0.911 m and 1.52 m resonance will be heard.
1
1 × 9.81 × 0.8
2 × 0.8
1.6 × 10−3
= 43.8 Hz
(d) f0 =
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cape physics - Unit 1
Question 7
(a) (i)
Conditions necessary for interference to be observed for two
sources of light:
– T
he two sources must be coherent, i.e., they must have the
same amplitude, velocity, frequency, wavelength and constant
phase difference.
– If polarized, they must be polarized in the same plane.
– The separation of the sources must be small (≈ 0.5 mm).
Monochromatic
source
(ii)
– They must meet.
S1
Area
of
interference
S
S2
D
METHOD:
OBSERVATION:
Unit I.indd 26
he apparatus is set up as shown in the
T
­diagram. Translucent paper can be used as
the screen.
series of bright and dark fringes will be
A
­observed on the screen.
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(iii) ∆x =
27
λD
d
∆x is fringe separation.
λ is wavelength of light used.
D is the distance from slits to the screen.
d is the slit separation.
(b) (i)
∆x =
λD
d
16
=
× 10−3 m
12
d = 0.55 × 10−3 m
D = 1.3 m
∆xd
λ=
D
= 5.64 × 10−7 m
This seems to be yellow (red is accepted).
(ii) I n the centre there will be a bright spot. The waves will meet in
phase since the path difference is zero in this case.
(iii) A
diffraction pattern will be observed as for one slit. The ­pattern
is ­similar to the interference pattern but will be less bright
­because there is less light ­coming. (Students can take the
­opportunity to read up on Huygen’s ­Principle!!).
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cape physics - Unit 1
(iv) A
ccording to the formula d Sin θ = nλ, different wavelengths
­(colours) will diffract by different amounts. At the centre when
n = 0, there will be a bright white spot. All other colours will
spread out to form a spectrum, with red spreading the most.
For higher orders overlapping of colours may take place.
R2
V2
R1
Diffraction
gravity
V1
White spot
(v) I nitially the fringes will be brighter. As the separation c­ ontinues,
there will be more and more overlapping and eventually no
fringes will be seen.
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29
Question 8
(a) (i) (a) Assumptions of the kinetic theory of gases:
– Gases are made up of many molecules.
– These molecules are moving randomly.
– The intermolecular forces are negligible.
– T
he total volume of the molecules is negligible compared
to the ­overall volume of the gas.
– All collisions are elastic.
– T
he duration of collisions is negligible compared to the
time between collisions.
– Newtonian mechanics apply.
(b) W
hen the molecules collide with the walls of the container
(or any surface), they experience a change of momentum.
The rate of change of momentum is force. Force divided by
area is pressure.
P=
F
A
v + 4v + 8v
3
= 4.3 v
(ii) Mean speed =
v 2 + ( 4v ) + (8v )
r.m.s speed =
3
= 5.21 v
2
(b) (i)
2
E = n C p ∆T
= 3 × 29 × 50
= 4350 J
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cape physics - Unit 1
V1 =
(ii)
n R T1
P1
3 × 8.31 × 270
5 × 105
= 13.462 × 10−3 m3
=
V2 =
3 × 8.31 × 320
5 × 105
= 15.955 × 10−3 m3
=
So
n R T2
P2
∆V = 2.49 × 10−3 m3
(iii) ∆W = P ∆V
= 5 × 105 × 2.49 × 10−3
= 1.25 × 103 J
(c) (i)
∆U = nCv ∆T
Cv = C p − R
So
(since C
p
− Cv = R
)
= 29 − 8.31
= 20.69 J/mol/k
∆T = 50
n=3
∆U = 3 × 20.69 × 50
= 3100 J
Alternately:
∆U = Q − W
= 4350 − 1250
= 3100 J
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31
(ii) C p − Cv = R
Cv = C p − R
= 29 − 8.31
= 20.7 J/mol/k
Alternately:
Using ∆u = n Cv ∆T and using ∆u from Q – W = 3100,
∆u 3100
=
n∆T 3 × 50
= 20.7 J/mol/k
Cv =
Unit I.indd 31
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cape physics - Unit 1
Question 9
(a) (i) (a) D
ensity is defined as mass per unit volume. It is a scalar
quantity and is measured in kg m–3.
(b) P
ressure is defined as force per unit cross-sectional
area. It is a scalar quantity and is measured in N m–2 or
­Pascals (Pa).
(ii) T
he difference in pressure between two depths in a fluid is
given by
Weight of fluid
Area
mg
=
A
Mass = Volume × ρ
And Volume = A × ∆h
A∆hρ g
So
∆p =
A
∆p = ∆hρ g.
∆p =
(iii) L
and and sea heat up at different rates due to their different
heat capacities (land has a lower heat capacity than sea). The air
above the land heats first and so expands. Thus the density falls
and this air rises. Cooler air from above the sea rushes in and so
a sea breeze is formed.
(b) T
he molecular separation in gases is much greater than that of solids.
Hence the same mass could occupy much greater volume in a gas.
(c) (i)
∆p = ∆hρ g
Pressure at 3 cm mark = 3 × 10−2 × 8.2 × 102 × 9.81
= 241 Pa
Pressure at 14 cm mark = 14 × 10−2 × 8.2 × 102 × 9.81
= 1126 Pa
Unit I.indd 32
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(ii)
(iii)
33
Force at 3 cm mark = 241 × 140 × 10−4
= 3.4 N
Force at 14 cm mark = 1126 × 140 × 10−4
= 15.8 N
Hence, net force = 15.8 − 3.4
= 12.4 N
Mass = Volume × Density
V = Ah = 11 × 10−2 × 140 × 10−6 m3
= 1.54 × 10−3 m3
Mass = 1.54 × 10−3 × 820
= 1.26 kg
∴ Weight = mg
= 1.26 × 9.81
So
= 12.4 N
(iv) Upthrust = weight of liquid displaced
= 12.4 N
Unit I.indd 33
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(2004) PAPER 1
Question 1
(a) Newtons second law:
The rate of change of momentum of a body is directly proportional to
the ­RESULTANT force and takes place in the direction of the force.
Newtons third law:
If body A exerts a force on body B, then body B exerts an equal and
opposite force on body A.
(b) (i)
roof of car
T
90°
40 g
Direction of
acceleration of car.
mg
(ii) The resultant force acts in the direction of acceleration.
So T Cos 29 = W and T Sin 29 = R.
So
Unit I.indd 34
R
= tan29
W
R = W tan29
= 0.04 × 9.81 × tan29
= 0.22 N
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35
(iii) Using F = ma,
F
m
0.22
=
0.04
= 5.5 ms −1
a=
(iv) T
he second force in the Newton’s third law pair of forces, which
includes the weight of the bob, will be the force of the bob on
the earth.
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cape physics - Unit 1
Question 2
(a) T
he ball does not reach its original height because energy is lost due
to resistive forces (drag forces). Also energy is lost in collision with
the ground.
(b) See graph page attached.
Note:
1. Slope must be the same.
2. Velocity decreases with each bounce.
(c) Acceleration is constant at 9.81 ms–2
(See graph page)
(d) T
he time of 30 s is an error. This will give a height of 4410 m !!!
­Impossible for a child!
Taking a time of 0.3 s,
1
s = ut + at 2
2
1
2
= 0 + (9.81)(0.3)
2
= 0.44 m ( more realistic!)
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37
(b)
V1
V2
V3
t
V3
V2
Unit I.indd 37
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38
cape physics - Unit 1
(c)
a/m/s 2
9.81
t /s
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39
Question 3
(a) (i) G
ravitational potential energy is energy a body possesses by
virtue of its ­position above some reference level (above the
ground)
G.P.E = mgh.
(ii) Kinetic energy is energy a body possesses by virtue of its
1
­motion, KE = mv 2 .
2
(iii) T
he law of conservation of energy states that energy cannot be
created nor destroyed but can be transformed from one form to
another.
(b) G
ravitational potential energy is converted to heat (by friction with
the medium).
Note: No gain in kinetic energy since speed is constant.
(c) (i) Friction.
(ii) Work done on plank (as heat) = F × d
= 150 × 8 = 1200 J
(iii) Change in G.P.E = mgh = 50 × 9.81 × 5 = 2450 J
So KE at deck = 2450 − 1200 = 1250 J
1
So mv 2 = 1250
2
v = 7.07 ms −1
Unit I.indd 39
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cape physics - Unit 1
Question 4
(a) (i) x = ACos ωt or x = –ACos ωt.
Note: x = A when t = 0.
(b)
(ii) V = –Aω Sin ωt or V = + Aω Sin ωt
Total energy constant
E
E total
Ep
Ek
t
o
m
K
2
4π m
11
Seconds
K=
T=
2
21
T
4π 2 (0.250)
=
2
 11 
 
21
(c) T = 2π
= 36.0 N m −1
(d) There will be no change since g is not in the expression.
Unit I.indd 40
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41
Question 5
(a) (i) R
efraction of light is the bending of light as it passes from one
medium to another. This is a consequence of change in speed.
(ii) R
efractive index is the ratio of sine of angle of incidence to sine
of angle of refraction, where the angle of incidence is taken in
the faster medium.
n=
Sin i
Sin r
(OR) In terms of Speed:
n=
Speed in vacuum
Speed in the medium
(b) Two conditions for total internal reflection to take place:
– T
he light must be travelling from the slower medium to the faster
medium. In this case, from potassium iodide to aniline.
– T
he angle of incidence in the potassium iodide must be greater
than the ­critical angle.
(c) Relative refractive index
1.67
= 1.05
1.59
1
1
= 0.95
Sin c = =
n′ 1.05
⇒ c = 72°
n′ =
Unit I.indd 41
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42
cape physics - Unit 1
(d) (i)
Analine
i r
Potassium
Iodide
l
l
Note: 1. Same wave lengths
2. ^i = ^r
(ii) S ince analine has the lower refractive index, the light will travel
faster in ­analine.
Unit I.indd 42
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43
Question 6
(a) r = 2f for thin lenses.
1
f
2
P=
r
P=
and
So
f must be in metres. P will be in dioptres.
(b)
r is radius of curvature, f is focal length and P is power.
F
F
(c) (i) Using the lens formula,
So
Unit I.indd 43
1 1 1
= +
f u v
1 1
1
=
−
v 20 10
1
=−
20
v = −20 cm
f = 20 cm
u = 10 cm
v=?
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cape physics - Unit 1
Magnification =
=
v
u
20
=2
10
So length of image = 2 × 1 cm = 2 cm
(ii) The image will be virtual.
Students should note the “real is positive” convention.
Unit I.indd 44
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45
Question 7
(a) (i) Tensile stress is the ratio of the force to the cross-sectional area.
Street =
F
A
(ii) Tensile strain is the ratio of the extension to the original length.
Strain =
x
l
(iii) Young’s modulus is the ratio of stress to strain.
Y=
Stress
Strain
(iv) H
ooke’s law states that the deformation of a material is directly
proportional to the force applied, provided that the limit of
­direct proportionality is not exceeded.
(Note the difference between elastic limit and limit of direct
­proportionality.)
(b) (i) At load 54 N, the graph deviates from Hooke’s law.
(ii) Stress =
F
54
=
A π 0.5 × 10−3
(iii) Strain =
∆l
l0
(
)
2
= 6.88 × 107 Nm −2
0.75 × 10−3
2.0
= 3.75 × 10−4
=
Stress
Strain
6.88 × 107
=
3.75 × 10−4
= 1.83 × 1011 Nm −2
(iv) Young’s modulus =
Unit I.indd 45
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cape physics - Unit 1
Question 8
(a) Either PV = nRT
or PV = NKT
where n is number of moles,
R is molar gas constant,
T is thermodynamic temperature,
N is number of molecules in the gas,
K is Boltzman’s constant.
1
(b) PV = Nmc 2 = NKT
3
2 1

N  mc 2  = NKT

3 2
(c) (i)
1
3
E k = mc 2 = KT
2
2
PV = nRT
PV
n=
RT
6 × 105 × 3.5 × 10−3
=
8.31 × 400
= 0.63 moles
(ii) Number of molecules = 0.63 × 6.02 × 1023
= 3.79 × 1023 molecules
∴ number of atoms = 2 × 3.79 × 1023
= 7.58 × 1023 Atoms
Unit I.indd 46
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47
3
(iii) E k = KT
2
3
= × 1.38 × 10−23 × 400
2
= 8.28 × 10−21 J/molecule
8.28
∴ E k per atom =
× 10−21 J
2
= 4.14 × 10−21 J/atom
Unit I.indd 47
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cape physics - Unit 1
Question 9
(a) T
he vacuum eliminates heat loss by conduction and convection since
these ­processes need a medium.
The cork stopper reduces heat loss by conduction.
The silvered surfaces reduce radiation from entering from outside.
(b) (i) (a) with scale.
TW
TA
1
21
x/mm
(b) without scale.
TW
TA
20
x/mm
(ii) From the formula
Q
∆θ
= KA
t
∆x
Q
If ∆x increases then
decreases. Now 1 mm of scale is
t
380
mm of copper.
­equivalent to
0.6
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49
So effective thickness of copper now is
633 + 20 mm
= 653 mm
Hence the rate of loss of heat is drastically reduced with the
scale.
(c) C
opper, being a metal with an abundance of free electrons, will have
as its ­mechanism of conduction, free electron movement as well
as lattice vibrations whilst the scale, not having free electrons, will
­conduct only by lattice vibrations.
Unit I.indd 49
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(2004) PAPER 2
Question 1
(a) Need to draw a tangent to the curve at the point where t = 0.5 s.
0.80 − 0.18
0.90 − 0.40
= 1.24 ms −1
Slope =
(b) Using the point (1.2, 1.8)
1
s = ut + at 2
2
1
1.8 = 0 + (a) × 1.44
2
a = 2.5 ms −2
(c) A
t impact kinetic energy would have been lost, so the speed in the
opposite ­direction will be smaller.
Note: Based on the question above, students can say that the ­gradient
before 1.25 is positive while the gradient after 1.25 is n
­ egative,
hence gradient before 1.25 is greater than gradient after 1.25.
(The ­question should say Magnitude of gradient in order to get an
energy consideration answer.)
(d) T
he displacement will not become zero again, i.e., the ball will not
reach back to its original height, because energy is lost on impact.
Unit I.indd 50
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51
Question 2
(a)
f/Hz
1/mm
1
f /Hz −1
1
f / Hz −1 × 10 −3
200
250
300
400
500
0.0050
0.0040
0.0033
0.0025
0.0020
5.0
4.0
3.3
2.5
2.0
406
322
264
194
153
Graph on graph page.
λ
(b)
=l+e
4
v
=l+e
4f
So

v
 Note λ = f 
1 4 4
= l+ e
f v
v
4
slope =
v
From graph, slope =
(4.90 − 1.75) × 10−3
(400 − 132)
= 1.18 × 10−5 Hz −1 mm −1
So
Unit I.indd 51
= 1.18 × 10−2 Hz −1 m −1
4
= 1.18 × 10−2
v
4
= 339 ms −1
v=
−2
1.18 × 10
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cape physics - Unit 1
I/f/HZ–1×10–3
5
4.9
4
each 2 mm ≡ .05 × 10–3
3.75
3
2
1.75
1
0
100
Unit I.indd 52
132
200
300
each 2 mm box ≡ 4 mm
400
1/mm
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53
(c) W
orking out the intercept using the point (300, 3.75 × 10–3) and the
slope as 1.18 × 10–5Hz–1 mm–1,
3.75 × 10–3 = 1.18 × 10–5 (300) + c
3.75 × 10–3 - 3.54 × 10–3 = c
c = 2.1 × 10–2Hz–1
4
4
But Intercept = l where is slope already worked out.
v
v
So
4
l = 2.1 × 10−4
v
2.1 × 10−4
l=
slope
2.1 × 10−4
=
1.18 × 10−5
= 17.8 mm
Unit I.indd 53
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cape physics - Unit 1
Question 3
θ/°C
(a) On graph page.
each 2 mm ≡ 0.8
80
71.2
(400, 71.2)
60
40
(115, 40)
20
0
100
200
300
400
450
t/s
115
each 2 mm box ≡ 5 s
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55
(b) The equation of the graph comes from
mc (θ – θ1) = Power × Time
where θ1 = 27.1°C
Pt
mc
Pt
θ=
+ 27.1
mc
P
Slope of graph ≡
mc
71.2 − 40
Slope =
400 − 115
= 0.11° C/s.
P
C=
So
m × 0.110
200
=
2 × 0.11
= 909 J kg −1 k −1
So
θ − 27.1 =
(Answer will depend on value of slope.)
(c) Heat capacity = Mass × Specific heat capacity
= 2 × 909
= 1820 J Kg–1
(d) Some suggestions as to how the experiment can be improved:
– Lag the block to reduce heat loss.
– Polish the block.
– Put the block to rest on an insulated surface.
Unit I.indd 55
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Question 4
(a) (i) F
or a body to undergo parabolic motion, the body must have a
constant ­velocity in one direction AND a constant acceleration
perpendicular to this velocity.
(ii) F
or a body to undergo circular motion, the body must
­experience an ­acceleration that is always perpendicular to
its velocity.
Note:
1.The magnitude of the acceleration will be constant and
always directed towards a fixed point (the centre of the
circle), but the acceleration will not be constant since the
direction will be changing continuously.
2.The velocity will not be constant but the speed will
be constant. Hence, there will be no change in kinetic
­energy.
(b) A
geostationary satellite is one that stays over the same point on the
earth all the time. It will have the same periodic time as the earth’s
rotation on its axis (24 hours).
For a satellite moving around the earth, the centripetal force is
provided by the gravitational force as given by Newton’s universal
gravitational law.
So
So
GME Msat Msat v 2
=
r2
r
GM
r = 2E
v
The Msat cancels out, so r is independent of Msat.
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57
(c) Force or tension in string provides the centripetal force.
At breaking point,
mv 2
r
2π r
where v =
T
F=
So
= 2π rf
(T is period)
1

 = f 
T
0.6 (2π rf )
60 =
r
60 = 0.6 × 4π 2rf 2
2
f 2 = 2.11 Hz2
f = 1.45 Hz
Alternatively:
mrω 2 = F
ω2 =
F
mr
ω = 83.3
now,
So
Unit I.indd 57
= 9.1 rad s −1
ω = 2π f
f=
ω
2π
= 1.45 Hz
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(d) (i) T
he centripetal force is the resultant towards the centre, of
weight and the ­tension
A
mg
T
T
B
mg
At the top, position A,
Centripetal force
So
At the bottom,
Centripetal force,
So
mv 2
= mg + T
r
mv 2
− mg
T=
r
mv 2
= T − mg
r
mv 2
+ mg
T=
r
Hence, the tension is greatest at the bottom and least at the top.
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59
(ii) The minimum tension = 2.1 N
This happens at the top.
So
So
mv 2
− mg = 2.1
r
0.6 × v 2
= (0.6 × 9.8) + 2.1
1.2
v 2 = 15.96 m2s −2
v = 4.0 ms −1
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Question 5
(a) (i) L
inear momentum is the product of the linear velocity of a body
and its mass. It is a vector quantity and is measured in N.S or
kg ms–1.
The law of conservation of linear momentum states that in
any collision or explosion, the total momentum before is equal
to ­total momentum after, ­provided that no external forces act
(i.e., it is a closed system).
(ii) (a) A
n inelastic collision is one in which momentum is
­conserved but kinetic energy is not conserved. A perfectly
elastic collision, on the other hand, is one in which both
momentum and kinetic energy are conserved.
(b) T
he laws of conservation apply in both cases, but for an
inelastic ­collision some kinetic energy is converted to other
forms of energy, mainly heat.
(iii) I mpulse of a force is the product of the force and the time
­duration through which the force acts.
Impulse = F × t
From Newton’s second law (F = ma),
mv − mu
t
Ft = mv − mu
F=
So impulse = change in momentum
(b) (i) For 1st collision:
Momentum before = (1.60 × 0.70) + (0.80 × 0.10)
= 1.20 N.S
Momentum after = (1.60 × 0.30) + (0.80 × 0.89)
= 1.19 N.S
The figures support the law of conservation of momentum.
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For 2nd collision:
61
Momentum before = (1.60 × 0.60) + (0.80 × 0.10)
= 1.04 N.S
Momentum after = (1.60 × 0.37) + (0.80 × 0.57)
= 1.05
Again the figures support the law.
(ii) For 1st collision:
1
Using KE = mv 2
2
Kinetic energy before = KE for 1.60 kg mass + KE for 0.80 kg mass
= (0.392 + 0.004)J
= 0.396 J
KE after
= (0.072 + 0.317)
= 0.389 J
This energy is reasonably close to say that KE is conserved.
For 2nd collision:
KE before = (0.288 + 0.004)
= 0.292 J
KE after = (0.110 + 0.130)
= 0.240 J
Hence KE is not conserved.
So first collision is elastic and second collision is inelastic.
(iii) B
eyond the immediate impact, external forces will act and so
slow down the trucks.
(iv) (a) The total momentum after the collision will be zero.
(b) T
otal momentum before collision is zero since the
­magnitude of the ­momentum of each track is the same
but moving in opposite directions.
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Question 6
(a) (i) R
efraction of sound waves is the bending (changing direction) of
the waves as they pass from one medium to another. This is as a
consequence of change of speed.
(ii)
Warmer air
Cool air
Cool air
Warmer
air
At night
Day time
(iii) T
he speed of sound in warm air is greater than the speed of
sound in cold air. Refraction of the sound occurs at night as
shown in the diagram above. At night, the air is cooler nearer
the ground than it is higher up.
During the day the air is warmer nearer the earth, so refraction
recurs as in the “day time” diagram above.
(b) T
wo consecutive antinodes are separated by a distance of
(1.0 – 0.6) m. But the distance between two successive antinodes
is half of a wavelength.
So
λ
2
= 0.4 m
λ = 0.8 m
v= fλ
= 440 × 0.8
= 352 ms −1
(c) (i) A
t the perpendicular bisector of the line between the two
­speakers, the waves would have travelled the same distance.
Hence the path difference is zero. Since the waves started 180°
out of phase, they will meet 180° out of phase, so cancellation
takes place. Hence the amplitude will be zero.
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63
(ii) First working out the wavelength:
v
330
=
f 4400
= 0.075 m
λ=
For the next “zero amplitude” position,
∆x =
λD
d
where D = 30 m,
λ = 0.075 m,
d = 0.50 m.
For the next loud sound, half this distance ∆x needs to be “travelled”.
So
Unit I.indd 63
∆x λ D
=
2 2d
0.075 × 30
=
2 × 0.5
= 2.25 m
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Question 7
(a) (i) A
motion is said to be simple harmonic if the acceleration is
proportional to the distance moved from a fixed point AND the
acceleration is always directed towards THAT fixed point.
(ii)
a = –ω2x
q
l
T
P
x
mg Sin q
mg
mg Sin q
Resultant force pulling the mass towards point P is mg Sin θ.
So
or
ma = –mg Sin θ.
a = –g Sin θ
If θ is small, then Sin θ ≈ θ.
θ=
x
l
g
x
l
This resembles a = −ω 2 x
So
where
a=−
ω2 =
T=
So
Unit I.indd 64
g
,
l
2π
⇒ω =
g
l
ω
T = 2π
l
g
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(b) (i)
T = 2π
= 2π
65
l
g
2.0
9.8
= 2.8s
(ii) I f the support accelerates, then the tension T in the string will
change, but the component mg Sin θ will not change. Hence the
period will not change.
(c) (i) x = 2 × 10–3 Sin 3πt
Compare this with x = x0 Sin ωt.
ω ≡ 3π rad/s
= 9.4 rad/s
(ii) T =
2π
ω
2π
=
3π
=
2
s
3
(iii) (a) At t = 0, Sin 3π (1) = 0
So x = 0
(b) v = + ωx0 Cos ωt
= 6π × 10–3 Cos 3π(1)
= –1.9 × 10–2 ms–1
(c) If x = 0, then a = 0.
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Question 8
(a) (i) C
ONDUCTION: When the colder molecules of the surroundings
come in contact with the hot body, there is a transfer of energy
from the higher kinetic energy molecules (hotter) to the lower
kinetic energy molecules.
CONVECTION: Transfer of energy takes place by the mass
­movement of ­hotter molecules.
RADIATION: Energy transfer takes place in the form of
­electromagnetic waves.
(ii) S pecific heat capacity of a substance is the amount of heat
needed to raise the temperature of one kilogram of a substance
by one kelvin (or one degree celsius).
Specific latent heat of fusion of a material is the amount of heat
needed to change one kilogram of the material from solid to
liquid without a change in temperature.
(b) T
he rate at which heat is received by the beaker and contents is given
by Stefan’s equation,
P = σAT4
= 5.67 × 10–3 × 1.0 + 10–4 × (1773)4
= 56 W
Energy will be needed to do the following:
1.
To heat the pyrex from 0°C to 20°C.
H1 = mc∆θ
= 20 × 10–3 × 840 × 20
2.
= 336 J
To melt 30 g of ice,
Hr = ML
= 30 × 10–3 × 3.34 × 105
= 10020 J
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3.
67
To raise the temperature of 80 g of water from 0°C to 20°C.
H3 = mc∆θ
= 80 × 10–3 × 4200 × 20
= 6720 J
So total energy required = 10020 + 6720 + 336
= 17076 J
17076
P
17076
=
56
= 305 s
So time =
(c) (i) Let the mass of the material be m kg,
So
P × t = mc∆θ
mc∆θ
P=
t
430 × 85
mW
=
50
= 731 m W
It takes 15 s to melt.
So heat supplied = 731 m × 15
So mLf = 731 m × 15
Lf = 731 × 15
(ii)
= 10960 J kg–1
P × t = mc ∆θ
731m × 55 = m × c × 25
731 × 55
c=
25
= 1610 J kg −1 k −1
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cape physics - Unit 1
Question 9
(a) (i) ∆u = ∆Q + ∆w
∆u = Change in internal energy.
∆Q = Heat added to the system.
∆w =Work done on the system.
(ii) T
he “mole” is that amount of a substance that contains the
­Avogadro’s number of molecules.
(iii) A
t constant pressure, heat is supplied to increase the internal
energy as well as do work on the surroundings whereas at
constant volume, energy is needed only to increase the internal
energy. Hence cp > cv.
(b) (i)
In each case, there is the same internal energy change.
P
4Po
Po
Vo
Unit I.indd 68
3Vo
V
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69
(ii) Using PV = nRT,
P0 V0
nR
P V P (3V )
T1 = 1 1 = 0 0 = 3T0
nR
nR
P V ( 4P0 )(3V0 )
T2 = 2 2 =
= 12T0
nR
nR
∆u = nCv ∆T
T0 =
= nCv (11T0 )
 3R   11P0V0 
= n  
 2   nR 
=
33
PV
2 0 0
(iii) (a) ∆w = p∆v
= p0 (3v0 − v0 )
= 2p0v0
= 2 × 3.039 × 105 × 4 × 10−3
= 2430J
(b) ∆Q = ∆u + ∆w
33
= (1215.6) + 2430
2
∆Q = 22500 J
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(2005) PAPER 1
Question 1
(a) (i) N
ewton’s first law states that a body will continue in its state
of rest or u
­ niform motion unless acted upon by an external
­resultant force.
(ii) N
ewton’s second law states that the rate of change of
­momentum of a body is directly proportional to the resultant
force acting upon it and is in the direction of the force.
(iii) N
ewton’s third law states that if body A exerts a force on body B,
then body B will exert an equal but opposite force on body A.
(b) (i) Horizontal component of 100 N force = 100 Cos 34
= 82.9 N
(ii) Horizontal component of 70 N force = 70 Cos 53
= 42.1 N
(iii) Resultant force on box = (82.9 + 42.1) − 30
= 95 N
F
Acceleration =
m
95
Acceleration =
80
= 1.2 ms −2
(iv) T
here is no resultant force in the direction perpendicular
to OX. That is, 100 Sin 34 is equal to 70 Sin 53 and opposite
in ­direction.
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71
Question 2
(a) S olar — In the Caribbean, there is an abundance of sun ­throughout
the year. This is a viable alternative source. It is also very easy
to set up in remote areas. It is useful for providing hot water for
­households.
Biomass — This will sustain the old sugar industry. Alcohol can also
be extracted from the Bagasse.
Wind — Abundance of wind in the Caribbean (N. E. trades).
Hydroelectric — for countries with water falls.
(b) – Use outdoor drying instead of electric dryers.
– Use natural lighting as far as possible.
– Air condition units can be designed and placed to reduce the load.
(c) Power on 1 m2 = 15% of 300 W/m2
= 45 W/m2
25 × 106
Area required =
45
= 5.6 × 105 m2
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Question 3
(a)
d/m
0
1
2
3
4
t/s
v/m/s
0
1
2
1
2
3
4
t/s
a/m/s2
0
3
4
t/s
acceleration is negative
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73
(b) At highest point,
v = u + at
o = u – gt
u = 9.8 × 2
= 19.6 ms–1
Hence the velocity with which the ball reaches back to the thrower
will be 19.6 ms–1 in the opposite direction (assuming no energy is lost
due to air resistance).
(c) v = u + at
= 19.6 – 9.8(3)
= –9.8 ms–1
Hence, after 3 s the ball will be travelling downwards with a “Speed”
of 9.8 ms–1.
(d) O
n the way up, the force of gravity and air resistance will be ­acting,
so the ­deceleration would increase. On the way down, air ­resistance
will be acting ­upwards, so acceleration down will be less than
­gravitational acceleration (g).
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Question 4
(a) (i)
F 0.4 × 9.81
=
0.125
x
= 31.4 Nm −1
K=
(ii) T = 2π
m
k
0.4
31.4
= 0.71 s
= 2π
Alternatively:
T = 2π
(b)
x
g
12.5 × 10−2
= 2π
9.81
= 0.71 s
x/cm
K/N/m
T/s
6.25
62.8
0.50
Parallel
Series
T11 = 2π
m
k
25
15.7
1.0
0.4
62.8
= 0.50 s
= 2π
Ts = 2π
m
k
0.4
15.7
= 1.0 s
= 2π
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75
Question 5
(a) (i) A
long the radius, the ray strikes the surface at 90°. So angle of
incidence is zero.
(ii) C
ritical angle is the value of θ1 (from the diagram), such that θ2
is 90°. If θ1 is larger than this, then total internal reflection will
take place.
(b) (i) When θ2 = 90°
θ1 ≈ 42°
(ii) n =
1
1
=
Sin c Sin θ1
1
Sin42
= 1.49
=
(c) T
o take angle of incidence θ1 as 45° would not make sense in the
­context of the question and the graph given.
So if 45° is taken as θ2, the angle in air,
Then n =
=
(d)
Sin θ2
Sin θ1
Sin45
= 1.48
Sin28.5
air
60°
Unit I.indd 75
60°
perspex
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Question 6
(a) S omeone who cannot see near objects clearly and can see far objects
better is suffering from long sight or hypermetropia.
(b)
(c) Using the “real is positive” convention,
u = 25 cm, v = –80 cm
(d)
1 1 1
= +
f u v
1
1
=
−
25 80
= 0.0275
f = 36.4 cm
1
Power =
0.364
= +2.75 D
80 cm
F
36.4
F
Image
He can focus onto the retina images of objects 80 cm or more away
without the help of glasses as seen from the diagram.
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77
Question 7
(a) (i)
F/N
Elastic limit exceeded
80
60
40
Area = work done
1
2
3
4
5
x/mm
(ii) On diagram
(iii) Work done = Area under graph
1
= F×x
2
1
= × 60 × 1.5 × 10−3
2
= 4.5 × 10−2 J
F
(iv) Y = A
x
l
2
πd2 π
= 0.43 × 10−3
A=
4
4
= 1.45 × 10−7 m2
60 × 2
Y=
1.45 × 10−7 × 1.5 × 10−3
= 5.5 × 1011 N m −2
(
Unit I.indd 77
)
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(b)
F
F
x
Rubber
Unit I.indd 78
x
Glass
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79
Question 8
(a) Radiation takes place in the form of electromagnetic waves.
The receiving object absorbs this energy and its molecules become
more agitated, thus increasing the temperature.
(b) R = σAT4
where R is rate of radiation in watts, σ is Stefan’s constant, A is
­surface area and T is temperature in kelvin.
(c) (i) Rate of absorption of energy by sphere = σAT4
A = 4πr2
T = 273 + 120
= 393 K
= 4π(15 × 10–2)2
= 2.83 × 10–1m2
R = 5.67 × 10–8 × 2.83 × 10–1 × 3934
= 3.83 × 102 W
(ii) Net rate of absorption = 2.83 × 10–1 × 5.67 × 10–8 (3934 – 3034)
= 16.046 × 10–9 × 1.54 × 1010
= 2.47 × 102 W
(iii) F
or the temperature to remain constant, the rate of heat
­radiated must be equal to rate of heat absorbed. This will
­happen when the temperature of the sphere reaches l20°C.
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Question 9
(a) (i)
Section
AB
BC
CD
DA
Description
Pressure increases, volume
­constant
Work done/J
0
Expansion at constant pressure
–16 × 105
Compression at constant
­pressure
4 × 105 J
Reduction in pressure at constant
volume
0
(ii) AB and CD, no work is done since volume is constant.
(Recall ∆W = P∆V.)
BC: W = – 4 × 105 × 4 = –16 × 105 J
(iii)
DA: W = +1 × 105 × 4 = 4 × 105 J
B
C
Area = Work done
A
(iv) ∆u = 0
D
Q = –w = 3 × 105 × 4
Unit I.indd 80
= 12 × 105 J
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(2005) PAPER 2
Question 1
(a)
Radius
r/mm
Time/s
Vel.
V/cms–1
1g
(V/cms–1)
1g
(r/mm)
1.00
44.8
1.79
0.252
0.000
7.08
0.850
0.305
16.00
1.204
1.49
20.1
3.98
7.2
11.11
2.02
11.3
2.99
5.0
2.51
Graph on graph page.
0.600
1.045
0.173
0.400
0.476
(b) log v = n log r + log K
(c) Slope of graph will be n.
1.2 − 0.3
0.475 − 0.03
= 2.02
Slope =
The most likely value for n is 2.
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cape physics - Unit 1
0.0
0.2
0.4
0.6
0.8
0.10
0.12
0.14
0.16
0.18
0.1
0.2
lg (r/mm)
0.3
0.4
0.5
82
lg (V/csm–1)
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Question 2
(a) On graph page.
11
10
9
8
Amplitude/cm
7
6
5
4
3
2
1
0
30
Unit I.indd 83
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
L/cm
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(b) T
hese points were taken so as to see where the peak would be. It is
always good practice to have more points near the turning point so
as to better locate the turning point exactly.
(c) T = 2π
L
g
0.40
9.81
= 1.27s
1
f=
T
1
=
1.27
= 0.79 Hz
= 2π
(d) T
he resonant frequency is equal to the natural frequency of the
heavy pendulum. That is, resonance occurs when the frequency of
the ­external driving force is equal to the natural frequency.
Hence, resonant frequency is 0.79 Hz.
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85
Question 3
(a)
Quantity
Mass, m
Initial temperature, θ1
Final temperature, θ2
P.D., V
Current, I
Time, t
Value
930 g
Uncertainty
±5g
28.0°C
± 0.5°C
9.8 V
± 0.1 V
51.8°C
4.3 A
500 s
± 0.5°C
± 0.05 A
Negligible
Note:Uncertainties should be taken as “half the smallest division” of
the ­instrument.
VIt
M ∆θ
9.8 × 4.3 × 500
=
0.93 × 23.8
= 952 Jkg −1 K −1
(b) S.H.C =
% error in S.H.C = % error in V + % error in I + % error in M + % in ∆θ
0.1
≡ 1.02%
% error in V =
9.8
0.05
≡ 1.16%
% error in I =
4.3
5
≡ 0.54%
% error in M =
930
(0.5 + 0.5) ≡ 4.20%
% error in ∆θ =
(51.8 − 28.0)
So total % error = 6.92%
(c) 1.
2.
Unit I.indd 85
Hence S.H.C = 952 ± 66 J Kg −1 K −1
Use lagging to insulate the cylinder.
Polish the metal to reduce heat loss by radiation.
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cape physics - Unit 1
Question 4
(a) (i) M
omentum of a body is defined as the product of the body’s
mass and its ­velocity. It is a vector quantity and is measured in
Kg ms–1 or N.S.
Note:Since only linear momentum is considered at this level,
then the ­velocity is the linear velocity.
(ii) I n an elastic collision, both momentum and kinetic energy are
conserved.
(iii) Momentum before = 2m(v) = 2mv
Momentum after = m(2v) = 2mv
So momentum is conserved.
1
(2m) v2
2
= mv 2
Kinetic energy before =
1
2
Kinetic energy after = m (2v )
2
= 2 mv 2
So kinetic energy is not conserved.
Hence, this outcome is not possible for an elastic collision.
(iv) Momentum before
Momentum after
= 2m(v) = 2mv
= 2m(v) = 2 mv
1
(2m) v2 = mv2
2
1
Kinetic energy after = (2m) v 2 = mv 2
2
Kinetic energy before =
( )
So both momentum and kinetic energy are conserved in this
case.
Unit I.indd 86
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87
(b) (i) Change in momentum = MV2 – MV1
= 1200 × (–1.5) – 1200(20)
= –1800 – 24000
= –25800 N.S
Note:Momentum is a vector quantity. The velocity is ­reversed
after the c­ ollision, so –1.5 ms–1.
(ii) Impulse is change in momentum.
(Ft = MV2 – MV1)
(iii)
So Impulse = 25800 N.S
F
Crash 2
t
(iv)
The area under the graph represents impulse or change in
­momentum.
So
Ft = 25 800
25 800
0.18
= 1.43 × 105 N
F=
(v) W
hen the car crumbles the collision will last longer. That is,
the car will take longer to come to rest. Hence the rate of
change of momentum will be ­reduced. Hence force will be less.
(Less chance of serious injury to ­passengers.)
Unit I.indd 87
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88
cape physics - Unit 1
Question 5
(a) (i) F
rom Newton’s second law, the force on the air is equal to the
rate of change of momentum.
From the third law, the plane pushes back the air, so the air
pushes the plane forward.
(ii) Force = rate of change of momentum
Change in momentum
Change in time
∆m
=
×v
∆t
∆m F
=
∆t v
1800
=
250
= 72 kgs −1
F=
So
(b) (i)
Lift
Resultant
Weight
Note: Lift force is always perpendicular to the wings.
Unit I.indd 88
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89
(ii) The resultant force is the centripetal force.
mv 2
r
mv 2
r=
F
F=
So
120 × 103
m/s
3600
= 33.3 m/s
v=
3000 × (33.3)
=
16000
= 208 m
2
(iii) Draw the vector diagram.
Fresultant
W
Lift
q
FResultant
16000
=
= 0.544
weight 3000 × 9.81
θ = 28.5°
tan θ =
Resultant force = Lift force × Sin θ
So lift force =
16000
Sin 28.5
= 3.35 × 104 N
Unit I.indd 89
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90
cape physics - Unit 1
(iv)
Push
The centripetal force on the passenger is provided by the
­horizontal ­component of the reaction force of the seat on the
passenger (i.e., towards the centre of the circle).
Unit I.indd 90
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91
Question 6
(a) D
iffraction takes place at each slit. Each colour will diffract
by a ­different amount according to the equation d Sin θ = n λ.
The ­diffracted waves will interfere, ­producing constructive
­interference in directions where the colours (wave length) meet
in phase. That is where the path difference is a whole number of
­wavelengths.
The zero order occurs where the path difference is zero. The first
order occurs where the path difference is one wavelength and the
second order occurs where the path difference is two wavelengths.
(Corresponding to n = 0, 1, 2 as in the formula.)
(b) (i)
Diffraction
grating
R2 Screen
2nd order
Y2
R1
Y1
1st order
Zero order
Symmetrical on both sides of
the zero order.
(ii) d Sin θ = nλ
nλ
d=
Sin θ
2 × 630 × 10−9
=
Sin43.9
= 1.82 × 10−6 m
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92
cape physics - Unit 1
1
1
=
d 1.82 × 10−6
= 5.5 × 105 lines per m
Number of lines per m =
So lines per mm = 5.5 × 105 × 10−3
= 550 lines per mm
1 × 630 × 10−9
(iii) Sin θr =
1.82 × 10−6
⇒ θr = 20.3°
1 × 570 × 10−9
Sin θ y =
1.82 × 10−6
⇒ θ y = 18.3°
(iv) For n = 3,
3 × 570 × 10−9
= 0.94
Sin θ y =
1.82 × 10−6
giving θ y = 70°; this is possible
Sin θr =
3 × 630 × 10−9
= 1.04
1.82 × 10−6
This is impossible since Sin θ must be less than or equal to 1.
Unit I.indd 92
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93
Question 7
(a) (i)
P
S2 Q
Y
q
q
aT
S1 R
Q
λN
D
From ∆ PTO, tan θ =
From ∆ QMR, sin θ =
y
D
λ
a
If θ is small, then sin θ ≈ tan θ.
So
y λ
=
D a
y=
λD
a
(ii)  T
he ∆ QRN is not truly a right-angled triangle if the path
­difference is to be λ.
 The formula holds only if θ is small.
Unit I.indd 93
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94
cape physics - Unit 1
(b) (i)
6m
P
0.75 m
2.5 m
X
Px2 = 62 + 3.252
Px = 6.82 m
Q
6m
1.75 m
X
Qx 2 = 62 + 1.752
Qx = 6.25 m
λ = Px − Qx = 0.57 m
(More significant figures will give λ as 0.574 m.)
(ii) y =
λD
a
ay
λ=
D
2.5 × 1.5
=
6
= 0.625 m
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95
(iii) T
here are no approximations in (b) (i). In (b) (ii) the assumption
that P and Q can be regarded as point sources is not valid.
(iv) f =
V
λ
330
0.574
= 575 Hz
=
(v) B
oth maxima will become minima since they will now be 180°
out of phase.
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96
cape physics - Unit 1
Question 8
(a) (i)
Q
∆θ
= −KA
t
∆x
Q
is rate of heat flow.
t
K is coefficient of thermal conductivity.
A is area of cross-section.
∆θ is temperature difference across a length ∆x.
(ii) F
or a good conductor, the length must be sufficient to give a
­measurable ­temperature difference and hence a measurable
temperature gradient, p
­ rovided the length is not too many
times greater than diameter (five to six times ­greater). Large
area will increase the heat flow but will make the ­difference in
­temperature small. Also the area must be such as to accomodate
the thermometers.
(iii) H
eat is lost through the sides. This heat loss can be reduced by
lagging.
(b) (i)
q /°C
30
2
0
Unit I.indd 96
1
2
3
x/cm
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97
(ii) The plywood having a conductivity of 0.24 Wm–1 k–1 and plastic
0.24
having a ­conductivity of 0.012 Wm–1 k–1 means that
cms
0.012
of plywood will be equivalent to 1 cm of foam.
So plywood equivalent would be 20 cm.
Alternately:
If all other terms are constant, then,
k1 k2
=
x1 x 2
x1 =
k1
× x2
k2
0.24
×1
0.012
= 20 cm
=
(iii) T
he box will be equivalent to 22 cm of plywood (1 cm each for
the sides plus 20 cm equivalent for the foam).
Q 0.24(0.6 × 0.4) × 28
=
22 × 10−2
t
= 7.3 W
A
Plywood
30°C
Unit I.indd 97
B
Plastic
q1
Plywood
q2
2°C
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98
cape physics - Unit 1
At boundary A, the drop in temperature is given by
0.24(0.6 × 0.4)∆θ
1 × 10−2
∆θ = 1.27°C
7.3 =
This will be the same temperature difference at junction B.
So temperature at A, θ1 = (30 − 1.27) = 28.7°C
and temperature at Junction B , θ2 = (2 + 1.27)
= 3.3°C
(Temperatures can be given to one decimal place.)
Unit I.indd 98
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99
Question 9
(a) (i) A
ll molecules in gas will not have the same speed because they
are constantly bombarding the walls of the container and each
other and so continuously exchanging momentum. Hence speed
will keep changing.
(ii) r.m.s. is root mean square speed.
Vrms
V12 + V22 + V32 VN2
=
N
(iii)
l
X
u
A
Consider a molecule of mass m moving in the x-direction
­towards face X of area A.
Momentum before collision = mu
Momentum after collision = –mu (if collision is elastic)
|change in momentum| = 2 mu
Time between collisions =
2l
u
2 mu

2
u
mu2
=

So rate of change of momentum =
If there are N molecules then Force on face X due to these
­molecules = rate of change of momentum
Unit I.indd 99
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100 c a p e p h y s i c s - U n i t 1
=
m 2 2 2
u1 + u2 + u3 +uN2 (1)
l
(
)
An average value of u12 + u22 uN2 can be found.
u12 + u22 uN2
N
2
2
u1 + u2 uN2 = Nu 2
u2 =
as
So
Sub in eq. 1
F=
Nm 2
u

Now consider the three directions (since the motion is random).
u 2 = v 2 = w2
and if c 2 is the mean square speed,
then
So
u 2 + v 2 + w2 = c 2
1
u2 = c 2
3
Hence force on area A will be
1 Nm 2
c
3 
F 1 Nm 2
P= =
c
A 3 A
A = V
1 Nm 2
So
P=
c
3 V
1
⇒ PV = Nmc 2
3
F=
Unit I.indd 100
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101
(b) (i) PV = nRT
PV 1.6 × 105 × 0.14
=
8.31 × 400
RT
= 6.7 moles
mass = n × molecular mass
= 6.7 × 4 = 27 g approximately
n=
1
(ii) PV = Nmc 2 where Nm = 27g
3
3
PV
c2 =
Nm
3 × 1.6 × 105 × 0.14
=
27 × 10−3
c 2 = 2.5 × 106
c = 1600 ms −1
(iii) Same temperature means same kinetic energy
1
1
2
2
m1 ( c1 ) = m2 ( c2 )
2
2
4 × 2.5 × 106
c22 =
32
= 313 × 103
The r.m.s. speed for oxygen will be 560 ms–1.
Unit I.indd 101
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(2006) PAPER 1
Question 1
(a) A
vector quantity has both magnitude and direction; e.g., velocity,
­acceleration. A scalar quantity has magnitude only; e.g., mass, time.
(b) (i)
Vector
x component
y component
P
Q
2.05
5.64
P–Q
–7.14
9.19
–7.71
13.35
x component of P = 6 Sin 20 = 2.05
y component of P = 6 Cos 20 = 5.64
x component of Q = 12 Cos 40 = 9.19
y component of Q = –12 Sin 40 = –7.71
x component of P – Q = (2.05 – 9.19)
(ii)
y component of P – Q = 5.64 – (–7.71)
y
13.35
P–Q
15.1 units
61.9˚
–7.14
Unit I.indd 102
X
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103
Question 2
(a)
x
v
u
t
t
1
Equation: x = ut + at 2 V = u + at
2
Gradient represents: velocity,
acceleration
(b) (i) For vertical motion:
y=
1 2
gt
2
1 2

using s = ut + at (1)
2

For horizontal motion:
x = ut
= 5t
x
t=
5 So
Sub 2 in 1
1
 x
y = (10)  
 5
2
=
Unit I.indd 103
x
5
(2)
2
2
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104 c a p e p h y s i c s - U n i t 1
(ii) When x = 10,
y=
100
= 20 m
5
Alternatively:
10
=2s
5
1
h = gt 2
2
1
= × 10 × 4
2
= 20 m
t=
Unit I.indd 104
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105
Question 3
(a) (i) N
ewton’s law of gravitation states that for any two bodies
in space, there is a force of attraction between them which
is ­directly proportional to the product of their masses and
­inversely proportional to the square of their distance apart.
F =G
M1 M2
r2
(ii) As the astronaut goes around the earth, his weight is given by
mV 2
=
− mg′ where m is mass and g′ is acceleration due to
W
r
gravity at that height.
If
V2
= g′ , then he feels “weightlessness”.
r
That is, if the acceleration of both bodies is the same, then there
is no resultant reaction force between them.
Note: “Weightlessness” is a sense of “feeling”.
(b) (i) A
geostationary orbit is one in which the satellite has the same
periodic time as the earth’s rotation on its axis (i.e., 24 hours).
Hence, the satellite appears over the same place all the time.
(ii) Angular velocity of a geostationary orbit is given by:
ω=
Unit I.indd 105
2π
24
=
= 7.27 × 10−5 rads −1
T 24 × 60 × 60
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106 c a p e p h y s i c s - U n i t 1
(iii) T
he centripetal force is provided by the gravitational force of
­attraction between the satellite and the earth.
So
mRω 2 =
R3 =
=
So
Unit I.indd 106
GME m
R2
GME
ω2
6.67 × 10−11 × 5.98 × 102 x
(7.27 × 10 )
= 7.55 × 1022
−5 2
R = 4.22 × 107 m
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107
Question 4
(a) (i)
So
∆x = 0.35 m
d = 0.08 m
D = SN ∴ SN 2 = 12 − 0.352
SN = 0.93675
∆x × d
λ=
D
0.35 × 0.08
=
0.93675
= 3.0 × 10−2 m
Note: If D in the formula is taken as 1.0 m, the answer for
λ may be the same, but will not be the correct method from
the ­diagram given.
(ii) At P, the waves meet out of phase so there is destructive
1

­interference, i.e., the path difference is  n +  λ where n = 0

2
in that case.
(iii) P will occur at a vertical height of 0.175 m above O.
So
⇒
0.175
0.93675
θ = 10.6°
tan θ =
(b) Z
eroth maximum occurs at O because the path difference is 0
(i.e., the paths are equal). With the glass in place, the path of one will
now increase (if t is the ­thickness of the glass and n is the refractive
index then path through glass is ­given by nt). So the equal paths will
now occur at a different spot.
Unit I.indd 107
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108 c a p e p h y s i c s - U n i t 1
Question 5
(a) Refractive index, n =
(b) (i)
Velocity in faster medium
Velocity in slower medium
l2
Medium 2
Index n2
q2
Medium 1
Index n1
q1
l1
l2 > l1
(ii) The waves will travel faster in medium 2.
(iii) n1 Sinθ1 = n2 Sinθ2
or
n1 Sin θ2
=
n2 Sin θ1
(c) (i) A
t each point of contact between the core and the cladding, the
angle of incidence in the core is greater than the critical angle
between the two media. So total internal reflection will take
place each time. Hence the light will continue through the core
until it emerges.
1.49
1.45
= 1.02758
1 1.45
∴Sin c = =
= 0.9732
n 1.49
c = 76.7°
(ii) Relative refractive index, n =
(iii) I f the cable is bent too much, then the angle of incidence on the
cladding from the core will be less than the critical angle and so
total internal reflection will not take place (i.e., light will escape).
Unit I.indd 108
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109
Question 6
(a) Wave length:
4 half wave lengths occupy 3.0 m.
3.0
= 1.5 m.
∴ 1 wave length will occupy
2
Velocity:
V = fλ
= 60 × 1.5
= 90 ms −1
(b) (i)
y/cm
4
8.3
16.7
t/ms
–4
1
f
1
=
× 103 ms
60
T=
(ii) ω = 2π f
= 60 × 2π
= 120π rads −1
= 377 rads −1
(Answer left in terms of π is accepted)
Unit I.indd 109
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110 c a p e p h y s i c s - U n i t 1
(iii) y/cm = 4 Sin 120 πt
(iv) Maximum acceleration is given by:
a = rω 2
= 0.04 × (120 π )
2
= 5.68 × 103 ms −2
Unit I.indd 110
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111
Question 7
(a)
Load/m
50
9.3
(b) K =
Extension/cm
F
x
50
9.3 × 10−2
= 538 Nm −1
=
(c)
When F = 30,
30
x=
= 0.0558 m
K
1
So energy stored = F × x
2
1
= × 30 × 0.0558
2
= 0.837 J
(d) Half of the “loss” in gravitational potential energy is dissipated as
1
1
heat in stretching the spring (i.e., mgx is stored and mgx is
2
2
­dissipated as heat). Also the load is applied steadily, so work done
is (average force) multiplied by distance.
1
So work done = F × x and not Fx .
2
Unit I.indd 111
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112 c a p e p h y s i c s - U n i t 1
Question 8
(a)
Temp
26
0˚C
Time
–16
(b) (i) Energy extracted = McΔθ
Energy extracted = Power × Time
Pt = Mc ∆θ
Mc ∆θ
t=
P
0.2 × 4200 × 26
=
80
= 273 s
So
(ii) Pt = mL
mL 0.2 × 3.3 × 105
=
80
P
2
= 8.35 × 10 s
t=
(c) D
uring the freezing process only the potential energy of the
­molecule changes (decreases). Kinetic energy does not change
since ­temperature remains constant.
Unit I.indd 112
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113
Question 9
(a) (i) C
opper is a good conductor of heat. When the heat is absorbed
by the black surface, it needs to be conducted to the water in the
pipes. Copper is best suited here.
(ii) B
lack is the best absorber of heat. The heat needs to be
­absorbed and then ­conducted to the water.
(iii) T
he glass cover makes use of the “green house effect”.
The ­shorter ­wavelength of infrared can come in, and the
­re-radiated longer wavelengths cannot leave, and so are
trapped.
(iv) T
he Styrofoam here reduces heat loss to the surroundings
through the base. Styrofoam is a poor conductor of heat.
(b) T
he water enters the tubes set on the base plate with the pressure
of the incoming mains. The base plate is placed at an angle to allow
gravity to enhance flow. The storage tank is placed slightly lower
than the outlet from the heater but higher than the taps to be used.
This is called “natural convection”.
60
Wm −2 is
(c) 6
0% efficiency in conversion means that 800 ×
100
­transferred to the ­water.
So
Unit I.indd 113
60
× 800 = 480 Wm −2
100
480 × A = 900
900
A=
480
= 1.88 m2
2/2/2016 5:21:20 PM
(2006) PAPER 2
Question 1
Unit I.indd 114
5
10
15
20
25
30
35
F
kN
40
50
100
150
200
250
300
350
400
450
500
550 t
ms
(a) See graph page.
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115
(b) (i) T
he area under the force-vs-time graph stands for impulse or
change in momentum.
(ii) One 1-cm square = 2.5 KN × 25 ms
So
≡ 2.5 × 103 × 25 × 10−3
≡ 62.5 N.S
186 cm2 ≡ 62.5 × 186
= 1.16 × 104 N.S
(c) Change in momentum = 1.1625 × 104 N.S
So
where v = 0,
mv − mu = 1.1625 × 104 NS
1.1625 × 104
1.2 × 103
= 9.69 ms −1
u=
(d) The maximum force experienced = 37.5 KN
F
So maximum acceleration =
m
37.5 × 103
=
1.2 × 103
= 31.3 ms −2
Unit I.indd 115
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116 c a p e p h y s i c s - U n i t 1
Question 2
(a) U
sing a travelling microscope, measure the height of the object
(the size of the slit, say). Using the travelling microscope, measure
the height of the image on the screen.
m=
(b) (i)
height of image
height of object
m=
=
m=
x +d
−1
f

x d
=  − 1

f f
d

1
( x ) +  − 1
f

f
This resembles y = mx + c,
Where
(ii)
Unit I.indd 116
1
is slope corresponding to m in the equation of y = mx + c.
f
When m = 0,
x +d
−1 = 0
f
x +d
So
=1
f
So
x +d = f
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(c) (i) The magnification, m, of the image, is given by m =
Below Figure is a plot of m versus x.
117
( x + d ) − 1.
f
1.4
Magnification m
1.2
1.0
0.80
0.60
0.40
0.20
0.0
0
20
30
40
x/cm
60
80
90
100
1.2 − 0
90 − 30
= 0.02 cm −1
(ii) (a) Gradient =
(b) f =
1
60
=
gradient 1.2
= 50 cm
(c) When m = 0, x = 30 cm
i.e., x + d = f
d = 50 − 30
= 20 cm
Unit I.indd 117
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118 c a p e p h y s i c s - U n i t 1
Question 3
(a) (i)
A
V
(ii) T
he apparatus is set up as shown with the thermistor in a water
bath. A ­thermometer is used to get the temperature of the water
bath. Vary the ­current via the rheostat and read I and V.
Precaution(s):– Heat in water bath for even heating.
– stirr.
(b) (i)
θ/°C
p.d./v
0.0
2.04
10.0
20.0
30.0
40.0
50.0
60.0
70.0
80.0
Unit I.indd 118
2.03
2.02
1.99
1.97
1.95
1.95
1.95
2.02
I/mA
R/Ω
2.74
745
3.50
5.05
6.75
8.90
11.9
15.9
19.9
28.0
580
400
295
221
164
123
0.98
72
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119
(ii) See graph page.
R/Ω
800
700
600
500
400
320
300
200
100
0
10
20
30
27°C
40
50
60
70
80
q./°C
(c) When R = 320 Ω ,
θ = 27°C.
Unit I.indd 119
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120 c a p e p h y s i c s - U n i t 1
Question 4
(a) (i) F
irst Law: A body will continue in its state of rest or uniform
motion unless acted upon by an external resultant force.
Second Law: The rate of change of momentum of a body is
directly ­proportional to the external resultant force and takes
place in the direction of the force: F = ma.
Third Law: If a body A exerts a force on body B, then body B will
exert an equal and opposite force on body A.
(ii) A
s the ball falls and accelerates downwards, the drag forces
­increases with velocity (F = 6πrηV – Stokes law). A stage is
reached where the net resistive forces (drag + upthrust) is equal
to the weight of the sphere. At this point no net force acts on
the body, so no acceleration. Hence constant velocity from here
onwards.
a/m/s 2
Shape – I
Intercepts – I
9.81
Terminal velocity is
reached at this
time
t /s
(b) (i)
upthrust
direction
of motion
r = 2.5 m
Fdrag
mg
Unit I.indd 120
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( 2 0 0 6 ) PAPE R 2
121
(ii) (a) Upthrust = Weight of air displaced
4
= πr3 × ρ × g
3
4
= × π × (2.5)3 × 1.29 × 9.81
3
= 828 N
(b)
Terminal velocity occurs when drag force + weight equals
upthrust.
So
1 2 2
π r ρv + mg = 828
2
1
2
π (2.5) × 1.29 × V 2 + (15 + 9.81) = 828
2
12.66 V 2 = 681
681
= 53.78
V2 =
12.66
V = 7.33 ms −1
(c)
Assume VT is reached quickly, then
t=
Unit I.indd 121
S 10 × 103
=
= 22.7 mins
V
7.33
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122 c a p e p h y s i c s - U n i t 1
Question 5
(a)
q
VB
r
r
S
A
VA
Let the body move from A to B in time t.
ω=
So
θ
t
and
S = rθ
S
θ
Linear velocity V = = r = rω
t
t
(1)
–VA
q
VB
∆V
V = VA = VB
∆V = VB − VA
∆V
θ
= V = Vω
t
t
V
ω=
r
2
V  V
a=V   =
r r
a=
From (1)
So
Unit I.indd 122
(2)
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( 2 0 0 6 ) PAPE R 2
123
(b) (i)
A
mg
T1
l
T2
B
mg
At position A, the forces acting (towards the centre of the circle)
will be the weight mg and the tension in the string T1

mv 2 
so
centripetal
force
=
+
=
.
mg
T
1

r 
At position B, the force acting will be mg downwards and
­tension T2 towards the centre

mv 2 
 So centripetal force = T − mg = r  .
(ii) The tension is max. at position B when tension is maximum
So
So
So
Unit I.indd 123
Tmax
mv 2
+ mg
T=
1
mv 2
− mg =
= mrω 2
1
ω=
Tmax − mg
ml
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124 c a p e p h y s i c s - U n i t 1
(iii) Tmax = mrω 2 + mg
(
)
= 0.5 1 × 42 + 9.81
(c) (i)
= 12.9 N
B
Parabolic path
h
(ii) (a) The horizontal velocity at this point is given by
v = rω = 1.0 × 4 = 4 ms–1.
So horizontal distance travelled = 4 × 0.5 = 2 m
(b) Vertical velocity is given by
v = u + gt
= 0 + 9.81 × 0.5
= 4.9 ms −1
Unit I.indd 124
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( 2 0 0 6 ) PAPE R 2
125
Question 6
(a) (i) A
ccommodation is the ability of the eye to change the thickness
of the lens, hence changing the focal length, so as to focus far
and near objects.
(ii) A
stigmatism is the uneven curvature of the cornea in the
­horizontal and ­vertical planes.
(iii) C
ataract is the “clouding” of the eye lens. The protein
­(protoplasm) that makes up the lens in the eye starts to “clump”
with age and so clouds the lens, ­leading to clouded vision.
(b) (i)
The woman suffers from “long sight”, i.e., she cannot see near
objects clearly.
≈ 25 cm
On putting the lens, the following happens.
≈ 25 cm
Unit I.indd 125
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126 c a p e p h y s i c s - U n i t 1
(ii)
1
1
=
= 0.4 m = 40 cm
D 2.5
1 1 1
= +
f u v
1 1 1 1
1
1000
= − =
−
= 67 cm.
so v =
v f u 40 25
15
f=
So the nearest distance she can read without her glasses is 67 cm.
(c) (i) 10–12 Wm–2
(ii) T
he scale is logarithmic because a change in dB level of 10 gives
a change in intensity by a factor of 10. A change by 20 dB
gives a change in intensity ­factor of 102 etc. So the response is
­logarithmic; i.e., a change in dB level of 10, 20, 30 gives a change
in intensity by 10, 100, 1000, etc.
(iii) T
he Bell is the standard unit. Since this is a large unit, the
­submultiple dB is usually used. So that we have
dB = 10 log
instead of B = log
I
I0
I
I0
Logarithmic, and matches human judgment of loudness
(i.e., every increase of 1 dB corresponds to same increase
in loudness).
(iv) dB = 10log
I
I0
6 × 10−2
10−12
= 10log 6 × 1010
= 10log
{
(
)
= 10 log6 + log1010
= 10{0.778 + 10}
}
= 108 dB
Unit I.indd 126
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( 2 0 0 6 ) PAPE R 2
(v) 20dB = 10log
127
I
I0
I
10−12
= log I + 12
log I = −10
2 = log
I = 10−10 Wm2
From the graph the person can hear between 100 Hz and
10 KHz.
Unit I.indd 127
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128 c a p e p h y s i c s - U n i t 1
Question 7
(a) (i) I f no damping, then total energy is constant. There will be
a ­continuous ­interchange between KE and PE.
TE
PE
KE
–r
+r
x
(ii) I f an oscillation is damped, then energy leaves the system during
each cycle in order to overcome the resistive forces. Hence, the
amplitude will decrease (the amplitude is a measure of the total
energy of the system, I = KA2).
(b) (i)
x0 = 0.05 m
3500
Hz = 58.3 Hz
60
ω = 2π f and a = −ω 2 x
f=
So
(ii)
a = (2π f ) (0.05) = 6.72 × 103 ms −2
2
V = ω x02 − x 2
When x = 0, V = x0ω
= 0.05 × 58.3 × 2π
= 2.92 × 2π ms −1
= 18.3 ms −1
Unit I.indd 128
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( 2 0 0 6 ) PAPE R 2
129
1
(iii) KE = mv 2
2
1
2
= × 0.45 × (18.3)
2
= 75.7 J ( using v as 2.92 × 2π )
(iv) The gain in KE = 76 J
The time taken to do this is
1
1
=
s
f 58.3
= 0.01715 s
T=
1
of the periodic time.
4
T
= 4.288 × 10−3 s
4
75.7
So power =
W
4.288 × 10−3
= 17600 W
= 17.6 kW
So
Unit I.indd 129
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130 c a p e p h y s i c s - U n i t 1
Question 8
(a) (i) T
hermal conduction in a metal can take place by free electron
movement and by lattice vibrations.
(ii) I n an insulator, there are no free mobile electrons, so thermal
conduction can only take place by lattice vibrations (electrons
firmly bound to atoms).
(iii) Conditions:
– Steady state must be reached.
– The material must be insulated.
– Area of cross-section must be constant.
– Linear flow.
(iv) T
he S.L.H of vaporization of a substance is the amount of heat
needed to change 1 kg of a substance at its boiling point to the
vapour state without a change in temperature. Units: J/kg.
(b) Heat supplied in 3 mins = 0.45 × Lv
= 0.45 × 2.26 × 106
= 1.02 × 106 J
1.02 × 106
180
= 5.67 × 103 J/s
So Heat/s =
So
5.67 × 103 = 50.2 × 0.15 ×
5.67 × 103 × 8.5 × 10−3
= ∆θ
50.2 × 0.15
6.4°C = ∆θ
∆x = 8.5 × 10−3
A = 0.15 m2
θ1 = 100°C
θ2 = ?
K = 50.2 Wm −1K −1
( ∆θ )
8.5 × 10−3
∴θ2 the temperature of Hotplate = 106.4°C
Unit I.indd 130
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( 2 0 0 6 ) PAPE R 2
131
IαP
(c) (i)
and P α T 4
∴
I α T4
V2
4.5
=
4
1500
22504
(ii)
4.5 × 22504
 2250 
= 4.5 
4
 1500 
1500
= 22.8µv
V2 =
1
r2
K
4.5 =
1.52
K
9= 2
r3
4
Iα
So
and
 1.5
(2) ÷ (1) :2 =  
 r 
r3 =
Unit I.indd 131
3
1.5
2
(1)
2
(2)
= 1.06 m
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132 c a p e p h y s i c s - U n i t 1
Question 9
(a) (i) Assumptions of K. T. of gases:
– All collisions are elastic.
– There are many molecules moving randomly.
– No inter moleculer forces.
– All motions obey Newton’s laws.
– T
he duration of a collision is negligible compared to the time
between ­collisions.
– T
he total volume of the gas molecule is negligible compared
to the overall volume of the gas.
1
1 Nm 2
c
(ii) P = ρcˆ 2 or P =
3
3 V
(iii)
So
Also
(b) (i)
1
Pv = Nmc 2 = nRT
3
1
3
Nmc 2 = nRT
2
2
1
1

Nmc 2 =  mc 2  N ⇒ The total KE of all the molecules.
2

2
1
3
Nmc 2 = NKT
2
2
nR
where K =
N
K is the Boltzmann constant.
PV = nRT
1.00 × 10 × V = 2 × 8.31 × 263
2 × 8.31 × 263
V=
1.00 × 105
= 0.0437 m3
5
Unit I.indd 132
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( 2 0 0 6 ) PAPE R 2
(ii)
133
V1 V2
=
T1 T2
0.0437 5 × 0.0437
=
263
T2
T2 = 263 × 5
= 1315 K
(iii) ∆W = P ∆V
= 1.0 × 105 × 4 × 0.0437
= 1.75 × 104 J
(iv) ∆u = nC p ∆T or mc ∆T , where m = 2.0 × 10−3 kg
3
= nR ∆T
2
3
= × 2 × 8.31 (1315 − 263)
2
= 26 226 J
(v) ∆Q = ∆u + ∆w
= 26 226 + 17480
= 43 706 J
(vi)
1
3
mc 2 = KT 2
2
3KT
c2 =
m
2.07 × 10−3
6.02 × 1023
= 0.33 × 10−26
m=
= 3.3 × 10−27 kg
3 × 1.38 × 10−23 × 1315
3.3 × 10−27
= 1.65 × 107
=
c = 4.06 × 103 ms −1
Unit I.indd 133
2/2/2016 5:21:28 PM
(2007) SPECIMEN PAPER 1
Multiple Choice
Qu. #
Key
1
B
Recall of base units.
3
D
4
C
The slope of the graph stands for velocity. The slope
starts of being negative. So the answer is either C or D.
The slope of the negative part is smaller in value than
the slope of the positive part, so D.
2
5
C
6
D
7
A
8
9
Unit I.indd 134
C
B
B
Explanations
Recall of vector diagrams and resultant.
The vertical component of acceleration (which is g) is
constant.
20 – F = m × a. So F = 20 – (0.80 × 5) = 16 N
Tan θ = Vertical velocity/Horizontal velocity.
­Horizontal velocity is constant. The higher the body
falls from, the greater is the final vertical velocity.
Momentum before = momentum after. The m
­ onkey
jumping ­vertically will not affect the horizontal
­momentum. Momentum before = (20 + 40) × 8.
­Momentum after = 40 × V. This gives V = 12 m/s.
For the ice floating, the weight of the ice is equal to
the weight of water displaced. The mass of water
­displaced will be 50 g. If the density of water is taken
as 1 g/cm3, then water displaced will be 50 cm3.
So new level will be 300 cm3.
Terminal velocity must be reached, so g must reach
zero. Acceleration does not decrease uniformly in
­falling through a fluid.
2/2/2016 5:21:28 PM
( 2 0 0 7 ) SPECI M EN PAPE R 1
Qu. #
Unit I.indd 135
Key
10
D
11
B
12
B
13
A
14
D
15
D
16
C
17
B
18
B
19
A
135
Explanations
If the masses are different, then
v12
v22
1
1
T = m1 = m2 . So m1v12 = m2v22
r
r
2
2
1
1
2
KE of A = m (2v ) = 2mv 2 . KE of B = (2m) v 2 = mv 2
2
2
The tension in the string does not remain constant
when moving in a vertical circle … lowest at the top
and highest at the bottom. Any answer with I is wrong.
That leaves B.
Definition of efficiency is output/input.
Zero error in the instrument is a systematic error.
Mass of dog, m = (m2 − m1 ) .
So
(m − m1 ) = 2 = 12%. Note the total error in
∆m
=∆ 2
m
(m2 − m1 ) 7
(m2 – m1) is the sum of the errors, i.e., 2 kg.
A longitudinal wave cannot be polarized.
Recall 1 α A2.
From graph 1, T = 20 s, so f =
λ = 10 m. V = f λ =
1
Hz. From graph 2,
20
1
× 10 = 0.5 m/s.
20
Air has a refractive index of 1. The smaller the
­refractive index, the faster the light will travel.
2/2/2016 5:21:28 PM
136 c a p e p h y s i c s - U n i t 1
Qu. #
20
A
21
C
22
B
23
A
24
D
25
26
27
Unit I.indd 136
Key
D
B
D
Explanations
Use d sin θ = nλ . So λ =
d sin θ
.
1
 1 
× 10−3 m.
d=

 2000 
Now
λ = 5 × 10−7 × 0.5 = 2.5 × 10−7 or 250 nm
So
Use of lens formula:
1 1 1
= + .
f u v
Recall range of wavelengths for visible region: longest
is red … 7.0 × 10–7 m to shortest, violet … 4.5 × 10–7 m.
In a given time, 6 full waves will fit between X and P
1
and 2 waves will fit between Y and P. So the waves
2
will meet 180° out of phase.
Recall I0 = 10–12 W m–2
The distance must be a whole number of half
­wavelengths.
1
1
wavelength. So 0.6 m = λ .
4
4
300
= 125 Hz. For the next
Hence λ = 2.4 m and f =
2.4
3
resonance, wavelengths will fit in 0.6 m. So λ = 0.8 m
4
and f = 375 Hz.
The fundamental gives
20 half wavelengths = 0.3 m. Therefore, 1 wavelength
= 0.03 m.:
f=
v
λ
=
3 × 108
= 1 × 1010 Hz.
0.03
2/2/2016 5:21:29 PM
( 2 0 0 7 ) SPECI M EN PAPE R 1
Qu. #
Key
28
A
29
C
30
31
32
Explanations
Acceleration α displacement and acceleration α force.
If ­displacement changes, then acceleration changes
and so force changes.
Recall. This can also be checked by using units
­analysis.
Maximum KE = Maximum PE. Maximum PE =
1 2

 recall energy stored = kx  .
2
1
(2k ) A2
2
The choices on the question paper were missing.
D
Recall zeroth law of thermodynamics.
34
D
35
D
First recognize that the temperature of the other side
Q
 ∆θ  gives Δθ as
will be less than 95°C. Using = kA 
 ∆x 
s
30°C. So temperature inside will be 65°C.
37
C
33
36
Unit I.indd 137
D
137
A
C
38
B
39
A
50 × 10–3 × c × 60 = 100 × 10–3 × 450 × 80 gives c as
1200 J/kg/k.
Change J kg–1k–1.... J → kg. m.s–2.m. This gives D.
3
Recall KE = KT . Substitute with T = 300 K.
2
Recall the assumptions of the kinetic theory of gases.
Use of PV = nRT. This gives n = 50 moles. Number of
atoms = n × Avogadro’s number.
Work is done ON the gas during a compression.
­Compression takes place in regions 1 to 2 and 2 to 3.
2/2/2016 5:21:29 PM
138 c a p e p h y s i c s - U n i t 1
Qu. #
40
41
Unit I.indd 138
Key
C
D
Explanations
More work is done BY the gas than ON the gas
(­ expansion at higher pressure). So net work done
on the gas will be negative. Net work done is area
­enclosed in the loop.
Recall that pressure due to a liquid is given by
P = depth × ρg. Depth = (x + h). Note: The total
­pressure will truly be this plus atmospheric pressure.
F 1
F
F
graph.
Y =   × , where   is the slope of the
 x A
 x
x
42
A
43
44
B
D
Recall.
45
C
X is definitely not ductile but brittle (no plastic
­region). Y is ­ductile and Z is polymeric .
A = 1 × 10–7m2.
Extension =
F
F 0.8 0.8
+
=
+
= 0.56 m.
K1 K2
2
5
2/2/2016 5:21:30 PM
(2007) EXAM PAPER 1
Multiple Choice
Qu. #
Key
1
C
2
C
3
C
4
C
5
C
Explanations
Power is watts. Watts is J/s. J = N.m. and N = kg m s–2.
Unit I.indd 139
C
7
B
= kg m2 s −3 .
X is equal in magnitude to the resultant of 7 N and 24 N
at 90°, R = 25 N.
Each graph is a disp/time graph. So slope stands for
velocity. No acceleration means constant velocity, which
means a straight line graph. This means A or C. Not A
­because the velocity is zero and the question says the
body is in motion.
F = ma. Acceleration is constant if force is constant for
constant mass.
Recall of a =
a=
6
W = kg m s −2 m s −1
So
F
F
and v 2 = u2 + 2as. If u = 0, v 2 = 2s ×   .
 m
m
So v =
F=
v2
and a = vω .
r
(2sF ) ×
1
m
. So v ∞
1
m
.
mv 2
F
v2
602
= =
= 0.73.
, Weight = mg. So
mg rg (500 × 9.81)
r
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140 c a p e p h y s i c s - U n i t 1
Qu. #
Unit I.indd 140
Key
8
C
9
B
10
B
11
D
12
B
13
A
14
D
15
D
16
D
17
B
Explanations
Impulse, Ft = change in momentum = mv – mu → area
under F/t graph. If initial momentum is zero, then final
momentum is area under graph.
Area =
1
(2 + 5) × 20 = 70 N.S.
2
Drag forces increase with increasing velocity.
So ­acceleration reduces to zero (terminal velocity)
(Stoke’s law).
Definition of “torque” → Product of one of the forces and
­perpendicular distance between them.
Constant velocity means zero acceleration. F = ma. Zero
­acceleration means zero resultant force.
Vertical acceleration does not change. It remains
­constant as “g”.
1 2 1
mv = Fx
2
2
−3 2
20 × 10 v = 20 × 7 × 10−2 → v = 8.4 ms −1
Precision means “small spread”. Accuracy means
­“average value is close to or equal to true value”.
Momentum is a vector. Momentum before
 m 
 1
= mV +   ( −v ) = mv.
 2 
 2
One is a sine function, i.e., d = 0 when t = 0 and the other
is a cosine function, d is maximum when t = 0. So they
are 90° out of phase or
π
2
radians.
3 half waves → 15 m. Therefore, 1 wave → 10 m.
V = fλ = 50 × 10 = 500 m s–1.
2/2/2016 5:21:31 PM
141
( 2 0 0 7 ) E X A M PAPE R 1
Unit I.indd 141
Qu. #
Key
Explanations
18
19
B
A
Longitudinal waves cannot be polarized.
20
D
21
D
nλ
shows that smaller wavelength
d
d
diffracts least. From ROYGBIV, blue has the shortest
wavelength of those given.
22
C
23
B
I ∞ A2, I1 = k × 302 and I2 = k × 102. So
24
A
25
C
At an open end there must be an antinode and at a closed
end there must be a node.
26
A
27
C
28
A
For S.H.M, velocity is maximum at the centre.
∆x =
λD
or sin θ =
Recall. Convex lens for long sight and concave lens for
short sight.
In the time taken for one wave to reach P from X, 1
1
2
waves will reach P from Y. So the waves will meet
out of phase at P. So ­amplitude is zero. (Destructive
­interference)
I1 302
=
=9
I2 102
For constructive interference, path difference must be
nλ, n = 0, 1, 2….
Using “real is positive” convention:
1 1 1 1 1 1
= + −
= + giving v = – 4, i.e., 4 cm on same
f u v 20 5 v
side of lens as object.
−
 I
dB = 10 log   . So 80 = 10 log  I−12  → I = 10 −4 W m −2 .
 I0 
 10 
The amplitude is 3 cm. So III is wrong. That means B, C, D
are wrong.
2/2/2016 5:21:31 PM
142 c a p e p h y s i c s - U n i t 1
Qu. #
29
30
Key
A
D
Explanations
T
 g
 I
 I
T1 = 2π   and T2 = 2π   . So 2 =   .
 g
 g′ 
 g′ 
T1
 g
If T1 = 1 s then T2 =   .
 g′ 
I is wrong. Resonance occurs when the frequency of the
external force is equal to the natural frequency of the
oscillating system. That means A, B, C are wrong.
( XT − X 0 ) × 100
( X100 − X 0 )
 (100) 
(100X 0 )
=
 × XT −
( X100 − X 0 )
 ( X 100 − X 0 ) 
θx =
31
B
→ y = mx + c
This is a straight line with positive slope and negative
intercept on the y-axis.
Stefan’s law P = σAT4.
32
Unit I.indd 142
D
33
C
34
A
Rate of loss of heat is P1 = σ AT14 .
Rate of gain of heat is P2 = σ AT24 .
(
So NET rate of heat loss is P1 − P2 = P = σ A T14 − T24
)
Note: T14 − T24 is not equal to (T1 – T2)4.
Recall: Amount of heat needed to raise the temperature
of entire body by 1 K.
∆W = P ∆V where ∆V = (140 − 16) × 10−3 m3
= 1750 × 103 × 124 × 10−3
= 217 kJ
2/2/2016 5:21:32 PM
( 2 0 0 7 ) E X A M PAPE R 1
Qu. #
Key
143
Explanations
P = hρ g
= 130 × 10−3 × 13600 × 9.8
35
C
36
37
C
A
Diamond is tetrahedral crystalline. Glass is amorphous.
38
C
39
B
Recall assumptions of kinetic theory of gases. Elastic
collision means “same speed” before and after. Also “no
forces” means “no acceleration”.
40
B
= 1.73 × 104 Pa
Air is a poor conductor of heat.
Recall: Energy lost is area enclosed by the hysteresis
loop.
3
KE = kT
2
(2KE )
T=
3k
2 × 8 × 10−21
=
3 × 1.38 × 10−23
(
(
)
)
= 386 K
41
B
In I, net work is done BY the gas. (PΔV for expansion
greater than PΔV for compression). This rules out C
and D. Similarly, in IV, expansion at higher pressure,
­compression at lower pressure means net work done
BY the gas.
Volume of all atoms in 1 mole is given by V =
V=
42
A
0.0635
, 1 mole has 6.02 × 1023 atoms.
8920
So volume of 1 atom =
(
0.0635
8920 × 6.02 × 1023
mass
density
)
= 1.18 × 10−29 m3
Unit I.indd 143
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144 c a p e p h y s i c s - U n i t 1
Qu. #
Key
43
D
44
45
Unit I.indd 144
B
B
Explanations
The pressure at the same depth in a fluid is the same.
Let volume of B be V, so volume of A will be 3 V.
Total mass = (3V × 8900) + (7800V ) = 34500V
Total volume = 4V
34500V
4V
= 8625 kg m −3
Net density =
Two springs in parallel require double force for the same
­extension. So k is doubled. Same mass will give only half
the extension.
2/2/2016 5:21:33 PM
(2007) PAPER 2
Question 1
(a) (i) L = 19.2 ± 0.4 cm
(ii) D = 1.92 ± 0.04 cm
(b) F
ind the mass of 10 balls and then divide by 10. The error of ±2.5 g
divided by 10 gives an error of ±0.3 g.
(c) Density =
Mass
Volume
m = 30.4 ± 0.3 g
1
V = π D3
6
=
1
× π × 1.923
6
= 3.71 cm3
∆V
∆D
=3
V
D
=
3 × 0.04
1.92
= 0.0625
∴∆V = 0.0625 × 3.71
= 0.232 cm3
Unit I.indd 145
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146 c a p e p h y s i c s - U n i t 1
30.4
= 8.91 g/cm −3 (3 sig figs )
3.71
∆ρ ∆m ∆v
=
+
ρ
m
v
Density =
0.3 0.232
+
30.4 3.71
= 0.072
=
So uncertainty in ρ is 7.2%.
Actual error in ρ = 0.072 × 8.19
= 0.59 × 103 kg/m −3 .
So ρ = (8.19 ± 0.59) × 103 kg/m–3.
Unit I.indd 146
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( 2 0 0 7 ) PAPE R 2
147
Question 2
(a)
f=
c
2π
A
Lv
This can be rearranged to give
 c
f =
 2π
A 1
×
L 
v
which looks like y = mx + c.
1
v/m3
v m
f/Hz
250 × 10–6
63.2
225
150 × 10–6
81.6
290
200 × 10–6
70.7
125 × 10–6
89.4
255
320
100 × 10–6
100.0
350
65 × 10–6
124.0
445
80 × 10–6
111.8
See graph page.
400
(b) Slope =
(430 − 250) = 180 = 3.6 Hz m 32
(120 − 70) 50
(c) Slope =
c
2π
A
L
c = 2π × 3.6 ×
= 342 ms −1
Unit I.indd 147
−3
2
L
A
L = 5.9 × 10−2
1
L = 2.42 × 10−1 m 2
A = 2.55 × 10−4
A = 1.60 × 10−2 m
2/2/2016 5:21:34 PM
148 c a p e p h y s i c s - U n i t 1
f/Hz
440
420
400
380
360
340
320
300
280
260
240
220
50
60
70
80
90
1
v
Unit I.indd 148
100
110
120
130
140
m–3/2
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149
Question 3
(a) On graph page.
F/N
45
40
35
30
25
20
18.5
15
10
5
.05
0
Unit I.indd 149
0
1
2
2.15
3
4
5
6
7
8
ext/mm
x
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150 c a p e p h y s i c s - U n i t 1
(b) Area up to 2.15 mm, i.e., 2.15 × 10–3 m.
1
× 2.15 × 10−3 × 20
2
= 2.15 × 10−2 J
A=
1
(20 + 30)(4 − 2.15) × 10−3
2
= 4.63 × 10−2
Area of appropriate trapezium =
So total energy = 6.78 × 10–2 J
(c) (i) The wire ceases to obey Hooke’s law when the load reaches 20 N.
(ii) Slope =
Y=
∆F
18.5 − 0
=
∆x (2.0 − 0) × 10−3
= 9.25 × 103 N/m
F l
×
x A
2
2.00 × 10−7
= 9.25 × 1010 Pa
= 9.25 × 103 ×
Unit I.indd 150
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( 2 0 0 7 ) PAPE R 2
151
Question 4
(a) (i) A scalar quantity is one that has magnitude only; e.g., time.
A vector quantity is one that has both magnitude and direction;
e.g., force.
(ii) Resolving vertically using upward direction as +ve.
Vertical component = 35 Cos 25 – 20 Sin 45 – 24 Sin 30
= 5.58 N
Resolving horizontally using to the right as +ve.
Horizontal component = 35 Sin 25 + 24 Cos 30 – 20 Cos 45
= 21.43 N
R
5.58
a
14.6°
21.43
R2 = 5.582 + 21.432 = 490.38
R = 22.14 N
5.58
21.43
α = 14.6°
tan α =
(b) (i) Conditions necessary for equilibrium:
1.
2.
Resultant force on body must be zero.
Resultant torque must be zero (moments).
W = P + Q(1)
Taking moments about pivot (Q).
W×
Unit I.indd 151
L
= P × L (2)
2
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152 c a p e p h y s i c s - U n i t 1
(ii) (a) The upward force R = downward force = 200 N
Taking moments about R.
(b) P × 6 = 200 × 1.5
200 × 1.5
P=
6
= 50 N
l = 62 + 32
(c) F = P = 50 N (Force to right = Force to left)
Resultant = 2002 + 502
= 206 N
200
tan α =
= 4.0
50
α = 76.0°
M
R 200
a
F 50
Unit I.indd 152
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153
( 2 0 0 7 ) PAPE R 2
Question 5
(a) (i)
t
V =0
a=
x
V=V
v
t
So F = ma = m
v
t
1
x = ut + at 2
2
1 v
1
=   t 2 = vt
2 t 
2
Work done on body = F × distance moved in the direction of force
v 1
= m × vt
t 2
1
= mv 2
2
But work done is energy gained.
1
So KE gained E k = mv 2 .
2
(ii) At this height, gravitational field strength will not be g.
g ′∞
1
where r is distance from center of earth.
r2
So g at this height would have been reduced by a factor
(rE + R )2
where rE and R are in metres, R being the height above the
­surface.
Unit I.indd 153
1
,
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154 c a p e p h y s i c s - U n i t 1
(iii) At the bottom, the PE will be converted to KE.
So total KE at bottom =
1
2
m (12) + mgh
2
= 72 m + 343 m
= 415 m J
So at bottom if v is the new velocity,
then
1 2
mv = 415 m
2
v 2 = 830
v = 28.8 ms −1
(b) (i)
∆PE = mg∆h
= 1200 × 35
= 4.2 × 104 J
(ii) P =
=
Work done
Time taken
4.2 × 104
24
= 1.75 kW
(iii)
70% of power input = 1.75 kW
1.75
× 100
70
= 2.5 kW
∴100% of power input =
Unit I.indd 154
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155
( 2 0 0 7 ) PAPE R 2
(c) (i)
disp
Taking position at
top as zero displacement
1
S = ut + at 2
2
disp
t/s
If disp. at top
is maximum at
t=0
t
V/m/s
V = u + at
t/s
(ii)
Unit I.indd 155
Assuming no air resistance.
1
S = at 2
2
1
35 = (9.8) t 2
2
2
t = 7.143
t = 2.67 s
2/2/2016 5:21:38 PM
156 c a p e p h y s i c s - U n i t 1
Question 6
(a) (i) A
wave can be made to reflect on to itself by hitting a ­reflecting
boundary. In this way the condition of coherence will be
­satisfied.
Also the distance from the source to the reflecting surface must
be a whole number of half wavelengths so that a node is at the
reflecting surface.
(ii) Amplitude at A is permanently at zero (node). Amplitude at
B is maximum (an antinode). Amplitude at C is between that
of A and B. Points B and C will be at their maximum ­amplitude
at this point at the top, so they will be in phase together “on their
way down”.
(iii) Loudness – this is related to amplitude.
Pitch – this is related to frequency.
Quality – this is related to number of overtones (i.e., number of
harmonics). Timbre is also related to this property of sound.
(b) (i)
λ=
=
v
f
340
1 × 103
= 3.4 × 10−1 m
= 0.34 m
(ii) dB = 10 log
I
I0
 10−2 
= 10 log  −12 
 10 
(
= 10 log 1010
)
= 100 dB
Unit I.indd 156
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( 2 0 0 7 ) PAPE R 2
120 dB = 10 log
(iii)
157
I
I0
I
10−12
= log I − log 10−12
= log I + 12
log I = 0
12 = log
So
Unit I.indd 157
I = 1 w/m2
1 × 10−2 x 2
= 2
1
20
2
x = 4 ∴ x = 2m
2/2/2016 5:21:38 PM
158 c a p e p h y s i c s - U n i t 1
Question 7
Sin i
where i is the angle in the faster
Sin r
­ edium that the ray makes with the normal.
m
(a) (i) Refractive index =
(ii)
Wave fronts
l1
Faster
Slower
l2
l1 > l2
(iii)
q0
air
A
q1 q1
n1
B
q2
n2
q2
C
q0
air
Consider a ray going through two media with absolute refractive
index n1 and n2 as shown.
At A,
n1 =
c Sinθ0
=
(1)
c1 Sinθ1
At C ,
n2 =
2 ÷1
So
Unit I.indd 158
c Sin θ 0
=
c2 Sin θ 2
n2 c1 Sin θ1
= =
= 1n2 n1 c2 Sin θ 2
(2)
n2 Sin θ 2 = n1 Sin θ1
2/2/2016 5:21:39 PM
( 2 0 0 7 ) PAPE R 2
159
(iv) W
hen a wave goes from one medium to another, the frequency
does not change.
λ1 v1
=
λ2 v2
(b) (i) For red light:
n=
1.51 =
c
c1
3.0 × 108
c1
3.0 × 108
1.51
= 1.99 × 108 ms −1
c1 =
For blue light:
3.0 × 108
1.55
= 1.94 × 108 ms −1
C2 =
(ii)
30
30
q1
q2
B
R
60
Angle of incidence for both is 30°.
Unit I.indd 159
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160 c a p e p h y s i c s - U n i t 1
For red light, find θ1.
Sin θ1
= 1.51
Sin 30
Sin θ1 = 1.51 × 0.5 = 0.755
θ1 = 49.02°
For blue light find θ2.
Sin θ2 = 1.55 × 0.5 = 0.775
θ2 = 50.81°
∴ θ2 – θ1 = 1.79°
⇒ angle between the two emergent rays.
1
× 10−3 m
1200
= 8.33 × 10−7 m
(c) d =
For red light,
nλ
for n = 1
d
6.78 × 10−7
=
8.33 × 10−7
= 0.8139
θr = 54.48°
Sinθr =
4.22 × 10−7
8.33 × 10−7
= 0.5066
θ b = 30.43°
Sinθ b =
∴ angle between colours = 54.48 –30.43
= 24.05°
Unit I.indd 160
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( 2 0 0 7 ) PAPE R 2
161
Question 8
(a) (i)
A constant volume gas thermometer can be used for ­measuring
temperatures in the range 3 K to 1800 K. For example, in
­measuring the temperature of liquid nitrogen. None of the other
thermometers can go lower than 20 K. This thermometer has
a wide range (–270°C to 1500°C). It is very accurate and very
sensitive.
(ii) (a)
A thermocouple can be used, since the temperature will be
over 1000°C. Also the small size of the “hot” junction allows
for measurement of temperature at a point.
(b)A thermocouple can be used here as well, since it is
­capable of measuring rapidly changing temperatures.
A ­thermocouple can measure temperature at a point.
(Note: In (a) modern radiation thermometers or
­pyrometers can also be used.)
(iii) –The resistance of a thermistor may increase or decrease
with temperature depending on negative or positive
­coefficient of resistance, whilst the resistance of the
­platinum increases with temperature.
(b) (i)
–The resistance of a thermistor does not vary linearly with
temperature and so its useful range in small, whereas the
resistance of the platinum varies linearly over a wide range.
Rice = 7360 Ω
R100 = 153 Ω
R = R0e
ln R =
Unit I.indd 161
B
T
B
+ ln R0
T
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162 c a p e p h y s i c s - U n i t 1
At ice point: 8.90 =
B
+ ln R0 (1)
273.15
At steam point: 5.03 =
B
+ ln R0 (2)
373.15
(
) (
)
3.87 = 3.66 × 10−3 B − 2.68 × 10−3 B
(1) − (2)
= 9.8 × 10−4 B
3.87
× 104
9.8
= 3.95 × 103 K
B=
Substitute in (1)
(ii)
(
)(
)
8.90 = 3.66 × 10−3 3.95 × 103 + ln R0
ln R0 = −5.56
When R = 2200 Ω ,
R0 = 3.85 × 10−3 Ω
3.95 × 103
− 5.56
T
3.95 × 103
13.26 =
T
3.95 × 103
T=
13.26
= 298 K
⇒ 298 − 273 = 25°C
7.696 =
(iii) (a) θ °C =
Pθ − P0 100
×
P100 − P0
1
2200 − 7360
× 100
153 − 7360
= 71.6°C
=
(b)This method assumes a linear relationship between
­resistance and temperature. The actual relationship is
an exponential one.
Unit I.indd 162
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( 2 0 0 7 ) PAPE R 2
163
Question 9
(a) (i) (a)
The cylinder will have many molecules of oxygen ­moving
randomly. Pressure is a measure of the frequency with
which the molecules bombard the walls of the container.
When the molecules bombard the walls of the container,
they will experience a change in momentum. The rate of
change of momentum is force. This force divided by the
area the molecules bombarded is pressure.
(b)When more oxygen is pumped in, there will be more
­molecules in the same space (moving at the same average
speed if temperature is constant), so there will be more
collisions per second with the walls. Hence greater rate
of change of momentum, greater force and hence greater
pressure.
(c)If the cylinder is left in the sun, the temperature increases.
Hence, the molecules gain kinetic energy and move faster,
thus colliding with the walls of the container more often.
Hence greater force, so greater pressure.
(ii) Case (1) P1V = n1RT1
4.5 × 105 V = 2.8 × R × 300
Case (2) P2V = n2RT2
P2V = (3.9 + 2.8) R × 320
(2) ÷ (1)
P2
6.7 × 320
=
5
4.5 × 10
2.8 × 300
P2 = 1.15 × 106 Pa
(1)
(2)
(b) (i) ∆Q = ∆u + P∆V
∆Q is heat energy supplied.
∆u is increase in internal energy.
P∆V is work done by the gas on the surroundings.
(P is constant pressure and ∆V is change in volume.)
Unit I.indd 163
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164 c a p e p h y s i c s - U n i t 1
(ii) (a) H = nCv ∆T
= 6.2 × 12.5 × 25
= 1940 J
(b)Since the volume is kept constant, then no work is
done on the surroundings. So all heat supplied go to
increase the internal energy. So increase in internal
energy will be 1940 J.
(c)
∆Q = ∆u + P ∆V
P ∆V = 3200 − 1940
= 1260 J
(d) From values given,
3200 = ∆u + P ∆v = nC P ∆T
3200
3200
=
n∆T 6.2 × 25
= 20.7 J/k/mol
CP =
From C P − Cv = R
C P = R + Cv
Unit I.indd 164
= 8.31 + 12.5
= 20.8 J/K/mol
2/2/2016 5:21:41 PM
(2008) Trinidad & Tobago
PAPER 2
Question 1
(a)
y
m
Distance,
±2 ms
0.600
0.342
0.800
1.000
0.0790
0.414
0.1714
0.500
1.400
0.534
1 2
gt
2
–
0.281
0.456
1.200
t2
s2
t
s
±2 mm
0.400
y=
Time,
0.1170
0.2079
0.2500
0.2852
So a graph of y – vs – t2 will have a slope of
So
g = 2 × slope
1
g.
2
1.200 − 0
0.2500 − 0
= 4.800 ms −2
∴ g = 2 × 4.800
Slope =
= 9.600 ms −2 .
Unit I.indd 165
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166 c a p e p h y s i c s - U n i t 1
(b)
a
9.6
t
v
t
y
t
(c)
v 2 = u2 + 2as
= 0 + 2(9.8)(0.90)
= 17.64
∴ v = 4.2 ms −1
Unit I.indd 166
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( 2 0 0 8 ) Tr i n i d a d & T o b a g o
PAPE R 2
167
l/m
1.7
1.6
1.5
1.4
1.3
1.2
1.1
1.0
.9
.8
.7
.6
.5
.4
.3
.2
.1
0
Unit I.indd 167
0.1
0.2
0.3
t 2/s 2
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168 c a p e p h y s i c s - U n i t 1
Question 2
(a) (i)
Diffraction is the spreading of waves as they pass through small
openings or around small objects.
(ii)
l
0
l
0
Note: Wavelength does not change.
(iii) On the diagram indicated by O and X.
Unit I.indd 168
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PAPE R 2
169
(iv)
d
d
t
d
t
d
t
t
d
d
t
Resultant
(b) (i)
t
Resultant is zero
displacement
Constructive interference
Destructive interference waves
waves meet in phase.
meet out of phase.
d = 600 m
v
λ=
f
3.0 × 108
9.00 × 105
= 333 m
nλ
Sin θ =
for n = 1
d
1 × 333
= 0.555
Sinθ =
600
θ = 33.7°
=
Unit I.indd 169
2/2/2016 5:21:43 PM
170
d = 600 m
v
λ=
f
c a p e p h y s i c s - U8 n i t 1
3.0 × 10
=
9.00 × 105
= 333 m
nλ
for n = 1
d
1 × 333
Sinθ =
= 0.555
600
θ = 33.7°
Sin θ =
(ii) A
t Q, the signals will meet out of phase. So destructive
­
interference will take place.
(iii) S ince the angles are big, Sin θ will not be approximately equal
to θ and so θ will not be proportional to n.
Using n = 1.5
1.5 × 333
= 0.8325
Sinθ =
600
θ = 56.4°
Unit I.indd 170
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PAPE R 2
171
Question 3
(a) (i) Stress is defined as force per unit cross-sectional area.
Stress =
F
N/m2 or Pa.
A
Strain =
x
∆
or


Strain is the ratio of the extension to the original length.
(ii)
F
Force
A
0
(iii)
∆l
Extension
Hooke’s law applies to the straight line part of the graph from
O to A.
F
F
∆l
No plastic region (brittle)
∆l
Hysteresis loop
(a) Glass(b) Rubber
Unit I.indd 171
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172 c a p e p h y s i c s - U n i t 1
(b) (i)
lm
T
mg
(ii) (a) At the lowest point,
mv 2
T − mg =
r
mv 2
r
= 0.5(9.81 + 1444)
T = mg +
So
(b) Stress =
(c)
F
A
= 727 N
727
=
3.14 × 10−6
= 2.31 × 108 Pa
A = πr2
(
= π 1 × 10−3
)
2
= 3.14 × 10−6 m2
Stress
Strain
Stress
So strain =
Y
2.31 × 108
=
= 1.155 × 10−3
11
2 × 10
Y=
So extension, ∆l = 1.155 × 10−3 × 1
= 1.16 mm
(c) T
he wire will not break because the stress is less than the breaking
stress; i.e., 2.31 × 108 Pa < 7.2 × 108 Pa.
Unit I.indd 172
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( 2 0 0 8 ) Tr i n i d a d & T o b a g o
PAPE R 2
173
Question 4
(a) U
sing a micrometer screw gauge, grip the wire between the jaws of
the gauge. Use the ratchet to tighten. Rotate the wire in the jaws to
cater for any “dents” in the wire. Note the reading. Do this for about
five other places along the length of the wire. Find the average of
these readings.
(b) P
recision deals with how much spread there is in the readings.
The less spread there is, the more precise, but not necessarily
­accurate.
Accuracy deals with how close the average value is to the correct
value, even though the spread may be wide (i.e., less precise).
f
Correct
value
True value
f
Precise but not accurate
x
x
Accurate but not precise
Pt = mL
Pt
L=
m
54 × 300
=
9.9 × 10−3
= 1.64 × 106 J/kg
(c)
∆L ∆P ∆t ∆m
=
+
+
L
P
t
m
2
2
0.1
=
+
+
54 300 9.9
= 0.0538
So
Unit I.indd 173
∴∆L = 0.0538 × 1.64 × 106
= 0.086 × 106 J/kg
L = (1.64 ± 0.09) × 106 J/kg
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174 c a p e p h y s i c s - U n i t 1
Question 5
(a)
boundary
l2
l1
a
l1 > l2
a = 90 – q
The waves will bend as shown, with smaller wavelength. This is
called refraction.
(b) (i)
(ii)
f=
λ
At
λ = 3 mm,
At
f=
69 × 10−3
= 23 Hz
3 × 10−3
λ = 8 mm,
f=
85 × 10−3
= 10.63 Hz
8 × 10−3
So decrease on f = 23.0 − 10.6
= 12.4 Hz
λ/mm
v/mm/s
v2/mm2/s2
1
λ
Unit I.indd 174
V
mm −1
1.0
1.2
1.4
1.6
1.8
7921
7056
6400
5776
5476
89
1.000
84
0.833
80
0.714
76
0.625
74
0.556
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175
See graph page.
Slope =
3
× 103
0.55
= 5.45 × 103
=
So
Unit I.indd 175
(8.2 − 5.2) × 103
(1.05 − 0.5)
k = 5.45 × 103 mm3s −2
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176 c a p e p h y s i c s - U n i t 1
v 2/mm 2/s 2
8.2
8
7
6
5.2
5
0.5
Unit I.indd 176
.6
.7
.8
.9
1.0
1.05
1.1
I/l/mm–1
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177
Question 6
(a) (i)
P = KA
(θ2 − θ1 )
x
where A is area and K is thermal conductivity of the material.
(ii) When x = 2.5 cm → 2.5 × 10−2m,
1
= 0.4 × 102 = 40
x
So P = 15 W
When x = 4.5 → 4.5 × 10−2 m,
1
= 0.22 × 102 = 22 m −1
x
So P = 9.0 W
(iii) The gradient of the graph will be K A (θ2 – θ1).
30
= 0.375
80
K (0.25)(35) = 0.375
Gradient =
So
So
(iv)
K = 0.0429 W m −1 k −1
⇒ 4.29 × 10−2 W m −1 k −1
p
x
Unit I.indd 177
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178 c a p e p h y s i c s - U n i t 1
(b) Net rate of heat loss is given by Stefan’s law.
(
)
Net R = α A T14 − T24 ,
(
T1 = 303 K
T2 = 268 K
= 5.67 × 10−8 × 2 3034 − 2684
)
= 3.27 × 109 × 5.68 × 10−5 × 2
= 371 W
Unit I.indd 178
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(2008) Other Islands
PAPER 2
Question 1
(a) (i)
v /m /s
5
4
3.25
3
2.85
2
1
0
0
Unit I.indd 179
1
2
3
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180 c a p e p h y s i c s - U n i t 1
(ii) T
he graph shows an initial acceleration which decreases with
time (since the slope of the graph decreases). The acceleration
eventually reaches zero (slope zero) at 2.8 seconds and beyond.
This is the time at which terminal velocity is reached.
The terminal velocity is 4.12 ms–1.
∆v
∆t
3.25 − 2.85
=
0.7 − 0.5
= 2.0 ms −2
(iii) Average acc =
K≡
(b) (i)
mg
6π rVt
≡ kg ms −2 m −1 m −1s
unit of K ≡ kg m −1 s −1 ( viscosity )
5 × 10−3 × 9.81
6π × 1.0 × 10−3 × 4.12
= 0.632 kg m −1s −1
(ii) K =
From the equation,
Vt =
mg
6π kr
if r′ = 2r and m is constant,
1
then v ′ = Vt ,
2
i.e., the terminal velocity will be half of what it was.
Unit I.indd 180
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181
Question 2
f/Hz
70
66
65
60
55
50
45
40
35
30
25
20
15
10
5
Unit I.indd 181
1
1.9 2
3
4
5
6
7
7.5
8
n
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182 c a p e p h y s i c s - U n i t 1
Question 2
(a) A
ny membrane that is made to vibrate will cause the medium (air in
this case) around it to vibrate and so cause sound. The frequency of
the sound produced in the case of a string depends on the tension in
the string, the length of the string and the mass per unit length of the
string.
(b) (i)
2.76 m
(ii) 2.76 m = 1.5λ
2.76
λ=
= 1.84 m
1.5
(iii) For n antinodes there will be n half wavelengths.
So
So
 λ
n  = L
 2
2L
n
V = fλ
λ=
f=
So
V
λ
=
(c) The slope of the graph ≡
V
n
2L
V
2L
66 − 15
7.5 − 1.9
= 9.11 Hz
V
∴ 9.11 =
2L
V = 2 × 2.76 × 9.11
So
Slope =
= 50.3 ms −1
Unit I.indd 182
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183
Question 3
(a) Thermometer
Liquid in glass
Thermocouple
Constant Volume gas
Advantage
Reads temperature
directly
Best for varying
­temperature
Very accurate for
wide range
Disadvantage
Not very accurate
Does not read
­temperature directly
Inconvenient to carry
(b) (i)
To keep the volume constant, adjust the right-hand side tube
up and down until the level in the left side tube (of the ruler) is
back to a specified mark on the ruler.
(ii) (a)
When the bulb is in pure melting ice the right arm is
­adjusted so that the left arm comes back to the designated
mark on the ruler. The difference in mercury levels is h0.
(b) At 100°C, again the difference in mercury levels will be h100.
(c)At temperature t°C, the difference in mercury levels will be
ht. In all cases, the left hand side must be brought back to
the designated standard mark.
(d)
h
16.8 − 5.0
× 100
20.0 − 5.0
11.8
=
× 100
15
= 78.7°C
(ii) The pressure of the gas in the bulb will be Atomospheric
­Pressure + ∆hρg
(c) (i)
t=
P = (0.76 + 0.168) 13600 × 9.81
= 1.24 × 105 Pa
Unit I.indd 183
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184 c a p e p h y s i c s - U n i t 1
Question 4
(a) (i)
If the body is moving in a circle, then the direction is
­continuously changing. So even if the speed is constant, the
velocity changes because direction changes (velocity is a vector,
speed is a scalar). Acceleration is rate of change of velocity, so if
velocity is changing, then there must be an acceleration.
v2
(ii) a =
r
This acceleration is always directed towards the centre of the
circle.
(iii) W
ork done is force multiplied by distance moved in direction of
force. The centripetal force is directed towards the centre of the
circle whereas the velocity (direction of motion at any instant)
is tangential. So, F is perpendicular to the direction of motion.
So no work is done by that force.
(b) (i)
30
T
30
mg
(ii) (a) Resolving vertically,
T Cos 30 = mg
1 × 9.81
T=
= 11.3 N
Cos 30
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185
(b) Finding the radius of the circle,
r
0.5
r = 0.25 m
Sin 30 =
So
mv 2
= T Sin 30
r
11.3 1
v2 =
× × 0.25 = 1.4125
1
2
∴ v = 1.19 ms −1
(c)The mass will move in a parabolic path with an initial
­vertical velocity of zero.
Looking from the side.
1.5 m
1
s = ut + at 2
2
1
1.5 = (9.81) t 2
2
2 × 1.5
t2 =
= 0.3058
9.81
t = 0.55 s
Unit I.indd 185
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186 c a p e p h y s i c s - U n i t 1
Question 5
(a) (i) Role of diffraction:
For each slit, spreading of the waves takes place.
Same l on both sides.
Diffraction takes place because the size of the slits is about the
size of the wavelength of the light waves.
Role of Interference:
These wave fronts now have a chance to meet and so ­interfere
with each other. In directions in which the wave fronts meet
inphase, there is constructive interference and the ­directions
in which the wave fronts meet out of phase there will be
­destructive interference.
Each colour diffracts by a different amount depending on the
wavelength as given by
d Sin θ = nλ.
The bigger the wavelength, the greater the diffraction.
Direction of constructive interference.
(ii) From the formula d Sinθ = nλ
or
Unit I.indd 186
Sinθ =
nλ
d
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187
The shorter the wavelength, the smaller the θ. Blue has the
shortest wavelength of the three, so it will be diffracted least.
C will be red, since red has the longest wavelength and B will be
yellow.
(iii) From the formula Sinθ =
nλ
d
For zero order, n = 0,
θ will be zero for all λ.
So all the colours will overlap at O.
(b) (i)
1
, m = 1.67 × 10−6 m
5
6 × 10
nλ
Sinθ =
≤ 1 using longer λ = 590 nm
d
d
So
n≤
d=
λ
1 × 10−5
≤
6 × 5.90 × 10−7
≤ 2.8
So max n is 2nd order.
(ii) When n = 2,
2 × 5.89 × 10−7
1.67 × 10−6
θ1 = 44.86°
Sinθ1 =
Also
2 × 5.90 × 10−7
1.67 × 10−6
θ2 = 44.96°
Sinθ2 =
So angular separation is 0.10°.
Unit I.indd 187
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188 c a p e p h y s i c s - U n i t 1
Question 6
(a) (i) ∆u ⇒ The increase in internal energy of the gas.
Q is the heat supplied to the gas.
W is the work done ON the gas by the surroundings.
(ii) At constant pressure,
Q = ∆u + w for expansion at constant pressure.
Q is heat supplied nCp ∆T
In this case, heat supplied goes to increase the internal energy
(hence temperature) plus do work on the surroundings.
If the volume is kept constant then W = 0, and so all the
­energy supplied nCv∆T goes to increase internal energy alone
(no change in volume) so less energy is needed to raise the
­temperature of one mole of the gas by one kelvin. Hence Cp > Cv.
(b) (i) Net work done is given by area enclosed by loop.
Net work done = P × V
= 1.01 × 105 × 0.0225
(ii)
P1 2P1
=
T1 T2
= 2.27 × 103 J
T2 = 2T1 = 2 × 273 = 546 K
(iii) From 1 → 2 no change in volume.
So
Unit I.indd 188
∆Q = ∆u = nCv ∆T
3
= 1 × R × (546 − 273)
2
= 3400 J
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189
From 2 → 3
∆Q = ∆u + P ∆V
= nCv ∆T + P ∆V
3
= 1 × R × (1092 − 546) + 2P × V
2
= 6800 + 2 2.27 + 103
(
)
= 11.34 × 103 J
(iv) Efficiency =
Useful work done
100
×
Total energy supplied
1
2.27 × 103 100
×
14.74 × 103
1
= 15.4%
=
Unit I.indd 189
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(2009) PAPER 2
Question 1
(a) (i)
F=
∆p
t
(ii) Impulse is the change in momentum.
Impulse = Ft = mv – mu
(b) (i) Initial weight = 1.8 × 9.8
= 17.64 N
(ii) Final weight = 0.40 × 9.8
= 3.92 N
(iii) Mass of fuel burnt = 1.8 – 0.4
= 1.4 kg
1.4
Time taken =
0.25
= 5.6 s
(b) (i) On graph page.
(ii) On graph page.
(iii) L
ift off just takes place when the lift force is just equal to the
total weight. This happens after 0.8 s.
(iv) Area under the curve shaded but above the line.
Unit I.indd 190
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Force
F/M
20.0
15.0
10.0
5.0
0
Unit I.indd 191
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
Time, t /s
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192 c a p e p h y s i c s - U n i t 1
(v) Change in momentum = Area under graph and above the line
Each box = 1 × 0.5 = 0.5 N.S
There are approximately 25 boxes under the curve and above
the line.
Therefore, estimated change in momentum = 25 × 0.5
≈ 13 NS (± 2.0 N.S)
Note: Change in momentum occurs only after take-off, so it is
NOT the total area under graph, but that area that corresponds
to motion after take off.
Unit I.indd 192
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193
Question 2
(a) (i) A periodic motion is one that repeats itself after some time period.
(ii) A
motion is simple harmonic if its acceleration is directly
­proportional to its displacement from some fixed point AND the
acceleration is always directed towards that fixed point.
(b) Practical procedures:
1.
2.
3.
4.
5.
a = –ω 2x
Get a rod of about 15 m long.
Put markings on it (graduations).
lace the rod in a section of the harbour which is calm,
P
i.e., sheltered from the waves.
Make sure the rod touches the bottom of the sea bed.
Take readings on the rod every hour over a 24-hour period.
(c) (i) Minimum depth of water = 9 – 5 = 4 m
(ii) Calculations for t1 and t2:
11.5 = 9 + 5 Sinωt
5 Sin ωt = 11.5– 9
= 2.5
2.5
Sinωt =
= 0.500
5
π
5π
ωt = or
6
6
So
t1 =
π
÷ 1.45 × 10−4
6
= 3.6 × 103 s
=1 hr
5π
t2 =
÷ 1.45 × 10−4
6
= 1.8 × 104 s
= 5 hrs
(iii) T
he length of time when the depth will be more than 11.5 m is
(5 – 1) = 4 hrs.
Unit I.indd 193
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194 c a p e p h y s i c s - U n i t 1
Question 3
(a)
ho = −50 mm
h100 = +220 mm
ht = +105 mm
θt =
=
ht − h0 100
×
h100 − h0
1
105 − ( −50) 100
×
220 − ( −50)
1
155
× 100
270
= 57.4°C
=
(b) (i)
By definition, the unit of temperature, the kelvin is defined
1
of this triple point temperature. This triple point is
as 273.16
­fundamental to all scales.
(ii) (a)
pT × 105 Pa
ptr × 105 Pa
pT
ptr
2.858
4.337
0.6590
2.294
3.480
0.6592
1.765
2.677
0.6593
1.195
0.6598
0.6595
0.6598
1.812
1.000
(b) 1.The pressures are determined by reading off the
heights of mercury. There can be errors in the read off.
2.Also, the pressure at the triple point should be
small. At higher pressure, the value of Ttr will not be
­constant. It may vary slightly (decreases).
3.
Unit I.indd 194
The gas is not ideal.
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195
(c) On graph page. A straight line accepted as well.
(d) When ptr = 0,
pT
= 0.6601
ptr
This means that at very low pressure for the triple point
Ttr = 273.16 k.
(e) From the equation
When
T PT
=
Ttr Ptr
PT
= 0.6601
Ptr
(f) °C = T − 273.15
= 180.31 − 273.15
= −92.84°C
Unit I.indd 195
T = 0.6601 × 273.16
= 180.31 K
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196 c a p e p h y s i c s - U n i t 1
PT
Ptr
0.6601
0.6600
0.6599
0.6598
0.6597
0.6596
0.6595
0.6594
0.6593
0.6592
0.6591
0.6590
0
1
2
3
4
4.5
Ptr × 105
Unit I.indd 196
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Question 4
(a) (i) Energy is the ability to do work.
Kinetic energy is energy a body possesses by virtue of its
1
­motion, KE = mv 2
2
Gravitational potential energy is energy a body possesses by
virtue of its height above some reference position.
G.P.E = mgh
(ii)
A
X
B
v=0
t=0
t=t
v=v
Consider a body of mass m moving (accelerating) uniformly
from A to B.
a=
v −0 v
=
t
t
So, force on body = ma =
mv
t
Distance moved, x, is given by
1
S = ut + at 2
2
1 v
= 0 +   t2
2 t 
1
= vt
2
Work done = Force × Distance moved in the direction of force
mv 1
× vt
2
t
1
= mv 2
2
=
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198 c a p e p h y s i c s - U n i t 1
(iii) But work done is energy acquired.
So KE acquired =
1 2
mv
2
(b) (i) If all the G.P.E is converted to KE at Q,
1
then mgh = mv 2
2
v = 2 gh
= 2 × 9.8 × 10
= 14 ms −1
(ii) V = r ω
V
ω=
r
14
=
= 1.4 rad s −1
10
(c) (i)
T
mg
mv 2
r

142 
= 75  9.8 +
10 

(ii) T = mg +
= 2200 N
Unit I.indd 198
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Question 5
(a) (i)
Accommodation is the ability of the eye to change the focal
length of the lens in the eye so as to see clearly far and near
­objects.
(ii) F
or a given accommodation, the eye is able to see clearly an
object slightly nearer and slightly further than some fixed point.
This variation in distance through which the eye can still see
clearly for a given accommodation is called “depth of focus”.
(b) (i) This defect is called “long sight” or “hypermetropia”.
(ii)
Normal
near point
(iii)
Apparent
near point
Use a convex lens
(c)
So
Unit I.indd 199
1
f
1
f = = 0.5 m = 50 cm
2
D=
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200 c a p e p h y s i c s - U n i t 1
(i)
1 1 1
= +
f u v
1 1 1
= −
v f u
1
1
1
=
−
=−
50 25
50
Therefore, v = 50 cm on same side of the lens as object. So near
point is 50 cm.
(ii)
1
1 1
=
+
50 40 v
1 1
1
=
−
v 50 40
4 −5
1
=
=
200 200
∴ v = 200 cm
Therefore, new near point is 200 cm.
(iii) Since the image is virtual,
1 1 1
= −
f u v
1
1
=
−
25 200
7
=
200
1
when f is in metres.
f
7
P=
× 102
200
= 3.5 D
P=
So
Unit I.indd 200
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201
Question 6
(a) (i) ∆Q = ∆u + ∆w
∆Q ⇒ Heat supplied to the system.
∆u ⇒ Increase in internal energy of the system.
∆w ⇒ Work done by the system on the surroundings.
(This format applies to an expansion with the +ve sign.)
(ii) A
t constant volume, all the energy supplied goes to increase the
internal energy whereas, at constant pressure, heat supplied
must increase the internal energy as well as do work on the
­surroundings. Hence CP > Cv.
CP – Cv = R
(b) (i) Using point P = 2 × 105 Pa, V = 0.005 m3
at T = 500 k (i.e., point B on graph),
PV = nRT
2.0 × 10 × 5 × 10−3 = n × 8.31 × 500
5
2.0 × 105 × 5 × 10−3
8.31 × 500
= 0.241 moles
n=
(ii) (a) From C to A, there is no change in volume.
So
Unit I.indd 201
∆Q = ∆u = nCv ∆T
3
= 0.241 × R × (500 − 200)
2
= 750 J
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202 c a p e p h y s i c s - U n i t 1
(b) From C to B,
∆Q = nC P ∆T
3

= 0.241 ×  R + R × (500 − 250)
2

5
= 0.241 × × 8.31 × 250
2
= 1300 J
(iii) Using the volume from (a) and (b),
1300 = ∆u + p∆v
where the amount of heat needed to raise the temperature of
the 0.241 moles of gas is 750 J (increase in internal energy for
temperature difference of 250 k).
So
p∆v = 1300 – 750
= 550 J
Alternately:
If ∆w = p∆v is used from values on the graph,
p∆v = 2.0 × 105 × (0.005 − 0.0026)
= 480 J
The graph lines in the graphs given in the question are too thick,
so tolerance will be given.
Unit I.indd 202
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(2010) PAPER 2
Question 1
(a) (i)
3.5
V/ms-1
4
3
2
1
0
−1
−2
−3
−4
0.5
0.1
1.5
2.0
t /s
Unit I.indd 203
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204 c a p e p h y s i c s - U n i t 1
(ii) T
he straight line graph shows the ball had constant acceleration
down the plane (and upwards also). It struck the block with a
velocity of 4.2 ms–1 and rebounded with a velocity of 3.4 ms–1.
(iii) (a)Acceleration down the plane = slope of graph while
moving down
3.5 − 0
1.0 − 0
= 3.5 ms −2
Slope =
(b) The length of the incline = Area under graph up to point of
contact with block
1
× 4.2 × 1.2
2
= 2.52 m
Area =
(c) Change in momentum = (0.6)( −3.4) − (0.6)( 4.2)
= 0.6 × ( −7.6) = −4.56 N.s
Time of contact = 0.05 Seconds
So Mean force =
−4.56
= −91.2 N
0.05
That is the force acts upwards on the ball.
(iv) T
he collision is not elastic since the rebound velocity is less
than the initial velocity at contact. That is kinetic energy is
not ­conserved.
Unit I.indd 204
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Question 2
(a) (i) Figure 2: Diffraction grating
R1
V1
White light
V1
R1
(ii) Figure 3: Prism
R
White
V
R
V
(iii) Figure 4: Glass block
White
White
Unit I.indd 205
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206 c a p e p h y s i c s - U n i t 1
(b) (i)
θ1
sin θ1
42.1°
0.67
31.0°
43.6°
0.52
θ2
sin θ2
75.2°
0.97
48.6°
0.69
90.0°
0.75
1.00
The value of the critical angle of the glass. 43.6°
(ii) T
otal Internal reflection takes place when angle of incidence in
slower medium is greater than critical angle.
0.65 − 0
0.45 − 0
= 1.44
(iii) n = gradient of graph =
Unit I.indd 206
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207
sin q2
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
Unit I.indd 207
0.1
0.2
0.3
0.4
0.5
0.6
0.7
sin q1
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208 c a p e p h y s i c s - U n i t 1
Question 3
(a) (i)
Load
Load
Load
Extension
Extension
Extension
Steel wire
Glass
Polymeric material
(ii) Stress is defined as force per unit cross-sectional area.
F
Stress = measured in Nm–2 or Pa.
A
Strain is the ratio of the extension to the original length.
e
No units. Stress = .
l
(b) (i)
Load, M/kg
0
Length, L/cm
Extension, ΔL/m
(ii) On graph page.
5.0
0
0.1
5.6
0.006
0.2
6.2
0.012
0.3
6.9
0.019
0.4
7.8
0.028
0.5
10.0
0.050
Mg × 5 × 10−2
∆L × 4.5 × 10−6
M 9.8 × 5 × 104
=
×
4.5
∆L
M
= 1.09 × 105
∆L
(iii) E =
(
Unit I.indd 208
)
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209
If S is slope
E = (1.09 × 105) × slope
(iv)
Slope =
So
Unit I.indd 209
0.23 − 0
(15 − 0) × 10−3
= 15.3 Kg m −1
E = 15.3 × 1.09 × 105
= 1.67 × 106 Pa
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210 c a p e p h y s i c s - U n i t 1
kg
0.5
0.4
0.3
0.2
0.1
0
Unit I.indd 210
10
20
30
40
50
∆L/M × 10−3
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211
Question 4
(a) (i)The conditions necessary for a body to be in equilibrium
are . . . 1. The resultant force on the body must be zero and
2. The ­resultant torque on the body must be zero.
(ii)
26°
T
F
150N
Resolving vertically:
T cos 26 = 150
150
T=
cos 26
= 167 N
Resolving Horizontally:
F = T sin26.
= 167 × ln26
= 73.2 N.
(b) (i)Vertical and Horizontal component of initial velocity is u sin θ
and u cos θ
Horizontal velocity = 1.6 cos 20 = 1.5 ms–1 (this stays constant)
Initial vertical velocity = 1.6 sin 20 = 0.55 ms–1
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212 c a p e p h y s i c s - U n i t 1
1
(ii) − H = u sin θ t − gt 2
2
1
⇒ −2.2 = 1.6 sin20°t − ⋅ 9.8 ⋅ t 2
2
⇒ −2.2 = 0.5472 t − 4.9t 2
(Taking g = 9.8 ms )
−2
⇒ 4.9t 2 − 0.5472t − 2.2 = 0
t=
0.5472 ±
( −0.5472)2 − 4 (4.9)( −2.2)
2 × 4.9
0.5472 ± 0.2994 + 43.12
9.8
0.5472 + 6.5893
0.5472 − 6.5893
or
=
9.8
9.8
= 0.7282
= −0.6165
as, time can't be negative, so these
≈ 0.73 sec
result can't be considered.
=
∴ the time taken by the boy to reach the ground is 0.73 sec.
(iii) the horizontal range (R) is given by,
R = u cosθ × t
= 1.6 × cos20° × 0.73
= 1.0976 m
So, the boy would land 1.0976 m horizontally far from the truck.
Unit I.indd 212
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213
Question 5
(a) (i)The threshold of hearing is the minimum intensity that can be
heard by a normal person. This is taken as 10–12 Wm–2.
The threshold of pain is the minimum intensity at which pain is
felt on the ear drum.
(ii) M
ost of the human senses machinery are logarithmic in
­response to an input (They obey Weber-Fetcher Law).
The ­decimal scale is a logarithmic scale and so is most suited.
I
(iii) β = 10 log  
 Io 
 3.82 × 10−3 
(iv) β = 10 log 
 10−12 
(
= 10 log 3.82 × 109
= 10{9 + 0.58}
= 95.8 dB.
Threshold of pain
Intensity (W/m2)
1
120
10−2
100
10−4
80
10−6
60
10−8
40
10−10
20
Threshold of hearing
10−12
20
50
100
200
Intensity level (dB)
(v)
)
0
500 1000 2000 5000 10000
Frequency (Hz)
Age related hearing loss (Presbycusis) occurs as a consequence
of deterioration in the middle and inner ear. This results in
­frequencies in the lower and upper ends of the hearing range
not able to be heard (reduced to about 16 KHz). Also the
­threshold of hearing will increase. That is the sound will have
to be louder for the aged person to hear.
Unit I.indd 213
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214 c a p e p h y s i c s - U n i t 1
(b) (i)The incident waves reflect off the wall and so set up standing
waves.
Where the incident waves and reflected waves meet
in half wavelength, nodes are formed. This gives minimum
­intensity. The reason why it is not an absolute silence is because
the reflected wave will not have the same intensity ­(amplitude/
loudness) as the incident wave. So there will not be total
­cancellation.
(ii) B
etween the speaker and the wall there will be 4 half wave
lengths.
So
So
1 
4  λ  = 2.25 m
2 
λ=
2.25
= 1.125 m
2
v= fλ
v
330
= 293 Hz.
f= =
λ 1.125
(iii) When f = 165 Hz and V = 330 ms–1
λ=
v 330
=
= 2 m.
f 165
The last maximum (antinode) from the wall will be 1.5 m away.
(In practice, at that frequency and distance, standing waves will
not be set up. The distance will have to be in 1 m increments.)
Wall
0.5 m
1.5 m
2.25 m
Unit I.indd 214
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215
Question 6
(a) (i)The hot body radiates electromagnetic waves in all directions.
A distant object will absorb this radiation (energy) and cause
the kinetic energy of the molecules of the receiving object to
increase. The temperature of a body is determined by the kinetic
energy of its molecules. Hence the body gets warm.
(ii) T
he green house effect: The shorter wavelength radiation
from the sun can penetrate the cloud of green house gases and
­enter the earth’s atmosphere. The earth gets warm. The ­longer
­wavelength re-radiated waves cannot penetrate the green
house gases on their way out, and so gets trapped in the earth’s
­atmosphere. Hence the earth becomes warmer and warmer.
(b) A = 4.6 m2
650°C
(i)
4.0 × 10⁻3 m
647°C
Q
∆θ
= KA
s
∆x
80.4 × 4.6 × 3
=
4.0 × 10−3
= 2.77 × 105 J/s
(
(ii) Rnet = σ A T14 − T24
)
(
= 5.67 × 10−8 × 4.6 9204 − 3034
)
= 1.85 × 105 J/s.
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216 c a p e p h y s i c s - U n i t 1
(iii) 2.77 × 105 Joules per second passes through the stove and
1.85 × 105 Joules per second net is lost by radiation. So the
heat lost per second by a combination of conduction and
­convection is
= 2.77 × 105 – 1.85 × 105
= 9.2 × 104 J/s
Unit I.indd 216
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(2011) PAPER 2
Question 1
(a) (i)
Fd
u
(ii) If upthrust is neglected
mg
Then Net force on body = (mg – F)
Net force
m
mg − bv n
=
m
bv n
a= g−
m
n
bv
⇒ g−a=
m
So acceleration, a =
Unit I.indd 217
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218 c a p e p h y s i c s - U n i t 1
(b) (i)
Acceleration
a/ms–2
Velocity
v/ms–1
g – a/ms–2
lg(g – a)
lg v
9.41
10
0.39
–0.41
1.00
8.24
20
1.56
0.19
1.30
8.91
15
7.36
25
6.28
30
5.00
(ii) On graph page.
35
0.89
–0.05
2.44
0.39
4.80
0.68
3.52
0.55
1.18
1.40
1.48
1.54
(iii) The equation of the line is
 b
log ( g − a) = n log v + log  
 m
So
0.70 − 0.0
= 2.0
1.55 − 1.20
n = 2.0
Slope =
(iv) At terminal velocity a = 0.
So
So
Unit I.indd 218
 0.251 
log g = 2 log v + log 
 78.5 
0.99 = 2 log v − 2.50
2 log v = 3.49
log v = 1.745
v = 55.6 m/s
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219
0.7
log ((g – a)/ms–2)
0.6
0.5
0.4
0.3
0.2
0.1
0.0
1.0
1.1
1.2
1.3
1.4
1.5
1.6
log (v/ms–1)
−0.1
−0.2
−0.3
−0.4
Unit I.indd 219
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Question 2
(a) (i)A loud sound would be heard. At this level resonance takes
place between the frequency of the AF generator and the
­natural frequency of that column of air. This leads to a maximum
­amplitude.
(ii) Resonance.
(b) (i)
Frequency of
Fork, f (Hz)
Length of
String, l (m)
 1
 
l
m −1
256
288
320
384
450
512
0.781
0.695
0.625
0.521
0.444
0.391
1.780
1.439
1.600
1.919
2.222
2.558
(ii) On graph page.
(iii)
1
= 2.400
0.417
f = 480 Hz
So
550 − 250
2.75 − 1.25
= 200 Hz m or ms −1
(iv) Slope =
(v) From equation slope =
T
= 200
4µ
T
= 40000
4µ
µ = 6.25 × 10−4 kg m −1
Unit I.indd 220
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221
700
650
Frequency,f(Hz)
600
550
500
480
450
400
350
300
250
200
1.0
1.25
1.5
2.0
2.5
2.75
3.0
1/Length (m-1)
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Question 3
(a) S pecific Latent heat of fusion of a substance is the amount of heat
needed to change 1 kg of the substance from solid to liquid without
a change in temperature.
(b) (i) Melting point 272 K, Boiling point 430 K.
(ii) Gradient P: Slope =
Gradient Q: Slope =
272 − 230
= 21 k min −1
2.0 − 0.0
430 − 272
= 11.3 k min −1
27 − 13
For solid 2 × 1.0 × 105 = 2 × c × (272 – 230)
c = 2381 J/kg/k
(iii) H = mc ∆ A
1.0 × 105 = 2 × c × slope. c = 4425 J/kg/K
For liquid
So SHC of liquid is greater.
(iv) SHC in liquid state
Heat supplied = p × ∆t = MC ∆θ
So
∆θ
∆t
C = 4425 J/kg/K.
1.0 × 105 = MC ×
∆θ
is slope
∆t
(v) Heat supplied during melting = (13 − 2) × 1.0 × 105
So
Unit I.indd 222
ML f = 1.1 × 106
= 1.1 × 106 J.
1.1
× 106
2
= 5.5 × 105 J/kg.
Lf =
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223
450
430
400
T/K
350
Q
300
272
P
250
230
210
200
0
5
2
Unit I.indd 223
10
15
13
20
25
t /min
30
35
40
45
27
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224 c a p e p h y s i c s - U n i t 1
Question 4
(a) (i)Kinetic energy is energy a body possesses by virtue of its
1
­motion. K.E = mv 2 .
2
Gravitational potential energy is energy a body possesses by
virtue of its position above some references G.P.E = mgh.
(ii) G
.P.E is taken from a reference level. Below is negative. KE is
­energy a body possesses once it is moving. Even if v is negative,
v2 will be +ve.
(b) (i)
1 2
mv = mg∆h
2
v = 2 g∆h = 2 × 9.80 × 12 = 15.3 m/s.
(ii) For vertical motion:
1
s = ut + at 2
2
1
10 = 0 + × 9.8 × t 2
2
20
= t 2,
t = 1.43 s
9.8
s = 10 m
u = 0 m/s
a = 9.8 m/s2
t =?
(iii) R = v × t since horizontal velocity remains constant
= 15.3 × 1.43
= 21.9 m.
(c) Assumptions:
Unit I.indd 224
1.
2.
3.
All change is G.P.E was converted to KE.
There is no friction between ski and ground.
Effects of air resistance is negligible.
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225
Question 5
(a) (i)Refraction is the bending of rays as they pass from one medium
to another. This is as a consequence of change in speed.
Diffraction is the spreading of waves as they pass though “small”
openings or around “small” obstacles.
air
l
l
glass
Refraction
(ii) Role of Diffraction
Diffraction
For each slit, spreading of the waves takes place.
l
l
Diffraction takes place because the size of the slits is about
the size of the wavelength of the light waves. Each slit acts as a
source of waves.
Role of interference:
The wave fronts now have a chance to meet and so interfere
with each other. In directions in which the wave fronts meet in
phase. There is constructive interference (bright fringes) and
the directions in which the waves meet out of phase. There will
be destructive interference (dark fringes).
Unit I.indd 225
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226 c a p e p h y s i c s - U n i t 1
(b) λ1 = 653.3 × 10–9 m
λ2 = 486.1 × 10–9 m
s = 6000 lines per cm n = 2.
(i)
1
× 10−2 m
s
1
=
× 10−2 m
6000
= 1.67 × 10−6 m
d=
(ii) For λ1:
2 × 6.567 × 10−7
= 0.7824
1.67 × 10−6
∴ θ1 = 51.5°
sin θ1 =
For λ2:
2 × 4.86 × 10−7
= 0.5821
1.67 × 10−6
∴ θ2 = 35.6°
sin θ2 =
(iii) Angular separation = (51.5 – 35.6) = 15.9°
Unit I.indd 226
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227
Question 6
(a)
T = 315 K
PV = nRT
n=
V1 = 2.90 × 10–4 m3,
PV 1.03 × 105 × 2.90 × 10−4
=
8.31 × 315
RT
= 0.0114 Moles
P1 = 1.03 × 105 Pa
(b) (i) From ∆u = ∆Q + ∆w for compression i.e. work done on the gas.
If ∆Q = 0, i.e. no heat added, then ∆u = ∆w. i.e. there is going to be
an increase in internal energy equal to the work done on the gas.
For an ideal gas increase in internal energy is totally k
­ inetic and
hence change (rise) in temperature.
(ii) C
ompressing the gas means applying a force on the molecules.
This force gives the molecules an acceleration and hence
­increase in velocity. Increase in velocity, and so increase in
­Kinetic Energy gives rise to an increase in Temperature.
(c)
P1V1 P2V2
=
T1
T2
P2 =
P1V1T2
V2T1
1.03 × 105 × 2.9 × 10−4 × 790
=
3.5 × 10−5 × 315
= 2.14 × 106 Pa
Unit I.indd 227
P1 = 1.03 × 105 Pa
V1 = 2.90 × 10−4 m3
T1 = 315K
P2 = ?
V2 = 3.5 × 10−5 m3
T2 = 790 K
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228 c a p e p h y s i c s - U n i t 1
(d) I f the cylinder is thermally isolated from the surroundings then
∆Q = 0.
∆u = ∆Q + ∆w
So
∴
∆w = ∆u
So increase in internal energy = 90 J.
(e) ∆u = nC ∆T
= .0114 × C × (790 − 315)
90
.0114 × 475
= 16.6 J mol −1 K −1
C=
Unit I.indd 228
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(2012) PAPER 2
Question 1
(a) (i)Displacement is distance moved in a specified direction.
This is a vector quantity. It is measured in metres.
(ii) V
elocity is the rate of change of displacement. This is a vector
quantity and is measured in ms–1, slope of a d-vs-t graph at an
instant in time.
(iii) A
cceleration is the rate of change of velocity. This is a vector
quantity and is measured in ms–2, slope of v-vs-t graph at an
instant in time.
(iv) Kinetic energy is the energy a body possesses by virtue of its
1
motion. This is a vector and is measured in Joules, KE = mc 2 .
2
(b) (i) (4, 40) and (1, 5)
(ii) (a)
So
and
Unit I.indd 229
t
X = x0  
 t0 
40 = x0 ( 4)
n
5 = x0 (1)n
40
= 4n
5
8 = 4n
n log 4 = log 8
n = 1.5
n
(1)
(2)
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230 c a p e p h y s i c s - U n i t 1
(b) x0:
5 = x0 (1)1.5
x0 = 5
x 
(iii) x =  1.50  × t 1.5
 t0 
=
V=
5 1.5
t
11.18
dx 1.5 × 5 0.5
t
=
dt 11.18
When t = 30 s, V = 3.67 ms–1
Unit I.indd 230
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Question 2
(a) T = 2π
m
K
(b) (i)
Unit I.indd 231
0
1
2
3
4
Amplitude, y(cm)
5
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
f /Hz
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232 c a p e p h y s i c s - U n i t 1
(ii) (a) When amplitude is maximum
f = 12.5 Hz
w = 2π f = 2π × 12.5
= 78.6 rad/s
1
f
1
=
12.5
= 0.08 s
(b) T =
K
(c) F
or springs in series K ′ = where K is the spring constant for one
2
spring.
So
T = 2π
m
k′
0.05 × 2
K
K = 617 N cm −1
0.08 = 2π
(d) (i) See graph page.
Note: The maximum amplitude occurs at a slightly lower
­frequency.
(ii) T
he phenomenon is called Resonance. This is advantageous
where the frequency of the external period force is equal to
(or a multiple of) the natural frequency of the swing.
Unit I.indd 232
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233
Question 3
(a) ∆Q = ∆u + ∆w
∆Q ⇒ is heat added to the system.
∆u ⇒ is increase in internal energy.
∆w ⇒ is work done by the system on the surroundings.
(b) (i) On graph page.
(ii) Work done = Area under graph
1
1
Each box = × 10−3 × × 105
4
4
= 6.25 J
These are approximately 100 boxes
So W.D.(Work Done) = 625 J
(Students can also use the trapezoidal Rule).
(iii) PV = nRT
Take P = 4.0 × 105 Pa
V = 1.0 × 10−3 m3
PV 4.0 × 105 × 1.0 × 10−3
n=
=
8.31 × 300
RT
= 0.16 moles
Unit I.indd 233
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Unit I.indd 234
0
0
1
2
3
4
P × 105/Pa
1
2
3
4
5
V × 10–3/m3
234 c a p e p h y s i c s - U n i t 1
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235
(iv) I f the temperature was kept constant then the change in internal
energy will be zero.
(v) From
If
and
∆Q = ∆u + ∆w
∆u = 0
∆w = 625 J
Then ∆Q, Heat supplied = 625 J.
Unit I.indd 235
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236 c a p e p h y s i c s - U n i t 1
Question 4
(a) N
ewton’s 2nd Law states that the rate of change of momentum of a
body is directly proportional to the net force applied and takes place
in the direction of the force.
(b) (i)
Fd
Fb
mg
(ii) A
viscous medium is one that offers resistance to motion as a
body moves through it.
(c) (i)
m = volume × density
= 0.002 × 2500
= 5 kg
(ii) mg = 5 × 9.81 = 49.1 N
So Buoyant force = 49.1 – 30 = 19.1 N
(iii) Initial acceleration =
Unit I.indd 236
30
= 6 ms–2
5
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237
(iv) Terminal velocity occurs when Fd + Fb = mg
So
(d) (i)
5VT + 19.1 = 49.1
5VT = 30
VT = 6 ms −1
V
VT
t
(ii)
a
t
Acceleration is slope of
v-vs-t graph
Unit I.indd 237
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238 c a p e p h y s i c s - U n i t 1
Question 5
(a) (i)Diffraction is the spreading of waves as they pass through
“small” openings or around “small” obstacles. Small means
about the size of the wavelength of the waves.
Refraction is the bending of waves as they go from one medium
to another. This is as a consequence of change in speed.
(ii) I f the width of the aperture is about the size of the wavelength of
the waves, then there is significant diffraction. If the width of the
aperture is very wide compared to the wavelength of the waves
then the diffraction is not significant.
Narrow
slit
l
Note: l is constant
Wide slit
(iii) When pass through a diffraction grating, the spreading takes
nλ
place at angles given by the equation sin θ =
, n and d are
d
constants and λ is the wavelength of the wave.
Red has the longest wavelength in the visible spectrum while
violet has the shortest wavelength, so when white light passes
through a diffraction grating, the colours separate out giving a
spectrum.
1
1
=
m
s 4.55 × 105
= 2.2 × 10−6 m
(b) d =
d sin θ = nλ
λv = 2.2 × 10 sin11.8
1
Unit I.indd 238
−6
−7
= 4.50 × 10 m
s = 4.55 × 105 lines/m
θv = 11.8°
1
θr = 15.8°
1
n = 1 for both
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239
λr = 2.2 × 106 sin 15.8
1
= 6.0 × 10−7 m.
(c) At θ = 54.8°, d sin θ = 1.7977 × 10–6 m
For Violet, n =
For Red, n =
1.7977 × 10−6
=4
4.5 × 10−7
1.7977 × 10−6
=3
6.0 × 10−7
This means that the 3rd order red will overlap with the 4th order
violet.
Unit I.indd 239
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240 c a p e p h y s i c s - U n i t 1
Question 6
(a) (i)The property is determined at the melting point of pure m
­ elting
ice (P0) and the property taken in steam from water boiling at
­normal atmospheric pressure (P100) and then the proper at the
­unknown temperature Pθ.
Then use the formula:
θ ° centrigrade =
Pθ − P0
× 100.
P100 − P0
(ii) T
he absolute thermodynamic scale is a theoretical scale with
one fixed point, the triple point of water. This scale does not
­depend on the physical property of any substance. The centigrade scale depends on the physical property of the ­substance
used, and so if the same temperature is measured ­using
­different properties, different answers can be obtained, since
the property may not vary in the same way or even ­linearly.
(iii) – Mercury does not wet the glass (cling to the sides).
(b) (i)
– M
ercury gives a wider range than most other liquids in the lab
(–39°C to 350°C).
950 − 3750
× 100
215 − 3750
−2800
=
× 100
−3535
= 79.2° centigrade
θ °C =
P0 = 3750 Ω
P100 = 215 Ω
Pθ = 950 Ω
(ii) T
he two properties do not vary the same way with temperature.
The variation of the property may not be linear.
(c) (i) P = IV = 120 × 10 = 1200 W
(ii) At steady state, all heat supplied goes to heat the water
Heat supplied/s = 1200 J/s
0.3
kg/s
Mass of water flowing/s =
60
Unit I.indd 240
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So
241
0.3
× 4200 × ∆θ
60
∆θ = 57.1°C
1200 =
So final temperature = 57.1 + 20 = 77.1°C.
(iii) A
t steady state the apparatus does not experience a change in
temperature.
i.e. Heat supplied = mc∆θ for water + C ∆ θ for a apparatus.
If ∆θ = 0 for apparatus, then C, Heat capacity for apparatus is not
needed.
Unit I.indd 241
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(2003)
(2013) PAPER 1
2
Question 1
(a) (i) This assumes falling with no resistance.
Using V = u + at
Velocity v/m s–1
Time t/s
0
0
9.80
1.0
29.40
3.0
19.60
39.20
(ii) On graph page.
2.0
4.0
(iii) Distance travelled = Area under graph
1
= × 4 × 39.2
2
= 78.4 m.
(b) (i) Vxi = Vi cos θ
Vyi = Vi sin θ
(ii) x = (vi cos θ) t(1)
1
y = (vi sin θ )t − gt 2 (2)
2
1
2
Using S = ut − gt 2 , using down as positive.
Unit I.indd 242
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243
( 2 0 1 3 ) PAPE R 2
V/ms–1
40
36
32
28
24
20
16
12
8
4
0
Unit I.indd 243
0
1
2
3
4
t/s
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244 c a p e p h y s i c s - U n i t 1
(iii) From (1) t =
Sub in (2),
x
v i cosθ
x
1  x 
− g
y = (vi sin θ )
vi cosθ 2  vi cosθ 
= x tan θ −
g
2(vi cosθ )
2
2
× x2
This resembles y = ax – bx2, which is a quadratic in x.
So a parabolic shaped graph.
Unit I.indd 244
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245
Question 2
(a) S imilarity: They both transfer energy from one point to another
­without matter working between the points.
Diff: Longitudinal waves the direction of vibration and ­direction
of travel are the some (parallel). For transverse they are
­perpendicular.
(b) Speed of sound at 10°C = 340 + (10 × 0.61)
= 346.1 m/s
2d = 346.1 × 8.3 = 2872.63
∴ d = 1436 m or 1440 m.
(c) (i)
d metres
t20 seconds
60
7.2
70
8.2
80
90
(ii) On graph page
t20
seconds
20
0.36
0.41
t
seconds
2
0.18
0.21
9.4
0.47
0.24
11.8
0.59
0.30
10.6
100
t=
0.53
0.27
(iii) The equation of the graph is
t
d = v 
 2
So slope of graph will be the speed of sound.
So
Unit I.indd 245
s=
(100 − 60)
(30 − 18) × 10−2
= 333 m/s.
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246 c a p e p h y s i c s - U n i t 1
d/m
100
95
90
85
80
75
70
65
60
55
50
15
Unit I.indd 246
20
25
30
t/2 × 10−2 /s
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247
Question 3
(a) LHS ≡ PV ≡ Nm −2 × m3 ≡ kg m s −2 × m
≡ kg m2 s −2 .
RHS ≡ kg (m s–1)2 ≡ kg m2 s–2
(b)
Same fundamental units.
PV =
So
K=
(c)
1
Nm c 2 = nRT
3
1
Nm c 2 = nRT
3
1
3 nR
3
mc 2 =
T = kT
2
2 N
2
nR
nR
R
⇒
=
⇒ Boltzmann constant.
N
nN A N A
c∝
So
cN =
co =
1
m
k
So
28
k
cN
32
=
co
28
= 1.07
32
(d) Using PV = nRT , V = nRT = 1.12 × R × 273
P
1.01 × 105
V = 0.025 m3
So
PV 1.33 × 104 × 0.025
=
8.61 × 223
RT
n = 0.179 moles.
For new case PV = nRT, n =
Unit I.indd 247
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248 c a p e p h y s i c s - U n i t 1
(e) W.D. = Area under graph
1
 1

=  (6 + 4) × 105 × 1 +  ( 4 + 2.95) × 105 × ( 4 − 2)
2
 2

5
5
= 5 × 10 + 6.95 × 10
= 1.20 × 106 J.
(f) From graph 3.0 m3.
P/N m–2
6
Graph of P vs V for a gas undergoing expansion
A
5
4
B
3
×10
3
C
2
1
0
0
1
2
3
4
V/m3
Unit I.indd 248
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249
Question 4
(a) T
he principle of moments states that if a system is in equilibrium
then the sum of the clockwise moments about any point is equal to
the sum of the anticlockwise moments about that point.
x
greater than 2x
100 kg
50 kg
If the 50 kg person sits greater than twice the distance from the
fulcrum than the heavier person, then this is possible. In this way
the clockwise moments in the diagram shown will be greater than
the anticlockwise moments.
(b) (i)The law of conservation of momentum states that in any
­collision or explosion, the total momentum before is equal to
total ­momentum after provided that no external forces act.
(ii)
m1
u1 = +4 m/s
Before
v1 = –1.5 m/s
After
m2
u2 = –3 m/s.
v2 = +5.5 m/s.
(a) Momentum before = 4m1 – 3m2
Momentum after = –1.5m1 + 5.5m2
If momentum is conserved then 4m1 – 3m2 = –1.5m1 + 5.5m2
4m1 + 1.5 m1 = 5.5 m2 + 3 m2
So
Unit I.indd 249
5.5 m1 = 8.5 m2
8.5
5.5
= 1.55.
m1 : m2 =
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250 c a p e p h y s i c s - U n i t 1
(b) If m1= 8.5 kg, then m2 =
8.5
= 5.48 kg.
1.55
1
1
(8.5)(4)2 + (5.48)( −3)2
2
2
= 68 + 24.66
= 92.7 J
Total K.E. before =
1
1
(8.5)( −1.5)2 + (5.48)(5.5)2
2
2
= 9.56 + 82.89
= 92.45 J
(c) Total K.E. after collision =
From the calculations some K.E. is lost so it will be an
­inelastic collision. However, since the K.E. lost is very small
from the calculations with respect to the values of the K.E.,
it can be considered an elastic collision. A 0.16% error in
the calculations is negligible.
Unit I.indd 250
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251
Question 5
(a) If L = 0.4 m, this represents four half wavelengths (i.e. 2 wavelengths)
so the wavelength of the wave is 0.2 m.
So
v= fλ
= 1200 × .2
= 240 m/s
(b) (i) v ∝ f and f ∝ T. (This relationship is not expected).
So if T is doubled then the velocity will increase by
So just “increase” should be accepted.
2 times.
(ii) If L is doubled then v will be doubled provided that the
­
frequency remains the same.
L is doubled then the two waves now fits in 0.8 m. So λ will be
0.4 m and
v = 1200 × 0.4
= 480 m/s.
(c)
 2π

y = A0 sin 
x + 2π ft 
 λ

Given equation y = 5 × 10–4 sin (740 x + 251300 t).
Comparing equations
(i) Amplitude = 5.0 × 10–4 m.
(ii)
λ
So
Unit I.indd 251
2π
= 740
λ=
2π
= 8.5 × 10−3 m.
740
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252 c a p e p h y s i c s - U n i t 1
(iii) 2π f = 251300
251300
f=
= 4.0 × 104 Hz.
2π
(iv) v = f λ
= 4.0 × 104 × 8.5 × 10−3
= 340 m/s.
(d) T
he hearing range is upto 20 kHz. This is 40 kHz so it is outside the
hearing range. This is a ultra sound. This can be used for imaging
­babies in the womb. Using a transducer, the vibrations of the baby
can be picked up and an image created.
Unit I.indd 252
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253
Question 6
(a)
Density r
h
A
Body X
Consider a body X of area A at a depth h below the surface of a liquid
of density I.
The volume of liquid pressing on area A is V = A h.
So the mass of liquid pressing on A is m = Ahρ.
(mass = volume × density).
Weight of liquid pressing on A is F = Ahρg.
Pressure is defined as P =
So pressure on A =
So
Unit I.indd 253
Force
Area
F Ahρ g
=
A
A
P = hρg.
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254 c a p e p h y s i c s - U n i t 1
(b) (i)
p = hρ g
= 2.5 × 103 × 1.04 × 103 × 9.80
= 2.55 × 107 Pa.
F
(ii) Young’s modulus = A .
e
l
F
= 2.55 × 107 Pa, l = 1 m
A
So
So
69 × 10
9
(2.55 × 10 ) .
Pa =
7
e
l
2.55 × 107
e=
69 × 109
= 0.369 mm.
(c) (i)At point A, the elastic limit is exceeded and the wire goes into
plastic deformation.
At B the wire breaks.
(ii) Strain energy (energy stored) = Area under graph
1
× 50 × 0.72 × 10−3
2
= 1.8 × 10−2 J.
=
Unit I.indd 254
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(2014)
(2003) PAPER 2
1
Question 1
(a) (i) VH = V cos α
= 300 × cos 40
= 230 ms–1
(ii) If total time of flight is 39 seconds.
Then range = 230 × 39
= 8,970 m
or 8.97 km.
(b) (i) 1
(ii) 4
(iii) Loss G.P.E. = mg∆h
= 2 × (0.7)
(c) On graph page.
(d) (i)
= 1.4 J.
0.65 − 0
1.1 − 0
= 0.59
Slope =
0.60 ± 0.05
So Hrebound = 0.59 × Hbefore.
Since the graph is a straight line through the origin.
(ii) when H before = 2.04
H r = 0.59 × 2.04
= 1.20 m.
Unit I.indd 255
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256 c a p e p h y s i c s - U n i t 1
Hrebound / m
1.1
1.0
.9
.8
.7
.65
.6
.5
.4
.3
.2
.1
0
Unit I.indd 256
0
.1
.2
.3
.4
.5
.6
.7
.8
.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7 Hbefore / m
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Question 2
(a) (i) From 20 Hz to 20 k Hz.
(ii) C
ontrolled variables – Same side head phone, constant
­amplitude (loud waves).
1.Put the headphone in the right ear and vary the frequency
up and down to determine the lower and upper cut off
­frequency of hearing.
Repeat for left ear.
I/W m–2
1k
fL
fU
f
frequency
(b) From v = f λ =
λ
T
So
Unit I.indd 257
=
T=λ
=
gλ
2π
gλ
1
and f = .
2π
T
2π
2πλ 2
=
gλ
gλ
2πλ
g
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258 c a p e p h y s i c s - U n i t 1
(c) T =
2π × 0.8
9.80
= 0.51291
= 0.716 S.
T=?
λ = 0.8 m
g = 9.80
(d) (i)
Radio Waves
Visible
X-ray
IR
λ1
uv
λ2
λ3
λ4
Decreasing wavelength
(ii) λ1(3.0 m normal radio) or 10–3 m or 10–2 m. Micro wave is
included.
λ2
λ3
λ4
Unit I.indd 258
7.0 × 10–7 m
4.5 × 10–7 m
1.0 × 10–10 m
(4.0 to 4.5) × 10–7 m
(accept 10–9 m to 10–11 m)
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259
Question 3
(a)
Temperature T
(°C)
Length of Gas
Column (mm)
Volume
V (mm3)
–35
83
498
–3
27
57
87
(b) (i) On graph page.
89
103
110
120
534
618
660
720
(ii) when v = 0, T = –265°C.
(c) No, this will mean infinite density, since the mass is not destroyed.
D=
m
.
v
(d) When length = 70 mm.
V = 6 × 70 = 420 mm3.
This corresponds to a temperature of –55°C.
So Temperature in kelvin
T
= −55 + 273
k
= 218 k ( ±10 k ) .
If –265°C is used as 0 k then
T
= 210 k.
k
Unit I.indd 259
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260 c a p e p h y s i c s - U n i t 1
(e) (i) PV = nRT = NkT.
 nR 
nR
(ii) V =   × T . If T is in kelvin then V ∝ T and slope is
.
 p
p
V=
−325 −300
nR
(θ + 273) .
p
−275
−250
−225
−200
−175
−150
−125
−100
−75
−50
−25
750
700
650
600
550
500
450
400
350
300
250
200
150
100
50
0
0
V/mm
25
50
75
100
T / °C
Unit I.indd 260
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261
Question 4
(a) (i)Acceleration is rate of change of velocity. If the speed is
­constant and the direction is changing then the velocity is
changing (a vector quantity). Hence the body is accelerating.
v2
(ii) a =
R
LHS ≡
F N
≡
≡ ms −2
M kg
(
RHS ≡ m s −1
)
−2
÷m
≡ m2 s −2 m −1
≡ m s −2 .
(b) During straight road
v 2 = u2 + 2as
= 0 + 2 × 6 × 150
= 1800
v = 42.4 ms −1
v=?
u=0
a = 6 m/s2
s = 150.
mv 2
Magnitude of force =
r
500 × 1800
=
200
= 4500 N
F
a
This force is directed towards the centre of the circle path.
Unit I.indd 261
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262 c a p e p h y s i c s - U n i t 1
(c) (i) (a)–(b)
Highest point
mg
T
T
Lowest point
mg
(ii) At the bottom (lowest point). At this point tension is greatest.
mv 2
mv 2
+ mg. At top T =
− mg.
T=
r
r
(iii) When the string breaks T = 20 N
So
20 v 2
= + g = rω 2 + g
m r
 20

− 9.8


0.5
ω2 =
0.5
= 60.4
ω = 7.77 rad s −1 .
Unit I.indd 262
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263
Question 5
(a) Coherent means:
1. Having the same frequency.
2. Same wave length.
3. Constant phase difference.
4. Same velocity.
5. Same periodic time.
(b) (i)
Screen
B
y
P
Q
q
q
O
d
R
l
N
D
From ∆BOQ tan θ =
From ∆PMR sin θ =
y
.
D
λ
.
a
For small θ sin θ ≈ tan θ .
So
Unit I.indd 263
y λ
λD
= ⇒ y=
.
D a
a
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264 c a p e p h y s i c s - U n i t 1
(ii) 1.4 × 10−3 =
λ × 0.5
a = 0.2 × 10−3 m
2.0 × 10−4
D = 0.5 m
1.4 × 10−3 × 2 × 10−4
λ=
−3
0.5
y = 1.4 + 10
= 5.6 × 10−7 m
= 560 nm.
(c) (i)
Screen
Grating
27.7
2
= 13.85°.
θ=
B1
q
q
B1
d sin θ = nλ
For n = 1
d=?
λ = 5.89 × 10−7 m
1 × 5.89 × 10−7
nλ
=
sin θ
0.2394
−6
= 2.46 × 10 m
d=
So number of slits/m =
1
= 4.065 × 105/m
d
So there will be 4.07 × 102 slits/mm or 407 lines/mm.
Unit I.indd 264
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265
(ii) For maximum number of orders.
sin θ =
So
nλ
≤ 1.
d
n≤
d
λ
2.46 × 10−6
5.89 × 10−7
≤ 4.18
≤
∴ Maximum number of orders is 4.
Unit I.indd 265
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266 c a p e p h y s i c s - U n i t 1
Question 6
(a) C
onduction: The copper pipes are good conductors of heat. So the
heat absorbed by the pipe is easily conducted to the water. The
­Styrofoam prevents conduction. Convection takes the water that
is heated to the storage tank (natural or by pump). Also heat is
­transmitted from hot water to colder water by convection.
Radiation: Heat reaches the blackened copper plates by radiation
from the sun. Black surfaces are best absorbers of heat.
Green house effect: The shorter wavelength infrared radiation from
the sun can penetrate the glass to pipes. Heat reradiated from the
black surface and the hot pipes are of a longer wavelength and
­cannot ­penetrate the glass to leave. Hence the heat is trapped giving
rise to the green house effect.
(b) (i)
Brick
Foam
Brick
Temp/°C
30°
q1
q2
–5°
K = 0.48
K = 0.016
x across the wall
(ii) The conductivity of the brick is 30 times better than that of the foam
 0.4p 

 . So it will take 30 times greater thickness of brick for same
0.016 
thickness of foam. So for 5 cm of foam, thickness of brick will be
30 × 5 = 150 cm of brick.
Unit I.indd 266
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267
(iii) The wall can be considered as made up of 170 cm of brick.
So
Q
∆θ
=k A
s
∆x
35
×1
170 × 10−2
Rate/m2 = 9.88 J/s/m2 .
Unit I.indd 267
= 0.48 ×
2/2/2016 5:22:27 PM
(2003)
(2015) PAPER 1
2
Question 1
(a) F
or any two bodies in space, there is a force of attraction between
them that is directly proportional to the product of their masses and
inversely proportional to the square of their distance apart.
(b) (i)
F =w=
GMm 6.67 × 10−11 × 6.42 × 1023 × 1.4
=
2
R2
3.395 × 106
(
)
= 5.20 N
d 6.79 × 106
=
2
2
= 3.395 × 106 m
R=
(ii) Using G.P.E = mg∆h ⋅ ( planet taken as point mass ) .
= 5.20 × 3.395 × 106
= −1.77 × 106 J
OR
G.P.E = –G M m/r
− 6.67 × 10−11 × 6.42 × 1023 × 1.4
3.39 × 106
= −1.77 × 106 J.
=
Unit I.indd 268
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269
(c) (i) On graph page.
(ii) Slope =
(1.0 − 0.2) × 105
(18.0 − 3.4) × 10−15
= 5.48 × 1018 Nm2
Formula relating F =
So
gME m
.
R2
slope = gMEm
m=
slope
5.48 × 1018
=
GME 6.67 × 10−11 × 5.6 × 1024
= 1.47 × 104 kg.
Unit I.indd 269
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270 c a p e p h y s i c s - U n i t 1
W × 105\N
1.2
1.1
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
1/R 2 × 10−15/m−2
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Unit I.indd 270
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271
Question 2
(a) (i) 1.The acceleration must be proportional to the distance from
a fixed point.
2.The acceleration must always be directed towards that
fixed point (a = –ω2 y).
(ii) M
aximum velocity (at equilibrium portion) is v = rω where r is
amplitude and ω = 2πf.
V = 3 × 10–3 × 2πf = 0.0848 ms–1
1
1
2
So Max KE = mv 2 = × 5.8 × 10−3 × (0.0848)
2
2
= 2.09 × 10−5 J.
(iii) a
∝ y i.e. acceleration increases with amplitude (the ­magnitude).
A stage is reached when the acceleration is equal to the
­acceleration due to gravity downwards. At this point the cube
will lose contact with the plate.
(iv) a = ω 2 y
y=
9.80
(9π )2
= 1.23 × 10−2 m
a = 9.80 ms–2
ω = 9π rad s–1
=12.3 mm.
(b) (i) On graph page.
(ii) f for maximum amplitude is 12.7 Hz (from graph).
So
Unit I.indd 271
ω = 2πf = 79.8 rad s–1.
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272 c a p e p h y s i c s - U n i t 1
0
1
2
3
4
5
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
f /Hz
y/cm
Unit I.indd 272
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273
Question 3
(a) 1.
2.
The properties vary differently with temperature.
The properties may not vary linearly with temperature.
(b) (i)It is the lowest theatrical temperature that can be reached
(–273°C → Ok).
(ii) I t is a theoretical scale with one fixed point, the triple point
of water. It does not depend on the physical properties of any
­substance.
(c) (i)
T
= 273.15 + 50
k
= 323.15 k.
(ii) ∆θ = 30°C
≡ 30 k ⇒ 30.00 k.
(d) Consider a liquid in glass thermometer (Mercury).
1.
2.
3.
4.
Unit I.indd 273
etermine the property (expansion) at the upper fixed
D
point p100.
Determine the property at the lower fixed point p0.
Determine the property at the unknown temperature pθ.
Use the formula θ ° Centigrade =
Pθ − P0
× 100.
P100 − P0
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274 c a p e p h y s i c s - U n i t 1
(e) (i)
R100 =
R0
{1 + (5.0 × 10 ) (100) }
= 0.667 R0
R80 =
(ii)
(
R0
2
−5
)
1 + 5.0 × 10−5 (80)
2
= 0.758 R0
Rθ = 0.758 R0
R0 = R0
R100 = 0.667 R0 .
θ=
0.758 R0 − R0
× 100
0.667 R0 − R0
= 72.7° Centigrade.
Unit I.indd 274
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275
Question 4
(a) Ax = A cos θ
Ay = A sin θ.
(b) (i)
A
X
35°
105
20°
14
5m
55°
Y
m
B
R
20°
RV
70°
0
x°
P
RH
C
(ii) Distance OX going upwards = 145 sin 70
= 136.25 m
Distance YB coming downwards = 105 sin 35
= 60.22
Net distance above OC = 136.25 − 60.22
= 76.3 m
or 76.0 m.
(iii) OP going east = 145 cos 70
= 49.59 m.
PC = AY going east = 105 cos 35
= 86.01 m
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276 c a p e p h y s i c s - U n i t 1
OC = 135.6 m
So
Hence OB = R = 135.6m2 + 762
= 155 m.
76.0
= 0.560
135.6
x = 29.2°.
tan x =
(c)
S≡
mx
(ms )
−2 x
≡ m x m− x s 2 x
s1 = s 2 x
So
Unit I.indd 276
1
2x = 1, x = .
2
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277
Question 5
(a)
Long sighted eye has difficulty
seeing near objects. The image
seems to form behind the retina.
This problem is corrected
using a convex lens of
suitable power.
(b)
Convex lens
F
F
Object placed inside F
Unit I.indd 277
eye
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278 c a p e p h y s i c s - U n i t 1
(c) (i)
1 1 1
= +
f u v
1
1 1
=
+
18 12 v
v=?
u = 12 cm
f = 18 cm
v = –36 cm. This means v is 36 cm away from the lens but on the
same side of the lens as the object.
(ii) Virtual
(iii) m =
v −36
=
= −3
u 12
The image is upright.
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279
Question 6
(a) Y
oung’s modulus is defined as the ratio of stress on a material to the
strain of the material.
stress
strain
F
= A.
e
l
Y=
(b) E
lastic deformation is one in which the material returns to its
­original size when the load is removed. This means that all the
­energy stored in this region is recoverable. This region extends
slightly beyond where Hooke’s law applies. Hooke’s law applies
only upto the region of direct proportionality.
Inelastic deformation (or plastic deformation) is one in which
the material does not return to its original size when the load is
­removed. All the stored potential energy is not recoverable (some is
converted into heat in the body itself). Hooke’s law does not apply
in this region.
Y = 1.8 × 1010 Pa
(c)
Maximum stress = 1.5 × 108 Pa
L = 0.47 m
e=?
stress 1.5 × 108
=
e
strain
0.47
8
1.5 × 10 × 0.47
e=
1.8 × 1010
= 0.392 × 10−2 m
= 3.92 mm.
Y=
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280 c a p e p h y s i c s - U n i t 1
Y = 2.4 × 1011 Pa (d) (i)
A = 1.3 × 10−4 m2
e
Strain, = 0.0010
l
F =?
Y=
Fl
Ae
 e
F = YA ×  
 l
= 2.4 × 1011 × 1.3 × 10−4 × 1.0 × 10−3
= 3.12 × 104 N.
(ii) 8 tonne ≡ 8000 kg ⇒ 8000 × 9.80 N ⇒ F
I = 8m, e = ?, Y = 2.4 × 1011 Pa
A = 1.3 × 10−4 m2
Fl
Ae
8000 × 9.80 × 8
Fl
e=
=
Ay 1.3 × 10−4 × 2.4 × 1011
= 2.0 cm.
Y=
Unit I.indd 280
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