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I
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Section III..................................................................................................................................................1
Chapter 1 Phases of matter........................................................................................................................1
1.1 Density ........................................................................................................................................1
1.2 Solids, liquids, gases ...................................................................................................................2
1.2.1 Solid .................................................................................................................................2
1.2.2 Liquid ...............................................................................................................................3
1.2.3 Gas ...................................................................................................................................4
1.2.1.1 The state parameters of gas ...................................................................................4
1.3 Pressure in fluids.........................................................................................................................4
1.3.1 Pressure ............................................................................................................................4
1.3.2 Hydrostatic Pressure.........................................................................................................5
1.3.3 Pascal Principle ................................................................................................................6
1.3.4 The Hydraulic Press .........................................................................................................6
1.3.5 Archimedes’s Principle.....................................................................................................7
1.4 Change of phase ..........................................................................................................................8
1.5 6 worked examples...................................................................................................................9
Chapter 2 Deformation of solids .............................................................................................................12
2-1 Deformation..............................................................................................................................12
2-2 tensile stress and tensile strain ..................................................................................................12
2-3 Hooke’s law ..............................................................................................................................13
2-4 Young’s modulus.......................................................................................................................13
2-5 Strain energy .............................................................................................................................14
2-6 Materials stretching ..................................................................................................................15
2-6-1 Stretching glass..............................................................................................................15
2-6-2 stretching a metal...........................................................................................................15
2.6 15 Worked examples ......................................................................................................................16
Chapter 3 Temperature............................................................................................................................30
3.1 Temperature and state of matter ................................................................................................30
3.2 Calorimetry ...............................................................................................................................30
3.3 Change of phase ........................................................................................................................30
3.4 12 Worked examples ......................................................................................................................30
Chapter 4 Ideal gases ..............................................................................................................................30
4.1 Boyle’s law................................................................................................................................30
4.2 The law of Charles and Gay-Lussac..........................................................................................30
4.3 The ideal gas law.......................................................................................................................30
4.4 The kinetic theory of gases........................................................................................................31
4.5 The kinetic-theory definition of temperature ............................................................................31
4.6 Internal energy of an ideal gas ..................................................................................................31
4.7 11 Worked examples ......................................................................................................................31
Chapter 5 Thermodynamics ....................................................................................................................31
5-1 Thermal equilibrium .................................................................................................................31
5-2 The first law of thermodynamics ..............................................................................................31
5-3 The carnot cycle and the efficiency of engines.........................................................................31
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...............................................................................32
II
5-4 Refrigerators and heat pumps ...................................................................................................31
5-5 The second law of thermodynamics..........................................................................................31
5-6 Entropy and the second law ......................................................................................................31
5-7 11 Worked examples......................................................................................................................31
Chapter 6 Waves .....................................................................................................................................31
6.1 Wave motion .............................................................................................................................31
6.1.1 Progressive waves ..........................................................................................................31
6.1.2 Wave features .................................................................................................................31
6.1.3 Definition of reflection, refraction and diffraction.........................................................31
6.2 Electromagnetic waves..............................................................................................................32
6.2.1 Electromagnetic waves...................................................................................................32
6.2.2 Polarization & Intensity .................................................................................................32
6.3 Interference ...............................................................................................................................32
6.3.1 Superposition and interference.......................................................................................32
6.3.2 Double-silt interference experiment
6.3.3 Diffraction ......................................................................................................................32
6.3.4 Diffraction by a single silt ..............................................................................................32
6.3.5 Diffraction grating..........................................................................................................32
6.4 Stationary waves .......................................................................................................................32
6.4.1 The superposition principle............................................................................................32
6.4.2 Standing waves...............................................................................................................32
6-5 35 Worked examples......................................................................................................................32
6-6 Thin lenses Locating images by ray tracing.........................................................................32
6-7 The thin-lens equation ..............................................................................................................32
6-8 Spherical mirrors ......................................................................................................................32
6-9 15 Worked examples......................................................................................................................33
Chapter 7 Sound......................................................................................................................................33
7.1 The speed of sound....................................................................................................................33
7.2 vibrating air columns ................................................................................................................33
7.3 Audible sound waves ................................................................................................................33
7.4 the Doppler Effect .....................................................................................................................33
7.5 9 Worked examples ........................................................................................................................33
Chapter 8 Oscillations.............................................................................................................................33
8-1 Simple harmonic motion Periodic motion .............................................................................33
8-2 Linking circular motion and SHM............................................................................................33
8-3 Graphical representations of SHM linking x, v, a and t ............................................................33
8-4 Mass—spring system................................................................................................................33
8-5 Simple pendulum......................................................................................................................33
8-6 The energy of SHM ..................................................................................................................33
8-7 Free and damped oscillations....................................................................................................33
8-8 Forced oscillations and resonance ............................................................................................33
8-9 41 Worked examples......................................................................................................................33
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Section III
Chapter 1 Phases of matter
1.1 Density
1. Weight density
The quantity which relates a body’s weight to its volume is known as its
weight density.
The weight density D of a body is defined as the ratio of its weight W to its
volume V. the SI unit is the Newton per cubic meter (N/m3).
WD
V W DV
2. Mass density
Since the weight of a body is not constant but varies according to location, a
more useful relation for density takes advantage of the face that mass is a
universal constant, independent of gravity.
The mass density of a body is defined as the ratio of its mass m to its
volume V.
m
V m V
SI unit of mass density is kilograms per cubic meter (kg/m3).
Note:
The relation between weight density and mass density is found by recalling
that , thus W mg
W mgD g
V V
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1.2 Solids, liquids, gases
1.2.1 Solid
1 Crystal and amorphism
Solids can be divided into crystals and amorphism. Crystal and
amorphism have differences in shape and physical properties. Generally
speaking, crystal has regular geometric shapes, a certain melting point and
shows anisotropy in its physical properties. But amorphism has an irregular
geometric form, no fixed melting point, and its physical properties show
isotropy.
Actually, some matter does show both crystal and amorphism, that is, it is
not absolute crystal or amorphism. For example, natural quartz is crystal, but
condensed crystal after melting is amorphous. A lot of amorphism can be
transformed into crystal under certain conditions. For instance, some glass
windows in ancient buildings become local crystals. It is known that almost
all materials can be transformed into amorphism under rapid cooling to a low
enough temperature.
2 Mono-crystal and poly-crystal
Crystal can be divided into mono-crystal and poly-crystal. If a body is a
perfect crystal, such as a snowflake, a particle of salt, etc, it is called
mono-crystal.
If an integral body is made of many crystals arranged at random, it is called a
poly-crystal.
3 The microstructure of solid
Why are there so many differences in shape and physical properties
between crystal and amorphism? In the 17th century, people put forward a
hypothesis about the regular arrangement of crystal grains according to their
crystalline anisotropy in shape and physical properties. Until the middle of
19th century, the theory of crystal structure was further developed. Many
people considered that grains in crystal were arranged regularly according to
a certain pattern. However, this theory could not be proved because of the
poor condition of science and technology at that time. So, it was still a
hypothesis. From 1912, people worked at the crystal structure with x-ray and
proved the hypothesis. Now people have seen the internal structure of a
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crystal and taken a lot of pictures by electron microscopes to confirm this
theory.
Particles (molecules, atoms or ions), made of crystal, are arranged
regularly in space according to certain patterns. The reaction between the
p[articles in a crystal is so strong that heat cannot overcome it to separate
them. The thermal motion of particles is shown as a vibration around the
balance position.
Some matter can form crystals of different kinds because their particles can
make different crystal structures.
1.2.2 Liquid
1 The microstructure of liquid
The property of liquid is between a solid and a gas. Like a solid, liquid
cannot be compressed. On the contrary like gas, it does not have a particular
shape, and is fluid. All these properties of liquid depend on its microstructure.
The volume of liquid expands a thousand times when it evaporates.
However, its volume reduces to 10% when it is condensed. This shows that
the arrangement of liquid molecules is similar to that of solid molecules. Like
a solid, the molecules in a liquid join together, so a liquid has a certain volum.
It is hard to compress. The interaction between liquid molecules, however, is
not as strong as a solid. The molecules of a liquid are arranged regularly in a
small region. This region is temporary as the boundary and the size of this
region change with time. Sometimes it breaks, and sometimes it reforms
again. Liquid consists of a great number of temporary small zones. These
small zones are random, so the liquid show isotropy.
The distance between liquid molecules is small, so the interactive force is
very strong. The thermal motion of liquid molecules is similar to solid, which
is mainly to vibrate around the balance position. However, liquid has no
permanent position. After vibrating around one balance position for a while,
the molecules shift to another balance position. That is, liquid molecules
move. This is the reason for liquid fluidity.
Diffusion in a liquid comes from the movement of liquid molecules. The
movement of molecules in a liquid is easier than in a solid. So diffusion in a
liquid is faster than in a soled.
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1.2.3 Gas
1.2.1.1 The state parameters of gas
1 Temperature
Temperature is a quantity, which describes how hot or cool a body is.
In SI system, the temperature, which is expressed by the thermodynamic
standard, is called thermodynamic temperature. Thermodynamic
temperature is one of the seven basic physical quantities, shown as KT . its
unit is Kelvin shown as K. it is also called an absolute temperature. We have
also learnt that the temperature can use the Celsius temperature
scale(shown as , and its unit is centigrade shown as ℃ ) or
Fahrenheit temperature scale(shown as , and its unit is ℉)
CT
FT
The relationship between Fahrenheit and Celsius temperatures is 5
( 32)9C FT T
The relationship between the Kelvin and Celsius temperature scales is 273.15K CT T
2 Volume
A certain amount of gas occupies a certain volume. The molecules of gas
move freely, so the gas always fills up the whole container. So the volume of
gas is equal to the volume of the container. In SI system, the volume is
expressed by V; its unit cubic meters and it is shown as m3. The unit of
volume also has liter, milliliter, shown as L, mL. The relationship between
cubic meter and litter as well as milliliter is as follows: 3 3
6 3
1 10
1 10
L m
mL m
1.3 Pressure in fluids
1.3.1 Pressure
To make sense of some effects in which a force acts on a body we have to
consider not only the force but also the area on which it acts. For example,
wearing skis prevents you sinking into soft snow because your weight is
spread over a greater area. We say the pressure is less.
Pressure is the normal force acting on unit area and is calculated from
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Where A is the area over which the perpendicular force F is applied.
The unit of pressure is the Pascal (Pa); it equals 1 Newton per square metre
(N/m2) and is quite a small pressure.
The greater the area over which a force acts, the less the pressure.
1.3.2 Hydrostatic Pressure
The fluid pressure (gauge pressure) at any point is directly proportional to the
density ( ) of the fluid and the depth (h) below the surface of the fluid (Fig.
1.1).
A
hP gh
Density
P gh
Fig. 1.1 Hydrostatic pressure
Note:
(i) The forces exerted by a fluid on the walls of its container are always
perpendicular.
(ii) The fluid pressure is directly proportional to the depth of the fluid and to
its density.
(iii) At any particular depth, the fluid pressure is the same in all directions.
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(iv) Fluid pressure is independent of the shape or area of its container.
1.3.3 Pascal Principle
The pressure discussed in the previous section is due only to the fluid itself.
Unfortunately, this is usually not the case. Any liquid in an open container,
for example, is subjected to atmospheric pressure in addition to the pressure
of the atmosphere is transmitted equally throughout the volume of the liquid.
This fact, first stated by the French mathematician Blaise Pascal, is called
Pascal’s principle. Generally, it can be stated as follows:
An external pressure applied to an enclosed fluid is transmitted uniformly
throughout the volume of the liquid.
Note:
Most devices which measure pressure directly actually measure the
difference between the absolute pressure and atmospheric pressure. The result
is called the gauge pressure.
Absolute pressure = gauge pressure + atmospheric pressure
atmAP gh P
1.3.4 The Hydraulic Press
The most universal application of Pascal’s law is found with the hydraulic
press, shown in Fig. 1.2. According to Pascal’s principle, a pressure applied
to the liquid in the left column will be transmitted undiminished to the liquid
in the column in the right. Thus, if an input force acts upon a piston of
area , it will cause an output force to act on a piston of area so
that
iF
iA oF oA
Input pressure = output pressure
i o
i o
F F
A A
The ideal mechanical advantage of such a device is equal to the ratio of the
output force to the input force. Symbolically, we write
o oI
i i
F AM
F A
A small input force can be multiplied to yield a much larger output force
simply by having the output piston much larger in area than the input piston.
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The output force is given by
o io
i
F FA
A
And, when the input force travels through a distance iF is while the output
force travels through a distance oF os , we can write
Input work = output work
o oi iF s F s
Thus, the ideal mechanical advantage can be also given by
i
o
oI
i
F sM
F s
1.3.5 Archimedes’s Principle
Archimedes’s Principle can be stated as follows:
An object which is completely or partly submerged in a fluid experiences an
upward force equal to the weight of the fluid displaced.
Buoyant force = weight of displaced fluid
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1.4 Change of phase
Fig. 1.3 shows what happens when a very cold solid (ice) takes in heat at a
steady rate. Melting and boiling are both examples of a change of phase.
A
BC
D E
0
100
Tem
per
atu
re/0 C
time
meltingSolidice
Liquidwater
boiling
GasWater vapour
(steam)
Fig. 1.3 change of phase A to B: The temperature rises until the ice starts to melt.
, the energy input is
re rises until the water starts to boil.
he energy input is
such as water, starts to turn to gas well below its boiling point. This
s a rapid type of evaporation in which vapour bubbles, forming in
B to C: Heat is absorbed, but with no rise in temperature
being used to overcome the attractions between the particles as the solid
changes into a liquid.
C to D: The temperatu
D to E: Heat is absorbed, but with no rise in temperature. T
being used to separate the particles as the liquid changes into a gas (water
vapour).
A liquid,
process is called evaporation. It happens as faster particles escape from the
surface.
Boiling i
the liquid, expand rapidly because their pressure is high enough to overcome
atmospheric pressure.
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The heat required to change a liquid into a gas (or a solid into a liquid) is
.5 6 worked examples
ater
called latent heat. When water evaporates on the back of your hand, it takes
the latent heat it needs from your hand. That is why there is a cooling effect.
Latent heat is released when a gas changes back into a liquid (or a liquid
changes back into a solid).
1
1. A cylindrical tank for w ( 31000 /kg m ) is 3 m long and 1.5 m in
nd the volume:
diameter. How many kilograms of water will the tank hold?
Solution:
First we fi2
2 1.5 33.14 3 5.32
V r h m
Substituting the volume and mass density into m V , we obtain
. A lady of weight 495 N, standing on the ground with the contact area of
the pressure of her shoes to the ground?
is the pressure？
3 3 31000 / 5.3 5.3 10m V kg m m kg
2
412 cm 2.
(i) What is
(ii) now she stands on the ground on one foot, what
Strategy:
(i) F = 495 N, S1 = 412 cm 2 = 0.0412 m 2
44951.2 10P Pa 1 0.0412
2 2 F = 495 N, S = 412/2 cm 2 = 0.0206 m(ii)
44952.4 10P Pa 2 0.0206
. A golf shoe has 10 cleats, each having an area of 0.01 in.2 in contact with
3
the floor. Assume that in walking, there is one instant when all 10 cleats
support the entire weight of a 180-lb person. What is the pressure exerted by
the cleats on the floor?
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Express the answer in SI units. ( 2 21 1000 / 0.145 / .kPa N m lb in )
area in contact with the floor is 0.1 in.2 ). Substitution
Solution:
The total ( 210 0.01 .in
into equation FP
A yields
22
180F lb1800 / .
0.1 .P lb in
A in
Converting to SI units, we obtain
2 11800 / . 1.
kPaP lb in 4
2 24 100.145 / .
kPalb in
. A tank is filled with water to a depth of 1.5 m. what is the pressure at the
mpute the pressure directly from equation
4
bottom of the tank due to the water alone?
Solution:
We can co P gh . The density of
The combination of units
water is approximately 1000 kg/m3, and the height h is 1.5 m. so
3 2 4 21000 / 9.81 / (1.5 ) 1.5 10 /P g h kg m m s m N m
2/N m is the Pascal (Pa). Thus, the pressure at the
completel t of the size and
. A nurse administers medication in a saline solution to a patient by infusion
so that th
iner must be hung high enough that the gauge pressure due to the
Pa
bottom of the tank due to the water is 41.5 10 Pa , or 15 kPa.
Note: the pressure due to the water is y independen
shape of the tank. And if we wish to know the total pressure on the bottom of
the tank, we must add atmospheric pressure to our answer here.
5
into a vein in the patient’s arm. The density of the solution is 1000 kg/m3, and
the gauge pressure inside the vein is 32.4 10 Pa . How high above the
insertion point must the container be hung ere is sufficient pressure
to force the fluid into the patient?
Solution:
The conta
liquid in the tube and container is at least as great as the gauge pressure
inside the vein:
2.liquidP gh 34 10
Solving for the height h yields
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3 2
32.4 10 2.4 10h Pa 30.24 24
1000 / 9.81 /m cm
g kg m m sPa
To actually establish a flow through the needle, the container would need to
. A scuba diver searches for treasure at a depth of 20.0 m below the surface
ric
be higher than this result.
6
of the sea. At what pressure must the scuba device deliver air to the diver?
Strategy: the pressure at the diver’s depth is greater than atmosphe
pressure because of the weight of the water above the diver. If the air
breathed in is not at the same pressure as the external pressure on the diver’s
chest, the excess pressure will collapse the chest. Thus, the breathing
apparatus must deliver air to the diver at the pressure of the surrounding
water. We can calculate this pressure from equation atmAP gh P
Solution:
The pressure on the diver is
atmP gh P
Where atmP is the pressure due to the atmosphere pressing down on the sea,
is the density of sea water, and h is the depth of the diver below the
face. And 101.3kPaatmPsur , 31030 kg / m . When the values for atmP , ,
and h are inse q3 3 2101.3 10 1030 / 9.81 / 20P Pa kg m m s
rted into the e ation, we get u
kPa
When expressed in terms of atmospheres, the pressure becomes
.0 303m
1 2.99 3303 atm atm atmP kPa 101.3kPa
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Chapter 2 Deformation of solids
solid may be atoms, or molecules, or ions. They are held
shape changes: deformation
stic, then the material returns to its original shape
tic, then the material does not return to its original
e strain
area A is under tension from a force F
2-1 Deformation
The particles of a
closely together by electric forces of attraction.
When external forces are applied to a solid, its
occurs. This alters the relative positions of its particles. There are two types
of deformation, as described on the right.
(i) Elastic deformation
If the deformation is ela
when the forces on it are removed.
(ii) Plastic deformation
If the deformation is plas
shape when the forces on it are removed. For example, Plasticine takes on a
new shape when stretched.
2-2 tensile stress and tensil
In Fig. 2.1, a wire of cross-sectional
(at each end). The tensile stress on the wire is defined like this:
l
sec
forcetensile stress
cross tional area
In symbol: F A
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The unit of tensile stress is the 2Nm .
The wire stretches so that its length increases by , called its extension. 0l l
The tensile strain is defined th like is:extension
tensile strainoriginal length
In symbol:
0l l
Tensile strain has no units.
ooke’s law states that the force needed to stretch a spring is directly
extension of the spring from its natural length.
2-3 Hooke’s law
H
proportional to the
Hooke’s law may be written as
force F k l
Where k is the spring constant (sometimes referred to as the stiffness constant)
a from its natural length .
is. The unit of k is
nd is the extension
n
l l
Note:
ⅰ. The greater the value of k, the stiffer the spri g 1Nm .
. The graph of F against l is a straight line of gradient k through the
Fig. 2.2)
ⅱ
origin. (
rce applied to it is removed.
ⅲ. If a spring is stretched beyond its elastic limit, it does not regain its initial
length when the fo
l
2-4 Young’s modulus
For a material which obeys Hooke’s law then, for an elastic deformation, the
o the stress. Which means stress/strain is a constant. strain is proportional t
This constant is called Young’s modulus, Y:
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Tensile stress ' mod tensile strYoung s ulus ain
Therefore,
Tensile stressYoung's modulus
In sym
tensile strain
bol:
Y
F
A ,
0
l
l And
0FlYSo
A l
e
is constant. If Y and A are also constant,
:Not
ⅰ. 0l l F . So the extension is
be used when the material is being
stic deformation, the graph of tensile stress against tensile strain
proportional to the stretching force.
ⅱ. The above equations can also
compressed.
ⅲ. For an ela
is a straight line of gradient Y through the origin. (Fig. 2.3)
2-5 Strain energy
Fig. 2.4
Force
Extension
Area = strain energy (for extension )l
l
F
Fig. 2.4 shows how the extension varies with the stretching force for a
material which obeys Hooke’s law. The work done for an extension l is
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given by the shaded area. The area of a triangle = 1
base height . 2
1So the work done
2F l .
As work is done on the material, energy is stored by the material. This is its
strain energy. So:
21Strain energy F l
1
2 2k l . Because F k l .
Note:
ke’s law is not obeyed, the work done is still equal to the area under a If Hoo
force-extension graph. However, the above equation does not apply.
2-6 Materials stretching
2-6-1 Stretching glass
Fig. 2.5
Elastic deformation
glass
Stress
Strain
A
Fig. 2.5 shows what happens if increasing tensile stress is applied to a glass
s varies with strain when a metal wire (steel) is
thread. Elastic deformation occurs until, at point A, a crack suddenly grows,
and the glass breaks. A material which behaves like this is said to be brittle.
The break is called a brittle fracture.
2-6-2 stretching a metal
Fig. 2.6 shows how stres
stretched until it breaks.
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Fig. 2.6
Note:
ⅰ. By convention, strain is plotted along the horizontal axis.
ⅱ. The sequence O to E is described in detail as follows
● From O to B the deformation of the wire is elastic, and the wire obeys
Hooke’s law up to point A.
● Point B: this is the elastic limit. Beyond it, the deformation becomes
plastic as layers of particles slide over each other. If the stress were
removed at, say, point D, the wire would be left with a permanent
deformation.
● Point C: this is the yield point. Beyond it, little extra force is needed
to produce a large extra extension. If a material can be stretched like
this, it is said to be ductile.
Point E: the wire develops a thin ‘neck’, and then a ductile fracture occurs.
The highest stress just before the wire breaks is called the ultimate tensile
stress (UTS) (sometimes is called breaking stress).
2.6 15 Worked examples
1. A vertical steel spring fixed at its upper end has an unstretched length of
300mm. its length is increased to 385 mm when a 5.0N weight attached to the
lower end is at rest. Calculate:
a. the spring constant,
b. the length of the spring when it supports an 8.0N weight at rest.
Solution:
a. Use F k l with 5.0N and mF 385 300 85 0.085l mm mm mm
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Therefore 15.059
0.085
F Nk Nm
l m
b. Use F k l with 8.0N and 159F k Nm to calculate l .
1
8.00.136
59
F Nl m
k Nm
Therefore the length of the spring = 0.300m + 0.136m = 0.436m
2. A steel elevator cable supports a load of 900 kg. the cable has a diameter of
2.0 cm and an initial length of 24 m. find the stress and the strain on the cable
and the amount it stretches under the load.
Young’s modulus for steel: 10 220 10 /Y N m
Solution:
We compute the stress directly from the definition:
2
F mgstress
A r
Where the force is the weight of the load and r is the radius of the cable: 2
7 22
(900 )(9.81 / )2.8 10 /
(0.010 )
kg m sstress N m
m
The strain can be computed from the value of Young’s modulus for steel: 7 2
410 2
2.8 10 /1.4 10
20 10 /
stress N mstrain
Y N m
Finally, we obtain the elongation of the cable from the strain:
4 30( )( ) (24 )(1.4 10 ) 3.4 10 3.4L L strain m m mm
3. A coil spring is used to support a 1.8-kg mass. If the spring stretches 2 cm,
what is the spring constant? What mass would be required to stretch the
spring 5 cm?
Solution:
For a force and displacement s, 2(1.8 )(9.8 / )
882 /0.02
F mg kg m sk N
s s m
When the spring was stretched 5 cm, the force now is given by (882 / )(0.05 ) 44.1F ks N m m N
Thus, the mass m is given by 2
44.14.50
9.81 /
F Nm k
g m sg
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4. A steel wire is 4.0 m long and 2 mm in diameter. How much is it elongated
by a suspended body of mass 20 kg? Young’s modulus for steel is 19600Mpa.
Solution:
Let be the elongation. Then, by Hooke’s law, L
0
F LY
A L
Where Y is Young’s modulus. The elongation is
300 9 2
1 20 9.8 41.273 10 1.273
196 10 (0.001)
mgLFL L m mm
Y A YA
5. An object of mass 0.15kg is attached to the lower end of a vertical spring
of unstretched length 300mm, which is fixed at its upper end. With the object
at rest, the length of the spring becomes 420mm as a result, calculate:
a. the spring constant
b. the energy stored in the spring
c. The weight that needs to be added to extend the spring to 600mm.
Solution:
a. Use F k l , with 0.15 9.8 1.47F mg N
And 420 300 120 0.12l mm mm mm m
So 11.4712.25
0.12
F Nk N
l mm
b. 2 21 1 1Energy stored=Strain energy 12.25 0.12 0.088
2 2 2F l k l J
c. when the length of the spring is 600mm,
600 300 300 0.3l mm mm mm m
Then 12.25 0.3 3.68F k l N
So the weight needs to be added, 3.68 1.47 2.21W F mg N N N .
6. (a) (i) State the difference between a scalar quantity and a vector quantity.
(a) (ii) State two examples of a scalar quantity and two examples of a vector
quantity.
Solution:
Scalar: quantity has direction only.
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Examples of scalar: mass, temperatures, volume, work…
Vector: quantity both has magnitude and direction
Examples of vectors: force, acceleration, displacement, velocity,
momentum…
(b) Fig. 6.1 shows a ship fitted with a sail attached to a cable. The force of
the wind on the sail assists the driving force of the ship’s propellers.
Fig. 6.1
The cable exerts a steady force of 2.8 kN on the ship at an angle of 35° above
a horizontal line.
(b) (i) Calculate the horizontal and vertical components of this force.
Solution:
(b) (ii) The ship is moving at a constant velocity of 8.3 m s-1 and the
horizontal component of the force of the cable on the ship acts in the
direction in which the ship is moving.
Calculate the power provided by the wind to this ship, stating an appropriate
unit.
Solution:
Review: Power and velocity
Above, the car’s engine provides a forward force F which balances the total
frictional force on the car. As a result, the car maintains a steady velocity v.
the displacement of the car is s in time intervals t . P is the power being
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delivered to the wheels.
So the work done (by F) = Fs
work done Fspower P
timetaken t
But s
vt
So P Fv42294 8.3 / 1.9 10hP F v m s W Therefore,
(c) The cable has a diameter of 0.014 m. Calculate the tensile stress in the
cable when it exerts a force of 2.8 kN on the ship, stating an appropriate unit.
Assume the weight of the cable is negligible.
Solution:
sec
forcetensile stress
cross tional area
In symbol: F
A
The unit of tensile stress is the 2Nm .
And the cross-sectional area of cable is given by 2 2
2 4 20.0143.14 1.54 10
2 2
dA r m
Therefore, 3
7 24
2.8 101.82 10
1.54 10
FNm
A
7. (a) State Hooke’s law.
Solution:
Hooke’s law states that the force needed to stretch a spring is directly
proportional to the extension of the spring from its natural length.
Hooke’s law may be written as
force F k l
Where k is the spring constant (sometimes referred to as the stiffness constant)
and is the extension from its natural length . l l
(b) A student is asked to measure the mass of a rock sample using a steel
spring, standard masses and a metre rule. She measured the unstretched
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length of the spring and then set up the arrangement shown in Fig. 7.1.
Fig. 7.1
(b) (i) Describe how you would use this arrangement to measure the mass of
the rock sample. State the measurements you would make and explain how
you would use the measurements to find the mass of the rock sample.
The quality of your written communication will be assessed in this question.
Solution:
(i) Use a metre rule to measure the original length ( ) of the spring. 0l
1m 1l
1 1 0l l l
2 3 4 5 6, ,l l l l l
(ii) When the spring supports a known mass ( ), measure the length ( ) of
the spring.
And the extension of the spring = length – unstretched length, thus
(iii) Repeat for different known masses ( , , , , ), and for each
known mass, the extension is
2m 3m 4m 5m 6m
(ⅳ) Plot graph of mass (M) against extension.
), and the extension now is (v) The spring supports the rock sample (mass sM
sL
(vi) Read off the mass M in the graph corresponding to the extension sL ,
whose value represents the mass of the rock sample. sM
(b) (ii) State and explain one modification you could make to the
arrangement in Fig. 7.1 to make it more stable.
Solution:
Adjust the stand so the spring is nearer to it, so the moment of the load is
reduced. And the weight of the load acts through the base.
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8. Fig. 8.1 shows a dockside crane that is used to lift a container of mass
22000 kg from a cargo ship onto the quayside. The container is lifted by four
identical ‘lifting’ cables attached to the top corners of the container.
(a) When the container is being raised, its centre of mass is at a horizontal
distance 32 m from the nearest vertical pillar PQ of the crane’s supporting
frame.
(a) (i) assume the tension in each of the four lifting cables is the same.
Calculate the tension in each cable when the container is lifted at constant
velocity.
Solution:
When the container is lifted at constant velocity, its resultant force is zero
(Newton’s first law). Thus the tension equals to the weight of the container.
Thus, the total tension 522000 9.81 2.16 10tT W mg N
Therefore, the tension in each cable is given by
5 51 12.16 10 0.54 10
4 4tT T N N
(a) (ii) Calculate the moment of the container’s weight about the point Q on
the quayside, stating an appropriate unit.
Solution:
Review:
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The moment of a force about any point is defined as the force × the
perpendicular distance from the line of action of the force to the point.
That is:
The moment of the force = F × d
Note: d is the line of action of the force to the point.
Unit of the moment of the force: Newton metre (Nm)
Therefore,
The moment of the weight = W × d = 5 62.16 10 32 6.9 10 N m
(a) (iii) Describe and explain one feature of the crane that prevents it from
toppling over when it is lifting a container.
Solution:
The counterweight provides a anticlockwise moment to that of the container
to prevent the crane toppling clockwise.
(b) Each cable has an area of cross-section of . 4 23.8 10 m
(b) (i) calculate the tensile stress in each cable, stating an appropriate unit.
Solution:
Review:
sec
forcetensile stress
cross tional area
In symbol: F
A
The unit of tensile stress is the 2Nm .
Therefore, 5
8 24 2
0.54 101.42 10 /
3.8 10
F NN m
A m
(b) (ii) Just before the container shown in Fig. 8.1 was raised from the ship,
the length of each lifting cable was 25 m. Show that each cable extended by
17 mm when the container was raised from the ship.
Young modulus of steel = 2.1×1011 Pa
Solution:
Review:
For a material which obeys Hooke’s law then, for an elastic deformation, the
strain is proportional to the stress. Which means stress/strain is a constant.
This constant is called Young’s modulus, E:
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Tensile stress = Young’s modulus × tensile strain
That is: Tensile stress
Young's modulustensile strain
In symbol:
E
And F
A ,
0
l
l
So 0FlE
A l
0l l
Where is the original length, is the extension.
Therefore, 5
011 4
0.54 10 250.017 17
2.1 10 3.8 10
Fll m mm
EA
9. Heavy duty coil springs are used in vehicle suspensions. The pick-up truck
shown in Fig. 9.1 has a weight of 14 000 N and length of 4.5 m. When
carrying no load, the centre of mass is 2.0 m from the rear end. The part of
the vehicle shown shaded in grey is supported by four identical springs, one
near each wheel.
(a) (i) Define the moment of a force about a point.
Solution:
The moment of a force about any point is defined as the force × the
perpendicular distance from the line of action of the force to the point.
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That is:
The moment of the force=F×d
Note: d is the line of action of the force to the point.
Unit of the moment of the force: Newton metre (Nm)
(a) (ii) State and explain which pair of springs, front or rear, will be
compressed the most.
Solution:
The center of mass is closer to the rear springs, thus the rear springs will be
compressed the most.
(a) (iii) By taking moments about axle B, calculate the force exerted on the
truck by each rear spring.
Solution:
The rear two springs have the same force on the truck, let the force and
; the front two springs also have the same force on the truck, let the force
1rF
2rF
1fF 2 and fF
1 2 12.5 2.5 2 2.5 1.4 1.4 14000 19600r r rF F F mg N m
And for equilibrium,
The sum of the clockwise moments=the sum of the anticlockwise moments
Gives
1
196003920
5rF N
(b) The spring constant for each of these springs is 100 000Nm-1.
Calculate the distance that each of these rear springs is compressed by this
vehicle as shown in Fig. 9.1.
Solution:
Using the equation , gives F k l
11
39200.0392 3.92
100000rF N
l m cmk Nm
0.065e m
(c) The springs must not be compressed by more than an additional 0.065 m.
Calculate the maximum load that could be placed at point X, which is
directly above the centre of the rear axle A, as shown in Fig. 9.1.
Solution:
The additional extension is
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Then the maximum force of the two rear springs is given by 2 100000 0.065 13000masF k e N
Thus, the maximum load is 13000 N.
10. (a) Describe how to obtain, accurately by experiment, the data to
determine the Young modulus of a metal wire.
The quality of your written answer will be assessed in this question.
Solution:
For a material which obeys Hooke’s law then, for an elastic deformation, the
strain is proportional to the stress. Which means stress/strain is a constant.
This constant is called Young’s modulus, Y:
Tensile stress = Young’s modulus × tensile strain
That is:
Tensile stressYoung's modulus
tensile strain
In symbol:
Y
And F
A ,
0
l
l
So 0FlY
A l
For an elastic deformation, the graph of tensile stress against tensile strain is
a straight line of gradient Y through the origin (Fig. 10.1).
(b) Fig. 10.2 is a plot of some results from an experiment in which a metal
wire was stretched.
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Fig. 10.2
(b) (i) Draw a best-fit line using the data points.
(b) (ii) Use your line to find the Young modulus of the metal, stating an
appropriate unit.
Solution:
(i)
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(ii)The gradient of the graph
710
3
5.1 1010
5.1 10
Pak Pa
10E k Pa
Thus, the Young’s modulus
10
(c) After reaching a strain of 37.7 10 , the wire is to be unloaded. On Fig.
10.3, sketch the line you would expect to obtain for this.
Solution:
What you draw on the Fig. 10.3:
The line Originates at last point and parallel to the first line, and then
touching the x axis:
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Fig. 10.3
11. A crane fitted with a steel cable of uniform diameter 2.3mm and length
28m is used to lift an iron girder of weight 3200N off the ground. Calculate
the extension of the cable when it supports the girder at rest.
The young modulus for steel = 2.1×1011 Pa
Solution:
From 0FlE
A l
, get 0Fl
lEA
With data: F=3200N, , cross-sectional area 0 28l m
221
2 4
DA D
So 2 3 21 13.14 (2.3 10 ) 4.15 10
4 4A D 6 2m
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Then 011 6
3200 280.103
2.1 10 4.15 10
Fll m
EA
12. A brass object consists of of copper and of zinc.
Calculate the mass and the density of this object. The density of copper =
8900 kgm
5 33.3 10 m 5 31.7 10 m
-3. The density of zinc = 7100kgm-3.
Solution:
The mass of the brass object is equal to the total mass of the copper and that
of zinc.
And the mass of the copper = density of copper×volume of copper.
= 3 5 38900 3.3 10 0.294kgm m kg
The mass of the zinc = density of zinc×volume of zinc
= 3 5 37100 1.7 10 0.121kgm m kg
So total mass, 0.294 0.121 0.415m kg
Total volume, 5 53.3 10 1.7 10 5.0 10V m 5 3
Density of alloy, 35 3
0.4158300
5.0 10
m kgkgm
V m
Chapter 3 Temperature
3.1 Temperature and state of matter
3.2 Calorimetry
3.3 Change of phase
3.4 12 Worked examples
Chapter 4 Ideal gases
4.1 Boyle’s law
4.2 The law of Charles and Gay-Lussac
4.3 The ideal gas law
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4.4 The kinetic theory of gases
4.5 The kinetic-theory definition of temperature
4.6 Internal energy of an ideal gas
4.7 11 Worked examples
Chapter 5 Thermodynamics
5-1 Thermal equilibrium
5-2 The first law of thermodynamics
5-3 The carnot cycle and the efficiency of engines
5-4 Refrigerators and heat pumps
5-5 The second law of thermodynamics
5-6 Entropy and the second law
5-7 11 Worked examples
Chapter 6 Waves
6.1 Wave motion
6.1.1 Progressive waves
6.1.2 Wave features
6.1.3 Definition of reflection, refraction and diffraction
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6.2 Electromagnetic waves
6.2.1 Electromagnetic waves
6.2.2 Polarization & Intensity
6.3 Interference
6.3.1 Superposition and interference
6.3.2 Double-silt interference experiment
6.3.3 Diffraction
6.3.4 Diffraction by a single silt
6.3.5 Diffraction grating
6.4 Stationary waves
6.4.1 The superposition principle
6.4.2 Standing waves
6-5 35 Worked examples
6-6 Thin lenses Locating images by ray tracing
6-7 The thin-lens equation
6-8 Spherical mirrors
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33
6-9 15 Worked examples
Chapter 7 Sound
7.1 The speed of sound
7.2 vibrating air columns
7.3 Audible sound waves
7.4 the Doppler Effect
7.5 9 Worked examples
Chapter 8 Oscillations
8-1 Simple harmonic motion Periodic motion
8-2 Linking circular motion and SHM
8-3 Graphical representations of SHM linking x, v, a and t
8-4 Mass—spring system
8-5 Simple pendulum
8-6 The energy of SHM
8-7 Free and damped oscillations
8-8 Forced oscillations and resonance
8-9 41 Worked examples
