LEARNING CONTENT
The content is divided into themes and topics as follow:
Mathematical requirements
Theme 1: General physics
1.1 Physical quantities and units
1.2 Measurement techniques
1.3 Kinematics
1.4 Dynamics
1.5 Forces, density and pressure
1.6 Work, energy and power
1.7 Deformation of solids
Theme 2: Waves
2.1 Progressive waves
2.2 Transverse and longitudinal waves
2.3 Determination of frequency and wavelength of sound waves
2.4 Doppler effect
2.5 Electromagnetic spectrum
2.6 Superposition
2.6.1 Stationary waves
2.6.2 Diffraction
2.6.4 Diffraction gratings
2.6.3 Interference, two-source interference
Theme 3: Electricity
3.1 Electric fields
3.2 Current of electricity
3.3 DC circuits
Theme 4: Modern physics
4.1 Atoms, nuclei and radiation
4.2 Fundamental particles
1
NSSCH Physical Science
Mathematical requirements
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add, subtract, multiply and divide
use averages, decimals, fractions, percentages, ratios and reciprocals
use a calculator to calculate pH (log) and refractive index (sin)
use direct and inverse proportion
use positive, whole number indices and exponents in calculations
make approximate evaluations of numerical expressions
use usual mathematical instruments (ruler, compasses, protractor, set square)
explain the meaning of angle, curve, circle, radius, diameter, square,
parallelogram, rectangle, diagonal
solve equations of the form x = yz for any one term when the other two are
known
recognise and use points of the compass (N, S, E, W), take bearing and apply
the rules for bearing taking
NSSCAS Physics
Mathematical requirements
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Arithmetic
Algebra
Geometry and trigonometry
Vectors
Graphs
Arithmetic
recognise and use expressions in decimal and standard form (scientific) notation
use an electronic calculator for addition, subtraction, multiplication and division
find arithmetic means, powers (including reciprocals and square roots), sines,
cosines, tangents (and the inverse functions)
take account of accuracy in numerical work and handle calculations so that
significant figures are neither lost unnecessarily nor carried beyond what is
justified
2
make approximate evaluations of numerical expressions (e.g. π ≈ 10) and use
such approximations to check the magnitude of calculated results
Algebra
Change the subject of an equation to the required variable. Most relevant
equations involve only the simpler operations but may include positive and
negative indices and square roots.
Solve simple algebraic equations. Most relevant equations are linear (e.g. y =
mx, y = mx + c) but some may involve inverse and inverse square relationships.
Linear simultaneous equations and the use of the formula (ax2 + bx + c or x =
−𝑏 ± √𝑏2 − 4ac
2𝑎
) to obtain the solutions of quadratic equations are required
2
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substitute physical quantities into physical equations using consistent units and
check the dimensional consistency (homogeneity) of such equations
• set up simple algebraic equations as mathematical models of physical situations,
and identify inadequacies of such models
• express small changes or uncertainties as percentages and vice versa
• understand and use the symbols <, >, ≤; ≥; ≪, ≫; ≈; /, ∝, ⟨X⟩, (= ̅
X), ∑, Δx, δx, √
Understand and use the symbols
<
- strict inequality, less than
>
- strict inequality, greater than
≤
- inequality, less than or equal to
≥
- inequality, greater than or equal to
≪
- much less than
≫
- much greater than
≈
- approximately equal, approximation
/
- division slash
∝
- proportional to
⟨X⟩
- average over all values in the set X
̅
(= X) - Average of the value
∑ - sigma, summation – sum of all values in range of series
Δx - Δx is about a secant line, a line between two points representing the rate of
change between those two points. That's a "differential" (between the two points).
δx - δx is about a tangent line to a partial derivative. That's a rate of change or
derivative in one direction, holding a number of other directions constant.
√ - surd, A number that can't be simplified to remove a square root (or cube root etc),
e.g. √2
e.g. linearization
Mathematical models of physical situations
Shape for y = mx graph y is directly proportional to x
Examples
V = IR
F = ke
𝑬𝒑 = mgh
3
Shape for y = mx graph
y is directly proportional to x …
Shape for y = mx + c graph
y is linearly proportional to x
Consider the equation V = E – Ir in line
with this graph:
y = V and x = I
This can be rearranged to give V = – Ir
+ E, since x = I, so in – Ir, m = – r so in
the form of y = mx + c, this gives V = –
rI + E
𝐬𝐡𝐚𝐩𝐞 𝐟𝐨𝐫 𝒚𝟐  x graph
4
𝐬𝐡𝐚𝐩𝐞 𝐟𝐨𝐫 𝒚𝟐  x graph …
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Consider the motion of a skydiver who
jumps from an airplane and falls vertically
through the air. The graph shows the
variation of the skydiver’s vertical velocity v
and time t.
•
top opening parabola
•
as x increases, y also increases; y increases at a
larger proportion compared to x
Shape for y  𝒙𝟐 graph
To change this into a y = mx or y = mx + c relationship,
square the values of x and plot a graph of 𝒚 versus 𝒙
𝟐
Shape for y  𝒙𝟐 graph
y = k𝒙𝟐
Consider the equation for elastic potential
energy as
𝟏
𝑬𝒑 = 𝟐 k𝒙𝟐
y = 𝑬𝒑
x = 𝒙𝟐
𝟏
Shape for y = 𝒙 graph
•
y is inversely proportional to x
To change this into a y = mx or y = mx + c
relationship, find the reciprocal of all the
𝟏
values of x and plot a graph of y versus 𝒙
5
𝟏
Shape for y = 𝒙 graph …
Consider the equation
E = IR + Ir
so E = I (R + r)
Since y = I, making I the subject of the formula gives
𝑬
I = 𝑹+𝒓
Since r is on the x = axis, if E and R are fixed, this means if r
𝟏
increases, I decreases and so I = 𝒓
Example of linearisation of position versus time for a rolling ball
Raw data
Time/ s
Length/ cm
0
00
1
3.1
2
12.2
3
27.0
4
47.9
5
75.2
6
108.3
7
146.8
8
192.1
Graph of position vs time from raw data
6
Linearisation of the data from y  𝒙𝟐 to y = mx
Time / s
Time𝟐 / 𝒔𝟐
length / cm
0
0
0.0
1
1
3.1
2
4
12.2
3
9
27.0
4
16
47.9
5
25
75.2
6
36
108.3
7
49
146.8
9
64
192.1
Graph of position vs time𝟐
7
Using the graph to determine gradient and y-intercept
Gradient =
(𝑦2 −𝑦1)
(𝑥2 −𝑥1
Unit for gradient =
=
(180−60)
(60−20)
𝑐𝑚
𝑠2
=
120
40
=3
= cm 𝑠 −2
y – intercept = 0, graph starts at true origin.
unit for y-intercept = unit on y axis = cm
Geometry and trigonometry
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calculate areas of right-angled and isosceles triangles, circumference and area of
circles, areas and volumes of cuboids, cylinders and spheres
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use Pythagoras’ theorem, similarity of triangles, the angle sum of a triangle
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use sines, cosines and tangents of angles (especially for 0°, 30°, 45°, 60°, 90°)
Area of right-angled triangle
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Area of isosceles triangle
Circumference of circle
Area of circle
Area of cuboid
Volume of cuboid
area of cylinder
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Volume of cylinder
area and volume of sphere
Pythagoras theorem
Similarity of triangles
Angles sum of a triangle
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Sines, cosines and tangents
Sine
Cosine
Tangent
0°
0
1
0
30°
0.5
0.899
0.577
45°
0.707
0.707
1
60°
0.866
0.5
0.732
90°
1
0
undefined
Vectors
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find the resultant of two coplanar vectors, recognising situations where vector
addition is appropriate
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obtain expressions for components of a vector in perpendicular directions,
recognising situations where vector resolution is appropriate
Pythagoras theorem
Resolution of vectors into components
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Graphs
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translate information between graphical, numerical, algebraic and verbal
forms
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select appropriate variables and scales for graph plotting
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determine the gradient, intercept and intersection of linear graphs
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choose, by inspection, a straight line which will serve as the line of best fit
through a set of data points presented graphically
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draw a curved trend line through a set of data points presented graphically, when
the arrangement of these data points is clearly indicative of a non-linear
relationship
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recall standard linear form y = mx + c and rearrange relationships into linear
form where appropriate
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sketch and recognise the forms of plots of common simple expressions like ; x2;
1
1
𝑥2
𝑥
−𝒙
; sin x; cos x; 𝒆
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draw a tangent to a curve, and understand and use the gradient of the tangent as
a means to obtain the gradient of the curve at a point
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understand and use the area below a curve where the area has physical
significance
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apply the conventions for labelling graph axes and table columns (see Annexe A,
A2, graph layout, plotting of points, trend)
Graph of y =
𝟏
Graph of y = x2
𝒙
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Graph of y =
𝟏
𝒙𝟐
graph of y = sin x and y = cos x
x/ °
0
90
180
270
360
Y = sinx/ unit
0
1
0
-1
0
Y= cosx/ unit
1
0
-1
0
1
Graph of y = sin x and y = cos x
13
Graph of y = 𝒆−𝒙
y = 𝒆𝒙
x
1
y = 𝒆−𝒙
= 0.0498
𝑒 −(−3) = 𝑒 3 = 20.0855
= 0.1353
𝑒 −(−2) = 𝑒 2 = 7.3891
-1
𝑒 −1 = = 0.3679
𝑒 −(−1) = e = 2.7183
0
𝑒 0 = 1.0000
𝑒 0 = 1.0000
1
e = 2.7183
𝑒 −1 = = 0.3679
2
𝑒 2 = 7.3891
𝑒 −2 =
3
𝑒 3 = 20.0855
𝑒 −3 =
-3
𝑒 −3 =
-2
𝑒 −2 =
𝑒3
1
𝑒2
1
𝑒
1
𝑒
14
1
𝑒2
1
𝑒3
= 0.1353
= 0.0498
NSSCAS Physics
Theme 1: General Physics
Theme 1 consists of 7 Topics
Topic 1.1 Physical quantities and units
Topic 1.2 Measurement techniques
Topic 1.3 Kinematics
Topic 1.4 Dynamics
Topic 1.5 Forces, density and pressure
Topic 1.6 Work, energy and power
Topic 1.7 Deformation of solids
Topic 1.1: Physical quantities and SI units
1.1.1 Physical quantities
1.1.2 SI units
1.1.3 Scalars and vectors
1.1.1 Physical quantities
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recall that all physical quantities consist of a numerical magnitude and a unit
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make reasonable estimates of physical quantities included within the syllabus
What is a physical quantity?
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A physical quantity is any quantity that can be measured. All physical quantities
consist of a numerical magnitude and a unit.
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Examples: mass (2.0 kg), length (1.9 m), time (0.15 s), current (1 600 μA),
temperature (273.0 K), amount of substance (0.05 mol)
Making reasonable estimates
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You are expected to be able to make reasonable estimates of physical
quantities included within the syllabus.
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When making an estimate, it is only reasonable to give the figure to 1 or at
most 2 significant figures since an estimate is not very precise.
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For example:
What is the approximate kinetic energy of an Olympic athlete when running at maximum
speed during a 100 m race?
A 400 J
B 4000 J
C 40 000 J
D 400 000 J
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Key = B
Reasonable mass estimate in the range 50kg to 100kg
Reasonable speed estimate around 10ms −1
1.1.2 SI units
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recall and use the following SI base quantities and their units: mass (kg), length
(m), time (s), current (A), temperature (K), amount of substance (mol)
•
express derived units as products or quotients of the SI base units and use the
named units listed in this syllabus as appropriate
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use SI base units to check the homogeneity of physical equations
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use the following prefixes and their symbols to indicate decimal submultiples or
multiples of both base and derived units: pico (p), nano (n), micro (μ), milli (m),
centi (c), deci (d), kilo (k), mega (M), giga (G), tera (T)
SI base quantities and their units
Quantity
mass
length
time
Electric current
Thermodynamic temperature
Amount of substance
Luminous intensity
Unit
kilogram
metre
second
Ampere (amp)
kelvin
mole
candela
Symbol
Kg
m
s
A
K
mol
cd
Derived units
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All quantities, apart from the base quantities, can be expressed in terms of
derived units.
•
Derived units consist of some combinations of the base units.
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Derived units are expressed when the SI base units are multiplied together or
divided by one another, but never added or subtracted (when the SI base units
are added or subtracted, they remain SI base units).
Derived quantities and units
Quantity
Frequency
velocity
Acceleration
Force
Energy
power
Electric charge
Potential difference
Electrical resistance
Specific heat capacity
unit
Hertz (Hz)
m s-1
m s-2
Newton (N)
Joule (J)
Watt (W)
Coulomb (C)
Volt (V)
Ohm (Ω)
J kg-1K-1
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Derived unit
s-1
m s-1
m s-2
kg m s-2
kg m2 s-2
kg m2 s-3
As
kg m2 s-3 A-1
kg m2 s-3 A-2
m2 s-2 K-1
Homogeneity of a physical equation
•
In any equation where each term has the same base units, the units on the left
hand side must be equal to the units on the right hand side of the equation,
the equation is said to be homogenous .
•
When an equation is known to be homogenous, then the balancing of base
units provides a means of finding the units of an unknown quantity.
•
Also known as dimensional consistency.
Using SI base unit to check homogeneity
1. Show that the SI base units of power are kg𝑚2 𝑠 −3. [3]
(i.e. show that 1 W = 1 kg𝒎𝟐 𝒔−𝟑)
Answer:
𝑠
𝑚
Velocity = 𝑡 = 𝑠 = 𝑚𝑠 −1
𝑚𝑠 −1
= 𝑚𝑠 −2
•
Acceleration =
•
Force = ma = kg × 𝑚𝑠 −2 = kg𝑚𝑠 −2
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Work = F s = kg𝑚𝑠 −2 × m = kg𝑚2 𝑠 −2
•
Power = 𝑡 =
𝑤
𝑠
𝑘𝑔𝑚2 𝑠 −2
𝑠
= kg𝑚2 𝑠 −3
OR Power = Fv = kg𝑚𝑠 −2 × 𝑚𝑠 −1 = kg𝑚2 𝑠 −3
𝒔
𝒎
•
Velocity = 𝒕 = 𝒔 = 𝒎𝒔−𝟏
•
Acceleration =
•
Force = ma = kg × 𝒎𝒔−𝟐 = kg𝒎𝒔−𝟐
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Work = F s = kg𝒎𝒔−𝟐 × m = kg𝒎𝟐 𝒔−𝟐
•
Power = 𝒕 =
𝒘
𝒎𝒔−𝟏
𝒔
𝒌𝒈𝒎𝟐 𝒔−𝟐
𝒔
= 𝒎𝒔−𝟐
= kg𝒎𝟐 𝒔−𝟑
OR Power = Fv = kg𝒎𝒔−𝟐 × 𝒎𝒔−𝟏 = kg𝒎𝟐 𝒔−𝟑
2. At temperatures close to 0 K, the specific heat capacity c of a particular solid is given
by c = bT3 where T is the temperature and b is a constant, characteristic of the solid.
The SI unit of specific heat capacity is J kg–1 K–1.
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What is the unit of constant b, expressed in SI base unity?
A. m2 s-2 K-3
B. m2 s-2 K-4
C. kg m2 s-2 K-3
D. kg m2 s-2 K-4
Answer
• c = bT3;
• b=
• =
•
𝑐
𝑇3
𝐽𝑘𝑔−1 𝐾 −1
𝐾3
𝒌𝒈𝒎𝟐 𝒔−𝟐 × 𝑘𝑔−1 × 𝐾 −1
𝐾3
•
𝑚2 𝑠 −2 𝐾 −4
•
Key = B
3. What is the number of SI base units required to express electric field strength and
power?
Electric field
power
A
3
3
B
3
2
C
4
2
D
4
3
Answer
Electric field strength
𝐹
𝑚𝑎
𝑘𝑔 × 𝑚𝑠 −2
•
E=
•
= 4 SI base units
𝑄
=
𝐼𝑡
=
𝐴×𝑠
= kg 𝑚𝑠 −3 𝐴−1
Power
𝑊
𝐹𝑠
𝑚𝑎𝑠
•
P=
•
= 3 SI base units
𝑡
=
𝑡
=
𝑡
=
kg 𝑚𝑠 −2 × 𝑚
𝑠
= kg 𝑚2 𝑠 −3
Key: D
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1.1.3 Scalars and vectors
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distinguish between scalar and vector quantities
– a scalar as a quantity which has a magnitude, but no direction
– a vector as a quantity which has both magnitude and direction
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state examples of scalar (e.g. mass) and vector (e.g. velocity) quantities
•
add and subtract coplanar vectors (vectors in the same plane)
•
represent a vector as two perpendicular components
Scalars and Vectors
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a scalar as a quantity which has a magnitude, but no direction
•
a vector as a quantity which has both magnitude and direction
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Examples of scalars and vectors
Quantity
Mass
Weight
Speed
Velocity
Acceleration
force
Pressure
temperature
scalar
√
Vector
√
√
√
√
√
√
√
Addition of vectors
•
Addition of vectors acting along the same line in the same direction or in opposite
direction.
Perpendicular vectors
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Vector triangles…
•
When two vectors do not act in the same or opposite directions, the resultant is
found by means of a vector triangle.
•
Each one of the two vectors V1 and V2 is represented in magnitude and direction
by the side of a triangle.
•
Both vectors must be in either clockwise or anticlockwise direction.
•
The combined effect (resultant) is given in magnitude and direction by the third
side of the triangle.
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•
If V1 and V2 are clockwise, R will be anticlockwise. The resultant may be
found by means of a scale diagram or solved using trigonometry.
•
For a vector triangle, the assumption is that the three forces are in an
equilibrium situation and should form a closed triangle.
Resolution of vectors
•
Recall: two vectors can be added together to produce a single resultant.
•
This resultant behaves in the same way as the two individual vectors.
•
It follows that a single vector may be split up, or resolved, into two vectors or
components.
•
The combined effect of the components is the same as the original vector.
•
The resolution of a vector into perpendicular components is a very useful means
of solving certain types of problems such as forces, momentum, velocity etc.
Fig: 1.19 resolving a vector into components
•
Consider a force of magnitude F acting at an angle of 𝜃 below the horizontal, Fig.
1.19.
•
A vector triangle can be drawn with a component 𝐹𝐻 in the horizontal direction
and a component 𝐹𝑉 acting vertically. Remembering that F, 𝐹𝐻 and 𝐹𝑉 form a
right-angled triangle, then:
•
𝐹𝐻 = F cos 𝜃
•
𝐹𝑉 = F sin 𝜃
•
The force F has been resolved into two perpendicular components, 𝐹𝐻 and 𝐹𝑉 .
•
This method applied to all types of vector quantities.
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End of topic activity
Multiple choice questions
Paper 12 February/ March 2019: 1, 2, 3
Paper 11 May / June 2019: 1, 2, 3
Paper 12 May / June 2019: 1, 2, 3, 4
Paper 13 May / June 2019: 1, 2, 3, 4
Structured questions
Paper 22 February/ March 2019: 1
Paper 23 May / June 2019: 1
24
Topic 1.2: Measurement techniques
1.2.1 Measurements
1.2.2 Errors and uncertainties
1.2.1 Measurements
•
Use techniques for the measurement of length, volume, angle, mass, time,
temperature and electrical quantities appropriate to the ranges of magnitude
implied by the relevant parts of the syllabus. In particular, learners should be able
to:
 measure lengths using rulers, calipers and micrometers
 measure weight and hence mass using balances
 measure an angle using a protractor
 measure time intervals using clocks, stopwatches and the calibrated timebase of a cathode-ray oscilloscope (c.r.o.)
 measure temperature using a thermometer
 use ammeters and voltmeters with appropriate scales
 use a galvanometer in null methods
 use a cathode-ray oscilloscope (c.r.o.)
 use both analogue scales and digital displays
 use calibration curves
Measurements
•
Theoretical ideas in physics are generally tested by experiment before being fully
accepted.
•
Experimental work is an important part of a physics course.
•
Make a sensible choice of the instrument to use to measure a particular
physical quantity.
•
There are sources of error and uncertainty in experimental work.
•
All experiments that are designed to obtain a quantitative result for a physical
quantity involve measurements.
•
These measurements must be of some combination of the base quantities
length, mass, time, temperature and current.
25
•
A need to understand the principles of the available methods, to make an
informed decision about the choice of a particular technique.
Methods of measuring length
1. Metre rule (precision to the nearest 1 mm)
𝟏
2. Vernier caliper (common calipers have a precision to the nearest 𝟏𝟎 or 0.1 mm)
𝟏
3. Micrometer screw gauge (precision to the nearest 𝟏𝟎𝟎 or 0.01 mm)
1. Metre rule
•
Simplest length measuring instrument.
•
Has a range of up to 1000 mm (1 metre ruler).
•
The smallest division is 1 mm.
•
Measurements to the nearest 1 mm (precision of instrument).
2. Vernier caliper
Reading a Vernier caliper
•
Fixed scale = 25.0 mm (the highest reading before the 0 on the vernier scale)
•
Vernier scale = 4 × 0.1 = 0.4 mm (where the fixed scale and vernier scale
coincide)
•
Total reading = 25.4 mm to the nearest 0.1 mm.
26
•
Versatile for measuring the dimensions of an object, the diameter of a hole,
depth of a hole.
•
Has a range of up to 100 mm.
•
Common Vernier calipers can be read to 0.1mm.
•
A sliding part B moves along the scale.
•
The slider has the Vernier scale engraved on it.
•
The zero of the Vernier corresponds with reference posts on the sliding part.
3. Micrometer screw gauge
•
The principle of the instrument is the magnification of linear motion using the
circular motion of a screw.
•
Consists of a u-shaped piece of steel with a fixed, plane, end-piece A.
•
Opposite this is a screw with a corresponding end piece B.
•
The position of the screw can be adjusted using the ratchet C which is
connected to the thimble D.
•
Most micrometer screw gauges measure up to maximum 5 cm = 50 mm.
•
Measures to the uncertainty of 0.01 mm.
27
•
There are graduations along the barrel of the instrument and round the
circumference of the thimble.
•
The purpose of the ratchet is to ensure that the same torque is applied to the
thimble for each reading.
•
One rotation moves by 0.5 mm. There are 50 divisions on the thimble so this
gives
0.5 mm
50
= 0.01
Example:
9.5 mm (highest reading that can be seen on the barrel)
36 × 0.01 = 0.36 mm on the thimble
Total reading 9.5 mm + 0.36 mm = 9.86 mm to the nearest 0.01 mm
Comparing metre rule, micrometer screw gauge and vernier caliper
Instrument
Metre rule
Micrometer screw gauge
Vernier caliper
range
1m
50mm
100mm
uncertainty
1mm
0.01mm
0.1mm
notes
Check zero, calibration errors
Check zero error
Versatile: inside and outside
diameters depth
Methods of measuring mass
•
Precision of mass measuring instruments vary and is usually quoted on the
instrument.
28
Electronic (top-pan) balance, precision 0.01 g
Measuring an angle –
spring balance
Measuring time
Protractor precision to the nearest 1 °
using a stop clock or a stop watch.
Precision 0.1 due to human reaction time.
Using a cathode-ray oscilloscope
•
A cathode ray oscilloscope (c.r.o) can be used to measure both the amplitude of
signals and short time intervals.
•
A potential difference applied to the y-input controls the movement of the trace
in a vertical direction.
•
A potential difference applied across the x-input controls the trace in the
horizontal direction.
•
The y-sensitivity is adjustable and is measured in volts per cm (V cm−1 ) or volts
per division (V div−1 ).
•
In the example in Figure 2.3 the y-sensitivity is set at 2 V div−𝟏 .
29
•
A d.c supply is applied across the y-input. No voltage is applied across the xinput. The trace appears as a bright spot.
In Fig. 2.3
•
Screen 1 shows the c.r.o with no input.
•
Screen 2 shows a deflection of 0.75 of a division. The voltage input across the yplates is 0.75 × 2 = 1.5 V.
•
Screen 3 shows a deflection of 1.5 divisions. The voltage input across the yplates is 1.5 × 2 = 3.0 V.
•
Screen 4 shows a deflection of − 0.75 divisions. The voltage input across the yplates is − 0.75 × 2 = − 1.5 V (in the opposite direction).
Using a c.r.o to measure time intervals
•
To measure time intervals, a time-base voltage is applied across the x-input (Fig.
2.4).
•
This drags the spot across the screen, before flying back to the beginning again.
•
The rate at which the time-base voltage drags the spot across the screen can be
measured either in seconds per division (s 𝑑𝑖𝑣 −1 ) or divisions per second (div
𝑠 −1 ).
•
You must check which method can be useful.
30
In Figure 2.4
•
Screen 1 – the spot moves slowly across the screen before flying back to the
beginning and repeating the process.
•
Screen 2 – with a higher frequency time base, the spot moves across the screen
more quickly. The fluorescence on the screen lasts long enough for a short tail to
be formed.
•
Screen 3 – with a much higher frequency, the fluorescence lasts long enough for
the spot to appear as a continuous line.
•
If successive pulses are applied to the y-plate while the time base voltage is
applied, the trace might appear as in Fig. 2.5.
•
The time interval between the pulses can be calculated by multiplying the
number of divisions between the two pulses by the time base.
Calibrated time-base of a c.r.o
•
A c.r.o has a calibrated time-base, measurement from the screen of a c.r.o which
can be used to give values of time intervals.
•
One application is to measure the frequency of a periodic signal.
•
You will learn more about this in Theme 2 topic 2.3 (Determination of
frequency and wavelength for sound waves).
31
Measuring temperature
Liquid-in-glass thermometer
Thermocouple
Liquid-in-glass thermometer
•
Based on the thermal expansion of a liquid.
•
Most experiments will involve temperatures between 0 ℃ and 100 ℃ .
•
Always allow time for thermometer to reach thermal equilibrium with its
surroundings before taking a reading.
•
Liquids must be thoroughly stirred before taking a reading (because of
convection currents).
•
Thermometers are fragile, handle with care.
Thermocouple thermometer
•
Consists of two wires made of different metals or alloys, joined at one end.
•
The other ends of the wires are connected to the terminals of a millivoltmeter.
•
This may be a digital instrument which is calibrated in ℃ .
•
The junction is placed in thermal contact with the object, the temperature of
which is required.
•
An e.m.f is created in the conductors due to differences in temperature. This
e.m.f is detected by the millivoltmeter and is interpreted into temperature for
example using a calibration curve.
Calibration curve of a thermocouple
•
You will need to draw a calibration curve, a graph of e.m.f against temperature
so that you can read off the temperature corresponding to an e.m.f. reading.
•
Note that this graph is often a curve rather than a straight line.
32
Voltmeters and ammeters
•
There are a selection of instruments for measuring current and potential
difference (voltage).
•
The two main types are analogue meters, in which a pointer moves over a scale
and digital, in which the value is displayed on the read-out consisting of a series
of integers.
Analogue meters
•
The normal analogue meter is restricted to the measurement of the relevant
quantity over a single range.
•
Example 0 – 1 A d.c. ammeter will measure direct currents in the range from 0 to
1 A.
•
A 0 – 30 V d.c. voltmeter will measure steady potential differences in the range
from 0 to 30 V.
•
Some analogue meters have a dual-range facility, with a common negative
terminal and two positive terminals, each of which is associated with a separate
scale on the instrument.
•
Thus, one scale might be 0 to 3 A and the other 0 to 10 A.
•
Each of the positive terminals is marked with the scale to which if refers.
•
Be careful to take the reading on the scale corresponding to the pair of terminals
you have selected.
33
An analogue ammeter
•
Analogue meters are subject to zero errors and parallax errors.
•
The uncertainty is normally taken ± the smallest division on the analogue scale.
A galvanometer
•
A galvanometer is a sensitive current-measuring analogue meter.
•
It may be converted into an ammeter by the connection of a suitable resistor
(shunt) in parallel with the meter.
•
The meter may be converted into a voltmeter by the connection of a suitable
resistor (multiplier) in series with the meter.
•
A galvanometer with a centre-zero scale shows negative currents when the
needle is to the left-hand side of the zero mark and positive currents when it is to
the right.
•
This type of meter is often used as a null indicator, that is, to detect when the
current in a path of a circuit is zero.
34
A galvanometer as an ammeter
A galvanometer as a voltmeter
A galvanometer
35
Digital meters
•
Digital meters may have a zero error.
•
Before switching on the circuit, check whether the reading is zero.
•
If it is not zero, make a note of the reading and take it into account when reading
the current or voltage.
•
The use of a digital meter may save you the trouble of selecting an instrument
with a suitable range for your application.
•
Most have an auto-ranging function, that is, the instrument selects the most
sensitive range.
•
The uncertainty in the reading of a digital meter is expressed in terms of the
overall uncertainty and the uncertainty in the last digit.
Fig: 2.33 Digital meter
Analogue scales and digital displays
•
A simple meter rule gives an analogue reading while a digital watch gives a
digital reading.
•
Do not assume that a digital reading is more accurate than an analogue reading.
•
Most digital readings come from analogue readings.
•
A digital thermometer, for example, will probably be using a thermistor as its
source of information.
•
The p.d. across the thermistor will be measured, this will be combined with a
calibration curve available from the manufacturer, which gives its resistance at
different temperatures.
•
The value of the p.d. will then be digitised and finally displayed.
Calibration curves
•
Many measuring devices are checked by the manufacturer against international
standards of length, temperature and electric current.
36
•
Details are produced on how any particular instrument’s accuracy is dependent
on external factors such as temperature.
•
The information will often be in the form of a calibration curve in which the
reading obtained under particular conditions, is plotted against a corrected value
under standard conditions.
•
You might use sensors whose output is not proportional to the quantity you are
attempting to measure.
Example of a calibration curve for measuring fuel
Example of a calibration curve for a thermocouple - deduce the temperature when
e.m.f = 0.250 mV.
37
1.2.2 Errors and uncertainties
•
explain the effects of systematic errors (including zero errors) and random errors
in measurements
•
distinguish between precision and accuracy
 precise measurements are all close to one another
 an accurate measurement is close to the true value
•
assess the uncertainty in a derived quantity by simple addition of absolute,
fractional or percentage uncertainties (a rigorous statistical treatment is not
required)
Accuracy
•
An accurate measurement is close to the ‘true value’.
•
Accuracy depends on:
 The equipment used
 The skill of the experimenter
 The techniques used
•
Reducing systematic error or uncertainty in a measurement improves its
accuracy.
Precision
•
Precise measurements are all close to one another.
•
Precision is determined by the size of the random error in the measurements.
•
Precision is that part of accuracy which is within the control of the experimenter.
•
The experimenter may choose different measuring instruments and may use
them with different levels of skill, thus affecting the precision of measurement.
•
E.g. to measure the diameter of a steel sphere or marble, we could use a metre
rule, a vernier caliper, or a micrometer screw gauge.
•
The choice of measuring instrument would depend on the precision with which
we want the measurement to be made.
•
A meter rule could be used to measure to the nearest mm, the vernier caliper to
the nearest 0.1 mm and a micrometer screw gauge to the nearest 0.01 mm.
•
Metre rule 1.2 ± 0.1 cm
•
Vernier caliper 1.21 ± 0.01 cm
38
•
Micrometer screw gauge 1.212 ± 0.001 cm
•
The degree of precision increases from the metre rule to vernier caliper to
micrometer screw gauge.
•
The number of significant figures quoted for the measurement increases as the
precision increases.
Precision and accuracy
•
T is the true value.
Random errors
•
Two-sided errors, the reading obtained is either above or below the true value.
•
Affect the precision of measurement.
Uncertainty
•
The uncertainty is the total range of values within which the measurement is
likely to lie.
•
E.g. a measurement of 46.0 ± 0.5 cm implies that the most likely value is 46.0
cm, but it could be as low as 45.5 cm or as high as 46.5 cm.
•
When writing down measurements, the number of significant figures indicates its
uncertainty.
•
The absolute uncertainty in this measurement is ±0.5 cm
Types of uncertainties
𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑢𝑛𝑐𝑒𝑟𝑡𝑎𝑖𝑛𝑡𝑦
•
The fractional uncertainty =
•
The percentage uncertainty =
𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑣𝑎𝑙𝑢𝑒
𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑢𝑛𝑐𝑒𝑟𝑡𝑎𝑖𝑛𝑡𝑦
𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑣𝑎𝑙𝑢𝑒
±1%
•
=
Uncertainties are always added.
39
0.5
46.0
×
= ± 0.01 or ±
100%
1
=
0.5
46.0
×
1
100
100%
1
=
 When you multiply or divide quantities, add % uncertainties.
 When you add or subtract quantities, add absolute uncertainties.
Examples of uncertainties
Table: 2.2 Examples of uncertainty
Instrument
uncertainty
Typical reading
Stopwatch with 0.1 s divisions
±0.1 s
16.2 s
Thermometer with 1 °C intervals
±0.5 °C
22.5 °C
Ammeter with 0.1 A divisions
±0.1 A
2.1 A
Systematic errors
•
One sided errors obtained when the reading obtained is always above or below
the true value
•
affect the accuracy of the measurement
•
Zero error
•
Wrongly calibrated scale
•
Reaction time of the experimenter
Zero error
•
A zero error is an example of a systematic error.
•
This is when the scale is not zero before a measurement is taken.
•
Zero errors can either be positive or negative. When it is negative, add it to the
reading you obtain. When it is positive, subtract it to the reading you obtain.
•
E.g. this ammeter has a zero error of about −0.2 A
Fig: 2.46 this ammeter has a zero error of about -0.2A
40
Wrongly calibrated scale
•
In school laboratories we assume that measuring devices are correctly calibrated
and you would not be expected to check the calibration in an experiment.
•
However, if there a doubts, you can check calibration of an ammeter by
connecting several in series in a circuit, or of a voltmeter by connecting several in
parallel in a circuit, or of thermometers by placing them in well-stirred water.
•
These checks will not enable you to say which of the instruments are calibrated
correctly, but they will show you if there is any discrepancy.
Reaction time of the experimenter
•
When timings are carried out manually, it must be accepted that there will be a
delay between the experimenter observing the event and starting the timing
device.
•
This delay, called the reaction time, may be as much as a few tenths of a
second.
•
To reduce this effect, you should arrange that the intervals you are timing are
much greater than the reaction time.
•
For example, you should time sufficient swings of a pendulum for the total time to
be of the order of at least ten seconds, so that a reaction time of a few tenths of a
second is less important.
Random errors
•
Results in readings being scattered around the accepted value.
•
These may be reduced by repeating a reading and averaging, and by plotting a
graph a drawing a best fit line.
•
Examples of random errors:
– Reading a scale
– Timing oscillations
– Taking readings of a quantity that varies with time
– Reading a scale from different angles
41
End of topic activity
Multiple choice questions
Paper 12 February/ March 2019: 4, 5
Paper 11 May / June 2019: 4, 5
Paper 12 May / June 2019: 5, 6, 7
Paper 13 May / June 2019: 5, 6
Structured questions
Paper 21 May / June 2019: 1 (b)
Paper 22 May / June 2019: 1
42
Topic 1.3 Kinematics
1.3.1 Equations of motion
1.3.1 Equations of motion
•
define and use distance, displacement, speed, velocity and acceleration
– distance as a measure of how far an object travels along a particular path
(without considering direction)
– displacement as a vector which has a magnitude equal to the shortest
distance between the initial and final points and a direction from the initial
to the final point
– speed as a rate of change of distance
– instantaneous velocity as a ‘rate of change of displacement’ or speed in
a given direction
– average velocity as the total displacement divided by the total time taken
– acceleration as the rate of change of velocity
•
use graphical methods to represent distance, displacement, speed, velocity and
acceleration
Use graphical methods to represent position
43
•
At the origin O, x =0 and t=0.
•
From O to A, the graph is a straight line, the particle is covering equal distances
in equal periods of time. This is uniform velocity.
•
The average velocity during this time is
(𝑥1 −𝑂)
𝑡1 −𝑂
. This is the gradient of the straight
line joining O and A.
•
Between A and B, the particle is slowing down, because the distances travelled
in equal periods of time are getting smaller. The average velocity during this
period is
(𝑥2 − 𝑥1 )
𝑡2 −𝑡1
.
•
On the graph this is represented by the gradient of the secant line joining A and
B.
•
At B, for a moment, the particle is at rest, and after B is has reversed its direction
and is heading back towards the origin.
•
Between B and C, the velocity is
(𝑥3 −𝑥2 )
𝑡3 −𝑡2
. Because 𝑥3 is less than 𝑥2 , the gradient
is negative, so the velocity is negative, indicating the reversal of direction.
•
Calculating the average velocity of the particle over the relatively long intervals
𝑡1 , (𝑡2 − 𝑡1 ) and (𝑡3 − 𝑡2 ) will not, however, give us the complete description of
the motion.
•
To describe the motion exactly, we need to know the particle’s velocity at every
instant. This is instantaneous velocity.
1.3.1 Equations of motion
•
determine displacement from the area under a velocity-time graph
•
determine velocity using the gradient of a displacement-time graph
•
determine acceleration using the gradient of a velocity-time graph
Use graphical methods to represent displacement
44
Displacement – time
Acceleration
45
Deducing acceleration
•
The gives the acceleration, a.
•
a =
•
The area under the graph gives displacement, s.
•
s=
∆𝑣
∆𝑡
1
2
=
20 −0
5 −0
= 4 𝑚𝑠 −2
× base × height =
1
2
× 5 × 20 = 50 m.
46
Equations of motion
•
recall and use the equations of uniformly accelerated motion:
𝑣−𝑢
v = u + at; a =
∆𝑡
;s=
𝑣+𝑢
2
t
•
derive, from the definitions of velocity and acceleration, equations that represent
uniformly accelerated motion in a straight line
•
solve problems using equations that represent uniformly accelerated (constant
acceleration) motion in a straight line, including the motion of bodies falling in a
uniform gravitational field without air resistance
•
describe an experiment to determine the acceleration of free fall using a falling
body
•
describe and explain motion due to a uniform velocity in one direction and a
uniform acceleration in a perpendicular direction
•
You should be able to recall and use the equations:
v = u + at; a =
•
𝑣−𝑢
∆𝑡
;s=
𝑣+𝑢
2
t
You should be able to use the equations (these will be provided under formulae
in the question paper for both Paper 1 and Paper 2).
1
s = ut + at2
𝑣 2 = 𝑢2 + 2as
2
Deriving equations of motion (suvat equations)
Equation 1: v = u + at
•
a=
This comes from the definition of acceleration, the rate of change of velocity,
which can be obtained from the gradient of a straight line graph.
𝑣−𝑢
∆𝑡
; make v the subject of the formula
v – u = a ∆𝒕
v = u + at
𝒗+𝒖
Equation 2: s = (
𝟐
)t
Average velocity, 𝑣𝑎𝑣𝑒𝑟𝑎𝑔𝑒 =
𝒗+𝒖
𝟐
and
𝒔
𝒕
this gives
𝒗+𝒖
𝟐
=
𝒔
𝒕
𝒗+𝒖
so s = (
This gives 2s = (v+u) t
47
𝟐
)t
Equation 3: s = 𝒖𝒕 +
𝟏
𝟐
𝒂𝒕𝟐
 Substituting equation 1:
 Substituting v = u + at into s =(
(𝒖𝒕+𝒖𝒕 + 𝒂𝒕𝟐 )
o s=
𝟐
𝟏
o s = 𝒖𝒕 +
𝟐
𝐯+𝐮
v = u + at into equation 2:
𝒗+𝒖
𝟐
(𝒖 + 𝒂𝒕+𝒖)
)t this gives
,
s=
s =( 𝟐 ) t
s=
(𝟐𝒖𝒕 + 𝒂𝒕𝟐 )
s=
𝟐
t,
𝟐
𝟐𝒖𝒕
𝟐
+
𝒂𝒕𝟐
𝟐
𝟐
𝒂𝒕
Equation 4: 𝒗𝟐 = 𝒖𝟐 + 2as
 From equation 1: v = u + at. When we make t the subject of the formula,
 this gives v – u = a𝒕,
 Substitute t =
v–u
 This gives s =
2
𝑎
𝑣+𝑢
2
𝒗–𝒖
𝒂
=
𝒂𝒕
𝒂
, so t =
𝒗–𝒖
𝒂
𝑣+𝑢
into equation 2, s = (
×
v–u
𝑎
=
(𝑣 + 𝑢)(v – u)
2𝑎
2
2
)t.
s=
𝑣 2 − 𝑢𝑣 +𝑢𝑣 − 𝑢2
2𝑎
=
𝑣 2 − 𝑢2
2𝑎
 𝑣 − 𝑢 = 2as
 𝑣 2 = 𝑢2 + 2as
We can find the first two equations from the velocity-time graph shown in Fig: 2.16. The
graph represents the motion of an object. It initial velocity is u. after time t, its final
velocity is v.
48
Fig: 2.16
Equation 1: The graph of Fig: 2.16 is a straight line, therefore the object’s acceleration
a is constant. The gradient (slope) of the line is equal to acceleration.
The acceleration is defined as:
a=
(𝑣+𝑢)
𝑡
Which is the gradient of the line. Rearranging this gives the first equation of motion
v = u + at
(equation 1)
Equation 2: displacement is given by the area under the velocity-time graph. Fig: 2.17
shows that the object’s average velocity is half-way between u and v. So the object’s
average velocity, calculated by averaging its initial and final velocities, is given by:
(𝑣 + 𝑢)
2
The object’s displacement is the shaded area in Fig: 2.17. This is a rectangle, and so
we have:
Displacement = average velocity × time taken
hence: s =
(𝑣+𝑢)
2
×t
(Equation: 2)
Fig: 2.17
49
Equation 3: From equation 1 and 2, we can derive equation 3:
(equation 1)
v = u + at
s=
(𝑣+𝑢)
2
×t
(equation 2)
Substituting v from equation 1 gives:
s=
(𝑢+𝑢+𝑎𝑡)
2
×t
1
𝑠 = ut + + 2t2
So
=
𝑎𝑡2
2𝑢𝑡
+ 2
2
(equation 3)
Looking at Fig: 2.16, you can see that the two terms on the right of the equation
correspond to the areas of the rectangle and the triangle which make up the area under
the graph. Of course, this is the same area as the rectangle in Fig: 2.17.
Equation 4: This equation is also derived from equations 1 and 2
(equation 1)
v = u + at
s=
(𝑣+𝑢)
2
×t
(equation 2)
Substituting for time t from equation 1 gives: s =
(𝑢+𝑣)
2
Rearranging this gives:
2as = (u + v)(v- u) = v2 – u2
Or simply:
v2 = u2 + 2as
+
(𝑣+𝑢)
𝑎
(equation 2)
(equation 4)
Solving problems using equations of motion
These equations can only be used:
•
for motion in a straight line
•
for an object with constant acceleration
•
Step 1 We write down the quantities which we know, and the quantity we want to
find.
•
Step 2 Then we choose the equation which links these quantities, and substitute
in the values.
•
Step 3 Finally, we calculate the unknown quantity
50
Worked examples
4. The rocket show in Fig: 2.12 lifts off from rest with an acceleration of 20 m s-2.
Calculate its velocity after 50 s.
u = 0 m s-1
Step 1: What we know:
a = 20 m s-2
t = 50 s
and what we want to know:
v=?
Step 2: The equation linking u, a, t and v is equation 1:
v = u + at
v = 0 + (20 × 50) = 1000 m s-1
Substituting gives:
So v = 1000 m s-1
So the rocket will be travelling at 1000 ms-1 after 50 s. This makes sense, since its velocity
increases by 20m s-1 every second, for 50 s.
You can use the same equation to work out how long the rocket would take to reach a velocity of
2000m s-1, or the acceleration it must have to reach a speed of 1000m s-1 in 40 s, and so on.
5. The car shown in Fig: 2.13 is travelling along a straight road at 8.0 m s-1. I
accelerates at 1.0 m s-2 for a distance of 18 m. How fast is it then travelling?
Fig: 2.13: For worked example5. This car accelerates for a short distance
as it travels along the road.
Step 1: What we know:
u = 8.0 m s-1
A = 1.0 m s-2
S = 18 m
What we want to know:
v =?
Step 2: The equation we need is equation 4:
Substituting: v2 = (8.0)2 + (2 ×1.0 × 18)
v = √ (100) = 10 ms-1
51
v2 = u2 + 2as
52
Experiment - determine the acceleration of free fall
53
Motion in two dimensions – projectiles
54
•
A multiflash photograph can reveal details of the path, or trajectory, of a
projectile.
•
Once the ball has left the child’s hand and is moving through the air, the only
force acting on it is its weight.
•
The ball has been thrown at an angle to the horizontal.
•
It speeds up as it falls – you can see that the images of the ball become further
and further apart.
•
At the same time, it moves steadily to the right. You can see this from the even
spacing of the images across the picture. The ball’s path forms a parabola.
•
After it bounces, the ball is moving more slowly. It slows down, or decelerates, as
it rises – the images get closer and closer together.
•
The vertical motion of the ball is affected by the force of gravity, that is, its
weight.
•
When it rises it has a vertical deceleration of magnitude g, which slows it
down, and when it falls it has an acceleration of g, which speeds it up.
•
The ball’s horizontal motion is unaffected by gravity. In the absence of air
resistance, the ball has a constant velocity in the horizontal direction.
•
We can treat the ball’s vertical and horizontal motions separately, because they
are independent of one another.
A stone is thrown upwards with an initial velocity of 20 m𝒔−𝟏
55
•
It is important to use a consistent sign convention here. We will take upwards as
positive, and downwards as negative. So the stone’s initial velocity is positive,
but its acceleration g is negative. We can solve various problems about the
stone’s motion by using the equations of motion.
How high?
•
How high will the stone rise above ground level of the cliff? As the stone rises
upwards, it moves more and more slowly – it decelerates, because of the force of
gravity.
•
At its highest point, the stone’s velocity is zero. So the quantities we know are:
How long?
•
How long will it take from leaving your hand for the stone to fall back to the cliff
top? When the stone returns to the point from which it was thrown, its
displacement s is zero.
•
Use the quadratic formula, the information is in the form of:
•
ax2 + bx + c = 0
x=
−𝒃 ± √𝒃𝟐 − 4ac
𝟐𝒂
56
57
Time/ s
0.00
0.04
0.08
0.12
0.16
0.20
0.24
0.28
Horizontal distance/ m
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
Vertical distance/ m
0.000
0.008
0.031
0.071
0.126
0.196
0.283
0.385
𝟏
You can calculate the distance s fallen using the equation of motion s = u t + at2. (The
𝟐
initial vertical velocity u = 0)
The horizontal distance is calculated using:
horizontal distance = 2.5 × t
The vertical distance is calculated using:
vertical =
𝟏
𝟐
× 9.81 × t2
Worked examples:
9. A stone is thrown horizontally with a velocity of 12 ms-1 from the top of a vertical cliff.
Calculate how long the stone takes to reach the ground 40 m below and how far the
stone land from the base of the cliff.
Step 1: Consider the ball’s vertical motion. It has zero initial speed vertically and travels
40 m with acceleration 9.81 ms-2 in the same direction.
𝟏
s = u t + at2
𝟐
𝟏
40 = 0 + × 9.81 ×t2
𝟐
Thus: t = 2.86 s
58
Step 2:
Velocity, 12 ms-1, as long as there is no air resistance
Distance travelled = u × t = 12 × 2.86 = 34.3m
You may find it easier to summarize the information like this:
Vertically:
s = 40m
u = 0 ms-1
u = 12 ms-1
Horizontally:
a = 9.81 ms-2
v = 12 ms-1
a = 0 ms-2
t =?
v =?
t =?
s =?
10. A ball is thrown with an initial velocity of 20 ms-1 at an angle of 30° to the horizontal
(Fig: 2.32). Calculate the horizontal distance travelled by the ball (its range).
Step 1:
Split the ball’s initial velocity into horizontal and vertical component:
Initial velocity = u = 20 ms-1
Horizontal initial velocity = u cos Ø = 20 × cos 30° = 17.3 ms-1
Vertical initial velocity = u sin Ø = 20 × sin 30° = 10 ms-1
Step 2: Consider the ball’s vertical motion. How long will it take to return to the ground?
In other words, when will its displacement return to zero?
u = 10 ms-1
a = -9.81ms-2
s = 0m
𝟏
Using s = u t + at2, we have
0 = 10 t - 4.905 t2
𝟐
This gives t = 0 s
t =?
or t = 2.04 s. so the ball is in the air for 2.04 s.
Components of the projective vectors
59
•
In order to understand how to treat the velocity in the vertical and horizontal
directions separately we start by considering a constant velocity.
•
If an aeroplane has a constant velocity v at an angle θ as shown in Figure 2.28,
then we say that this velocity has two effects or components, 𝑣𝑁 in a northerly
direction and 𝑣𝐸 in an easterly direction.
•
These two components of velocity add up to make the actual velocity v. This
process of taking a velocity and determining its effect along another direction is
known as resolving the velocity along a different direction.
•
In effect splitting the velocity into two components at right angles is the
reverse of adding together two vectors – it is splitting one vector into two
vectors along convenient directions.
•
To find the component of any vector (e.g. displacement, velocity, acceleration) in
a particular direction, we can use the following strategy:
Step 1 Find the angle θ between the vector and the direction of interest.
Step 2 Multiply the vector by the cosine of the angle θ. So the component of an object’s
velocity v at angle θ to v is equal to v cos θ (Figure 2.28).
End of topic activity
Multiple choice questions
Paper 12 February/ March 2019: 6, 7
Paper 11 May / June 2019: 6, 7
Paper 13 May / June 2019: 7
60
Topic 1.4: Dynamics
1.4.1 Momentum and Newton’s laws of motion
1.4.2 Non-uniform motion
1.4.3 Linear momentum and its conservation
1.4.1 Momentum and Newton’s laws of motion
•
state that mass is the property of a body that resists change in motion
•
recall the relationship F = ma and solve problems using it, appreciating that
acceleration and resultant force are always in the same direction
•
define and use linear momentum as the product of mass and velocity
(recall p = mv)
•
define and use force as rate of change of momentum
•
state and apply each of Newton’s laws of motion:
– Newton’s first law (the law of inertia): an object at rest continues in a
state of rest or if moving continues moving with constant velocity unless it
is acted on by a resultant force
– Newton’s second law: the resultant force exerted on a body is directly
proportional to the rate of change of linear momentum of that body; and
recall and use (F =
∆𝒑
∆𝒕
)
– Newton’s third law: when two bodies interact, they exert forces on each
other, these forces have the same magnitude but are in opposite
directions
Force, mass and acceleration
•
The mass of an object is a measure of its inertia, a property that resists change
in its motion.
•
Recall and use F = ma
•
Define and use p = mv, unit = kg 𝐦𝐬 −𝟏
•
Define and use F =
•
(𝒎𝒗 −𝒎𝒖)
∆𝒑
𝒕
𝒕
Recall and use (F =
;F=
∆𝒑
∆𝒕
)
61
•
State and apply Newton’s laws of motion:
•
Newton’s first law (the law of inertia): an object at rest continues in a
state of rest or if moving continues moving with constant velocity unless it
is acted on by a resultant force
•
Newton’s second law: the resultant force exerted on a body is directly
proportional to the rate of change of linear momentum of that body;
•
Newton’s third law: when two bodies interact, they exert forces on each
other, these forces have the same magnitude but are in opposite
directions
Characteristics of Newton’s third law pair of forces
•
They are forces of the same type.
What does it mean to say that the forces are ‘of the same type’?
We need to think about the type of interaction which causes the forces to appear.
•
Two objects may attract each other because of the gravity of their masses –
these are gravitational forces.
•
Two objects may attract or repel because of their electrical charges – electrical /
electrostatic forces.
•
Two objects may attract or repel because of their magnetic fields – magnetic
forces.
•
Two objects may touch – contact forces.
•
Two objects may be attached by a string and pull on each other – tension
forces.
•
They are forces of the same type.
•
They act on different objects (bodies).
•
They are equal in magnitude / size.
•
They are opposite in direction.
•
They are in the same line of action.
•
They act for the same time.
62
Example of Newton’s third law pair of forces
1.4.2 Non-uniform motion
•
describe and use the concept of weight as the effect of a gravitational field on a
mass and recall that the weight of a body is equal to the product of its mass and
the acceleration of free fall
•
describe and explain qualitatively the motion of bodies falling in a uniform
gravitational field with air resistance (including reference to terminal velocity)
•
recall that acceleration may be constant even when the motion is non-uniform
63
•
If an object has two or more forces acting on it, we have to consider whether or
not they are ‘balanced’.
•
Forces on an object are balanced when the resultant force on the object is
zero.
•
The object will either remain at rest or have a constant velocity.
•
We can calculate the resultant force by adding up two (or more) forces which act
in the same straight line.
•
We must take account of the direction of each force.
Weight
•
W = mg (relates to F = ma)
•
When a body (e.g rain drops, a parachutist) falls in a uniform gravitational field, air
resistance plays an important role and it is taken into consideration to describe
and explain the motion.
•
When a body falls through a resistive fluid (liquid or gas), as the speed increases,
the resistive force (air or water resistance) also increases. When this resistive
force reaches a value equal and opposite to the weight of the falling body, the
body no longer accelerates and continues at uniform velocity.
•
At the beginning of falling, the acceleration = g but decreases to zero at the time
when the terminal velocity is reached.
•
This is a case of motion with non-uniform acceleration.
64
•
acceleration may be constant even when the motion is non-uniform
•
Skydivers are rather like cars – at first, they accelerate freely. At the start of the
fall, the only force acting on the diver is his or her weight. The acceleration of the
diver at the start must therefore be g.
•
Then increasing air resistance opposes their fall and their acceleration
decreases.
•
Eventually they reach a maximum velocity, known as the terminal velocity. At
the terminal velocity the air resistance is equal to the weight, so the resultant
force = 0 and at this point the diver moves with constant velocity.
•
The idea of a parachute is to greatly increase the air resistance. Then terminal
velocity is reduced, and the parachutist can land safely.
•
Terminal velocity depends on the weight and surface area of the object. For
insects, air resistance is much greater relative to their weight than for a human
being and so their terminal velocity is quite low.
•
Insects can be swept up several kilometres into the atmosphere by rising air
streams.
•
Later, they fall back to Earth uninjured.
•
It is said that mice can survive a fall from a high building for the same reason
1.4.3 Linear momentum and its conservation
•
define impulse as F∆t
•
relate impulse to change in momentum (F∆t = ∆p)
65
•
use the relationship between impulse and change in momentum to calculate the
force exerted, time for which the force is applied and change in momentum for a
variety of situations involving the motion of an object in one dimension
•
apply the concept of impulse to safety considerations in everyday life, e.g.
airbags, seatbelts and arrestor beds
•
state the principle of conservation of momentum, that when bodies in a system
interact, the total momentum remains constant (momentum is always conserved)
provided that no external force acts on the system
•
apply the principle of conservation of momentum to solve simple problems,
including elastic and inelastic interactions between bodies in both one and two
dimensions (knowledge of the concept of coefficient of restitution is not required)
– in elastic interactions, kinetic energy is conserved
– in inelastic interactions, kinetic energy is not conserved
•
recognise that, for a perfectly elastic collision, the relative speed of approach is
equal to the relative speed of separation
•
explain that, while momentum of a system is always conserved in interactions
between bodies, some change in kinetic energy may take place
Impulse
•
impulse = F∆t, unit for impulse = Ns
•
impulse = change in momentum; F∆t = ∆p
•
The unit for momentum is = kg 𝐦𝐬 −𝟏 , however since F∆t = ∆p and the unit for
F∆t = Ns, the unit for momentum can also be regarded as Ns.
•
Recall: N = kg 𝐦𝐬 −𝟐 × s = kg 𝐦𝐬 −𝟏
•
use the relationship F∆t = ∆p to calculate:
– force exerted,  F =
∆𝒑
∆𝒕
– time for which the force is applied, ∆𝒕 =
∆𝒑
F
– change in momentum, ∆p = F∆t
Impulse and safety considerations
•
A very important application of impulse is improving safety and reducing
injuries.
66
•
In many cases, an object needs to be brought to rest from a certain initial
velocity.
•
This means there is a certain specified change in momentum.
•
If the time during which the momentum changes can be increased then the
force that must be applied will be less and so it will cause less damage.
•
This is the principle behind arrestor beds for trucks, airbags, seatbelts and
bending your knees when you jump off a chair and land on the ground.
F =
∆𝒑
∆𝒕
•
Airbags, seatbelts and arrestor beds ensure that smaller forces are exerted over
longer periods of time. This reduces the impact.
•
From the equation F = ∆𝒕 , the resultant force F is inversely proportional to ∆𝒕.
∆𝒑
Thus, the longer the ∆𝒕, the lesser the F .
•
This is the principle of application of impulse to safety considerations in everyday
lives such as in road safety.
A seatbelt in a vehicle
An airbag in a vehicle
Namibia’s first arrestor bed at Usakos
An arrestor bed is a patch of ground that is softer than the road.
67
•
state the principle of conservation of momentum, that when bodies in a system
interact, the total momentum remains constant (momentum is always conserved)
provided that no external force acts on the system
• 𝑚1 𝑣1 + 𝑚2 𝑣2 = 𝑚1 𝑣1 + 𝑚2 𝑣2
•
in elastic interactions, kinetic energy is conserved, the total kinetic energy
before interaction = total kinetic energy after interaction
•
1
•
for a perfectly elastic interaction, the relative speed of approach is equal to the
relative speed of separation
•
𝒖𝟏 + 𝒖𝟐 = 𝒗𝟏 + 𝒗𝟐
•
in inelastic interactions, kinetic energy is not conserved
•
while momentum of a system is always conserved in interactions between
bodies, some change in kinetic energy may take place
2
1
1
1
2
2
2
𝑚1 𝑢1 2 + 𝑚2 𝑢2 2 = 𝑚1 𝑣1 2 + 𝑚2 𝑣2 2
End of topic activity
Multiple choice questions
Paper 12 February/ March 2019: 8, 9, 10
Paper 11 May / June 2019: 9, 10, 11
Paper 12 May / June 2019: 8, 9, 10, 11
Paper 13 May / June 2019: 8, 9, 17
Structured questions
Paper 22 February / March 2019: 3
Paper 21 May / June 2019 : 2
9702/13/MJ/19
17. Step 1: Find velocity
𝑀𝐴 𝑣𝐴 + 𝑀𝐵 𝑣𝐵 = 𝑀𝐴 𝑣𝐴 + 𝑀𝐵 𝑣𝐵
(500 × 20) + 0 = v (95 + 5)
1000 = 100v
So v = 10
68
Step 2: Find height
𝑣 2 = 𝑢2 + 2as
0 = 102 + 2 (−9.81) s
0 = 100 − 19.62 s
− 100 = − 19.62 s
s = 5.1 m
•
(a) 𝑀𝑥 𝑣𝑥 + 𝑀𝑦 𝑣𝑦 = 𝑀𝑥 𝑣𝑥 + 𝑀𝑦 𝑣𝑦
•
𝑀𝑥 (3.0) + 2.5 (−9.6 cos 60) = 0
•
3.0 𝑀𝑥 = 12
•
𝑀𝑥 = 4.0 kg
•
(b) 2.5 (9.6 sin 60) = (4.0 + 2.5) v
•
2.5 (9.6 sin 60) = 6.5 v
•
v = 3.2 𝑚𝑠 −1
1
2
1
(c) 𝐸𝑘 = 2 × 2.5 × (9.6) − 2 × 4.0 × (3.0)
2
69
Topic 1.5: Forces, density and pressure
1.5.1 Types of forces
1.5.2 Turning effects of forces
1.5.3 Equilibrium of forces
1.5.4 Density and pressure
1.5.1 Types of forces
•
describe the force on a mass in a uniform gravitational field and on a charge in a
uniform electric field
•
explain the origin of the up-thrust acting on a body in a fluid (due to the
difference in hydrostatic pressure)
•
explain frictional forces and viscous forces including air resistance (no treatment
of the coefficients of friction and viscosity is required)
•
apply the concept that the weight of a body may be taken as acting at a single
point known as its centre of gravity
Identifying forces
70
71
•
Weight of a body is an example of a force acting on a mass in a uniform
gravitational field.
•
Near the surface of the earth, the gravitational field is approximately constant
(uniform). This means that in calculations we take the same value for g = 9.81
𝒎𝒔−𝟐 whatever the position on the surface of the Earth or for a short distance
(compared with the Earth’s radius) above it.
•
An electric field is another example of a field of force. An electric charge
experiences a force in an electric field for example a charged particle ( + or − )
between two parallel plates (+ plate and − plate) (Theme 3 section 3.1.1 and
Theme 4 section 4.1).
•
An object in a fluid (liquid or gas) experiences a frictional forces (viscous force)
caused by the viscosity of the fluid. Examples of viscous forces are air
resistance and water resistance.
•
An object immersed in a fluid appears to weigh less than when in a vacuum, this
is due to upthrust acting on the body in a fluid.
•
The origin of upthrust is due to the difference in hydrostatic pressure
(hydrostatic pressure depends on depth and increases with depth).
72
•
The weight of a body may be taken as acting at a single point known as its
centre of gravity.
•
The centre of gravity of an object is defined as the point where all the weight of
the object may be considered to act.
•
An object may have two or more forces acting on it and, since these are vectors,
we must use vector addition find their combined effect (their resultant).
•
There are several forces acting on the car (Figure 4.2) as it struggles up the
steep hill.
•
They are:
•
its weight W (= mg)
•
the contact force N of the road (its normal reaction)
•
air resistance D
•
the forward force F caused by friction between the car tyres and the road.
1.5.2 Turning effects of forces
•
define moment as the product of force and perpendicular distance through the
line of action from the pivot
•
apply the moment of a force to everyday examples such as crowbar,
wheelbarrow, pliers, scissors, tweezers or tongs
•
state that a couple is a pair of (equal but opposite) forces (acting along parallel
but different lines) that tends to produce rotation only
•
define and apply the torque of a couple (torque as the product of the magnitude
of one of the forces and the distance of separation)
73
Moment of a force
•
moment = force × perpendicular distance through the line of action from the pivot
•
The moment of a force depends on two quantities:
– the magnitude of the force (the bigger the force, the greater its moment)
– The perpendicular distance of the force from the pivot (the further the
force acts from the pivot, the greater its moment).
Couple of forces
•
a couple is a pair of (equal but opposite) forces (acting along parallel but
different lines) that tends to produce rotation only
A torque of a couple
•
a torque of a couple (a pair of equal but opposite forces) is the product of the
magnitude of one of the forces and the distance of separation
•
Pure turning effect. When we calculate the moment of a single force, the result
depends on the point or pivot about which the moment acts.
•
The further the force is from the pivot, the greater the moment.
•
A couple is different; the moment of a couple does not depend on the point about
which it acts, only on the perpendicular distance between the two forces.
74
•
A single force acting on an object will tend to make the object accelerate (unless
there is another force to balance it).
•
A couple, however, is a pair of equal and opposite forces, so it will not make the
object accelerate.
•
This means we can think of a couple as a pure ‘turning effect’, the size of which
is given by its torque.
•
For an object to be in equilibrium, two conditions must be met at the same time:
•
The resultant force acting on the object is zero.
•
The resultant moment is zero.
75
1.5.3 Equilibrium of forces
•
state and apply the principle of moments
•
recall and apply the principle that, when there is no resultant force and no
resultant torque, a system is in equilibrium
•
use a vector triangle to represent three coplanar forces in equilibrium
•
The principle of moments states that: For any object that is in equilibrium, the
sum of the clockwise moments about any point provided by the forces acting on
the object equals the sum of the anticlockwise moments about that same point.
•
We can use the idea of the moment of a force to solve two sorts of problem:
•
We can check whether an object will remain balanced or start to rotate.
•
We can calculate an unknown force or distance if we know that an object is
balanced (in equilibrium).
•
We can use the principle of moments to solve problems.
•
Note that, for an object to be in equilibrium, we also require that no resultant
force acts on it.
76
Vector triangle to represent three coplanar forces in equilibrium
•
The spider shown is hanging by a thread. It is blown sideways by the wind. The
diagram shows the three forces acting on it:
•
Weight acting downwards
•
Tension in the thread acting diagonally
•
The push of the wind acting from west.
77
•
If the wind blew a little harder, there would be an unbalanced force on the spider
and it would move to the right.
•
If we work out the resultant force on an object and find that it is zero, this tells us
that the object is in equilibrium.
•
If we know that an object is in equilibrium, we know that the forces on it must add
up to zero.
•
We can use this to work out the values of one or more unknown forces.
Components of forces
•
If we consider the spider in equilibrium example, even though three forces are
acting on it, the spider is in equilibrium. The tension has an effect like that of a
diagonal in a right angled-triangle.
•
The weight of the spider is the vertical component of the tension and the wind
blowing from west is the horizontal component of the tension.
•
The principle of resolution of vectors is also applicable to a vector triangle of
coplanar forces in equilibrium.
1.5.4 Density and pressure
•
define and use density (density as the mass per unit volume)
•
define and use pressure (pressure as the perpendicular force per unit area)
•
derive, from the definitions of pressure and density, the equation Δp = ρgΔh
•
use the equation for hydrostatic pressure Δp = ρgΔh
78
𝐹
𝑚𝑔
•
define: p =
•
define : 𝜌 =
•
recall: v = l × b × h
•
recall: A = l × b
•
substitute: p =
•
so p = 𝜌𝑔ℎ
𝐴
=
𝑚
𝑣
𝐴
and m = 𝜌v
𝐹
𝐴
=
𝑚𝑔
𝐴
=
𝜌v𝑔
𝐴
=
𝜌 ×𝒍 ×𝒃 ×ℎ ×𝑔
𝒍 ×𝒃
End of topic activity
Multiple choice questions
Paper 12 February/ March 2019: 11, 12, 13, 14
Paper 11 May / June 2019: 12, 13, 14, 15, 16
Paper 12 May / June 2019: 12, 14, 15, 16
Paper 13 May / June 2019: 11, 12, 13, 14
Structured questions
Paper 22 February/ March 2019: 2
Paper 22 May / June 2019 : 2 (a), (b)
Paper 23 May / June 2019 : 3
79
Topic 1.6: Work, energy and power
1.6.1 Energy conversion and conservation
1.6.2 Work and efficiency
1.6.3 Potential energy and kinetic energy
1.6.4 Power
1.6.1 Energy conversion and conservation
•
give examples of energy in different forms, its conversion and conservation, and
apply the principle of conservation of energy to simple examples (e.g. the kinetic
energy changing to potential energy in a pendulum and the sum of the two is
constant if air resistance is negligible)
Forms of energy
•
Gravitational potential energy – energy of a mass due to its position in a
gravitational field.
•
elastic potential energy – energy stored in an object as a result of of
reversible (elastic) deformation
•
Electrostatic potential energy – energy due the position of a charge in an
electric field.
•
Nuclear potential energy – energy associated with particles in the nuclei of
atoms.
•
Chemical potential energy – energy (stored in chemical bonds) released during
chemical reactions.
•
Kinetic energy – energy due to motion.
•
Sound energy – energy transferred from particle to particle associated with
sound wave.
•
Electromagnetic radiation – energy associated with waves in the
electromagnetic spectrum.
•
Solar energy – electromagnetic radiation from the sun.
•
Internal energy – random kinetic and potential energy of the molecules in an
object.
•
Thermal energy – energy transferred due to temperature difference (also called
heat energy).
80
•
Electrical energy – energy associated with moving charge carriers due to a
potential difference.
Energy conversion
•
When chemical energy is used, the energy is transformed into other forms of
energy, some of which are useful and some of which are not.
•
For example, when petrol is burned in a car engine, some of the chemical
energy is converted into kinetic energy of the car and some is wasted as heat
(thermal) energy.
•
When the car stops, its kinetic energy is converted into internal energy in the
brakes. The temperature of the brakes increases and heat energy is released
into the atmosphere.
•
When energy changes from one form to another, this is called energy
conversion.
Example 1: diver jumping off a diving board
•
The diver uses his gravitational potential energy to do work in bending the
diving board.
•
The work done is stored as elastic potential energy, which is then converted
into kinetic energy of the diver as he is pushed upwards and off the diving
board.
•
At the same time, some of the elastic potential energy is lost as heat and
sound due to dissipative forces in the diving board.
Example 2: hammering a nail into a wooden block
•
A person uses the chemical energy in his muscles to work against the
gravitational pull in order to lift the hammer.
•
The work done is converted into the gravitational potential energy of the
hammer in its raised position.
81
•
As the hammer falls, its gravitational potential energy is converted into kinetic
energy.
•
When the hammer hits the nail, its kinetic energy is used to do work in driving
the nail into the wooden block, producing sound energy in the air and thermal
energy in the block, nail and hammer.
Example 3: bouncing ball
•
As the ball falls, its gravitational potential energy is converted into its kinetic
energy.
•
When the ball hits the ground, the ball is deformed during the collision. Its kinetic
energy is converted into elastic potential energy. Some kinetic energy may be
lost as thermal energy or sound energy.
•
The elastic potential energy is converted back into kinetic energy as the ball
regains its original shape.
•
The kinetic energy is converted into gravitational potential energy as the ball
bounces upwards, until it reaches its highest position.
•
During the flight, presence of air resistance will cause kinetic energy to be
dissipated as thermal energy, thus reducing the total energy in the ball and its
subsequent height after each bounce.
Example 4: burning of fossil fuel
•
When fuels such as oil, coal and wood are burnt, the chemical energy stored in
these materials is converted into thermal energy (heat) and light energy.
Example 5: falling plasticine
•
As the plasticine is falling, the gravitational potential energy is converted to
kinetic energy.
•
During impact, all kinetic energy is converted into thermal and sound
energies as the plasticine is permanently deformed.
Energy conservation
•
As energy conversions take place, the outcome is that the total energy present in
the universe remain constant.
•
All the energy conversions are governed by the principle of conservation of
energy which states that energy cannot be created or destroyed, it can only
be converted from one form to another.
82
•
The total energy of a closed system is constant. This means that energy
cannot be created or destroyed but it can be changed from one form to
another.
Pendulum example of energy conversion
•
A 2.0 kg metal ball is suspended from a rope as a pendulum. If it is released from
point A and swings down to the point B (the bottom of its arc):
– show that the velocity of the ball is independent of its mass,
– Calculate the velocity of the ball at point B.
Step 1: Analyse the question to determine what information is provided
•
The mass m of the metal ball is m = 2 kg
•
The change in height going from point A to point B is h = 0.5 m.
•
The ball is released from point A so the velocity at point, 𝑣𝐴 = 0 m s-1
•
All quantities are in SI units.
Step 2: Analyse the question to determine what is being asked
• Prove that the velocity is independent of mass.
• Find the velocity of the metal ball at point B.
Step 3: Apply the Law of Conservation of Mechanical Energy to the
situation (mechanical energy = potential energy + kinetic energy)
• Since there is no friction, mechanical energy is conserved. Therefore:
• 𝑬𝒎𝟏 = 𝑬𝒎𝟐 this means 𝑬𝒑𝟏 + 𝑬𝒌𝟏 = 𝑬𝒑𝟐 + 𝑬𝒌𝟐
• 𝑚𝑔ℎ1 +
1
2
m𝑣1
2
= 𝑚𝑔ℎ2 +
1
2
m𝑣2
83
2
1
• 𝑚𝑔ℎ1 + 0 = 0 +
• 𝑚𝑔ℎ1 =
1
2
m𝑣2
2
m𝑣2
2
2
• The mass of the ball m appears on both sides of the equation so it
can be eliminated so that the equation becomes:
1
• 𝑔ℎ1 = 𝑣2
2
2
• 2𝑔ℎ1 = 𝑣2
2
• This proves that the velocity of the ball is independent of its mass.
Step 4: Calculate the velocity of the ball at point B
• We can use the equation above, or do the calculation from “first
principles”:
2
• 𝑣2 = 2𝑔ℎ1 = (2) (9.81) (0.5)
• 𝑣2
2
= 9.81
• 𝑣2 = √9.81
• 𝑣2 = 3.13 ms-1
• It does not matter what its mass is, it will always have the same
velocity when it falls through this height.
• We could use the same equation to calculate the speed of an object
falling from height h. An object of small mass gains the same speed
as an object of large mass, provided air resistance has no effect.
Alternatively this can be summarised as:
𝑬𝒎𝟏 = 𝑬𝒎𝟐 this means 𝑬𝒑𝟏 + 𝑬𝒌𝟏 = 𝑬𝒑𝟐 + 𝑬𝒌𝟐
• 𝑚𝑔ℎ1 +
1
2
m𝑣1
• 𝑚𝑔ℎ1 + 0 = 0 +
2
1
2
= 𝑚𝑔ℎ2 +
m𝑣2
1
2
m𝑣2
2
84
2
•
1
2
m𝑣2
• 𝑣2
2
=
2
= 𝑚𝑔ℎ1
𝑚𝑔ℎ1
𝑚
=
(2) (9.81)(0.5)
2
• 𝑣2 = √9.81
• 𝑣2 = 3.13 ms-1
• NB: The sum of 𝐸𝑝 and 𝐸𝑘 is constant if air resistance is negligible.
1.6.2 Work and efficiency
•
explain the concept of work in terms of the product of a force and displacement
in the direction of the force
•
calculate the work done in a number of situations including the work done by a
gas that is expanding against a constant external pressure: W = p Δ V
•
recall and apply that the efficiency of a system is the ratio (which can be
expressed as percentage) of useful energy output from the system to the total
energy input
•
discuss the implications of energy losses in practical devices and use the
concept of efficiency to solve problems
•
The work done by a force is defined as the product of the force (F) and the
distance (s) moved in the direction of the force.
•
W (in Joule) = F (in Newton) × s (in metre)
•
1 joule = 1 newton × 1 metre; 1 J = 1 N m
•
The size of the force F – the bigger the force, the greater the amount of work you
do.
•
Work done = energy transferred.
•
1 joule is also the amount of energy transferred when a force of 1 newton moves
a distance of 1 metre in the direction of the force.
Force, distance and direction
• It is important to appreciate that, for a force to do work, there must be
movement in the direction of the force.
85
• Both the force F and the distance s moved in the direction of the
force are vector quantities, so you should know that their directions
are likely to be important.
• To illustrate this, we will consider three examples involving gravity.
• In the equation for work done, W = F × s , the distance moved s is
thus the displacement in the direction of the force.
•
Suppose that the force F moves through a distance s which is at an angle θ to F,
as shown in Figure 5.6. To determine the work done by the force, it is simplest to
determine the component of F in the direction of . This component is F cos θ, and
so we have: work done = (F cos θ) × s Or simply: work done = Fs cos θ Worked
example 1 shows how to use this.
86
A gas doing work
•
Gases exert pressure on the walls of their container. If a gas expands, the walls
are pushed outwards – the gas has done work on its surroundings. In a steam
engine, expanding steam pushes a piston to turn the engine, and in a car engine,
the exploding mixture of fuel and air does the same thing, so this is an important
situation.
87
•
Figure 5.8 shows a gas at pressure p inside a cylinder of cross-sectional area A.
•
The cylinder is closed by a moveable piston.
•
The gas pushes the piston a distance s.
•
If we know the force F exerted by the gas on the piston, we can deduce an
expression for the amount of work done by the gas.
•
From the definition of pressure (pressure = area ), the force exerted by the gas
force
on the piston is given by: force = pressure × area; F = p × A ; and the work
done is force × displacement: W = p × A × s
•
A × s is the same as l×b×h which is the increase in volume (l × b × h) of the
gas; that is, the shaded volume in Figure 5.8. We call this ∆V, where the ∆
indicates that it is a change in V.
•
Hence the work done by the gas in expanding is:
W = p∆V
•
Notice that we are assuming that the pressure p does not change as the gas
expands.
•
This will be true if the gas is expanding against the pressure of the atmosphere,
which changes only very slowly.
Efficiency
•
the efficiency of a system is the ratio (which can be expressed as percentage) of
useful energy output from the system to the total energy input
•
efficiency =
•
Since energy cannot be created, efficiency can never be greater than 100%
useful energy output
total energy input
100%
×(
1
)
88
•
Machines are used to change energy from one form to some other more useful
form. In most energy changes, some energy is wasted as heat (thermal) energy
due to resistive forces such as friction.
•
For practical devices to work, energy input is needed. Most modern practical
devices run on electrical energy (e.g. television, computer) or chemical energy
(e.g. vehicle).
•
When a practical device works, it converts the energy input into both useful
energy output and wasted energy output.
•
If the output is intended in our design or operation of the device, then it is useful.
If it is not intended, then it is considered as wasted energy output.
•
Efficiency of a practical device is a measure of how much useful work that device
produces from a given amount of energy input.
•
Its value depends on what energy output we consider as useful.
•
Efficiency is dimensionless and can be expressed as a ratio or percentage.
•
We can never make a practical device with 100% efficiency because:
•
We have limited control over physical processes (e.g. a filament bulb must heat
up before it produces light, but the heat produced becomes wasted energy);
•
Dissipative forces tend to convert useful energy into heat, which is a form of
wasted energy. Mechanical devices are especially susceptible to this.
1.6.3 Potential energy and kinetic energy
1
•
derive, from the equations of motion, the formula for kinetic energy 𝐸𝑘 = 𝑚𝑣 2
•
recall and apply the formula 𝐸𝑘 = 𝑚𝑣 2
•
distinguish between gravitational potential energy and elastic potential energy
2
1
2
– gravitational potential energy as energy of a mass due to its position in a
gravitational field
– elastic potential energy as energy stored in an object as a result of
reversible (elastic) deformation
•
apply the relationship between force and potential energy in a uniform field to
solve problems
•
derive, from the defining equation W = F s, the formula ΔEp = mgΔh for
gravitational potential energy changes near the Earth’s surface
89
•
recall and use the formula Δ𝐸𝑝 = mgΔh for gravitational potential energy
changes near the Earth’s surface
Derivation of 𝐸𝑘 equation
•
recall W = Fs;
•
recall: 𝒗𝟐 = 𝒖𝟐 + 2as, so 𝒗𝟐 − 𝒖𝟐 = 2as
•
note: if u = 0 then 𝒗𝟐 − 𝒖𝟐 = 𝒗𝟐
•
Thus 𝒗𝟐 = 2as;
•
If we makes the subject of the formula: s =
•
recall: F = ma;
•
recall W = Fs; thus W = mas and W = mgs
•
since s = 2a therefore W =
𝑣2
W = 𝐸𝑘 =
•
•
•
𝑚𝑣 2
2
ma
1
×
𝒗𝟐
𝟐𝒂
𝑣2
2a
1
= 𝑚𝑣 2
2
Potential energy (𝐸𝑝 ) is the energy possessed by a system by virtue of the
relative positions of its component parts.
gravitational potential energy (𝐸𝑝 ) is energy of a mass due to its position in a
gravitational field
elastic potential energy (𝐸𝑘 ) is energy stored in an object as a result of
reversible (elastic) deformation
Relationship between force and potential energy in a uniform field
•
In a uniform field (such as a uniform gravitational, electrical or magnetic field),
a body experiences the same force F at all points.
•
For example a body in a uniform gravitational field experiences the same force
(weight) due to uniform gravitational acceleration such as g = 9.81 ms−𝟐 on or
near the surface of the Earth.
•
If this force F moves a body along a distance s in its direction, then the work
done W = F ∆s.
•
Using the principle of conservation of energy, this work done must be
compensated for by a decrease in potential energy − ∆𝐸𝑝 .
•
Thus W = Fs = − ∆𝐸𝑝 so if F∆s = − ∆𝐸𝑝
90
−∆𝐸𝑝
•
then F =
•
F is the force acting on the point mass / charge placed at that particular point in
the field.
•
∆𝑠
−∆𝐸𝑝
is the change in the potential energy of a point mass / charge with a
∆𝑠
variation of the distance from the source of the field.
Answer
(a) F =
−∆𝐸𝑝
∆𝑠
so − ∆𝐸𝑝 = F ∆𝑠 and ∆𝐸𝑝 = − F ∆𝑠
−24 × 5.0 = −120 (decreasing ∆𝐸𝑝 )
(b) ∆𝐸𝑝 = − F ∆𝑠 = −24 × − 5.0
= 120 (increasing ∆𝐸𝑝 )
(c) ∆𝐸𝑝 = − F ∆𝑠 = −24 × 0 = 0
Derivation of 𝐸𝑝 = mgh
•
Suppose we want to lift an object of mass m to a height h above the ground, so
that its velocity remains constant.
91
•
to do so, we must apply a force F that is equal but opposite to the weight mg of
the object, where g is the acceleration of free fall, recall: if v is constant, resultant
force = 0
•
recall: W = Fs
•
s is distance which can be expressed as height h
•
recall: F = ma and if F = weight, F = mg
•
The work done W by force F on the object is:
•
W = Fs; W = (mg)s; W = mgh
•
Since the object’s velocity is constant, its kinetic energy is also constant. Hence,
by conservation of energy, the work done W on the object must be equal to the
gain in 𝑬𝒑 of the object.
•
Change in 𝐸𝑝 near Earth’s surface; 𝐸𝑝 = mgh
1.6.4 Power
•
define power as work done per unit time
•
derive power as the product of force and velocity
•
recall and use the relationships P =
𝑊
𝑡
and P = Fv
Power
•
Machines such as wind turbines or engines do work for us when they change
energy into a useful form.
•
However, not only is the availability of useful forms of energy important, but also
the rate at which it can be converted from one form to another.
•
The rate of converting energy or using energy is known as power.
•
Power =
•
Power is the work done per unit time, or the rate of work done, or the rate of
energy conversion.
•
Power, like energy, is a scalar quantity.
•
Consider a force F which moves a distance x at constant velocity v in the
direction of the force, in time t. The work done W by the force is given by W = Fx.
𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒
𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
;P=
𝑊
𝑡
92
𝑊
•
Recall P =
•
Recall v =
•
Thus P = Fv
𝑡
; so P = F
𝑥
𝑡
𝑥
𝑡
End of topic activity
Multiple choice questions
Paper 12 February/ March 2019: 15, 16, 17, 18
Paper 11 May / June 2019: 17, 18, 19
Paper 12 May / June 2019: 17, 18, 19
Paper 13 May / June 2019: 15, 16, 18
Structured questions
Paper 21 May / June 2019 : 3
Paper 23 May / June 2019 : 2
93
Topic 1.7: Deformation of solids
1.7.1 Stress and strain
1.7.2 Elastic and plastic behaviour
1.7.1 Stress and strain
•
outline that deformation is caused by a force and that, in one dimension, the
deformation can be tensile or compressive
•
use the terms load, extension and compression
•
explain and use the terms limit of proportionality, elastic limit, yield point and the
spring constant (i.e. force per unit extension)
•
obtain and draw force-extension, force-compression, and tensile/compressive
stress-strain graphs
•
recall and use Hooke’s law (F = kx)
•
define and use the terms stress, strain and the Young modulus:
– stress as the force per unit area of a material
– strain as extension per unit length
– Young modulus as the ratio of stress to strain
•
describe an experiment to determine the Young modulus of a metal in the form
of a wire
Tensile and compressive deformation
•
deformation is caused by a force, in one dimension, the deformation can be
tensile or compressive
•
a pair of forces is needed to change the shape of a spring. If the spring is being
squashed and shortened, we say that the forces are compressive. More usually,
we are concerned with stretching a spring, in which case the forces are
described as tensile (Figure 7.4).
94
Load, extension and compression
Fig. 8.4 shows a spring without any weight attached.
Fig. 8.4 and Fig. 9.1 show weights attached to the lower ends of the springs. As the
weight is increased, the spring becomes longer. The increase in length of the spring is
the extension and the weight attached is the load. When the load is removed,
compression occurs.
Force-extension graph
95
Extension-force graph
Force-compression graph
A force-extension, extension-force and force compression graph
•
Specific for an object made of a certain material with certain dimensions and
NOT to the type of material in general.
•
Different springs for example with different dimensions would have different
graphs, even though they are made up of the same material.
•
Similarly, a spring and straight wire made up of the same material would have
different graphs.
96
Limit of proportionality and the spring constant
•
Hooke’s law states that, provided the proportionality limit is not exceeded, the
extension of a body is proportional to the applied load.
•
The law can be expressed in the form of an equation: fore F ∝ extension e.
•
This gives F = ke where k is a constant known as the spring constant (or force
constant).
•
The spring constant is the force (in N) per unit extension (in m) = ( ).
•
The unit of the constant is newton (N) per metre (m) (Nm−1 ).
•
On a force-extension graph, the spring constant is given by the gradient of the
𝑵
𝒎
straight line (k =
∆𝑭
∆𝒙
).
97
Stress and strain
•
stress (𝜎 sigma symbol) is the force per unit area of a material and does not take
the dimensions of the material into account but depends on the area, force per
unit area.
𝜎=
•
𝐹
𝐴
=
𝑁
𝑚2
; unit is N𝑚−2
strain (𝜀 epsilon symbol) is extension per unit length, it is independent of the
original dimensions of the material.
𝑥
𝑚
𝐿
𝑚
𝜀 = = ; strain has NO unit
Stress-strain graph
•
A stress-strain graph is the same for all objects of the same material regardless
of their dimensions.
•
A stress-strain graph can be used to investigate the behaviour of a material
under tensile force.
•
The tensile stress and tensile strain of a material are used to describe the
behaviour of a material independent of its original dimensions.
98
Tensile/compressive stress-strain graph
•
The point on a stress-strain graph where the stress is proportional to the strain is
the limit of proportionality.
•
The elastic limit is the maximum force that can be applied to a wire/ spring such
that the wire/ spring returns to its original length when the force is removed.
•
Beyond the elastic limit, the spring/ wire becomes permanently deformed
(plastic deformation).
•
The yield point is the point on a stress-strain curve that indicates the limit of
elastic behaviour and the beginning of plastic behaviour.
•
Prior to the yield point, a material will deform elastically and will return to its
original shape when the applied stress is removed.
Stress, extension and Young modulus
99
•
Young modulus (modulus of elasticity, E) is the ratio of stress to strain.
•
From a stress-strain graph within the limit of proportionality, the gradient of the
∆𝜎
graph is given by m =
and it is a constant gradient.
∆𝜀
•
This is used to determine the Young modulus of a material which is a
characteristic physical property of the material (just like density, resistivity, etc).
•
E=
•
Thus, the Young modulus is a measure of the stiffness of a material.
•
The stiffer the material, the greater its Young modulus and the steeper the
gradient on a stress-strain graph.
𝜎
𝜀
=
N𝑚−2
𝑛𝑜 𝑢𝑛𝑖𝑡
; unit = N𝑚−2
Determining Young modulus
•
As the long wire is stretched, the position of the sticky tape pointer can be read
from the scale on the bench.
•
Why do we use a long wire? Obviously, this is because a short wire would not
stretch as much as a long one. We need to take account of this in our
calculations, and we do this by calculating the strain produced by the load.
•
A travelling microscope is placed above the wire and focused on the sticky tape
pointer. When the pointer moves, the microscope is adjusted to keep the pointer
at the middle of the cross-wires on the microscope.
•
Metals are not very elastic. In practice, they can only be stretched by about 0.1%
of their original length. Beyond this, they become permanently deformed.
100
•
As a result, some careful thought must be given to getting results that are good
enough to give an accurate value of the Young modulus.
•
First, the wire used must be long. The increase in length is proportional to the
original length, and so a longer wire gives larger and more measurable
extensions.
•
Typically, extensions up to 1 mm must be measured for a wire of length 1 m
(1000mm). To get suitable measurements of extension there are two
possibilities: use a very long wire, or use a method that allows measurement of
extensions that are a fraction of a millimetre.
•
The distance that the pointer has moved can then be measured accurately from
the scale on the microscope.
•
In addition, the cross-sectional area of the wire must be known accurately. The
diameter of the wire is measured using a micrometer screw gauge. This is
reliable to within ±0.01 mm (precision of micrometer screw gauge). Once the
wire has been loaded in increasing steps, the load must be gradually
decreased to ensure that there has been no permanent deformation of the wire.
•
Other materials such as glass and many plastics are also quite stiff, and so it is
difficult to measure their Young modulus.
•
Rubber is not as stiff, and strains of several hundred percent can be achieved.
However, the stress–strain graph for rubber is not a straight line. This means the
value of the Young modulus found is not very precise, because it only has a
very small linear region on a stress–strain graph.
1.7.2 Elastic and plastic behaviour
•
distinguish between elastic and plastic deformation of a material
– elastic deformation being reversible when the stress is removed
– plastic deformation being permanent as a result of dislocations
•
relate the area under the force-extension graph to the work done (the area under
the force-extension graph = work done)
•
determine the elastic potential (strain) energy in a deformed material from the
area under the force-extension graph
•
recall and use 𝐸𝑝 =
1
2
Fx =
1
2
kx2 for a material deformed within its limit of
proportionality
Elastic and plastic deformation
•
elastic deformation is reversible when the stress is removed
101
•
plastic deformation is permanent as a result of dislocations
•
Whenever you stretch a material, you are doing work. This is because you have
to apply a force and the material extends in the direction of the force (recall:
Work = Force × distance).
•
Similarly, when you push down on the end of a springboard before diving, you
are doing work. You transfer energy to the springboard, and you recover the
energy when it pushes you up into the air.
•
We call the energy in a deformed solid the elastic potential energy or strain
energy.
•
If the material has been strained elastically (the elastic limit has not been
exceeded), the energy can be recovered.
•
If the material has been plastically deformed, some of the work done has gone
into moving atoms past one another, and the energy cannot be recovered
(wasted as other forms of energy).
•
The material warms up slightly. We can determine how much elastic potential
energy is involved from a force–extension graph, see Figure 8.18.
•
We need to use the equation that defines the amount of work done by a force.
That is: work done = force × distance moved in the direction of the force
102
•
Elastic potential energy = work done = area under the force-extension graph
(triangle), E𝑝 =
1
2
Fx.
•
The work done in stretching or compressing a material is always equal to the
area under the graph of force against extension.
•
This is true whatever the shape of the graph, provided we draw the graph with
extension on the horizontal axis. If the graph is not a straight line, we cannot use
the
1
2
Fx relationship, so we have to resort to counting squares or some other
technique to find the answer.
•
However, the elastic potential energy relates to the elastic part of the graph (i.e.
up to the elastic limit), so we can only consider the linear section of the force–
extension graph.
•
There is an alternative equation for elastic potential energy, E𝑝 =
•
According to Hooke’s law the applied force F and extension x are related by F =
kx, where k is the force constant (obtained from the gradient of the F versus x
graph).
•
Substituting for F into E𝑝 =
•
E𝑝 =
1
2
Fx =
So:
1
2
1
2
Fx gives:
× kx × x
E𝑝 =
1
2
𝑘𝑥 2
End of topic activity
Multiple choice questions
Paper 12 February/ March 2019: 19, 20, 21
Paper 11 May / June 2019: 20, 21
Paper 12 May / June 2019: 20
Paper 13 May / June 2019: 19, 20
Structured questions
Paper 21 May / June 2019 : 3
Paper 22 May / June 2019 : 2 (c)
103
1
2
Fx.
Theme 2: Waves
Theme 2 consists of 6 Topics
2.1 Progressive waves
2.2 Transverse and longitudinal waves
2.3 Determination of frequency and wavelength of sound waves
2.4 Doppler effect
2.5 Electromagnetic spectrum
2.6 Superposition
Part 1
2.1 Progressive waves
2.2 Transverse and longitudinal waves
2.4 Doppler effect
2.5 Electromagnetic spectrum
Topic 2.1 Progressive waves
•
describe what is meant by wave motion (propagation), an oscillation which
transfers energy from one place to another without any net movement of the
medium, as illustrated by vibration in ropes, springs and by experiments using
water waves
•
describe and use the terms displacement, amplitude, phase difference, period,
frequency, wavelength and speed
•
derive, from the definitions of speed, frequency and wavelength, the wave
equation v = fλ
•
recall and use the equations v = fλ and f = 𝑇
•
describe that energy is transferred by a progressive wave
•
recall and use the relationship intensity = 𝑎𝑟𝑒𝑎 and intensity ∝ (amplitude)2
1
𝑝𝑜𝑤𝑒𝑟
Wave motion (propagation)
•
Waves that move through a material (or a vacuum) are called progressive
waves.
•
A progressive wave transfers energy from one position to another without the
transfer of matter.
104
•
A wave motion (propagation) is an oscillation which transfers energy from one
place to another without any net movement of the medium.
•
This can be illustrated by vibration in ropes, springs and by experiments using
water waves.
Illustration of progressive waves by vibration in ropes
Illustration of progressive waves by vibration in springs
Variation of the displacement of particles with distance along a transverse wave
See: https://www.youtube.com/watch?v=AUBAMlMoI1g
105
Illustration of progressive waves by vibration in springs
Illustration of progressive waves by experiments using water waves
106
Displacement, amplitude, phase difference, period, frequency, wavelength and
speed
Displacement, amplitude, wavelength
•
Displacement (x) – the distance, in a specified direction, of a particle / point on
the wave from its equilibrium position.
•
Amplitude (A) – the maximum displacement of any point on the wave from its
equilibrium position.
•
Wavelength () – distance between two corresponding points in successive
waveforms such as two successive crests or troughs.
Phase and phase difference
107
•
Phase difference – is the amount (in degrees or seconds) by which one
oscillation leads or lags behind another. All points along a wave have the same
pattern of vibration. However, different points do not necessarily vibrate in step
with one another. As one point on a stretched string vibrates up and down, the
point next to it vibrates slightly out-of-step with it. We say that they vibrate out
of phase with each other – there is a phase difference between them.
A central angle has a measure of 1 radian if it is
subtended by an arc whose length = radius of the
circle
Degrees vs radians
Period, frequency and wave speed
•
Period (T) – time taken for one complete oscillation of a point in a wave. It is the
time taken for a point to move from one particular position and return to that
same position, moving in the same direction. It is measured in seconds (s).
108
•
Frequency (f) – the number of complete oscillations in one second. Unit Hz =
𝑠 −1 .
•
Wave speed (v)– the speed (rate of change of distance) with which energy is
transmitted by a wave is. The unit is 𝑚𝑠 −1 . The wave speed for sound in air is
about 340 𝑚𝑠 −1, while the wave speed for light in a vacuum it is almost 300 000
000 𝑚𝑠 −1 .
Derive, from the definitions of speed, frequency and wavelength, the wave
equation v = fλ
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
•
recall: speed =
•
for one oscillation of the wave:
𝑡𝑖𝑚𝑒
– distance = wavelength (λ)
– time = period (T)
𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆
•
from speed =
•
recall: f =
•
thus v = λ÷ 𝒇; v = λ × 𝟏
•
Hence v = fλ
1
𝒕𝒊𝒎𝒆
and T =
𝑇
𝟏
we get speed =
𝒘𝒂𝒗𝒆𝒍𝒆𝒏𝒈𝒕𝒉 (λ)
𝒑𝒆𝒓𝒊𝒐𝒅 (𝑻)
1
𝑓
𝒇
Wave energy
•
It is important to realise that, for both types of mechanical waves, the particles
that make up the material through which the wave is travelling do not move
along – they only oscillate about a fixed point.
•
It is energy that is transmitted by the wave.
•
Each particle vibrates; as it does so, it pushes its neighbour, transferring energy
to it.
•
Then that particle pushes its neighbour, which pushes its neighbour.
•
In this way, energy is transmitted from one particle to the next, to the next,
and so on down the line.
Wave speed and intensity
•
recall and use the equations:

v = fλ
109
 f=
•
recall and use the relationship:

•
1
𝑇
intensity =
𝑝𝑜𝑤𝑒𝑟
𝑎𝑟𝑒𝑎
=
𝑤𝑎𝑡𝑡, 𝑊
𝑚2
= 𝑊𝑚−2
The intensity of a wave is defined as the rate of energy transmitted (POWER)
per unit area at right angles to the wave velocity.
Intensity and amplitude
•
The intensity of a wave generally decreases as it travels along. There are two
reasons for this:
– The wave may ‘spread out’ (e.g. light spreading out from a light bulb) and
since intensity =
𝒑𝒐𝒘𝒆𝒓
𝒂𝒓𝒆𝒂
, when the area increases, the denominator
increases and so the intensity decreases.
– the wave may be absorbed or scattered (as when light passes through the
Earth’s atmosphere)
•
As a wave spreads out, its amplitude decreases.
– This suggests that the intensity I of a wave is related to its amplitude A.
– Intensity ∝ (amplitude) 2 so I ∝ A𝟐 (y  𝒙𝟐 relationship).
– The relationship I ∝ A𝟐 also implies that for a particular wave I = kA𝟐 this
means
I
A𝟐
= k.
– From I = kA𝟐 , when the amplitude A decreases, the intensity I also
decreases.
•
If the waves from a point source spread out equally in all directions, we have
what is called a spherical wave.
•
As the wave travels further from the source, the energy it carries passes through
an increasingly large area.
•
Since the surface area of a sphere is 4𝒓𝟐 (recall from Mathematical
𝑝𝑜𝑤𝑒𝑟 (𝑊)
requirements), the intensity is 𝑎𝑟𝑒𝑎 (4𝑟 2 ), where W is the power of the source.
•
The intensity of the wave thus decreases with increasing distance from the
source.
110
Topic 2.2 Transverse and longitudinal waves
•
compare transverse and longitudinal waves
– for a transverse wave, the oscillations are perpendicular to the direction of
travel of the energy of the wave
– for a longitudinal wave, the oscillations are parallel to the direction of travel
of the energy of the wave
•
analyse and interpret graphical representations of transverse and longitudinal
waves
Longitudinal and transverse waves
•
Both longitudinal and transverse can be demonstrated using a slinky spring
lying along a bench.
– for a transverse wave, the oscillations are perpendicular to the direction
of travel of the energy of the wave
– for a longitudinal wave, the oscillations are parallel to the direction of
travel of the energy of the wave
•
Sound waves are an example of a longitudinal wave.
•
Light and all other electromagnetic waves are transverse waves.
•
Waves in water are quite complex. Particles of the water may move both up and
down and from side to side as a water wave travels through the water. You can
investigate water waves in a ripple tank.
Graphical representations of transverse and longitudinal waves
111
•
Figure 14.8 shows how we can represent longitudinal and transverse waves.
•
The longitudinal wave shows how the material through which it is travelling is
alternately compressed and expanded. This gives rise to high and low pressure
regions respectively.
•
However, this is rather difficult to draw, so you will often see a longitudinal wave
represented as if it were a sine wave (Mathematical requirements).
•
The displacement referred to in the graph is the displacement of the particles in
the wave.
•
We can compare the compressions and rarefactions (or expansions) of the
longitudinal wave with the peaks and troughs of the transverse wave.
Sine wave - Mathematical requirements
112
Topic 2.4 Doppler effect
•
explain that when a source (of waves) moves relative to a stationary observer,
there is a change in observed frequency
•
use the expression 𝑓𝑜 =
𝑓𝑠 𝑉
(𝑣 ±𝑣𝑠 )
for the observed frequency when a source of
sound waves moves relative to a stationary observer
•
explain that Doppler shift is observed with all waves, including sound and light
Doppler effect
•
The whistle of a train or the siren of a police car appears to increase in
frequency as it moves towards a stationary observer.
•
When a source (of waves) moves relative to a stationary observer, there is a
change in observed frequency.
•
The frequency change due to the relative motion between a source of sound
or light and an observer is known as the Doppler effect.
•
When the observer and source of sound are both stationary, the number of
waves per second reaching the observer will be the same frequency as the
source (see Figure 14.13).
•
Recall: the velocity of a wave depends on the medium in which the wave travels
113
•
When the source moves towards the observer the effect is to shorten the
wavelength of the waves reaching the observer (see Figure 14.14).
•
Let v be the speed of sound in air (recall that v depends on the medium in which
the wave travels). A source of sound has a frequency 𝒇𝒔 and wavelength λ.
•
The source moves towards an observer at a speed 𝒗𝒔 .
•
The period of oscillation (T = 𝒇 ) of the source of sound is T (=
•
In the time of one oscillation the source moves towards the observer a
distance 𝒗𝒔 T (recall: distance = speed × time).
•
Hence the wavelength is shortened by this distance 𝒗𝒔 T
•
The wavelength of the sound received by the observer is λ − 𝒗𝒔 T
•
Recall f =
•
recall T =
•
from v = fλ, λ = 𝒇
•
If 𝒇𝒐 =
•
𝒇𝒐 = ÷ ( -
𝟏
𝒗
λ
, hence the frequency observed 𝒇𝒐 =
𝟏
𝒇
𝒗
𝟏
𝒇
𝒇𝒔
and λ = so 𝒗𝒔 T = 𝒗𝒔
𝒗
(λ - 𝒗𝒔 T)
𝒗
= 𝒔
𝒇𝒔
𝒗
𝒗
𝒔
, so 𝒇𝒐 =
(λ - 𝒗𝒔 T)
𝒗
𝒗
𝒗𝒔
𝟏
𝒇𝒔
𝒇𝒔
𝒗
𝒗 𝒗 [split this into two fractions]
(𝒇 - 𝒇 𝒔 )
𝒔 𝒔
) [notice that the denominator is the same]
114
1
𝒇𝒔
).
𝒇𝒐 =
𝒗
𝟏
÷
𝒗 −𝒗𝒔
𝒇𝒔
[multiply the first fraction by the reciprocal of the second one]
𝒗
𝒇𝒔
𝟏
𝒗 −𝒗𝒔
𝒇𝒐 = ×
SO 𝒇𝒐 =
𝒇𝒔 𝒗
(𝒗 - 𝒗𝒔 )
•
The source could move away from a stationary observer at position P on the lefthand side of Figure 14.14. The observed wavelengths would lengthen.
•
For a source of sound moving away from an observer the observed frequency
𝒇𝒔 𝒗
𝒗
𝒗
can be shown to be 𝒇𝒐 =
= 𝒗 𝒗𝒔
𝒇𝒐 =
(λ + 𝒗𝒔 T)
(𝒗 + 𝒗𝒔 )
( + )
𝒇𝒔
𝒇𝒔
•
The frequency is increased when the source moves towards the observer (due
to shortened wavelengths) and the frequency is decreased when the source
moves away from the observer (due to lengthened wavelengths).
•
The expressions apply only when the source of waves is sound. However, a
change of frequency (Doppler shift) is observed with all waves, including light.
•
In astronomy, the wavelength tends to be measured rather than the frequency. If
the measured wavelength of an observed spectral line is less than that measured
for a stationary source, then the distance between the source (star) and detector
is decreasing (blue shift). If the measured wavelength is greater than the value
of a stationary source, then the source is moving away from the detector (red
shift).
115
•
The blue and red shifts are referred to in this way as red has the longest
wavelength and blue the shortest wavelength in the visible spectrum
•
(recall: dispersion of light through a triangular prism – red, orange, yellow,
green, blue, indigo, violet)
Topic 2.5 Electromagnetic spectrum
•
state that all electromagnetic waves are transverse waves that travel with the
same speed in free space
•
recall the orders of magnitude of the wavelengths of the principal regions of the
electromagnetic spectrum from radio waves to gamma rays
Electromagnetic spectrum
•
Visible light is just a small region of the electromagnetic spectrum.
•
All electromagnetic waves are transverse waves, consisting of electrical and
magnetic fields which oscillate at right angles to each other and to the
direction in which the wave is travelling.
•
Electromagnetic waves show all the properties common to wave motions: they
can be reflected, refracted, diffracted, obey the principle of superposition
and produce interference patterns.
•
In a vacuum, all electromagnetic waves travel at the same speed, 3.0 × 𝟏𝟎
𝒎𝒔−𝟏 .
•
The complete electromagnetic spectrum is divided into a series of regions based
on the properties of electromagnetic waves in these regions.
•
It should be noted that there is no clear boundary between regions (the
regions overlap).
116
𝟖
•
Recall the mnemonic RaMIVUXY to recall the orders of the electromagnetic
waves: Radio waves, Micro waves, Infrared radiation, Visible light, Ultraviolet
radiation, X-rays, Gamma (𝜸) rays.
End of topic activity
Cambridge International AS and A level question papers
Multiple choice questions
Paper 12 February/ March 2019: 22, 23, 25, 26
Paper 11 May / June 2019: 22, 23, 25, 26
Paper 12 May / June 2019: 23, 24, 26, 27
Paper 13 May / June 2019: 21, 22, 24, 25
117
Topic 2.6 Superposition
2.6.1 Stationary waves
2.6.2 Diffraction
2.6.3 Interference, two-source interference
2.6.4 Diffraction gratings
Part 2
2.6.3 Interference, two-source interference
2.6.1 Stationary waves
2.6.3 Interference, two-source interference
•
define the terms interference and coherence
– coherence - when two waves both have the same frequency (and
wavelength) and a constant phase difference
– interference - when two or more waves overlap/superpose, the resultant
displacement is the sum of the displacements of each wave
•
describe experiments that demonstrate two-source interference using water
ripples, light (monochromatic light source e.g. laser) and microwaves
•
discuss the conditions required if two-source interference fringes are to be
observed
•
recall and solve problems using the equation λ = 𝐷 for double-slit interference
𝑎𝑥
using light
Interference
Fig: 15.1
118
•
In Fig. 15.1, the fisherman is going to experience the effects of interference.
•
The amplitude of oscillation of his boat will be significantly affected by the two
approaching waves and their interaction when they reach his position.
•
If two or more waves overlap, the resultant displacement is the sum of the
individual displacements.
•
Remember that displacement is a vector quantity.
•
The overlapping waves are said to interfere.
•
This may lead to a resultant wave of either a larger or a smaller displacement
than either of the two component waves.
Two-source interference using water ripples
Constructive interference
119
Destructive interference
•
If two waves arrive at a point at the same time in phase, that is if their crests or
troughs arrive at exactly the same time, they will interfere constructively.
•
A resultant wave will be produced which has crests much higher than either of
the two individual waves, and troughs which are much deeper.
•
If the two incoming waves have the same frequency and equal amplitude A,
the resultant wave produced by constructive interference has an amplitude 2A.
The frequency of the resultant is the same as that of the incoming waves.
•
If two waves are in antiphase (have a phase difference of 180 °) the crests of
one wave arrive exactly the same time as the troughs from the other wave and
they interfere destructively.
•
The resultant wave will have smaller amplitude.
•
If the incoming waves have equal amplitudes, the resultant wave has zero
amplitude.
The principle of superposition of waves
•
The situation of constructive and destructive interference of waves is an example
of the principle of superposition of waves.
•
The principle of superposition states that, when two or more waves meet at a
point, the resultant displacement at that point is equal to the sum of the
displacements of the individual waves at that point.
•
Because displacement is a vector quantity, we must remember to add the
individual displacements taking into account of their directions.
•
Superposition applies to all types of waves.
120
Two-source interference using monochromatic light
Double slit
Two-source interference using monochromatic light
•
A monochromatic light has one colour, hence one wavelength.
121
•
A simple arrangement to show the interference effects produced by light involves
directing a monochromatic light source from a laser through two slits.
•
A monochromatic light has one colour, hence one wavelength.
•
The slits are two clear lines on a black slide, separated by a fraction of a
millimetre.
122
•
Light from the monochromatic source is diffracted a the slit, producing two
lights sources at the double slit.
•
Because these two light sources originate from the same primary source, they
are coherent (have the same frequency, wavelength and a constant phase
difference).
•
Where the light falls on the screen, a series of equally spaced dots of light are
seen (Fig. 15.21).
Interference ‘fringes’
•
These bright dots (maxima) are referred to as interference ‘fringes’, and they
are regions where light waves from the two slits are arriving in phase with each
other, i.e. constructive interference.
•
The dark regions (minima) in between are the result of destructive
interference.
•
In order to observe interference, we need two coherent sources of waves
(hence the use of a monochromatic light source).
Two-source interference using microwaves
Fig: 15.17 Microwave can also be used to show interference effects.
•
Using 2.8 cm wavelength microwave equipment (Fig. 15.17), you can observe an
interference pattern.
•
The microwave transmitter is directed towards the double gap (similar to a
double slit) in a metal barrier.
123
•
The microwaves are diffracted at the two gaps so that they spread out into the
region beyond, where they can be detected using the probe receiver.
•
By moving the probe around, it is possible to detect regions of high intensity
(constructive interference) and low intensity (destructive interference), now
always zero intensity.
•
The probe may be connected to a meter, or to an audio amplifier and
loudspeaker to give an audible output.
Rays from the two slits travel different distances to reach the screen
Fig: 15.22 Rays from the two slits travel different distances to reach the screen.
•
Point A: this point is directly opposite the midpoint of the two slits (1 and 2). Two
rays of light arrive at A, one from slit 1 and the other from slit 2. Point A is
equidistant from the two slits, and so the two rays of light have travelled the
same distance. The path difference between the two rays of light is zero. If we
assume that they were in phase (in step) with each other when they left the
slits, then they will be in phase when they arrive at A. Hence they will interfere
constructively, and we will observe a bright fringe at A.
•
Point B: this point is slightly to the side of point A, and is the midpoint of the first
dark fringe. Again, two rays of light arrive at B, one from each slit. The light
from slit 1 has to travel slightly further than the light from slit 2, and so the two
rays are no longer in step. Since point B is at the midpoint of the dark fringe, the
two rays must be in antiphase (phase difference of half a wavelength/ have a
period = 180°) so the two rays interfere destructively.
•
Point C: this point is the midpoint of the next bright fringe with AB = BC. Again,
ray 1 has travelled further than ray 2; this time, it has travelled an extra distance
equal to a whole wavelength λ. The path difference between the rays of light
is now a whole wavelength. The two rays are in phase at the screen. They
interfere constructively and we see a bright fringe.
124
Recall and use the equation λ =
•
𝒂𝒙
𝑫
for determining wavelength λ
The double-slit experiment can be used to determine the wavelength λ of light.
The following three quantities: slit separation a; fringe separation x and Slitto-screen distance D have to be measured.
1. Slit separation a
•
This is the distance between the centres of the slits, though it may be easier to
measure between the edges of the slits. (It is difficult to judge the position of the
centre of a slit. If the slits are the same width, the separation of their left-hand
edges is the same as the separation of their centres.) A travelling microscope
is suitable for measuring a.
2. Fringe separation x
•
This is the distance between the centres of adjacent bright (or dark) fringes. It
is best to measure across several fringes (say, ten) and then to calculate
later the average separation. A metre rule or travelling microscope can be
used.
3. Slit-to-screen distance D
This is the distance from the midpoint of the slits to the central fringe on the screen.
It can be measured using a metre rule or a tape measure. Once these three quantities
have been measured, the wavelength λ of the light can be found using the relationship:
𝒂𝒙
λ=
𝑫
Work example
1. In a double-slit experiment using light from a helium-neon laser, a student obtained
the following results: width of 10 fringes 10x = 1.5 cm
Separation of slits a = 1.0 mm
Slit-to-screen distance D = 2.40 cm
Determine the wavelength of the light.
Step 1: work out the fringe separation: fringe separation 𝑥
=
1.5 ×102
10
=1.5 ×10-3 m
Step 2: substitute the valves of a, x and D in the expression for wavelength λ:
λ=
𝒂𝒙
𝑫
=
𝟏.𝟎 ×𝟏𝟎−𝟑 ×𝟏.𝟓 ×𝟏𝟎−𝟑
𝟐.𝟒𝟎
= 6.3 × 10-7 m
Don’t forget to convert all the distances into metres.
125
Topic 2.6.1 Stationary waves
•
explain and use the principle of superposition in simple application
•
describe experiments that demonstrate stationary waves using microwaves,
stretched strings and air columns
•
explain the formation of a stationary wave using a graphical method and identify
nodes and antinodes
The principle of superposition of waves
•
The situation of constructive and destructive interference of waves is an example
of the principle of superposition of waves.
•
the principle of superposition states that, when two or more waves meet at a
point, the resultant displacement at that point is equal to the sum of the
displacements of the individual waves at that point.
•
Because displacement is a vector quantity, we must remember to add the
individual displacements taking into account of their directions.
•
Superposition applies to all types of waves.
RECAP: progressive waves
126
Stationary waves in music
progressive vs stationary waves: https://www.youtube.com/watch?v=b3j1RE7EGG0
Stationary waves
•
The waves we have considered so far have been progressive waves; they start
from a source and travel outwards, transferring energy from one place to
another.
•
A second important class of waves is stationary waves (standing waves).
•
A stationary wave is set up by the superposition of two progressive waves of
the same type, amplitude and frequency travelling in opposite directions.
•
A stationary (or standing) wave is one in which some points are permanently at
rest (nodes), others between these nodes are vibrating with varying amplitude,
and those points with the maximum amplitude (antinodes) are midway
between the nodes.
•
These can be produced using a spring or a slinky spring. A long rope or piece
of rubber tubing will also do. Lay it on the floor and fix one end firmly.
•
Move the other end from side to side so that transverse waves travel along the
length of the spring and reflect off the fixed end (Figure 16.2).
•
If you adjust the frequency of the shaking, you should be able to achieve a
stable pattern like one of those shown in Figure 15.7.
•
Alter the frequency in order to achieve one of the other patterns.
127
Demonstrating stationary waves using a slinky string
Fig: 15.17 First four modes of vibration of a string
Progressive vs Stationary
Demonstrating stationary waves using a slinky spring
128
Stationary waves using a slinky spring
Fig: 15.16 A stationary wave is created when two waves travelling in opposite directions
interfere.
Fundamental mode also known as first harmonic – lowest frequency known as
fundamental frequency
•
The fundamental mode or first harmonic of a standing wave is the wave which
has the lowest frequency and fewest number of nodes and antinodes.
•
This wave pattern has a single (one) loop.
•
At the end of the string there is no vibration. These points are called nodes (no
motion). At the centre, the amplitude is a maximum. A point of maximum
amplitude is the antinode.
•
Nodes and antinodes do not move along the string. The wavelength  = 2L.
From v = f  , f =
v

so f =
𝒗
2L
129
•
There are points along the spring that remain (almost) motionless while points
on either side are oscillating with the greatest amplitude.
•
The points that do not move are called the nodes and the points where the
spring oscillates with maximum amplitude are called the antinodes.
•
We normally represent a stationary wave by drawing the shape of the spring in
its two extreme positions.
•
If two points are on either side of a node, they are in antiphase (out of phase)
1
1
and will have a phase difference of 180 ° (2 wavelength/ 2 period).
•
The spring appears as a series of loops, separated by nodes.
The second and third harmonic
𝒗
•
For the second harmonic wave, the wavelength  = L. From v = f  , f = .
•
For the third harmonic wave, the wavelength
L
=
𝟐𝑳
𝟑
from v = f  , f =
𝒗
1
÷
𝟐𝑳
𝟑
,f=
𝟑𝒗
𝟐𝑳
1st, 2nd , 3rd and 4th harmonic
130
131
132
General expression for the frequency of the nth mode
𝑓𝑛 =
𝑛𝑣
2𝐿
where n is the number of loops on the stationary wave. Examples:
1 loop: 𝒇𝟏 =
𝟏𝒗
𝟐𝑳
2 loops: 𝒇𝟐 =
3 loops: 𝒇𝟑 =
4 loops: 𝒇𝟒 =
5 loops: 𝒇𝟓 =
6 loops: 𝒇𝟔 =
𝟐𝒗
𝟐𝑳
=
𝒗
𝑳
𝟑𝒗
𝟐𝑳
𝟒𝒗
𝟐𝑳
=
𝟐𝒗
𝑳
𝟓𝒗
𝟐𝑳
𝟔𝒗
𝟐𝑳
=
𝟑𝒗
𝑳
Use superposition to explain stationary waves
•
The diagram shows the formation of a stationary wave by superposition of
two progressive waves of equal amplitude and frequency travelling in opposite
directions.
•
The red-dashed line is travelling from left to right, and the blue-dashed line is
going from right to left.
•
When the string is clamped (at the right-hand end), the effect of the clamp is to
change the phase of the reflected wave by 180°.
•
In this graph, the waves meet at an instant at the same time, in phase.
Superposition gives the purple curve, which has twice the amplitude of either
of the progressive waves. The two waves interfere constructively.
133
•
In this graph, the waves meet a quarter of a period (a quarter of a cycle) later
in opposite directions.
•
This has brought the two waves to a situation where the movement of one wave
relative to the other is half a wavelength, so that the waves are exactly out of
phase.
•
The resultant, obtained by superposition, is zero everywhere.
•
In this graph, half a period (half a cycle) from the start, the waves are again in
phase, with maximum displacement for the resultant.
•
In this graph, the waves meet three-quarters of a period (cycle) later in
opposite directions.
•
This has brought the two waves to a situation where the movement of one wave
relative to the other is half a wavelength, so that the waves are exactly out of
phase.
•
The resultant, obtained by superposition, is zero everywhere.
134
•
In this graph, one period from the start, the waves are again in phase, with
maximum displacement for the resultant.
•
In the previous diagrams, we saw how there are some positions, the nodes N,
where the displacement of the resultant is zero throughout the cycle.
•
The displacement of the resultant at the antinodes A fluctuates from a
maximum value when the two progressive waves are in phase to zero when
they are out of phase.
Demonstrate stationary waves using air columns
•
Fig. 15.22 shows an experiment to demonstrate the formation of stationary
waves in air.
•
A fine, dry powder (such as cork dust or lycopodium powder) is sprinkled evenly
along the transparent tube. A loudspeaker powered by a signal generator is
placed at the open end.
•
The frequency of the sound from the loudspeaker is gradually increased.
•
At certain frequencies, the powder forms itself into evenly spaced heaps along
the tube.
135
•
A stationary wave has been set up in the air, caused by the interference of the
sound wave from the loudspeaker and the wave reflected from the closed
end of the tube.
•
At nodes (positions of zero amplitude) there is no disturbance, and the powder
can settle into a heap.
•
At antinodes, the disturbance is at a maximum, and the powder is dispersed.
Stationary wave in a closed pipe
•
In Fig. 15.23 (b) a fundamental mode (first harmonic) is illustrated.
•
The wavelength of this stationary wave (ignoring the end-correction) is 4L, where
L is the length of the pipe.
•
Using the wave equation v = f , the frequency f𝟏 of the fundamental mode is
v
given by f𝟏 = 4L where is the speed of sound in air.
136
•
Fig. 15.24 and Fig. 15.25 show how other modes of vibration (second harmonic
and third harmonic respectively) are possible.
•
Their corresponding wavelengths are
•
Their corresponding frequencies are f𝟐 =
•
The general equation for the frequency f𝒏 of the 𝑛𝑡 ℎ mode of vibration of the air
in the closed tube (the 𝑛𝑡 ℎ harmonic is
• f𝒏 =
4𝐿
3
and
3v
4L
4𝐿
5
and f𝟑 =
(2n - 1) v
4L
Stationary wave in an open pipe
137
5v
4L
Demonstrate stationary waves using microwaves
•
Stationary waves can be demonstrated using microwaves.
•
A source of microwaves faces a metal reflecting plate, as shown in Fig. 15.26.
•
A small detector is placed between source and reflector.
•
The reflector is moved towards or away from the source until the signal picked up
by the detector fluctuates regularly as it is moved slowly back and forth.
•
The minima are nodes of the stationary wave pattern, and the maxima are
antinodes.
•
The distance moved by the detector between successive nodes is half the
wavelength.
End of topic activity
Multiple choice questions
Paper 12 February/ March 2019: 24, 27, 28
Paper 11 May / June 2019: 24, 27, 29
Paper 12 May / June 2019: 25, 28, 30, 31
Paper 13 May / June 2019: 23, 26, 28, 29
138
Part 3
2.3 Determination of frequency and wavelength of sound waves
2.6 Superposition
2.6.2 Diffraction
2.6.4 Diffraction gratings
Topic 2.3: Determination of frequency and wavelength of sound waves
•
determine the frequency of sound using a calibrated cathode-ray or PC
oscilloscope
•
determine the wavelength of sound using stationary waves (e.g. use of
sonometer, resonance tubes, tuning forks)
Determine the frequency of sound using a calibrated cathode-ray or PC
oscilloscope
139
•
A cathode ray oscilloscope (c.r.o.) with a calibrated time-base may be used to
determine the frequency of sound.
•
A signal generator and loudspeaker are used to produce a note of a single
frequency. The microphone is connected to the Y-plates of the c.r.o.
•
The microphone detects the sound and a trace on the c.r.o. can be obtained by
adjusting the Y-plate sensitivity and the time-base settings.
•
The distance between peaks or troughs () is measured using the scale on the
c.r.o. display.
•
The time-base setting is used to determine the period T and frequency of the
sound. The calculated value can be compared with that shown on the signal
generator.
Determine the wavelength of sound using stationary waves (e.g. use of
sonometer, resonance tubes, tuning forks)
•
Since we know that adjacent nodes (or antinodes) of a stationary wave are
separated by half a wavelength, we can use this fact to determine the
wavelength λ of a progressive wave.
•
If we also know the frequency f of the waves, we can find their speed v using the
wave equation v = f λ
140
•
In Fig. 16.16, a loudspeaker sends sound waves along the inside of a tube. The
sound is reflected at the closed end.
•
When a stationary wave is established, the dust (fine powder) at the antinodes
(maximum amplitude) vibrates violently. It tends to accumulate at the nodes
(zero amplitude), where the movement of the air is zero. Hence the positions of
the nodes and antinodes can be clearly seen.
•
The loudspeaker produces sound waves, and these are reflected from the
vertical board. The microphone detects the stationary sound wave in the
space between the speaker and the board, and its output is displayed on the
oscilloscope. It is simplest to turn off the time base of the oscilloscope, so that
the spot no longer moves across the screen.
141
•
The spot moves up and down the screen, and the height of the vertical trace
gives a measure of the intensity of the sound (recall: I ∝ A𝟐 ) .
•
By moving the microphone along the line between the speaker and the board,
it is easy to detect nodes and antinodes.
•
Since we know that adjacent nodes (or antinodes) of a stationary wave are
separated by half a wavelength, we can use this fact to determine the
wavelength λ of a progressive wave.
Determining wavelength using microwaves
•
Stationary waves can be demonstrated using microwaves.
•
A source of microwaves faces a metal reflecting plate, as shown in Fig. 15.26.
•
A small detector is placed between source and reflector.
•
The reflector is moved towards or away from the source until the signal picked up
by the detector fluctuates regularly as it is moved slowly back and forth.
•
The minima are nodes of the stationary wave pattern, and the maxima are
antinodes.
•
The distance moved by the detector between successive nodes is half the
wavelength.
142
Determination of the wavelength of sound using stationary waves using a
sonometer
What is a sonometer?
•
A sonometer consists of a hollow rectangular wooden box of more than one
meter length.
•
It has a hook at one end and a pulley at the other end.
•
One end of a string is fixed at the hook and the other end passes over the
pulley.
•
A weight hanger is attached to the free end of the string.
•
Two adjustable wooden bridges (A and B) are put over the board, so that the
length of string can be adjusted.
•
If a string which is stretched between two fixed points is plucked at its center (R),
vibrations are produced and it move out in opposite directions along the string.
•
Because of this, a transverse wave travels along the string.
143
Materials required to do the sonometer experiment
•
a sonometer
•
a set of tuning forks of known frequency
•
0.5kg weight hanger
•
some 0.5kg slotted weights
•
rubber pad
•
paper rider
•
Take a tuning fork of known frequency. Make it vibrate by striking its prong
with a rubber pad. Bring it near the ear.
Procedure for the sonometer experiment
144
To find the relation between frequency and length
•
Place the sonometer on the table.
•
Make sure that the pulley is frictionless. If you feel any friction, oil them.
•
Stretch the wire by placing a suitable maximum load on the weight hanger.
•
Move the wooden bridges outward, so that the length of wire between the
bridges is maximum.
•
Pluck the sonometer wire and leave it to vibrate.
•
Compare the sounds produced by tuning fork and sonometer wire.
•
(Sound which has low pitch has less frequency).
•
Gently adjust the bridges for decreasing the length of wire, till the two sounds
appear alike.
•
Put an inverted V shaped paper rider on the middle of the wire.
•
Vibrate the tuning fork and touch the lower end of its handle with sonometer
board. The wire vibrates due to resonance and the paper rider falls.
145
•
Measure the length of wire between the bridges using a meter scale. It is the
resonant length and record it in the ‘length decreasing’ column.
•
Now, bring the bridges closer and then slowly increase the length of the wire till
the paper rider falls.
•
Measure the length of wire and record it in ‘length increasing’ column.
•
Repeat the above steps with tuning forks of other known frequencies, and find
resonant length each time.
To find the relation between length and tension
1. Select a tuning fork of known frequency
2. Set the load in the weight hanger as maximum.
3. Repeat the steps in the previous section to find out the resonant length.
4. Now, remove 0.5 kg weight from the weight hanger and find resonant length with
same tuning fork.
5. Repeat the experiment by removing slotted weights one by one in equal steps of
0.5 kg.
6. Record the observations each time.
146
147
•
If a string of length l having mass per unit length m is stretched with a tension T,
the fundamental frequency of vibration f is given by; f =
•
𝟏
𝟐𝒍
√
𝑻
𝒎
Law of length: the frequency of vibration (f) of a stretched string is inversely
proportional to its resonating length (l), provided its mass per unit length (m)
and tension remain (T) constant, f 𝜶
𝟏
𝒍
.
Law of tension: the frequency (f) of vibration of a stretched string is directly
proportional to the square root of its tension (√𝑻), provided its resonating length (l) and
mass per unit length (m) of the wire remains constant), f 𝜶 √𝑻
•
From the law of length f 𝜶
𝟏
𝒍
, it follows that
f × l = constant.
𝟏
1
•
A graph between f and
•
From the equation for frequency f =
•
A graph between T and 𝒍𝟐 (y = 𝒙𝟐 ) will be a straight line.
𝒍
(y = ) will be a straight line.
𝑥
𝟏
𝟐𝒍
𝑻
√ , it follows that
𝒎
√𝑻
= constant.
l
Resonance tubes
•
One of our best models of resonance in a musical instrument is a resonance
tube (a hollow cylindrical tube) partially filled with water and forced into vibration
by a tuning fork. The tuning fork is the object that forced the air inside of
the resonance tube into resonance.
148
Determining the wavelength of sound using stationary waves using resonance
tubes and tuning forks
Apparatus
1000 ml graduated cylinder, resonance tube, set of tuning forks in the frequency range
256 Hz to 512 Hz, vernier callipers, metre stick, stand (longest upright type), clamp and
wooden block.
5. Calculate the wavelength λ as described below:
• L1 =
1λ
4
and
• L = L2 - L1 =
L2 =
3λ
4
-
1λ
4
3λ
4
=
λ
𝟐
149
6. Once λ is determined, equation v = f λ may be used to find the measured value
of v. Of course, f is the frequency of the tuning fork.
2.6.2 Diffraction
•
explain the meaning of the term diffraction
•
describe experiments that demonstrate diffraction, including the qualitative effect
of the gap width relative to the wavelength of the wave, for example diffraction of
water waves in a ripple tank
2.6.4 Diffraction gratings
•
recall and solve problems using the formula dsin 𝜃 = nλ
•
describe the use of a diffraction grating to determine the wavelength of light (the
structure and use of the spectrometer are not included)
Diffraction
•
When waves pass through a narrow gap, they spread out.
•
This spreading out is called diffraction.
•
The extent of diffraction depends on the width of the gap compared with the
wavelength.
•
Diffraction is most noticeable if the width of the gap is approximately equal to
the wavelength.
150
Diffraction can be demonstrated in a ripple tank
•
The extent to which ripples spread out depends on the relationship between their
wavelength and the width of the gap.
•
In a, the width of the gap is very much greater than the wavelength and there is
hardly any noticeable diffraction.
•
In b, the width of the gap is greater than the wavelength and there is limited
diffraction.
•
In c, the gap width is equal to the wavelength and the diffraction effect is
greatest.
•
When diffraction occurs, the wavelength () does NOT change.
151
Diffraction gratings
•
A diffraction grating is a plate on which there is a very large number of
parallel, identical, very closely equally spaced slits ruled on a glass or plastic
slide.
•
A transmission diffraction grating is similar to the slide used in the double-slit
experiment, but with many more slits than just two slits.
•
If monochromatic light is incident on this grating, a pattern of narrow bright
interference fringes is seen.
•
As waves pass through the slits, they spread into a geometric shadow.
•
Each line is capable of diffracting the incident light.
•
There may be as many as 10 000 lines per centimetre.
•
A diffraction grating can be used to make a spectrometer and a spectrometer
is a device that measures the wavelength of light.
Reflection diffraction grating
•
In a reflection diffraction grating, the lines are made on a reflecting surface so
that light is both reflected and diffracted by the grating.
•
The shiny surface of a compact disc (CD) or DVD is an everyday example of a
reflection diffraction grating.
•
Hold a CD in your hand and twist it so that you are looking at the reflection of
light from a lamp. You will observe coloured bands (Figure 15.24). A CD has
thousands of equally spaced lines of microscopic pits on its surface; these carry
the digital information.
•
It is the diffraction from these lines that produces the coloured bands of light from
the surface of the CD.
152
Fig: 15. 34 Arrangement for obtaining a fringe pattern with a diffraction grating
How a diffracting gratings work?
•
Monochromatic light from a laser is incident normally (perpendicularly) on a
transmission diffraction grating.
•
In the space beyond, interference fringes are formed.
•
These can be observed on a screen, as with the double slit. However, it is usual
to measure the angle θ at which they are formed, rather than measuring their
separation (Figure 15.25).
•
With double slits, the fringes are equally spaced and the angles are very small.
•
With a diffraction grating, the angles are much greater and the fringes are not
equally spaced.
•
The fringes are also referred to as maxima.
The fringes are also referred to as maxima
•
The central fringe is called the zeroth-order (n = 0) maximum, the next fringe is
the first-order maximum, and so on.
•
The pattern is symmetrical, so there are two first-order maxima (n = 1 and n=
−1), two second-order maxima (n = 2 and n= −2), etc.
153
Explaining the diffraction grating experiment
•
The principle is the same as for the double-slit experiment, but here we have light
passing through many slits.
•
As it passes through each slit, it diffracts into the space beyond. So now we have
many overlapping beams of light, and these interfere with one another.
•
There is a bright fringe, the zeroth-order, 0th order maximum, in the straightthrough direction (θ = 0°).
•
Because all of the rays here are travelling parallel to one another and in phase,
so the interference is constructive.
•
To form first-order maximum of light emerge from all of the slits; to form a bright
fringe, all the rays must be in phase.
Many slits are better than two
•
It is worth comparing the use of a diffraction grating to determine wavelength with
the Young two-slit experiment.
– With a diffraction grating the maxima are very sharp.
– With a diffraction grating the maxima are also very bright. This is because
rather than there being contributions from only two slits, there are
contributions from a thousand or more slits.
– With two slits, there may be a large inaccuracy in the measurement of the
slit separation a. The fringes are close together, so their separation may
also be measured imprecisely.
–
With a diffraction grating, there are many slits per centimetre, so d can be
measured accurately. Because the maxima are widely separated, the
angle θ can be measured to a high degree of precision.
154
•
So an experiment with a diffraction grating can be expected to give
measurements of wavelength to a much higher degree of precision than a simple
double-slit arrangement.
Diffracting white light
Diffracting white light
•
A diffraction grating can be used to split white light up into its constituent
colours (wavelengths).
•
This splitting of light is known as dispersion (similar to dispersion of light using
a triangular prism – refraction vs diffraction of white light)
•
A beam of white light is shone onto the grating.
•
A zeroth-order, white maximum is observed at θ = 0°, because all waves of
each wavelength are in phase in this direction.
•
On either side, a series of spectra appear, with violet (shortest ) closest to the
centre and red (longest ) furthest away.
•
We can see why different wavelengths have their maxima at different angles if
n
we rearrange the equation d sin 𝜽 = n to give: sin 𝜽 = d
•
The greater the wavelength λ, the greater the value of sin θ and hence the
greater the angle θ.
•
Red light is at the long wavelength end of the visible spectrum, and so it appears
at the greatest angle.
155
Determining wavelength λ with a grating using d sin 𝜽 = n
•
Fig. 15.35 shows a parallel beam of light incident normally (at 90 °) on a
diffraction grating in which the spacing between adjacent slits is d.
•
Consider first rays 1 and 2 which are incident on adjacent slits.
•
The path difference between these rays when they emerge at an angle 𝜽 is d sin
𝜽.
•
To obtain constructive interference in this direction from these two rays, the
condition is that the path difference should be an integral number of wavelengths.
•
The path difference between rays 2 and 3, 3 and 4, and so on, will also be d sin
𝜽.
•
The condition for constructive interference is the same.
d sin 𝜽 = n
•
Thus, the condition for a maximum of intensity at angle 𝜽 is d sin 𝜽 = n where
 is the wavelength of the monochromatic light used, and n is an integer.
End of topic activity
Multiple choice questions
Paper 12 February/ March 2019: 29
Paper 11 May / June 2019: 28, 30
Paper 12 May / June 2019: 29
156
Paper 13 May / June 2019: 23, 30
Structured questions
Paper 22 February/ March 2019: 5
Paper 22 May / June 2019: 2(d) and 4
Paper 21 May / June 2019: 5
Paper 23 May / June 2019: 5
157
THEME 3: Electricity
3.1 Electric fields
3.2 Current of electricity
3.3 DC circuits
3.1 Electric fields
3.1.1 Concept of an electric field
3.1.2 Uniform electric fields
3.1.1 Concept of an electric field
• define and use electric field strength as force per unit positive charge (point
𝑭
charge) (E = 𝑸)
•
•
•
•
•
•
•
explain the concept of an electric field as an example of a field of force (a region
in which an electric charge experiences a force due to another charge)
represent an electric field by means of field lines
Prior-knowledge – static electricity in everyday life
A balloon rubbed with wool will stick on a wall
A TV screen attracts dust
Dry hair crackles (and may actually spark) when brushed
You may feel a shock when you touch the metal door-handle of a car on getting
out after a journey in dry weather
A plastic ruler picks up pieces of paper when rubbed
Prior-knowledge – electric charge
• Insulated objects gain electric charge by friction – that is, by being rubbed
against other objects.
• Insulators that are charged by friction with attract other objects.
• There are two kinds of electric charge. Polythene becomes negatively charged
when rubbed with wool. Cellulose acetate becomes positively charged, also
when rubbed with wool.
• A material becomes positively charged when it loses electrons and negatively
charged when it gains electrons.
Prior-knowledge
Static electricity in everyday life
• A balloon rubbed with wool will stick on a wall
• A TV screen attracts dust
• Dry hair crackles (and may actually spark) when brushed
• You may feel a shock when you touch the metal door-handle of a car on getting
out after a journey in dry weather
• A plastic ruler picks up pieces of paper when rubbed
158
Electric charge
• Insulated objects gain electric charge by friction – that is, by being rubbed
against other objects.
• Insulators that are charged by friction will attract other objects.
• There are two kinds of electric charge. Polythene becomes negatively charged
when rubbed with wool. Cellulose acetate becomes positively charged, also
when rubbed with wool.
• A material becomes positively charged when it loses electrons and negatively
charged when it gains electrons.
Charging by friction
Friction can transfer electrons from one material to another
Attraction and repulsion
Attraction and repulsion between electric charges
This is the basic law of force between electric charges
Like charges repel,
unlike charges attract.
Prior-knowledge – types of fields
Different types of fields
• electric fields - act on objects with electric charge
• magnetic fields - act on magnets (and on electric currents)
• gravitational fields - act on objects with mass
159
•
•
•
•
•
Electric field
An electric field is an example of a field of force and it is defined as a region in
which an electric charge experiences a force due to another charge.
An electric field is represented by electric field lines.
The field lines tells us the direction as well as the strength of the electric field.
The direction of the electric field is defined as the direction in which a positive
charge would move if it were free to do so, and it is represented by an arrow,
from positive to negative.
The strength is shown by the closeness of the field lines (the closer the lines are
to each other, the stronger the field is).
Investigating electric fields
Electric field lines
Around a point
between two points
between two unlike points
Between sphere and charge plate
160
•
Electric field strength
The electric field strength at a point is defined as the force per unit charge
acting on a small positive charge placed at that point
•
Equation: 𝐄 = 𝐐
•
The unit for electric field strength, E is NC
•
∆𝑽
𝐅
-1
3.1.2 Uniform electric field
•
•
•
•
•
•
recall and use 𝑬 = ∆𝒅 to calculate the field strength of the uniform field between
charged parallel plates in terms of potential difference and separation
calculate the forces on charges in uniform electric fields
describe the effect of a uniform electric field on the motion of charged particles
Uniform electric fields
A uniform electric field can be set up between two parallel plates by connecting
to a high-voltage power supply
A uniform electric field is constant at every point, this means the field strength
does not differ from place to place.
In a uniform electric field the filed lines are parallel.
Field strength of uniform field
If the electric plates are connected to a voltage then the electric field strength will
depend on:
– The potential difference between the plates (the higher the voltage,
the stronger the field: E ∝ V)
– The separation between the two plates (the greater the separation,
𝟏
•
the weaker the field: E ∝ 𝒅)
For a uniform field, the two factors can be combined to give the equation: electric
∆𝑉
•
field strength 𝐸 given by: 𝐸 = ∆𝑑
The unit for uniform field strength is Vm-1
Force on electric charge
∆𝑽
𝑭
•
The force F on charge Q can be calculated by combining 𝑬 = ∆𝒅 and 𝑬 = 𝑸
•
From 𝑬 = 𝑸, F = EQ
•
Substituting 𝑬 = ∆𝒅 into F = EQ, this gives
F=
𝑭
∆𝑽
−𝑸𝑽
𝒅
161
•
For an electron with a charge −𝒆 this becomes
𝒆𝑽
F= 𝒅
Examples
1. Two metal plates 5.0 cm apart have a potential difference of 1000 V between
them. Calculate
∆𝑉
1000 V
a) The strength of the electric field between the plates. 𝐸 = ∆𝑑 = 0.05 𝑚 =
2.0 × 104 𝑉𝑚−1
b) The force on a charge of 5.0 nC between the plates
𝑭
𝐄 = 𝑸 so
•
•
F = EQ
= (𝟐. 𝟎 × 𝟏𝟎𝟒 𝐕𝐦−𝟏 ) × (𝟓 × 𝟏𝟎−𝟗 𝐂)
= 𝟏. 𝟎 × 𝟏𝟎−𝟒 𝑵
Effect of a uniform electric field on the motion of a charged particle
A charge parallel to the field
– The charge will accelerate directly to the plate of opposite charge
A charge perpendicular to the field
– Will follow a curved path toward the opposite charged plate
Questions
1. An oil droplet has a charge –q and is situated between two horizontal metal
plates as shown in the diagram.
The separation of the plate is d. The droplet is observed to be stationary when the
upper plate is at potential +V and the lower plate is at potential –V
For this to occur, what is the weight of the droplet?
𝑉𝑞
A. 𝑑
B.
2𝑉𝑞
𝑑
C.
𝑉𝑑
𝑞
D.
2𝑉𝑑
𝑞
2. A horizontal beam of electrons is passed between two horizontal parallel plates, 2.0
cm apart, as shown.
162
The upper plate has an electrical potential of +4.0 V, and the lower plate has an
electrical potential of – 4.0 V.
What is the force on each electron when between the plates?
A. 3.2 × 10–17 N downwards
B. 3.2 × 10–19 N upwards
C. 6.4 × 10–19 N downwards
D. 6.4 × 10–17 N upwards
3.2 Current of electricity
3.2.1 Electric current
3.2.2 Potential difference and power
3.3.3 Resistance and resistivity
•
•
•
•
•
3.2.1 Electric current
explain that electric current is a flow of charge carriers
recall that the charge on charge carriers is quantised
define the coulomb as the SI unit of electric charge, equal to the quantity of
charge conveyed in one second by a current of one ampere
recall and use Q = It
derive and use, for a current-carrying conductor, the expression I = Anvq, where
A is the cross sectional area, n is the number density of charge carriers (number
of charge carriers per unit volume), v is drift velocity and q is the charge carried
by the individual charge carrier
Electricity???
•
•
•
Electricity in everyday life, its meaning may be rather vague – sometimes we use
it to mean electric current; at other times, it may mean electrical energy or
electrical power.
In this chapter and the ones which follow, we will avoid using the word electricity
and try to develop the correct usage of more precise scientific terms.
Electric current
The direction of the current is from the positive terminal of the cell, around the
circuit to the negative terminal. This is referred to as conventional current
163
Electric current - what is going on inside the wire?
•
•
•
•
•
•
•
•
•
A wire is made of metal. Inside a metal, there are negatively charged electrons
which are free to move about.
We call these conduction (free) electrons, because they are the particles which
allow a metal to conduct an electric current.
The atoms of a metal bind tightly together; they usually form a regular array, as
shown in Fig. 10.6. In a typical metal such as copper or silver, one electron from
each atom breaks free to become a conduction electron. The atom remains as a
positively charged ion. Since there are equal numbers of free electrons
(negative) and ions (positive), the metal has no overall charge – it is neutral.
When the cell is connected to the wire, it exerts an electrical force on the
conduction electrons that makes them travel along the length of the wire.
Since electrons are negatively charged, they flow away from the negative
terminal of the cell and towards the positive terminal.
This is in the opposite direction to conventional current.
Note that there is a current at all points in the circuit as soon as the circuit is
completed.
We do not have to wait for charge to travel around from the cell. This is because
the charged electrons are already present throughout the metal before the cell is
connected.
We can use the idea of an electric field to explain why a current flows almost
instantly.
164
•
•
•
•
•
•
•
•
•
•
•
•
•
Connect the terminals of a cell to the two ends of a wire and we have a complete
circuit.
The cell produces an electric field in the wire; the field lines are along the wire,
from the positive terminal to the negative.
This means that there is a force on each electron in the wire, so each electron
starts to move and the current exists almost instantly.
Charge carriers
Sometimes a current is a flow of positive charges – for example, a beam of
protons produced in a particle accelerator. The current is in the same direction
as the particles.
Sometimes a current is due to both positive and negative charges – for
example, when charged particles flow through a solution (electrolyte ).
An electrolyte contains both positive and negative ions. These move in opposite
directions when the solution is connected to a cell
Any charged particles which contribute to an electric current are known as
charge carriers;
Charge carriers can be protons, electrons or ions
Charge carriers are particles that move freely in a material and carry an
electric charge.
Electric current is thus a flow of charge carriers
The charge on charge carrier is quantised, meaning it is a multiple of integer
electrons
The SI unit for electric charge is coulombs, and is equals to the amount of
charge conveyed in 1s by a current of 1A
𝑄 = 𝐼𝑡
Charged particles – elementary charge
• As we have seen, current is the flow of charged particles called charge carriers.
• But how much charge does each particle carry?
• Electrons each carry a tiny negative charge of
≈ −1.6 × 10-19 C.
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•
•
•
•
This charge is represented by −e. The magnitude of the charge is known as the
elementary charge.
Protons are positively charged, with a charge +e. This is +1.6 × 10-19 C which
equal and opposite to the elementary charge of an electron.
Ions carry charges that are multiples of +e and −e.
Current in a conductor
A current I in a conductor of cross-sectional area A depends on:
– The number of conduction electrons per unit volume is called the number
density n
– the length of the wire
– The average drift velocity
– The charge of each charge carrier
•
Expression for a current-carrying conductor
The number of electrons per unit volume n (number density) is a characteristic
property of a material.
•
So 𝒏 =
•
•
•
•
•
•
•
•
number of electron in a conductor, 𝑵 = 𝒏 ×V
For a conductor of length 𝒍 and cross-section area A then
𝑵 = 𝒏 × 𝑨 × 𝒍 (note: 𝑨 × 𝒍 = V)
The amount of charge (Q) in a particle or electron is:
𝑸= 𝑵×𝒆 ∴𝑸= 𝒏×𝑨×𝒍×𝒆
𝒏×𝑨×𝒍×𝒆
But 𝑸 = 𝑰𝒕, so 𝑰𝒕 = 𝒏 × 𝑨 × 𝒍 × 𝒆 hence 𝑰 =
𝒕
As the electron moves across the conductor they move with a certain average
velocity (drift velocity - velocity attained by a particle because of an electric
field), v.
𝒔
𝒍
recall velocity, 𝒗 = 𝒕 so 𝒕 = v
•
𝑰 = 𝒏𝑨𝒗𝒆 𝒔𝒐, 𝑰 = 𝒏𝑨𝒗𝒒
𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒆𝒍𝒆𝒄𝒕𝒓𝒐𝒏𝒔 (𝑵)
𝒗𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒘𝒊𝒓𝒆 (𝑽)
(𝒏 =
𝑵
𝑽
)
Questions
1. Which of the following quantity of electric charge is possible?
A. 6.0 × 10−19
B. 8.0 × 10−19
C. 9.0 × 10−19
D. 10.0 × 10−19
166
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•
•
•
•
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•
•
3.2.2 Potential difference & power
define potential difference and the volt
– potential difference (p.d) [V] as energy transferred (work done) per unit
charge
– the volt (the SI unit for of both potential difference and electromotive force)
as the ratio of joule to coulomb
recall and use V = 𝑊÷𝑄
recall and use 𝑃=𝑉𝐼, 𝑃=𝑉2÷𝑅 and 𝑃=𝐼2𝑅
Potential difference (V)
Energy transfers as 1 C of charge flows round a circuit.
The voltmeter readings indicate the energy transferred to the component by each
unit of charge.
The voltmeter placed across the power supply measures the e.m.f. of the supply.
the voltmeters placed across the resistors measure the potential difference
(p.d.) across these components
•
The term potential difference is used when charges lose energy by transferring
electrical energy to other forms of energy e.g. thermal energy in a component.
It is the energy transferred per unit charge.
•
𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒, 𝑉 =
•
•
Potential difference is measured in volts (V).
One volt is defined as energy consumption of one joule per electric charge of
one coulomb. 1 V = 1 J / C
𝑒𝑛𝑒𝑟𝑔𝑦 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟𝑒𝑑
𝑐ℎ𝑎𝑟𝑔𝑒
𝑊
(𝑉 = 𝑄 ).
•
Electric power
Power is the rate at which energy is transferred.
The rate at which energy is transferred in an electrical component is related to
two quantities:
– the current I in the component
– the potential difference V across the component
P = W/t and W = VΔQ
•
So, 𝑃 =
•
P = VI
•
•
𝑉∆𝑄
𝑡
𝑄
= 𝑉𝐼 since 𝑡 = I
167
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•
•
•
•
3.2.3 Resistance and resistivity
define resistance of a conductor as the ratio of the potential difference across it
to the current through it
define the ohm (the SI unit for of electrical resistance) as the ratio of volt to
ampere, transmitting a current of one ampere when subjected to a potential
difference of one volt
recall and use V = IR
sketch and discuss the I-V characteristics of a metallic conductor at constant
temperature, a semiconductor diode and a filament lamp
explain that the resistance of a filament lamp increases as current increases
because its temperature increases
explain that the resistance of a light-dependent resistor (LDR) decreases as the
light intensity increases
explain that the resistance of a thermistor decreases as temperature increases
(negative temperature coefficient [NTC] thermistor only)
state and use Ohm’s law
define resistivity of a material as a product of the resistance and cross-sectional
area per length of the specimen
recall and use 𝜌 = 𝑅𝐴÷𝑙, where R is the resistance, ρ is the resistivity of the
material, l is the length of the conductor and A is the cross sectional area
Resistance and the ohm
Resistance of a conductor is the ratio of the potential difference across it to the
current through it.
The resistance of a conductor depends on:
– The potential difference across the conductor or component (Resistance
is direct proportional to the potential difference, 𝑹 ∝ 𝑽)
– The current flowing through the component (Resistance is inversely
𝟏
proportional to the current, 𝑹 ∝ 𝑰 )
𝑽
•
The two factors can be combined to give the equation 𝑹 = 𝑰 which can be re-
•
•
arranged as V = IR.
The unit for resistance is the ohm, and can be derived from the equation above.
The ohm is equivalent to 1 volt per ampere (1Ω = 1 V/A), the transmission of a
current of 1 ampere when subjected to a potential difference of one volt.
Resistance & electric power
•
𝑃 = 𝑉×𝐼
But 𝑉 = 𝐼 × 𝑅
So 𝑃 = 𝐼 × 𝑅 × 𝐼 = 𝐼 2 × 𝑅
𝑉
• Also, 𝑃 = 𝑉 × 𝐼, but 𝐼 = 𝑅
𝑉
So 𝑃 = 𝑉 × 𝑅 =
𝑉2
𝑅
168
•
Power dissipated in a resistor is given by:
𝑉2
𝑃 = 𝐼 2 𝑅 or 𝑃 = 𝑅
I – V characteristic graphs
Metallic conductor
Straight line
passing through
the origin.
Current is directly
proportional to
voltage
diode
•
•
•
•
•
•
•
•
Filament bulb
The line pass through the
origin
For small current the line is
roughly a straight line and
for higher voltage is a curve
Resistance increases and
depends on the
temperature of the filament
A diode only allows current to flow in one direction through
it (forward biased), when the current tries to flow the other
way (reverse biased) no current is allowed to flow through
the diode.
Diode has a threshold voltage of 0.6 V
Light-dependent resistor (LDR)
The resistance of an LDR decreases as the light intensity increases.
LDR’s are used in circuits which automatically switch on lights when it gets dark,
for example street lighting.
Thermistor: NTC
Thermistors have resistance which change rapidly with temperature.
Two types:
– Negative temperature coefficient (NTC)
– Positive temperature coefficient (PTC)
The resistance of a NTC thermistor decreases as it’s temperature increases.
Thermistors can be used as thermostats, the thermistor is used in circuits which
monitor and control the temperature of rooms, freezers & fridges etc.
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•
•
•
Ohm’s law
Ohm’s law states that for a conductor at constant temperature, the current in the
conductor is proportional to the potential difference across it.
Resistivity
The resistivity (𝝆) of a material is the product of the resistance and crosssectional area per length of the specimen
resistivity (𝝆) is a characteristic property of a material.
Resistance of a wire depends on:
– Length (𝑅 ∝ 𝑙)
1
– Cross-section area (𝑅 ∝ 𝐴)
𝒍
– The two factors can be combined to give the relationship 𝑹 ∝ 𝑨 where R
𝒍
= k 𝑨 where k is a constant (resistivity).
𝒍
– The resistivity (𝝆) depends on R, l and A: R = 𝜌 𝑨
𝑅𝐴
– Thus, 𝜌 = 𝑙
Example
1. Calculate the resistance per metre at room temperature of a constantan wire of
diameter 1.25 mm. The resistivity of constantan wire at room temperature is 5.0 ×
10−7 𝛺𝑚.
Answer: 0.41𝛺𝑚−1
3.3 DC circuits
3.3.1 Practical circuits
3.3.2 Kirchhoff’s laws
3.3.3 Potential dividers
•
•
3.3.1 Practical circuits
recall and use appropriate circuit symbols (see Annexe C)
draw and interpret circuit diagrams containing sources, switches, resistors,
ammeters, voltmeters, and/or any other type of component referred to in the
syllabus
170
•
•
•
•
define electromotive force (e.m.f.) of a source as energy transferred per unit
charge in driving charge round a complete circuit
distinguish between e.m.f. and potential difference
discuss the effects of the internal resistance of a source of e.m.f. on the terminal
potential difference and output power
recall and use the equation V = E – Ir, where V is the p.d., E is the e.m.f, I is the
current and r is the internal resistance
Appropriate circuit symbols (see Annexe C)
171
You should be able to draw and interpret circuit diagrams containing:
Sources
Switches
Resistors
Ammeters
Voltmeters
and/or any other type of component referred to in the syllabus
•
•
Electromotive force (e.m.f) and potential difference (p.d)
e.m.f. is the total work done (energy transferred) from chemical to electrical
energy per unit charge when the charge flows around a complete circuit.
Potential difference between two points is the energy transferred from electrical
to thermal energy per unit charge as the charge moves from one point to the
other.
•
𝑒. 𝑚. 𝑓. =
•
𝑝. 𝑑 =
𝑒𝑛𝑒𝑟𝑔𝑦 𝑐𝑜𝑛𝑣𝑒𝑟𝑡𝑒𝑑 𝑓𝑟𝑜𝑚 𝑜𝑡ℎ𝑒𝑟 𝑓𝑜𝑟𝑚 𝑡𝑜 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙
𝑐ℎ𝑎𝑟𝑔𝑒
𝑒𝑛𝑒𝑟𝑔𝑦 𝑐𝑜𝑛𝑣𝑒𝑟𝑡𝑒𝑑 𝑓𝑟𝑜𝑚 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑡𝑜 𝑜𝑡ℎ𝑒𝑟 𝑓𝑜𝑟𝑚
𝑐ℎ𝑎𝑟𝑔𝑒
172
Internal resistance
• All sources of e.m.f have an internal resistance (r).
• Internal resistance is the inherent resistance to the flow of current within the
source itself
• It can be due to:
– the wire or components inside (for power supply).
– chemicals within (for a cell)
– Any power source has two fundamental parts:
• e.m.f. and internal resistance
• The internal resistance increases as a battery depletes.
• If the internal resistance is present, the terminal potential difference is always
lower than the electromotive force (e.m.f.)
• Consider the circuit Fig. 20.2 with e.m.f E and internal resistance r.
• It delivers a current I when connected to an external resistor of resistance
(load) R.
• 𝑽𝑹 is the potential difference across the load, and 𝑽𝒓 is the potential difference
across the internal resistance r.
Using conservation of energy: E = 𝑽𝑹 + 𝑽𝒓
•
From Fig. 20.2
𝑽𝑹 = IR and 𝑽𝒓 = Ir
• Since E = 𝑽𝑹 + 𝑽𝒓 , so E = IR + Ir
• The potential difference 𝑽𝑹 across the load is thus given by 𝑽𝑹 = E − 𝑽𝒓
• 𝑽𝑹 is called the terminal potential difference
• From 𝑬 = 𝑰𝑹 + 𝑰𝒓
E = 𝑰(𝑹 + 𝒓)
• Hence 𝑰𝑹 = 𝑬 − 𝑰𝒓
V = 𝑬 − 𝑰𝒓
Terminal p.d = e.m.f. – (lost volt)
(The ‘lost volts’ indicates the energy transferred to the internal resistance of the supply.)
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•
•
Internal resistance and power
From the circuit on the right:
– VRI = power dissipated in the load
𝑉
– 𝐼𝑅 = load resistance
The source delivers maximum power to a circuit when the load resistance of the
circuit is equal to the internal resistance of the source.
Example
1. Three cells each of emf 1.5 V are connected in series to a 15Ω light bulb.
The current in the circuit is 0.27 A. Calculate the internal resistance of each cell.
Answer: 0.56 Ω
•
•
•
•
3.3.2 Kirchhoff’s laws
recall Kirchhoff’s first law and appreciate the link to conservation of charge
recall Kirchhoff’s second law and appreciate the link to conservation of energy
derive, using Kirchhoff’s laws, a formula for the combined resistance of two or
more resistors in series
solve problems using the formula for the combined resistance of two or more
resistors in series
174
•
•
•
•
•
•
•
•
•
derive, using Kirchhoff’s laws, a formula for the combined resistance of two or
more resistors in parallel
solve problems using the formula for the combined resistance of two or more
resistors in parallel
apply Kirchhoff’s laws to solve simple circuit problems
Kirchhoff’s first law
Kirchhoff’s first law states that the sum of the currents entering any point in a
circuit is equal to the sum of the currents leaving that same point.
– ∑ 𝐼𝑖𝑛 = ∑ 𝐼𝑜𝑢𝑡
– i.e. Algebraically sum of all the current entering and leaving the node
must be equal to zero
• I1 + I2 + (-I3 + -I4) = 0
Kirchhoff’s first law is an expression of the conservation of charge.
Kirchhoff’s first law – series circuit
In a series circuit, the components are connected one after another, forming
one complete loop.
The current in a series circuit is the same at each point.
Kirchhoff’s first law – parallel circuit
In a parallel, the current takes alternative routes in different loops. The current
divides at a junction, but the current entering the junction is the same as the
current leaving it.
The fact that the current does not get ‘used up’ at a junction is because current is
the rate of flow of charge, and charges cannot accumulate or get ‘used up’ at a
junction.
175
•
–
–
–
–
•
•
•
•
In Fig. 20.9, the current:
A = 175 mA
A𝟏 = 175 mA
A𝟐 = 75 mA − 25 mA = 50 mA
A𝟑 = 175 mA − 75 mA = 100 mA
Kirchhoff’s second law – conservation of energy
Charge flowing round a circuit gains electrical energy on passing through the
battery and loses electrical energy on passing through the rest of the circuit.
From the law of conservation of energy, we know that the total energy must
remain the same.
The consequence of this conservation of energy is known as Kirchhoff’s
second law.
Kirchhoff’s second law states that the sum of the e.m.fs in a closed circuit is
equal to the sum of the p.ds
– ∑𝐸 = ∑V
176
•
– Or: the sum of voltages around a closed conducting loops must be equal
to zero
Kirchhoff’s second law is an expression of the conservation energy.
•
•
Fig. 20.12 shows a circuit containing a battery, lamp and resistor in series.
Applying Kirchhoff’s second law, the e.m.f in the circuit is the e.m.f E of the
battery.
• The sum of the p.d’s is the p.d. 𝑉1 across the lamp plus the p.d. 𝑉2 across the
resistor thus, E = 𝑽𝟏 + 𝑽𝟐 .
• If the current in the circuit is I and the resistance of the lamp and resistor are 𝑅1
and 𝑅2 respectively, then pd.s can be written as:
𝑽𝟏 = I 𝑹𝟏 and 𝑽𝟐 = I 𝑹𝟐 .
•
•
Both e.m.f and potential difference have direction.
Direction must be considered when working out the equation for Kirchhoff’s
second law. For example in Fig. 20.13, two cells have been connected in
opposition. Here, the total e.m.f in the circuit is 𝐸1 − 𝐸2 , and by Kirchhoff’s
second law,
𝑬𝟏 − 𝑬𝟐 = 𝑽𝟏 + 𝑽𝟐 = 𝑰𝑹𝟏 + 𝑰𝑹𝟐 .
177
Derive, using Kirchhoff’s laws, a formula for the combined resistance of two or
more resistors in series
•
•
•
•
•
•
•
Resistors in series
Fig. 20.14 shows two resistors of resistances 𝑹𝟏 and 𝑹𝟐 connected in series, and
a single resistor of resistance R equivalent to them.
The current I in the resistors, and in their equivalent single resistor, is the same.
The total potential difference V across the two resistors must be the same as that
across a single resistor.
If 𝑽𝟏 and 𝑽𝟐 are the potential differences across each resistor, V = 𝑽𝟏 + 𝑽𝟐 .
But since the potential difference is given by V = IR, then IR = IR𝟏 + IR𝟐 .
If we divide all by current I then R = R𝟏 + R𝟐 .
The equation R = R𝟏 + R𝟐 can be extended so that the equivalent resistance of
several resistors connected in series is given by the expression R = R𝟏 + R𝟐 +
R𝟑 ….
Derive, using Kirchhoff’s laws, a formula for the combined resistance of two or
more resistors in parallel
•
•
•
In Fig. 20.15, two resistors of resistances R𝟏 and R𝟐 are connected in parallel.
The current through each resistor will be different, but they will each have the
same potential difference.
The equivalent single resistor of resistance R will have the same potential
difference across it, but the current will be the total current through the separate
resistors.
By Kirchhoff’s first law: I = 𝑰𝟏 + 𝑰𝟐 .
•
From R = 𝑰 , it follows that I = 𝑹
•
𝑽
𝑽
178
𝑽
𝑽
𝑽
𝟏
𝟐
•
Thus 𝑹 = R + R
•
Dividing the equation by V gives 𝑹 = R + R
•
𝟏
𝟏
𝟏
𝟏
𝟏
𝟐
𝟏
𝟏
𝟏
𝟐
The equation 𝑹 = R + R can be extended so that the equivalent resistance of
𝟏
𝟏
𝟏
𝟏
𝟏
𝟐
𝟑
several resistors connected in parallel is given by the expression 𝑹 = R + R + R
•
•
•
•
…
The reciprocal of the combined resistance of the resistors in parallel is the sum of
the reciprocals of all the individual resistors.
Applying Kirchhoff’s laws to solve simple circuit problems
To solve simple circuit problems using Kirchhoff’s laws, consider the way in
which the source of e.m.f. is connected as well as the direction of current.
E.g. in Fig 11.11, calculate the current in each of the resistors in the circuit.
•
•
•
•
Step 1: Mark the currents flowing in the circuit.
See 𝑰𝟏 , 𝑰𝟐 and 𝑰𝟑 in Fig. 11.11.
Step 2 : Apply Kirchhoff’s first law.
At point P, this gives: 𝑰𝟏 + 𝑰𝟐 = 𝑰𝟑 (equation 1)
Step 3: Choose a loop and apply Kirchhoff’s second law.
Recall: E = 𝑽𝟏 + 𝑽𝟐 and 𝑽𝟏 = I 𝑹𝟏 and 𝑽𝟐 = I 𝑹𝟐 .
This means for the upper loop E = (𝑰𝟑 R) + (𝑰𝟏 R)
so 6.0 V = (𝑰𝟑 × 30 ) + (𝑰𝟏 × 10 ). (equation 2)
Step 4: Repeat Step 3 around the other loops until there are the same number of
equations as there are unknown currents.
Around the lower loop 2.0 V = (𝑰𝟑 × 30 ) (equation 3)
We now have three equations with three unknowns (the three currents)
Step 5: Solve the three equations as simultaneous equations.
𝑰𝟏 + 𝑰𝟐 = 𝑰𝟑 (equation 1)
6.0 V = (𝑰𝟑 × 30 ) + (𝑰𝟏 × 10 ). (equation 2)
2.0 V = (𝑰𝟑 × 30 ) (equation 3)
•
From equation 3, 𝟑𝟎 = 𝟑𝟎 𝟑 so 𝑰𝟑 = 0.067 A
•
•
•
𝟐.𝟎
𝟑𝟎 𝑰
179
•
•
•
•
Substituting the value for 𝑰𝟑 into equation 2, this gives 6.0 V = (0.067 A × 30 )
+ (𝑰𝟏 × 10 ) so
𝑰𝟏 = 0.40 A.
Substituting 𝑰𝟏 and 𝑰𝟑 into equation 1 gives
𝑰𝟏 + 𝑰𝟐 = 𝑰𝟑 so 0.40 A + 𝑰𝟐 = 0.067 A ,
Thus 𝑰𝟐 = −0.333 A (notice that 𝑰𝟐 is negative)
•
•
3.3.3 Potential dividers
explain the principle of a potential divider circuit as a source of variable p.d
𝑅
recall and use 𝑉𝑜𝑢𝑡 = 𝑉𝑖𝑛 (𝑅 +2𝑅 )
•
•
•
•
•
•
•
•
•
•
•
1
2
solve problems using the principle of the potentiometer as a means of
comparing potential differences
explain the use of thermistors (negative temperature coefficient [NTC]
thermistors only), light-dependent resistors (LDR) in potential dividers to provide
a potential difference that is dependent on the temperature and illumination
respectively
Potential dividers
Two resistors connected in series with a cell each have potential difference.
They may be used to divide the e.m.f of the cell.
This is illustrated in Fig. 20.19. The current in each resistor is the same, because
they are in series.
Thus 𝑉1 = 𝐼𝑅1 and 𝑉2 = 𝐼𝑅2
𝑉
𝑅
Dividing the first equation by the second gives 𝑉1 = 𝑅1.
2
2
The ratio of the voltages across the two resistors is the same as the ratio of their
resistances.
If the potential difference across the combination were 12 V and 𝑅1 were equal to
𝑅2 then each resistor would have 6V across it.
If 𝑅1 were twice the magnitude of 𝑅2 then 𝑉1 would be 8V and 𝑉2 would be 4V.
180
•
•
•
•
•
•
Potential divider is a device or a circuit that uses two or more resistors or a
variable resistor (potentiometer) to provide a fraction of the available voltage (p.d.)
from the supply.
The potential difference from the supply divides across the resistors is direct
proportion to their individual resistances.
It is used in volume control knobs as well as in sensory circuits using LDR and
thermistors.
The diagrams shows two potential divider circuits.
The high-voltage resistance voltmeter measures the voltage across the resistor.
This is the output voltage (Vout)
Circuit (b) contains a single variable resistor. By sliding the contact, the V out
changes (between 0.0 V and the value of Vin)
• V depends on the relative values of R and R .
out
•
1
2
The value of output voltage can be calculated using:
𝑅
1
𝑉𝑜𝑢𝑡 = 𝑉𝑖𝑛 (𝑅 +𝑅
)
2
1
Vin is the total voltage across the two resistors
Example
Calculate the output p.d, Vout from the potential divider circuit shown.
Answer
𝑅1
(𝑅2 + 𝑅1 )
10
𝑉𝑜𝑢𝑡 = 20𝑉 ×
15 + 10)
𝑉𝑜𝑢𝑡 = 8 𝑉
𝑉𝑜𝑢𝑡 = 𝑉𝑖𝑛
181
Potentiometer
•
•
•
•
•
•
•
•
•
•
A potentiometer is a continuously variable potential divider.
Potentiometer is a resistor that we can attach a wire to at different parts, splitting
it into two resistors with one part in series with the attached wire and the other in
parallel with the attached wire.
It is a three terminal variable resistor used for comparing potential differences.
It is connected as a potential divider to give a continuous variable output voltage.
NTC thermistor and LDR in potential dividers
If the variable resistors a thermistors or LDRs, whose value depends on an
external conditions, the potential difference across it will vary depending on those
conditions and a control circuit can be created.
An LDR is a semiconducting resistor whose value decreases with light. It can be
used to create a sensing circuit which outputs a varying potential difference
across it in response to light conditions.
As the light on the LDR is increased its resistance falls and the potential
difference across it would also fall. This would cause the bulb to dim.
If the LDR was completely covered, its resistance would be at its maximum
value, the potential difference would also be at the maximum value and the bulb
would shine brightly.
If the position of the two resistors was reversed and the fixed resistor was placed
in parallel with the bulb the opposite effect would be observed.
If the LDR in a circuit is swapped with a thermistor, the exact
temperature/brightness controls would be adjusted by adjusting the ratio of the
variable resistor and the resistance of the thermistor at the desired temperature.
Example
Calculate the reading on the meters shown below when the thermistor has a
resistance of
(a) 1 kΩ warm condition
(b) 16 kΩ cold condition
182
answers
𝟒
(a) Vout = 𝟓 × 𝟏+𝟒 = 4 V
𝟒
(b) Vout = 𝟓 × 𝟏𝟔+𝟒 = 1 V
End of topic activity
Multiple choice questions
Paper 12 February/ March 2019: 36, 37, 38
Paper 11 May / June 2019: 35, 36, 37, 38
Paper 12 May / June 2019: 35, 36, 37
Paper 13 May / June 2019: 34, 35, 36, 37, 38
Structured questions
Paper 22 February/ March 2019: 6
Paper 21 May / June 2019: 6
Paper 22 May / June 2019: 5, 6 (a) – (b)
Paper 23 May / June 2019: 6
183
Theme 4: Modern physics
4.1 Atoms, nuclei and radiation
4.2 Fundamental particles
4.1 Atoms, nuclei and radiation












describe and explain the simple structure of the nucleus
recall that radioactive decay is the random and spontaneous emission of
particles and/or electromagnetic radiation from an unstable nucleus
recall the nature and properties of of α, β and γ radiations (both β– and β+ are
included)
distinguish between proton number and nucleon number ( mass number) and
proton number ( atomic number) and use standard nuclide notation ((_Z^A)X)
 proton number (atomic number), denoted by Z
 nucleon number (mass number), denoted by A
state that an element can exist in various isotopic forms each with a different
number of neutrons
describe and explain the transformation of nuclei when they emit radiation
appreciate that nucleon number proton number and energy are all conserved in
nuclear processes
234
230
represent simple alpha and beta decay by equations of the form
𝑈→
Th
92
90
4
214
214
+ α
Pb →
Bi + 0 β
82
83
2
−1
deduce the mass number and proton number of the daughter and granddaughter
products in a decay series
recall that during beta decay that beta particles are emitted with a range of
kinetic energies
recognise the effects of a uniform electric field on the path of alpha and beta
particles and gamma rays
calculate the force on alpha and beta particles when passing through a uniform
∆𝑉
electric field (e.g using F = EQ; E = ∆𝑑)



use the unified atomic mass unit and/or the mass of an electron in calculations
involving forces on alpha and beta particles (e.g. using F = ma and equations of
motion)
deduce from the results of the α particle scattering experiment the existence and
small size of the nucleus
state that (electron) antineutrinos and (electron) neutrinos are produced during
β– and β+ decay
184
•
•
Atoms
The atoms of all elements are made up of three particles called protons,
neutrons and electrons.
The protons and neutrons are at the centre (nucleus) of the atom. The electrons
orbit the nucleus.
Protons and neutrons both have a mass of about one atomic mass unit u.
1 u = 1.66 × 10−27 kg.
•
By comparison, the mass of an electron is very small, about
•
•
u
2000
1
2000
of 1 u, which is
.
•
The vast majority of the mass of the atom is therefore in the nucleus.
•
Electrons carry a charge of −1e, where e = 1.6 × 10-19 C known as the
elementary charge.
Protons carry a charge of +1e.
•
Atoms – the structure of the atom
•
•
•
•
•
•
•
•
Atoms and ions
Atoms are uncharged (electrically neutral) because they contain equal numbers
of protons and electrons, and the charge of one electron is equal and opposite to
the charge of one proton.
If an atom loses one or more electrons, so that it does not contain an equal
number of protons and electrons, it becomes charged and it called an ion.
For example, if a sodium atom loses one of its electrons, it becomes a positive
sodium ion (cation).
Na → 𝑁𝑎 + + 𝑒 − .
If an atom gains an electron, e.g Cl2 + 2𝑒 − → 2𝐶𝑙 − , it becomes a negatively
charged ion (anion).
Proton number and nucleon number
The number of protons in the nucleus of an atom is called the proton number
(atomic number) Z.
The number of protons together with the number of neutrons in the nucleus is
called the nucleon number (or mass number) A.
A nucleon is the name given to either a proton or a neutron in the nucleus.
185
•
•
The difference between the nucleon number (A) and the proton number (Z) gives
the number of neutrons in the nucleus.
The standard nuclide notation 𝑨𝒁𝑿
If the chemical symbol of an element is X, a particular atom of this element, a
nuclide, is represented by the notation
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
𝒏𝒖𝒄𝒍𝒆𝒐𝒏 𝒏𝒖𝒎𝒃𝒆𝒓
𝑿
𝒑𝒓𝒐𝒕𝒐𝒏 𝒏𝒖𝒎𝒃𝒆𝒓
= 𝑨𝒁𝑿.
The element changes for every Z number and the symbol X changes.
A nuclide is the name given to a class of atoms whose nuclei contain a specified
number of protons and a specified number of neutrons.
The nucleus of one form of sodium contains 11 protons and 12 neutrons.
Therefore its proton number Z is 11 and the nucleon number A is 11 + 12 = 23.
This nuclide can be shown as 𝟐𝟑
𝟏𝟏𝑵𝒂.
All atoms with nuclei that contain 11 protons and 12 neutrons belong to this class
and are the same nuclide.
Isotopes
Nuclides with nuclei which have the same atomic number Z but a different
nucleon number A are isotopes of the same element.
Sometimes atoms of the same element have different numbers of neutrons in
their nuclei.
The most abundant form of chlorine contains 17 protons and 18 neutrons in its
nucleus, giving it a nucleon number of 17 + 18 = 35. This is often called chlorine35.
Another form of chlorine contains 17 protons and 20 neutrons in the nucleus,
giving it a nucleon number of 37. This is chlorine-37.
Chlorine-35 and chlorine-37 are said to be isotopes of chlorine.
Isotopes are different forms of the same element which have the same number of
protons but different numbers of neutrons in their nuclei.
Radioactivity
Some elements have nuclei which are unstable. That is, the combination of
protons and neutrons in the nucleus is such that the forces acting on the
nucleons do not balance.
In order to become more stable, they emit particles and/or electromagnetic
radiation.
The nuclei are said to be radioactive, and the emission is called radioactivity.
Radioactive decay is the random and spontaneous emission of particles
and/or electromagnetic radiation from an unstable nucleus.
Tracks of 𝜶 − 𝐩𝐚𝐫𝐭𝐢𝐜𝐥𝐞𝐬 in a cloud chamber
The emissions are invisible to the eye, but their tracks were first made visible in a
device called a cloud chamber.
186
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•
•
•
•
•
•
•
•
The nature and properties of α, β and γ radiations
Investigations of the nature and properties of the emitted particles or radiation
show that there are three different types of emission.
The three types are α-particles (alpha-particles), β-particles (beta-particles)
and γ-radiation (gamma radiation).
All three emissions originate from the nucleus.
α-particles (alpha-particles)
α-particles are like a helium nucleus.
An α-particle contains two protons and two neutrons, and hence carries a charge
of +2e (recall e = 1.6 × 10-19 C ).
α-particles travel at speeds up to about 107 𝑚𝑠 −1 (about 5% of the speed of light).
α-particle emission is the least penetrating of the three types of emission. It can
pass through very thin paper, but is unable to penetrate thin card.
Its range in air is a few centimetres.
Because α-particles are positively charged (recall that there are no electrons in
α-particles hence the positive charge), they can be deflected by electric and
magnetic fields.
An α-particle is identical to the nucleus of a helium atom and is represented as
4
2He.
•
•
•
•
•
•
•
•
•
•
As α-particles travel through matter, they interact with nearby atoms causing
them to lose one or more electrons.
The ionised atom and the dislodged electron are called an ion pair.
The production of an ion pair requires the separation of unlike charge, and
consequently they are efficient ionisers.
They may produce as many as 105 ion pairs for every centimetre of air through
which they travel.
Thus, they lose energy relatively quickly, and have low penetrating power.
When the nucleus of an atom emits an α-particle, it is said to undergo α-decay.
The nucleus loses two protons and two neutrons in this emission.
In α-decay, the proton number of the nucleus decreases by two, and the nucleon
number decreases by four.
α-decay causes one element to change into another (transmutation).
The original nuclide (parent nuclide) changes into a daughter nuclide.
187
•
For example, uranium-234 (the parent nuclide) may emit an α-particle, the
daughter nuclide is thorium-230, energy is also released.
•
•
234
230
4
92U → 90Th + 2He + energy
•
The atomic mass of the decay product is less than the atomic mass of the parent
nuclide ( 234
92U).
The energy equivalent of the difference in the mass appears as kinetic energy of
•
•
the α-particles and recoiling daughter nuclide ( 230
90Th) and a γ-photon.
Therefore, mass-energy is conserved.
Linear momentum is also conserved in this type of nuclear reaction.
•
•
•
The same amount of energy is released in the decay of each nucleus of 234
92U.
The α-particles emitted from a particular radioactive nuclide have the same
kinetic energy.
β-particles (beta-particles)
−
A radioactive nucleus that decays by β-decay may emit a negative (β ) or
+
positive (β ) electron.
+
The positive electron (β ) is also known as a positron or an antielectron (𝑒̅ ).
−
+
β-particles are fast moving electrons, β , or positron β .
β-particles have speeds in excess of 99% of the speed of light.
These particles have half the charge and very much less mass than α-particles.
Consequently, they are much less efficient than α-particles in producing ion pairs.
They are thus far more penetrating than α-particles, being able to travel up to
about a metre in air.
They can penetrate card and sheets of aluminium up to a few millimetres thick.
Their positive or negative charge means that they are affected by electric and
magnetic fields.
There are however important differences between the behaviour of α- and β- in
these fields.
β-particles may carry a negative charge or a positive charge, and thus may be
deflected in the same direction or opposite direction to the positively charged αparticles.
β-particles experience a much larger deflection when moving at the same speed
as α-particles, this is because the mass of a β-particle is much less than that of
an α-particle.
A β-particle may be emitted from a lead-214 nucleus (the parent nuclide).
The daughter nuclide is busmuth-214, in addition, energy is released.
The emission is represented by the nuclear equation:
•
0
0
214
214
82Pb → 83Bi + −1𝑒 + 0𝑣̅ + energy
•
•
A β particle may be emitted from a phosphorus-30 nucleus (the parent nuclide).
The daughter nuclide is silicon-30 and energy is also released.
•
•
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•
•
•
•
•
•
•
•
•
+
188
•
The emission is represented by the nuclear equation
•
30
30
0
0
15P → 14Si + 1𝑒 + 0𝑣 + energy
0
0
0𝑣 represents a neutrino and 0𝑣̅ represents an antineutrino.
•
•
•
•
•
•
•
•
•
•
These particles have no electrical charge and little or no mass and are emitted
from the nucleus at the same time as the β-particle.
Recall that the nucleus contains protons and neutrons. The β-particles (both
−
+
β and β ) also originate from the nucleus.
−
The process for this type of decay is that, just prior to β emission, a neutron in
the nucleus forms a proton, a negative electron and an antineutrino ( 00𝑣̅ ).
The ratio of protons to neutrons in the nucleus is changed and this makes the
daughter nucleus more stable.
Free neutrons are known to decay as follow:
0
0
1
1
0𝑛 → 1𝑝 + −1𝑒 + 0𝑣̅ + energy.
•
A similar process happens in the nucleus.
In β-decay, a negative electron and antineutrino 𝑣̅ are emitted from the nucleus.
This leaves the nucleus with the same number of nucleons (protons + neutrons)
as before, but with one extra proton and one fewer neutron.
+
In β emission, a proton in the nucleus forms a neutron, a positive electron and a
neutrino.
This process again changes the ratio of protons to neutrons in the nucleus and
makes the daughter nucleus more stable.
+
In β decay the proton is considered to transform itself as follows:
•
0
0
1
1
1𝑝 → 0𝑛 + 1𝑒 + 0𝑣 + energy.
•
In β decay, the positive electron and neutrino are emitted from the nucleus. This
leaves the nucleus with the same number of nucleons (protons + neutrons) as
before, but with one extra neutron and one fewer proton.
−
In β decay (negative electron), a daughter nuclide is formed with the proton
number increased by one, but with the same nucleon number.
+
In β decay, (positive electron), a daughter nuclide is formed with the proton
number decreased by one, but with the same nucleon number.
+
The antimatter particle, the positive electron (β ) very quickly meets its
equivalent matter particle, the negative electron.
The two particles annihilate (‘destroy’) each other to produce γ-radiation.
This makes the positive electron difficult to detect.
The atomic mass of the decay products is less than the mass of the parent
nucleus.
The energy equivalent of the difference in the mass is shared between the kinetic
energy of the β-particle and the recoiling daughter nucleus and the energy of the
neutrino or antineutrino.
•
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•
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•
•
•
•
+
189
•
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•
•
•
•
•
•
•
•
Therefore mass-energy is conserved.
The same amount of energy is released in the decay of each particular parent
nucleus.
However the electrons emitted from a particular radioactive nuclide have varying
amounts of kinetic energy.
The amount depends on the way the total energy available is shared between
the electron and the neutrino.
The sum of the electron’s energy and the neutrino’s energy is constant for the
decay of a particular nuclide.
−
+
β and β decay
Kinetic energy of the sub-atomic particles
The SI unit of energy is the joule.
The energies in nuclear reactions are very small compared to the joule.
A more convenient unit to use is the electron-volt.
This is the work done (energy gained) by an electron when accelerated through a
potential difference of one volt.
Since work done = potential difference × charge, one eV is equivalent to 1.602 ×
10−19 J.
γ-radiation (gamma radiation)
γ-radiation is part of the electromagnetic spectrum with wavelengths between
10−11m and 10−13 m
Since γ-radiation has no charge, its ionising power is much less than that of
either α- or β-particles.
γ-radiation penetrates almost unlimited thicknesses of air, several metres of
concrete or several centimetres of lead.
α- and β-particles are emitted by unstable nuclei which have excess energy. The
emission of these particles results in changes in the ratio of protons to neutrons,
but the nuclei may still have excess energy.
190
•
•
The nucleus may return to its unexcited (or ground) state by emitting energy in
the form of γ-radiation.
In γ-emission, no particles are emitted and there is, therefore, no change to the
proton numbers or nucleon number of the parent nuclide.
For example, when uranium-238 decays by emitting an α-particle, the resulting
nucleus of thorium-234 contains excess energy (it is an excited state) and emits
a photon of γ-radiation to return to the ground state.
This process is represented by the nuclear equation
•
234
234
∗
90𝑇ℎ → 90𝑇ℎ + γ (the asterisk * shows that Th is in an excited state).
•
•
Summary of the nature and properties of α, β and γ radiations
Property
α-particles.
β-particles.
γ-radiation
Mass
4u
About u/2000
0
Charge
+2e
-e or +e
0
Nature
Helium nucleus
Negative or positive Short-wavelength
electron
electromagnetic waves
Speed
Up to 0.05c
More than 0.99c
C
Penetrating power
Few cm of air
Few mm of
Few cm of lead
aluminium
4
Relative ionising
10
102
1
power
Affects
Yes
Yes
Yes
photographic film?
Deflected by
Yes see figure 26.3 Yes, see figure 26.3 No
electric, magnetic
fields?
191
Conservation of proton number, nucleon number and energy in nuclear
processes
• Notice that in all radioactive decay processes (and in fact in all processes of
nuclear reactions), the nucleon number and proton number are conserved.
• Hence, for all equations representing nuclear reactions, the sum of the nucleon
numbers and well as proton numbers on the left hand side of the equation is
equal to the sum of the nucleon numbers on the right hand side of the equation.
• This shows that proton number, nucleon number and energy are all conserved in
nuclear processes.
Radioactive decay series
• When a parent nuclide decays into a daughter nuclide, the daughter nuclide may,
itself, be unstable and so may emit radiation to give another different nuclide
(granddaughter) nuclide.
• This sequence of radioactive decay from parent nuclide through succeeding
daughter nuclides to granddaughter nuclides is called radioactive decay series.
• The series ends when a stable nuclide is reached.
• You should be able to deduce the mass number and proton number of the
daughter and granddaughter products in a decay series.
Effect of a uniform electric field on the path of α, β and γ
Effect of a magnetic field on the path of α, β and γ
Force on α- and β- particles when passing through a uniform electric field
• When α- and β- particles pass through a uniform electric field, because they are
charged, they experience a force and are accelerated towards the oppositely
charged plates.
192
•
•
The acceleration depends on the mass of the particle and the force exerted on
the particle.
You should be able to use the unified atomic mass unit or mass of an electron /
proton in calculations involving forces on alpha and beta particles. These are
given in the data section of the syllabus and will be provided in Paper 1 and
Paper 2.
 Unified atomic mass unit
1u = 1.66 × 10-27 kg
 Rest mass of electron
me = 9.11 × 10-31 kg
 Rest mass of proton
mp = 1.67 × 10-27 kg
You should be able to use the equations to calculate information related α- and βparticles in a uniform electric field such as:
Strength of force exerted on α- and β- particles using F = EQ
∆V
Strength of the electric field between two plates using E = ∆d
using F = ma and equations of motion (see Theme 1)
where:
F = Force (N) exerted on the particle
−1
E = Electric field strength (NC ) of the uniform electric field
Q = Charge (C) of the particle
∆V = potential differences between the two parallel plates (V)
∆d = distance between the two parallel plates
m = mass (kg) of the particle
a = acceleration (ms−2) of the particle
∆V
Example of questions using F = EQ, E = ∆d,
F = ma and equations of motion
−1
An electron is placed in a uniform electric field of 4.5 NC created by two parallel
plates that are 5 cm apart. The electron accelerates from rest at the negative plate.
Calculate:
a) The force on the electron.
b) The acceleration on the electron.
c) The final speed of the electron after travelling a distance of 5 cm.
d) The final kinetic energy of the electron of the electron after travelling a
distance of 5 cm.
e) The time taken for the electron to travel a distance of 5 cm.
(a) force
−1
F = EQ = (4.5 NC ) × (−1.6 × 10−19 C)
= 7.2 × 10−14 N
(b) acceleration
193
𝐹
F = ma, a = 𝑚 =
7.2 × 10−14 N
9.11×10−31 𝑘𝑔
= 7.9 × 1016 Nkg−1
(c) final speed
𝒗𝟐 = 𝒖𝟐 + 2as
𝒗𝟐 = 0𝟐 + 2 (7.9 × 1016 Nkg−1) (0.05 m)
15
√𝑣 = √7.9 × 10
v = 8.89 × 107 𝑚𝑠 −1
(d) final kinetic energy
1
1
𝐸𝑘 𝑓𝑖𝑛𝑎𝑙 = 2 𝑚𝑣 2 = 2 (9.11 × 10−31 𝑘𝑔)(8.89 × 107 𝑚𝑠 −1 )
2
3.6 × 10−15J
OR
W = ∆𝑬𝒌 = 𝑬𝒌 𝒇𝒊𝒏𝒂𝒍 − 𝑬𝒌 𝒊𝒏𝒊𝒕𝒊𝒂𝒍
Since the electron accelerated from rest, 𝑬𝒌 𝒊𝒏𝒊𝒕𝒊𝒂𝒍 = 0
Thus, W = 𝑬𝒌 𝒇𝒊𝒏𝒂𝒍 = Fs = 7.2 × 10−14 N × 0.05 m
•
•
•
= 3.6 × 10−15J
(e) Time
v = u + at
8.89 × 107 𝑚𝑠 −1 = 0 + (7.9 × 1016 Nkg−1) t
t = 1.125 × 10−9 s
Alpha particle scattering experiment
In 1911, Ernest Rutherford and two of his associates Hans Geiger and Ernest
Marsden did an experiment in which they fired a beam of 𝛼-particles to a very
thin piece of gold foil.
A circular fluorescent screen surrounded the gold foil to detect the direction in
which the 𝛼-particles travelled after striking the gold foil.
They observed that:
 The vast majority of the 𝛼-particles passed through the foil with very little
or no deviation from the original path.
 A small number of particles were deviated through an angle of more than
about 10 °
 An extremely small number of particles (on in ten thousand) were
deflected through and angle greater than 90 °
194
From the observations, the following conclusions were drawn:
• The majority of the mass of an atom is concentrated in a very small volume at
the centre of the atom (the nucleus). Most 𝛼-particles would, therefore, pass
through the foil undeviated / undeflected.
• The centre (nucleus) of an atom is charged. 𝛼-particles, which are also
charged, passing close to the nucleus will experience a repulsive force, causing
them to deviate.
• Only 𝛼-particles which pass very close to the nucleus, almost striking it head-on,
will experience large enough repulsive forces to cause them to deviate through
angles greater than 90° the fact that so few particles were deflected in this way
confirms that the nucleus is very small and that most of the atom is empty space.
Idea of size of nucleus
4.2 Fundamental particles
•
•
•
appreciate that protons and neutrons are not fundamental particles since they
consist of quarks
describe a simple quark model of hadrons in terms of up, down and strange
quarks and their respective antiquarks
describe protons and neutrons in terms of a simple quark model
195
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
appreciate that there is a weak interaction between quarks, giving rise to β
decay
describe β– and β+ decay in terms of a simple quark model
appreciate that electrons and neutrinos are leptons
Fundamental Particles
In the 19th century, the atom was considered to be the fundamental particle form
which all matter was composed.
This idea was used to explain the basic structure of all elements.
Experiments performed at the end of the 19th century and beginning of the 20th
century provided evidence for the structure of an atom.
The conclusions were that all atoms have a nucleus containing protons which is
surrounded by electrons and that the nucleus was very small compared with the
size of the atom.
The neutron was introduced to explain the discrepancy between the mass of the
atom and the mass from the number of protons (positive charges).
In 1932 James Chadwich discovered the neutron and the fundamental particles
were then considered to be the proton, the neutron and the electron.
The particles in the atom must experience forces in order to maintain its
structure.
The forces were the gravitational force that acts between all masses and the
electrostatic force that acts between charged objects.
The electrostatic force of repulsion is approximately 1036 times greater than the
gravitational force of attraction between protons.
Another attractive force must keep the protons together in the nucleus. This force
is known as the strong force and acts between the nucleons. The strong force
does not seem to have any effect outside the nucleus and is therefore and is,
therefore, considered to be very short range (a little than the diameter of nuclei,
10−14 m).
There appears to be a limiting spacing between nucleons which is similar in
different nuclei and this suggests that the force is repulsive as soon as the
nucleons come close together.
The strong force does not act on electrons.
196
Fundamental particles (also called elementary particles) are particles that cannot
be broken into anything smaller.
• They are the smallest building blocks of the universe.
• They have no internal structure. ...
• Particles that make up all matter, called fermions.
• Particles that carry force, called bosons.
• Protons and neutrons are NOT fundamental particles because they consist of
quarks.
Quarks are tiny particles held together by gluons
Hadrons and leptons
• The discovery of antimatter is cosmic radiation supported the theory developed
from the special theory of relativity and quantum theory that all fundamental
particles have a corresponding antimatter particle.
• The matter and antimatter particles have the same mass but opposite charge.
• The following particles were required to support the theory: the antiproton, the
antineutron and the antielectron.
• The symbols used for the antiparticles are 𝑝̅ (antiproton), 𝑛̅ (antineutron), 𝑒̅
(antielectron).
• The antielectron (positive electron / positron) was introduced in 𝛽-particle decay.
• Many other particles were discovered in cosmic radiation throughout the
twentieth century, giving support for the idea that the electron, proton and
neutron were NOT the only fundamental particles.
• The numerous types of particles are placed into two main categories. Those
affected by the strong force are called hadrons, for example protons and
neutrons, and those not affected by the strong force are called leptons, for
example electrons and positrons.
• The many different particles have been produced in high-energy collisions using
accelerators such as those at Standford in Califonia and CERN in Switzerland
during the second half of the twentieth century.
• A vast number of collisions were carried out and a large number of hadrons were
produced. Two of the conclusions to these reactions were:
• The total electrical charge remains constant
• The total number of nucleons normally remains constant
The quark model of hadrons
• The problem of what were considered to be fundamental particles was resolved
by the quark model for hadrons.
• In this model, the hadrons are made up of three smaller particles called quarks.
The types of quark, called flavours of quark, are up (u), down (d) and strange (s).
• The quark flavours have charge and strangeness as shown in the table.
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•
•
There are three antiquarks, 𝑢
̅, 𝑑̅ and 𝑠̅. These have the opposite values of
charge and strangeness.
Protons and neutrons consists of three different quarks.
•
Leptons
Leptons are particles that are not affected by the strong force. The electron and
neutrino and their antimatter partners, the positrons and antineutrino, are
examples of leptons.
These types of particle do not appear to be composed of any smaller particles
and are, therefore, considered to be fundamental particles.
During the decay of a neutron in the nucleus, a proton is formed and an electron
and antineutrino emitted.
In terms of the fundamental particles, quarks, the reaction can be shown as:
•
0
0
1
1
0𝑛 → 1𝑝 + −1𝑒 + 0𝑣̅
•
•
•
•
•
u
u
d
u
d
d
The quark flavour is not conserved as down quark has changed to up quark.
The reaction cannot be due to the strong force. The 𝛽-decay must be due to
another force. This force is called the weak force or weak interaction.
The total lepton number before a reaction is equal to the total lepton number after
the reaction.
The lepton number is +1 for the particle and −1 for the antiparticle.
The total lepton number before the reaction is zero in the 𝛽 − -decay given.
The lepton numbers for the particles after the reaction are +1 for the electron and
−1 for the antineutrino, giving a total of zero.
•
•
•
•
•
•
•
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Protons and neutrons in terms of quark model
Protons and neutrons in terms of quark model
•
•
A proton consists of 2 ups and 1down (uud)
If we add the quarks we see the proton has a charge of +1
•
( + )−
•
•
A neutron consists of 2 downs and 1 up (udd)
If we add the quarks we see the neutron has no charge
•
•
•
•
2 2
1
3 3
3
2
3
4
1
3
3
= −
1
1
0
3
3
3
=
3
3
= +1
− − = =0
Antinutrons and antiprotons are made up of antiquarts(antiquark has a charge
opposite to the quark)
In beta+ decay a up quark changes to a down quark and a positron and electron
neutrino are formed
In beta- decay a down quark changes to an up quark and an electron
antineutrino are emitted.
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