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Measurements SI Units Errors and uncertainties Scalars and
vectors Learning Outcomes Candidates should be able to: (a) recall
the following base quantities and their units: mass (kg), length
(m), time (s),
current (A), temperature (K), amount of substance (mol).
(b) express derived units as products or quotients of the base
units.
(c) Show an understanding of and use the conventions for
labelling graph axes and table columns as set out in the ASE
publication SI units, Signs, Symbols and Abbreviations, except
where these have been superseded by Signs, Symbols and Systematics
( The ASE Companion to 5-16 Science, 1995 )
(d) use the following prefixes and their symbols to indicate
decimal sub-multiples or multiples of both base and derived units:
pico (p), nano (n), micro (), milli (m), centi (c), deci (d), kilo
(k), mega (M), giga (G), tera (T).
(e) make reasonable estimates of physical quantities included
within the syllabus.
(f) show an understanding of the distinction between systematic
errors (including zero errors) and random errors.
(g) show an understanding of the distinction between precision
and accuracy.
(h) assess the uncertainty in a derived quantity by simple
addition of actual, fractional or percentage uncertainties
(i) distinguish between scalar and vector quantities, and give
examples of each.
(j) add and subtract coplanar vectors.
(k) represent a vector as two perpendicular components.
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(a) recall the following base quantities and their units: mass
(kg), length (m), time (s), current (A), temperature (K), amount of
substance (mol).
Physical Quantities & SI units A physical quantity defines
some measurable feature of many different items. It
consists of a number and a unit.
Example: Area of the school compound, A = 5 000 m2
Numbers are not physical quantities. Without a unit, numbers
cannot be a measure of any physical quantity.
Physical quantities are the building blocks of Physics in terms
of which the laws of Physics are expressed.
There are 2 types of physical quantities: (1) base (fundamental)
quantities (2) derived quantities
Base Quantities & Units
A base quantity is chosen and arbitrarily defined rather than
being derived from a combination of other physical quantities.
Base unit is the standards assigned to each of these basic
quantities and to no others. It is not derived from other units,
i.e. independent of other units.
The seven base quantities and their SI units are:
Base Quantity / Symbol Unit Symbol SI Unit
Length, l Mass, m Time, t electric current, I temperature, T
amount of substance, n *luminous intensity, Iv
m kg s A K
mol cd
metre kilogram
second ampere
kelvin mole candela
* luminous intensity is not in the syllabus
Physical Quantity Numerical magnitude
Unit
Base units are units from which all other units can be
defined
All physical quantities consist of a numerical magnitude and a
unit.
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(b) express derived units as products or quotients of the base
units.
A derived quantity is defined based on combination of base
quantities and has a derived unit that is the product and/or
quotient of these base units.
Example
timentdisplaceme velocity =
Unit of velocity = unit of distance / unit of time
= m / s = m s1
Example Force = mass x acceleration
unit of F = unit of m x unit of a = kg m s-2
Example Potential Energy = mgh
unit of PE = unit of (m x g x h) = kg m s-2 m
so, SI unit of energy, J, can be expressed as kg m2 s-2 Worked
Example 1 (J93/I/2, J83/II/28) What are the SI base units of
specific heat capacity? Solution Specific heat capacity, c, of a
substance is defined as the thermal energy required by a unit mass
of the substance (e.g.1 kg) to experience a unit rise in
temperature (e.g. by 1 K).
Unit of c
= =2 -2unit of thermal energy J kg m s =
(unit of mass) x (unit of temperature change) kg . K kg . K
2 -2 -1= m s K More on homogeneity of equations
A physical equation is true irrespective of the system of units
used for the physical quantities mentioned in the equation. Note
that: - Each term of the equation has the same units - Only
quantities of the same units can be added, substracted or equated
in an
equation - The equation is said to be homogeneous or
dimensionally correct if all the terms
in it has the same base units.
Derived quantity
Base quantities
Derived unit
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- Consider the following equation
21s = ut + at2
Unit for s = Unit for ut = Unit for at2 = Note that all the
three terms have the same unit. Therefore, we can conclude that the
equation is homogeneous. An equation which is not homogeneous must
be physically wrong. On the other hand, if the units for the
various terms in an equation are the same, it does not imply that
the equation is physically correct. Cases where an equation can be
homogeneous and yet incorrect are: a. Incorrect Coefficient (s)
lT = 2g
is dimensionally correct and physically correct, but lT = 3g
is
dimensionally correct but physically wrong. b. Missing terms
The relationship between s, u, t and a may just be written as
21s = at2
which is
incomplete and wrong even though the equation is homogeneous c.
Extra Terms
An equation may be wrongly written with an extra term which has
the same units
as the other terms for e.g. +l lT = 2g g
The correctness of a physical equation is determined
experimentally. To test for the homogeneity of physical equations,
the units of the terms on the right hand (RHS) of the equation must
be equal to the units of the terms on the LHS.
Worked Example 2 (N82/II/29) At Temperatures close to 0 K, the
specific heat capacity of a particular solid is given by c = aT3,
where T is the thermodynamic temperature and a is a constant. Find
the units of a in terms of SI based units. Solution c = aT3 unit of
c = unit of (aT3) m2 s-2
K-1 = unit of a unit of K3
unit of a = m2 s-2 K-4
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Worked Example 3 The experimental measurement of the specific
heat capacity c of a solid as a function of temperature T is to be
fitted to the expression c = T + T3. What are the possible units of
and expressed in SI base units? Solution
c = T + T3
Base unit of c = base unit of T = base unit of T3
m2 s-2 K-1 = (unit of ) (unit of T ) unit of = m2 s-2 K-2
Also base unit of c = base unit of T3
m2 s-2 K-1 = (unit of ) (unit of T3 ) unit of = m2 s-2 K-4
Worked Example 4 Which of the following units are not equivalent to
that of force? A. Pa m2 B. W m1 s C. J m s1 D. kg m s2 Solution
Working Base units A) Pa m2 kg m-1 s-2 x m2 kg m s-2 B) W m-1 s
kg m2 s-3 x m-1 s kg m s-2 C) J m s-1 kg m2 s-2 x m s-1 kg m3 s-3
D) kg m s-2
Answer is C.
For the equation above to be dimensionally correct or
homogeneous, the base units of the terms on the L.H.S. of the
equation must be the same
as the base units of the terms on the R.H.S.
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Worked Example 5 The critical density of matter 0 in the
universe may be written as
What are the SI base units of H ?
Solution
3G8H 0=
unit of H = ( kg m-3 )( N m2 kg-2 ) = ( kg m-3 )( kg m s-2 m2
kg-2 )
= s-2 Worked Example 6 The energy E of a damped oscillator of
mass m varies with time t is given by What are the units of k and
b?
Solution
e-bt/2m is a number and hence has no unit
Power has no unit and hence 2btm
= 1 =(unit of b).s 1kg
unit of b = kg s-1
unit of E = (unit of A2 ) ( unit of k )
( unit of k ) = J m-2 = ( kg m2 s-2 ) m-2 = kg s-2
G8H
3
0 =H = Hubble constant G = gravitational constant (unit: N m2
kg2)
G8H
3
0 =
2mbt
2ekAE
= A = amplitude (unit: m) k, b = constants
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(c)
Signs, Symbols and Systematics ( The ASE Companion to 5-16
Science, 1995 )
Show an understanding of and use the conventions for labelling
graph axes and table columns as set out in the ASE publication SI
units, Signs, Symbols and Abbreviations, except where these have
been superseded by
as the leading column in a table, e.g. pre-selected physical
quantity such as length.
readings recorded from experiment or calculated values from
previous columns. e.g.
Time taken for 20 oscillat ns
A table of ordered pairs is used to represent a relationship. In
a table, it is conventional to have the independent variable to be
recorded
The second and subsequent columns should be the dependent
variable(s), e.g.
ioPendulum length
L
Period, T = ( t1 + t2 ) / 40
/ cm
t1 / s
t2 / s
T / s 20.0 10.3 10.5 0.520 40.0 15.7 15.8 0.788 60.0 20.9 21.2
1.05
Independent
ces (d.p.) according to the type of the measuring instrument
i.e. the etre rule.
count of the human reaction time of ~ .2 s in starting and
stopping the stopwatch.
ll have the ame number of significant figures (s.f.) as its raw
score ( raw data ).
e horizontal axis ( x-axis ) and the dependent variable on the
vertical axis ( y-axis ).
tercept of a graph, y = 0 is substituted into the graph or
equation
tercept of a graph, x = 0 is substituted into the graph or
equation to solve for y.
variable
Notice the pendulum length readings are recorded with the same
number of decimal plam The time taken by a stopwatch however was
only recorded to 1 d.p. throughout instead of the usual 2 to 3 d.p.
as indicated in the time instrument. This is because the time
readings recorded must take into ac0 The period of one oscillation
is a calculated value ( processed data ) taken from records of the
time taken for 20 oscillations, hence, the values recorded wis A
graph is an infinite set of order pairs plotted on a Cartesian
plane, representing a relationship between the elements of the
ordered pair. It is conventional to plot the independent variable
on th
To find the x-in
to solve for x. To find the y-in
Dependent variable
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The selected scales of the axes should be easy to read and large
enough to enable the curve(s) (Curve refers to both curves and
straight lines) to occupy at least half of the graph grid. The
selected scale should usually be in multiples of 1, 2, 5 or 10
unit(s). A false origin is sometimes used in order to accommodate
the curve into the graph.
When determining the gradient of the curve at a particular point
or straight line, the chosen reference triangle should fall within
the recorded data points and be at least half the size of the
curve. The chosen points should not coincide with any of the
recorded data points from the experiments. Label these chosen
points for easy checking.
(d) use the following prefixes and their symbols to indicate
decimal sub-multiples or
multiples of both base and derived units: pico (p), nano (n),
micro (), milli (m), centi (c), deci (d), kilo (k), mega (M), giga
(G), tera (T).
Prefixes are used to simplify the writing of very large or very
small orders of
magnitude of physical quantities.
Fraction / multiple Prefix Symbol 10-12 pico p 10-9 nano n 10-6
micro 10-3 milli m 10-2 centi c 10-1 deci d 103 kilo k 106 mega M
109 giga G 1012 tera T
L / cm
Reference triangle
Independent variable
Dep
ende
nt v
aria
ble
False origin
T / s
10.0
15.0
20.0
25.0
30.0
0.0 20.0 40.0 60.0
x
x
xx
x x
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Examples
1500 m = 1.5 x 103 m = 1.5 km 0.00077 V = 0.77 x 10-3 V = 0.77
mV 100 x 10-9 m3 = 100 x (10-3)3 m3 = 100 mm3
(e) make reasonable estimates of physical quantities included
within the syllabus.
The following are examples of the estimated values of some
physical quantities:
Diameter of an atom ~ 10-10 m Diameter of a nucleus ~ 10-15 m
Air pressure ~ 100 kPa Wavelength of visible light ~ 600 nm
Resistance of a domestic lamp ~ 1000 Mass of a car ~ 2000 kg
Maximum speed of a car on an expressway ~ 90 km h-1
(f) show an understanding of the distinction between systematic
errors (including
zero errors) and random errors.
Measuring any physical quantity requires a measuring instrument.
The reading will
always have an uncertainty. This arises because a) Experimenter
is not skilled enough b) Limitations of instruments c)
Environmental fluctuations
As a result, measurements can become unreliable if we do not use
good measurement techniques. Some of the common practices to
minimise the errors made in measurements are as follows: a) Taking
average of many readings b) Avoid parallax errors c) Take readings
promptly
Analogue & Digital displays
Often, when we measure a quantity with an instrument, we can
make an estimate of the uncertainty by using the following
rule:
Uncertainty of a Reading = Half the smallest scale division
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For example, take a look at the following analogue scales:
Reading = 5.7 Reading = 2.36
Even when instruments with digital displays are used, there are
still uncertainties present in the measurements. So for example, a
digital ammeter may show the current to be 358 mA. This does not
mean that the current is 358 mA exactly. For digital displays, the
uncertainty is usually provided by the manufacturer.
Errors & Uncertainties
Errors or uncertainties fall generally into 2 categories: a)
Random errors b) Systematic errors
Random Errors
The readings are equally likely to be higher or lower than the
MEAN value. Random errors are of varying sign (higher or lower than
average) and magnitude and cannot be eliminated. Averaging repeated
readings is the best way to minimize random errors.
Examples: Measuring the diameter of a wire due to its
non-uniformity
x x x
xx
xx
x
xx
5
7
6
2.2
2.4
2.3
Uncertainty = 0.1 Uncertainty = 0.01
Random error is one that occurs without a fixed pattern
resulting in a scatter of readings about a mean value.
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Systematic errors
The readings are consistently higher or lower than the ACTUAL
value. Systematic error is consistent in both magnitude and sign
and results in readings taken being faulty in one direction.
Systematic errors cannot be reduced or eliminated by taking the
average of
repeated readings. It could be reduced by techniques such as
making a mathematical correction or correcting the faulty
equipment.
Examples: zero error of a measuring instrument, a clock running
too fast or too slow.
(h) assess the uncertainty in a derived quantity by simple
addition of actual, fractional or percentage uncertainties
If we denote the uncertainty or error of P as P, then we write
the measured quantity as
P + P
Fractional error of P = PP
Percentage error of P = %100PP
x x
x x
xx
x x
x
xx
xx
x x
x x
x
Systematic error is one that occurs with a fixed pattern
resulting in a consistent over-estimation or under-estimation of
the actual value.
Absolute error
Mean Value
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Worked Example 7
The length of a piece of paper is measured as 297 1 mm. Its
width is measured as 209 1 mm. (a) What is the fractional
uncertainty in its length? (b) What is the percentage uncertainty
in its length? Solution Note : 297 + 1 mm
(a) Fractional uncertainty in its length = 1
297 = 0.00337
(b) Percentage uncertainty in its length = 1 x 100%
297 = 0.337 %
Note:
Rule of thumb: answers should always be rounded off to 3
significant figures (3.S.F.) except for absolute errors, which are
always rounded off to 1.S.F. MARKS will be deducted for expressing
the answers to an incorrect number of significant figures!
The average value is always rounded off to the number of decimal
figures of the absolute error when expressed in scientific
notation.
(g) show an understanding of the distinction between precision
and accuracy.
Measurements are often described as precise or accurate. In
layman terms, these 2 terms are used interchangeably to mean the
same thing. In Physics, it is possible to have precise but
inaccurate measurements and accurate measurements that are not
precise.
Suppose we do some experiments to find g, the acceleration of
free fall. The expected result is 9.81 m s-2. The results obtained
are shown below:
Absolute error
Mean Value
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8.63, 8.78, 8.82, 8.59, 8.74, 8.88 9.76, 9.79, 9.83, 9.85, 9.88,
9.90
( precise, not accurate ) ( accurate & precise )
9.64, 9.81, 9.95, 10.02, 9.77, 9.68 7.65, 8.92, 10.00, 9.12,
8.41, 9.45
( accurate but not precise ) (neither precise nor accurate)
To summarise, Precision: A set of measurements is precise if a)
the measurements have a small spread or scatter b) there are small
random errors in the measurements Accuracy: A set of measurements
is accurate if a) the measurements are close to the actual value b)
there are small systematic errors in the measurements
Determining Uncertainty in Derived Quantities
No. of readings, n
Value of reading, x Expected 9.81
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Addition & Subtraction
sured with uncertainties A and B respectively. C = A + B, find
the uncertainty in C.
If D = A B, find the uncertainty in D.
Multiplication and Division in E.
If F =
Suppose A and B are mea If
If E = AB, find the uncertainty
AB
, find the uncertainty in F.
orked Example 8
W
cm and Y = (5.0 0.2) cm, find D and the associated uncertainties
if ) D = X + Y
D = X + Y = 6.0 cm
D = X + Y = 0.2 + 0.2 = 0.4 cm Therefore, D = (6.0 +
Given X = (1.0 0.2) (a
0.4) cm
) D = Y X
(b
D= A + B
Generalising:
If A = Bn, then BBn
AA =
If A = Bm Cn , then CCn
BBm
AA +=
If A = Bm / Cn , then CCn
BBm
AA +=
BB
AA
EE +=
BB
AA
FF +=
C = A + B
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D = Y X = 4.0 cm
D = Y + X = 0.2 + 0.2 = 0.4 cm Therefore, D = (4.0 +
0.4) cm
) D = 2Y 3X D = 2Y 3X = 7.0 cm D = 2(Y) + 3(X) = 2(0.2) + 3(0.2)
= 1.0 cm
(c 1 cm (to 1 sig. fig)
Therefore, D = (7 + 1) cm
D =
(d) 4
X-Y
D = 4
X-Y = 1.0 cm
D = (Y + X) = 0.1 cm
Therefore,
= (0.2 + 0.2)
D = (1.0 + 0.1) cm (e) D = XY
YY
XX
DD +=
0.2 0.2= + 1.0 5.0
= 0.24
Average D = X Y = (1.0) (5.0) = 5.0 cm2
DDDD =
= 0.24 x 5.0 = 1.2 cm2 (to 1 sig. fig)
Therefore, D = (5 +
21 cm
1) cm
) D = 4XY
2
(f
YY
XX
DD +=
= 0.52.0
0.12.0 + = 0.24
= 4(1.0)(5.0) = 20 cm2 D
D = 0.24 x 20 = 4.8 cm2 25 cm (to 1 sig. fig)
= (20 5) cm2
D
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(g) D = X2Y
YY
XX
DD += 2
0.2 0.2= 2( ) + 1.0 5.0
= 0.44
Average D = X2 = (1.0)2 3
Y (5.0) = 5.0 cm
DDDD =
= 0.44 x 5.0 = 2.2 cm3 (to 1 sig. fig)
Therefore, D = (5 +
32 cm 2) cm3
ple 9Worked Exam experiment to determine g the equation used is
where
ind the value of g and its uncertaint
olution
In a simple pendulumT = (2.16 0.01) s and l = (1.150 0.005) m. F
y.
S
Note: T was the original subject of the equation but since we
are interested in g, before
we form the error equation, we have to make g the subject of the
equation first.
glT 2=
TT
ll
gg += 2
22
Tl4g =
16.201.02
150.1005.0 +=
gg
216.2)150.1(24g = 0136.0
gg =
g = 9.73 m s-2 g = 0.132 = 0.1 m s-2 (to 1 sig. fig)
g = (9.7 0.1) m s-2
glT = 2
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Worked Example 10 (N89/II/2) The length of a piece of paper is
measured as 297 1 mm. Its width is measured as 209 1 mm. What is
the area of one side of the piece of paper? State your answer with
its uncertainty. Solution
Area of the paper = 297 x 209 = 6.21 x 104 mm2
A lA l b
b = + 209
1297
1 += 410x 6.21A
A = 506 = 0.0506 x 104 mm2 (to 1 sig. fig) 4 20.05 x 10 mm
Therefore, A = (6.21 + 0.05 ) x 104 mm2
Worked Example 11 In an experiment to measure the Youngs modulus
of a wire, E, the following measurements were made:
Length of wire, l 3.025 0.005 m Diameter of wire, d 0.84 0.01 mm
Mass, M 5.000 0.002 kg Extension, e 1.27 0.02 mm Acceleration of
free fall, g 9.81 0.01 m s-2
If E is calculated as 24
edMglE = , calculate E with its associated uncertainty.
Solution
24
edMglE =
dd
ee
ll
gg
MM
EE ++++= 2
0426298.084.001.02
27.102.0
025.3005.0
81.901.0
000.5002.0
EE =
++++=
E = 0.0426298 x 2.11 x 1011
11 2 -3 -3 2 2.11 x 10 Pa
4 4(5.000)(9.81)(3.025)(1.27 x 10 )(0.84 x 10 )
MglEed == =
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E = 8.78 x 109 Pa
E = 0.09 1011 Pa (to 1 sig. fig) E = (2.11 0.09) x 1011 Pa
Note (REPEATED):
Rule of thumb: non-exact answers should always be rounded off to
3 significant figures (3.S.F.) except for absolute errors, which
are always rounded off to 1.S.F. MARKS will be deducted for
expressing the answers to an incorrect number of significant
figures!
The average value is always rounded off to the number of decimal
figures of the absolute error when expressed in scientific
notation.
(i) distinguish between scalar and vector quantities, and give
examples of each.
Scalars and vectors are both physical quantities (they have both
a number and a unit).
Examples: mass and charge
Examples: displacement and force Examples
Scalars Vectors Distance displacement speed velocity time
acceleration frequency force density momentum (encountered in the
topic of
Dynamics)
Note:
A vector can be placed anywhere in a diagram as long as it keeps
its same length and direction.
Two vectors with the same length but different directions are
different. Directions for vectors must be given clearly without
ambiguity.
A scalar quantity consists of a magnitude only.
A vector quantity consists of a magnitude and a direction.
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3 different ways to give directions clearly are: i) Compass
points, e.g. due east, 75o north of west, 20o east of south ii)
Bearings, e.g. bearing of 090o, 345o, 160o
iii) X-Y plane, e.g. positive x-axis, 75o above the negative
x-axis, 70o below the positive x-axis
(j) add and subtract coplanar vectors.
Dealing with Vectors
When vectors are added however, the result is NOT just the sum
of the numbers (magnitudes). The directions of the vectors must be
considered, especially when they point in different directions.
Vectors Addition by Scaled Drawing
Examples
75o
A
B
R
i) Due East ii) Bearing of 090o
i) 75o north of west ii) Bearing of 345o iii) 75o above the -ve
x-axis
i) 40o south of east or ii) Bearing of 130o iii) 40o below the
+ve x-axis 40o
A
B
R
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Vectors Addition by Triangle Law
Worked Example 12 Find R.
Solution
180 84 96 = = 96 40
120sin sin
R =
R2 = 1002 + 1202 2(100)(120) cos 40 R = 77.6 m
R is 77.6 m at an angle of 84.0o north of east (other possible
answers?)
Vectors Subtraction Example : P - Q = P + (- Q)
(k) represent a vector as two perpendicular components.
When 2 perpendicular vectors are added, they give a resultant as
shown:
P+(-Q)
P
- Q
P
Q
V + H = R
H
V R
100 m
120 m R
40o 84o
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Finding the components of a vector is the reverse process of
vector addition. Instead of combining 2 vectors into one, a vector
can be split into 2 components, the horizontal component and the
vertical component.
Worked Example 13 Find the horizontal and vertical components of
the forces A, B & C. Solution
Vector Horizontal / N Vertical / N A - 400 cos 25o 400 sin
25o
B 500 cos 35o 500 sin 35o
C - 200 cos 20o - 200 sin 20o
Total - 141 387
Note:
Check that your calculator is in the correct mode i.e. radian or
degree if you cannot get the final answer for the computation of
trigonometric functions.
Worked Example 14 (N87/I/3, J94/I/1) Which of the following
pairs contains one vector and one scalar quantities? A.
displacement; acceleration B. force; kinetic energy C. power; speed
D. work; potential energy Ans : B
Rx = H = R cos Ry = V = R sin
tan = VH
R
H
V
35o25o
20o
B (500 N)
C (200 N)
A (400 N)
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Worked Example 15 (N90/I/1) Which list contains only scalar
quantities? A. mass, acceleration, temperature, kinetic energy B.
mass, volume, kinetic energy, temperature C. acceleration,
temperature, volume, electric charge D. moment, velocity, density,
force Ans : B
Worked Example 16 (N90/I/1) (a) Two vectors A and B are at right
angles to each other. Draw a vector diagram to
show how the sum of the vectors could be found. (b) A car
changes its velocity from 30 m s1 due East to 25 m s1 due
South.
(i) Draw a vector diagram to show the initial and final
velocities and the change in velocity.
(ii) Calculate the change in speed. (iii) Calculate the change
in velocity.
Solution
bii) Change in speed = 25 30 = -5 m s-1
biii) Change in velocity = ( 252 + 302 )0.5
= 39.1 m s-1
angle = tan-1 (25 / 30) = 39.8
The change in velocity is 39.1 m s-1 at an angle of 39.8 south
of west.
(other possible answers?)
fvJJK
ivJK
bi)
fvJJK
ivJJJK
( )f iv v+ JJK JK
AJK
BJK
A B+JK JK
a)
-
2011 H1 PHYSICS (8866) Measurements
JJ Physics Dept Page 23 of 23
Note (IMPORTANT):
Change in a physical quantity = Final physical quantity Initial
physical quantity: For scalars, one can simply take the difference
in the magnitude; For vectors, one need to consider the directions
involved!
In the previous example, speed is a scalar whereas velocity is a
vector!
Link to mathematical requirements found in O Level 4016
Mathematics (2010) Syllabus:
Acknowledgements - Updated by Chong K.W. , 2011
Analogue & Digital displaysRandom Errors Systematic
errors
Worked Example 7Precision: Accuracy: