Demonstrate understanding of aspects of mechanics 15
Mass, weight, gravity, pressure
Mass, m, is the amount of matter in an object and remains the same wherever the object is (Earth, Moon,
outer space). Units are grams (g), kilograms (kg).
1.1
exam
ial
notes
Essent
Weight is the force with which an object pushes down on whatever it is resting on. Weight is caused by the
pull of gravity on an object (gravity force on an object is often called weight force). Unit is newton (N).
Weight of an object is calculated by multiplying mass by g, the strength of gravity. On Earth, g is 10 newtons
per kilogram, 10 N kg–1. This means every object pushes down with a force of 10 N for each kg of its mass;
therefore, to find the weight of an object in newtons, multiply its mass (in kg) by 10.
The value of g is also called acceleration due to gravity as a falling object accelerates downwards at 10 m s–2,
i.e. the same value as g.
1
1
that on Earth; therefore, objects on the Moon weigh about
of their
6
6
weight on Earth. Outer space has no gravity, so objects are effectively weightless.
Gravity on our Moon is about
When an object is at rest or is travelling along any surface, the downwards push from the object’s weight is
balanced by the upwards push from the surface. Upwards force from the surface is called a reaction force
because it occurs only when an object pushes down.
F
F
Pressure, P, is the amount of force an object generates per unit area: P =
A
P
A
Units are N m–2 (N/m2); other units are N mm–2 (N/mm2). Pressure may also be given in
–2
pascals (Pa). One pascal is one newton per square metre (1 Pa = 1 N m ).
Pressure can change the shape of the object it is acting on – the smaller the area the force is acting on, the
greater the change in shape. For example, a person wearing football boots with ‘sprigs’ on the bottom is
more likely to leave an impression in the surface of the ground than when the person is wearing running
shoes (the person’s weight force is spread over a wider area when wearing running shoes). Softer the
surface, more likely it will undergo a change of shape.
Sir Isaac Newton (1642–1727) – his work formed the basis for mechanics
ESA Publications (NZ) Ltd, Freephone 0800-372 266
AMEy11 Sci book.indb 15
14/05/14 11:15 am
Year 2013 Ans. p. 102
practic
e
NCEA
16 Achievement Standard 90940 (Science 1.1)
Questions
Mass, weight, gravity, pressure
Question One: Fun in the snow
A family decides to spend a day at a snow field.
The father hires a snowboard for himself and a
pair of skis for his daughter.
Assume the snowboard and skis are
rectangular in shape.
The father and snowboard have a combined
mass of 80 kg.
1.75 m
1.75 m
1.6 m
0.08 m
0.25 m
0.08 m
0.25
a. Calculate the pressure exerted by the father and snowboard on the snow.
Your answer should include:
• an area calculation
• a calculation of the pressure.
 ESA Publications (NZ) Ltd, Freephone 0800-372 266
AMEy11 Sci book.indb 16
14/05/14 11:15 am
Demonstrate understanding of aspects of mechanics 17
The daughter and the skis have a combined mass of 58 kg.
Explain why the daughter on her skis sinks further into the snow than her father on his snowboard.
In your answer you should:
• calculate the pressure exerted by the daughter and her skis on the snow
• compare the pressure exerted by the daughter and father (from a.) on the snow
• explain the difference in pressure in terms of force AND area
• explain how pressure relates to how far the person will sink in the snow.
1.1
b. The father notices that his daughter on her skis has sunk further into the snow than he has on his
snowboard.
ESA Publications (NZ) Ltd, Freephone 0800-372 266
AMEy11 Sci book.indb 17
14/05/14 11:15 am
18 Achievement Standard 90940 (Science 1.1)
Year 2012 Ans. p. 102
Question Two: The tractor
A woman drives her tractor down a sandy beach to pick up her friend’s boat. The distance-time graph
below shows part of the journey.
Distance
(metres)
Section A
600
Section B
120
0 10 20 30 40 50 60 70 80 90 Time
(seconds)
a. U
se the information from the graph to calculate the average speed of the tractor during the
90 seconds.
average speed =
m s–1
b. Describe the motion of the tractor in section B, and explain what this tells us about the forces acting
on the tractor during this time.
c. The total mass of the tractor and driver is 1 660 kg.
Calculate the speed of the tractor at the end of section A, and then calculate the net force acting on
the tractor during section A of the graph.
net force =
N
 ESA Publications (NZ) Ltd, Freephone 0800-372 266
AMEy11 Sci book.indb 18
14/05/14 11:15 am
AMEy11 Sci book.indb 35
Li
6.9
Fr
223
87
Cs
133
55
Rb
85.5
37
K
39.1
19
Na
23.0
11
3
1
Be
9.0
103
Lr
262
Lu
175
71
Y
88.9
39
Sc
45.0
21
3
Actinide Series
Lanthanide Series
Ra
226
88
Ba
137
56
Sr
87.6
38
Ca
40.1
20
12
Mg
24.3
4
2
Ac
227
89
Th
232
90
Ce
140
58
57
La
139
105
Db
262
Ta
181
73
41
Nb
92.9
V
50.9
23
5
104
Rf
261
Hf
179
72
Zr
91.2
40
Ti
47.9
22
4
Pa
231
91
Pr
141
59
106
Sg
263
W
184
74
42
Mo
95.9
Cr
52.0
24
6
U
238
92
60
Nd
144
107
Bh
264
Re
186
75
Tc
98.9
43
25
Mn
54.9
7
H
1.0
93
Np
237
61
Pm
147
108
Hs
265
Os
190
76
Ru
101
44
Fe
55.9
26
8
Pu
239
94
62
Sm
150
109
Mt
268
Ir
192
77
Rh
103
45
Co
58.9
27
9
Atomic Mass
95
Am
241
Eu
152
63
Pt
195
78
Pd
106
46
Ni
58.7
28
10
96
Cm
244
64
Gd
157
Au
197
79
Ag
108
47
Cu
63.6
29
11
Bk
249
97
Tb
159
65
Hg
201
80
Cd
112
48
Zn
65.4
30
12
B
10.8
Cf
251
98
Dy
163
66
Tl
204
81
In
115
49
Ga
69.7
31
Al
27.0
13
5
13
C
12.0
Es
252
99
Ho
165
67
Pb
207
82
Sn
119
50
Ge
72.6
32
Si
28.1
14
6
14
N
14.0
100
Fm
257
Er
167
68
Bi
209
83
Sb
122
51
As
74.9
33
P
31.0
15
7
15
O
16.0
101
Md
258
69
Tm
169
Po
210
84
Te
128
52
Se
79.0
34
S
32.1
16
8
16
F
19.0
102
No
259
Yb
173
70
At
210
85
I
127
53
Br
79.9
35
Cl
35.5
17
9
17
18
He
4.0
Rn
222
86
Xe
131
54
Kr
83.8
36
Ar
40.0
18
Ne
20.2
10
2
exam
ial
Atomic Number 1
Periodic Table of the Elements
notes
Essent
Achievement Standard 90944
SCIENCE
Demonstrate understanding of aspects
of acids and bases
1.5
Externally assessed 4 credits
Periodic Table
14/05/14 11:15 am
36 Achievement Standard 90944 (Science 1.5)
Table of ions
You must know the names of the ions. Following is the Table of Ions that you will be given in the 90944
exam but with spaces for you to write the names of the ions.
Ca2+
Al3+
O2–
OH–
NO3–
NH4+
CO32–
HCO3–
Cl–
Cu2+
SO42–
S2–
K+
Pb2+
Fe3+
Ag+
Fe2+
Mg2+
H+
Ba2+
Na+
Li+
+1
1.5
Zn2+
+2
+3
–2
F–
–1
 ESA Publications (NZ) Ltd, Freephone 0800-372 266
AMEy11 Sci book.indb 36
14/05/14 11:15 am
Demonstrate understanding of aspects of acids and bases 37
exam
ial
Atomic structure
notes
Essent
Atomic number = no. protons. Mass number = no. of protons + no. neutrons (electrons are so small their
mass is negligible in contributing to the overall mass of an atom).
Example: sodium, Na: atomic no. 11 (11 protons); mass no. 23 (11 protons + 12 neutrons).
Atoms of an element that have different numbers of neutrons are called isotopes.
Example: Oxygen, O: atomic no. 8 (8 protons) has two common isotopes:
• mass no. 16 (8 protons + 8 neutrons) is written as
16
8
• mass no. 18 (8 protons + 10 neutrons) is written as
1.5
Elements are made up of only one type of atom. Atoms are made up of protons (positively charged),
electrons (negatively charged), and neutrons (no charge). Protons and neutrons form the nucleus,
electrons orbit the nucleus.
O
18
8
O
Atoms are neutral (have no overall charge) – the number of (positive) protons
equals the number of (negative) electrons. Example: sodium, Na, atomic no. 11,
has 11 protons and also 11 electrons.
Electrons orbit the nucleus in energy levels (‘shells’). For the first 20 elements,
first shell (closest to nucleus) holds maximum of 2 electrons; 2nd and 3rd shells
hold up to 8 electrons; any remaining electrons are found in the 4th shell. The
electron arrangement (or electron configuration) describes the number of electrons
in each shell. Example: sodium (11 electrons: 2 electrons in 1st shell, 8 electrons
in 2nd shell, 1 electron in 3rd shell) electron arrangement is written Na: 2,8,1
The sodium atom
n n
+
Na: 2,8,1
Ions form when atoms gain or lose electrons – becoming ‘charged’ particles (no
longer neutral). Atoms react to get a complete outer shell of electrons. For the first 20 elements:
• If outermost shell has < 4 electrons, atom tends to lose electrons, becoming a positive ion (now more
protons than electrons). Examples: Na: 2,8,1 reacts to lose the one electron in its outermost shell
becoming Na+ ion with electron arrangement 2,8; magnesium, Mg: 2,8,2, loses two electrons from
outermost shell becoming Mg2+ 2,8. With a complete outer shell, ions are more stable and less reactive
than the parent atom.
• If outermost shell has > 4 electrons, atom tends to gain electrons, becoming a negative ion (now fewer
protons than electrons). Examples: fluorine, F: 2,7 reacts to gain one more electron in its outermost shell
becoming F– ion with electron arrangement 2,8; oxygen, O: 2,6 gains two electrons in its outermost
shell becoming O2–: 2,8
• Atoms that already have a complete outermost shell are very stable and do not react in normal
laboratory conditions (Group 18, inert gases – e.g. He, Ne).
• Hydrogen, H, has a first shell of just one electron. It reacts to lose its electron and form the hydrogen
ion, H+, effectively becoming just a proton.
Metals, found on the left and middle of the periodic table, react by losing electrons to form positive ions.
Group 1 metals (e.g. Li, Na, K) react to lose one electron; group 2 metals (e.g. Mg, Ca) react to lose two
electrons.
Non-metals, found on the right side of the periodic table, react by gaining electrons to form negative ions;
group 17 (e.g., F, Cl) all react to gain one electron.
Polyatomic ions are made of more than one atom – e.g. ammonium, NH4+ ; hydroxide, OH– ; carbonate,
CO32– ; sulfate, SO42 ; nitrate, NO3–
Metals reacting with non-metals form ionic compounds. The attraction of unlike changes (+ metal ion
and – non-metal ion) holds the ions together forming an ionic bond. Examples: sodium fluoride, NaF;
ESA Publications (NZ) Ltd, Freephone 0800-372 266
AMEy11 Sci book.indb 37
14/05/14 11:15 am
38 Achievement Standard 90944 (Science 1.5)
magnesium oxide, MgO; copper chloride, CuCl2
To write a formula for an ionic compound: (1) Select the required ions from the Table of Ions. (2) Put the
positive ion first, negative ion second. (3) Numbers of positive and negative charges must balance as the
compound is overall neutral. If the charges are not equal for the two ions, ‘drop and swap’ the values of the
ions – the value of the charge becomes the subscript of the other ion. Examples follow.
Magnesium oxide: Mg forms Mg2+ ion, O forms O2– ion. Values of charges the same (both 2), so formula MgO
1.5
Magnesium chloride: Values of charges (Mg2+ and Cl–) different, so value of Mg ion (2) becomes subscript
for Cl (2) and the formula is MgCl2 (Value of Cl– ion is 1; no need to put a 1 either in front of the charge or
as a subscript for Mg)
Magnesium hydrogencarbonate: hydrogencarbonate is the polyatomic ion HCO3–; brackets must be used
before the subscript is included, e.g. Mg(HCO3)2
Write the formulae of the ionic compounds on the following table.
Formulae of ionic compounds
Name
Formula
Name
Formula
sodium hydroxide
zinc oxide
silver chloride
ammonium hydroxide
calcium sulfide
ammonium carbonate
copper chloride
iron(II) sulfate
copper hydroxide
iron(III) sulfate
barium hydroxide
aluminium hydroxide
lead oxide
aluminium oxide
lead nitrate
sodium hydrogencarbonate
potassium oxide
magnesium hydrogencarbonate
For all reactions, word equations and balanced symbol equations need to be written. To write a balanced
symbol equation: (1) Write the chemical formula for each of the reactants and products using the ion table.
(2) Count to see if there is the same number of each kind of atom on both sides of the equation. (3) If needed,
put a whole number in front of the compound. For example, reaction of sodium hydroxide with sulfuric acid:
Word equation:
sodium hydroxide
NaOH
Symbol equation:
Balanced symbol equation:
2NaOH
+ sulfuric acid
+
H2SO4
+
H2SO4
→
→
sodium sulfate
Na2SO4
+
+
water
H2O
→
Na2SO4
+
2H2O
2 is needed in front of the NaOH in the reactants as there are two Na’s in Na2SO4 in the products. 2NaOH
gives two O and two H. Therefore, a 2 is also needed in front of H2O in the products. Counting up the
atoms shows the equation is now balanced:
Na
Reactants
Products
AMEy11 Sci book.indb 38
O
H
S
two Na
six O
four H
one S
(from 2NaOH)
(two from 2NaOH,
four from H2SO4)
(two from 2NaOH and
two from H2SO4)
(from SO4)
two Na
six O
four H
one S
(from Na2SO4)
(four from Na2SO4,
two from H2O)
(from 2H2O)
(from SO4)
 ESA Publications (NZ) Ltd, Freephone 0800-372 266
14/05/14 11:15 am
Demonstrate understanding of aspects of acids and bases 39
practic
e
NCEA
Questions
Atomic structure
Question One – Atomic structure
Year 2013 Ans. p. 106
F , Ne, and Mg have the same electron arrangement.
–
2+
a. Complete the table, using the periodic table on page 35.
Number of protons
Number of
electrons
Electron
arrangement
1.5
Atomic number
F–
Ne
Mg2+
b. Compare the atomic structure of F–, Ne, and Mg2+. In your answer you should:
• describe the difference between an atom and an ion
• explain the charges on F–, Ne, and Mg2+ in terms of electron arrangement and number of protons
• r elate the position of F–, Ne, and Mg2+ on the periodic table to the charges and electron
arrangement
• explain why all three have the same electron arrangement.
ESA Publications (NZ) Ltd, Freephone 0800-372 266
AMEy11 Sci book.indb 39
14/05/14 11:15 am
40 Achievement Standard 90944 (Science 1.5)
Year 2012 Ans. p. 106
Question Two: Atoms and ions
a. Complete the table below for ions formed by Ca, F, and Cl.
Atom
Atomic number
1.5
Ca
20
F
9
CI
17
Electron
arrangement of
atom
Electron
arrangement of
ion
Ion symbol
b. Explain the charges on ALL three ions, in terms of electron arrangement and number of protons.
c. Use their positions on the periodic table to explain why two of the atoms form ions with the same
charge, AND two of the atoms form ions with the same electron arrangement.
 ESA Publications (NZ) Ltd, Freephone 0800-372 266
AMEy11 Sci book.indb 40
14/05/14 11:15 am
Answers and explanations
Achievement Standard 90940
(Science 1.1): Demonstrate understanding
of aspects of mechanics
Thrust = Air friction
support
thrust
air friction
thrust = air friction
Runner at constant speed.
In all answers involving calculations:
gravity
•
give the relevant formula, e.g. F = ma
•
if required, rearrange the formula to find the correct
quantity,
e.g. m =
•
F
a
substitute the correct values, e.g. m =
Not possible to say
10
2
support
air friction
(remember to convert any non-standard values to standard
values first – e.g. kilometres to metres, grams to kilograms –
as required in the formula)
•
gravity
do the calculation and give the correct unit, e.g. 5 kg.
1.1 Straight-line motion and force
or
p. 4
support
Question One: The runner
thrust
a. Section A: Runner has constant acceleration for 8 seconds,
indicated by straight line upwards (has constant gradient) on
the speed-time graph.
v
10 – 0
a=
=
= 1.25 m s–2
t 8.0 – 0
Section B: Runner has constant speed of 10 m s−1 for
7 seconds, indicated by horizontal straight line on the speedtime graph.
gravity
or
support
thrust
Section C: Runner has constant deceleration for 5 seconds,
indicated by straight line downwards (has constant gradient)
on the speed-time graph.
10
v 0 – 10 m s–1
deceleration =
=
=
= –2.0 m s–2
5
t
20 – 15 s
Section D: Runner has constant speed of 0 m s−1 for 5 seconds,
so is stopped / stationary, indicated by the speed-time graph
running along the x-axis (i.e. horizontal straight line at
0 m s−1).
b. Thrust > Air friction
gravity
AMEy11 Sci book.indb 99
comparative
size of arrows
irrelevant
See answer to c. for explanation to Not possible to say.
c. The net force is the combination of all the forces acting on the
runner and determines the motion of the runner.
When the forces are balanced (equal and opposite), they
combine to give a net force of zero and the motion is
constant. This occurs in Section B, where the forces of thrust
and air friction are equal and opposite so the runner is at a
constant speed (10 m s−1).
When the forces are unbalanced, the speed will change.
air friction
thrust > air friction
Runner is accelerating.
air friction
gravity
support
thrust
air friction
• In Section A, the forces are opposite but force of thrust
is greater than force of air friction, therefore the runner
is accelerating (speeding up). Net force is in forward
direction.
14/05/14 11:16 am
100 Answers and explanations
• In Section C, for deceleration to happen, there must be
a net force in the opposite direction to the motion. A
decelerating runner is using their legs to slow their motion,
so the thrust force is in the opposite direction to the
motion. Air friction is still in the opposite direction to the
motion, and so both forces combine to give a net force
that is against the motion.
It is not air friction that is mainly slowing the person, but
the person’s leg muscles. The question uses a person/
runner not an inanimate object (e.g. a skateboard), hence
other things come into play in terms of slowing down.
As the forces of thrust and air friction are opposite, there is no
change in the runner’s direction of movement. As the runner
is running horizontally, the forces of gravity and support are
equal and opposite so their net force is zero therefore they do
not affect the runner’s motion.
ii.
= (9 × 2) + (8 × 9) + (1 × 2) + (9 × 2)
= 9 + 72 + 1 + 18
= 100 m
Area under Tama’s graph
= (10 × 5) + (10 × 5) + (2.5 × 2) + (10 × 2)
= 25 + 50 + 2.5 + 20
= 97.5 m
As the length of the race was 100 m, it was Sam who finished
the race at 12 seconds.
Section A: Distance =
Answers
Section D: Distance is 0 m, as runner has stopped/is stationary.
Therefore, total distance travelled is:
40 + 70 + 25 + 0 = 135 m
of acceleration in b. and describes Sam’s and Tama’s speed and acceleration and makes
a correct comparison of speed or acceleration in c. and attempts calculation of distance
using graph in d.; M – as for A but explains why Sam has greater acceleration in a. and
attempts work calculation in b. but contains error (may be carried over error) and describes
(and since t = 5 s)
P=
W
t
P=
150
5
∆v = a∆t = 0.2 × 5 s = 1.0 m s−1
Therefore, at end of section Y, the boy is running at 3 m s–1
He is running at 2 m s–1 during section X, then
accelerates during section Y at 0.2 m s–2. As section
Y lasts 5 s, his speed increases by 1 m s–1 so his
speed at the end is 3 m s–1
M – as for A and correctly calculates motion for each section in a. and correctly describes
relation of the forces in all sections of b. and explains link between net force and motion in
E – a., b. and c. all correct (no carried-over errors) and in c. correctly and comprehensively
explains and compares the relationship between net force and motion, including direction
of motion)
p. 6
a. The slope/gradient of the speed-time graph gives
acceleration. The gradient of Sam’s graph is higher/greater,
so he has the greater acceleration in the first two seconds.
v
9
= = 4.5 m s–2
2
t
W = Fd = 270 × 9 = 2 430 J
c.i.
Sam’s speed and acceleration: Sam accelerates at
4.5 m s–2 (from b.) for the first two seconds and reaches
a speed of 9 m s–1. He runs at a constant speed of 9 m s–1
for the next 8 seconds until 10 seconds have elapsed.
Tama’s speed and acceleration: Tama accelerates at
2 m s–2 ( ) for the first 5 seconds and reaches a speed
of 10 m s–1. He runs at a constant speed of 10 m s–1 for
the next 5 seconds until 10 seconds have elapsed.
Comparison of Sam and Tama: Sam accelerates faster
than Tama (4.5 m s–2 compared with 2 m s–2) but does
not accelerate for as long as Tama (2 s compared with
5 s). Sam reaches and runs at a lower constant speed
(9 m s–1 compared with 10 m s–1) during the first 10
seconds.
10
5
v
t
it to motion in c. and correctly calculates distance travelled in two sections in d.;
c. and correctly calculates total distance travelled in d. but may have a carried-over error;
= 30 W
a=
labels arrows for the forces in all sections of b. and describes the net force and correctly links
F = ma = 60 × 4.5 = 270 N
p. 8
c.i.
(A – correctly describes motion of runner in all four sections in a. and correctly draws and
2
b. W = Fd = 12 × 12.5 = 150 J
1
2
a. Fnet = ma = 60 × 0.2 = 12 N
1
2 × base × height gives area of the triangle in Section C.
b. a =
1
Question Three: Running
1
1
(20 – 15) × 10 = (5 × 10) = 25 m
2
2
Question Two: 100 metre race
2
throughout with needed comparisons correctly made)
Section B: Distance = (15 – 8) × 10 = 7 × 10 = 70 m
Section C: Distance =
1
2
correct calculation of distance for both runners in d.; E – as for M but all calculations correct
1
1
(8 × 10) = × 80 = 40 m
2
2
base × height gives the area of the rectangle in Section B.
1
Sam’s and Tama’s speed and acceleration with correct comparison in c. and has essentially
1
2 × base × height gives area of the triangle in Section A.
Area under Sam’s graph
(A – correct identification of Sam having greater acceleration in a. and correct calculation
d. Distance travelled by the runner is equal to the area under
the speed-time graph.
Area under a speed-time graph equates to distance.
ii.
Speed
(m s –1 )
3
2
1
15
20 Time (s)
(A – correctly calculates F in a. and calculates W correctly in b. and gives correct speed in
c. i. or graph drawn has correct shape in c. ii.; M – as for A and calculates P correctly but
with no/incorrect unit and/or carried over error from incorrect calculation of W in b. and
correctly calculates speed in c. but errors omissions in graph; E – all calculations correct in all
sections and graph complete and correct in c.)
Question Four: Parachuting
a. vv ==
p. 9
∆d
= 2 400 = 40 m s−1
60
∆t
 ESA Publications (NZ) Ltd, Freephone 0800-372 266
AMEy11 Sci book.indb 100
14/05/14 11:16 am