Download SPH3U Exam Solutions Lisa Di Lorenzo - ped4126-2010

Page 1 of 16
OTTAWA VALLEY COLLEGIATE
SPH3U PHYSICS FINAL EXAMINATION
Instructor: Lisa Di Lorenzo
2.0 h (1:30 pm to 3:30 pm)
Date: _________________
NAME:__________________________________
General Instructions – Rulers, pens, pencils, protractors, non-programmable calculators and one
8 ½ x 11 crib sheet (supplied) are the only aids permitted. Remember to watch your significant
digits and accuracy at ALL times! Be sure to show your work when required. Also remember,
IF I CAN’T READ IT, I CAN’T MARK IT!
A. Multiple Choice Questions
Question #
Marks allotted to
Knowledge/
Understanding
Marks allotted to
Thinking/
Investigation
Marks allotted to
Communication
1
1
2
1
3
1
4
1
5
1
6
B. Extended Answer
Questions
7
8
9
10
Total
Marks
4
5
5
5
5
26
5
2
2 marks overall will be allotted to communication: this includes
significant digits, accuracy, clarity, and legibility.
TOTAL:
/32
If you have used additional pages to record your responses, please indicate
here the number of additional pages submitted: ______________________
PART A: MULTIPLE CHOICE. Circle the most appropriate answer. When
finished, record your answers in the answer grid below. (1 mark each)
Using CAPITAL letters only, record your multiple choice answers here.
1
2
3
4
5
Page 2 of 16
/1
KU
1.
Which of the following is an example of non-uniform linear motion?
a)
b)
c)
d)
e)
A skydiver at terminal velocity.
A ball rolls along a table without changing velocity
A jogger runs 50 m along a straight track at a constant speed.
An elevator moves vertically upward at zero acceleration.
All of the above are examples of uniform motion.
Solution: e) They are all examples of uniform linear motion. Uniform motion is that
with zero acceleration. Otherwise, motion is non-uniform; and in this case it will either
be motion with constant acceleration or motion with variable acceleration. Expectation
B3.2 sees students with the understanding to distinguish between and provide examples
of scalar and vector quantities as they relate to uniform and non-uniform linear motion.
Students will have to know that terminal velocity is when the (net force is zero and the)
acceleration is zero. They will have had exposure to this through having met expectation
B2.4.
/1
TI
2.
Stephanie exerts a constant force of 1.5 N to pull a 2.0 kg object at
constant velocity along a level surface on the moon (g = 1.6 N/kg [down]).
The coefficient of kinetic friction for this situation is:
a)
b)
c)
d)
e)
0.077
0.47
0.75
1.3
2.1
Solution: b) If Stephanie exerts a constant force of 1.5 N just to keep the object moving
at a constant speed (without the result of it speeding up or slowing down), then we can
assume that it is only the force of friction that she is ‘working against’. We can assume
that 1.5 N is equal to the constant force of friction:
Ff = 1.5 N
Students should know the relationship between the force of friction, Ff , the coefficient of
kinetic friction, k , and the normal force, Fn :
Ff = k Fn
and they should know to arrive at the normal force by: |Fn | = |FG|.
FG = maG = 2.0 kg (1.6 N/kg) [down] = 3.2 N [down]
Fn = 3.2 N [up]
Page 3 of 16
Therefore,
Ff = k Fn
1.5 N = k (3.2 N)
k = 0.46875 = 0.47
Answer h) is what one would get if they took mass as weight and neglected to multiply
the mass of the object by the acceleration due to the gravity of the moon to get the force
of gravity. It would hopefully be clearly wrong to neglect to do this. Answer d) is the
mass of the object divided by the acceleration due to gravity, m/a, and this should clearly
be wrong!
Answer j) would be what students would get if they did everything correctly except they
recalled the relationship incorrectly as Fn = k Ff so it is meant as a possible distractor.
Answer a) is what they would get if they accidentally used the g force of Earth, 9.8 m/s2,
but did everything else correctly! (Another distractor response).
This question fits in with the expectations C2.4 for students to be able to analyse the
relationships between acceleration and applied forces such as the force of gravity, normal
force, force of friction, and the coefficient of kinetic friction.
I am marking it as a thinking/inquiry question as it involves some thought to arrive at the
solution from the known relationships – students must first ask themselves what
relationships are relevant to the answering the question, and they must reason that the
force that Stephanie is pushing the object with is equal in magnitude to the force of
friction in order for the velocity to remain constant.
/1
KU
3.
Thermal Radiation is the method of heat transfer that:
f)
g)
h)
i)
j)
travels at the speed of light.
requires that molecules circulate in ‘cells’.
can only occur within physical forms of matter.
is radioactive.
is only emitted from very hot bodies, such as the Sun.
Solution: The correct answer is a). Strand D is all about ‘Energy & Society’. I
particularly designed this question as I feel that ‘radiation’ is given a misnomer in general
society as exclusively referring to ionizing radiation, and I expect a Grade 11 physics
student to be educated on radiation referring to all wavelengths of EMR. The curriculum
expectations also do, as indicated in Expectation D3.8, that students be able to distinguish
between and provide examples of conduction, convection, and radiation. Moreover, the
general expectations D2 and D3 see students investigating energy transformations and
demonstrating an understanding of work, various storage forms of energy, including
Page 4 of 16
thermal energy, and the transfer of heat energy. Within this context, I expect students to
know that all forms of electromagnetic radiation travel at the speed of light. In my
teaching of this material, I plan to emphasize energy stores vs. energy carriers, and I will
expect students to know that thermal radiation travels at the speed of light, just like other
forms of EMR. This serves Expectation E3.1, that students should be able to describe
and explain a variety of energy transfers and transformations.
It would hopefully be clear that c) is wrong as this is referring to conduction. Likewise,
d) is clearly wrong as this refers to the common misnomer of radiation being used to
soley refer to ionizing radiation. Students will have learned about radioactivity and
ionizing ability in D3.10 and D3.11.
Choice b) refers to convection and the word ‘cells’ will likely make this an obvious
incorrect response to students, and students should know that radiation does not require
the specifically circular movement of molecules, although it does require the movement
of charged particles within the molecules to generate EMR, so this may be a distractor for
some students. Choice e) might throw some students off, but they should know that all
objects (with a temperature above absolute zero) emit thermal radiation.
/1
KU
4.
When a crest traveling in one direction on a rope meets a crest traveling in
the opposite direction, the result is:
k)
l)
m)
n)
o)
a standing wave.
a loop.
constructive interference.
destructive interference.
both a) and c).
Solution: The correct answer is c) constructive interference. The principle of
superposition states that when two waves intersect, the resulting displacement is the sum
of the individual displacements of the wave. Answers b) and d) hopefully would appear
clearly wrong to students, as they would recognize b) as outside the experience of
theoretical and experiential phenomena that they are familiar with, and d) destructive
interference applies to the superposition of a crest and a trough, rather than two crests as
we have in this question. Answers a) and d) are meant to be distractors. Initially, a
student must know what a standing wave is. They will likely recognize c) as a correct
response, but must further understand that a standing wave can arise as the result of
interference between two waves traveling in opposite directions.
Expectation E3.4 targets the properties of standing waves and the conditions for them to
occur, and E2.1 realizes understanding of superposition, and constructive and destructive
interference. Students will specifically have had exposure to understanding the principle
of superposition with respect to standing waves from expectation E3.3. Knowledge and
Understanding are required to correctly answer this question.
Page 5 of 16
/1
KU
5.
A transformer with 500 turns in the primary coil and 60 turns in the
secondary coil is connected to a 120 V A.C. electric potential source.
What is the electric potential of the secondary coil, and what kind of
transformer is this?
a)
b)
c)
d)
e)
14.4 V; step-up
14.4 V; step-down
14.4 kV; step-up
14.4 kV; step-down
120 V; step-down
Solution: The correct answer is b). Expectation F2.6 sees students solving problems
involving energy, power, potential difference, current, and the number of turns in the
primary and secondary coils of a transformer. This is knowledge and understanding that
students demonstrate by recalling the relationship:
Vs / Vp = Ns / Np
where Vp and Vs are the potential differences in volts in the primary and secondary coils,
respectively, and Np and Ns are the number of turns in the primary and secondary coils,
respectively.
Therefore,
Vs / Vp = Ns / Np
Vs / Vp = 60 / 500
At this point, students must know to assess that the 120 V electric potential source
denotes a potential difference across the primary coil, whereby:
Vs / 120 V = 60 / 500
Vs = 14.4 V
This is a step-down transformer, of the type that would be used in the home for doorbells
or small electric motors, such as in a toy or humidifier or an adaptor for electronic
equipment.
Lots of distractors here, particularly a), as it is the correct potential difference, but
erroneously indicates a ‘step-up’ transformer. Also, e) correctly indicates a step-down
transformer, but might make the student pause to re-check their understanding of the
primary and secondary potential differences. Hopefully, the answers on the order of kV
are clearly wrong to the student from the math.
Page 6 of 16
PART B: EXTENDED ANSWER QUESTIONS.
Answer the following questions in the space provided. Be sure to show ALL your
work. Watch your accuracy and precision.
Note that all of the marks in the extended answer questions fall under thinking and
investigation; while knowledge and understanding are required, it is primarily their
problem-solving skills that students are demonstrating in answering these questions.
6.
/5
TI
A 2.0 kg mass is moving north at 32 m/s when a force of 16.0 N [W]
acts on the mass for 4.0 seconds. What is the final velocity of the mass?
Include a free-body diagram.
Solution:
m = 2.0 kg
v1 = 32 m/s [N]
t = 4.0 s
Fa = 16.0 N
In order to calculate v2 we need a 3rd item  “a”
Fnet = ma
16.0 N [W] = (2.0 kg) a
a = 8.0 ms-2 [W]
1 mk
N
1/2 mk
E
a = Fnet /m
1 mk
v2 = v1 + at
= 32 m/s [N] + 8.0 ms-2 [W] (4.0 s)
= 32 m/s [N] + 32 m/s [W]
1 mk
Draw a vector diagram in order to help in proceeding to calculate the magnitude and
direction of v2:
32 m/s
vf
1/2 mk
32 m/s

vf2 = (32 m/s)2 +(32 m/s)2
vf = 45.25… m/s
tan  = 32m/s/ 32m/s
 = 45
1 mk
1 mk
 v2 = 45 m/s [NW]
Page 7 of 16
Justification of Question and Marking:
This is a multi-strand question that requires students to apply some of their learning
from both Kinematics and Forces strands. Initially, they must recognize the need for
Newton’s 1st Law to find the acceleration due to applied force, as per expectation C3.3
that they know and have some understanding of when to apply Newton’s laws. After this
the question is solved in terms of kinematic equations learned in Strand B, specifically
addressing the expectations of B2.7 to solve problems involving uniform linear motion in
two dimensions. From B2.6, it is evident that students have some knowledge of vector
diagrams and components and using them to solve problems in uniform acceleration
equations.
I allotted half a mark for the initial free body diagram and defining the coordinate axes
directions. One mark is then for knowing to find and arriving at a value for the
acceleration. One mark is for knowing to evaluate final velocity in terms of v 1 , a, and t,
and writing the correct formula for this (if the formula is not explicitly written, but it is
obvious from steps taken that students had to have used this formula then that is
acceptable). One mark for arriving at the values for the velocity vector components. One
mark for determining the magnitude of the final velocity from information shown on the
vector diagram; also a half mark given for vector diagram properly set up – if vector
diagram isn’t included, then this mark goes towards the calculations leading towards the
final response for the velocity vector magnitude and direction. One mark for finding and
for properly stating the compass orientation [NW] of the direction of the final velocity.
7.
A peach that weighs 2.0 N is accelerated by a net force of 8.0 N [E]. the
force of friction on the peach is 3.0 N [W].
/2.5
a)
Find the force that a squirrel applies to the peach to overcome
friction and accelerate the peach. Include a free body diagram with
relevant information labeled.
/1
b)
What is the mass of the peach?
/1.5
c)
A groundhog sees the squirrel pushing the peach from a distance of
10.0 m away, and begins to run towards the peach at a constant
speed of 5.5 m/s. Assuming the squirrel continues to push the
peach with a net force of 8.0 N [E], and assuming that the
groundhog is initially due west of the squirrel and peach, will the
groundhog reach the squirrel and the peach? Answer yes or now,
but back up your answer with calculations. You can also assume
that the squirrel was traveling at 4.0 m/s [E] at the moment the
groundhog began to run towards him.
/1
d)
Is this situation realistic? Why?
Page 8 of 16
Solution:
Given:
Fnet = 8.0 N [E]
Ff = 3.0 N [W]
Required to find F, the force that the squirrel applies.
W
Ff = 3.0 N
-
E
1/2 mk
Fnet = 8.0 N

F=?
1 mk
F = Fnet + (-Ff )
= 8.0 N [E] – (3.0 N [W] )
= 8.0 N [E] + 3.0 N [E]
= 11 N [E]
1 mk
The squirrel applies a force of 11 N [E] to overcome a friction force of 3.0 N [W] and
give a net force of 8.0 N [E] to the peach.
b)
Fg = mg
2.0 N = m (9.81 ms-2)
m = 0.20387… kg
m = 200 grams
1 mk
The mass of the peach is approximately 200 grams
c)
1 mk
Given: v1 squirrel +peach = 4.0 m/s
d initial groundhog to peach = 10.0 m
vgroudnhog = 5.5 m/s
v=d/t
d/v=t
10 m / 5.5 m/s = 1.82 s
The groundhog will reach the spot where the peach initially is in 1.82 seconds.
However! The squirrel is pushing the peach at a rate of 4.0 m/s and then this
speed is increasing by a = F/m = 8.0 N/0.20 kg = 0.40 m/s2. Therefore in 1.82 seconds,
when the groundhog reaches this spot, the squirrel will have already pushed the peach a
further distance away:
d
= v1 t + ½ a (t2)
= (4.0 m/s) (1.82 s) + ½ (40 m/s2) (1.82 s)2
+
Page 9 of 16
= 7.28 m + 66.25… m
= 74 m
1/2 mk
Assuming the squirrel maintains a constant applied force of 11 N to result in the
peach continuing to accelerate at 8.0 N, it is not likely that the groundhog will catch up to
the squirrel – his peach is safe! By the time the groundhog runs towards the location that
the peach initially was, the squirrel has already pushed the peach a further 74 m away.
This is not realistic because in reality, the squirrel would not be able to continue
increasing its velocity indefinitely with an acceleration of 40 m/s2, but there’d be some
upper limit as the squirrel’s little squirrely muscles fatigue.
1 mk
Justification of marking and question:
I intentionally included the weight of the peach as the first bit of information as
extraneous information for the purposes of solving part a) of this question. Students must
be aware that the weight is irrelevant in finding the force that squirrel applies.
Successfully solving this problem relies on students correctly understanding positive and
negative signs and having set up an initial coordinate axes related to east and west. ½
mark is allotted for the initial free body diagram correctly labeled. 1 mark for the correct
initial statement of the relationship between F, Fnet and Ff. 1 mark for correctly fitting
the values of Fnet and Ff into this equation with proper signage and for knowing to reverse
the sign of Ff due to the convention for west being negative set up by the coordinate axes,
(or if they have otherwise defined their axes as west being positive, marks allotted
accordingly), which brings them to the correct solution for Fnet . 1 mark for finding the
mass of the peach. ½ mark for the position that the groundhog will not be able to catch
up to the squirrel and the peach and 1 mark for calculations supporting this argument. 1
mark for explaining how realistically, the squirrel would likely fatigue from exerting this
constant force, and his and the peach’s velocity would not rise without limit.
This question asks students to use reasoning that meets expectations C2 to investigate in
qualitative and quantitative terms, net force, acceleration, and mass, and solve related
problems. They are doing so using free-body diagrams and algebraic equations as per
expectation C2.4. They must also recall and build on knowledge from Strand B.
Kinematics in answering part c), using relations between distance, time, velocity, and
acceleration.
8.
A 300 g piece of iron (c = 4.5 x 102 J/kg °C) at 450°C is submerged in
200 g of water (c = 4.18 x 103 J/kg °C) at 10 °C to be cooled.
/3
a)
Determine the final temperature of the iron & water. (Assume 2
significant digits).
/1
b)
It is suggested that if just a small amount of steam is produced
Page 10 of 16
from the immersion of hot iron into the water, then the water
vapour in the air will pose problems for some of the delicate
machinery in the room. What temperature would the iron have to
be initially for this to begin to be a concern, that is to just get the
liquid water up to 100°C, but not to cause a change of state?
/2
c)
It is found that for liquid water at 100°C, the additional immersion
of an unknown given mass of iron at 450°C (cooled by the
immersion to 100°C) results in 2.0 g of water undergoing a change
of state to steam. Given that the specific latent heat of
vaporization of water is 2255 kJ/kg, find the mass of the iron that
was the source of this heat transfer that resulted in 2 g of water
being vaporized.
Solution:
Iron:
mi = 0.300 kg
ci = 450 J/ kg °C
Ti = 450 °C
T2i = Tf = ?
Water: mw = 0.200 kg
cw = 4180 J/ kg °C
Ti = 450 °C
T2w = Tf = ?
1 mk
a)
Qiron + Qwater = 0
mici∆Ti + mwcw∆Tw = 0
(0.300 kg)(450 J/ kg °C)(Tf - 450°C ) + (0.200 kg)(4180 J/ kg °C)(Tf - 10°C) = 0
(135 J/°C) Tf – 60750 J + (836 J/°C) Tf - 8360 J = 0
(971 J/°C) Tf = 69110 J
Tf = 71.17..
1/2 mk
Tf = 71 °C
1/2 mk
b)
It is necessary to recognize that some amount of heat is required to raise the
temperature of the water to 100°C, and then some heat is also required to change
its state from liquid to water vapour. In this part of the questions, students will
find the amount needed to bring the temperature of the liquid water up to 100°C.
Qiron + Qwater = 0
mici∆Ti + mwcw∆Tw = 0
(0.300 kg)(450 J/ kg °C)(100°C – Tiron) + (0.200 kg)(4180 J/ kg °C)(100°C - 10°C) = 0
13500 J/°C – 135 J/°C Tiron + 75240 J = 0
657.3… °C = Tiron
1 mk
Tiron = 660°C
1 mk
Page 11 of 16
c)
Using Q = mlv, students can now find how much heat is necessary to vaporize
0.0020 kg of water. We define Qvap as the heat gain to the water to affect this
change of state.
Q = mlv
Qvap = (0.0020 kg)(2255 kJ/kg)
= 4.51 kJ
1 mk
To find the mass of iron that would result in this vaporization, we define Qi vap as
the heat transferred to the iron as undergoes heat transfer with the water (and results in 2
grams of vaporization). Note that this heat transfer will be negative as the iron is actually
losing thermal energy to the water. Also note that:
Qvap = - Qi vap
1/2 mk
1/2 mk
- Qiron vap = - ( mi vap ci vap ∆Ti vap )
- 4 510 000 J = - (mi vap (450 J/kg °C) (100°C – 450°C) )
28.63 kg = mi vap
Therefore, 28 kg of iron at 450°C would be responsible for vaporizing 2 g of water
(initially at 100°C).
Justification of marking and question:
Students must apply the law of conservation of energy (Expectation D3.1) to arrive at a
starting point, viewing the iron & water as a closed system [1 mark for Qiron + Qwater = 0].
They must use their knowledge of heat and energy equivalence and the relationship for
heat transfer Q = mc∆T to move forward in solving the problem. Also, an understanding
that the final temperature of the iron will be equal to that of the water is implied in the
question, but students must be aware of this implication. Therefore, ½ mark is given for
setting up the equation with conservation of energy in terms of m, c, and ∆T and then 1
mark for correctly substituting in values to this equation. ½ mark for arriving at the
correct final answer for the initial temperature of the iron.
In part b) students are applying the same understanding as in part a), expect now they are
working backwards to find the initial temperature of the iron in this scenario. 1 mark is
allotted for correct application of theory, calculations, and final answer.
In part c) students are now demonstrating expectation D2.10, in solving problems
involving changes in temperature and changes of state. They are assumed to be
comfortable working with problems involving latent heat, as per expectation D2.1 and
D3.3. Correctly using the relation involving specific latent heat to find the amount of
heat transferred resulting in the vaporization of 2.0 g of water is allotted 1 mark. ½ mark
is allotted for recognizing that the heat gain to the water is equal to the heat loss of the
iron, and then ½ mark for correctly making their calculations based on this and arriving at
the final answer for the mass of iron.
Page 12 of 16
9.
A 440 Hz tuning fork is sounded together with the A string on a guitar,
and 21 beats are heard in 3.0 seconds.
/3
a)
What are the possible frequencies of the guitar string?
(/2 KU, /1 TI)
/2
b)
When an elastic band is wrapped tightly around one prong of the
tuning fork (which lowers the pitch), a new beat frequency of
2.0 Hz is heard. What is the frequency of the guitar string?
Explain. (2 KU)
Solution:
a) Knowns: ƒo = 440 Hz
Nb = 21
tb = 3.0 s
ƒb = Nb / tb
= 21/3.0 s
= 7 Hz
1 mk
1 mk
Since ƒb , beat frequency = |ƒ2 – ƒ1 |, there are two possible frequencies
of the guitar string:
ƒ2 – ƒ1 = 7 Hz
ƒ2 = ƒ1 + 7 Hz
= 440 Hz + 7 Hz
= 447 Hz
or
ƒg1 = 433 Hz;
ƒ2 – ƒ1 = -7 Hz
ƒ2 = ƒ1 - 7 Hz
=440 Hz–7 Hz
= 433 Hz
ƒg2 = 447 Hz
1 mk
b) If an elastic lowers the pitch and a smaller beat frequency is heard, this
means that the guitar string must be the lower of the two possible
1 mk frequencies (the closer the frequencies, the smaller the beat
frequency.)
ƒguitar = 433 Hz 1 mk
Page 13 of 16
Justification of question and marking:
This question addresses overall expectations E2. and E3. related to the properties of
mechanical waves and sound and the demonstration of understanding of the principles
underlying their production, transmission, interaction, and reception. Specifically, I am
looking for students to first know and understand beat frequency as the number of beats
per unit time, but then also to demonstrate understanding that beat frequency is equal to
the difference in frequency between two sources [1 mark for finding beat frequency as
number of beats per time; 1 mark for beat frequency = |ƒb – ƒo|; students can demonstrate
this knowledge by explicitly writing out this equation or by arriving at the answer for the
possible guitar string frequencies based on this knowledge]. Qualitatively, students will
have to have at least some basis of an understanding that the tuning fork and the A-string
of the guitar both have their own frequencies, but beats will be heard at moments in time
where their sound waves are constructively interfering, and that the beat is the
superposition of both of these waves at their loudest. [1 mark given for stating the two
possibilities for the guitar string frequency – this was the only mark allotted to thinking
and inquiry on the basis of the absolute value reasoning that they did to arrive at this final
answer].
In correctly answering part b), students understand that pitch is determined by sound
wave frequency, and that notes with lower-pitch have lower frequencies. When the
frequency of the tuning fork becomes lower than 440 Hz, the student must frame the
question: is a tuning fork with frequency lower than 440 Hz now closer or farther from
the one true frequency of the guitar A-string? If the guitar A-string is 433 Hz, then the
frequencies of the guitar string and tuning fork are now closer; if the guitar A-string is
447 Hz, then the frequencies are now farther apart. The student must then be able to use
an understanding that as two frequencies become closer together, the beat frequency
becomes smaller. [n part b), 1 mark is given for knowing that closer the frequencies,
smaller the beat frequency, and 1 mark is given for the final answer for the guitar
frequency].
10.
/2.5
Electricity & Magnetism: DO ONLY TWO out of the following three
questions. If you do more than two questions, I will only mark the
first two: a and b, unless you CLEARLY indicate the two out of three
you’d like me to grade.
a)
Explain how Lenz’s law for induced currents can be related to
conservation of energy. (CHOOSE ONLY TWO OUT OF THE
THREE QUESTIONS IN Question #10 to ANSWER)
Solution:
a)
Lenz’s law is an extension of the law of conservation of energy to nonconservative forces in electromagnetic induction. The law of conservation of energy is
exclusively for conservative forces. Lenz’s law extends the principles of energy
Page 14 of 16
conservation to situations such as the following: when one moves a magnet towards the
face of a closed loop of wire (eg. a coil or solenoid) , an electric current is induced in the
wire, because the electrons in it are subject to an increasing magnetic field as the magnet
approaches. This produces an EMF that acts upon them. The direction of the induced
current depends on whether the north of south pole of the magnet is approaching. It can
be used to give the direction of the induced electromotive force (EMF) and current
resulting in electromagnetic induction.
In short, Lenz’s law states that the magnetic field of an induced current always opposes
the change in magnetic field that is causing the induced current. The induced current
must have received energy in order to begin to flow. This iinduced current can then
transfer energy into forms such as light or heat. But energy cannot be created from
nothing. The energy is provided by the action causing the induction. The energy that
was expended in pushing the magnet, for example, that was responsible for the change in
magnetic field in the first place.
Lenz’s Law is part of expectations F2.7, F3.4 and F3.5. F2.7 specifically involves an
investigation into Lenz’s law and the law of conservation of energy. This question
targets the understanding from these expectations, tying in nicely with the theme of the
law of conservation of energy from the previous Strand.
/2.5
b)
Explain the operation of an electric motor OR a generator,
including the roles of their respective components. (CHOOSE
ONLY TWO OUT OF THE THREE QUESTIONS IN Question
#10 to ANSWER)
Solution:
b)
This question is actually verbatim expectation F3.6.
- An electric motor operates by a current-carrying wire within a magnetic field resulting
in a force on the wire which causes the physical movement of the wire. Thus electrical
energy is converted to mechanical energy. An electric motor consists of:
1. a current-carrying conductor called an armature or rotor – in the case of the simplest
motor, this will be a square-shaped flat coil (typically wire wrapped around a permeable
ferromagnetic core). The coil is free to rotate in a magnetic field.
2. external field magnets are present to create this magnetic field.
3. a split-ring commutator allows current to flow in and out of the coil even while the
coil is rotating. In a DC motor, the commutator usually has brushes that are at the ends
of electric leads; they come into contact with the commutator, yet allow it to turn freely
without the leads getting twisted as the motor runs. In an AC motor, slip rings usually
allow current to flow from the rotating armature to an external circuit.
Page 15 of 16
- A generator or dynamo is a device used to produce electric current from mechanical
(kinetic) energy. It works on the principle of electromagnetic induction.
1. In the simplest AC generator, an alternating electromotive force is induced in the
armature (consisting of a coil of wire wrapped around a permeable core) as it rotates
between the poles of a..
2. field magnet which produces the necessary magnetic field. As the armature rotates,
the magnetic field experienced by the electrons in the wire is constantly changing, and so
this changing magnetic field induces a current in the wire.
3. A generator for DC has a commutator, as on an electric motor, which helps to ensure
that the current always flows in the same direction; while an AC generator has slip rings
that likewise allow current to flow from the rotating armature to an external circuit.
½ mark each for each of the three components, and 1 mark overall for describing the
operation, with respect to the individual components.
/2.5
c)
Distinguish between alternating current (AC) and direct current
(DC), and explain why alternating current is presently used in the
transmission of electrical energy. (CHOOSE ONLY TWO OUT
OF THE THREE QUESTIONS IN Question #10 to ANSWER)
Solution:
c)
This question is actually verbatim expectation F3.7.
A periodic reversal of the direction of current is an alternating current. Direct current
always travels in the same direction and does not reverse. In North America, the current
from generating plants alternates with a frequency of 60 hz.
When Thomas Edison and his associates set up the first DC power station in NYC in
1882, they had not way of stepping the voltage up or down. Long-distance transmission
would have required a copper wire as thick as one’s arm to avoid large power loses and
voltage drops along the way. Edison’s solution was to install a power station every
couple of kilometers. Nikola Tesla arrived in NYC in 1884 with four cents in his pocket
and a design for AC motors and generators. He eventually was able to compete with DC
power stations, as AC is easier to distribute than DC, because AC can be raised or
lowered in voltage steps by a transformer. This allows very high voltages to be used in
long distance travel of electricity, which is then stepped down several times before being
supplied to the end user. As the voltage is increased in a circuit, the current decreases for
a given load. This allows the use of lighter wire to transmit large amounts of electricity.
Page 16 of 16
½ mark for difference between AC and DC.
1 mark for discussing the lighter wire for AC and why.
1 mark for discussing the ease of stepping up/down AC voltage vs. DC..