Physics 2021

Any questions
smitha@ignatius.vic.edu.au
To view presentation:
https://www.slideshare.net/secret/cVUaf5V3uyDYsf

The examiners aren’t set out to trick you.
Make sure that you are familiar with the
setting and layout of the 2019 exam. This
will be more useful than 2020.
Also 2021 NHT exam is a good guide and
available now.

No marks are awarded for writing a correct
formula.
You need to show correct substitution into
the formula.
NO formula or NO substitution equals NO
working marks. However, is there is some
attempt at working out and the answer is
correct you will generally get full marks.
No marks have been taken for incorrect
decimal places or significant figures in the
recent past. However, the practical questions
will require correct uncertainty responses.

Some general hints:
Approximately 30% will be written responses.
Pay attention to how many marks are awarded
b.
Use dot points in written response:
ie. 2 clear points two marks: role and operation:
The best responses are concise and address the
question. Try to link your ideas together.
• The role of the commutator is to keep the motor
spinning in the same direction.
• It does this by reversing the direction of current in the
coil every 180o.
• When the orientation of the coil is vertical.

Study score 30
C+ - B 80/130
62%
Study score 40
A+ 114/130
88%

Practice exams
VCAA exams back to about 2000 with
solutions
(The best resource)
iTute
Great resource for past exams detailed
solutions.
Their exams tend to be difficult.

Unit 3: How do fields explain motion and electricity?
Outcome 1
On completion of this unit the student should be able to analyse gravitational, electric and
magnetic fields, and use these to explain the operation of motors and particle accelerators
and the orbits of satellites.
Key knowledge
Fields and interactions
• describe gravitation, magnetism and electricity using a field model.
• investigate and compare theoretically and practically gravitational, magnetic
and electric fields, including directions and shapes of fields, attractive and
repulsive fields, and the existence of dipoles and monopoles

• investigate and compare theoretically and practically gravitational fields and
electrical fields about a point mass or charge (positive or negative) with reference
to:
– the direction of the field
– the shape of the field
– the use of the inverse square law to determine the magnitude of the field
– potential energy changes (qualitative) associated with a point mass or charge
moving in the field

Field Lines never touch and never cross
These are monopole fields
Gravity is missing but it can only be attraction

Attraction Repulsion
Dipole negative
charges will be
similar with
arrows going
towards the
charges.

Note that Magnetism is an example of a
dipole force field.
There are NO monopole magnetic fields.

Question 1
A. B.
C. D.
Must be showing repulsive forces (like charges)
Field lines are away from the sources
Correct C

r
i
t
e
i
n
t
h
i
s
a
r
e
a
Question 2
X Y Z
A.
B.
C.
D.
Question 3
G × 10–11 2 –2
A. –3
B. –5
C.
X can be B
Y can be E or B or G
Z can be E or B only
C only option

You need to realise the assump1on that H has one proton and that
the electron is also an elementary par1cle
For some reason many
students forgot to square
the radius in the
denominator of the
frac1on.

The radius is the orbital radius of the atom as given in part a. This
connec1on needs to be made. Also circular mo1on.
Think about what the electron would
look like! It is travelling about 10,000
1mes the speed of a jet aircraF and is
circling a H atom. It would just be a
complete blur and you would never
really be able to locate its posi1on. This
is what the quantum world states the
electron would look like.

Question 3
G × 10–11 2 –2
A. –3
B. –5
C.
D.
Question 4
X
A.
B.
F =
GMm
r2
F =
6.67 ×10−11
( ) 1
( ) 100
( )
0.1
( )2
F = 6.67 ×10−7
N
C do not use cm as a measurement

Question 5
× 10–8
–9
k 9
A.
B.
C.
D. –5
Question 6
A.
B.
C.
D.
Question 7
F =
kQ1Q2
r2
F =
9 ×109
( ) 1×10−8
( ) 1×10−9
( )
0.3
( )2
F = 1.0 ×10−6
N
B do not use cm as a measurement

Question 1
23
G –11 2 –2
a.
–2
g =
GM
r2
g =
6.67 ×10−11
( ) 6.4 ×1023
( )
3.4 ×106
( )
2
g = 3.69 N / kg

b.
U
EK
energy (J)
height (m)
10
0
0
1
2
3
4
5
6
7
8
9
8
7
6
5
4
3
2
1
This question is different to others in the past:
Energy v Height graphs are usually satellites, so
the H is large and the graph is hyperbolic in shape.
In this case H is small (10m) and the graph will be
approx. linear i.e. E=mgh, m and g are constant.
NOTE: E = mgh = 20 x 3.7 x 10 = 740J

EK
energy (J)
height (m)
10
0
0
1
2
3
4
5
6
7
8
9
8
7
6
5
4
3
2
1
J x100
Gravitational Potential Energy
Kinetic Energy
Energy v Height on mars surface

This type of question students need to:
• adjust the scale of the axis given.
• provide a key for the graphs.
• Give a heading for the graph

• investigate and apply theoretically and practically a vector field model to magnetic
phenomena, including shapes and directions of fields produced by bar magnets, and
by current-carrying wires, loops and solenoids
• identify fields as static or changing, and as uniform or non-uniform.
d
o
n
o
t
w
r
i
t
e
i G × 10–11 2 –2
A. –3
B. –5
C.
D.
Question 4
X
A.
B.
C.
D.
Use Right
Hand
Grip rule. B is
into the coil
at X. Correct
response is A

Effects of fields
• analyse the use of an electric field to accelerate a charge, including:
– electric field and electric force concepts: and
– potential energy changes in a uniform electric eld: W = qV,
– the magnitude of the force on a charged particle due to a uniform electric eld:
F = qE
E = k
Q
r2
F = k
q1 ⋅q2
r2
E =
V
d

• analyse the use of a magnetic field to change the path of a charged particle,
including:
– the magnitude and direction of the force applied to an electron beam by a
magnetic field: F = qvB, in cases where the directions of v and B are perpendicular
or parallel
– the radius of the path followed by a low-velocity electron in a magnetic field:
qvB = m⋅v2
r
• model the acceleration of particles in a particle accelerator (limited to linear
acceleration by a uniform electric field and direction change by a uniform
magnetic field).

In General:
Electrons fired into B Fields will have a
perpendicular force towards the centre of
the path. (circular motion)
Electrons travelling through an E field will
have constant speed in one direction and
perpendicular force in another.
(parabolic / projectile motion)

Question 2
Data
–31
–19
region of uniform
magnetic field
plate plate
10 000 V
0.10 m
Figure 1

region of uniform
magnetic field
plate plate
10 000 V
0.10 m
Figure 1

a.
b.
–1
E =
V
d
E =
10,000
0.1
= 100,000 V / m or N /C
E = 10,000 eV
E = 1.6 ×10−15
J
v =
2E
m
=
3.2 ×10−15
9.1×10−31
= 5.93×107
m / s
E must be in Joules and NOT
eV.

The Magnetic force acting on a wire carrying a current
F = B⋅ I ⋅l
Writing I as charge per second
F = B⋅
Q
t
⋅l
Change position of t as being under length
F = B⋅Q⋅
l
t
Length divided by time is speed
F = B⋅Q⋅v
Force on a moving charge in a B field

t
h
i
s
a
r
e
a
c.
d.
Bqv =
mv2
r
r =
m⋅v
B⋅q
r =
9.1×10−31
( )⋅ 5.93×107
( )
0.02
( )⋅ 1.6 ×10−19
( )
= 0.017 m
The magnetic force on the electrons is equal to the
centrifugal force acting on them.

d.
Figure 2a Figure 2b
In the electric field the deflection will be from negative
towards positive. The shape will be parabolic in nature. a ↓
v →, parabolic
In B field the direction of Force is always perpendicular to
the velocity. The shape will be circular. RH slap rule. Note
Direction is - current.

• For circular motion.
• Force is perpendicular to the direction of motion.
• Force MUST be constant magnitude.

Application of field concepts
• apply the concepts of force due to gravity, Fg, and normal reaction force, FN,
including satellites in orbit where the orbits are assumed to be uniform and
circular
• model satellite motion (artificial, Moon, planet) as uniform circular orbital
motion: a =
v2
r
=
4π2
r
T 2

• analyse the use of gravitational fields to accelerate mass, including:
– gravitational eld and gravitational force concepts: and
– potential energy changes in a uniform gravitational eld: Eg = mgΔh
– the change in gravitational potential energy from area under a force-distance
graph and area under a field- distance graph multiplied by mass.
F = G
m1 ⋅m2
r2
g = G
M
r2

For Gravitational Motion write these formulae on your Cheat
Sheet
DO NOT try to manipulate the equations in the exam.
v =
GM
r
r =
GM ⋅T 2
4π2
3
T =
4π2
⋅r3
GM
Use T is seconds not
minutes, hours or days
Use v in m/s not km/h
Know how your calculator
uses scientific notation

This is oFen confused with Force, the ques1on is asking for Field
Strength. The units are N/kg and on earth it is equivalent to the
accelera1on.

You should have a formula sheet that will give the correct formula.
Trying to derive the correct formulas under the pressure of the exam
is a possible disaster.
Students oFen transpose these equa1ons incorrectly or make
mistakes like forgeKng to cube the radius or such.

The small m (mass of satellite) does not appear in any of the equa1ons for
period, radius or velocity of the satellite around a body. As such satellites
with different mass will travel at exactly the same period and velocity if they
are in the same orbit. Melissa is the only correct student.
For 3 marks you should approach the ques1on as:
Rick is incorrect because…..
Melisa is correct because…
Nam is incorrect because…..
This follows the instruc1on to “Evaluate these three opinions”.

• describe the interaction of two fields, allowing that electric charges, magnetic
poles and current carrying conductors can either attract or repel, whereas masses
only attract each other
• investigate and analyse theoretically and practically the force on a current
carrying conductor due to an external magnetic eld, F = nIlB , where the
directions of I and B are either perpendicular or parallel to each other
• investigate and analyse theoretically and practically the operation of simple DC
motors consisting of one coil, containing a number of loops of wire, which is free
to rotate about an axis in a uniform magnetic eld and including the use of a split
ring commutator

Question 3
10–2
A +
–
F G
E H
N S
6.0 V
Figure 3
a.
Common Question
ZERO
I is parallel to B.

A Commutator:
-Enables the direction of the current in the
loop to be reversed.
-Every half turn or 180o.
-This reverses the direction of the force on
the sides of the motor loop.
*To keep the motor spinning in the same
direction.
S
b.

Area of Study 2
How are fields used to move electrical energy? The production, distribution and use
of electricity has had a major impact on human lifestyles. In this area of study
students use empirical evidence and models of electric, magnetic and
electromagnetic effects to explain how electricity is produced and delivered to
homes. They explore magnetic fields and the transformer as critical to the
performance of electrical distribution systems.

Generation of electricity
• calculate magnetic flux when the magnetic field is perpendicular to the area, and
describe the qualitative effect of differing angles between the area and the field:
ΦB = B⊥ A
• investigate and analyse theoretically and practically the generation of
electromotive force (emf) including
AC voltage and calculations using induced emf: , with reference to:
– rate of change of magnetic flux
– number of loops through which the flux passes
– direction of induced emf in a coil
Ε = −N
ΔΦ
Δt
• explain the production of DC voltage in DC generators and AC voltage in
alternators, including the use of split ring commutators and slip rings
respectively.

This is an easier ques1on than where the frequency is given. The
loop will make 4 area changes in one complete period. In this
ques1on the t will be 1/16 of a second for a quarter turn. This is the
main difficulty with this ques1on.

The area is not required in this ques1on as the magne1c flux is given
and NOT the magne1c field strength. This is an example when
reading the ques1on carefully is important. A li]le bit of a trick
ques1on but this happens all of the 1me.
This is a good example of how the exam ques1ons can just change a
li]le and confuse students from year to year.
Read the ques1ons carefully and take in all the informa1on.

Question 4
N
S
oscilloscope
Figure 4
2
a.

There are tricks to this question:
-The area changes x4 per rotation so while
f=10 Hz
the time for the area will be 1/40 sec
ξ = n⋅
B⋅ΔA
Δt
ξ = 20⋅
0.5⋅0.02
1/ 40
= 8.0 V

SECTION B – conti
b.
time
voltage output
0
The period will get larger and larger
while the amplitude gets smaller and
smaller.

Transmission of electricity
• compare sinusoidal AC voltages produced as a result of the uniform rotation of
loop in a constant magnetic field with reference to frequency, period,
amplitude, peak-to-peak voltage (Vp–p ) and peak-to-peak current (Ip–p )
• compare alternating voltage expressed as the root-mean-square (rms) to a constant
DC voltage developing the same power in a resistive component
• convert between rms, peak and peak-to-peak values of voltage and current

Question 5
Figure 5
a.
Vpeak = 30 V VRMS = 21.2 V

b.
f = 1/T
T= 80 ms = 0.08
sec
f = 12.5 Hz

• analyse transformer action with reference to electromagnetic induction for an
ideal transformer:
• analyse the supply of power by considering transmission losses across
transmission lines
• identify the advantage of the use of AC power as a domestic power supply.
NP
NS
=
VP
VS
=
IS
IP

This is the deriva1ve of the graph shown. There is a nega1ve sign
but in general the nega1ve that represents the concepts of Len’s law
and can be leF out. The only possible answer is D. In the past, when
drawing EMF graphs that have a posi1ve and nega1ve constant value
it has not ma]ered which is posi1ve and nega1ve, only that there is
a dis1nct difference.

Question 10
250 VRMS AC
to town
transformer
primary
coils
secondary
coils
alternator
20 km
power plant
transmission line
50000 VRMS AC
Figure 11
a.

b.
V
Ploss = I2
R
Ploss = 152
× 40 = 25,000 W
VOUT = VIN − VDROP
VOUT = 50,000 − 15 × 40
VOUT = 49,400 V

c.
To transmit Power (P=VI)
If V ↑ , then I ↓ For the same amount of P.
Power loss = I2 R, therefore at HIGH V and
LOW I
In order to change the voltage in the power
lines transformers are needed.
Transformers required a △ Flux to
operate.
This can only be supplied by AC and not
DC.
Voltage is stepped UP at transmission and
stepped DOWN closer to the customer.

Part A is fairly straight forward.

Part B
This is a ra1o problem that strong students could possibly do with no
working out.
-If the voltage is halved, the current is doubled to keep the power
the same.
-As Ploss = I2R, doubling current will increase power loss by x4.
Just by wri1ng x4 with no working or jus1fica1on (1 mark).

. Area of Study 3
How fast can things go?
In this area of study students use Newton’s laws of motion to analyse relative motion,
circular motion and projectile motion. Newton’s laws of motion give important
insights into a range of motion both on Earth and beyond. At very high speeds,
however, these laws are insufficient to model motion and Einstein’s theory of special
relativity provides a better model. Students compare Newton’s and Einstein’s
explanations of motion and evaluate the circumstances in which they can be applied.
They explore the relationships between force, energy and mass.

• investigate and analyse theoretically and practically the uniform circular motion of an
object moving in a horizontal plane: , including:
– a vehicle moving around a circular road
– a vehicle moving around a banked track
– an object on the end of a string
FNET =
m⋅v2
r
• investigate and apply theoretically Newton’s second law to circular motion in a
vertical plane (forces at the highest and lowest positions only)

Question 12
–1
L
Figure 14a
4.00 m

L
Figure 14b
a.
b.
The arrow must be pointing toward the
centre of the circle as it is travelling in a
circle and the FNET is always towards
the centre.

SECTION B – continued
a.
b.
F =
mv2
r
∑
⇒ FN − mg =
mv2
r
⇒ FN = mg +
mv2
r
⇒ FN = 5⋅10 +
5⋅52
4
= 81.3 N

• investigate and analyse theoretically and practically the motion of projectiles near
Earth’s surface, including a qualitative description of the effects of air resistance
No calculations about air resistance:
-Path will no longer be parabolic/
symmetrical.
-Air resistance will reduce the range and
max. height of projection.

Question 13
–1
15 m
20 m s–1
30°
cliff
sea
Figure 15
vertical component only
u = 20sin30 = 10ms−1
v = −
a = −10 ms−2
s = −15 m
t = ?
s = ut +
1
2
at2
−15 = 10t − 5t2
t2
− 2t − 3 = 0
t = 3.0 sec

• investigate and apply theoretically and practically the laws of energy and momentum
conservation in isolated systems in one dimension.
Key knowledge Newton’s laws of motion
• investigate and apply theoretically and practically Newton’s three laws of motion
in situations where two or more coplanar forces act along a straight line and in
two dimensions

Relationships between force, energy and mass
• investigate and analyse theoretically and practically impulse in an isolated system
for collisions between objects moving in a straight line: FΔt = mΔv
• investigate and apply theoretically and practically the concept of work done by a
constant force using:
– work done = constant force × distance moved in direction of net force
– work done = area under force-distance graph

• analyse transformations of energy between kinetic energy, strain potential energy,
gravitational potential energy and energy dissipated to the environment
(considered as a combination of heat, sound and deformation of material):
– kinetic energy at low speeds: ; elastic and inelastic collisions with
reference to conservation of kinetic energy
– strain potential energy: area under force-distance graph including ideal springs
obeying Hooke’s Law:
– gravitational potential energy: Eg = mgΔh or from area under a force-distance
graph and area under a eld- distance graph multiplied by mass
E =
1
2
m⋅v2
E =
1
2
k ⋅ x2

Make sure that you read the ques1on carefully and you are not just
looking at the picture and trying to figure out what the ques1ons
states from that. The “car comes to rest when spring is compressed
to 0.5 m” statement is cri1cal.
Q12. k = gradient of the graph (400 N/m)

Q13. E= ½ k x2 or the area under the graph = 50 J.
It was a common error to give the area under the en1re graph which
is 200 J.

Question 11
–1
–1
A 6.0 kg
Before After
B 2.0 kg
u = 2.0 m s–1 u = 0 m s–1
A 6.0 kg B 2.0 kg
v = 1.0 m s–1 v = ? m s–1
Figure 12
a.
–1
pbefore = pafter
12 = 6 + 2v
v = 3 ms−1

SECTION B – Question 11
b.
KEbefore =
1
2
mv2
=
1
2
6 2
( )2
= 12J
KEafter =
1
2
mv2
+
1
2
mv
=
1
2
6 1
( )2
+
1
2
2 3
( )2
= 3 + 9 = 12 J
Therefore it is Elastic.
In general if object stick together
INELASTIC
You must follow direction to show
working.

w
r
i
t
e
i
n
t
h
i
s
a
r
e
a
c.
t
Figure 13a
t
Figure 13b
Pat is correct.
Some kinetic energy stored as elastic
energy while the spring is deformed.

Question 14
k
70
60
50
40
unstretched
spring
ruler (cm) stretched
spring
masses added
30
20
10
0
Figure 16
Number of 0.50 kg masses Length of spring
0

Number of 0.50 kg masses Length of spring
0
1
2
3
5
The readings are approximately to the
nearest cm, by the 53, 57 and 63 cm
reading. Therefore ± 0.5 cm

a.
Error bars of approx 2.5 cm each way.
You must follow the instructions:

SECTION B – Question 14
TURN OVER
b. part a. k
–1
There would be a range of correct answers, but you
will need to justify your answer.
± 0.5 cm over approx. 50 cm ±1%
0.5 kg masses ± 20g ±4%
Gravity ± 2% Total error 7%
k =
1
Gradient
k = 81 N / m ± 7%

c.
The 40 cm mark is 30cm above the rest position.
When released the masses will go 30cm below the
rest position. This will be 60cm below where they
are released. However the question requires
some working .
The lost GPE when falling is gained as EPE.
mgh =
1
2
k ⋅ x2
2.5 ×10 × h =
1
2
81⋅h2
h = 0.62 m

The mistake in this ques1on is to confuse the point where the spring
comes to permanent rest and the point where it is at rest
momentarily.
For permanent rest
If you understand springs very well, the momentarily at rest point is
double this.
You can also show this through the conserva1on of energy. But you
could also simply state the correct answer, if you have the equa1on
from above correct.

Conserva1on of Energy

For the full 4 marks list the three posi1ons and the total energy.
At top: GPE=max EPE=0 KE=0
At mid: GPE=some EPE=some KE=some
At bo]om: GPE=0 EPE=max KE=0
Total E is always constant

Question 12
A.
B.
C.
D.
Question 13
Transverse wave Longitudinal wave
A.
B.
C.
D.
Question 14
The proper length of an object needs to be measured
at rest to the observer.
B
The same reference frame could be moving at a
constant velocity compared to the object. D is the
distractor.

. Einstein’s theory of special relativity
• describe Einstein’s two postulates for his theory of special relativity that:
– the laws of physics are the same in all inertial (non-accelerated) frames
of reference
– the speed of light has a constant value for all observers regardless of
their motion or the motion of the source
• compare Einstein’s theory of special relativity with the principles of classical
physics
• describe proper time (t0) as the time interval between two events in a
reference frame where the two events occur at the same point in space

• interpret Einstein’s prediction by showing that the total ‘mass-energy’ of an
object is given by:
Etot = Ek + E0 = γ mc2 where E0 = mc2, and where kinetic energy can be
calculated by: Ek = (γ – 1)mc2
• describe how matter is converted to energy by nuclear fusion in the Sun, which
leads to its mass decreasing and the emission of electromagnetic radiation.

• describe proper length (L0) as the length that is measured in the frame of
reference in which objects are at rest
• model mathematically time dilation and length contraction at speeds approaching
c using the equations:
t = t0γ and L =
L0
γ
whereγ = 1−
v2
c2
⎛
⎝
⎜
⎞
⎠
⎟
−1
2

• explain why muons can reach Earth even though their half-lives would suggest that
they should decay in the outer atmosphere.
The muons are moving at great speed compared to
the earth.
The height of the upper atmosphere is smaller from
the muons perspective and the time they take to
reach the ground is longer than for a stationary
observer on earth.

Question 15
c
muons
top of mountain
ground level
2627 m
Figure 17

SECTION B – Question 15
Figure 17
a.
t0 = 2.2 ×10−6
s
v = 0.995 c
t =
t0
1−
v2
c2
=
2.2 ×10−6
1− 0.9952
= 2.2 ×10−5
Don’t use c = 3.0 x 108 .
If measured in term of c they will cancel.
Note: v = 0.995c has factor of x10
v = 0.99 c has factor of x 7
v =v 0.9 c has factor of x 2.3
Time Dilation will be greater

b. c
c.
Length contraction will be smaller
l0 = 2627 m
v = 0.995 c
l = l0 1−
v2
c2
=
2627
10
= 263 m
Note : use previous result

c.
According to classical physics (2627 m @ 3 x 108)
the time will be 8.8 x 10-6 sec. (more than a half
life 2.2 x10-6 sec)
From the muons perspective they are travelling
263 m at 0.995 c.
This will take 8.8 x 10-7 sec. (much less than half
life not so many muons disintegrate)
More strike the surface of earth than predicted by
classical physics.
Make sure that you compare the two.

Unit 4:
How can two contradictory models explain both light
and matter?
Outcome 1
On completion of this unit the student should be able to apply wave
concepts to analyse, interpret and explain the behaviour of light.

Key knowledge
Properties of mechanical waves
• explain a wave as the transmission of energy through a medium without the
net transfer of matter
• distinguish between transverse and longitudinal waves
• identify the amplitude, wavelength, period and frequency of waves
• calculate the wavelength, frequency, period and speed of travel of waves
using: v = f λ = λ/T

Waves transfer energy without transfer of
mass.
Transverse Waves
The particles vibrate perpendicular to the motion.
Think of T for perpendicular to help remember.
Easy to draw
Water waves are an easy example.

Longitudinal Waves
The particles vibrate parallel to the motion.
Difficult to draw
Sound waves
Gases cannot support shear forces. The particles
must
BUMP into each other.

Question 13
Transverse wave Longitudinal wave
A.
B.
C.
D.
Question 14
A.
B.
C.
D.
A is Correct

• investigate and analyse theoretically and practically constructive and destructive
interference from two sources with reference to coherent waves and path
difference: nλ and (n-1/2)λ respectively

n is the number of the antinodal line from the centre of the pattern
P is the point in question
S1 and S2 are the sources of the waves
d(PS1) is the distance from P to S1.
S1 S2
trough
crest
A
A
A
N
N
n = 0 n = 1
n = 1 n = 2
n = 2
central antinodal line
PN
PA
Similaranalysisshowsthat,forapointonanodalline,thedifferenceindistance

Nodal lines are lines where
destructive interference occurs
on a surface, resulting in no
displacement of the surface.
source, and vice versa. Any point on a
minimum, because of the minimum dis
Between the nodal lines are regions w
The centres of these regions are called
antinodal line, a crest from one sourc
from the other source, or a trough from
as a trough from the other source, and s
sometimes called a local maximum, bec
occurs there.
When the two sources are in phase, a
page, the interference pattern produced
line. Note that any point on the central a
each source. Since the sources produce
two sources will arrive at any point on the
Similar analysis will show that, for an
either side of the centre of the pattern,
exactly one wavelength further from o
means that crests from one source stil
although they were not produced at the
Point PA is on the first antinodal line fro
seen that PA is 4.5 wavelengths from S1 a
A way to establish whether a point is a
Antinodal lines are lines where
constructive interference occurs on
a surface.
The path difference The path difference
nλ n −
1
2
⎛
⎝
⎜
⎞
⎠
⎟ λ

Question 18
S1
S2
P
X
laser
double slits screen
central bright band
Figure 22
1 2
a.

1 2
a.
nm
b.
2 1
nm
a.
nm
b.
2 1
nm
X is the second dark fringe and this represents
a path difference of 1.5 waves.
One wave is 500 nm
P is the second bright fringe and this represents
a path difference of 2 waves. i.e. 1000 nm
Both average 1.2 / 2 marks

• explain qualitatively the Doppler effect
If the ambulance is
stationary both observers
hear the same frequency.
The person moving
towards the ambulance
(relatively) hears a higher
frequency. The person
moving away from the
ambulance hears lower
frequency.
The Doppler Effect explains red shift and is evidence of
an expanding universe.

Moving towards Moving Away
f =
f0 ⋅vS
vS − v
( )
f =
f0 ⋅vS
vS + v
( )
f0 is the stationary frequency
vs is the speed of sound ( 340 ms-1)

D.
Question 14
A.
B.
C.
D.
The speed of waves in a particular medium is
constant.
Doppler Effect will explain the change in
wavelength or distance between crests.
D is correct.

• explain resonance as the superposition of a travelling wave and its reflection, and
with reference to a forced oscillation matching the natural frequency of vibration
• analyse the formation of standing waves in strings fixed at one or both ends
Fixed end reflection reverses the orientation of
the wave

Open end reflection does not reverse the
orientation of the wave.
Standing waves are due to superposition
of waves
that have the SAME wavelength.

Resonance occurs when a particular
wavelength matches the item
generating the sound. This is a
musical instrument.

Resonance occurs in many objects.
You may have seen images/film of
Tacoma Narrows bridge. Build in the
1930’s before the engineers knew about
resonance. Particular wind speeds
matched the length of the bridge and
caused the bridge to fail.
Electrons shells or
energy levels are
another example of
resonance.

Superposition of waves and other
examples of wave interference is
evidence for the wavelike nature of
something. In particular light.
Only waves can produce an interference
pattern.
Both Transverse and Longitudinal waves
can produce interference. Longitudinal
are difficult to draw.
Only particular frequencies can be
supported by resonance. The term is the
“Natural frequency of vibration.”

Use the following information to answer Questions 15 and 16.
ends of guitar string
S
Question 15
A.
B.
C.
D.
Question 16
immediately after
Standing waves are formed by coherent waves.
Travelling in opposite directions in a medium.
Due to opposing sources or reflections.
In 2017 exam a simple diagram like above was enough.

Question 15
A.
B.
C.
D.
Question 16
immediately after
A.
B.
C.
D.
Question 17
A.
B.
C.
D.
In this case the wave is a Transverse
wave.
The wave must be travelling in opposite
directions. B is Correct

SECTION A – continu
Question 16
immediately after
A.
B.
C.
D.
Question 17
A.
B.
C.
D.
The particles DO NOT move left or right
only UP or DOWN.
At the extreme position as shown. The
particle is at it’s highest point. Only
possible direction is DOWN. D is correct.

• investigate and explain theoretically and practically diffraction as the directional
spread of various frequencies with reference to different gap width or obstacle
size, including the qualitative effect of changing the λ/w ratio.

heard other than from directly in front of the speaker cone.
Diffraction of water waves
The diffraction of sound can be modelled with water waves in a ripple tank.
The next figure shows the way water waves diffract in various situations. The
diagrams apply equally well to the diffraction of sound waves.
aves:
round
velength
ct,
hrough
ngth
,
round
nd
ound the
ier
(c)
shadow
shadow
shadow
(a)
(e)
shadow
(d)
(b)
(f)
shadow
e tank
tion
t summary
ctice
ns

Some rules about diffraction:
Longer waves diffract more than short waves.
(This is the DOOF DOOF effect).
If the wavelength is smaller than the gap width there
will be minimal diffraction.
Maximum diffraction is when the wavelength is equal
to or larger than the gap width
The width of a diffraction pattern is proportional to
λ
w

35
c. w
y
laser
single slit
screen
w y
Figure 23
w
A.–D.
A. y
B. y
C. y
D.
1
2
y

If the slit width get larger the diffraction width will be
smaller.
OR even better
Spacing is proportional to
D is correct
λ
w

. Light as a wave
• describe light as an electromagnetic wave which is produced by the
acceleration of charges, which in turn produces changing electric fields
and associated changing magnetic fields
• identify that all electromagnetic waves travel at the same speed, c, in a
vacuum
• compare the wavelength and frequencies of different regions of the
electromagnetic spectrum, including radio, microwave, infrared, visible,
ultraviolet, x-ray and gamma, and identify the distinct uses each has in
society

In 1865 James Clerk
Maxwell proposed
mathematically that
light is a combination
of electric and
magnetic waves self
propagating each other.
He used maths to
suggest the speed was
3.0 x108 ms-1
A number of scientists including Foucault and
Michelson-Morley have since verified the speed of light.

Question 8
Longest Shortest
Radio micro Infra-red visible UV X-ray

• explain polarisation of visible light and its relation to a transverse wave model
Polarising filters have a lens structure that
only allows light of one orientation to pass,
either vertically or horizontally.

Polarising filters are evidence of the
transverse nature of electro-magnetic
radiation.

• investigate and analyse theoretically and practically the behaviour of waves
including:
– refraction using Snell’s Law: n1 sin(θ1) = n2 sin(θ2) and n1v1 = n2v2
– total internal reflection and critical angle including applications:
n1 sin(θ) = n2 sin(90°)

More generally this would be expressed as follows:
n1 sin θ1 = n2 sin θ2.
f
he light
use the
he ray
tion.
medium 1
refractive index n1
medium 2
refractive index n2
boundary
normal
θ2
θ1
n1 sin θ1 = n2 sin θ2
Sample problem 10.1
A ray of light strikes a glass block of refractive index 1.45 at an angle of inci-
dence of 30°. What is the angle of refraction?
nair = 1.0, θair = 30°, nglass = 1.45, θglass = ?
1.0 × sin 30° = 1.45 × sin θglass (substitute values into Snell’s Law)
sin
sin 30
θ =
°
olution:
Angles of incidence and refraction are always
measured from the NORMAL.
Light is bent at the boundary as the light changes
velocity
Slowing down light bends towards the normal.
Speeding up light bends away from the normal.

The refractive index of a material is related to the
speed of light in the material.
n1 =
c
v1
If the refractive index of a material is 1.5 (water),
then the speed of light in the material is:
n1 =
c
v1
1.5 =
c
vw
vw =
2
3
c = 2.0 ×108
m / s

Total internal reflection ONLY occurs going from
slower medium to faster medium i.e. water into
air.
For total internal reflection Ør = 90o.
θc = sin−1 n2
n1
⎛
⎝
⎜
⎞
⎠
⎟
If going into air (which is common)
θc = sin−1 1
n1
⎛
⎝
⎜
⎞
⎠
⎟
If ø is greater than øc
then the light will have
Total Internal Reflection.

Question 7
a.
45°
25° glass slab
air
n = 1.00
laser
Figure 8 n1 sinθi = n2 sinθr
1⋅sin45 = n2 sin25
n2 = 1.67

r
i
t
e
i
n
t
h
i
s
a
r
e
a
incoming ray
air
n = 1.00
water
n = 1.33
Figure 9
b.
c. i
θc = sin−1 1
n1
⎛
⎝
⎜
⎞
⎠
⎟
θc = sin−1 1
1.33
⎛
⎝
⎜
⎞
⎠
⎟
θc = 48.60

c. i
incoming ray
air
n = 1.00
water
n = 1.33
ic
The incoming ray has angle of
incidence greater than critical angle.
Therefore, Total Internal Reflection
will occur.

The diagram should show the par1ally reflected and par1ally
refracted rays, this is typical of what happens in the real world.

Total internal reflec1on occurs.
Therefore, no light leaves the block and the observed cannot see the
laser.
It was very common for students to make the first point and not
con1nue to explain why the observer could not see it.

• investigate and explain theoretically and practically colour dispersion in
prisms and lenses with reference to refraction of the components of white light
as they pass from one medium to another

ROYGBIV
The dispersion (spectrum) is a result of the
relative refractive index of each colour and
ultimately the speed of each colour in the
glass. While the speed of all light in a
vacuum will be the same. Each colour will
have slightly different speed in glass. Hence
the angle of refraction will be slightly
different resulting in dispersion.

• explain the results of Young’s double slit experiment with reference to:
– evidence for the wave-like nature of light
– constructive and destructive interference of coherent waves in terms of
path differences: nλ and (n − 1/2) λ
– effect of wavelength, distance of screen and slit separation on interference
patterns: ∆x = λL/d
Young’s double slit experiment relies on
coherent light.
It is irrefutable evidence of wavelike properties
of light:
It is an interference pattern.
ONLY light is capable of an interference
pattern.
If light was a particle there would be only two
maxima in the double slit experiment.

Question 18
a.
b.

There is a number of components to Young’s
experiment:
-A colour filter to allow one wavelength/
frequency.
-A small single slit for the waves to diffract
through. This is to ensure the waves are
coherent.
-The double slits for the coherent waves to
pass through and produce the interference
pattern.

Key knowledge
Behaviour of light
• investigate and describe theoretically and practically the effects of varying the
width of a gap or diameter of an obstacle on the diffraction pattern produced
by light and apply this to limitations of imaging using light
Very similar dot point covered earlier.
However, if using optical instrument such
as microscope. It is difficult to see
objects smaller than one wavelength as
the light will diffract around the object.
This is a limitation. Hence, higher
resolution in electron microscopes as the
DeBroglie wavelength is much smaller.

Question 6
a.
–1
b.
P
v = λ ⋅ f
v = 0.1× 5
v = 0.5 ms−1
Don’t use cm as a
measurement

b.
Y X
P
Figure 6
As P is the second min. the
path difference is 1.5 λ. As
XP is 16 cm, YP must be
31cm.

c.
Figure 7
wavefront meets a small gap, e
which means the next wavefron
Now let us investigate what h
ling at an angle θ such that:
θ
λ
=
w
sin .
Point sources in a diffraction
gap
Diffraction patterns change
with gap width. As the gap
width gets smaller, coming
down the figure, the pattern
spreads out more.
Diffraction spreading
If gap larger than
wavelength not a lot of
diffraction
If gap equal or smaller
than wavelength then a
lot of diffraction
Diffraction spreading is at a maximum when the gap is equal
or smaller than the wavelength. Increasing the frequency
will reduce the wavelength and reduce the angle of
diffraction.

Elli is assuming that with two speakers will produce sound that is
twice as loud as one speaker, this is incorrect.
Sam is correct in that an interference pa]ern will be produced
between the speakers and this will have areas of loudness
(construc1ve interference) and low volume (destruc1ve
interference).

The rule for the spacing between fringes is:
Student 2 will hear a high volume being at a point of construc1ve
interference.
Student 5 will hear low volume being at a point of destruc1ve
interference.
It is also possible to use Pythagora’s Theorem with this ques1on and
arrive at correct answers.
It is not enough to say Student 2 will hear high volume and Student 5
low without any jus1fica1on.

You could draw a diagram to show the first point or state:
λ
2
= 4 m
λ = 8m

This is a li]le bit of a trick ques1on need to find fo:
fo = 30 Hz this is the fundamental freq.
For a string the second-lowest resonance is 60 Hz.
All mul1ples of the fundamental are possible in strings and open
pipes.

The waves are travelling in opposite direc1ons or have been
reflected from the ends of the string.
The waves need be coherent. (same wavelength and amplitude)
The waves interfere and cause the observed pa]ern.

Area of Study 2
How are light and matter similar?
In this area of study students explore the design of major experiments that have led
to the development of theories to describe the most fundamental aspects of the
physical world – light and matter.
Outcome 2
On completion of this unit the student should be able to provide evidence for the
nature of light and matter, and analyse the data from experiments that supports
this evidence.

• analyse the photoelectric effect with reference to:
– evidence for the particle-like nature of light
– experimental data in the form of graphs of photocurrent versus electrode
potential, and of kinetic energy of electrons versus frequency
– kinetic energy of emitted photoelectrons: Ek max = hf – φ, using energy units
of joule and electron-volt
– effects of intensity of incident irradiation on the emission of photoelectrons
There are two values for Planck’s constant:
6.63 x 10-34 J.s (S.I. units)
4.14 x 10-15 eV.s (non S.I. units)
Know when to use each.

• describe the limitation of the wave model of light in explaining experimental
results related to the photoelectric effect.
The major points are:
Classical Theory means waves
Quantum Theory means particles
-Classical theory would predict that waves
of any frequency would eventually eject
electrons from the surface of the metal as
the energy builds up.
-Classical theory would predict that the
stopping voltage is also related to the
intensity of the light as the wave energy is
related to the amplitude.
These observations are not made in PE.

b.
The photo-electric effect shows how a single
photon will behave like a particle.
However, when a single photon is used in
Young’s experiment the photon splits into a
wave of potential, passes the slits and
interferes with itself causing a interference
pattern.
Only waves can cause interference patterns.
The photon is both wave and particle.
Q 18

c.
If light is a wave then:
1 The more intense the light, the more energy the electrons will have
when the fly off the plate.
2 If the light is very feeble, one may have to expose the source plate for
several seconds or minutes until enough waves strike it to knock
electrons loose.
3 Waves of any frequency ought to knock electrons free.

1 The energy of the electrons does NOT depend on the intensity of the light.
◦ Each electron absorbs only one photon at a time. If the absorbed energy
is large enough to expel the electron from the metal, it leaves. If not, the
electron dissipates its energy in collisions with nearby electrons and
atoms before it can absorb another photon. And yes, this implies that the
time it takes for an electron to lose the energy gained in one absorption is
much smaller than the interval between absorptions. Under ordinary
circumstances, it is.
2 The electrons always appear AS SOON AS the light reaches the plate.
◦ As soon as a single photon containing sufficient energy strikes the source
plate, it will knock an electron free. There is no need to wait for multiple
waves to build up enough energy.
3 NO electrons are produced if the frequency of the light waves is below a
critical value.
◦ Since the energy of each photon is
E=h f
below some critical frequency, no photon has enough energy to knock an
electron free.

... We shall assume that in leaving the body, each electron must perform an
amount of work W characteristic of the substance. The ejected electrons
leaving the body with the largest normal velocity will be those that were
[located exactly on] the surface. The kinetic energy of such electrons is
given by h*f - W.
Albet Einstein

Quantum mechanics is all about par1cles as waves so the possible
answers are C or D.
Electron diffrac1on explains interference pa]erns seem by electrons.
Electron shells (energy levels) are explained by the standing
wavelength pa]erns. Therefore D.

Question 17
A
V
metal plate
collector electrode
light source
filter voltmeter
variable
DC voltage
source
photocell
+
–
ammeter
Figure 19

a.
I ( A)
V (volts)
X
O
Figure 20
The point X is the stopping voltage.
The voltage X stops the most energetic
photoelectron that is ejected from the plate and
the current falls to zero.

The difficulty with the previous question is that the
diagram usually used is like the one below. Students
did not recognise the difference and were unable to
apply their knowledge of the situation.

Frequency (Hz) Stopping voltage (Vs)
× 10
b.
2.0
2.5

b.
0
0.5
1.0
1.5
2.0
2.5
4.0 5.0 6.0
frequency (Hz × 1014)
7.0 8.0
EK max of
photoelectrons
(eV)
c. part b. h
+
+
+
+
+

SECTION B – Question 17
c. part b. h
h =
rise
run
h =
2
3.75 ×1014
= 5.3×10−15
eVs
You can not just quote a known value for h. The
instructions are “from the graph plotted…”.

d.
e. one
f.
2
3
EK max of
photoelectrons
(eV)
The Frequency of light used
Stopping voltage OR max E of photoelect
The target metal would be controlled.
Different metals have a different work function.

f.
0
1
2
3
frequency (Hz × 1014)
EK max of
photoelectrons
(eV)

SECTION B – continued
TURN OVER
Draw in the error bars that have the same length as
the one already shown.
Draw a straight line that can pass through the errors
bars.
If you can then a straight line is appropriate

f.
0
1
2
3
frequency (Hz × 1014)
EK max of
photoelectrons
(eV)

To get the answer in eV you will need to use the correct value of
Planck’s constant.

. Matter as particles or waves
• interpret electron diffraction patterns as evidence for the wave-like nature of
matter
• distinguish between the diffraction patterns produced by photons and
electrons
• calculate the de Broglie wavelength of matter: λ = h/p

w
r
i
t
e
i
n
t
h
i
s
a
r
e
a
Question 16
–31
–1
crystal
e–
fluorescent
screen
electron gun
Figure 18a Figure 18b
a.

SECTION B – Question 16
a.
b.
λ =
h
p
λ =
6.63×10−34
9.1×10−31
×1.5 ×107
λ = 4.86 ×10−11
m
The momentum will increase.
Therefore the wavelength will be smaller.
The observed pattern will become smaller as well.

c.
–1
If it has the same pattern it must have the same
wavelength.
However X-rays and electrons will have different
energies.
E =
h⋅c
λ
E =
6.63×10−34
⋅3.0 ×108
4.86 ×10−11
E = 4.09 ×10−15
J

. Similarities between light and matter
• compare the momentum of photons and of matter of the same wavelength
including calculations using: p = h/λ
• explain the production of atomic absorption and emission line spectra,
including those from metal vapour lamps
• interpret spectra and calculate the energy of absorbed or emitted photons:
ΔE = hf

d
o
n
o
t
w
r
i
t
e
i
n
t
h
i
s
Question 20
Data
h –15
c 8 –1
A. –19
B.
C.
D.
E =
h⋅c
λ
E =
4.14 ×10−15
⋅3.0 ×108
4.14 ×10−7
E = 3.0 eV

w
r
i
t
e
i
n
t
h
i
s
a
r
e
a Question 9
ionisation energy
n = 6
n = 5
n = 4
n = 3
n = 2
n = 1 (ground state)
0
10.2
12.1
12.8
13.1
13.2
13.6
energy (eV)
Figure 10
n
12.8, 2.6, 0.7 12.1, 1.9 10.2
Arrows MUST
go DOWN

• analyse the absorption of photons by atoms, with reference to:
– the change in energy levels of the atom due to electrons changing state
– the frequency and wavelength of emitted photons: E = hf = hc/λ

• describe the quantised states of the atom with reference to electrons forming
standing waves, and explain this as evidence for the dual nature of matter

• interpret the single photon/electron double slit experiment as evidence for the
dual nature of light/matter

• explain how diffraction from a single slit experiment can be used to illustrate
Heisenberg’s uncertainty principle
• explain why classical laws of physics are not appropriate to model motion at
very small scales.

d
o
n
o
t
w
r
i
t
e
i
n
t
h
i
s
a
Question 10
y
O
x
y
electron’s
path
barrier with a slit
A.
B.
C. y
D. x
D from the previous video

. Production of light from matter
• compare the production of light in lasers, synchrotrons, LEDs and
incandescent lights.
Light
Amplification
by
Stimulated
Emmission
of
Radiation
Laser light is
COHERENT
All photons are the
same wavelength and
are all in phase with
each other.

Incandescent light globes work by heating
the electrons in the wire. Electrons move
into higher energy levels and release energy
as photons.
Incandescent light globes produces
INCOHERENT light of many colors
(frequencies) and are not in phase.

LED’s are semi conductor materials with a
P-N junction.
Electrons move from the conductive band
to the valence band and give extra energy
as photons.
The light is INCOHERENT although there
will be discrete colors the light will NOT be
in phase.

Fluorescent lights absorb energy into a gas
vapour and then release the energy in the
form of photons.
There will be many frequencies of light due
to the discrete energy levels. Think of the
Sodium lamp or the energy rungs of a
Hydrogen atom.

A synchrotron is a particle accelerator.
Beams of particles (usually electrons) are
accelerated and bent in Magnetic fields.
They are capable of making a very wide
range of photon wavelengths. However,
usually make very very high intensity X-
rays. The intensity can be x100,000’s the
brightness of the sun.

e
i
n
t
h
i
s
a
r
e
a
Question 18
A.
B.
C.
D.
Question 19
A.
B.
C.
D.
Question 20
Data
h –15
c 8 –1
t
h
i
s
a
r
e
a
Question 18
A.
B.
C.
D.
Question 19
A.
B.
C.
D.
Question 20
Data
D is the only correct statement
C LED’s move electrons from the conduction band
to the valence band.

Area of Study 3
Practical investigation
A student-designed practical investigation related to waves, fields or motion is
undertaken either in Unit 3 or Unit 4, or across both Units 3 and 4. The
investigation relates to knowledge and skills developed across Units 3 and 4 and is
undertaken by the student through practical work.

• the characteristics of scientific research methodologies and techniques of
primary qualitative and quantitative data collection relevant to the selected
investigation, including experiments and/or the construction and evaluation of
a device; precision, accuracy, reliability and validity of data; and
the identification of, and distinction between, uncertainty and error

• methods of organising, analysing and evaluating primary data to identify
patterns and relationships including sources of uncertainty and error, and
limitations of data and methodologies
• models and theories, and their use in organising and understanding observed
phenomena and physics concepts including their limitations
• the nature of evidence that supports or refutes a hypothesis, model or theory
• the key findings of the selected investigation and their relationship to concepts
associated with waves, fields
and/or motion

• the conventions of scientific report writing and scientific poster presentation,
including physics terminology and representations, symbols, equations and
formulas, units of measurement, significant figures, standard abbreviations and
acknowledgment of references.

Key knowledge
• independent, dependent and controlled variables
• the physics concepts specific to the investigation and their significance, including
definitions of key terms, and physics representations
A.
B.
C.
D. –5
Question 6
A.
B.
C.
D.
Question 7
The investigator changes the value of the
independent variable. A

Question 7
A.
B.
C.
D.
Question 8
A.
B.
C.
D.
D.
Question 7
A.
B.
C.
D.
Question 8
A.
B.
C.
D.
The dependent variable changes with the
independent D
The control variables are fixed by the investigator D

i
t
e
i
n
t
h
i
s
a
r
e
a
Question 9
–2
–2
–2
–2
–2
–2
A. –2
B. –2
C. –2
D. –2
Question 10
y
O
y
Highest value = 9.84
Lowest value = 9.78
Difference 0.06
Uncertainty = ± 0.03 A

Q18.
Accurate: is close the expected value.
Precise: the measured values are close to each other. A