2014 St Josephs Lecture Physics

1. VCE physics exam revision
• Physics examination questions that received <60% score for
2010-2013.

6. 2013 Exam all topics

8. 20 N

11. 20 J Read from graph

12. 5 J missing

13. They have not taken into account that at Q
the spring is extended 0.5m and at P it is
extended 1.5 m
Giving a change in energy of 10J as required.
They have not taken into account that at Q

14. 24 hours = 86,400 sec

15. The correct term is apparent weightlessness. While in
orbit the astronaut is constantly falling.
Weight is the product of mass and gravity. This applies in
this situation.
True weightlessness only occurs when in very deep space
where the action of gravity is negligible. This does not
apply in this case.

19. Resistor changes electrical energy into thermal energy
LED changes electrical energy into light
NOTE: LED’s do not generate heat. That is why they are
efficient.

20. An increase in the input voltage causes the output voltage to
decrease.
Negative gradient (1 mark only)

21. Make sure you use your RIGHT hand. If you are right handed
put your pen/pencil down.

22. To generate EMF in the resistor the transformer needs a
change in the flux (Faraday’s Law)
The constant current produced by the battery generates a
constant flux in the transformer. Hence NO EMF is
generated.
The only time a change in flux is observed is when the switch
is opened or closed.
Write about the Change in Flux and NOT about magnetic
fields. This shows a lack of understanding about how
electricity is made.

24. When the gradient is ZERO
t=0.5, 1.0, 1.5 sec

25. HARD QUESTION (AVERAGE 0.7/4)
As the ring falls down the B field is increasing and upwards
inside the ring.
The EMF generated will produce a B to oppose the change in
flux {Lenz’s Law} OR Keep flux at ZERO in the ring.
The induced B field will be going down inside the ring.
The current must be CLOCKWISE for this to occur.
You could draw a diagram to show the same thing.

26. Tricky: Longest wavelength has the smallest energy
E=3.19-2.11 = 1.08 eV
ie. First excited state to ground state

28. The students have chosen a frequency that is below the
threshold frequency for the target metal.
Therefore, no photoelectrons are being ejected.

29. The intensity will be at a MAXIMUM.
The path difference is ZERO.
Constructive interference is occuring
Lower frequency
Longer wavelength
More width in the pattern
D

30. INCORRECT
-Young’s experiment supports the wave model.
-only waves can produce constructive/destructive interference
pattern.
-The particle model would predict just two bright bands on
the screen.

32. A is CORRECT
The patterns will have the same fringe spacing if the
wavelengths are the same.
The deBroglie wavelength is a function of the momentum
NOT energy.

33. Newtons motion

34. Consider the trailer by itself:
F = m x a
F = 2000 x 0.5 = 1000N

35. T1 pulls both masses. It will need to overcome the friction of both logs.
800

36. Still need to overcome friction and acceleration,
but only for 2nd log.
F = 300 + 400 = 700 N
d= 20m v2=u2+2ad
u=4 ms-1 v2=16+2(0.5)(20)
v=? v=6 ms-1
a=0.5 ms-2
t=

37. åF = ma
1.0 = (0.1+ 0.4)a
a = 2ms-2

38. d = 1.0m v2 =u2 + 2ad
u = 0ms-1 v2 = 0 + 2 ×2 ×1
v= ? v2 = 4
a = 2ms-2 KE = 1
2
mv2 = 1
2
(0.4)4 = 0.8J
t =

39. 1.5 N
The Normal reaction force supplied by block C onto B will be
equal to the combined weight of blocks A and B.
The most common mistake was to forget about block A and
give the answer as 1.0 N.

40. Not a great question as the meaning is ambiguous.
The reaction force is the Tension in cable A.
It has a value of 30 N. The combined weight of both masses.
It acts up away from the centre of the earth.

42. If traveling in a circle at a constant speed the Net Force (centripetal
force) will be constant, acting towards centre of the circle.
At the bottom of the circle the Tension in the string and the weight
are acting in opposite directions. To maintain the same Net Force
the tension must be greater to overcome this.
At the top of the circle the Tension and weight force act in the
same direction and add up to the same Net force.
You could draw a diagram with some Vectors to show the same
idea and still get full marks.

43. Fnet
Must be clear that the arrow is pointing horizontally.
By vector addition of the Weight force and Normal
reaction force the Fnet is horizontal.

44. q = Tan−1 v2
rg
⎛⎝ ⎜
⎞⎠ ⎟
θ = Tan−1 102
100 ⋅10
⎛⎝ ⎜
⎞⎠ ⎟
θ = 5.7o

45. E = 1
2
kx 2
m gh = 1
2
kx 2
1´10 ´ 0.3 = 1
2
k (0.3)2
k (0.3)2 = 6
k = 67N /m

46. E = 1
2
kx2
Þ5.4 = 1
2
k(0.08)2
k = 1.69 ´103 N /m
Must use units as m not cm.

47. åF = 0
Þmg = kx
k = mg
x
= 2 ×10
0.4
=50N /m

48. The system gains Elastic and loses GPE
DEPE =
1
2
kx2 =
1
2
50(0.4)2 = 4 J
ΔGPE = mgh = 2 ⋅10 ⋅0.4 = 8 J
ΔEnergy = 4 J

50. A: The KE is returned to the system as it is elastic but some is
stored in the spring for a short period as EPE during the collision.
B: The total momentum is ALWAYS constant.

51. C: Not all of the KE is returned to the system. Some is lost as
heat or sound during the collision.

52. The easiest way to show this is using vectors:
If right is positive and 1.2 kg block has x kgm/s of momentum before the collision
and -y kgm/s after the collision. The 2.4 kg block will need to have x+y, to the
right for momentum to be conserved.

53. d = 200m v2 =u2 + 2ad
u = 0 v2 = 0 + 2(10)(200)
v= v= 63.2ms-1
a = 10ms-1
t =

54. 10

55. g = GM
r2
M = gr2
G
=
10 ´ 3´104 ( )
2
6.67 ´10-11 ( )
= 1.35 ´1020 kg

56. T = 4p 2r3
GM
=
4p 2 1´109 ( )
3
6.67 ´10-11 ( ) 5.7 ´1025 ( )
= 3.2 ´106 sec

57. T = 4p 2r3
GM
T 2 = 4p 2r3
GM
r = GMT 2
3 =
4p 2
6.67 ´10-11 ( )(7.36 ´1022 )(7200)2
4p 2
3
r = 1.86 ´106 m
h = 1.86 ´106 -1.74 ´106 = 1.21´105 m

58. The term weightless applies in situations where there
is only micro gravity or ZERO gravity. Apparent
weightlessness is the absence of a Normal reaction
force. In situations such as an orbit “Apparent
weightlessness” would be the correct term to apply.

59. F = GMm
r2
F=
6.67 ´10-11 ( ) 5.98 ´1024 ( ) 3.04 ´105 ( )
6.72 ´106 ( )
2 = 2.69 ´106N

60. T = 4p 2r3
GM
T =
4p 2 6.72 ´106 ( )
3
6.67 ´10-11 ( ) 5.98 ´1024 ( )
T = 5480sec
Same

61. Electronics & photonics

62. 2013 exam

64. Resistor changes electrical energy into thermal energy
LED changes electrical energy into light
NOTE: LED’s do not generate heat. That is why they are
efficient.

65. An increase in the input voltage causes the output voltage to
decrease.
Negative gradient (1 mark only)

66. 2012-2010 exams

67. From Q1: Use V=IR
10 = I x 3.3
Itotal = 3 Amp
Resistance through C,D is
twice that of B
IB= 2 Amp

68. P=I2R
P=12 . 2 = 2 W

69. Voltage across the 200 Ohm resistor is
1V and across 500 Ohm resistor is 5V
10mA through the 500 Ohm resistor but
only 5mA through the 200 Ohm resistor.
The other 5mA will go through the diode.
But still only 10mA through the battery and
hence ammeter.

70. 12
12

71. V=iR
24=ix2
i=12 Amp

73. 10luxÞ10,000W
ÞV out = 4V
ÞV R = 8V
R = 20,000W

74. -As sunset approaches the lux decreases,
therefore the R increases as read from the graph.
-Vout will take a greater proportion of the 12 V
available and will increase.
-For the light to come on earlier that means
more lux and less resistance.
-To maintain the 2:1 ratio the value of R should
also decrease.

75. -Compare Vin and Vout during the linear section (say 5 ms)
-Pay attention to the scales used on the graph.
-Gain is 50
-There is never a negative gain.

77. Must have battery, thermistor and resistor connected
in series. The most basic concept.
Does the switching circuit go across the resistor or the
thermistor? What size resistor is needed?
Then T rises, Rt falls and Vt also falls therefore VR will
increase. The circuit must be across the resistor.
Rt=1500 Ω RR=? must be 4500 Ω
Vt=1.5 V VR= 4.5 V
4500 Ω
6 V
Thermistor

78. Note: As the Resistance
increases, the current
decreases and the
brightness of the LED will
decrease.

79. The brightness of the LED
will never be ZERO
The variation of the Resistance in the circuit is
synchronised with the variation of the brightness of
the LED OR variation of intensity of one wave (R) is
matched with the variation in intensity the other
(Brightness)

80. You must answer THE question not simply waffle on
about clipping and distortion.
-When the magnitude of input is greater than 10 mV.
-The maximum output will have a magnitude of 4 V.

81. Note: The different threshold voltages
of the diodes.

82. 8 - 3 = 5V
V=IR V=IR
6=Ix100 5=Ix150
I=0.06 A I=0.033 A
93

84. At 2.5 lux, RLDR =3000 Ω.
For output of 6V or more, Rr= 2000 Ω. For 3:2 ratio.
As light decrease RLDR increases. Switching unit across RLDR.
2000 Ω
10 V
Marking scheme
1 mark for all components in a diagram.
1 mark for 2000Ω resistor
1 mark for switching circuit across the LRD.

85. -The modulation device combines the low frequency
input signal with the high frequency carrier wave.
-The light beam carries the modified signal to the
demodulation device.
-The demodulation device removes the carrier wave
to recover the input signal.
DO NOT write about changing electricity to light,
this is what a transducer does not a
modulation/demodulation system.

86. materials & structures

88. Solve Q8 and Q9 together as simultaneous equations:
120,000 N
Fb
Fs
Take torque about the bolt
20 Fs=30 x 120,000
Fs=180,000 N
Add vertical forces
Fb + 120,000 = Fs
Fb=180,000 - 120,000
Fb=60,000 N
Q8 D
Q9 B

89. Place the steel cables where the board will be in tension and NOT compression:
All along the top of the board.
ANSWER A

90. e = Dl
lo
⇒ 0.01 = Δl
2m
∴ Δl = 0.02m = 2cm

91. StrainEnergy / m3 = Area
= 1
2
3´108 ( ) 1.5 ´10-3 ( ) = 2.25 ´105 J
V olume = 10-3
StrainEnergy= 2.25 ´105 ´10-3 = 225J

92. Electricity generation and
supply

93. 2013 EXAM

95. When the switch closes this causes the
current to go from ZERO to ON.
This will cause the flux in the coil/core to
change from ZERO to MAX
By Faraday’s Law this will induce an EMF in
the coils and hence current in the resistor.
When the current is ON there is no Change
in flux thus NO EMF.

97. When the change in flux is ZERO ie.
Gradient is ZERO
T=0.5, 1.0, 1.5 sec (full marks if include t=0
and 2.0 and other times are correct)

98. Link ideas together:
CLOCKWISE
As the ring moves down the magnetic field
increases in an upwards direction relative to the
ring.
Lenz’s Law states the induced current will
oppose the change in flux (ie. try to keep the
magnetic field the same.)
This means the induced magnetic field will
increase in the down direction.
To do this the induced current must be
clockwise.

99. 2012 EXAM

101. P =V I
P= 900 ´ 50 = 45kW

102. R T = 7 +18 = 25W
V = IR
1000 = I ×25
I = 40Amp

103. No the motor will not work correctly
Ploss = I 2R
= 402 ×25 = 40,000W
Thismeansonly5,000W reachesthemotor
at40Amp, 125V
It is also possible to show that
the current is 40 Amp while
the motor operates correct at
50 Amp
Use thicker wires or change the material that the wires are made of. This
will reduce the resistance and hence the power loss.
Use a series of step up and step down transformers. This will enable the
electricity to be distributed at a higher voltage and lower current that will
reduce the power loss as Ploss=I2R.

104. Step down transformer.
5
To induce a current in the secondary coil Faradays’s Law implies a
change in flux is needed. The DC will supply a constant flux and will
be unable to induce a current in the secondary coil

105. Initially the flux is a maximum and decreases to a
minimum after 1/4 turn. After a further 1/4 turn the
magnitude is a max but the direction of the flux
relative to the coil is negative. Before returning to
zero and then a max again.
Students obtain full marks for a single cos graph.

106. x = n ΔΦ
t
ξ = n
B × Area
t
3.6 = n
0.03×(0.3× 0.4)
0.125
n = 125
Best answer was with a sketch:
Slip rings give a sin function
Commutator with mod(sin)

107. As the loop moves from position 2 to position 3, the flux is out of the page and decreasing.
The induced current will produce a magnetic field out of the page to oppose the decrease.
Therefore, the current must flow anticlockwise through the loop.
It was common for students to omit reference to the initial flux and how it was changing.
Students also seem to have had problems deciding whether the direction of the current was X
to Y or Y to X through the loop.

108. 2011 EXAM

110. The direction must be
exactly right as the S pole
and earth will cancel each
other

111. The moving magnet supplies a source of changing
magnetic flux. This induces an EMF in the loop.

112. Or something similar, first
direction is not important

115. 2010 EXAM

116. ZERO
The loop is parallel to the magnetic field.

121. 4W &2VÞ2Amp
Rglobe = 1W
RT = 5WÞV =10

122. Ploss = I 2R
= 22 ´ 4 = 16W

123. The globe needs 2 V and 2 Amp. The primary side of the
transformer will have 20 V and 0.2 Amp
Ploss = I 2R
= 0.22 ´ 4 = 0.16W

124. theories of light and matter

126. E = hf -W
E = 4.14 ´10-15 ( ) 7.5 ´1014 ( )- 2.28
E = 3.105 - 2.28 = 0.83eV

145. recording and reproduction of
sound