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Trial Examination 2014

VCE Physics Units 3&4
Written Examination

Suggested Solutions

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TEVPHYU34_SS_2014.FM

VCE Physics Units 3&4 Trial Examination Suggested Solutions

SECTION A – CORE Area of study – Motion in one and two dimensions Question 1 (10 marks) a. v v = 12 sin 60° = 10.4 m s b. t air = 2 × t top 0 – 10.4 t top = ------------------ = 1.04 – 10 t air = 2 × 1.04 = 2.1 1 2 s = -- at 2
2 1 = -- ( 10 ) ( 1.04 ) 2 –1

1 mark

1 mark 1 mark Note: Consequential on part a.

c.

1 mark 1 mark

= 5.4 m d. Gravity is 10 m s ∴ 10 m s e.
–2 –2

down 1 mark

1 2 KE = -- mv 2 v h = 12 cos 60° =6ms
–1

1 mark

2 1 KE = -- ( 80 ) ( 6 ) 2

= 1440 J f. R = v h t air = ( 6 ) ( 2.08 ) = 12.5 m

1 mark

1 mark 1 mark Note: Consequential on part e.

2

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VCE Physics Units 3&4 Trial Examination Suggested Solutions

Question 2 (6 marks) a. P before = ( 5 × 10 ) ( 15 ) – ( 10 × 10 ) ( 15 ) ˜ 4 –1 = – 7.5 × 10 kg m s – 7.5 × 10 V after = -----------------------4 1.5 × 10 ˜ = –5 m s ∴ 5.0 m s
–1 –1 4 3 3

LEFT

2 marks 1 mark for magnitude 1 mark for direction

b.

3 2 1 3 2 1 KE before = -- ( 5 × 10 ) ( 15 ) + -- ( 10 × 10 ) ( 15 ) 2 2

= 1 687 500 J 1 3 2 KE after = -- ( 15 × 10 ) ( 5 ) 2 = 187 500 J ∆KE = 1 500 000 J = 1.5 MJ c. It is lost as heat and sound. 1 mark Note: Consequential on part a. 1 mark 1 mark 1 mark

Question 3 (3 marks) a. ΣF = ma 30 = ( 2 + 4 ) a a=5ms b.
–2 –2

1 mark

Block A accelerates at 5 m s . ΣF A = 2 ( 5 ) = 30 – F B on A = 10 N to the right = 30 – F B on A ∴ FB = 20 N (to the left) 1 mark 1 mark

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VCE Physics Units 3&4 Trial Examination Suggested Solutions

Question 4 (3 marks) a. mv F c = --------R Fc R v = --------m ( 2800 ) ( 80 ) = --------------------------1400 = 12.6 m s b.
–1 2

1 mark 1 mark

F

c

1 mark Question 5 (3 marks) a. mv ΣF = --------R
2

mv T – mg = --------R mv T = mg + --------R
2

2

1( 5 ) = 1 ( 10 ) + ------------0.5 = 60 N b. 0J As the force is always perpendicular to the direction of travel, no work is done.

2

1 mark 1 mark 1 mark

Question 6 (3 marks) F ∆t = m ∆v Area under graph gives F ∆t. F ∆t = 400 × 10 × 10 =4 4 ∆v = ---------------------–3 60 × 10 = 66.7 m s
–1 –1 –3

1 mark

1 mark 1 mark
TEVPHYU34_SS_2014.FM

= 240 km h
4

VCE Physics Units 3&4 Trial Examination Suggested Solutions

Question 7 (4 marks) a. F = kx F k = -x 2000 = ----------10 = 200 N m
–1

2 marks 1 mark for correct value 1 mark for correct unit

b.

1 2 PE stored = -- kx 2
2 1 = -- ( 200 ) ( 8.6 ) 2

1 mark

= 7396 J = 7.4 × 10 J
3

1 mark Note: Consequential on part a.

Question 8 (3 marks) The trainee astronauts still have weight, as they are still in a strong gravitational field and W = mg. 1 mark Weightlessness only occurs in deep space away from all masses. This does not apply here. Because the centripetal acceleration at the top of the circle is given exactly by gravity, they feel ‘apparent weightlessness’ or apparent zero-g. Question 9 (5 marks) GMT R = 3 ---------------2 4π
2

1 mark 1 mark

a.

1 mark
– 11 24 2

( 6.67 × 10 ) ( 5.98 × 10 ) ( 86 400 ) = 3 -------------------------------------------------------------------------------------------2 4π = 4.2 × 10 m b. GM F = mg = m -------2 R
7

1 mark 1 mark

( 1432 ) ( 6.67 × 10 ) ( 5.98 × 10 ) = -----------------------------------------------------------------------------------7 2 ( 4.2 × 10 ) = 3.24 × 10 N
2

– 11

24

1 mark

1 mark Note: Consequential on part a. Subtract one mark if answer not given to three significant figures.

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VCE Physics Units 3&4 Trial Examination Suggested Solutions

Area of study – Electronics and photonics Question 10 (12 marks) a. From graph, at 25°, RTDR = 500 Ω V OUT R2 ------------ = -----------------V IN R1 + R2 4 500 -- = -------------------9 R 1 + 500 R 1 = 625 Ω b. V P = ----R
2

1 mark

1 mark 1 mark

4 = -------500 = 3.2 × 10 c.
–2

2

1 mark W 1 mark

P TDR = VI 3.2 × 10
–2

=4×I
–3

I = 8.0 × 10 A Vvariable resistor = 9.0 – 4.0 – 1.4 = 3.6 V V 3.6 R variable resistor = -- = ------------------------ = 450 Ω I 8.0 × 10 –3 Hence the value of the variable resistor must decrease. d. As temperature decreases, the resistance of thermistor increases, so VOUT increases. When the temperature DECREASES this would turn ON a device connected in parallel with the thermistor so could control the heater. OR When the temperature DECREASES it could turn OFF a device connected in parallel with the variable resistor so could control an air conditioner. e. The LED is an opto-electrical device. As the current through the LED increases, its brightness will also increase, so the LED converts electrical energy to light energy.

1 mark

1 mark 1 mark 1 mark 1 mark 1 mark

1 mark 1 mark 1 mark 1 mark

Question 11 (2 marks)

30 Ω 30 Ω 30 Ω 30 Ω

30 Ω

2 marks

6

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VCE Physics Units 3&4 Trial Examination Suggested Solutions

Question 12 (5 marks) a. When VIN = 150 mV, VOUT = –30 V 30 magnitude of gain = --------0.15 = 200 b. 1 mark 1 mark

VOUT

30 20 10 –300 –200 –100 –1 –2 –3 100 200 300 VIN (mV)

3 marks 1 mark for correct scale 1 mark for clipping region shown 1 mark for inverted (negative gradient) Question 13 (4 marks) Position Waveform (A, B, C or D) P C Q A R D S C

4 marks 1 mark for each correct answer

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VCE Physics Units 3&4 Trial Examination Suggested Solutions

Area of study – Electric power Question 14 (7 marks) a.

X

variable power supply

closed switch
1 mark

If the north pole is repelled then there must be an induced north pole at the left-hand end of the coil. Use the right-hand grip rule to predict the current flow. b.

1 mark 1 mark

The correct answer is B. 1 mark Looking from X, the current is flowing in a clockwise direction. Using the right-hand grip rule, this means the induced field is into the page. Note: Consequential on part a. current direction: from B to A As the south pole approaches, the induced field will oppose the increasing flux. Using the right-hand grip rule this will mean the current flows downwards in the coil and from B to A in the galvanometer. 1 mark 1 mark 1 mark

c.

Question 15 (16 marks) a. F = NBIl = 30 × 0.35 × 2.5 × 0.1 = 2.6 N b. c. There will be no force (NF written next to side XY) as the current is parallel to the field. 1 mark 1 mark 1 mark

The force on side YZ is out of the page so side YZ would initially move upwards. 1 mark 1 When YZ has passed -- of a turn (the coil is horizontal) the force will still be out of the page, 4 so this will cause the coil to rotate back in the opposite direction. 1 mark The coil will oscillate around this horizontal position and eventually stop. 1 mark 1 mark

d.

The correct answer is B. The split-ring commutator will change the current direction in the coil each half-turn, allowing continual rotation. flux = BA = 0.35 × ( 0.10 × 0.15 ) = 5.3 × 10
–3

e.

1 mark 1 mark

Wb

8

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VCE Physics Units 3&4 Trial Examination Suggested Solutions

f.

N ∆Φ EMF = ----------∆t 30 × ( 5.3 × 10 – 0 ) 0.15 = -------------------------------------------------∆t ∆t = 1.1 s
–3

1 mark 1 mark Note: Consequential on part e.

g.

R

P Q Z

W

X

Y
1 mark 1 mark

The flux is upwards and decreasing. Lenz’s law tells us that the induced field will oppose this change, i.e. it will be in the same direction (up). The right-hand grip rule predicts current will flow ZYXW, which means it flows down through R. h. The correct answer is D.

1 mark 2 marks

1 The time from position A to B is -- of the time from position B to C. Hence the EMF for A to B 4 N ∆Φ will be four times the EMF for B to C, since EMF = ----------- and N ∆Φ is constant. ∆t 2 V Since P = ----- , the power dissipated as the coil moves from A to B will be 16 times the power R dissipated as the coil moves from B to C.

Question 16 (13 marks) a. 1 f = -T 1 = -----------------------–2 ( 5 × 10 ) = 20 Hz b. 120 Equivalent DC voltage is V RMS = -------2 V = 85 V 1 mark 1 mark 1 mark 1 mark

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VCE Physics Units 3&4 Trial Examination Suggested Solutions

c.

For the step-up transformer, VIN = 120 V, so VOUT = 120 × 20 = 2400 V P I = -V 500 = ----------2400 = 0.21 A P loss = I R = 0.21 × ( 5 × 400 ) = 88 W
2 2

1 mark

1 mark

1 mark

d.

V drop = IR = 0.21 × 2000 = 420 V V delivered = 2400 – 420 = 1980 V 1980 After the step-down transformer, V = ----------- = 99 V. 20 1 mark 1 mark 1 mark 1 mark 1 mark 1 mark

e.

The transformer needs AC to operate as the current in the secondary coil is induced by a changing current and magnetic flux in the primary coil. A DC battery will not provide this changing current so the pump will receive no power/voltage and will not operate.

10

TEVPHYU34_SS_2014.FM

VCE Physics Units 3&4 Trial Examination Suggested Solutions

Area of study – Interactions of light and matter Question 17 (7 marks) a. first nodal line ⇒ path difference = -2 532 -------- = 266 nm 2 b. c. d. The lines of the pattern will be closer together and the spacing will be smaller. As λ RED > λ GREEN , the spacing will be larger as the pattern spreads out. 1. The interference pattern is created by constructive and destructive interference. 2. This is explained by the path difference between S1P1 – S2P2, requiring it to be reliant on a whole number of wavelengths and a whole number plus half wavelengths. 3. This is well-modelled by wave theory. Thus if the light has a wavelength it has wave properties. Question 18 (2 marks)

λ

1 mark 1 mark 1 mark 1 mark 1 mark 1 mark 1 mark

1 mark If the orbit is an exact multiple of the wavelength of the electron, a standing wave is formed. This means that electrons of those particular wavelengths can exist in the atom. Since different wavelengths correspond to specific energies, only those energy states are possible for the atom. The diagram above shows an example of exactly 3λ .

1 mark

Question 19 (8 marks) a. Planck’s constant is given by gradient. 1.8 h = -----------------------14 4.4 × 10 = 4.09 × 10
– 15

1 mark eVs 2 marks 1 mark for correct value 1 mark for correct units Subtract one mark if answer not given to three significant figures.

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VCE Physics Units 3&4 Trial Examination Suggested Solutions

b.

total energy of photon = 3.0 + 1.8 = 4.8 eV hc E = -----

1 mark

λ

( 4.14 × 10 ) ( 3 × 10 ) λ = --------------------------------------------------------4.8 = 2.56 × 10
–7

– 15

8

m

1 mark Subtract one mark if answer not given to three significant figures.

c.

KEmax (eV) 4.0 3.0 2.0 1.0 0.0 –1.0 –2.0 –3.0
2 marks 1 mark for intercept KEmax = –3.0 eV. 1 mark for same gradient as metal A. (Students may dot line below KE = 0 axis.)

metal A 4.4 × 10
14

Hz

metal B f (× 10
14

Hz)

1 2 3 4 5 6 7 8 9 10 11 –1.8 eV (intercept)

Question 20 (4 marks) a. 3.61 eV – 2.10 eV = 1.51 eV photon

ionisation 3.75 eV 3.61 eV 3.19 eV ∆E = 1.51 2.10 eV

0 eV
2 marks

12

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VCE Physics Units 3&4 Trial Examination Suggested Solutions

b.

Longest λ is the smallest energy change. ∆E = 3.75 – 3.61 = 0.14 eV 1 mark c E
– 15 8

λ = h --

( 4.14 × 10 ) ( 3 × 10 ) = --------------------------------------------------------0.14 = 8.8 × 10
–6

m

1 mark

Question 21 (6 marks) a.

λ = --

h p
– 34

6.63 × 10 = -------------------------------------------------------------– 31 6 ( 9.11 × 10 ) ( 6.8 × 10 ) = 1.07 × 10 b.
– 10

1 mark 1 mark 1 mark 1 mark

m

This experiment does demonstrate the wave nature of electrons because there will be a significant diffraction occurring as λ ≈ d (the atomic spacing). λ = ------------------------ as p = 2 ( KE )m h 2 ( KE )m

c.

λ neutron me 1 ------------------ = ------ = --------λ electron m n 42.8
( λ neutron = λ electron ) Thus since ----------------------------------------------- , then λ neutron < atomic spacing and so diffraction will not 42.8 be observable, therefore the answer is NO. Question 22 (2 marks) Nia is correct.

1 mark

1 mark

1 mark

For electrons and X-rays to produce the same diffraction pattern they must have the same de Broglie h wavelength. Since for both electrons and photons the de Broglie wavelength is given by λ = -- , if p the wavelength is the same then the momentum must be the same as well. 1 mark

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VCE Physics Units 3&4 Trial Examination Suggested Solutions

Detailed study 5 – Photonics Question 1 A

The term incoherent when applied to an incandescent light source means that the light emitted is out of phase. Question 2 D

Laser light is monochromatic (one wavelength) and coherent (in phase). Question 3 c = λf c f = -C

λ

3 × 10 = -----------------------–7 5.2 × 10 = 5.8 × 10 Question 4 E = hf = ( 4.14 × 10 = 2.4 eV Question 5 C
– 15 14

8

Hz B
14

) ( 5.8 × 10 )

Light is emitted by the LED is as a result of the movement of electrons from the conduction band to the valence band of a semiconductor. Question 6 B

Modal dispersion in optical fibres is characterised by different modes taking different paths while travelling through the optical fibre. Question 7 D

Modal dispersion can be minimised using a graded-index fibre. Question 8 B

An optical-fibre sensor system measures the stress forces in a plane via the modulation of the intensity of the infrared light, which can then be read at the sensor. Question 9 1.41 sin θ c = --------1.48 C

θ c = 72.3°

22

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VCE Physics Units 3&4 Trial Examination Suggested Solutions

Question 10
–1 2

B
2

α = sin ( 1.48 – 1.41 ) = 26.7° Question 11 C

When placing the optical fibre in water (n = 1.33) the critical angle stays the same but the acceptance angle becomes smaller.

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Trial Examination 2013

VCE Physics Units 3 & 4
Written Examination

Suggested Solutions

VCE Physics Units 3 & 4 Trial Examination Suggested Solutions

SECTION A – CORE Unit 3 Area of study 1 – Motion in one and two dimensions Question 1 (7 marks) a. p i = 0.058 × 20 = 1.164 kg m s
–1 –3 –1

1 mark (in opposite 1 mark

∆p = area under graph = 27 squares × ( 5 × 20 × 10 ) = – 2.75 kg m s direction to initial motion) ∆p = p f – p i – 2.75 = p f – 1.164 p f = – 2.75 + 1.164 = – 1.59 kg m s p v f = --m 1.59 –1 = ------------ = 27.3 m s 0.058 b. range = ut cos θ = 15 × 1.9 × cos 28° = 25.2 m The ball lands outside the baseline. Question 2 (9 marks) a. u = 0, x = 1.0, t = 1.2, v = ? 0+v 1 u+v x ----------- = - so ----------- = ------ 2 1.2 2 t v = 1.7 m s b.
–1 –1

(in opposite direction to initial motion)

1 mark

1 mark 1 mark 1 mark 1 mark

1 mark 1 mark

1 2 At the bottom of the ramp E k = -- mv 2 1 2 = -- × 1.5 × 1.7 = 2.2 J 2 work done to overcome friction = Fx = 2.9 × 1 = 2.9 J total energy at top of ramp = 2.2 + 2.9 = 5.1 J U g = mgh 5.1 = 1.5 × 10 × h so h = 0.34 m 1 mark 1 mark 1 mark 1 mark

2

TEVPHYU34_SS_2013.FM

VCE Physics Units 3 & 4 Trial Examination Suggested Solutions

c.

25 –1 spring constant k = --------- = 2500 N m 0.01 1 2 E k = -- mv 2 1 2 = -- × 1.5 × 1.7 so U e = 2.2 J 2 1 2 U e = -- kx 2 1 2 2.2 = -- × 2500 × x so x = 0.04 m 2

1 mark

1 mark

1 mark

Question 3 (5 marks) a. The forces acting on the car are the weight force (N = mg) and N. These should both be shown on the diagram as they are below.

N

21˚ W
1 mark As the normal reaction force (N) has a component towards the centre of the circle, the centripetal force is provided which enables him to travel in a circle. The component of the normal reaction is greater than the road tyre friction value, which enables the car to travel at a greater speed around the curve. b. v tan θ = ---rg v tan 21° = ---------------------( 20 × 10 ) v = 8.8 m s
–1 2 2

1 mark 1 mark

1 mark 1 mark

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VCE Physics Units 3 & 4 Trial Examination Suggested Solutions

Question 4 (5 marks) a. If the car is about to lose contact with the track at point Y, then N = 0. mv ΣF = -------- = mg + N r mv –1 So if N = 0 then -------- = mg and v = rg = ( 0.15 × 10 ) = 1.22 m s . r At point Y, total energy = E k + U g 1 2 =  -- × 0.08 × 1.22  + ( 0.08 × 10 × 0.3 ) = 0.30 J 2  Therefore at X, total energy = 0.3 = U g 0.3 = 0.08 × 10 × h h = 0.38 m b. mv ΣF = -------- = mg – N r mv So N = mg – -------r ( 0.08 × 1.9 ) = ( 0.08 × 10 ) – -----------------------------0.5
2 2 2 2 2

1 mark

1 mark

1 mark

1 mark 1 mark

= 0.22 N Question 5 (11 marks) a. R = ( 6.4 × 10 ) + ( 900 × 10 ) = 7.3 × 10 m 2πr v = -------T ( 2 × π × 7.3 × 10 ) = --------------------------------------------( 96.2 × 60 ) = 7.9 × 10 m s b. 4π r M E = ------------2 GT ( 4 × π × ( 7.3 × 10 ) ) = --------------------------------------------------------------------– 11 2 ( 6.67 × 10 × ( 96.2 × 60 ) ) = 6.75 × 10
24 2 6 3 2 3 3 –1 6 6 3 6

1 mark

1 mark 1 mark

1 mark 1 mark

kg

4

TEVPHYU34_SS_2013.FM

VCE Physics Units 3 & 4 Trial Examination Suggested Solutions

c.

total energy provided = kinetic energy in orbit + energy to overcome gravitational field 1 2 E k = -- mv 2 1 3 2 9 = -- × 85 × ( 7.9 × 10 ) = 2.7 × 10 J 2 energy per kg = area under graph = 12.5 squares × ( 2 × 0.3 × 10 ) = 7.5 × 10 J kg total U g change = area under graph × mass = 7.5 × 10 × 85 = 6.37 × 10 J
6 8 6 6 –1

1 mark 1 mark

1 mark
9

total energy required = 2.7 × 10 + 6.37 × 10 = 3.3 × 10 J
9 8

1 mark 1 mark 1 mark

d.

The apparent weight of Sputnik I is 0 N. The only force acting on the satellite is gravity, hence it is in free fall and has no ‘reaction force’, and hence there is no apparent weight.

.

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VCE Physics Units 3 & 4 Trial Examination Suggested Solutions

Area of study 2 – Electronics and photonics Question 1 (4 marks) a. Resistors X and Y are in a series and therefore have a total resistance of 60 Ω, which in turn is in parallel with the 30 Ω resistor Z. Using the parallel resistance formula gives 1 1 1 -- = ----- + ----R 60 30 R = 20 Ω b. Resistors X and Y are in series and therefore form a voltage divider circuit. As both resistors are of the same value (30 Ω), each must have a potential difference of 10 V. Using V = IR gives a current through resistor X of 0.33 A. Question 2 (5 marks) a. forward bias This can be deduced from the direction of the current through the diode. b. The graph shows that the voltage drop across the LED is 1.2 V. Therefore V OUT = 1.2 V c. The voltage drop across the LED is 1.2 V. This means that there is a 10.8 V drop across the 150 Ω resistor, as the resistor and the diode form a voltage divider circuit. V 150 10.8 current = --------- = --------- = 0.072 A = 72 mA R 150 150 Using V = IR gives a current through the 150 Ω resistor of 0.072 A = 72 mA. Question 3 (5 marks) a. The resistance axis scale is logarithmic and therefore • • b. c. R at 0°C = 10 kΩ R at 25°C = 1.0 kΩ 1 mark 1 mark 2 marks 1 mark 2 marks 1 mark 1 mark 1 mark 1 mark

1 mark 1 mark

( 1 kΩ ) ( 12 V ) Using the voltage divider formula V OUT = ------------------------------------ = 3.0 V ( 3 kΩ + 1 kΩ ) C. The resistance of the thermistor increases as the temperature of the room decreases. Applying the voltage divider formula means V OUT will be larger.

1 mark

Question 4 (8 marks) a. inverting amplifier The gradient of the V OUT versus V IN graph is negative. 1 mark 1 mark

6

TEVPHYU34_SS_2013.FM

VCE Physics Units 3 & 4 Trial Examination Suggested Solutions

b.

V OUT gain = ----------V IN 3V = ---------------– 6 mV = 500 1 mark 1 mark

c.

The output signal is shown below

VOUT

0 –1 –2 –3

10

20

t (ms)

clipping at –3 V
4 marks 1 mark for correct vertical axis scale 1 mark for correct period 1 mark for correct orientation (inverted) 1 mark for showing correct clipping Question 5 (3 marks) a. reverse bias This can be deduced from the direction of the current through the photodiode when light shines on it. b. When Φ = 0.3 W m , the current through the photodiode is 30 µA.
–2

1 mark 1 mark 1 mark

Using Ohm’s law, V = IR = 30 × 10 Question 6 (2 marks)

–6

× 1.0 × 10 = 3.0 V

5

The representation of the intensity modulated wave produced by the encoder is shown below.

brighter information

dimmer

encoder intensity modulated carrier

carrier wave
2 marks

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VCE Physics Units 3 & 4 Trial Examination Suggested Solutions

UNIT 4 Area of study 1 – Electric power Question 1 (4 marks) a. As there are two identical bar magnets the resultant magnetic field B will be shown as

X N S B S Y N
2 marks b. The magnitude of the combined magnetic field of these two bar magnets will be given by the theorem of Pythagoras as 40 2 mT = 56.6 mT 2 marks

P

Question 2 (19 marks) a. Chris is correct – a DC motor needs a split-ring commutator to work properly. 1 mark

A split-ring commutator allows the current to reverse every half-cycle, which keeps the motor turning in the same direction. 1 mark b. F = NIlB = ( 100 ) ( 2.0 ) ( 0.1 ) ( 1.0 ) = 20 N The magnitude of the force acting on side CD = 20 N Note: Use V = IR to determine the current through the loop. c. D. The direction of the force acting on side CD is given by the left-hand FBI rule or the right-hand slap rule and is out of the page. A, B, D. To make the motor turn faster requires a larger force ( F = NIlB ). This can be achieved by an increase in the number of coils of wire, an increase in the strength of the magnetic field or an increase in the voltage supply. An increase in the resistance of the coils of wire will result in a smaller current and a slower speed. e. The magnetic flux (Φ) when the coil is at right angles is given by Φ = BA = ( 1.0 ) ( 0.1 ) ( 0.05 ) Wb = 5.0 × 10
–3

2 marks

2 marks

d.

2 marks

Wb

3 marks 2 mark for correct magnitude of magnetic flux 2 1 mark for correct unit (Wb or T m )

8

TEVPHYU34_SS_2013.FM

VCE Physics Units 3 & 4 Trial Examination Suggested Solutions

f.

The average emf (ε) produced by the coil during one-quarter of a revolution is given by ∆NBA ( 100 ) ( 5.0 × 10 ) ε = -------------- = ------------------------------------------ = 40 V ∆t ( 0.0125 )
–3

2 marks

100 ( e. ) Note: consequential on answer to part e., i.e. -------------------( 0.0125 ) g.

emf (ε)

0

25

50

t (ms)

3 marks 2 marks for a sine graph (polarity does not matter) 1 mark for correct period h. .

emf (ε)

0

25

50

t (ms)

3 marks 2 marks for correct modulus sine graph (polarity does not matter) 1 mark for correct period Question 3 (4 marks) a. The RMS current being drawn by the 2400 W electric kettle is given by P 2400 I = -- = ----------- = 10.0 A V 240 b. V peak = 240 2 = 339 V . 2 marks 2 marks Only award 2 marks if working is shown

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VCE Physics Units 3 & 4 Trial Examination Suggested Solutions

Question 4 (10 marks) a. The RMS current of the 20 kW generator can be determined using P = IV. P 20 000 I = -- = ---------------- = 83.3 A V 240 b. The number of turns in the secondary coil of step-up transformer T1 is 200. This can be determined • • N1 V1 from the transformer ratio formula  ----- = -----  , or  N2 V2  by using a symmetry argument with the transformer T2 at the other end. 2 marks c. P 20 000 I = -- = ---------------- = 8.3 A V 2400 The RMS current in the transmission wires is 8.3 A. d. P = I R = ( 8.3 ) ( 5.0 ) = 345 W The power lost in the transmission wires is 345 W. 2 marks Note: If the current is not rounded down to 8.3 A the answer may be slightly higher to receive full marks. e. A. There will be a voltage drop across the transmission wires, therefore the input voltage to T2 will be less than 2.4 kV and T2 will transform it to less than 240 V. 2 marks
2 2

2 marks

2 marks

10

TEVPHYU34_SS_2013.FM

VCE Physics Units 3 & 4 Trial Examination Suggested Solutions

Area of study 2 – Interactions of light and matter Question 1 (6 marks) a. W = hf0 = 4.14 × 10
– 15

× 5.4 × 10 = 2.24 eV
– 15

14

1 mark 1 mark

E k ( max ) = hf – W = ( 4.14 × 10

× 7.0 × 10 ) – 2.24 = 0.66 eV
14

The stopping voltage is equal to the value of E k ( max ) in eV, so stopping voltage is equal to 0.66 V. 1 mark b. D. Increasing the intensity of the light does not change the energy of the photons or the behaviour of the metal, hence there will be no change in stopping voltage, work function or threshold voltage. As there are more photons incident on the metal, there will be more photoelectrons ejected and a greater photocurrent. W 4.3 14 f 0 = ---- = --------------------------------- = 10 × 10 Hz h ( 4.14 × 10 –15 ) This will be the x-intercept of the graph. The gradient of the graph will be the same as the graph for potassium, as this is equal to Planck’s constant.

1 mark

c.

maximum kinetic energy (eV)

3 2.5 2 1.5 1 0.5 0 2 4 6 8
14

10 Hz)

12

14

frequency (×10

2 marks 1 mark for each of cutoff frequency and gradient correctly shown on the graph Question 2 (10 marks) a. The band marked X will be a dark band. 1 mark

The central band (on the dotted line) is shaded and the central band is always a maxima, as constructive interference occurs here. As the band at X is not shaded it must be a minima or a dark band. 1 mark

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VCE Physics Units 3 & 4 Trial Examination Suggested Solutions

b.

1 path difference =  n – --  λ  2 1 700 =  2 – --  λ  2 λ = 4.7 × 10
8 –7

1 mark m 1 mark 1 mark

v 3 × 10 14 f = -- = ----------------------- = 6.4 × 10 Hz λ 4.7 × 10 –7 c. ( number of photons × energy of photon ) P = -----------------------------------------------------------------------------------------------time E = hf = ( 6.63 × 10
18 – 34

) × ( 6.4 × 10 ) = 4.2 × 10
– 19

14

– 19

J

1 mark 1 mark 1 mark

( 1.5 × 10 × 4.2 × 10 ) P = -----------------------------------------------------------60 = 0.01 W d. As the frequency is lowered, the wavelength will be longer. Since spacing of bands is λL proportional to the ratio ------ , the bands will spread further apart. d The greater intensity of light will mean that the bright bands will be brighter.

1 mark 1 mark

Question 3 (5 marks) Since maximum diffraction occurs when the gap width is equal to the wavelength, the spacing of the atoms will be approximately equal to the de Broglie wavelength of the electron. electron energy = 15 eV = 15 × 1.6 × 10 1 2 E = -- mv 2 So 2.4 × 10
– 18 – 19

1 mark 1 mark

J = 2.4 × 10

– 18

J

1 – 31 2 = -- × 9.1 × 10 × v 2
6 –1

v = 2.3 × 10 m s h λ = -----mv ( 6.63 × 10 ) = ---------------------------------------------------------– 31 6 ( 9.1 × 10 × 2.3 × 10 ) = 3.2 × 10
– 10 – 34

1 mark

1 mark 1 mark
– 10

m m.

So atom spacing is approximately equal to 3.2 × 10

12

TEVPHYU34_SS_2013.FM

VCE Physics Units 3 & 4 Trial Examination Suggested Solutions

Question 4 (7 marks) a. hc photon energy = ----λ ( 4.14 × 10 ) × ( 3 × 10 ) = -------------------------------------------------------------–9 ( 344 × 10 ) = 3.61 eV The missing level is –5.13 + 3.61 = –1.52 eV; as this is between the values for n = 3 and n = 5, this is the energy level for n = 4. The 0.42 eV photon will be ejected when the electron falls from n = 4 to n = 3, as the energy difference is equal to 0.42 eV. The 1.5 eV photon must be ejected when electron falls from n = 4 to n = 2, so the energy level for n = 2 is ( – 1.52 – 1.5 ) = – 3.02 eV . 1 mark b. hc λ = ----E ( 4.14 × 10 × 3.0 × 10 ) = ------------------------------------------------------------3.02
– 15 8 – 15 8

1 mark 1 mark

= 4.1 × 10

–7

m
– 34

1 mark 1 mark 1 mark 1 mark

h ( 6.63 × 10 ) – 27 –1 p = -- = --------------------------------- = 1.6 × 10 kg m s –7 λ ( 4.1 × 10 ) c. Yes, the 3.02 eV photon would be absorbed by the ion. If an electron were in n =2 it could absorb the photon and move to n = ∞ (ionisation).

TEVPHYU34_SS_2013.FM

13

VCE Physics Units 3 & 4 Trial Examination Suggested Solutions

Detailed study 5 – Photonics Question 1 C

The reason why a hand-held torch with a filament light globe produces incoherent light is due to the thermal motion of electrons. Question 2 D

Light emitted by the LEDs in the torch is a result of the movement of electrons from the conduction band to the valence band of a semiconductor. Question 3 hc λ = ----E ( 4.14 × 10 ) ( 3.0 × 10 ) = -----------------------------------------------------------2.0 = 6.21 × 10 = 621 nm The wavelength of the light produced by the LED shown in Figure 1 is closest to 620 nm. Question 4 A
–7 – 14 8

C

m

Safety precautions need to be observed when using green laser pointers as they are a very intense non-diverging light source which can be very harmful if shone in the eyes. Question 5 A

Rayleigh scattering is the scattering of light due to variations in impurities in the fibre. Question 6 D

In order of having the most modal dispersion to having the least modal dispersion: step-index (significant modal dispersion); graded index (medium modal dispersion); single mode (which has no modal dispersion). Question 7 A

The optical fibre does not work as intended because the construction with the higher refractive index cladding does not allow total internal reflection to occur in the fibre. Question 8 1.42 sin θ c = --------1.46 D

θ c = 76.56°
The critical angle for this optical fibre is 76.56°.

TEVPHYU34_SS_2013.FM

21

VCE Physics Units 3 & 4 Trial Examination Suggested Solutions

Question 9

C
2 2

The numerical aperture (NA) = 1.46 – 1.42 = 0.339

θ a = sin ( 0.339 )
–1

= 19.84° The acceptance angle for this optical fibre is 19.84°. Question 10 B

Only imaging fibre bundles used in key-hole surgery need to be coherent. Question 11 hc E = hf = ----λ ( 6.62 × 10 ) ( 3.0 × 10 ) = -----------------------------------------------------------7 6.0 × 10 = 3.31 × 10
– 19 – 34 8

D

J
–3

5.0 mW = 5.0 × 10 ( 5.0 × 10 ) N = ---------------------------– 19 3.31 × 10 = 1.5 × 10
16 –3

Js

–1

The number of photons per second produced by the laser is closes to 1.5 × 10 .

16

22

TEVPHYU34_SS_2013.FM

Trial Examination 2012

VCE Physics Unit 3
Written Examination

Suggested Solutions

Neap Trial Exams are licensed to be photocopied or placed on the school intranet and used only within the confines of the school purchasing them, for the purpose of examining that school’s students only. They may not be otherwise reproduced or distributed. The copyright of Neap Trial Exams remains with Neap. No Neap Trial Exam or any part thereof is to be issued or passed on by any person to any party inclusive of other schools, non-practising teachers, coaching colleges, tutors, parents, students, publishing agencies or websites without the express written consent of Neap. Copyright © 2012 Neap
ABN 49 910 906 643

96–106 Pelham St Carlton VIC 3053

Tel: (03) 8341 8341 Fax: (03) 8341 8300

TEVPHYU3_SS_2012.FM

VCE Physics Unit 3 Trial Examination Suggested Solutions

SECTION A – CORE

Area of study 1 – Motion in one and two dimensions Question 1 a. N = mg cos θ = 1800 × 10 × cos ( 18 ) N = 1.7 × 10 N b. ΣF = ma = T – friction – mg sin θ Since speed is constant ΣF = 0 0 = T – 300 – ( 700 × 10 × sin ( 18 ) ) T = 2.5 × 10 N c. For whole system, ΣF = 0 = F D – 850 – 300 – ( 2500 × 10 × sin ( 18 ) ) F D = 8.9 × 10 N 15 W Fd 3 Power = ---- = ------ = ( 8.9 × 10 ) × ----t t 8 Power = 1.7 × 10 W Question 2 a. r = 0.8 sin ( 15 ) = 0.21 m 2πr ( 2 × π × 0.21 ) v = -------- = --------------------------------T 1.7 v = 0.77 m s b.
–1 4 3 3 4

1 mark 1 mark

1 mark 1 mark 1 mark

1 mark 1 mark 1 mark

1 mark 1 mark 1 mark

Vertically, ΣF = 0 so mg = T cos θ m × 10 = 5.6 cos ( 15 ) m = 0.54 kg mv ( 0.54 × ( 0.75 ) ) F c = -------- = -------------------------------------r 0.21 (Note: give consequential mark if mass calculated is used here) F c = 1.45 N OR F c = T sin 15° = 5.6 sin 15° = 1.45 N 1 mark
2 2

1 mark 1 mark 1 mark

2

TEVPHYU3_SS_2012.FM

VCE Physics Unit 3 Trial Examination Suggested Solutions

c.

The correct answer is D. mg T = ------------------ so if m increases, T will increase. sin ( 15 ) mv F c = -------- so if m increases, F c will increase. r
2

2 marks

d.

At the top, when the ball is travelling at its minimum speed T = 0 mv ΣF = -------- = mg r v 50 × ------ = ( 50 × 10 ) 0.8 v = 2.8 m s OR At top of flight, v = rg = ( 0.8 × 10 ) v = 2.8 m s
–1
–1 2 2

1 mark 1 mark

1 mark 1 mark

Question 3 a. Kinetic energy of ball = elastic potential energy stored in spring E k = 0.5mv = 0.5 ( 0.02 ) ( 12 ) = 1.4 J U e = 0.5kx so 1.4 = 0.5k ( 0.08 ) K = 450 N m b.
–1 2 2 2 2

1 mark 1 mark 1 mark

Vertically to the highest point: u = 12 sin ( 35 ) , v = 0 , a = – 10 , x = ? v = u + 2ax 0 = ( 12 sin ( 35 ) ) + ( 2 × ( – 10 ) × x ) x = 2.4 m
2 2 2

1 mark 1 mark

c.

To find the speed of ball when it lands: u = 12 sin ( 35 ) , a = – 10 m s , x = 1.2 m , v = ? ( v = u + 2ax = ( 12 sin ( 35 ) ) ) + ( 2 × – 10 × 1.2 ) v = 4.8 m s
–1 2 2 2 –2

1 mark
–1

or – 4.8 m s

–1

so on the way down it will be – 4.8 m s

1 mark

To find the time of flight: v = u + at – 4.8 = 12 sin ( 35 ) + ( – 10 )t so t = 1.17 seconds Note: the time of flight can also be found by solving a quadratic equation but this method is not required. To find the range, x = ut = 12 cos ( 35 ) × 1.17 = 11.5 m 1 mark 1 mark

TEVPHYU3_SS_2012.FM

3

VCE Physics Unit 3 Trial Examination Suggested Solutions

Question 4 a. Pi = Pf ( 600 × 5 ) + ( m × – 2 ) = 1 ( m + 600 ) 1 mark 3000 – 2m = m + 600 m = 800 kg b. m∆v = ΣFt 600 × ( 1 – 5 ) = ΣF × 0.2 ΣF = 1.2 × 10 N c. The change in kinetic energy for each car is a fixed quantity, this is equal to the work done.
4

1 mark

1 mark 1 mark 1 mark

The rigid metal bumper bar will compress less than the rubber bumper bar, hence the car will stop in a shorter distance. 1 mark Since W = Fx , if W is constant and x is decreased, the Force (F) will increase. 1 mark

Note: do not accept answer if it is in terms of momentum change and time rather than energy change and distance. Question 5 a.
3

r GM ---- = -----------2 2 T ( 4π ) r ( 6.67 × 10 ) × ( 6.37 × 10 ) ----------------------------------- = ----------------------------------------------------------------------2 2 ( 24.6 × 3600 ) 4π r = 2.04 × 10 m Altitude = r – radius of Mars = ( 2.04 × 10 ) – ( 3.43 × 10 ) Altitude = 1.7 × 10 m
7 7 6 7 – 11 23

3

1 mark 1 mark

1 mark 1 mark

b.

Apparent weight = 0 N

The only force acting on the satellite is gravity, so it is in ‘freefall’ and the apparent weight is zero (there is no ‘normal’ force which is the apparent weight). 1 mark c. GM ( 6.67 × 10 ) × ( 6.37 × 10 ) g = -------- = ----------------------------------------------------------------------2 6 2 r ( 3.43 × 10 ) g = 3.6 N kg
–1 – 11 23

1 mark 1 mark

Note: no marks to be given unless calculations are shown.
.

4

TEVPHYU3_SS_2012.FM

VCE Physics Unit 3 Trial Examination Suggested Solutions

Area of study 2 – Electronics and photonics Question 1 a. 4.0 Ω Resistors Y and Z are in series and equivalent to 12.0 Ω . The YZ resistor combination is in parallel with resistor X which is 6.0 Ω Using the parallel resistance formula, 1 1 1 1 ----- = ----- + -- = -- , - - R p 12 6 4 which means R p = 4.0 Ω . b. 1.0 A Resistors Y and Z are in series and equivalent to 12.0 Ω and the current through each resistor will be the same as they are in series. The total voltage drop across the YZ combination is 12 V. V 12 I = -- = ----- = 1.0 A R 12 c. 24 W The power dissipated in resistor X is given by P = IV . V 12 I = -- = ----- = 2.0 A R 6 P = IV = 2.0 × 12 = 24 W . Question 2 a. b. c. YES Because the LED is in forward bias. 22.5 mA From the graph the voltage across the LED is 1.5 V. Therefore the voltage across the resistor is 6.0 – 1.5 = 4.5 V. The current through the LED will be the same as the current through the resistor. V 4.5 I = -- = -------- = 22.5 mA R 200 d. A If the 200 Ω resistor is replaced with a 300 Ω resistor, the current through the LED-resistor series combination will be smaller and therefore the LED will glow less brightly than before. 2 marks Question 3 a. 5.0 k Ω The value of the resistance of the thermistor when the temperature in the oven is 100 °C is read directly from the graph. b. 750 Ω The value of the resistance of the thermistor when the temperature in the oven is 400 °C is read directly from the graph (allow a range of 700 Ω –800 Ω ). 1 mark

1 mark 1 mark

2 marks

1 mark 1 mark

1 mark 1 mark

1 mark

1 mark

1 mark

TEVPHYU3_SS_2012.FM

5

VCE Physics Unit 3 Trial Examination Suggested Solutions

c.

6.0 V RT V OUT =  ------------------ V IN  R T + R V 5 =  -----------  12 V  5 + 5 = 6.0 V 2 marks

d.

10.4 V RV V OUT =  ------------------ V IN  R T + R V 5 =  ------------------- 12 V  5 + 0.75 = 10.4 V 2 marks Note that there will be a small allowable range based on consequentials from Question 3.

Question 4 a. C V OUT 3 The gain of the amplifier is given by ----------- = --------- = 100 . V IN 0.03 b. c. Inverting amplifier. The gradient of the graph is negative. The graph of the output voltage versus time graph is marked as follows: 1 mark for correct voltage values and correct period. 1 mark for showing inverting nature of amplifier. 1 mark for showing clipping at the +2 V and –2 V marks.
VOUT
3 2 1

2 marks Note 30 mV = 0.03 V 1 mark

10 –1 –2 –3

20

30

40

t (ms)

3 marks

6

TEVPHYU3_SS_2012.FM

Trial Examination 2012

VCE Physics Unit 4
Written Examination

Suggested Solutions

Neap Trial Exams are licensed to be photocopied or placed on the school intranet and used only within the confines of the school purchasing them, for the purpose of examining that school’s students only. They may not be otherwise reproduced or distributed. The copyright of Neap Trial Exams remains with Neap. No Neap Trial Exam or any part thereof is to be issued or passed on by any person to any party inclusive of other schools, non-practising teachers, coaching colleges, tutors, parents, students, publishing agencies or websites without the express written consent of Neap. Copyright © 2012 Neap
ABN 49 910 906 643

96–106 Pelham St Carlton VIC 3053

Tel: (03) 8341 8341 Fax: (03) 8341 8300

TEVPHYU4_SS_2012.FM

VCE Physics Unit 4 Trial Examination Suggested Solutions

SECTION A Area of study 1 – Electric power Question 1 a. The answer is P. The field lines point in the direction of North. Therefore P is North and Q is South.

P I I

Q

1 mark b. The answer is A. The two electromagnets attract each other as a South meets a North where they meet each other. 2 marks

Question 2 a. Each string of 6 globes in each of the parallel strings has 240 V across it. Therefore one globe has a 40 V potential difference (the 6 globes in the string being in series). 1 mark Using P = IV 10 W = I × 40 V Therefore I = 0.25A b. The resistance of globe G can be calculated using V = IR 40 V = 0.25 × R Therefore R = 160 Ω . 1 mark 1 mark Note: a sensible consequential based on an incorrect answer to Question 2a, but correct working for Question 2b, can be awarded 2 marks. 1 mark 1 mark 1 mark

c.

Each string of six globes in each of the parallel strings draws 0.25 A. Therefore the current being drawn from the mains is I = 3 × 0.25 = 0.75 A .

d.

The answer is D. When globe G fails and no longer works, the bottom row does not work (as this is a series circuit with the other five globes in series with globe G) but the other two rows are the same brightness as before (as they still are in parallel with the 240 V AC supply). 2 marks

2

TEVPHYU4_SS_2012.FM

VCE Physics Unit 4 Trial Examination Suggested Solutions

Question 3 a. b. 0N Side AD of the coil is parallel to the magnetic field therefore no force. The magnitude of the force acting on side AD is given by F = NIlB To determine I use V = IR 12 = I × 4.0 I = 3.0 A F = ( 50 ) ( 3.0 ) ( 0.1 ) ( 2.0 ) = 30 N c. d. e. f. The answer is B. The direction of the force acting on side AD is down. The direction of the force is given by the right-hand slap rule or the left-hand FBI rule. When the switch is closed the DC motor rotates a quarter of a cycle anticlockwise (as seen from the front) and then comes to a stop standing up vertically. 2 marks 2 marks 1 mark 1 mark 2 marks

The student-built DC motor could be made to work properly if a split-ring commutator was placed between the power supply and the motor. 2 marks The purpose of the curved magnets in a DC motor is to provide a maximum torque for a longer period of time as the magnetic field is perpendicular to the current for a longer period of time than with a straight magnet (see diagram below).

straight pole pieces

curved pole pieces
2 marks

Question 4 a. The magnetic flux Φ B for exactly one quarter of a rotation after the coil is parallel to the magnetic field will be the maximum magnetic flux given by Φ = BA = ( 2.0 ) ( 1.0 ) = 2.0 Wb
2

2 marks 1 mark Note: must have the correct unit Wb or T m for the full 3 marks

TEVPHYU4_SS_2012.FM

3

VCE Physics Unit 4 Trial Examination Suggested Solutions

b.

The magnitude of the average emf generated by the generator in one quarter of a revolution is given by ∆Φ ε = – N ------∆t ( 100 ) ( 2.0 ) = – ------------------------1  -----   40  = – 8000 V ε = 8.0 kV 1 mark 1 mark 2 marks

c. d.

The answer is B. The voltage variation of the DC generator with a split-ring commutator is rectified AC.

The answer is D. The voltage variation of the generator with two slip rings will be alternating current (AC). 2 marks

Question 5 a. The current in the transmission wires can be calculated using P = IV ( 1.0 × 10 ) I = -------------------------250 = 40 A 1 mark b. The power lost in the transmission lines can be calculated using p=I R = ( 40 ) ( 2.0 ) = 3200 W 1 mark 1 mark c. There is a voltage drop given by V = IR = ( 40 ) ( 2.0 ) = 80 V across the transmission lines. Therefore the voltage available at the lighting system is 250 V – 80 V = 170 V . d. 1 mark 1 mark
2 2 4

1 mark

The scientists can use the 2.4 km of the same type of wire to create 4 parallel strands of wire in each direction (three new ones plus the one already there) with an equivalent total resistance of 0.5 Ω . Use the parallel resistance formula for four 2.0 Ω resistors). 1 mark This then means the power loss will be only 800 W instead of the 3200 W with the original wiring, therefore making more power available for the penguin lighting system. 1 mark

_____________________________________________________________________________________

4

TEVPHYU4_SS_2012.FM

VCE Physics Unit 4 Trial Examination Suggested Solutions

Area of study 2 – Interactions of light and matter Question 1 a. W = hf0 = 4.14 × 10 W = 1.7 eV b. ∆E k = hf – W = ( 4.14 × 10
– 15 – 15

× 4.0 × 10

14

1 mark 1 mark

× 6.0 × 10 ) – 1.7
– 19

14

1 mark
– 19

∆E k = 0.828 eV = 0.828 × 1.6 × 10

J = 1.3 × 10

J

1 mark 1 mark 1 mark 2 marks

1 1 2 – 19 – 31 2 ∆E k = -- m v so 1.3 × 10 = -- × ( 9.1 × 10 ) × v 2 2 v = 5.4 × 10 m s c.
5 –1

The correct answer is C. When the intensity of light is reduced but the frequency remains the same, there will be less photons to absorb and hence less photoelectrons ejected each second. As the frequency has not changed, the energy provided by each photon, and hence the maximum kinetic energy of the photons, will not change.

d.

Each photon of light has a discrete amount of energy, E = hf . If the photon energy is less than the minimum energy required for an electron to escape from the surface of the metal (the work function) no photoelectron will be emitted. If light were a continual wave the electron could collect energy from light of any frequency until it had sufficient energy to escape, hence a wave model cannot explain this phenomena.

1 mark 1 mark

1 mark

Question 2 a. path difference 1.75 × 10 ----------------------------------- = -------------------------- = 2.5 therefore destructive interference occurs. –9 wavelength 700 × 10 1 mark  n – 1  = 2.5 so n = 3, P is on the third dark band on the left side (close to S ). -1  2 P drawn correctly on diagram as shown below. 1 mark 1 mark
–6

P

Note: subtract 1 mark if P is on the right of the central band rather than the left. b. The correct answer is B. λL ∆x = ------ so if the wavelength is increased, the spacing of the bands will increase. This means d that both light and dark bands will get wider. 2 marks

TEVPHYU4_SS_2012.FM

5

VCE Physics Unit 4 Trial Examination Suggested Solutions

c.

h 6.63 × 10 p = -- = ---------------------------λ 700 × 10 – 9 p = 9.5 × 10
– 28

– 34

1 mark
–1

kg m s

1 mark

d.

An individual photon/particle cannot cancel itself out or destructively interfere with itself, so a particle model cannot be used to explain the destructive interference. 1 mark Interference can only be explained by a wave model so this is evidence for the wavelike nature of light. 1 mark

Question 3 The correct answers are A and C. 2 marks

For X-rays and particles to produce the same diffraction pattern, the de Broglie wavelength of the particle h must be the same as the wavelength of the X-rays. Since for both particles and photons, λ = -- , if the p wavelength is the same then they must have the same momentum also.

Question 4 a. hc ∆E = ----λ 4.14 × 10 × 3 × 10 So 10.4 = ---------------------------------------------------λ λ = 1.2 × 10 b.
–7 – 15 8

1 mark 1 mark

m
– 15

∆E = hf = ( 4.14 × 10

) × ( 5.0 × 10 ) = 2.1 eV

7

1 mark

This corresponds to the energy difference between n = 4 and n = 3 so the energy transition is from n = 4 to n = 3 . 1 mark

6

TEVPHYU4_SS_2012.FM

VCE Physics Unit 4 Trial Examination Suggested Solutions

Question 9

D

The undulator rapidly changes the direction of electrons so that they emit energy in the form of electromagnetic radiation. Question 10 2d sin ϑ = nλ 2 × d × sin ( 17 ) = 2 × 1.5 × 10 d = 5.1 × 10
– 10 – 10

C

m B

Question 11

Other maxima will be detected for angles where n = 1, 2, 3 etc. in the equation 2d sin ϑ = nλ When n = 1, ϑ = 8.5° , when n = 3, ϑ = 26.2° , when n = 4, ϑ = 36° so the only option with two of these angles is option B, 8.5° and 26.2° . Question 12 hc E = ----λ 4.14 × 10 × 3.0 × 10 For the incident photons, 1.5 = -------------------------------------------------------λ λ = 8.3 × 10
– 10 – 15 8

D

m

Since Compton scattering occurs, the collision is inelastic and energy is lost, so the scattered photons will have less energy and a longer wavelength than the incident photons. Thus the wavelength must be greater than 8.3 × 10
– 10

m , i.e. option D.

Detailed study 2 – Photonics (24 marks) Question 1 B

On the list given only the laser produces coherent light. Question 2 C

On the list given only the laser produces monochromatic light. Question 3 B

An LED produces its light via the spontaneous emission of photons.

8

TEVPHYU4_SS_2012.FM

VCE Physics Unit 4 Trial Examination Suggested Solutions

Question 4

D

hc The energy gap for blue LED is given by E = ----λ ( 4.14 × 10 ) ( 3.0 × 10 ) E = ------------------------------------------------------------ = 2.86 eV –7 4.35 × 10 Note: Make sure the correct value for Planck’s constant is used. Question 5 C
– 15 8

When the current in the LED is 20 mA this means that the voltage drop across the LED is 2.0 V. Therefore there is a 4.0 V voltage drop across the resistor. Use V = IR to get R = 200 Ω . Question 6 A

Both LEDs will work but each one will shine less than L mW. These two LEDs in series will have a combined voltage drop of 4.0 V across them. Therefore the resistor R will now only have a 2.0 V voltage drop across it. Using V = IR the current will now be half of what it was originally (that is it will go from 20 mA to 10 mA) and hence the LEDs will shine less brightly than before. Question 7 D

The optic fibre shown is a stepped-index fibre. Question 8 1.37 sin θ c = --------1.41 θ c = 76.3° Question 9 D B

The critical angle for this one step-index multimode optic fibre is given by

The acceptance angle for this one step-index multimode optic fibre is given by sin α = 1.41 – 1.37 α = 19.5° Question 10 A
2 2

Attenuation in an optic fibre communication system is the loss of optical power of the signal along the optical fibre. Question 11 B
2.5

The optical fibre loses half of its signal strength for every 2.0 km travelled along the optical pipe. At a 5.0 km distance it will be 40 mW (0.5) = 7.07 mW .

TEVPHYU4_SS_2012.FM

9

VCE Physics Unit 4 Trial Examination Suggested Solutions

Question 12

D

Light travelling down a graded-index fibre bends as it travels through the different layers of the core. It does not reflect sharply at an intersection between the core and cladding as happens with step-index fibres, therefore P is a step index multimode fibre and Q is a graded-index multimode fibre.
.

Detailed study 3 – Sound (24 marks) Question 1 B

The sound produced is a longitudinal wave, which means that the particle at point P will oscillate parallel to the direction of wave motion (left to right) at the same frequency as the wave. Question 2 A

v 330 λ = - = -------- = 0.6 m f 550 Question 3 B

 2.5 × 10 –7 I dB = 10log 10  --- = 10log 10  -----------------------  = 54 dB – 12  I 0  10  Question 4 D

If the intensity is halved, the dB level is decreased by 3 dB. If it is halved again (so quartered in total) the dB level will decrease by another 3 dB, resulting in a total decrease of 6 dB. Question 5 D

The dynamic loudspeaker operates due to electromagnetic principles. A changing current in the coil produces a changing magnetic field. This coil is wrapped around a magnet attached to the speaker cone, and the interaction of the two magnetic fields causes the speaker cone to move in and out. Question 6 A

The baffle eliminates destructive interference between the waves produced by the front and back of the speaker cone, as they are out of phase. The ports produce a path difference so that for low frequencies the waves produced by the front and back of the cone meet in phase and so interfere constructively. This improves the frequency response of the speaker for low frequencies. Question 7 B

nv For a closed end pipe, f n = ----- . 4L n × 336 360 = ----------------4 × 0.7 n = 3 so this is f 3 or the first overtone for a closed end pipe (it can only produce the odd harmonics).

10

TEVPHYU4_SS_2012.FM

Trial Examination 2011

VCE Physics Unit 3
Written Examination

Suggested Solutions

Neap Trial Exams are licensed to be photocopied or placed on the school intranet and used only within the confines of the school purchasing them, for the purpose of examining that school's students only. They may not be otherwise reproduced or distributed. The copyright of Neap Trial Exams remains with Neap. No Neap Trial Exam or any part thereof is to be issued or passed on by any person to any party inclusive of other schools, non-practising teachers, coaching colleges, tutors, parents, students, publishing agencies or websites without the express written consent of Neap. Copyright © 2011 Neap
ABN 49 910 906 643

96–106 Pelham St Carlton VIC 3053

Tel: (03) 8341 8341 Fax: (03) 8341 8300

TEVPHYU3_SS_2011.FM

VCE Physics Unit 3 Trial Examination Suggested Solutions

SECTION A – CORE

Area of study 1 – Motion in one and two dimensions Question 1 At 5.0 m, speed is constant so ΣF = 0 . Therefore F d (driving force) is equal to the frictional force at 5.0 m, and so is equal to 20 N. At 2.0 m, ΣF = 20 – 9 = 11 N . ΣF = ma 11 = 2.0a a = 5.5 m s Question 2 ∆E k = work done on car – work done to overcome friction Work done to overcome friction = area under graph = 53 squares × 1 = 53 J (accept 52.5 – 53.5 J) 1 2 ∆E k = -- mv = ( Fxd ) – 53 2 1 -- × 2 × v 2 = ( 20 × 5 ) – 53 2 v = 6.9 m s Question 3 Centripetal force will act towards the centre of the circle.
Remote control car
–1 –2

1 mark 1 mark

1 mark

1 mark

1 mark 1 mark

1 mark

0.8 m

Question 4 mv F c = -------r
2

(2 × v ) 4.5 = -----------------0.8 v = 1.3 m s
–1

2

1 mark 1 mark

2

TEVPHYU3_SS_2011.FM

VCE Physics Unit 3 Trial Examination Suggested Solutions

Question 5 mv Since F c = -------- by the maximum speed at which the car can travel in a circle of radius r can be r increased by increasing F c . If the curve is banked, there is a component of the normal force acting towards the centre of the circle. Centripetal force will be equal to the sum of the sideways frictional force and the component of the normal towards the centre of the circle, and hence increase the speed at which the car can travel in the circular path. Question 6 Apparent weight is equal to the normal force acting on the car. 2 mv At the top of the speed hump, F c = -------- = mg – N . r 2 ( 2 × 2.5 ) N = ( 2 × 10 ) – ----------------------1.5 N = 12 N Question 7 Conservation of momentum: p i = p f ( 2.5 × 2 ) = ( – 1.2 × 2 ) + ( 1 × m ) m = 7.4 kg Question 8 m ∆v = ΣFt 2 ( – 1.2 – 2.5 ) = ΣF ( 0.05 ) ΣF = 148 N = 1.5 × 10 N Direction is West Question 9 1 2 Initial kinetic energy = -- mv 2 1 2 = -- × 2 × 2.5 2 = 6.25 J 1 1 2 2 Final kinetic energy =  -- × 2 × 1.2  +  -- × 7.4 × 1  2  2  = 5.14 J Since kinetic energy has been lost, the collision is inelastic. Therefore Matilda is not correct. 1 mark 1 mark 1 mark
2 2

1 mark 1 mark

1 mark

1 mark 1 mark

1 mark 1 mark

1 mark 1 mark 1 mark

TEVPHYU3_SS_2011.FM

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VCE Physics Unit 3 Trial Examination Suggested Solutions

Question 10 Horizontally, x = ut cos θ = 2 × 0.43 × cos ( 37° ) x = 0.67 m Question 11 Vertically to the top of the flight, u = 2 sin ( 37° ) m s , v = 0 and a = – 10 m s . v = u + 2ax 0 = ( 2 sin ( 37° ) ) + ( 2 × – 10 )x x = 0.07 m Question 12 Vertically for the whole flight, u = 2 sin ( 37° ) m s , t = 0.43 sec and a = – 10 m s . 1 2 x = ut + -- at 2 1 2 = ( 2 sin ( 37° ) × 0.43 ) +  -- × – 10 × 0.43  2  x = – 0.4 m , therefore the height of the ramp is 0.4 m. Question 13
–1 1 From the graph, k = --------- = 100 N m . 0.01 –1 –2 2 2 2 –1 –2

1 mark 1 mark

1 mark 1 mark

1 mark

1 mark 1 mark

1 mark

1 2 1 2 U e = E k hence -- kx = -- mv . 2 2 1 1 -- × 100 × ( 5.0 × 10 –3 ) = -- × 0.3 × v 2 2 2 –1 v = 0.09 m s Question 14 T = 2 × 60 × 60 = 7200 seconds GM r ---- = -------2 2 T 4π r - ( 6.67 × 10 × 5.98 × 20 ) ------------- = -----------------------------------------------------------------2 2 7200 4π r = 8.1 × 10 m Altitude = r – radius of earth = ( 8.1 × 10 ) – ( 6.37 × 10 ) = 1.7 × 10 m 1.7 × 10 m = 1.7 × 10 km
6 3 6 6 6 6 3 – 11 24 3

1 mark 1 mark

1 mark 1 mark

1 mark 1 mark

4

TEVPHYU3_SS_2011.FM

VCE Physics Unit 3 Trial Examination Suggested Solutions

Question 15 The apparent weight of the satellite is equal to 0 N. 2 marks This is because the only force acting on the satellite in orbit is the force of gravity, and so the satellite is in ‘free fall’ and its apparent weight is zero. Question 16 A

The area under the graph shows the increase in gravitational potential energy and corresponding decrease in kinetic energy as the object moves from position A to position B. 2 marks
.

Area of study 2 – Electronics and photonics Question 1 An ammeter is a device that has a very low internal resistance. The ammeter is connected in series and not connected in parallel with the LED and resistor R as it will short circuit the LED and resistor R combination when placed in parallel. This will normally burn out the fuse in the ammeter. 2 marks Question 2 15 mA The LED has 1.5 V maximum across it (read information from the graph) and therefore the resistor has 4.5 V across it – the supply voltage minus voltage across LED (6.0 V – 1.5 V). 1 mark Using V = IR
–2 4.5 I = -------- = 1.5 × 10 A = 15 mA 300

1 mark

Question 3 0.0 mA When the LED is reversed in the circuit, the current through the LED will be zero. This is because the LED is now in reverse-bias. Question 4 40 kΩ The resistance of the LDR when the light intensity is 100 Lux is 40 kΩ (read the information directly from the graph). Question 5 12.0V When the light intensity is 300 Lux, the resistance of the LDR is 10 kΩ (read the information directly from the graph). As the variable resistor and the LDR form a voltage divider circuit we can use the voltage 20 divider formula. V OUT = ----------------------------------- = 12.0 V ( 20 + 10 )18 V 1 mark 1 mark 1 mark 1 mark

1 mark

TEVPHYU3_SS_2011.FM

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VCE Physics Unit 3 Trial Examination Suggested Solutions

Question 6 The output voltage V OUT decreases when the light intensity decreases to 100 Lux. The resistance of the LDR when the light intensity is 100 Lux is 40 kΩ . 20 V OUT = ----------------------------------- = 6.0 V ( 20 + 40 )18 V When the light level reaches 100 Lux, the change in voltage (from 12.0 V to 6.0 V) can be used to trigger a circuit to make the lighting turn on. Question 7 300 The gain of the amplifier is given by the gradient of the graph in the linear region. V OUT 9V ----------- = ------------------- = – 300 V IN – 30 mV (Note: the gain is normally given as the magnitude of the gradient of the graph.) Question 8 The amplifier is an inverting amplifier as the gradient is negative. Question 9 In reference to amplifiers clipping means the flattening of the output signal when the input signal is too large. When the amplitude of the input signal is too large, the amplifier is not physically capable of giving the normal voltage gain. 1 mark For the amplifier shown in Figure 6 the input voltage ranges which would be clipped would be those shown as flat straight lines from –30 mV to –50 mV and +30 mV to +50 mV. 1 mark The most common effect of clipping is that the output wave signals do not match the input wave signals to the amplifier and hence there is distortion of the sound output compared to the sound input. 1 mark Question 10 The process of amplitude modulation is the changing of the amplitude of a carrier wave by superimposing the waveform of the input signal. 1 mark Waveform of the input signal + Carrier wave = Amplitude modulated signal 1 mark 2 marks 1 mark 1 mark

1 mark

1 mark Note: Some acceptable drawing of how an amplitude modulated signal is created is required for the full two marks.

6

TEVPHYU3_SS_2011.FM

VCE Physics Unit 3 Trial Examination Suggested Solutions

Question 11 12 Ω Call the resistance of resistor R X . 1 1 1 1 Then, using the parallel resistor formula: ----- = ------ + -- = -- Rp RX 6 4 1 1 ------ = ----R X 12 R X = 12 Ω Question 12 12W V 2 ( 12 ) The power dissipated by resistor R X is given by P = ----- = ------------- = 12 W . R 12 (Note: Only 1 mark for the correct answer without any working and 2 marks for a fully worked consequential from the answer to Question 11.) Question 13 A
2

1 mark

1 mark

2 marks

The power dissipated by resistor R 2X is given by V P = ----R ( 12 ) = -----------24 =6 W 2 marks
2 2

TEVPHYU3_SS_2011.FM

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Trial Examination 2011

VCE Physics Unit 4
Written Examination

Suggested Solutions

Neap Trial Exams are licensed to be photocopied or placed on the school intranet and used only within the confines of the school purchasing them, for the purpose of examining that school's students only. They may not be otherwise reproduced or distributed. The copyright of Neap Trial Exams remains with Neap. No Neap Trial Exam or any part thereof is to be issued or passed on by any person to any party inclusive of other schools, non-practising teachers, coaching colleges, tutors, parents, students, publishing agencies or websites without the express written consent of Neap. Copyright © 2011 Neap
ABN 49 910 906 643

96–106 Pelham St Carlton VIC 3053

Tel: (03) 8341 8341 Fax: (03) 8341 8300

TEVPHYU4_SS_2011.FM

VCE Physics Unit 4 Trial Examination Suggested Solutions

SECTION A Area of study 1 – Electric power Question 1 The magnetic field lines between the poles of a bass loudspeaker magnet run from North to South as shown.

S

S

N

S

S

1 mark

Question 2

F

The direction of the magnetic force acting on the current carrying wire is out of the page (use right hand slap or left hand FBI rule). 2 marks Question 3 F = ( N ) ( I )l ( B ) The length of one loop of wire l = ( 2π ) ( 3.0 × 10 ) = 0.188 m F = ( 200 ) ( 0.1 ) ( 2π ) ( 3.0 × 10 F = 3.8 N Question 4 A, B and C
–2 –2

1 mark 1 mark 1 mark

m ) ( 1.0 )

The designer of the bass loudspeaker requires a larger force to act on the speaker cone. As F = ( N ) ( I )e ( B ) , all three of the changes suggested would achieve this: Increase the current in the coil; Increase the number of turns of wire; Increase the strength of the magnetic field. Question 5 The RMS current being drawn by the electric kettle can be determined by using the Electric Power equation. P = IV 2400 W = I × 240 V I = 10.0 A 2 marks 2 marks

2

TEVPHYU4_SS_2011.FM

VCE Physics Unit 4 Trial Examination Suggested Solutions

Question 6 The resistance of the electric kettle can be determined using Ohm’s law, V = IR . 240 V = 10 A × R R = 24 Ω Note: No consequential marks from the wrong current determined in Question 5 as the resistance ----can also be determined from the primary data and the formula P = V . R Question 7 The magnitude of the force acting on side PQ is found using F = ( N ) ( I )l ( B ) . The first step is to determine the current using V = IR which gives I = 6.0 A . F = ( 30 ) ( 6.0 ) ( 0.1 ) ( 1.0 ) = 18 N Question 8 B 1 mark 1 mark
2

2 marks

The direction of the force acting on side PQ is: down (use right hand slap or left hand FBI rule). Note that the current needs to be determined from the battery terminals (the current runs from P to Q). Question 9 The function of the brushes in a DC motor is to provide a sliding electrical contact between the battery terminals and the split-ring commutator. (Carbon brushes are often used as they both conduct electricity and also minimise frictional contact between the slip-rings and the brush.) Question 10 The purpose of the curved magnets in the DC motor is to: 1. Provide a magnetic field for the current carrying coils of wire. 2. Create a magnetic field that is perpendicular to the current carrying coils of wire for a longer time period than a straight magnet. This improves the overall torque.

2 marks

2 marks

2 marks Question 11 The magnetic flux ΦB is given by Φ B = BA = 1.8 Wb or Tm
2

2 marks

Note: 1 mark for correct numerical answer and 1 mark for correct unit. Question 12 The purpose of the slip-rings in an AC generator is to allow the alternating current produced by the generator to be connected to an external circuit. 2 marks Question 13 D 2 marks
3

The voltage variation of the AC generator as seen at V OUT gives a sinusoidal graph.
TEVPHYU4_SS_2011.FM

VCE Physics Unit 4 Trial Examination Suggested Solutions

Question 14

B

The voltage variation of the generator with a single split-ring commutator as seen at V OUT gives a rectified DC output. 2 marks Question 15 The current through the carbon fibre wire when the electric blanket is set on maximum heat is P = IV 60 W = I × 24 V I = 2.5 A Question 16 0.25 A The current drawn from the mains electricity when the electric blanket is set on maximum heat can be determined by transformer radio which in this case is 24 V : 240 V or 1:10. 2 marks Question 17 The total resistance of the carbon fibre wire when the electric blanket is set on maximum heat can V be determined by using P = ----- . R V R = ----P
2 2

2 marks

( 24 ) = ------------ = 9.6 Ω 60 Question 18 D

2

2 marks

For Yao to turn his electric blanket down from the maximum heat setting (HIGH) 60 W to the lowest heat setting (LOW) 15 W, the total resistance of the carbon fibre wire has to be increased by 2 V a factor of four using P = ----- . 2 marks R Question 19 The current in the transmission wires for a 500 MW system transmitting at 500 kV is calculated using P = IV . 5.0 × 10 = I × 5.0 × 10
3 8 5

I = 1.0 × 10 A = 1.0 kA Question 20 4:1 The power-loss ratio for the 500 MW electrical power being transmitted at 250 kV rather than 2 500 kV is calculated using P LOSS = I R . The resistance of the transmission network remains constant. As the voltage is halved, the current doubles for the same amount of power to be transmitted and therefore the P LOSS increases by a factor of 4.

2 marks

2 marks

4

TEVPHYU4_SS_2011.FM

VCE Physics Unit 4 Trial Examination Suggested Solutions

Area of study 2 – Interactions of light and matter Question 1 f h p = -- = h c λ = ( 6.63 × 10 = 1.7 × 10 Question 2 v λ=f
–7 3.0 × 10 = ----------------------- = 3.9 × 10 m 14 7.8 × 10 8 – 34

( 7.8 × 10 ) ) × ---------------------------8 ( 3.0 × 10 )
–1

14

1 mark 1 mark

– 27

kg m s

1 mark

Path difference = ----------------------- = 1.5 so path difference = 1.5 λ ----------------------------------- 5.8 × 10 –7 λ 3.9 × 10 Thus the point on the graph is the 2nd nodal line which is point A. Question 3 The correct answer is A. (See Question 4 below for explanation.) Question 4 Interference can only be described using a wave model. When the wave passes the slits, it diffracts and the two resultant circular waves interfere, causing constructive interference if they meet in phase, or destructive interference when they meet out of phase. Even with only one photon in the apparatus at a time, the wave-like nature of light will enable the interference pattern to be produced. Question 5 E k(max) = qV 0 = 1.6 × 10 = 1.6 × 10
– 19 – 19

–7

1 mark 1 mark

1 mark

1 mark

1 mark 1 mark

J
– 19

1 mark = ( 6.63 × 10
– 34

E k(max) = hf – hf 0 so 1.6 × 10 f 0 = 5.4 × 10 Question 6
14

) ( ( 7.8 × 10 ) – f 0 )

14

1 mark 1 mark

Hz

The stopping voltage will remain the same. Increasing the intensity of the light increases the number of photons and hence the number of photoelectrons emitted. But the energy of individual photoelectrons remains unchanged so the stopping voltage will be the same.

1 mark 1 mark

1 mark
TEVPHYU4_SS_2011.FM

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VCE Physics Unit 4 Trial Examination Suggested Solutions

Question 7 1 2 E k = qV = -- m v 2 ( 1.6 × 10
– 19

1 3 – 31 2 ) ( 1.2 × 10 ) = -- ( 9.1 × 10 ) ( v ) 2
7 –1

1 mark 1 mark

v = 2.1 × 10 m s Question 8

For the X-ray photons and the electrons to produce the same diffraction pattern, they must have the same wavelength. For the electrons (and hence the photons): h( 6.63 × 10 ) λ = ------ = --------------------------------------------------------mv ( 9.1 × 10– 31 ) ( 2.1 × 107 ) λ = 3.6 × 10 Question 9
– 11 – 34

1 mark 1 mark

m (Note: consequential on Question 7) A

hc If the energy of the electrons decreases, since E = hf = ----- , then their wavelength would increase. λ An increase in wavelength will lead to an increase in the amount of diffraction.

2 marks

Question 10 Photon energy: E = hf = hc ----λ ( 4.14 × 10 ) ( 3.0 × 10 ) = ------------------------------------------------------------ = 3.54 eV –7 ( 3.5 × 10 )
– 15 8

1 mark

For the photon to be absorbed by the sodium ion in its ground state, there must be an excited energy level in the sodium ion which is 3.54 eV higher than the ground state. 1 mark As there is not an energy state at this level (–1.59 eV), then the photon cannot be absorbed. Question 11 The energy difference between the two energy levels that the photon moves between must be equal to 1.08 eV. 1 mark From the diagram, this corresponds to n = 3, to n = 2, (3.02 – 1.94 = 1.08), so the ion was originally in the n = 3 state. 1 mark 1 mark

6

TEVPHYU4_SS_2011.FM

VCE Physics Unit 4 Trial Examination Suggested Solutions

Question 9

B

The undulator produces the brightest light for selected wavelengths – up to 1 million times that produced by a bending magnet, because of the constructive interference that occurs. Question 10 B

For Bragg diffraction to occur, constructive interference must occur between the waves that have been reflected from subsequent layers of atoms in the crystal. Question 11 D 2dsinθ = nλ Assume n = 1 to start with, 2 ( 1.2 × 10
– 10

)sin ( 28 ) = 1λ λ = 6.5 × 10
– 11

m – which is not one of the options.

For n = 2, λ = 5.6 × 10 For n = 3, λ = 3.8 × 10 For n = 4, λ = 2.8 × 10

– 11 – 11 – 11

(which is option C) (which is option B) (which is option A)

Hence all three options (A, B and C) could produce Bragg diffraction in this sample. Question 12 C

When Thomson scattering occurs, the collision between the photon and a particle is elastic. As a result, the wavelength of the scattered photon is the same as the wavelength of the incident photon.

Detailed study 2 – Photonics (24 marks) Question 1 C

The particular light source spectrum shown in Figure 1 is monochromatic (one specific wavelength) and therefore most likely to be produced by a laser. Question 2 A

The predominant colour produced by the particular light source spectrum shown in Figure 1 is at 632.8 nm, which is in the red part of the visible spectrum. Question 3 C

An LED produces its light via the spontaneous emission of photons. Question 4 E = ( 4.14 × 10 = 1.91 eV Note: Make sure you use the correct value for Planck’s constant.
– 15

B ) ( 4.62 × 10 )
14

The energy gap for this tail light LED is given by E = hf.

8

TEVPHYU4_SS_2011.FM

VCE Physics Unit 4 Trial Examination Suggested Solutions

Question 5

A

The physics principle of how an optic fibre works is total internal reflection. Question 6 C

A high-order mode makes many internal reflections compared to a low-order mode. Question 7 --------sinθ C = 1.48 1.51 θ C = 78.6° Question 8 A C

The critical angle for this one step-index multimode optic fibre is given by:

The acceptance angle for this one step-index multimode optic fibre is given by: sinα = ( 1.51 ) – ( 1.48 ) α = 17.4° Question 9 D
2 2

Attenuation in an optic fibre communication system is the loss of optical power of the signal along the optical fibre. Question 10 B

The purpose of a coherent optic fibre bundle as used for medical imaging is the ability to create a precise image of the site under medical examination. Question 11 C

The most likely reason for a significant loss of signal when an optical fibre sensor is placed under such excessive unintentional stress is that, although it does not break, it has excessive bending. The optical fibre cannot totally internally reflect all of the light as some of the incoming light has exceeded the critical angle. Question 12 hc N ----λ P = --------t N = Pλt ------hc = 1.28 × 10
.

C

16

photons per second

TEVPHYU4_SS_2011.FM

9

Trial Examination 2010

VCE Physics Unit 3
Written Examination

Suggested Solutions

Neap Trial Exams are licensed to be photocopied or placed on the school intranet and used only within the confines of the school purchasing them, for the purpose of examining that school's students only. They may not be otherwise reproduced or distributed. The copyright of Neap Trial Exams remains with Neap. No Neap Trial Exam or any part thereof is to be issued or passed on by any person to any party inclusive of other schools, non-practising teachers, coaching colleges, tutors, parents, students, publishing agencies or websites without the express written consent of Neap. Copyright © 2010 Neap
ABN 49 910 906 643

96–106 Pelham St Carlton VIC 3053

Tel: (03) 8341 8341

Fax: (03) 8341 8300

TEVPHYU3_SS_2010.FM

VCE Physics Unit 3 Trial Examination Suggested Solutions

SECTION A – CORE

Area of study 1 – Motion in one and two dimensions Question 1 Momentum is conserved and needs to be treated first in order to find the common speed after the collision. Σmomentum before = Σmomentum after ( Let motion to the right be positive ) P before = P combined after ( 60.0 × 2.00 ) + ( 12 × – 1.00 ) = ( 60.0 + 12.0 ) V common 120 – 12 = 72V common
–1 –1 108 V common = --------- = 1.50 m s = 1.50 m s right 72.0

2 marks for correct method

2 marks 1 mark for correct velocity 1 mark for correct direction

Question 2 The total momentum of the system is conserved independently of whether the collision is elastic or inelastic. Thus, John’s statement is incorrect. Momentum is conserved before the collision is determined to be elastic or inelastic. Thus, Selena’s statement is also incorrect as momentum is conserved before the collision is determined to be elastic or inelastic. In any case, the collision is inelastic which contradicts Selena’s statement. Question 3 1 mark 1 mark

1 mark

Towards the centre of the circle

1 mark Question 4 F net = m4π r --------------2 T 60.0 × 4π × 10.2 F net = -----------------------------------------2 ( 5.11 ) F net = 925 N or 9.25 × 10 N
2 2 2

1 mark 1 mark

TEVPHYU3_SS_2010.FM

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VCE Physics Unit 3 Trial Examination Suggested Solutions

Question 5 At the highest point on the track, Method 1 ΣForces ( horizontally ) = Net Force = Answer Q4 Tension × sin(θ ) = 925 N Tension × sin ( 57 ) = 925 N 925 Tension = -----------------sin ( 57 ) Tension = 1103 N T = 1.10 × 10 N ( consequential answer to Question 4 ) Method 2 ΣForces ( vertically ) = 0 Tension × cos ( θ ) – ( m × g ) = 0 Tension × cos ( 57 ) – ( 60.0 × 10 ) = 0 600 Tension = -------------cos 57 Tension = 1102 N Tension = 1.10 × 10 N Question 6 Work done by net force = net force × distance × cos ( θ ) W = F net × s × cos ( θ ) F net = mass × net acceleration F net = m × a To find acceleration: given u = 0, t = 4.59 sec, s = 10 m s = ut + 0.5at
2 2 3 3

1 mark for method

1 mark for answer

10 = 0 + 0.5 × a × 4.59 a = 0.949 m s
2

1 mark

F net = 30.0 × 0.949 = 28.5 N W = F net × s × cos ( θ ) W = 28.5 × 10 × cos ( 0 ) W = 285 J or 2.85 × 10 J
2

1 mark

1 mark

TEVPHYU3_SS_2010.FM

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VCE Physics Unit 3 Trial Examination Suggested Solutions

Question 7 Apparent Weight = size of normal reaction. ΣForces (vertically) = 0, let up be positive – Force Fred – Weight + Normal reaction = 0 – F Fred – W + N = 0 – 140 sin ( 23 ) – ( 30 × 10 ) + N = 0 N = 140 sin ( 23 ) + 300 N = 355 N N = 3.5 × 10 N Question 8 1 2 Energy stored = area beneath force–extension curve up to 15 cm, or use E = -- kx (k = gradient) 2 200 k = -------- = 2000 N/m 0.1 E stored = 0.5 × 2000 × ( 0.15 ) E stored = 22.5 J E stored = 23 J Question 9 The maximum height reached occurs when the total energy stored in the slingshot converts to gravitational potential energy. Total mechanical energy before firing = Total mechanical energy after firing. 1 2 -- kx = mgh 2 22.5 = 0.020 × 10 × h (consequential to Question 8) 22.5 h = ------------------------10 × 0.020 h = 112.5 h = 113 m or 1.1 × 10 m
2 2 2

1 mark

1 mark

1 mark

1 mark

1 mark

1 mark

1 mark

1 mark

TEVPHYU3_SS_2010.FM

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VCE Physics Unit 3 Trial Examination Suggested Solutions

Question 10 Time of flight = 2 × time taken to get to the maximum height Vertically, let upwards be positive a = – 10 v=0 s = 18 Use s = vt – 0.5at
2 2 2

18 = 0 – 0.5 × 10 × t t = 3.6 t = 1.897 sec

1 mark 1 mark 1 mark

∴time in air = 2 × 1.897 = 3.8 sec Question 11 Horizontally, u cos ( θ ) × time of flight = range 54.0 = u cos ( θ ) × 3.79 54.0 u cos ( θ ) = --------- = 14.248 3.79 Vertically, v = u sin ( θ ) + at 0 = u sin ( θ ) – 10 × 1.897 u sin ( θ ) = 18.97 u sin ( θ - 18.97 ) thus ------------------- = --------------u cos ( θ ) 14.248 tan ( θ ) = 1.331 θ = INVtan ( 1.331 ) = 53° Question 12 (u = initial speed)

1 mark consequential upon Question 10

1 mark

1 mark

1 mark

At position A, the ball experiences friction, which opposes its velocity and it experiences its weight force. Thus the forces acting are Thus the answer is D. 2 marks and . The net force is the sum of these + =

TEVPHYU3_SS_2010.FM

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VCE Physics Unit 3 Trial Examination Suggested Solutions

Question 13 The feeling of weightlessness occurs when the human body makes no contact with any surfaces and so that the normal reaction equals zero. The occupants will feel weightless if the normal reaction acting on them at the top of the ride becomes zero. Let’s take the downwards towards the centre of the circular track as positive. Σ Forces (vertically) = centripetal force Weight – Normal reaction = centripetal force. normal reaction

1 mark

weight

1 mark
2

-------W – N = mv r

-------N = mg – mv r

2

-----------------------N = ( 60 × 10 ) – ( 60 × 12 ) 14.4 N = 600 – 600 = 0 Yes, the occupants will feel weightless. Question 14 Given: Period = T = 102 × 60 = 6120 seconds Mass of the Earth = 5.98 × 10
6 24

2

1 mark

1 mark

kg
– 11

Universal gravitational constant = 6.67 × 10 Radius of Earth = 6.38 × 10 m
3 r GM Earth Use ----- = ----------------2 2 T 4π

Nm kg

2

–2

r 6.67 × 10 × 5.98 × 10 ------------------ = ------------------------------------------------------------2 2 ( 6120 ) 4π
– 11 24

3

– 11

24

1 mark
2

3 6.67 × 10 × 5.98 × 10 × ( 6120 ) r = --------------------------------------------------------------------------------------2 4π

r = 3.78 × 10

3

20 1 -20 3

1 mark

r = ( 3.78 × 10 ) r = 7.23 × 10 m
6

altitude = orbital radius – radius of Earth altitude = 7.23 × 10 – 6.37 × 10 altitude = 8.53 × 10 m altitude = 8.53 × 10 km or 853 km
.

1 mark

6 5 2

6

1 mark
TEVPHYU3_SS_2010.FM

6

VCE Physics Unit 3 Trial Examination Suggested Solutions

Area of study 2 – Electronics and photonics Question 1 The two resistors are in parallel, so the parallel resistance formula is used 1 1 1 3 ----- = ------ + ------ = -- = 0.5 - R P 6.0 3.0 6 R P = 2.0 Ω Question 2 As the two resistors are in parallel with the battery, they both have a 6.0 V drop across them. Therefore the current through the 3.0 Ω resistor can be determined using Ohm’s Law V = IR 6.0 I = -----3.0 = 2.0 A Question 3 The power dissipated in the 6.0 Ω resistor is found using the power formula: P = IV. As the two resistors are in parallel with the battery, they both have a 6.0 V drop across them. Therefore the current through 6.0 Ω resistor is 1.0A (V = IR) P = IV = ( 1.0 ) ( 6.0 ) = 6.0 V Question 4
A LED

1 mark 1 mark

1 mark

1 mark

1 mark

1 mark

6.0 V R

2 marks 1 mark for identifying correct component (circled) 1 mark for label (LED or Light Emitting Diode) Question 5 From the current–voltage characteristic graph the voltage drop across the LED is 1.5 V. Therefore the voltage drop across the resistor, R = 6.0 – 1.5 = 4.5 V V = IR 4.5 = I ( 450 ) I = 1.0 × 10 = 10 mA

1 mark

–2

A 1 mark
7

TEVPHYU3_SS_2010.FM

VCE Physics Unit 3 Trial Examination Suggested Solutions

Question 6 6.0 V When the optoelectronic device is reversed in the circuit it means that no current flows through the circuit. This means that the potential difference across the resistor is 0.0 V and across the diode 6.0 V. Question 7 The value of the resistance of the thermistor when the temperature in the room is 30°C is 100 Ω (read from graph). Question 8 R1 and the thermistor (R2) form a voltage divider circuit. R2 V OUT = V IN ---------------------( R1 + R2 ) 9.0 × 100 = ---------------------------( 200 + 100 ) = 3.0 V Question 9 When the room reaches 18°C the thermistor resistance is 200 Ω (read from graph). R2 V OUT = ---------------------- × V IN ( R1 + R2 ) 200 = ---------------------------- × 9.0 ( 200 + 200 ) = 4.5 V Question 10 The gain of the amplifier (taken from the linear region of the graph) is given by V OUT 3 –3 V Gain = ----------- = ------------- = ------------------ = 500 6 mV 6 × 10 –3 V IN 1 mark 1 mark 1 mark 1 mark 1 mark 1 mark 1 mark

1 mark

2 marks

TEVPHYU3_SS_2010.FM

8

VCE Physics Unit 3 Trial Examination Suggested Solutions

Question 11
VOUT (V) 6.0
4.0 2.0 0 –2.0 –4.0 –6.0 t (ms)

10

20

30

40

3 marks 1 mark for indicating the inverting nature of the amplifier 1 mark for the correct minimum and maximum values of VOUT (3 V and –3 V respectively) 1 mark for showing that the wave form has been clipped Note: Dashed line is not required

Question 12 The beam’s brightness varies through a range of values in carrying the information – this is a characteristic of analogue information. 1 mark

TEVPHYU3_SS_2010.FM

9

VCE Physics Unit 3 Trial Examination Suggested Solutions

SECTION B – DETAILED STUDIES (2 marks for each correct answer) Detailed study 1 – Einstein’s special relativity Question 1 B

Travelling at 0.9 c there would be significant contraction of length in the direction of travel so the square would appear as in B. Question 2 C

1 The Lorentz factor, γ = ----------------2 v 1 – ---2 c 1 = -----------------------------2 ( 0.9c ) 1 – ---------------2 c = 2.3 Question 3 B

Postulate one was: No law of physics can identify a state of absolute rest. Question 4 A

Postulate two was: The speed of light is independent of the motion of the light source or observer. Question 5 B

The purpose of the Michelson–Morley Experiment was to determine the existence of the ether. Question 6 A

The results of the Michelson–Morley Experiment demonstrated the ether does not exist. Question 7 D

1 The Lorentz factor, γ = ----------------2 v 1 – ---2 c 1 = -----------------------------------------2 ( 0.99999c ) 1 – ---------------------------2 c = 707 Question 8 D
–1

Although the space shuttle travels at 28 000 km h us, it is slow relative to the 300 000 km s
–1

in its orbit around the earth and seems relatively fast to

speed of light.

TEVPHYU3_SS_2010.FM

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Trial Examination 2010

VCE Physics Unit 4
Written Examination

Suggested Solutions

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TEVPHYU4_SS_2010.FM

VCE Physics Unit 4 Trial Examination Suggested Solutions

SECTION A Area of study 1 – Electric power Question 1 The magnetic field lines between the poles of the magnet are as shown below. Note: the direction must be correct (North to South).

S

N
1 mark

Question 2

B

2 marks

The direction of the magnetic force acting on the current carrying wire is down (use the right hand slap rule or the left hand FBI rule). Question 3 F = IlB = ( 5.0 ) ( 0.1 ) ( 1.2 ) = 0.6 N Question 4 P = IV 2400 I = ----------240 = 10 A Question 5 The peak voltage for the kettle is 240 2 = 339 V . Question 6 P=I R 2400 R = ----------- = 24 Ω 100 2 marks
2

1 mark 1 mark

2 marks

1 mark

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VCE Physics Unit 4 Trial Examination Suggested Solutions

Question 7 F = NIlB = ( 200 ) ( 2.0 ) ( 0.1 ) ( 1.0 ) = 40 N Question 8 G 1 mark 1 mark 2 marks

There is no force acting on side BC as the current runs parallel to the field lines. Question 9 B 2 marks

DC motors require a split-ring commutator to work properly. Question 10 F 1 mark

The direction of the magnetic field passing through the loop is out of the page (arrow convention). Question 11 Φ B = BA = ( 1.5 × 10 ) ( 0.1 ) ( 0.04 ) = 6.0 × 10 Question 12 ) ε = – ( BA -------------t = ( 6.0 × 10 Wb ) --------------------------------------( 0.2 ) = 3.0 × 10
–4 –5 –5 –2

1 mark 1 mark

Wb

1 mark 1 mark

V

– Ans ( Q11 ) consequential: --------------------------( 0.2 ) Question 13 A 2 marks

As the loop is pulled out of the magnetic field, the direction of the induced current runs from P to Q (use Lenz’s law). Question 14 AC The voltage shown on the graph alternates between positive and negative values of voltage, hence producing an alternating current (AC) through a load resistor. 1 mark 1 mark

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VCE Physics Unit 4 Trial Examination Suggested Solutions

Question 15 X Y Z 17.6 –17.6 0.025 Volts Volts seconds 3 marks 1 mark per correct answer Question 16 NP VP ----- = ----NS VS VP 500 = ----- = 240 -------- -------N S V S 12 N S = 25 turns Question 17 Ideal transformer so that P P = P S P = IV 12 = I ( 240 ) I = 0.05 A Question 18 A 1 mark 2 marks 1 mark

1 mark

1 mark

The transformer is warm because P primary > P secondary . Question 19 P = IV 2.0 × 10 I = --------------------- = 40.0 A 3 5 × 10 Question 20 P=I R = ( 40 ) ( 5 ) = 8.0 kW 2 marks
2 2 5

2 marks

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VCE Physics Unit 4 Trial Examination Suggested Solutions

Area of study 2 – Interactions of light and matter Question 1 B 2 marks

When light passes through each slit it initially undergoes diffraction. Question 2 B 2 marks

Maximum diffraction of light occurs when the slit width is approximately equal to the wavelength of the light. Question 3 Path difference = 1750 nm, wavelength = 700 nm Path difference 1750 ----------------------------------- = ----------- = 2.5 wavelength 700 1 Since destructive interference occurs when PD =  n – --  λ ,  2 n = 3 and so X will be the third dark band from the centre. Marking on diagram the location of point X closer to slit A. 1 mark

1 mark 1 mark

X

A monochromatic light source B bright band

screen

Question 4 Bright bands occur when light waves undergo constructive interference because they are in phase when they meet at the screen. Dark bands occur when light waves undergo destructive interference because they are out of phase when they meet at the screen. This cannot be explained by the particle model as it is not possible for particles to meet and cancel each other out, hence the dark bands cannot be explained. Question 5 W = hf 0 = ( 4.14 × 10 W = 2.5 eV Question 6 V 0 is the stopping voltage. This is the voltage required to stop the most energetic photoelectrons which are ejected from the metal.
TEVPHYU4_SS_2010.FM

1 mark 1 mark 1 mark

– 15

) × ( 6.0 × 10 )

14

1 mark 1 mark

1 mark 1 mark
5

VCE Physics Unit 4 Trial Examination Suggested Solutions

Question 7 E k ( max ) = hf – W = ( 4.14 × 10 0.4 eV = 0.4 × ( 1.6 × 10
– 19 – 15

× 7.0 × 10 ) – 2.5 = 0.4 eV
– 20

14

1 mark 1 mark

) J = 6.6 × 10

J

1 1 2 – 20 – 31 2 E k = -- mv so 6.6 × 10 = -- ( 9.1 × 10 )v 2 2 v = 3.8 × 10 m s
5 –1 – 34

1 mark 1 mark

5 –9 h - ( 6.63 × 10 ) λ = ------ = --------------------------------- × ( 3.8 × 10 ) = 2.0 × 10 m – 31 mv ( 9.1 × 10 )

Question 8

current (mA)

–V0

–V0

0

voltage (V)
1 mark 1 mark

Higher energy photon means more energetic photoelectrons so the magnitude of V 0 should be greater. The intensity is less so less photons are incident on the metal, resulting in less photoelectrons ejected and a lower photocurrent. Question 9 hc E = ----λ ( 5.5 – 1.6 ) = ( 4.14 × 10 λ = 3.2 × 10 Question 10 -- 6.63 × 10–7 p = h = ---------------------------λ 3.2 × 10 = 2.1 × 10
– 27 – 34 –7 – 15

( 3 × 10 ) ) × --------------------λ

8

1 mark 1 mark

m

1 mark
–1

kg m s

1 mark

Question 11 Since the energy of the photon (12 eV) is greater than the ionisation energy of the atom (10.4 eV), the photon will be absorbed. As a result of this, an electron would be ejected from the atom and ionisation would occur. 1 mark 1 mark

TEVPHYU4_SS_2010.FM

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VCE Physics Unit 4 Trial Examination Suggested Solutions

Question 11

C
– 10

For second order diffraction, n = 2 . 2d sin θ = nλ so 2 × ( 2.4 × 10 θ = 32.7° Question 12 B ) × sin ( θ ) = 2 × 1.3 × 10
– 10

When Thompson scattering occurs there is no change in the wavelength as the collision is elastic. When Compton scattering occurs the collision is inelastic and the emerging photons have lower energy, hence lower frequency and longer wavelengths. Question 13 D

In Bragg diffraction, photons are diffracted off a layer of atoms and no energy is transferred. In Compton scattering, a photon-electron collision occurs and some (but not all) of the photon energy is transferred to the electron. In Thompson scattering there is no transfer of energy to the electron. In the photoelectric effect the photon is absorbed and all of its energy is transferred to the electron.

Detailed study 2 – Photonics (26 marks) Question 1 B

The spectrometer would show that the light emitted by a 20 W incandescent light globe is polychromatic (creating a rainbow-like spectrum). Question 2 D

The light emitted by a 20 W incandescent light globe is out of phase (incoherent). Question 3 B

A laser produces light by the stimulated emission of photons. Question 4 D

Laser light is coherent, monochromatic and in phase. Question 5 B

The need for extreme caution when using lasers in the physics laboratory relates to the intensity of the laser light which is extremely concentrated and dangerous for human eyes. Question 6 C

An LED produces red light by spontaneous emission of photons. Question 7 B

The band energy gap is smaller for a red LED compared to a green LED as the red light has a lower frequency (and E = hf).

TEVPHYU4_SS_2010.FM

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VCE Physics Unit 4 Trial Examination Suggested Solutions

Question 8 hc E = hf = ----λ

B

The energy gap required for the production of the green light from the LED is

) = ( 4.14 × 10 ) ( 3.0 × 10 -----------------------------------------------------------–7 ( 5.15 × 10 ) = 2.41 eV Question 9 Using Snell’s law n 1 sin θc = n 2 sin 90° 1.50 θc = sin  ---------   1.52 = 80.7° Question 10
–1 –1 –1

– 15

8

C

B

Using the formula for the acceptance angle θa = sin ( ( n1 ) – ( n2 ) )
2 2

= sin ( 0.2457 ) = 14.2° Question 11 D

Rayleigh scattering in relationship to an optic fibre is best explained as the scattering of light due to variations in impurities in the fibre. Question 12 C

A coherent optic fibre bundle as used for medical imaging purposes is designed so that the individual fibres at both ends of the bundle are in the same position relative to each other to give an exact image. Question 13 C

The 400 optical fibres bundle will give a more detailed image as, in a sense, these individual fibres behave like pixels in the formation of the image. The greater number of pixels per unit area, the greater the detail of the image.
.

TEVPHYU4_SS_2010.FM

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