VCE physics exam revision

• Physics examination questions that received <60% score for 2010-2013.

2013 Exam all topics

20 N

20 J Read from graph

5 J missing

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 the spring is extended 0.5m and at P it is extended 1.5 m

24 hours = 86,400 sec

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.

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.

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

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

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.

When the gradient is ZEROt=0.5, 1.0, 1.5 sec

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.

Tricky: Longest wavelength has the smallest energyE=3.19-2.11 = 1.08 eV

ie. First excited state to ground state

The students have chosen a frequency that is below the threshold frequency for the target metal.

Therefore, no photoelectrons are being ejected.

The intensity will be at a MAXIMUM.The path difference is ZERO.Constructive interference is occuring

Lower frequencyLonger wavelengthMore width in the patternD

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.

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

Newtons motion

Consider the trailer by itself:F = m x aF = 2000 x 0.5 = 1000N

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

800

Still need to overcome friction and acceleration, but only for 2nd log.F = 300 + 400 = 700 N

d= 20m v2=u2+2adu=4 ms-1 v2=16+2(0.5)(20)v=? v=6 ms-1

a=0.5 ms-2

t=

F = µ α∑1.0 = 0.1+ 0.4( )αα= 2 µ σ−2

d = 1.0 µ ϖ2 = υ2 + 2αδυ = 0 µ σ−1 ϖ2 = 0 + 2 ⋅2 ⋅1ϖ= ? ϖ2 = 4

α= 2 µ σ−2 ΚΕ = 12

µ ϖ2 = 12

0.4( )4 = 0.8 ϑ

τ=

1.5 NThe 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.

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.

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.

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.

θ = Tan−1v2

rg

⎛

⎝⎜⎞

⎠⎟

θ = Tan−1102

100 ⋅10

⎛

⎝⎜⎞

⎠⎟

θ = 5.7o

E = 12

κξ2

µ γη = 12

κξ2

1×10 × 0.3 = 12

κ 0.3( )2

κ 0.3( )2 = 6κ = 67 Ν / µ

E = 12

κξ2

⇒ 5.4 = 12

κ 0.08( )2

κ = 1.69 ×103 Ν / µMust use units as m not cm.

F∑ = 0

⇒ µ γ = κξ

κ = µ γξ

= 2 ⋅100.4

= 50 Ν / µ

The system gains Elastic and loses GPE

∆EPE =1

2kx2 =

1

250 0.4( )

2= 4 J

ΔGPE = mgh = 2 ⋅10 ⋅0.4 = 8 J

ΔEnergy = 4 J

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.

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

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.

d = 200 µ ϖ2 = υ2 + 2αδυ = 0 ϖ2 = 0 + 2 10( ) 200( )

ϖ= ϖ= 63.2 µ σ−1

α= 10 µ σ−1

τ=

10

g = ΓΜρ2

Μ = γρ2

Γ=

10 × 3×104( )2

6.67 ×10−11( )= 1.35 ×1020 κγ

T = 4π 2ρ3

ΓΜ

=4π 2 1×109( )

3

6.67 ×10−11( ) 5.7 ×1025( )

= 3.2 ×106 σεχ

T = 4π 2ρ3

ΓΜ

Τ 2 = 4π 2ρ3

ΓΜ

ρ= ΓΜΤ 2

4π 23 =

6.67 ×10−11( ) 7.36 ×1022( ) 7200( )2

4π 23

ρ= 1.86 ×106 µη= 1.86 ×106 −1.74 ×106 = 1.21×105 µ

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.

F = ΓΜµρ2

Φ=6.67 ×10−11( ) 5.98 ×1024( ) 3.04 ×105( )

6.72 ×106( )2 = 2.69 ×106 Ν

T = 4π 2ρ3

ΓΜ

Τ =4π 2 6.72 ×106( )

3

6.67 ×10−11( ) 5.98 ×1024( )

Τ = 5480σεχ

Same

Electronics & photonics

2013 exam

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.

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

2012-2010 exams

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

P=I2RP=12 . 2 = 2 W

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.

1212

V=iR24=ix2i=12 Amp

10 lux ⇒10,000Ω⇒ ς ουτ = 4ς⇒ ς Ρ = 8ςΡ = 20,000Ω

-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.

-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.

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 RΩ R=? must be 4500 ΩVt=1.5 V VR= 4.5 V

4500 Ω

6 V

Thermistor

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

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)

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.

Note: The different threshold voltages of the diodes.

8 - 3 = 5V

V=IR V=IR6=Ix100 5=Ix150I=0.06 A I=0.033 A

93

At 2.5 lux, RLDR =3000 . ΩFor output of 6V or more, Rr= 2000 . For 3:2 ratio.ΩAs light decrease RLDR increases. Switching unit across RLDR.

2000 Ω10 V

Marking scheme1 mark for all components in a diagram.1 mark for 2000 resistorΩ1 mark for switching circuit across the LRD.

-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.

materials & structures

Solve Q8 and Q9 together as simultaneous equations:

120,000 N

Fb

Fs

Take torque about the bolt

20 Fs=30 x 120,000Fs=180,000 N

Add vertical forces

Fb + 120,000 = FsFb=180,000 - 120,000Fb=60,000 N

Q8 D

Q9 B

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

ε = ∆l

lo

⇒ 0.01 =Δl

2 m

∴ Δl = 0.02 m = 2cm

Strain Energy / m3 = Αρεα

= 12

3×108( ) 1.5 ×10−3( ) = 2.25 ×105 ϑ

ς ολυµ ε = 10−3

∴ΣτραινΕνεργψ= 2.25 ×105 ×10−3 = 225 ϑ

Electricity generation and supply

2013 EXAM

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 MAXBy 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.

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

Link ideas together:CLOCKWISEAs 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.

2012 EXAM

P = ς ΙΠ= 900 × 50 = 45 κΩ

R T = 7 +18 = 25Ως = ΙΡ1000 = Ι ⋅25Ι = 40 Αµ π

No the motor will not work correctlyPloss = Ι 2Ρ= 402 ⋅ 25 = 40,000 Ω

Τηισµ εανσονλψ5,000Ω ρεαχηεστηεµ οτορατ40 Αµ π, 125ς

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.

It is also possible to show that the current is 40 Amp while the motor operates correct at 50 Amp

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

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.

ξ = nΔΦt

ξ = nB × Area

t

3.6 = n0.03× 0.3× 0.4( )

0.125n = 125

Best answer was with a sketch:Slip rings give a sin functionCommutator with mod(sin)

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.

2011 EXAM

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

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

Or something similar, first direction is not important

2010 EXAM

ZEROThe loop is parallel to the magnetic field.

4W &2V ⇒2 Αµ πΡγλοβε = 1ΩΡΤ = 5Ω ⇒ς =10

Ploss = Ι 2Ρ= 22 × 4 = 16 Ω

The globe needs 2 V and 2 Amp. The primary side of the transformer will have 20 V and 0.2 Amp

Ploss = Ι 2Ρ= 0.22 × 4 = 0.16 Ω

theories of light and matter

E = ηφ− Ω

Ε = 4.14 ×10−15( ) 7.5 ×1014( ) − 2.28

Ε = 3.105 − 2.28 = 0.83ες

recording and reproduction of sound