Mid Year Exam Student Guide

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2011 J2 H2 Mid-Year Physics Exam

Student Guide

Section A – answer key

Pg 3 Pg 4 Pg 5 Pg 6 Pg 7 Pg 8 Pg 9 Pg 10

1 5 10 13 17 21 24 28

A C C C A A B A

2 6 11 14 18 22 25 29

D B D D C D D C

3 7 12 15 19 23 26 30

A B B C A D B C

4 8 16 20 27

B B C A C

9

C

Section A – key ideas to selected questions (apparently more difficult questions)

(Some of you MAY have got the correct answer BY CHANCE)

1 I = k(a)2

I60o

= k(acos60o)

2

Solve the two equations

3 Two particles on either side of a node are 90o out of phase with each other.When X has downwards displacement, Y has upwards displacement

4 Distance between successive antinodes = half a wavelength

Speed of microwave = speed of light

5 Mass submerged in water, tension in string decreases. Velocity of wave decreases. No

change in frequency. Wavelength decreases.

6 Node at eardrum, Antinode at ear opening

Wavelength = 3.5 x 4 cm

7 At closed end change in amplitude is zero displacement node

pressure antinode change in pressure is maximum

At open end change in amplitude is max displacement antinode pressure node

change in pressure is zero

8 Magnetic north on the left of diagram to produce the magnetic field

9 Electric Field :

Electric force acting on electron

When there is electric force on electron, there is acceleration of electron.

Hence there is a change in the speed of electron

Magnetic force on electron is always perpendicular to the electron’s velocity. Hence no

component of magnetic force in the direction of motion of electron. Therefore, there is no

change in the speed of electron.11 Magnetic force (Bevsinθ) on electron in the direction normal to the field lines, provide

the centripetal force for the electron to move in a circular path.

There is no magnetic force (no resultant force) on electron in the direction along the field



lines.

Motion of electron in the direction along field lines is due to the vertical component of

electron’s velocity

12 Isotopes nuclei of the same element with different mass number (different number of

neutrons)

Since Bqv = mv2 /r

r = mv/Bq

r is directly proportional to m , assuming speed (v) is constant (ions are passed through

velocity selector), and charge (q) is the same

The purpose of the mass spectrometer is to sieve the isotopes according to their masses

16 When conductor rod moves to the right, the mobile electrons in the rod moves to end A

(lower potential). This means that an emf is generated between ends A & B and induced

current will flow in the direction DCBA. Magnetic force on induced current carrying rod

will be to the left (Fleming’s LHR)

17 Average induced emf = -(change of magnetic flux linkage / time)

Change of magnetic flux linkage= final flux linkage – initial flux linkage

= NBA – 0 = NBA

Thus, -2 x 10-3

= - (NBA / t)

18 Peak Power = Vo2 /R

But Vo2

= (Vrms2) x 2

19 (Analogy)

Direct current:

Mean Power α I2

AC:

Mean Power α Irms2

Direct Current:

Force F between a pair of parallel coils

α I2

AC

Mean force Fmeanα Irms2

So for mean force Fmean = F

Irms = I = 1 A

21 Energy of emitted photon with freq f 1

= energy of emitted photon with freq f 2 + energy of emitted photon with freq f 3

∴hf 1 = hf 2 + hf 3

22 In this practical scenario, the atmosphere is like the discharge tube of gaseous atoms (inthe notes)

Emission spectrum is observed by the astronaut outside the planet

Absorption spectrum is observed by the astronaut on the planet

23 Gain in kinetic energy = eV

λ = h/mv

Derive an equation that relates wavelength (λ) and potential difference (V )

24 Speed of electron increased de broglie wavelength decreases

from dsinθ=nλ When λ decreases, θ decreases

Spacing between concentric rings decreases

25 Take note that y-axis is not stopping potential but max KE

Photoelectric equation:



hf = φ + KEmax

Rearrange,

KEmax = hf - φ

26 When energy of incident electrons increased, the maximum energy of x-ray photon

increases λmin decreases

No change in wavelength of the characteristic peaks since target element is not changed

27 Accelerating PD is reduced maximum energy of x-ray photon decreases λmin

increases

Overall intensity across all wavelength decreases (do you know why?)

No change in wavelength of the characteristic peaks

29 As single slit gets narrower uncertainty in the location of photon at the slit decreases

From uncertainty principle, when Δx decreases, Δp of photon at the screen increases,

resulting in larger spreading of light (larger diffraction)

30 Work function of metal U – E where U is represented by height of potential barrier

and E is the energy of the electron

When work function of metal is small, there is higher probability of electron tunnellingthrough the potential barrier (higher current)

Distance of conducting tip to metal surface represented by the width of the potential

barrier

When distance is small, there is higher probability of electron tunnelling through the

potential barrier (higher current)

Section B – solutions

1(a) The two waves have a constant phase difference with time (or constantphase relationship with time) hence they are coherent

(b)(i) Intensity is directly proportional to (amplitude)2

IA = k(3)2

and IB = k(2)2

Dividing the two equations

IB = 4IA/9 = 4I/9

(ii) Resultant amplitude at point P= (3.0 x 10-4 cm) – (2.0 x 10-4 cm)= 1.0 x 10

-4cm

IA = k(3)2 and IRes = k(1)2


IRes = IA/9 = I/92(a) Interference is when two or more waves meet (or superpose)

At different points where the waves meet, the resultant displacement (or resultant intensity) can be higher (constructive interference) or lower (destructive interference) than the displacement of the individual waves.The resultant displacement can be determined by the principle of superposition

(b)(i) dsinθ=nλ

(10-3

/550)sin90o=n x 644 x 10

-9

n=2.8

Hence, number of orders on each side of zero order is two(ii) When n is larger, Δθ is larger

So greater separation in second order (n=2)

(c) Light is not normal to the diffraction grating (or screen not parallel todiffraction grating)



3(a) Wavelength = 2 x 0.75 m= 1.50 m

(b) speed =f λ

speed = 360 x 1.50

= 540 m s−1

(c) When string is plucked at the centre, two travelling waves are sent to thefixed ends

The two travelling waves are then reflected and superposed to form a

stationary waveThe speed calculated in (b) refers to the speed of the travelling waves

(d) To increase the frequency of vibration, we can increase the speed of thewave on the string by lowering the mass of string (or by increasing thetension of string)[Think of a guitar]

4(a)(i) As wire moves, it cuts magnetic field (or there is change of magnetic flux)

By Faraday’s law, rate of cut of flux (or change of flux)is directional proportional to induced emf

Hence there is emf induced in the wire

(ii) Area of magnetic field swept by the wire is increasing and decreasing with

timeMagnetic flux is increasing and decreasing with time

Hence induced emf changes direction

as wire changes direction

(b) V=Vosinωt (or cosine equation)Find V0 Amplitude = (7.5 squares/ 5 squares) x 1 mV = 1.5 x 10

-3V

Find ω

ω = 2π/ TPeriod (30 squares/5 square) x 0.5 ms = 3 ms

ω = 2094 rad s-1

V = 1.5 x 10

-3sin 2094 t

5(a) To prevent magnetic flux losses (or to improve magnetic flux linkagebetween primary and secondary coils)

(b)(i) Graph must have same frequency as Fig. 5.2 and sinusoidal

At t=0 s, emf has a negative maximum value

(iii) 90o

or π/2 rad

(c)(i) Magnetic flux is constant throughout the core assuming no magnetic fluxloss

Coil C is wound on the part of core which has the smallest cross-sectional.

Since flux = BA, and flux is constantCoil C will have the maximum flux density

(ii) To step up, output coil needs to have increased number of coils (or morenumber of coils) than input coil B

Hence either Coil C or D can be used

6(a) It is the value of the direct current that produces the same average (or mean) power in the same resistor

(b) power is equal to I2R

so power is independent of current direction (whether current is positive or negative)

Note: possible misconception would be that since average current is zero

for an AC, then power should be zero (no power converted in the resistor)Note: Current is not a vector, current direction refers to its direction of flowin an electrical circuit

(c) average power = IRMS2R = (I0 / √2)

2

maximum power = I02R




Hence ratio is 0.5

7(a) Work function refers to the minimum energy of the incident photon that isrequired to release an electron from a metal surface

(b)(i) Emitter:FTarget: E

Reason: when wavelength λ1 is used, there is saturation current whenmetal E is positive potential (higher potential with respect to metal F). This

shows that photoelectrons are from metal F and accelerating towardsmetal E.

(ii) λ2 is smaller than λ1

φE is greater than φF

Note: When λ2 is used, the incident light has higher frequency. It is thenable to emit photoelectrons from metal E which has a higher work function

than metal F. Previously, when λ1 is used, no photoelectrons are emittedfrom metal E.

(c) Important note: Question mentioned “Light of constant intensity…”So when frequency is increased, number of incident photons per unit time(n / time) must decrease since

Intensity =timearea

nhf

×

At higher frequency, there are fewer photons per unit time for same(constant) intensity

Since one incident photon causes one electron to be emitted from metalsurface, the number of electrons emitted per unit time decreases

(d)1. Relation: Energy of photon =h x f

where a photon is a discrete amount of energy of EM radiation withfrequency f

2.Wavelength =

p

h

Every particle has a wave nature with a de broglie wavelength

And p is the momentum of the particle

8(a) When an atom is excited, the electron is at a higher energy level

When the electron goes to a lower energy level, a photon with adiscrete/specific energy and frequency is emitted

The line spectrum shows discrete/specific frequencies of photons

This means that the electron must transit between energy levels that arediscrete in order to have emitted photons of discrete energies andfrequencies

(b) Four transitions correct (seediagram)

All transitions are from higher tolower energy level And USE OFRULER

9(a) Electrons with high KE bombard the metal target

When all KE of an electron is lost during a single collision with the targetatoms, one X-ray photon with the maximum energy is produced

Since the accelerating potential is the same for both tubes,



the X-ray photon produced from either tube has the same maximumenergy and same frequencysince energy of x-ray photon = hf

Hence the minimum wavelength is the samesince speed of x-ray = frequency of x-ray x minimum wavelength)

(b) At lower accelerating potential, the energy of electrons (25 keV) is notsufficient

to knock out the innermost electrons from the tungsten atom

10(a) De Broglie wavelength = 6.63 x 10

-34

/ 0.06 x 20= 5.53 x 10-34

m

De broglie Wavelength is much smaller than the 1m gap

Hence no significant diffraction

(b)

mv

h=λ

mE

h

2=λ

meV

h

2=λ

Show correct substitutionV = 67 V

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Mid Year Exam Student Guide