v3-sg.AL.themis - H2 Physics - Yearly Solutions - 2008 ...

2008 Nov

2008 − 1

2008nov

advanced

physicscomplete yearly solutions

Advanced Physics solutions - yearly © themis

2008 − 2

2008 Nov Paper 1

MCQs Answer all questions.

1. [Measurement] Solution

Velocity is a vector. Time is a scalar.

∴ To define velocity with respect to time, the measurement of distance travelled must be of vector-form.

Distance travelled in unit time

Distance travelled per unit time

Speed in a particular direction

Displacement per unit time

(A) (ans) ≡≡≡ themis ≡≡≡


Realistic estimate:

The average kinetic energy of a bus (say, m = 3000kg) travelling on an expressway (say, v =

100km/h) is 212 mv ≈ 1 MJ.

the kinetic energy of a bus travelling on an expressway is 30,000 J

The average power of a domestic light is 12 – 100W.

the power of a domestic light is 300 W

The average temperature of a hot oven is 300°C.

the temperature of a hot oven is 300 K

The average volume of air in car tyre is say,

1.0m (tube) × 0.2m (height) × 0.2m (wide)

= 0.04m3 (close)

the volume of air in a car tyre is 0.03 m3

(D) (ans) ≡≡≡ themis ≡≡≡


Specification of a digital voltmeter is:

“accuracy ± 1 % with an additional uncertainty of ± 10 mV”

The meter reads 4.072 V.

∴ Recorded reading = (4.072 ± 1% ± 10 mV)

= (4.072 ± 1% ± 0.010)

= (4.072 ± 0.04072 ± 0.010)

= (4.07 ± 0.05) V (error has only 1sf and the main reading follows the decimal place of the error)

Recorded reading, together with its uncertainty:

(4.07±0.01) V

(4.07±0.04) V

(4.072±0.052) V

(4.07±0.05) V


4. [Kinematics] Solution

Air resistance is negligible.

A metal ball is at rest and dropped over a bed of sand for 1.0 s before making a depth of 8.0 mm.

∴ Just before hitting the sand, the metal ball (mass m) from rest (initial speed u = 0) has a final speed

(v) = u + gt = (0) + (10)(1) = 10 ms-1.

Kinetic energy before impact is 212 mv

= ( )212 10m = 50m ⎯

Work done against sand

= Ffrictions = mas ⎯

where a is the deceleration of ball in sand and s is its depth of displacement.

= :

50m = mas ⇒ 50

0.0080a =

⇒ 3 -26.3 10 msa = × (2sf) (close)

The average deceleration of the ball:

6.0 × 102 m s-2

1.2 × 103 m s-2

2008 Nov

2008 − 3

1.2 × 104 m s-2

6.0 × 103 m s-2

(C) (ans) ≡≡≡ themis ≡≡≡

5. [Kinematics] Solution

A cyclist accelerates down a hill, then constant speed, before decelerating up another hill.

∴ To deduce the graph of distance s moved by the cyclist against time taken, t:

• When the cyclist goes down slope, for every tΔ , there would be an increasing sΔ , giving an upwardly increasing graph.

• When the cyclist is travelling at a constant speed, for every tΔ , there would be a constant sΔ , giving a constant increasing graph.

• When the cyclist goes up another slope, for every tΔ , there would be a decreasing sΔ , giving an

downwardly decreasing graph.

Since the cyclist is moving further away from his original position, the graph of s is ever increasing.

The shape of graph:


6. [Dynamics] Solution

Trolley 1:

Mass = 6.0 kg, Velocity = 5.0 ms-1 Direction: to the right

Trolley 2:

Mass = 10 kg, Velocity = 3.0 ms-1 Direction: to the left

The trolleys had a head-on collision that lasted 0.20 s. After the collision, the trolleys stick together.

∴ Taking ( )→ to be positive,

Total momentum ( ) ( ) ( ) ( )6.0 5.0 10 3.0= + −

= 0 kgms-1

Since objects stick together,

Total kinetic energy = 0 J

Average force = change in momentum per unit time of impact

= ( ) ( )6.0 5.0 0.20 = 150 N

Total momentum of the two trolleys

before the collision (kgms-1)

The average force acting on each

trolley during the collision (kgms-1)

0 300

60 150

60 300

0 150


7. [Forces] Solution

A ladder of weight W rests against a vertical wall.

Friction exists at each contact point prevents the ladder from slipping.

∴ According the Lami’s theorem, the lines of action of the 3 co-planar forces must meet at a common point.

6.0 kg

pin cork

10 kg

5.0 ms-1 3.0 ms-1

W

s

t 0

0

s

t 0

0

s

t 0

0

s

t 0

0


2008 − 4

The diagram that shows the directions of the forces on the ladder:


8. [Forces] Solution

A uniform rod with a rubber section and wooden section is made to balance as shown.

∴

Taking moment about the pivot:

( ) ( )w r4.00 1.00 2.00 2.10 2.10 0.5W l W l+ − − = −

w r0.90 1.60W W× = × ⎯

: ( )W V g AL gρ ρ= =

w w r r0.90 1.60AL g AL gρ ρ× = ×

r

w

0.90 4.001.60

l l ρρ

⎛ ⎞× = ⎜ ⎟

⎝ ⎠

The ratio, ( )

( )r

w

density of rubber

density of wood

ρρ

= 2.25

The ratio, =density of rubberdensity of wood

1.71 2.50 3.27 2.25

(B) (ans) ≡≡≡ themis ≡≡≡

9. [Work, Energy & Power] Solution

A car of mass m has a driving force F.

In a time t, it travels a distance s and its speed increases from u to v.

∴ Useful work done by the car engine

= final energy − initial energy

= 2 21 12 2mv mu− =

( )2 2

2

m v u−

The useful work done by the car engine:

Fst

, it has a unit equivalent to power

( )-m v u , it has a unit equivalent to momentum

Ft , it has a unit equivalent to impulse

( )2 2

2

m v u−, it has a unit equivalent to energy


10. [Electromagnetic Induction] Solution

A square loop of wire is placed in a region of uniform magnetic field.

Direction of field = perpendicular to plane

The loop is pulled out of the field at a uniform speed v.

∴ Initially when the loop is pulling, the loop experienced no change in magnetic flux linkage, i.e., no work done.

When the loop is partially out of the magnetic field, the loop experienced a change in magnetic flux linkage, i.e., work done needs to be done to oppose that change. Since the change of flux is constant, the work needs to be done is also constant.

magnetic field

loop of wire

v

2.10 l

l 4.00 l

Ww Wr 2.00 l

2.10 l

l 4.00 l rubber handle

wooden section

W

W

W

W

2008 Nov

2008 − 5

The graph shows work done W versus the speed v:


11. [Motion in a Circle] Solution

The rotor of an engine is rotating at 3000 rpm (revolutions per min); radius = 8.0 cm.

∴ Centripetal acceleration

= 2rω — in its usual notations

= ( )2

2 3000 28.0 1060

π− ×⎛ ⎞× ⎜ ⎟⎝ ⎠

= 7900 m s-2

The centripetal acceleration:

25 ms-2

3.1×104 ms-2

7.2×107 ms-2

7900 ms-2


12. [Motion in a Circle] Solution

A pendulum has:

Bob mass = 1.27 kg Radius of oscillation = 0.600 m

Velocity (horizontally) = 0.575 ms-1, at the centre of its motion when the string is vertical.

∴ Tension in string = cW F+ — in its usual notations

= 2vmg m

r+ =

20.5751.27 9.810.600

⎛ ⎞+⎜ ⎟⎜ ⎟

⎝ ⎠

= 20.5751.27 9.81

0.600⎛ ⎞

+⎜ ⎟⎜ ⎟⎝ ⎠

= 13.2 N

Tension in the string:

11.8 N

12.5 N

13.7 N

13.2 N


13. [Gravitational Field] Solution

A satellite has:

Mass = m

Orbit = Circular, initially at radius 1r around the

Earth, then moved to a new circular orbit of radius 2r , as shown.

Mass of the Earth = M

Gravitational constant = G

∴ Change in potential energy of the satellite, ΔΦ

= m φΔ = 2 1m φ φ−⎡ ⎤⎣ ⎦

= 2 1

1 1m GM GMr r

⎡ ⎤⎛ ⎞⎛ ⎞ ⎛ ⎞− − −⎢ ⎥⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎝ ⎠⎣ ⎦

= 1 2

1 1GMmr r⎡ ⎤

−⎢ ⎥⎣ ⎦

(note: +ve)

Increase in the potential energy of the satellite:

2 1

1 1GM

r r

⎛ ⎞−⎜ ⎟

⎝ ⎠

1 2

1 1GM

r r

⎛ ⎞−⎜ ⎟

⎝ ⎠

satellite mass m C Earth

mass Mr1

r2

W

T

a

0 0 v

W

0 0 v

W

0 0 v

W

0 0 v

W


2008 − 6

2 1

1 1GMm

r r

⎛ ⎞−⎜ ⎟

⎝ ⎠

1 2

1 1GMm

r r

⎛ ⎞−⎜ ⎟

⎝ ⎠



Earth is orbiting round the Sun.

Radius of the Earth's orbit = 111.50 10 m×

Period = 365 days

∴ Centripetal force, 2 S EE 2

GM mm rr

ω = —

— in its usual notations

: 2 S2

GMrr

ω =

⇒ Mass of sun, 3 2

SrM

Gω=

23 2r

TG

π⎛ ⎞⎜ ⎟⎝ ⎠=

( ) ( )

2311

11

21.50 10

365 24 60 60

6.67 10

π

−

×× × ×

=×

⎛ ⎞⎜ ⎟⎝ ⎠

302.01 10= × kg

The mass of the sun:

6.4×1029 kg 1.16×1033 kg

3.31×1033 kg 2.01×1030 kg


15. [Oscillations] Solution

Defining equation for a particle moving in simple harmonic motion is

2a xω= −

where, acceleration of the particle = a , displacement = x , and angular frequency = ω

How a varies with x for a particle moving in simple harmonic motion, is as shown.

∴ Deduce from graph, amplitude of motion = 5.0 cm

gradient = 210

5.0 10−− × = 2ω =

22Tπ⎛ ⎞

⎜ ⎟⎝ ⎠

⇒ Period of motion, 0.44T = s

Amplitude (cm) Period (s)

5.0 14

10 0.44

10 14

5.0 0.44



An object is said to undergo lightly-damped oscillations.

∴Although the object is undergoing lightly-damped oscillations, i.e., its amplitude will reduce slowly, but its period remains unchanged.

Diagram shows the displacement, y of that object against time, t:


y

t

y

t

y

t

y

t

5

-5

-10

x / cm

a / m s-2

0

10

2008 Nov

2008 − 7

17. [Thermal Physics] Solution

A container contains a hot vapour and is allowed to cool and loses heat to its surroundings.

The graph shows how its temperature changes with time.

∴ The length of the line Q is greater than the length of the line S, implies it needs to lose more heat in liquefaction than freezing.

The part of the graph shows that the specific latent heat of vaporization of the substance is greater than its specific latent heat of fusion:

The gradient of the graph at P is greater than the gradient at R.

The gradient of the graph at T is greater than the gradient at R.

The value of X is greater than the value of Y.

The length of the line Q is greater than the length of the line S.



A fixed amount of an ideal gas:

Pressure = p

Volume = V

The graph shows the variation of 1 p with

V at a constant temperature.

The amount of gas and the thermodynamic temperature are both doubled.

∴ Original Ideal gas equation:

pV nRT= ⇒ 11 p V

nRT⎛ ⎞= ⎜ ⎟⎝ ⎠

—

New Ideal gas equation:

( ) ( )2 2pV n R T= ⇒ 14

11 p VnRT

⎛ ⎞= ⎜ ⎟⎝ ⎠

—

The new gradient would be 14 that of the original

graph.

Line represents the new gas:



An electric kettle contains:

500 g of water at 15°C Power = 2.2 kW

Specific heat capacity of water = 4.2 × 103 J kg-1 K-1

The temperature of the water is raised to 100°C.

∴ Recall, Pt mc θ= Δ — in its usual notations

Time taken, mct

PθΔ=

( )0.500 4200 100 15

2200× × ° − °

= 81= s (2sf)

Time taken to raise the temperature of the water to 100°C:

22 s 95 s 8.1×104 s 81 s


temperature

time0

0

R X

Y

Q P

S T

1/p

V

0.4

0

0.1

0.2

0.3

0 1 2 3 4

1/p

V

0.4

0

0.1

0.2

0.3

0 1 2 3 4

position of original graph


2008 − 8

20. [Wave Motion] Solution

A point source produces waves is made to pass through an area that is 2.0 cm wide, as shown.

Within this area, the intensity of the waves is I and their amplitude is A. The waves reach a second area of width 16 cm.

∴ Recall, 1

AreaI ∝ — and 2I A∝ —


: Area 64↑ = × ⇒ 64I↓ = × ⇒ 64I

: 64I↓ = × ⇒ 8A↓ = × ⇒ 8A

Intensity at 2nd area Amplitude at 2nd area

8I

4A

64I

8A

256I

16A

64I

4A


21. [Quantum Physics] Solution

Energy E of a photon and its wavelength λ is

KEλ

= , where K is a constant.

∴ If E is measured in electronvolts ( )'E and λ in

nanometres ( )'λ , then:

Recall, cE hf hλ

⎛ ⎞= = ⎜ ⎟⎝ ⎠


If measured in the other units, then

( ) ( )19

9' 1.6 10

' 10cE h

λ−

−

⎛ ⎞⎜ ⎟× =⎜ ⎟⎝ ⎠

• ( ) ( )9 191''10 1.6 10

hcEλ− −

⎛ ⎞⎜ ⎟=⎜ ⎟×⎝ ⎠

• ( ) ( )9 1910 1.6 10hcK

− −

⎛ ⎞⎜ ⎟=⎜ ⎟×⎝ ⎠

( ) ( )

( ) ( )34 8

9 19

6.63 10 3.00 10

10 1.6 10

−

− −

× ×=

×

= 1240 (3sf)

Numerical value of K:

3.18×10-53 3.18×10-35

1.24×10-15 1.24×103


22. [Superposition] Solution

Diagrams show air particles moving in a column.

A stationary wave is formed.

The first diagram shows the displacement of some particles at one instant and the second diagram shows the displacement of some particles half a cycle later.

∴ Deduce from diagrams that 34 Lλ = .

Maximum varying pressures occur at points with maximum particle movements (anti-nodes).

16 cm

2.0 cmpoint

source of waves

L

node node antinode antinode

2008 Nov

2008 − 9

Length L in terms of wavelength A

Position within the column that pressure change by the largest

amount

34 λ node

32 λ node

32 λ antinode

34 λ antinode



A beam of incident monochromatic light is shined normally to grating:

Wavelength = 600 nm

Diffraction grating = 53.0 10× lines per metre

∴ Recall, θ λ=sind n —

For maximum images, set sin 1θ = :

: maxdnλ

= 5

91 3.0 10600 10−

×=×

5.5=

⇒ 5 images, maximum

Total number of images produced by light transmitted through this grating:

8 9 11 5


24. [Electric Fields] Solution

Electric field strength is defined:

“force per unit positive charge on a small test charge”.

∴ The key definition of electric field strength came from Coulomb. Hence, it is necessary for the test charge to be small,

so that the test charge does not distort the electric field

so that the force on the test charge is small

so that the test charge does not create any forces on nearby charges

so that Coulomb's law for point charges is obeyed



In an electric field, the electric potentials at points 1P

and 2P in free space are 1V and 2V respectively.

A point electric charge q− is brought from 1P to 2P

by an external agent.

∴ Work done on the charge = ( )2 1q V V− −

= ( )1 2q V V−

Work done on the charge:

( )2 1q V V+ ( )2 1q V V− −

( )2 1q V V− ( )1 2q V V−


26. [Current of Electricity] Solution

A wire has:

Resistivity = 81.3 10 m−× Ω

Diameter = 0.50 mm Length = 30 times round an insulating rod Insulating rod’s diameter = 1.5 cm

∴ Recall, resistance, RA

ρ= —


: ( )

( )2

823

30 2 1.5 10 21.3 10

0.5 10 2R

π

π

−−

−

× × × ÷= × ×

× ÷

29.4 10−= × Ω

Resistance of the wire:

11.1 10× Ω 47.0 10−× Ω

54.7 10−× Ω 29.4 10−× Ω


wire woundround rod 30

timesinsulating rod diameter 1.5 cm


2008 − 10


A battery, connected to an external resistor, has:

e.m.f. = 6.0 V ( )E

Internal resistance = 0.40 Ω ( )r

External resistor = 2.90 Ω ( )R

∴ Recall, current ( )Ei

R r=

+ ( )

6.02.90 0.40

=+

1.82= A

Power dissipated 2P i R= ( )21.82 2.90=

9.6= W

Power supplied to the external resistor:

1.3 W 5.3 W

12.4 W 9.6 W


28. [D. C. Circuits] Solution

The resistance R of a thermistor varies during part of a day is shown.

Thermistor is connected to the potential divider circuit as shown.

At 0730, an output of 6.O V is required.

∴ If the output is to be set at 6.0 V, then across the other section (A), the voltage drop needed to be 3.0 V.

At 0730, the resistance of the thermistor = 1.5 kΩ

The current flowing through section (A):

( )3.0 V

1.5 k 1.5 ki =

Ω + Ω 0.001= A

Similarly, the same current flows through the section (B):

( ) ( )V6.0 V 1.5 k 0.001 AR= + Ω

• v 4.5 kR = Ω

Value of the variable resistor be set:

1.5 kΩ 3.0 kΩ

6.0 kΩ 4.5 kΩ


29. [D. C. Circuits] Solution

The circuit consists:

A battery of e.m.f. 6.0 V, of negligible internal resistance;

Three resistors, each of resistance R, and a variable resistor T, as shown.

The resistance of T changes from R to 5R.

time of day

R/ kΩ

4.0

3.0

2.0

1.0

0.00500 0600 0700 0800 0900

0 V

output

+9.0 V

1.5 kΩ

1.5 kΩ A

BRv

6.0 V R

R R

T

V +6.0 V +0.0 V

X

Y

2008 Nov

2008 − 11

∴ When variable resistor, T is set to R: Potential X = +3.0 V Potential Y = +3.0 V Potential difference, 0.0 VVΔ =

When variable resistor, T is set to 5R: Potential X = +1.0 V Potential Y = +3.0 V Potential difference, 2.0 VVΔ =

Change in the reading of the high-resistance voltmeter

= 2.0 V 0.0 V 2.0 V− =

Change in the reading of the high-resistance voltmeter:

zero 4 V 5 V 2 V



A charged particle, in vacuum, travels in a straight line, enters a uniform field. The particle then travels in a curved path that is not the arc of a circle.

∴ Only electric field influences a charged particle in a non-circular way.

For the charged particle to go on a subsequent non-parallel path, the field direction has to be non-parallel.

Type of field Initial direction of the particle

Electric parallel

magnetic parallel

magnetic perpendicular

electric perpendicular


31. [Electromagnetism] Solution

Different particle beams are made to enter a region between two metal plates in which there are uniform electric and magnetic fields.

∴ By applying Fleming’s left hand rule and electric field direction, the forces acting on the particle beams

must exactly opposed in order for the beams to emerge undeflected.

The arrangement that would allow the beam to pass through undeflected:



A coil has area A and n turns.

A uniform magnetic field of flux density B acts at an angle θ to the plane of the coil, as shown.

pivot

magnetic flux density B

θ

coil area A

electrons

−

+

protons

−

+

magnetic field

electrons

+

−

protons

+

−


2008 − 12

∴ Initial flux linkage i sinBAn θΦ =

The coil is rotated by an angle of θ .

Final flux linkage f 0Φ =

Change in flux linkage sinBAn θΔΦ =

Change in magnetic flux linkage when the coil rotates so that the angle θ is reduced to zero:

cosBAn θ 2 cosBAn θ

2 sinBAn θ sinBAn θ



The magnetic flux linking a coil, with respect to time is as shown.

∴ The induced e.m.f. of a coil, varies with the rate of change of flux linkage, i.e., the gradient of the flux-time graph.

Further increasing or decreasing flux linkage induces differently signed e.m.f..

The corresponding e.m.f. induced in the coil, with respect to time:


34. [Alternating Currents] Solution

An alternating current with a square waveform is shown.

Its peak value for the current is ol .

∴ Consider a complete period, T.

Squaring the wave to give the 2oI⎡ ⎤

⎣ ⎦ waveform.

Take its average over its period, 2

2 2oo o

I TI IT×

= = .

Take the square root of this average value,

2 2rms o o oI I I I= = =

Root-mean-square value for the current:

0 o

2I

o

2I

oI



A beam of electrons is made to pass through a thin carbon film, velocity, v, as shown.

flux

time 0 0 t 2t

current

time

square wave-form Io

- Io

alternating regions of bright and dark zones

graphite plate fluorescent

screen

electron beam

current

time

Io

- Io

Io2

e.m.f.

time00 t 2t

e.m.f.

time0 0 t 2t

e.m.f.

time0 0 t 2t

e.m.f.

time00 t 2t

2008 Nov

2008 − 13

The electrons produce a pattern of concentric circles on the fluorescent screen.

∴ According to de Broglie’s principle of wave-particle duality, the electrons exhibited wave-like property, i.e., diffraction.

From hmvλ

= — in its usual notations

When ( )mv ↑ ⇒ ( )λ ↓

From θ λ=sind n — in its usual notations

When ( )λ ↓ ⇒ ( )θ ↓

• More fringes and diameters of circles decrease.

Cause of the pattern Changes to pattern when the velocity is

increased

diffraction

diameters of circles increase

refraction

diameters of circles increase

refraction

diameters of circles decrease

diffraction

diameters of circles decrease



Sketch shows how the wave function ψ of an electron

varies with position.

∴ We usually interpret the absolute square of the wavefunction ( ),r tΨ as the probability density for the

particle to be found at each point in space. In other

words, ( ) 2 3,r t d rΨ is the probability, at time t, of

finding the particle in the infinitesimal region of

volume 3d r surrounding the position, r .

Graph shows the probability of finding an electron at each position:

(B) (ans)

≡≡≡ themis ≡≡≡

37. [Lasers and Semiconductors] Solution

The symbols + + + and − − − represent the majority carriers in the p-type and n-type sides of a p-n junction.

∴ As interpreted in standard text.

0 distance

ψ

0 distance

probability

0 distance

probability

0 distance

probability

0distance

probability


2008 − 14

Pair of diagrams illustrates how a p-n junction acts as a rectifier:



In a helium-neon laser:

• Helium atoms collide with neon atoms and excite them

• It produces a population inversion which allows stimulated emission.

∴ The spontaneous infra-red emission is low energy.

The stimulated emission of red light is high energy.

The total excitation energy should equal its total emissions.

Neon energy level diagram that correctly shows the excitation of the neon atoms by the helium atoms, the spontaneous infra-red emission from the neon, and the stimulated emission of red light:


39. [Nuclear Physics] Solution

A detector has:

• Ionising radiation • Background count rate = 24 counts per minute

A radioactive source has:

• Reading = 532 counts per minute

∴ True initial reading 532 24 508= − =

After 2 half-lives,

the final true reading 508 127

4= =

We now have to add back the background count,

the final meter reading 127 24 151= + =

Reading after two half-lives of the source:

127 133 157 151



A nucleus of bohrium xyBh decays to mendelevium

255101Md by a sequence of three α-particle emissions.

bohrium xyBh → dubnium + α

→ lawrencium + α

→ mendelevium 255101Md + α

∴ Rewriting, xyBh → 255

101Md + 3α —

: ( )255 3 4 267x = + =

( )101 3 2 107y = + =

• no. of neutrons 267 107 160= − =

No. of neutrons in a nucleus of xyBh :

267 261 154 160


+ −

+ + + + + + + + + + +

− − − − − − − − − − −

p

n

conventional current

voltagesource

+ − p

n no current

+ + + +

− − − −

+ −

p

n


voltagesource

+ + + + + + + + + + +

− − − − − − − − − − − +

− p

n no current

+ + + +

− − − −

+ −

p

n


voltagesource

+ + + + + + + + + + +

− − − − − − − − − − − +

− p

n no current

+ + + +

− − − −

+ −

+ + + + + + + + + + +

− − − − − − − − − − −

p

n


voltagesource

+ − p

n no current

+ + + +

− − − −

spontaneous emission

stimulated emission excitation

stimulated emission

spontan. emission excitation

spontaneous

emission


spontaneous emission


2008 Nov

2008 − 15

[2008N P1 MCQ Key]

Q. Key Q. Key Q. Key Q. Key

1. A 11. B 21. D 31. B

2. D 12. C 22. B 32. B

3. D 13. D 23. A 33. C

4. C 14. B 24. D 34. D

5. D 15. A 25. D 35. B

6. C 16. D 26. B 36. B

7. C 17. C 27. C 37. D

8. B 18. D 28. C 38. D

9. D 19. B 29. B 39. C

10. D 20. B 30. B 40. C ≡≡≡ themis ≡≡≡


2008 − 16

2008 Nov Paper 2

Questions

Answer all questions. 1. [Kinematics] Solution

At the top of a cliff of height 32m, a stone of mass 130 g is thrown horizontally as shown.

Air resistance is negligible.

The stone hits the sea at speed of -134 ms .

(a) At point of impact,

(i) Let the velocity of impact be v .

Consider its vertical descent, 2 2 2y yv u gs= + — in its usual notations

• ( ) ( )2 20 2 9.81 32yv = +

• -125.1 msyv = (the vertical component of

the velocity of the stone) (ans) [2]

(ii) Deduce from the final velocity,

2 2 2y xv v v= + ⇒ 2 2 234 25.1 xv= +

• -123.0 msxv =

The angle θ to the horizontal of the stone’s path

= 1tan y

x

vv

− ⎛ ⎞⎜ ⎟⎝ ⎠

1 25.1tan23.0

− ⎛ ⎞= ⎜ ⎟⎝ ⎠

47.5= ° (3sf) (ans) [2]

(b) Upon impact, the speed of the stone is reduced

from -134 ms to -12.0 ms in a time of 0.95 s.

∴ Using momentum considerations,

Impulse, ( )averageF t m v u= − — in its notations

• Average force acted on the stone during this time,

( )average

0.130 2.0 340.95 s

F−

=

4.38 N= (ans) [2]

(c) If the stone causes a big splash on hitting the sea,

using energy considerations, the high kinetic energy from the stone is not only transferred to its immediate vicinity but over a wide area. Hence, the whole water surface and column is made to resist the impact. Together with a louder sound, the stone will be slowed down in a shorter time (shorter distances) than when its impact is restricted to a small water column, where no splash is produced. (ans) [2]



The diagram showed a flat horizontal plate vibrating in a vertical plane.

The plate’s the variation with displacement x of the acceleration a is as shown.

32 m

path of stone

sea

θ A

plate

vertical oscillations

x / mm-8 -6 -4 -2 2 4 6 8 0

a / m s-2 20

10

0

-10

-20

v

vx A

vyθ

2008 Nov

2008 − 17

(a) From the straight-line graph, the negative gradient gives a relationship i.e., a x∝ − (oppositely directed).

It is also given that it is a periodic oscillation, i.e., the graph does not extend itself indefinitely.

By definition, the oscillations of the plate are simple harmonic (SHM) as it is a periodic motion in which acceleration (a) is proportional to, but oppositely directed to, the displacement (x) from its equilibrium position. (ans) [3]

(b) Some sand is introduced onto the plate.

Amplitude of vibration of the plate is gradually increased from zero.

At one particular amplitude, the sand is seen to lose contact with the plate.

(i) When the sand particle is resting

on the vibrating plate, the external forces acting on the particle are as shown in the free body diagram (upward positive).

ma R W= − —

When the sand particle loses contact, the reaction force, R become zero.

: 0ma W= − ⇒ ma mg= −

⇒ a g= − 9.81= − m s-2 (ans)

At a g= − (upward +), the plate is travelling downward from its top displacement position. At this point, the acceleration due to gravity ( )g and

the acceleration due to the vibrating plate is the same, hence, if the plate were to accelerate faster than g (downwards), the sand particle will not be able to “catch-up” and stay in contact, thereby losing contact with the plate. (ans) [3] (ii) From graph, when a g= − , the amplitude of

vibration of the plate at which the sand first loses contact is at 3.8x = mm. (ans) [1]



A planet, with a mass M and radius pR , has an

orbiting satellite of mass m and the orbit is R .

(a) In terms of M, m and R:

(i) For the satellite to be in a stable orbit, it implies that the attractive gravitational force that

provides the centripetal acceleration is perfectly balanced by the satellite’s kinetic urge to leave the orbit.

⇒ 2

2GMm vm

RR= — in its usual notations

⇒ 212 2

GMmmvR

= = kinetic energy of the

satellite (ans) [2]

(ii) Gravitational potential energy GMm

R= − — in its

usual notations

gravitational potential energy of satellite

kinetic energy of satellite

2

GMmR

GMmR

−=

= −2 (QED) [1]

(b) The variation with orbital radius R of the gravitational potential energy of the satellite is shown.

(i) The variation with orbital radius of the kinetic energy of the satellite (The line should extend from

p1.5R R= to p4R R= ):

∴ Relating the ratio in (a)(ii) with the actual given K.E. values in the graph. We get,

12K.E. of satellite G.P.E. of satellite= −

energy / 109 J

-2.0

-4.0

-6.0

-8.0

-10.0

10.0

8.0

6.0

4.0

2.0

0R Rp 2Rp 3Rp 4Rp

reaction, R

weight, W

acce

lera

tion sand grain,

mass m


2008 − 18

(ans) [2]

(ii) The radius of the orbit of the satellite is changed from p4R R= to p2R R= .

Given the mass m of the satellite is 1600kg.

∴From 212K.E. mv= ,

2 K.E.v

m×= —

When p4R R= , 9K.E. 1.25 10 J= ×

: 92 1.25 10

1600v × ×= 1250= m s-1

When p2R R= , 9K.E. 5.00 10 J= ×

: 92 5.00 10

1600v × ×= 2500= m s-1

Change in orbital speed of the satellite

2500 1250 1250= − = m s-1 (ans) [5] ≡≡≡ themis ≡≡≡


An electrical circuit (as shown) has:

• A variable resistor R connected • A battery of e.m.f. E with internal resistance r .

The resistance of resistor R can be varied.

Take the potential difference across R to be denoted as V and the power dissipated in R as P .

The graph thus show the variation with V of P .

(a) For the maximum value of P,

(i) P is maximum at 5.62 W.

Its corresponding potential difference across the resistor is 4.50 V.

The current in the circuit = current in the resistor

PiV

= 5.62W4.50 V

= 1.25 A= (3sf) (ans) [2]

(ii) Correspondingly,

Resistance in R Vi

= 4.50 V1.25 A

=

= 3.6Ω (QED) [1] (b) Set the resistor R to 2.03 Ω, the current in the

circuit is 1.60 A.

∴ When 2.03R = Ω and 1.60 Ai = ,

Let E be the e.m.f. of the battery,

• ( )E i R r= +

From (a), ( )1.25 3.6E r= + —

energy / 109 J

-2.0

-4.0

-6.0

-8.0

-10.0

10.0

8.0

6.0

4.0

2.0

0R Rp 2Rp 3Rp 4Rp

kinetic energy of satellite

voltmeter

switch

ammeter

resistor R

A

V

− +

battery e.m.f. E

internal resistance r

current i

p.d. V

P / W5.8

5.6

5.4

5.2

5.0V / V3.0 3.5 4.0 4.5 5.0 5.5

2008 Nov

2008 − 19

From (b), ( )1.60 2.03E r= + —

Substituting E of into ,

( ) ( )1.25 3.6 1.60 2.03r r+ = +

( ) ( )3.6 1.28 2.03r r+ = +

3.6 2.5984 1.28r r+ = +

1.0016 0.28r= ⇒ r = 3.58Ω

Internal resistance r of the battery = 3.58Ω (this is a proof of maximum power theorem) (ans) [3]



Using the band theory of conduction, explain why the electrical resistance of an intrinsic semiconductor material decreases as its temperature rises.

Draw a diagram if needed. [4]

In semiconductors, the conduction and valence bands are spaced closely enough together (≈1 eV) that, at room temperature, a nontrivial number of electrons is found in the conduction band. These materials have significant conductivity that is highly temperature-sensitive.

• At low temperatures, semiconductors behave like insulators. The valence electrons have no adjacent energy levels to transit into, and do not have sufficient energy to the energy gap.

• At high temperatures, semiconductors behave like conductors. The valence electrons have sufficient energy to cross over the small energy gap to assist in the conduction. (ans)



Strontium-90 is a radioactive nuclide.

(a) A substance is said to be radioactive if it comprised unstable nuclei that will disintegrate into more stable configurations by the emission of alpha-particles (helium nuclei), beta-particles (electrons or positrons) and/or gamma radiation (electromagnetic waves of short-wavelengths). (ans) [2]

(b) A sample of Strontium-90 has a mass of 82.40 10−× g. The average activity of this sample

during a period of 1 hour is found to be 51.26 10× Bq.

(i) The decay constant (λ) of a radioactive nuclide is

defined as the constant of proportionality relating its activity to the number of undecayed nuclei. (ans) [2]

(ii) From (i), AN

λ = — in its usual notations —

: The decay constant of Strontium-90,

( )5

8 3 27

1.26 10 Bq

2.40 10 10 kg 90 1.66 10 kgλ

− − −

×=

× × × ×

107.84 10−= × s-1 (3sf) (ans) [3]

(c) For nuclides that have relatively small decay constants, it will have a relatively large half-life. It is thus suitable to accurately measuring its changes in mass and activity to determine its decay constant; whereas if the half-life is short, its mass and activity measurements would not be too accurate. (ans) [1]


7. [Thermal Physics] [Data Response] Solution

A serious hazard for fire-fighters is the explosion of containers of `liquefied gas' (butane) that have been heated in a fire. When the butane suddenly burns in an explosion, the fire spreads very rapidly in the form of a spherical fireball of increasing radius that is at very high temperature.

In order to study such fireballs, a series of experiments

is carried out. Some butane of volume 3 312.5 10 m−×

is put in a sealed container and is then heated until it

conduction band

valence band

energy of electrons,

E

small energy gap

energy bands in semiconductors


2008 − 20

explodes. The variation with time t of the radius R of the fireball is determined. The results are shown in the diagram.

(a) Use the diagram to

(i) Without any calculation,

• From t = 0 → 2 ms, initially the rate at which the radius of the fireball increases is very high.

• From t = 2 → 25 ms, the rate reduces.

• From t = 25 ms onwards, the rate is reduced further to an approximate constant (still positive). (ans) [2]

(ii) In a room of length 12 m, width 5 m and height 3 m, such an explosion would be very hazardous. From the graph, it will take less than t = 25 ms for the explosion to reach every part of the room. This is a very short reaction time.

The space is relatively small and enclosed, the blast and sound will add sudden pressure onto the ears and may confuse the victim subsequently.

The debris created by the exploding gas will behave like bullets from a shotgun thus harming the victim quickly. (ans) [3]

(b) It is thought that, for a fixed volume of butane, the radius R of the fireball varies with time t according to the expression

n mR k t= ,

where n and m are integers and k is a constant.

Some corresponding values of lg t and lg R for

the data in the diagram are plotted on the graph as shown following.

(i) On the diagram,

(1) At t = 40 ms, R = 14.6 m

⇒ lg R = 1.16 and lg t = 1.60

The point corresponding to time t = 40ms is plotted at (1.60,1.16). (ans) [1]

(2) The best-fit line for all the plotted points is

drawn. (ans) [1]

(ii) From the best-fit line drawn in (i) part (2),

The gradient 1.24 0.9351.79 1.00

−=−

0.386= (ans)

[2]

(iii) Recall, n mR k t=

• lg lg lgn R m t k= +

• 1lg lg lgmR t kn n

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

: 0.386 0.4mn

⎛ ⎞ ≈ ≈⎜ ⎟⎝ ⎠

Given that n and m are integers, making m as small as possible, implies 5n = and 2m = (ans) [3]

(c) The experiment is repeated using similar

containers but with different volumes of butane. The results are shown in the next diagram.

t / ms0 10 20 30 40 50

R / m16

12

8

4

0

20

1.4

1.3

1.2

1.1

1.0

0.9

lg (R

/ m

)

×

lg (t / ms)

1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7

×

×

× ×

1.4

1.3

1.2

1.1

1.0

0.9

lg (R

/ m

)

×

lg (t / ms)1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7

×

×

× ×

× (b)(i)1.

(b)(i)2.

2008 Nov

2008 − 21

Without drawing a further graph,

Proof: 5R cV= , where c is a constant

⇒ 5 lg lg lgR V c= +

• 1 15 5lg lg lgR V c= + —

From the graph at t = 40ms,

R / m V / 10-3 m3 R5 / 105 5R V / 107

14.6 12.5 6.63 5.30

14.0 10.0 5.38 5.38

13.1 7.5 4.01 5.35

12.2 5.0 2.70 5.41

10.6 2.5 1.34 5.35

average 5.36

Within limits of 1.1% error, it is found that 5R

V is

a constant; hence, the relationship

R5 = cV is true. (QED) [3]

(d)

(i) The equation in (c) may also be applied to other exploding gases.

∴ One suggested physical quantity on which the constant c will depend is its surrounding pressure. (ans) [1]

(ii) The data were collected for butane in a container in a room.

One other situation where the theory developed predicts a high level of hazard for fire-fighters is fire-fighting onboard ships where flammable gases build up in enclosed small cavities onboard the ship. (ans) [1]


t / ms 0 10 20 30 40 50 60

20

16

12

8

4

0

R /

m 12.5 × 10-3

volume of container / m3

10.0 × 10-3 7.5 × 10-3 5.0 × 10-3

2.5 × 10-3


2008 − 22

2008 Nov Paper 3

Questions Section A

Answer all questions in this section.

1. [Work, Energy and Power] Solution

(a) At constant temperature, for a given fixed mass of

gas, the variation of volume V of the pressure p is as shown (A).

(i) Using values from (A), using the axes drawn, plot in diagram below (B) to show that p is inversely proportional to V .

(ans) [2]

(ii) Within limits of errors, inverse proportion is demonstrated by the plotted graph, since the plot of 1 p against V is approximated to a

straight line passing through the origin, p is shown to be inversely proportional to V . (QED) [1]

(b) F and d are in the same direction.

The variation with displacement d of the force F applied to an object is as shown below (C).

Draw a graph showing the variation with d of the work done in (D).

Work done is area under the force − displacement graph.

(ans) [4]

V / cm3 0 100 200 300 400

3

2

1

0

p /

105 P

a

(A)

d / m 0 1.0 2.0 3.0 4.0

(D)

wor

k /

J

V / cm3 0 100 200 300 400

(B)

3.0

2.0

1.0

0.0

1/p

/ 1

0-5 P

a-1

×

×

× ×

×

0 0

(B)

30

20

10

0

F /

Nd / m

0 1.0 2.0 3.0 4.0

(C)

60

40

20

0

wor

k /

J

d / m 0 1.0 2.0 3.0 4.0

(D)

2008 Nov

2008 − 23

(c) Measurements are made of the Earth's gravitational field strength g for different distances r from the centre of the Earth. Diagram (E) shows the variation of lg r against lg g .

(i) The gradient of the graph (E) =

( )0.73 0.85

7.5 6.7

− −−

= 1.98− (3sf) (ans) [3]

(ii) Within limits of errors, since the plot of lg g

against lg r is approximated to a straight line, lg g

is shown to have a linear relationship with lg r

From (i), the gradient 2≈ − or lg g relates to 2 lg r−

• g relates to 2

1

r

• The gradient relates g to vary inversely with

the square of the distance 2

1

r⎛ ⎞⎜ ⎟⎝ ⎠

from the

centre of the earth. (ans) [2] ≡≡≡ themis ≡≡≡

2. [Electromagnetism] Solution

(a) A simple generator has:

• A coil with a large number of turns • Rotates at a constant rate in a uniform

magnetic field, as shown.

(i) Faraday's law of electromagnetic induction states that the magnitude of the induced e.m.f. in a conductor is directly proportional to the rate at which magnetic flux is cut by the conductor or flux linkage is changed in a coil.

Hence, when the coil rotates, its flux linkages changes with the magnetic field resulting in an e.m.f. being generated between the ends of the coil. (ans) [2]

(ii) Two factors that affect the magnitude of the maximum e.m.f.:

• the number of turns in the coil • the field magnetic field • (the coil is rotated at a relatively faster speed)

(ans) [2]

(iii) When the coil is at its horizontal position, its direction of motion is perpendicular to the magnetic field. Its induced e.m.f. is therefore at its maximum.

When the rotation of the coil reaches its vertical position, its direction of motion is parallel to the magnetic field. No e.m.f. would be induced.

Since the motion of the coil is circular, its induced e.m.f. follows a sinusoidal curve. (ans) [2]

(b) The output from a similar generator is connected to the input of an ideal transformer:

• primary coil = 30 turns • secondary coil = 600 turns • input e.m.f. = 72 V r.m.s.. • output is connected to a resistor of resistance

160 Ω, as shown in diagram.

(i) The peak input e.m.f.

= rms 2V ⋅ = 72 2⋅ = 102 V (3sf) (ans) [1]

lg (r / m)

6.7 6.8 6.9 7.0 7.1 7.2 7.3 7.4

(E)

1.0

0.5

-0.5

-1.0

lg (g

/ m

s-2)

0.0

N

S A

B

C

D

P

Q permanent magnet

transformer input

30turns

600 turns 16

0 Ω

time/ t

e.m.f / V

Voltage output-time graph

A B

A

B

B

A B A A B

T/2 T

one revolution

0

+Eo

-Eo


2008 − 24

(ii) The r.m.s. value of the p.d. across the resistor =

outputinput

input

NV

N× =

60072

30×

= 1440 V (3sf) (ans) [1] (iii) The r.m.s. value of the current in the resistor

= rmsVR

= 1440160

= 9.00 A (3sf) (ans) [1]

(iv) The mean power dissipated in the resistor,

P = 2rmsI R = ( )29.00 160

= 13.0 kW (3sf) (ans) [1] (v) The r.m.s. value of the current from the

generator

= rms, input

P

V =

1296072

= 180 A (3sf) (ans) [2] ≡≡≡ themis ≡≡≡


Explain the meaning of each of the following terms as applied to waves. (a)

A standing wave, also known as a stationary wave, as opposed to progressive wave, is a wave that remains in a constant position, i.e., their waveforms do not move. (ans) [2]

(b) Diffraction is the bending and

spreading of waves when they meet an obstruction or gap into regions where a shadow might be expected. (ans) [2]

(c) Sources are said to be coherent, if the waves

leaving them bear the same phase relationship to each other at all times. The same phase relationship implies the same frequency, and therefore the same wavelength. (ans) [2]

(d)

Polarization, also called wave polarization, is an

expression of the orientation of the lines of electric flux in an electromagnetic field. (ans) [1]


4. [2008N P3] [Thermal Physics]

Solution

(a) Define:

(i) Every minute particle (such as ions, electrons, atoms or molecules) in a body has random

potential energy ( )pE , due to their state and

position, and, random kinetic energy ( )kE , due to

their motion. Collectively, the sum of these randomly distributed energies is termed the internal energy ( )U of the body, i.e.,

p kU E E= + . (ans) [2]

(ii) The first law of thermodynamics is defined as the

heat ( )q absorbed by a system either raises the

internal energy ( )UΔ of the system and/or does

work by the system on the environment ( )byw ,

i.e., in byq U w= Δ + . — (ans) [1]

(b) An ideal gas undergoes a cycle of changes

A B C A→ → → , as shown in the diagrams

below.

unpolarised wave

plane- polarized

Polaroid

lens

light

gas

pA

frictionless piston

area A

cylinder

xΔ

volume V / cm3

pressurep / 105 Pa

1

5

B

20

A

C

7

2008 Nov

2008 − 25

(i) The work done by a gas during the change C → A expanding against an external pressure p is given by area under p V− graph or avep V⋅ Δ

( ) ( )5 6 3by 1 10 Pa 20 - 5 10 mw −⎡ ⎤= × ×⎣ ⎦

1.5 J= (2sf) (ans) [2]

(ii) The figure below is a table of energy changes

during one cycle.

section of cycle

heating supplied to

gas / J

work done on gas / J

increase in internal

energy / J

A B→ zero 4.2

B C→ −8.5

C A→

section of

cycle heating

supplied to gas / J

work done on gas / J

increase in internal

energy / J

A B→ zero 4.2 4.2

B C→ −8.5 zero −8.5

C A→ 5.8 −1.5 4.3

Rearranging , in onU q wΔ = + — Process A B→ :

: 0 4.2UΔ = + ⇒ 4.2UΔ = (ans)

Process B C→ :

( )on 0.0 0.0w p= = (ans)

: 8.5 0UΔ = − + ⇒ 8.5UΔ = − (ans)

Process C A→ :

cycle A B B C C A 0U U U U→ → →Δ = Δ + Δ + Δ =

• ( ) ( ) C A0 4.2 8.5 U →= + − + Δ

⇒ C A 4.3U →Δ = (ans)

: in4.3 1.5q= − ⇒ in 5.8q = (ans) [4] ≡≡≡ themis ≡≡≡

Section B Answer two questions in this section.


(a)

(i) The electric field strength, E , or electric field intensity or electric field is defined by the ratio of the electric force on a charge at a point in free space to the magnitude of the charge placed there, i.e.,

electric field strength, electric forcecharge

E =

(ans) [1]

(ii) The diagram below shows a uniform electric field of electric field strength E where a charge q+ is placed at point.

The charge at X is moved to Y through a distance d .

Using the definition in (i), Since the charge is positive, positive work done has to be

W.D. EF d= ⋅ qE d= ⋅ (ans) [1]

(iii) The potential difference between X and Y is V.

Using the answer from (ii),

The electric potential (V) at a point in free space in an electric field is defined as the work done in bringing a unit positive charge from infinity to the point, i.e.,

electric potential, work done

chargeV = —

: XYXY

work donecharge

V VΔΔ = =

qEdq

=

Ed= (ans) [2]

(b) An X-ray tube has:

• a vacuum • anode and cathode

uniform fieldE Y X

+q

d


2008 − 26

• electrons are accelerated from rest through a potential difference of 60 kV between the cathode and the anode.

• The current in the tube is 8.6 mA.

(i) The number of electrons passing through the tube in one second

electron

currentcharge

= -3

19

8.6 10 A1.6 10 C−

×=×

165.38 10= × (n) (3sf) (ans) [2]

(ii) Work done on the electron qV=

This work done is used to accelerate the electron to a high speed – converted to pure kinetic energy,

212qV mv= — in its usual notations

⇒ the speed of electrons arriving at the anode,

2qV

vm

=19 3

31

2 1.6 10 60 109.11 10

−

−

× × × ×=×

81.45 10= × m s-1 (3sf) (ans) [4] (iii) Work done by the electrons per second

= Power supplied by the electrons hitting the anode

nqV=

16 19 35.38 10 1.6 10 60 10−= × × × × ×

= 516 W (3sf) (ans) [2]

(c) X-ray production:

• uses a negligible fraction of the power reaching the anode in (b)

• has to be cooled by passing a coolant through the anode.

• The coolant has specific heat capacity of 3500 J kg-1 K-1 and the temperature rise is 30 K.

∴ The heat that needed to be removed = (iii)

Rate at which the coolant must be pumped

through the anode work done

c θ=

Δ

5163500 30

=×

34.91 10−= × kg s-1 (ans) [3]

(d) A conducting sphere has:

• radius = 0.10 m • charge = +0.060 μC

The electric field around the sphere is as shown in the diagram below.

(i) By referring to the above diagram, If the conducting sphere is replaced by a point charge of the same value, beyond the boundaries of the original conducting sphere, it will produce exactly the same field pattern and strength as that of the conducting sphere at points A, B and C. Hence, to points A, B and C, it appears as if the charge is concentrated at the centre of the sphere. (ans) [1]

(ii) Electric field strength at the surface of the

sphere,

2o

14

QErπε

= — in its usual notations

πε

−×=6

2o

1 0.060 104 0.10

= × 45.39 10 N C-1 (ans) [2] (iii) Point A and B are 0.40 m and 0.50 m from the

centre of the sphere respectively.

Potential difference (V) for AB = V for BC.

∴ πε

⎛ ⎞= −⎜ ⎟

⎝ ⎠AB

o B A

1 14

QVr r


Similarly, πε

⎛ ⎞= −⎜ ⎟

⎝ ⎠BC

o C B

1 14

QVr r

.

Since =AB BCV V , rearranging

⎛ ⎞⎛ ⎞

− = −⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠B A C B

1 1 1 1r r r r

⇒ ⎛ ⎞⎛ ⎞− = −⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠C

1 1 1 10.50 0.40 0.50r

Distance from the centre of the sphere to C, =C 0.67r m (ans)


field of isolated

point charge

AB

C

+0.060 μC

2008 Nov

2008 − 27


(a) Frequency ( )f is defined as the number of

complete oscillations produced per unit time, is distinct from angular frequency ( )ω which is

defined as the magnitude of the vector quantity angular velocity, that is defined as the rate of change of angular displacement with respect to time. (ans) [2]

(b) A spring has:

• An unstretched length = 0.650 m • Attached to a fixed point • A mass = 0.400 kg is attached to the spring

and gently lowered until equilibrium is reached.

• The spring has then stretched elastically by a distance of 0.200 m.

For the stretching of the spring,

(i) The loss in gravitational potential energy of the mass

mgh= ( )( )( )0.400 9.81 0.200=

= 0.785 J (ans) [1]

(ii) The elastic potential energy gained by the spring

12 Fx= ( )( )( )1

2 0.400 9.81 0.200=

= 0.392 J (ans) [2]

(c) The two answers to (b) are different because the spring−mass system is a rather inefficient mechanical energy storage system. It only manages to store half of the energy supplied, while the rest are lost through heat, sound and other forms of energy. (ans) [2]

(d) The load on the spring is now set into simple

harmonic motion of amplitude 0.200 m.

(i)

At its equilibrium position,

kx mg=

⇒ mg

ke

= 0.400 9.81

0.200×= 19.6= N m-1

At its lowest point of its movement, the resultant force on the load

'T W= − ( )k x e mg= + −

19.6 0.400 0.400 9.81= × − ×

3.92= N (ans) [2] (ii) Angular frequency of the oscillation,

km

ω = 19.60.400

=

7.00= rad s-1 (ans) [2] (iii) Maximum speed of the mass,

2 20v x xω= ± − 27.00 0.200=

1.40= m s-1 (ans) [1] (e) Complete the table below which is a list of

energies of the simple harmonic motion.

gravitational potential energy / J

elastic potential energy / J

kinetic energy / J

total energy / J

lowest point 0

equilibrium position

highest point

By observation,

gravitational potential energy / J

elastic potential energy / J

kinetic energy / J

total energy / J

lowest point 0 1.57 0 1.57

equilibrium position 0.785 0.392 0.392 1.57

highest point 1.57 0 0 1.57

(ans) [5] spring

e+x equilibrium

position

W

simple mass-spring system

e

W

T

T’


2008 − 28

(f) Based on the axes below, sketch four graphs to show the shape of the variation with position of the four energies. Label each graph.

By conservation of energies:

(ans) [3] ≡≡≡ themis ≡≡≡


(a) In the past, experiments were performed to

provide evidence for a small charged nucleus in the atom.

Rutherford and his students conducted an experiment on alpha particles scattering. Alpha-particles from a radium source are made to impinge on a very thin sheet of gold foil, approximately 10-6 m thick.

Observations:

• Most of the incident alpha particles pass straight through the foil, to hit the fluorescent screen.

• There are few particles, however, which suffer deviations in the forward and backward directions.

• A very few particles even retrace their original path, backwards.

Conclusions:

• The observation can be explained by proposing that the atom is made up of a very small, positively charged nucleus surrounded by a cloud of electrons. The atom is mainly empty space, so that most alpha particles pass through the foil with practically no deviation.

• If the alpha particles, however, come too close to the nucleus, the strong Coulomb repulsion between the nucleus and the positively charged alpha particle will cause the alpha particle to deviate from its original direction. The repulsive force indicates that the nucleus must be of the same nature as the alpha-particle.

• The retracing of a few alpha-particles resolved that the atomic nucleus must be real, massive, hard and physical. (ans) [4]

(b) Uranium-235 nucleus:

• at rest, and • absorbs a slow neutron and undergoes fission.

• Sometimes the fission don’t emit any neutrons.

One such fission:

235 1 9892 0 52 yU n Te Zrx+ → +

The masses of these particles are

uranium 23592U 235.0439 u

tellurium 52Tex 137.9603 u

zirconium 98y Zr 97.9197 u

neutron 10n 1.0087 u.

(i) Nucleons: 235 1 98x+ = +

⇒ 138x = (ans)

Protons: 92 0 52 y+ = + ⇒ 40y = (ans) [2]

KE

0 xox

EPE

TE

energy

–xo

GPE

fluorescentscreen

gold foil

radium source

α -particles

highest point lowest point equilibrium position

energy

gold atom

alpha nuclei

2008 Nov

2008 − 29

(ii) Assume all particles are initially at rest.

Energy released in the fission = mass excess equivalent

= (235.0439 u + 1.0087 u − 137.9603 u − 97.9197 u) × 2c = (0.1777 u) × 2c

= ( )227 80.1777 1.66 10 3.00 10−× × ×

= 112.65 10−× J (ans) [4]

(iii) Part of the energy released, 112.3 10−× J were used to become kinetic energy of the tellurium and zirconium nuclei.

∴ The remaining energy released may be in a form of photon(s). (ans) [1]

(iv) 1. Initial momentum is zero, hence final momentum

must also be zero

⇒ Zr Zr Te Tem v m v= (oppositely directed)

∴ the ratio speed of zirconium nucleusspeed of tellurium nucleus

Zr

Te

vv

= Te

Zr

mm

= 137.9603 u97.9197 u

=

1.41= (3sf) (ans) [2]

2. The ratio kinetic energy of zirconium nucleus

kinetic energy of tellurium nucleus

Zr

Te

K.E.

K.E.=⎛ ⎞⎜ ⎟⎝ ⎠

21

Zr Zr221

Te Te2

m vm v

= 2

Zr Zr

Te Te

m vm v

⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

21

Te Zr

Zr Te

m vm v

−⎛ ⎞⎛ ⎞

= ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

Zr

Te

vv

⎛ ⎞= ⎜ ⎟⎝ ⎠

1.41= (ans) — [2]

(v) 11Zr TeK.E. K.E. 2.3 10−+ = × —

Combining and :

11ZrZr

K.E.K.E. 2.3 10

1.41−+ = × ⇒

11ZrK.E. 1.346 10−= × ⇒

2 111Zr Zr2 1.346 10m v −= × ⇒

( )27 2 111Zr2 97.9197 1.66 10 1.346 10v− −× × = ×

⇒ Speed of the zirconium nucleus,

7Zr 1.29 10v = × m s-1 (ans) [3]

(vi) Two assumptions made in the calculation in (iv) part 1.:

1. The fission takes place in isolation.

2. Newtonian laws of motion are obeyed at atomic level. (ans) [2]



2008 − 30

Notes:

v3-sg.AL.themis - H2 Physics - Yearly Solutions - 2008 ...

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