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1 SAJC 2010 Prelims/9646 Solutions [Turn Over 2010 SAJC H2 Physics Prelims Examiner’s Report Paper 1 : Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 B A B A D B D A A A 35.1% 52.0% 91.3% 45.3% 70.4% 68.5% 37.7% 32.6% 54.2% 32.4% Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 B C D C A D A D D C 62.0% 45.8% 71.3% 80.1% 82.2% 51.1% 74.1% 69.8% 50.5% 54.8% Q21 Q22 Q23 Q24 Q25 Q26 Q27 Q28 Q29 Q30 A B C A C B B A D B 58.9% 62.2% 32.4% 23.4% 51.3% 62.9% 88.5% 89.7% 36.1% 79.6% Q31 Q32 Q33 Q34 Q35 Q36 Q37 Q38 Q39 Q40 B C B B B D C A A B 80.1% 61.3% 94.1% 40.3% 55.2% 44.2% 87.2% 41.4% 68.8% 83.2% Average score = 24/40, figures in % refers to % of cohort who got answer correct. 1 B A typical saloon car, like a Nissan Sunny, is about 1500 kg. Mrs Quek’s car has an unladen mass of 1410 kg. A Mercedes Benz weighs about 2250 kg. An old SAF military truck used to weigh 3 tonnes (which is why it was called a 3-tonner but these days the new version is 5 tonnes (ie. 5000 kg) The cultural centre has a ground floor area of 709.5 m 2 . A yellow bunsen flame is typically above 1000 ºC and a blue flame can go as high as 1500 ºC. The upthrust is ρgV. Density of water, ρ = 1000 kg m -3 . g is 9.81 and the volume of a tube float (totally submerged) is estimated by taking the diameter of its cross-section to be 15 cm and the diameter of the whole tube (top-view) to be 70 cm. This will give a volume of 0.0389 m 3 and an upthrust of 381 N. 2 A Identify the formula => v 2 = u 2 + 2as, u = 0, hence, v 2 = 2as Calculating v => v = (2as) = (2)(23.1)(100) = 67.97 Fractional error of v => v v = ½ a a + ½ s s Uncertainty of v, v => 1.075 v in one sig. fig => 1 v in same dec. pl => 68 Hence, (v ± v) => (68 ± 1) m s -1
26
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Page 1: 2010 Sajc h2 Prelim Solutions Print

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SAJC 2010 Prelims/9646 Solutions [Turn Over

2010 SAJC H2 Physics Prelims Examiner’s Report Paper 1 :

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10

B A B A D B D A A A 35.1% 52.0% 91.3% 45.3% 70.4% 68.5% 37.7% 32.6% 54.2% 32.4%

Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20

B C D C A D A D D C 62.0% 45.8% 71.3% 80.1% 82.2% 51.1% 74.1% 69.8% 50.5% 54.8%

Q21 Q22 Q23 Q24 Q25 Q26 Q27 Q28 Q29 Q30

A B C A C B B A D B 58.9% 62.2% 32.4% 23.4% 51.3% 62.9% 88.5% 89.7% 36.1% 79.6%

Q31 Q32 Q33 Q34 Q35 Q36 Q37 Q38 Q39 Q40

B C B B B D C A A B 80.1% 61.3% 94.1% 40.3% 55.2% 44.2% 87.2% 41.4% 68.8% 83.2%

Average score = 24/40, figures in % refers to % of cohort who got answer correct.

1 B A typical saloon car, like a Nissan Sunny, is about 1500 kg. Mrs Quek’s car has an unladen mass of 1410 kg. A Mercedes Benz weighs about 2250 kg.

An old SAF military truck used to weigh 3 tonnes (which is why it was called a 3-tonner but these days the new version is 5 tonnes (ie. 5000 kg) The cultural centre has a ground floor area of 709.5 m2. A yellow bunsen flame is typically above 1000 ºC and a blue flame can go as high as 1500 ºC. The upthrust is ρgV. Density of water, ρ = 1000 kg m-3. g is 9.81 and the volume of a tube float (totally submerged) is estimated by taking the diameter of its cross-section to be 15 cm and the diameter of the whole tube (top-view) to be 70 cm. This will give a volume of 0.0389 m3 and an upthrust of 381 N.

2 A Identify the formula => v2 = u2 + 2as, u = 0, hence, v2 = 2as

Calculating v => v = √(2as) = √(2)(23.1)(100) = 67.97

Fractional error of v => ∆vv

= ½∆aa

+ ½∆ss

Uncertainty of v, ∆v => 1.075 ∆v in one sig. fig => 1 v in same dec. pl => 68 Hence, (v ± ∆v) => (68 ± 1) m s-1

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3 B Option A => g will always be overstated (error in one direction) Option B => The experimenter’s timing could be below or above the true

value, thus, its error can be in either direction. This constitutes random error.

Option C => timing will always be understated (error in one direction) Option D => measure will either be always overstated or always understated

(error in one direction) 4 A

The ball rebounds at 0.6 s. Area under trapezium in first 0.6 s where the ball drops to the ground = 3.6 m. The ball reaches a maximum height at 1.5 s. Hence, area of triangle from 0.6 s to 1.5 s is the upward trajectory of the ball before it changes direction, and it is also 3.6 m. Therefore, the vertical displacement between the starting point and highest point is zero.

5 D

The equation that links the height and time is sy = uyt + ½gt2

In this case, for both x and y, they have the same sy (height) and uy (both zero) and g (constant at 9.81 m s-2), Hence, the time will be same for x and y.

6 B For the ball travelling to X from either Y or Z, there is no friction or other resistive forces. Hence, the speed at X comes purely from the loss in GPE from Y or Z.

Loss in GPE = Gain in KE mgh = ½mv2

Hence, v α √(h) => v'v

= √(2hh

) => v’ = √(2) v = 1.41v

7 D

Note that answer cannot be C because the air resistance cannot cause an object to change direction horizontally. This is because when an objected afflicted by air resistance slows down to zero speed, air resistance will then cease to act on the object, as air resistance is proportional to velocity (or square of velocity). Therefore, the air resistance acting horizontally will drop to 0 when the horizontal component of the velocity is 0.

8 A

If the resultant force acting on charge X is horizontal to the left, the force that charge at Z acts on X (ie FZX) must be repulsive, hence, Q must be a negative charge (since X has a negative charge). Also, for vertical equilibrium =>

FXY = vertical component of FZX

( )2

04 r

qq

πε =

( )2

0 24 r

Qq

πεsin45° ,where r = side of the square

Q = 2.8q

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9 A When the passenger feels weightless, there is no normal reaction force on him (ie. N = 0). To do this question, form an equation for if the vehicle is at the top of the track (ie. options A or B) and when the vehicle is at the bottom of the track (ie. options C or D). The option where N would have been negative is the answer.

At the top => mg – N = mv2

R => N = mg -

mv2

R

In option A, v > √(gR), and N would have been negative. Hence, option A is the answer.

10 A

Since person is moving in a circle, the resultant force acting on the person must be pointing towards centre of circular motion path (ie. to the right). For that to be possible, there must be a frictional force (upwards) to counter the weight. Hence, the forces exerted by the cage are normal reaction and friction and the resultant of these two will be arrow A. The weight is not included as it is exerted by the Earth and not the cage.

11 B

The slope is at an incline of 20°. Since car is travelling at a constant speed, the resultant force = 0 (because there is no acceleration). T – mgsin20° - 500 = 0 T = mgsin20° + 500 = 3855.22 N Power, P = Tv = (3855.22)(8) = 30842 W

12 C

Force is a vector and depending on the direction declared, the force can be either positive or negative with respect to the declared direction. However, potential energy U is a scalar and gravitational potential energy is always negative, hence, answer is between options B and C.

Since F α 1r2 and U α

1r ,

Graph C gives the more accurate relative shapes for U and F.

thrust, T

weight, mg Friction, 500 N

friction

normal reaction

weight

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13 D If the gravitational potential at X, ΦX, = -8 kJ kg-1, then ΦY would be -4 kJ kg-1,

according to the formula, Φ = - GM

r .

Work done = m∆Φ = (1)[ΦY – ΦX] = (1)[(-4) – (-8)] = 4 kJ Also note that the work done has to be positive because it is by an external agent and energy is required to bring an object against the gravitational field.

14 C

Kinetic energy depends also on mass (ie. KE = ½mv2) so if the mass is different, the KE would be different even if the velocity is the same.

15 A

Since pV = nRT, at constant volume and no change in gas quantity, both V and n as well as R are constants, and p α T. Hence, the graphs that shows a direct relationship between p and T is A or B. However, the temperature scale is in Celsius, and at t = 0°C or 273 K, the pressure could not possibly be zero, thus, the answer is option A.

16 D With reference to the 1st Law of Thermodynamics, ∆U = ∆Q + ∆W,

the temperature at both A and B are similar, meaning there is no change in internal energy from A to B and ∆U = 0. Also, there is expansion from A to B so ∆W is negative, which means ∆Q must be positive, indicating that heat is supplied to the gas.

17 A Work done by gas refers to an expansion. Work done on gas refers to a compression.

From Y to Z, it is a compression, hence, the work done by gas should be negative, and the area under YZ is 300 J. Thus, work done by gas from Y to Z = -300 J. The net work done by the gas is the area of the rectangle WXYZ and it is 600 J. It is positive because the work done through expansion process from W to X is larger than the work done through compression from Y to Z.

18 D Option A => speed at P is zero as it is a turning point for particle

Option B => displacement at Q is zero only at that instant, not always. Option C => at R where the speed is zero, the KE is zero. Option D => S is a point of maximum displacement, hence, its acceleration is

maximum, afterall, a = -ω2x and a is max. when x is max. (Note that each particle vibrates vertically in SHM)

19 D

Critical damping requires the object to return to equilibrium immediately and in the case of option D, the effect took place so fast that the passenger hardly noticed it, hence, critical damping.

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20 C Let the maximum number of bright fringe on either side be nmax. dsinθ = nλ, but for nmax, sinθ = 1

nmax = dλ =

0.001400

550 x 10-9 = 4.545.

n must be an integer so nmax = 4, the maximum on either side. But the total number of bright fringes that can be observed includes the central bright fringe and all the fringes on the other side, so the total number is 4 + 1 + 4 = 9

21 A The path difference between S1X and S2X is 1 m or 2λ.

If the two sources are in phase, the path difference is in the form of nλ so it would be constructive interference. In this case, the two sources have a phase difference of π radians, meaning they are in anti-phase, hence, the outcome is destructive interference.

22 B The diameter of the kink is 1.5 cm or 0.015 m, hence, the radius is 0.0075 m.

ε = -∆NBA∆t

= (1)(2.5 x 10-2)(π0.00752)

0.050 = 8.8369 x 10-5

P = ε2

R = 4.8806 x 10-9 W

23 C

To do this question, trace on the diagram the path of the current in one cycle and then using Fleming’s LHR, figure out that the magnetic force acting on P is downwards. Now trace the current round the circuit in the other direction and you will realise that the force is also acting downwards.

24 A

Bev = mv2

r => Be =

mvr

=> Be = mrω

r =>

Bem

= ω = 2πT

Hence, T = 2πmBe

and independent of velocity.

25 C

Electric field at a point is its potential gradient at that point, or dVdx

.

Hence, if the potential around the point is constant, its gradient there would be zero and the electric field there would be zero.

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26 B

Bx is parallel to the direction of the current on wire Y, so there is no force on P due to Bx. The magnetic force on P is contributed by By (downwards) and IY (to the right). Hence, by Fleming’s LHR, the force is into the page.

27 B

F = Q1Q2

4πεor2 => F α

1r2 =>

F'F

= (rr' )2 => F = (

12 )2(1.5) = 0.375 N

28 A

Electric field always points from higher to lower potential, hence, EPX and EXQ are both to the right.

Also, we can see that the vertical components of the electric fields at Y would cancel out, leaving only the horizontal components pointing to the right. The magnitudes of EQY and EPY would be the same since Y is equidistant from P and Q.

Potential at a point due to P = VP = 2µ

4πεor . Also, VQ =

-2µ4πεor

VTotal = VP + VQ is always zero because rPX = rQX and rPY = rQY. 29 D

Option A => RBC is indeed 0.5 Ω. Option B => If another resistor, say 1 Ω, is connected in parallel to AC, the

resultant resistance is 0.462 Ω. Option C => RAD is indeed 1 Ω. Note that resistor between B and C serves no

purpose as no current will go through there because the potential at B and C are equal. Since VCD = 0, IBC = 0.

Option D => This statement contradicts the factual statement in option C.

P Q

X

Y

2 µC - 2 µC

EQY

EPY

P Bx

By B

B = B-field due to wire X acting on point P Bx = horizontal component of B-field By = vertical component of B-field

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30 B

R = ρLA

= ρL

πd2

4 ; Hence, R α

Ld2

Since L and d are both doubled, resistance is halved, and current is doubled. 31 B Flux linkage is the main function of the soft-iron core. 32 C For steady (dc) current : P = I2R

For alternating current : P = (Io√2

)2R4

Equating them : Io = √8 I = 2√2 I 33 B This is one of the four observations of the Photoelectric Effect Experiement. 34 B

The probability of the electron tunnelling the barrier depends on the transmission coefficient, T = exp(-2kd), where d = barrier thickness, L

and k = 2

2

h

)EU(m8 −π

Hence, for X=> T α exp(-√(5E – E) L) = exp(-2EL) for Y=> T α exp(-√(17E – E) 2L) = exp(-8EL) for Z=> T α exp(-√(2E – E) 3L) = exp(-3EL)

Thus, X has the greatest T value (transmission coefficient) followed by Z and then Y.

35 B It is an electron (not photon) incident on the gas atom, so the ‘all-or-nothing- rule does not apply. 20.0 Xx 10-19 J of mechanical energy can raise the atom to the -5.4 x 10-19 J level or the -2.4 x 10-19 J level. If it is raised to the -2.4 x 10-19 J level, the remaining energy which becomes the kinetic energy is 0.6 x 10-19 J, which is not one of the four answer options given. Hence, consider that the atom is raised to the -5.4 x 10-19 J level, which means that -5.4 x 10-19 J – (-21.8 x 10-19 J) = 16.4 x 10-19 J is being absorbed by the atom. Hence, the ke transferred to the free electron is 20.0 x 10-19 J - 16.4 x 10-19 J = 3.6 x 10-19 J

36 D

It is a fact that heating and cutting is usually takes place at the energy level of the infrared region.

37 C This is a reiteration of this fact which many students got wrong in BT2.

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38 C During reverse bias condition of a p-n junction, the p-type semiconductor becomes more (rather than less) negative. The depletion region widens, as the holes in the p-type semi-conductor migrate to the positive terminal of the battery. Hence, statement A is false.

39 A

Equation : U23592 + n1

0 Cs14355 + Rb90

37 + 3 n10

Binding Energy : 235(7.6) + 0 + 195 = 143X + 90X + 0 X = 8.502 MeV

40 B

After 16 years,

NNo

= (12 )n

31250500000

= (12 )n

116

= (12 )n

(12 )4 = (

12 )n

n = 4

Therefore, 16 years is 4 half-lives. t1/2 = 4 yrs

After 12 years, NNo

= (12 )n

M

500000 = (

12 )12/4

M

500000 = (

12 )3

M = 62500

t / years N 0 500000 4 250000 8 125000 12 62500 16 31250

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Paper 2 : Tutor’s comments/Common Mistakes Section A

1 Generally, students must: • explain what is happening,

• give a sensible example which illustrates how the graph can arise

Graph A : - The body travels at a constant speed, suddenly reverses and travels with a lower constant speed in the opposite direction. - example : ball bouncing horizontally against wall / ball hit by racquet - This occurs when there is a collision. The lower speed after the collision shows that KE is lost. A poor example: a car moving in one direction and suddenly turn around and move in the opposite direction. Due to its large mass, a car cannot possibly achieve such a stunt in an extremely short time.

Graph B : - acceleration going from zero to a high value. - example : dropping an object from rest - acceleration changes suddenly as the force holding the object in place is

removed A poor example: a boy pushing a supermarket trolley at constant speed and suddenly push it with a constant acceleration. It is difficult for anybody to maintain a trolley at constant speed or acceleration.

Graph C : - constant force being applied drops to zero suddenly.

- example taking foot off car accelerator/ceasing to pedal bicycle - so zero subsequent acceleration A common wrong example is a body/car accelerating and suddenly stopped by an obstacle. In this case, there is a large negative force experienced by the body before it remains at rest with zero resultant force.

2 (a) GPE∆

( ) ( )A A B Bm g h m g h= ∆ + ∆

2.00 9.81 0.500sin40 5.00 9.81 ( 0.500sin50 )= × × ° + × × − °

12.481 J= − 12.5 J (3 sig fig)= −

(b) Loss in GPE = Gain in KE + Work done against friction

12

mv2 = 12.481 – 0.500 x 3.00 x 2

12

(2.00 + 5.00) v2 = 9.481

-11.65 m s (3 sig fig)v =

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Alternatively :

ΣF = ma => 5gsin50° - 2gsin40° - 3 x 2 = (5 + 2)a a = 2.709 m s-2 v2 = u2 + 2as => v2 = 02 + 2(2.709)s v = 1.65 m s-1

(c)

Ek = 12

mv2 = 12

m (u + at)2 = 12

m (0 + at)2 = 12

ma2t2 ⇒ Ek ∝ t2.

Wf = F x d = F x (ut + 12

at2) = F x (0 + 12

at2) = 12

Fat2 ⇒ Wf ∝ t2.

The three graphs should add up to 12.5 J. So, Ep should curve in the way shown above. The question should have defined Wf as work done against frictional force to obtain a positive work done. Hence, students were not penalized for drawing a (- kt2) graph as the question mentioned work done by frictional force.

3 (a) (i) ω = dθdt

= 2π

24 x 60 x 60 = 7.27 x 10-5 rad s-1

(ii) (Gravi force) – (Tension) = m r ω2 Tension = 19.66 – (2.00)(6.37x106)(7.27x10-5)2 = 19.59 N

(iii) Part of the gravitational force on the mass supplies the centripetal force for the mass to move in circular motion.

(b) • Astronaut and spacecraft are both accelerating at the same value towards the Earth.

• No (normal) reaction force experienced by the astronaut. Both astronaut and spacecraft are falling at the same rate towards the centre of the Earth and hence, are not pushing against each other.

Energy / J

Time 0

Ep

Wf (∝ t2)

Ek (∝ t2)

12.5

9.5

3.0

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4 (a) The magnetic flux density is defined as the force experienced by a conductor of unit length and carrying unit current, placed at right angles to the magnetic field.

(b) (i) pointing upwards (ii) F = BIL

B = 16000

10 x 0.30 = 5330 T

(iii) Advantage: no wear and tear due to moving parts, low noise

Two-word answer such as “environmentally friendly” will not be accepted. Students should at least state why it is environmentally friendly such as no fossil fuel is needed and therefore no emission of harmful gases.

Disadvantage: high magnetic field is required

“High current needed” is not a good answer as 10 A is not considered large for such a huge vehicle. In fact, students could have mentioned that it is an inefficient vehicle compared to propeller-powered crafts.

(c) (i)

f(x)=5*sin(2*x)*exp(-x/5)

x

y

The graph above assumes that only the South-pole oscillates in and out of the loop. In this case, one cycle of SHM corresponds to one cycle of the graph.

(ii) Changing direction: As the south pole approaches, the induced current produced in the coil flows in such a way that a repulsive force is produce. When the South pole withdraws from the coil, the induced current flows in the opposite direction so as to produce an attractive force.

The induced current flowing in the coil will generate heat in the coil. The energy lost is not recovered by the system. Hence, the amplitude decreases with time.

Changing directions Decreasing magnitude

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However, if it interpreted that the whole magnet goes right through the loop, then the graph may look like the one shown below.

In this second case, one cycle of SHM corresponds to 2 cycles of the graph.

5 (a) A.C. can be easily stepped up or down using a transformer

Some students only mention “stepped-up”. It is important to mention “stepped-down” as well as a.c. can be stepped-down to desired values.

(b) To reduce power loss as the transmission current is reduced. (c) (i) Given: Rc = 6 kΩ, Vload = 80 kW, Power delivered = 250 kW To show: Ic = 2.61 A By principle of conservation of energy/power, Pgen = Power Output + Power Loss in cables, Pc ie, 250 kW = IcVload + Ic

2Rc = Ic (80 x 103) + Ic

2(6 x 103) Solving, Ic = 2.61A (ii) Total voltage drop across cables = Ic Rc = 2.61 x 6 x 103 = 15.7 kV (iii) Power dissipated in cables = Ic

2 Rc = 2.612 x 6 x 103 = 41 x 103 W or, = Ic x Ans in (ii)

(iv) Power delivered to load = Ic x Vload = 2.61 x 80 x 103 = 2.1 x 105 W

or, = Pgen - Ploss = 250 kW – Ans in (iii)

I

t

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6 (a) The conduction band is the lowest energy band that is empty or partially filled when the atoms are unexcited. The valence band is the highest energy band that is completely populated (by electrons). (b) (i) The energy gap between the valence and conduction band in an insulator may be greater than the photon energy of all the given wavelengths; These photons will pass through the insulator without being absorbed. (ii) Energy levels of electrons in the conduction band are finely spaced; Photons of all visible light wavelengths have sufficient energy to excite an electron in the conduction band to a higher energy state; Photons are re-emitted from the surface as reflected light. (c) The impurities in a p-type semiconductors are trivalent atoms while the impurities in a n-type semiconductors are pentavalent atoms. The majority charge carriers in a p-type semiconductors are holes while the majority charge carriers in a n-type semiconductors are electrons.

It is wrong to say that p-type semiconductors are positively charged while n-type, negatively charged. If fact, both types are electrically neutral. It is also wrong to simply say that p-type has more holes while n-type has more electrons. Furthermore, students should avoid using positive or negative ion sites of a p-n junction to distinguish the two types of semiconductors.

7 (a) (i) F = GMm

r2 Or F = - GMm

r2

where F is the gravitational force M and m are two point masses r is the separation between the point masses

Common error: not realizing that M and m should be point masses.

(ii) GMm

r2 = m r ω2

GM = r3 (2πT

)2

T2 = 4π2r3

GM

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(b) (i)

moon

period

T/days

Mean distance from centre of Jupiter r/109m

log10(T/days) log10(r/m)

Sinope 758 23.7 2.88 10.37

Leda 239 11.1 2.38 10.05

Callisto 16.7 1.88 1.22 9.27

Lo 1.77 0.422 0.248 8.63

Metis 0.295 0.128 -0.530 8.11

(ii)

• Graph should show 5 plots

• Graph should show a straight line, passing almost perfectly through all the 5 plots

(10.45, 3.00)

(7.80, -1.00)

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(c) (i) gradient = 3.00 - (-1.00)10.45 - 7.80

= 1.51 (accept 1.43 to 1.59)

(ii) T2 = 4π2r3

GM

2 lg T = lg ( 4π2

GM ) + 3 lg r

Show: lg T = ½ lg( 4π2

GM ) +

32 lg r

Conclusion: Data support the relation because

• straight line graph with non-zero intercept is obtained

• gradient of about 1.5 is obtained

(d) When T = 7.16 days, lg T = 0.85 From the graph, lg r = 9.02 r = 1.05 x 109 m (accept 0.95 – 1.16 x 109 m)

(e) • No, because a different planet has a different mass M.

• Although the gradient remains the same, the y-intercept changes.

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8 Suggested diagram: Diagram and Procedure Marks

Labelled diagram of workable (with high voltage source) experimental arrangement Note : Battery as voltage source is not acceptable

1

Measure pressure p using pressure gauge (instrument shown on diagram is acceptable)

1

Measure high voltage V using a voltmeter/CRO (instrument shown on diagram is acceptable)

1

Increase/decrease pressure using air/vacuum pump

1

Keep air gap constant, or Keep type of gas constant (if specifically mentioned) Temp. need not be kept constant as the pressure gauge will always give the pressure within the chamber, with or without temperature change.

1

Keep pressure constant (or set p to a certain value), vary voltage V slowly until sparks occur.

1

Repeat experiment several times using different pressure

1

Analysis: Plot a graph of lg V against lg p.

1

Good design/experimental details

1) perform experiment in the dark hence enhance visibility of spark 2) increase V slowly from an initial low value until sparks are seen 3) describe a suitable pressure chamber (eg air-tight container with

toughened glass sides) 4) repeat reading for minimum voltage and take the average 5) use goggles to view spark possible danger of uv radiation 6) use safety screen around pressure chamber in case of

explosion/implosion 7) use rubber gloves when touching the circuit 8) repeat measurement of voltage V, and taking the average

1

mark each,

max. 4

Total 12

vacuum pump

digital voltmeter

tap

glass pressure chamber

pressure gauge

spark gap, d V

variable high voltage suppy

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Paper 3 : Key: ( ) optional but good to include; Tutor’s comments

1 (a) (i) Energy of each alpha particle in joules = 5 x 106 x 1.6 x 10-19 J = 8.0 x 10-13 J [ 1 ] (Power of α beam = No of α emitted per sec x Energy of each α = Activity x Energy of each α)

Thus activity required = 20

8.0 x 10-13 = 2.5 x 1013 Bq [ 1 ]

(ii) Using λ =

21

2ln

t =

0.69388 x 365 x 24 x 3600

= 2.5 x 10-10 s-1 [ 1 ]

(i) Number of nuclei required, N = Aλ

[ 1 ]

= 2.5 x 1013

2.5 x 10-10

= 1.0 x 1023 [ 1 ]

( Since sample mass, M ∝ No of nuclei in sample, N

Molar mass, Mr ∝ NA )

Thus Mass of sample required = N Mr

NA =

1.0 x 1023 x 0.238 6.02 x 1023

= 0.040 kg [ 1 ] Molar mass Mr = mass number expressed in g = 238 g =0.238 kg

(iv) Rocket may burn up in atmosphere during launch or re-entry [ 1 ] The plutonium would be vaporized; it could then be ingested by people. Once inside the body, α-particles become hazardous.

[ 1 ] Note that α-particles are not hazardous to us when they are outside our body as they cannot penetrate through our skin.

(b) A= λN =

2

1

2ln

t

N

Since masses are similar & the mass no are also approx the same, the number of active nuclei , N, in the 2 samples are about the same. Thus

A ∝∝∝∝

21

1

t.

Since half-life of Iodine >> that of caesium, the initial activity of Iodine << that of caesium. [ 1 ]

Thus the number of active Iodine nuclei decreases much more slowly

that that of caesium. Given that both their initial activities are high enough to be hazardous, the Iodine waste would remain hazardous for a longer time, before the no of active nuclei decreases to a safe level and hence should be stored in a secure place for a longer time before it can be allowed to be released into the open. [ 1 ]

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2 (a) (i) Amplitude = 0.02 nm = 2×10-11 m [ 1 ] Frequency = 1/T = 500 Hz [ 1 ]

(ii) 2 19

0

1Max KE 2.4 10

2mv −= = ×

2 19

0

1(2 ) 2.4 10

2m fxπ −= × [ 1 ]

11 2 191(2 500 2 10 ) 2.4 10

2m π − −× × × = × [ 1 ] for correct v0

41.22 10 kg (3 sig fig)m −= × [ 1 ]

(iii) [ 1 ] for elliptical shape Some are still confused about the v-x & v-t graphs [ 1 ] for value of amplitude

(b) Since the Vuvuzela can be considered as a tube with two open ends, at the fundamental frequency,

0.702

λ= A diagram to illustrate will be helpful

330

236 Hz1.4

vf

λ= = = [ 1 ]

The next 2 harmonics will be integer multiples of the fundamental

frequency, ie, 236×2 = 472 Hz and 236×3 = 708 Hz. [ 1 ] A diagram to illustrate will be helpful By generating sound at these resonant frequencies and with similar amplitudes [ 1 ] but 180º out of phase, [ 1 ]

destructive interference occurs which causes both sound waves to cancel each other out. A lot of explanations lacked the key ideas!

0.02 −0.02

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3 (a) (For that given light intensity,) all electrons ejected are already collected by the collector electrode even for a low positive voltage ie none has managed to “escape”; thus increasing to higher positive V values will make no difference to the saturation current. [ 1 ]

(b) The decreasing current in the region of negative pd is due to a

decelerating force opposing the motion of the electrons (which have a range of ke) towards the collector. This force increases with the negative pd. [ 1 ]

As the pd becomes more and more negative, this opposing force

increases to a point where even the fastest electron is prevented from reaching the collector. The current is thus reduced to zero. [ 1 ]

(c) Power P = hft

N)( ,

t

N= rate of incidence of photons.

Thus when f increases at constant power, → t

N must decrease.

Since current is proportional to t

N, current must decrease. [ 1 ]

Since eVs = hf - φ , when f increases for a given metal (ie constant φ ),

Vs must increase in magnitude [ 1 ] (d) (i) Given: Photon energy = 4.7 eV, Power = 3.8 mW, I = 0.8 x 10-8 A Thus, energy of each photon = 4.7 x 1.6 x 10-19 = 7.52 x 10-19 J [ 1 ]

Since power P = hft

N)( ,

Rate of incidence of photons t

N=

hf

P =

3.8 x 10-3

energy of photon

= 5.1 x 1015 s-1 [ 1 ]

V ext

I

0 -Vs

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(ii) Max rate of electron emission, nt =

Ie I =

Qt = e

t

n)(

= 0.8 x 10-8

1.6 x 10-19

= 5.0 x 1010 s-1 [ 1 ] Quite a number of candidates are confused about the rate of incidence of photons & the rate of emission of electrons

(ii) Why rate of electron emission<<rate of incidence of photons 1. Not every photon would hit an electron; most are reflected by

the metal or miss hitting any electron.

2. On the way out to the metal surface, an electron may lose its kinetic energy to ions and other electrons it encounters along the way. This energy loss prevents it from overcoming the work function & so such electrons are absorbed by the metal. [ 2 ]

One mark per underlined, maximum two marks 4 (a) (i) (Current in the 2 identical lamps ie same resistance in parallel

must be the same, ie = 0.5 A. Thus, by Kirchoff’s 1st law, ) Current through A1, Itot = 0.50 + 0.50 = 1.0 A [ 1 ] Applying Kirchoff’s 2nd law to the left loop, E.m.f. = PD across L1 + Itot r PD across L1 = 2.0 V – 1.0(0.2) = 1.80 V [ 1 ]

(ii) Connecting L3 in parallel to L1 and L2 will lower the effective/total resistance of the whole circuit

Some thought there will be no change. [ 1 ]

hence, total current (A1) increases Itot = Emf/Effective resistance [ 1 ]

(b) (i) VAB = RTotal

RAB x PD across Total R potential divider

= ( 2.0

8.0 + 2.0 ) 2.00 = 0.40 V = 400 mV [ 1 ]

Some candidates gave their answer in terms of volts 0.40 V but the Examiner requires the answer in mV! You need to be

mindful of the specified unit of all calculated quantities and abide by it!

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(ii) At balance, e.m.f. of Cell Y = PD across balance length

= AB

JB

R

Rx VAB

= AB

JB

L

Lx VAB since R=

ρl

A , R ∝∝∝∝ length

= 100

)30100( −x 400 mV

= 280 mV [ 2 ]

(iii) If contact J is closer to A, ⇒balance length is to be made longer; many incorrectly thought that the balance length is given by AJ, when it is actually JB.

(Since, at balance, Emf of cell under test (ie Y) = PD per unit

length of potentiometer wire x Balance length), to increase the balance length, we can:

• Replace the resistance wire with one of lower resistivity,

or, use a thicker wire of the same resistivity R= ρl

A [ 1 ]

• Replace Cell X with one of lower e.m.f. [ 1 ]

• Replace the 8.0 Ω resistor with one of higher resistance. [ 1 ] Changing the cell Y to one of smaller emf is considered an invalid (ie illegal) modification due to the “context of the question”, which is to determine the emf of cell Y. Additional Question: Suggest why in an expt to determine the emf of an unknown source (like Y in this case,) it is desirable that the balance length be long rather than short.

5 (a) (i) Newton’s 1st law: Every object continues in its state of rest or

uniform motion/const speed in a straight line, unless a net (external) force acts on it to change that state. [ 1 ]

The law implies that a “force” is that quantity which is required to change the velocity of an object. [ 1 ]

Most did not know how this law leads to the concept of force

(ii) Mathematically, resultant force, F ∝ dt

mvd )( = k

dt

mvd )( [ 1 ]

(where the value of the proportionality const k would depend on the definition of the unit of force.)

Since the definition of the newton was chosen as “the force which causes a mass of 1 kg to have an acceleration of 1 m s-2,

thus, 1 N = k x 1 kg x 1 m s-2

⇒ k = 1 [ 1 ]

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Thus, F = 1dt

mvd )(

= m

dvdt

(since m = const) [ 1 ]

= ma (shown)

(b) As the stone accelerates towards Earth, its downward momentum is

increasing from an initial value of zero momentum. [ 1 ]

Since there is no net force acting on the system of stone & Earth, the total momentum of the stone-Earth system must remain unchanged, ie = 0. [ 1 ]

Thus, the Earth must accelerate upwards towards the falling stone. [ 1 ]

Assume air resistance is negligible (c) (i)

[ 1 ] (ii) Given KEα = 9.2 x 10-13 J Thus ½ mαvα

2 = 9.2 x 10-13 J, where mα = 4u = 4x1.66x10-27 kg [ 1 ]

→ Vα = 27

13

1066.14

102.92−

xx

xx

= 1.67 x 107 = 1.7 x 107 m s-1 (shown) [ 1 ] (iii) Since no net force is involved in radioactive decay, & since total momentum before decay = zero (as Ra is initially stationary), [ 1 ] by principle of conservation of momentum, 0 = mαvα - mRnvRn scalar form of the principle [ 1 ]

Thus, vRn = mαvαmRn

= (4u)(1.7 x 107)

220u

= 3.1 x 105 m s-1 [ 1 ] (d) (i) Centre of gravity of an object is defined as that point through which the entire weight of the object may be considered to act. [ 1 ]

(ii) Principle of Moments:

For a body to be in (rotational) equilibrium, the sum of all the anticlockwise moments about any point must be equal to the

sum of all the clockwise moments about that same point. [ 1 ] The condition, “For a body to be in equilibrium” must not be omitted!

HeRnRa 4

2

220

86

224

88 +→

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(iii)

Moment exerted by the weight about the base = mg x (2.3cos450) [ 1 ] = 180 x 9.81 x 2.3cos450 [ 1 ] = 2.9 x 103 N m (shown) (iv) Applying the principle of moments about the base, T x (4.0 sin350) = 2.9 x 103 [ 1 ] → T = 1.26 x 103 = 1.3 x 103 N [ 1 ] 6 (a) (i) In general, internal energy = sum of p.e. + k.e. of molecules [ 1 ] For ideal gas, with no force of attraction between molecules, the molecular pe = 0 [ 1 ]

Since ke of molecules ∝ its absolute temperature, ½ m <c2> ∝ T,

internal energy ∝ absolute temperature ie in Kelvin [ 1 ]

(ii) Since the minimum k.e. of molecules is 0 assuming classical physics is applicable; recall that according to Heisenberg Uncertainty Principle, the molecules cannot be at rest, [ 1 ]

the corresponding minimum absolute temperature is also 0, referred to as the absolute zero or, cannot be negative. [ 1 ] (b) “heated steadily” means thermal energy is being supplied at const

rate; internal energy is increasing ALL THE TIME but what is happening to the ke & the pe, bearing in mind that U = ke + pe for this non-ideal gas system? Recall that pe depends on the intermolecular force which increases as the intermolecular

separation increases while ke ∝ T. The above facts & the deduction of the phase at the various sections

of the temp-time graph were not well understood by the majority. (i) AB: As temp of the pure liquid H2O rises from room temp to its bp,

its internal energy rises k.e. increases with the temperature [ 1 ]

separation, and thus p.e. remains constant

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If you consider the expansion as not negligible, then the pe can be said to be increasing very slightly. [ 1 ]

(ii) BC:

Thermal energy supplied is being used to vaporise the liquid to gaseous state

k.e. remains constant as temperature is constant [ 1 ] separation, and thus p.e increases greatly as volume expands during vaporisation [ 1 ]

(iii) CD: Temp of the gas increases within the sealed container, ie

volume remains const k.e. increases with temperature [ 1 ]

As separation remains constant, pe remains constant [ 1 ]

(c) (i) Amount of gas in moles, n = mM

, where m is the mass of the

gas In a question which requires candidates to “show”, any new symbol introduced by the candidate - m, in this case - MUST be defined.

Density ρ = mV

; p = nRT

V =

(mV

)RT

M [ 1 ]

= ρRTM

[ 1 ]

Note that the symbol M should have been defined as the mass

PER mole of gas. (ii) This question proved to be one of the most challenging

questions because candidates did not know how to use the data to verify the following 2 characteristics of an exponentially decreasing function. Recall the radioactive decay law

The exponential function can be represented by: p = p0e

-kh where k & p0 are positive constants. Thus, ln p = lnp0 – kh. A graph of lnp vs h must then be a straight line with a

negative gradient; so candidates could tabulate at least 4 sets of lnp & h values to verify this.

Alternatively, using the concept of half-life, show for 3 values of p, that the time for each value to decrease to half its value is the same or approx equal

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(iii) p at 8 km = 3.5 ± 0.3 x 104 Pa [ 1 ]

ppo

= ρ

ρo since R, T & M are constants

ρ = 0.35 x 1.3 = 0.455 kg m-3 [ 1 ]

(iv) pρT

= constant

105

1.3 x 293 =

3 x 104

ρ x 250 [ 1 ]

At summit, ρ = 0.457 kg m-3 [ 1 ] Some converted the temp in 0C to Kelvin incorrectly

7 (a) (i) Time taken = 80 x 10-3

2.50 x 107 = 3.2 x 10-9 s [ 1 ]

(ii) tan 30º = Vy

2.50 x 107 tan θ =

x

y

v

v [ 1 ]

vy = tan 30º x 2.5 x 107 = 1.44 x 107 m s-1 [ 1 ]

(iii) (Ave vert) Acceleration = t

Vy 0− = 4.51 x 1015 m s-2 [ 1 ]

Quite a number used vy

2 = uy2 + 2aysy . Substituting ½ x40x10-3 m

for sy is incorrect as Sy can’t be assumed to be = ½ the plate separation.

(iv) a = eEm

(assuming mg<<qE for electron here)

= e(V/d)

m [ 1 ]

V = adm

e = 1.02 x 103 V [ 1 ]

(b) (i) Pointing perpendicularly into the page [ 1 ]

Apply FLH rule with FB (thumb) pointing vertically down the page & current finger pointing from right to left horizontally

(ii) electric force = magnetic force (since there is no deflection) ie eE = Bev v= velocity as electron as it enters plates[ 1 ]

B = Ev

= Vdv

= 1.02 x 103

40 x 10-3x 2.5 x 107

= 1.03 x 10-3 T [ 1 ]

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(iii) The most energetic X-ray photon is produced when a bombarding electron loses ALL its KE in one single collision with the tungsten target.

Thus ½ mv2 = hc

λ [ 1 ]

λ = 6.99 x 10-10 m [ 1 ]

Why is it incorrect to use λ = mv

h here as some did?

(c) (i) Since the electron is moving horizontally from left to right, by FLH rule, FB will be vertically down the page, causing a circular path as shown. The arc must begin right from the point of entry into the B field, ie, there must not be a hint of a linear path within the magnetic field! All diagrams used to illustrate a key point must not be ambiguous.

[ 1 ]

(ii) Bevsin900 = mv2

r [ 1 ]

r = 0.138 m [ 1 ]

(iii) No change to speed because the magnetic force is always perpendicular to the velocity recall FLH rule and hence cannot change the speed/ cannot do any work on the electron. [ 2 ] In all such questions, the gravitational force is typically considered negligible

(d) (i) Each electron behave like a wave. [ 1 ] The electron wave diffracts upon passing the micro-grid [ 1 ] The diffracted waves interfere/superpose with each other to produce an interference pattern. [ 1 ]

(ii) a “wider pattern” means that θ in dsin θ = nλ is larger

Using electrons with longer wavelength. [ 1 ]

Since λ = mv

h an alternative answer is: reduce the speed of the

electrons v, or, by reducing the accelerating potential difference that is applied to the electrons.

End of Solutions