2007 H2 Prelims Structured Questions
Ans – 1m Solutions to AJC 2007 Physics Prelim Paper 21(a)(i) As the apple is released from the balloon, it is traveling at the same speed
of the balloon at the moment of release. Apple’s speed at moment of release = 2.0 m s–1
Ans – 1m (ii) Apple’s acceleration = 9.81 m s–2 (experiencing free fall) (iii) Using s = ut + ½ a t2,
Subs –1m Conversion – 1mAnswer – 1m
(↓) 200 = – 2.0 t + ½ (9.81) t2 t2 – 4.0/9.81 t – 400/9.81 = 0
Using a2
ac4bbx
2 −±−= , we get t = 6.59 s or –6.18 s (N.A)
Incorrect for (a 1 & 2) Correct for (a 1 & 2) Initial speed of apple
0 m s–1 2.0 m s–1
Acceleration of apple
4.0 m s–2 9.81 m s–2
Part (a 3) marking scheme
(0, 4) (2, 9.81) (0, 9.81) or (2, 4) or (0, any other values)
Any wrong values subs
√ subs × subs × subs × subs √ convrs √ convrs √ convrs √ convrs × ans √ ans × ans × ans Marks 2 2 1 1 1(b) At constant speed, net F acting on object is 0 N.
Mthd – 1m Weight of object = resistive force
(78) (9.81) = k (62) 2 Ans – 1m
Mthd – 1m
k = 0.199 Base unit of F = (Base units of k) (Base units of v2) Base units of k = (kg m s–2) / (m2 s–2) = kg m–1 Ans – 1m
2007 AJC H2 Solutions
2 (a) The direction of the object changes continuously and hence there is a change in velocity of the
object. [1] Based on Newton’s first law, this means that there must be a force acting on the object. [1] Since the speed remains unchanged, this force cannot have a component tangential to the circle. This implies that the force is perpendicular to the velocity/displacement of object (so towards centre of circle). [1]
(b) (i)
Length of vector L must be longer than vector W and vertical components of L must equal that to W [1]
Horizontal component of L must be directed towards centre [1] L
× centre
W W vertically downwards from CG of plane [1]
(ii) Resultant force = centripetal force = mv2/r = (5.0 × 104)×(0.20 × 103)2 / (15 × 103) = 1.3 × 105 N [1] Direction = towards centre of (horizontal) circle [1]
3(a) Two sources are said to be coherent if the waves emitted from them have the same frequency and have a zero or constant phase difference. [1]
(b)
(i) slit separation ≤ 1.0 mm [1] Use fringe separation, x = λD/d x = 1.04 mm [1]
(ii)
1. the bright fringes would be less intense [1] 2. the bright fringes would be less intense [1] and the dark fringes would not be completely dark
[1] 3. fringe pattern will appear and disappear at every half cycle of rotation [1] 4. fringe separation would be reduced [1]
colour of fringes would be yellow [1]
2007 AJC H2 Solutions
4(a) I = 2.0/(1.0 + 5.0) = 0.33 A [1]
(b)(i) Vxy = ( 0.20.10.5
0.5+
) = 1.67 V [1] XP = 1.5/1.67 = 0.90 m [1]
(ii) V10Ω = ( 5.1 )r10
10+
= r10
15+
[1]
75.00.1
VV
10
XY =Ω
( )
75.00.1
15r1067.1=
+ [1]
r = 1.97 Ω = 2.0 Ω [1]
2007 AJC H2 Solutions
0 0.05 0.10 0.15 0.20 0.25
t/s
φ/ wb 0.01
5(a)(i)
t1.0t)0.1)(10x0.5)(0.2()vt(BL 2 ===Φ −
Max φ is reached when t = (10 x 10-2)/1.0 = 0.10 s
wb01.00.1
10x101.0va1.0
2
max =⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟
⎠⎞
⎜⎝⎛=Φ
−
Time interval for constant magnetic flux = (15.0 -10.0) x 10-2 / 1.0 = 0.05 s 1 mark for general shape of graph, 1 mark for the correct interval (0.10 s to 0.15 s) for constant magnetic flux 1 mark for labelling the Max φ
(ii)
E / V
V 1.010.001.0
tE ==
ΔΔΦ
=
1 mark for general shape of graph, 1 mark for labelling the Max E (b)
• When the rod QR cuts the magnetic field, an induced current flows in it. • the magnetic force on QR is in opposite direction (by Fleming’s left-hand rule) and equal
in magnitude to the applied force. [3]
(c) External force on QR, Fext = FB = BIL
A 00625.016
1.0REI === [1]
Hence Fext = BIL
= (2.0)(0.00625)(5.0 x 10-2) [1] = 6.25 x 10-4 N = 6.3 x 10-4 N [1]
0 0.05 0.10 0.15 0.20 0.25
0.1
t/s
-0.1
2007 AJC H2 Solutions
6(a) a quantum of electromagnetic energy [1]
(b) - not traveling in direction of crystal - some pass through the crystal
[1] for a sensible suggestion (c) (i) 1. 1.3 – 1.5 cm [1] 2. smooth curve drawn through points [1]
acceptable line as tangent, length at least half length of curve [1] gradient = 0.07 cm-1 ± 0.01 [1] correct unit [1]
(ii) Q takes the same distance to halve in value [1] (d) (i) fraction of total number of photons directed towards crystal [1]
becomes smaller as x increases [1]
(ii) for low photon energies, all photons directed towards crystal are absorbed [1] high energy photons can penetrate crystal [1]
(e) (i) when S is in contact with the crystal [1] only half of all emissions move towards the crystal [1] (ii) if S is at the bottom of the well [1] some photons directed ‘backwards’ will be detected [1] so efficiency will increase [1]
2007 AJC H2 Solutions
2007 Prelim H2 Paper 3 Solutions 1 a) (i) Work is done when a force moves its point of application in the direction of the force. The work done by a force is the product of the force and the distance moved in the
direction of the force. (ii) power is the work done (or energy converted) per unit time. [2] (b) (i) lead particles fall and lose potential energy (as tube is inverted), this is converted to
kinetic energy. Kinetic energy converted into heat energy on impact. [2]
(ii) 1. Total change in potential energy = 50 x mgh = 50 x 0.025x 9.81 x 1.2 = 14.7 J [1]
2. [1] θΔ=Δ mcQ
14.7 = 0.025 x c x 4.5 c = 130.7 J kg-1 K-1 [1]
2 (a) Maximum possible upthrust = (1.0 X 103)g(0.15) (1) (b) Weight of dinghy and n children on it = (5.0 + 30n)g, where n is the number of
children in the dinghy (1) To stay afloat, Weight of dinghy and n children ≤ Maximum possible upthrust (5.0 + 30n)g (1.0 X 103)g(0.15) (1) ≤ n ≤ 4.83 n = 4 (1) 3(a)(i) The gravitational force exerted on a 1kg (or a unit) mass on the surface (or near to the surface) of Mars is 3.7 N [1] and is an attractive force. [1] (ii) By Newton’s 2nd Law, F = ma where F = Gravitational force on mass m and a = acceleration of free fall For 1 kg mass, 3.7 = (1)a [1]
a = 3.7 m s-2 (b)(i) Area between horizontal axis and graph [1]
1
2007 AJC H2 Solutions
(ii) estimated no of squares = 206 (accepted range: 202-210) 1 square : (0.5 × 106) × 0.1 = 5 × 104 J kg-1 [1] estimated energy required for a mass of 1 kg to escape from the gravitational field of Mars = estimated area = 206 × 5 × 104 [1] = 1.0 × 107 J (iii) k.e. at surface of planet = estimated energy required for a mass of 1 kg to escape from the gravitational field of Mars Hence (1/2)(1) v2 = 1.0 × 107 [1] Escape speed v = 4.5 × 103 m s-1 [1] (iv) By conservation of energy, kinetic energy of mass is converted to gravitational potential energy as the mass travels further away from Mars. To escape from gravitational field of Mars, the mass has to travel to a large distance away from Mars, Hence, k.e. at surface of Mars + G.P.E at surface of Mars = G.P.E. + k.e at infinity
------ idea for conservation of energy [1] (1/2)mv2 + (- GMm/r) = 0 v = (2GM/r)1/2 Escape speed is independent of mass Escape speed = 4.5 × 103 m s-1 [1] (c) 1. Gravitational pull from Earth is much stronger than that from Mars. So less fuel needed for rocket to accelerate towards Earth on the return journey. [1] 2. Escape speed from Mars is less than that from Earth. So less fuel needed for rocket to reach an initial speed required to leave Mars than to leave Earth. [1] 4ai) The total kinetic energy of both vehicles is conserved.
Ans – 1m OR: The relative speed of approach is equal to the relative speed of separation.
Ans – 1m 4aii) The subsequent motion of the two vehicles are collinear.
4b) Relative spped of separation = relative speed of approach
Ans – 1m Mthd – 1m
v’c – v’t = – (vc – vt) = – [(–188) – 40.5] = 228.5 m s–1
2
2007 AJC H2 Solutions
4c) By conservation of momentum, (→) mt vt + mc vc = mt v’t + mc v’c
Mthd – 1m (3200)(40.5) + (200)(–188) = (3200)v’t + (200) v’c ----(1) 92000 = 3200v’t + 200 v’c
Mthd – 1m 460 = 16v’t + v’c From (a), velocity of separation = v’c – v’t = 228.5 ----(2)
Ans – 1m + 1m Solving the 2 equations, v’c = 242.1 = 242 m s–1 & v’t = 13.62 = 13.6 m s–1
OR 4c) By Conservation of Kinetic Energy
Mthd – 1m ½ (3200)(40.5)2 + ½ (200)(188)2 = ½ (3200)( v’t)2 + ½ (200)( v’c)2 61588 = 16 v’t2 + v’c2
Mthd – 1m From (a), velocity of separation = v’c – v’t = 228.5 ⇒ v’c = 228.5 + v’t
Ans – 1m + 1m Substituting, we get v’c = 242.1 = 242 m s–1
& v’t = 13.62 = 13.6 m s–1 5 (a) At low temperature, valence band remains full, conduction band empty and the semiconductor is an insulator. [1]
Conductivity increases with temperature because some valence electrons acquire thermal energy greater than the energy gap[1] and hence move into the conduction band to become free (conducting) electrons. [1]
(b) n-type silicon is doped with phosphorus (pentavalent atom) n-type silicon has free electrons (rather than positive holes) n-type silicon has a lattice of positive ions
(any TWO of the above, 1 mark each) [2] DO NOT accept n-type is negative, p-type is positive
(c) In junction, movement of electrons to holes (and holes to electrons) leaving the n-type positive (and the p-type negative). This depletion layer is widened when the n-type is made positive with a battery, so no current but decreased when the n-type is made negative allowing continual removal of electrons from the p-type region and supply of electrons to the n-type region. [4]
OR
3
2007 AJC H2 Solutions
In forward bias, the positive terminal of the battery is connected to the p-type of the junction & the negative terminal to the n-type. The height of the potential barrier at the junction will be lowered (by the applied forward voltage). Holes cross the junction to the n-type and electrons (cross the junction) to the p-type and constitute a current. The width of the depletion region will decrease. [2]
In reverse bias, the negative terminal of the battery is connected to the p-type of the junction & the positive terminal to the n-type. The height of the potential barrier at the junction will be raised (by the applied reverse voltage). Hence majority carriers cannot cross the junction resulting in no current. The width of the depletion region will increase. [2]
6.(a)(i) 1. Simple harmonic motion is a motion where the acceleration is propotional to the displacement and directed opposite to it. [2] 2
acceleration displacement [1] (ii) 1. v =ω( xo
2 – x2 )1/2
0.3 = (2π/T) ( 0.102 – 0.062 )1/2 [1] T = (2π/0.3) (0.0064)1/2 = 1.676 = 1.7 s [1] 2. vmax = ωxo = (2π/1.676)(0.10) [1] = 0.375 = 0.38 m s-1 [1] (iii) Fnet = ma W – N = ma N = W – ma = m (g-a) [1]
4
2007 AJC H2 Solutions
When N = 0 as the mass losses contact with the platform, [1] a = g [1] g = ω2xo
xo = g/ω2 = 9.81 / (2π/1.676)2 = 0.6976 = 0.70 m [1]
(b) (i) When the frequency of wave generated by the earthquake is equal to the natural frequency of the building, resonance occurs where the vibration amplitude is the greatest. [1] [1]
amplitude of vibration of the building
fo natural frequency of the building
frequency of earthquake
(ii) The maximum aplitude will be lowered. The resonant frequency may be shifted to the left. All amplitudes at the different frequencies will be lowered. [3] (c) (i) dsinθ = nλ 1/400000 (sinθ) = (1)(570 × 10-9) [1] θ = 13.179o tan θ = x/2 tan 13.179o = x/2 x = 0.468 = 0.47 m [1] (ii) Now there are 200 lines per mm, thus d will increase. With increase in d, each of the nth orders found on the screen will be closer to the central maximum/ more orders can be seen [1] and the intensity maximums are less bright. [1] 7 (a) (i) 1. Electric field strength E: force per unit positive charge (at the point in the field)
[1] 2. Electric potential V: work done per unit positive charge in bringing the charge
from infinity to the point in the field. [1]
5
2007 AJC H2 Solutions
(ii) E = dxdV
− where dxdV is called the potential gradient (in the x-direction) [1]
Negative sign indicates that the direction of electric field strength is in the
direction of the decreasing potential. [1] (b) (i) 1, 2, 3. [1] [1] [1]
+Q
A
y
EA
r
B
C
-Q
Fig. 7.1
r
EB
ER x
(ii) EA = EB = 20 r
Q4
1πε
[1]
ER = 2 ( 20 r
Q4
1πε
cos 60°) [1]
= 8.99 x 109 2rQ [1]
(iii) Potential at C = rQ
41
0πε +
rQ
41
0
−πε
[1]
= 0 [1]
(c) (i) E = dVΔ [1]
6
2007 AJC H2 Solutions
50,000 = 3u
1012V300−×
− [1]
Vu = - 300 V [1] (ii) 1. ke lost = EPE gained
21 m v2
min = e V [1]
21 (9.11 x 10-31) v2
min = (1.6 x 10-19) x 600 [1]
vmin = 1.45 x 107 m s-1 [1] 2. F = e E = (1.6 x 10-19)(50,000) [1] = 8.0 x 10-15 N Direction: vertically downwards. [1] or directed vertically towards the lower plate 8(a) (i) — electron in an atom have specific energy [1]
— electron falls from one energy level to a lower energy level with release of electromagnetic wave of certain frequency ( or loss of energy during
transition release a photon of frequency f = E / h). [1] — certain frequencies only therefore line spectra [1]
(ii) atoms must be in gaseous form [1] electrons at high energy states implies gas at high temperature [1] (b) (i) 1. ΔE = 0.65 eV [1] 2. ΔE = 0.65 eV = 1.6 x 10-19 J x 0.65 = 1.04 x 10-19 J [1] f = ΔE / h [1] f = 1.6 x 1014 Hz [1] (ii) 1. five absorption lines [1] 2. n = 2 to n = 1 only [1] (c) (i) Speed is well below speed of light, hence momentum can be found from its
nonrelativistic formula: p = px = m v = (9.11 x 10-31 kg)(2.05 x 106) = 1.868 x 10-24 kg ⋅ m s-1 [1] Δp / p = 0.5/100 ⇒ Δp = 9.34 x 10-27 kg ⋅ m s-1 [1]
7
2007 AJC H2 Solutions
[1]m1065.5
1035.922/1063.6
p2xhence 9
27
34−
−
−
×=××
π×=
Δ≈Δ
h (ii) Δx is about 100 atomic diameters [1] which implies the electron do not have a
sharply defined position and that it is found with equal probability [1] anywhere along the x-axis.
(d) (i) ( ) 19
234
19312
2
2
1067.6)1063.6(
)106.1)(1.58.6)(1011.9(88k −−
−−
×=×
×−×=
−= m
hEUm ππ
[1]
8
60.10-2kd 1045eeT −− ×==≈ [1]
(ii) In the case of the electron, its transmission coefficient means for every 1
million electrons, 45 will tunnel through [1] but the transmission coefficient for the proton is so enormously reduced that for every 1 million protons practically none [1] will be able to tunnel through.
2007 AJC H2 Solutions
CJC H2 Prelim Solutions 2007
1
CJC Physics Preliminary Examinations 2007
Paper 1
1 C v = √(0.62 + 0.82) = 1.0 m s-1, s = v t = (1)(5) = 5 m
2 C Assume 1 heartbeat per second, 1 hour = 3600 sec ⇒ 3600 heartbeats ⇒ 103
3 C
Component of weight down the slope = Mg sin θ Net force = Mg sin θ = Ma When M = m, mg sin θ = ma, hence a = g sin θ When M = 2m, 2mg sin θ = 2ma, hence a = g sin θ. There is no change in the acceleration a. a will only change when θ changes.
4 B Distance traveled before brakes are applied = 13 x 0.7 = 9.1 m
Using ( )
m 8.185.42
130
a2
uvs ,as2uv
22222 =
−−
=−
=+=
Total distance traveled = 9.1 + 18.8 = 27.9 m. Hence he stops 27.9 – 25 = 2.9 m beyond the line.
5 B The free-body diagrams on A and B are as below.
Since A is pulled with constant velocity, A and B does not accelerate. Hence for body B, T = 12 N For body A, F + T + 12 = 40, F = 40 – 12 – 12 = 16 N
6 D Area under F – t graph gives us the impulse acting on the body. Total impulse from t = 75 ms to t = 150 ms is equal to area of two trapeziums, = ½ (60 +80) (25) + ½ (80 + 40 ) (50) = 4750 N ms = 4.75 N s F ∆t = m ∆v, 4.75 = 0.2 x (v – 15), v = 38.9 m s-1
7 C Vertical height of liquid above point X = 2 x l sin 45O = l √2
ρgl = P ρg(l √2) = P’, hence P’ = P √2
B
A
12 N T
12 N T F
40 N
θ Mg
θ
CJC H2 Prelim Solutions 2007
2
8 D F t = m (v – u) ⇒ (6)(4) = (2) (v – 0) ⇒ 24 = 2v ⇒ v = 12 m s-1 Kinetic energy = ½ m v2 = ½ (2) (12)2 = 144 J
9 B By definition, gravitational field strength = gravitational force per unit mass
10 B GMm/r2 = mω2r ⇒ GM/r2 = (2π/T)2 r ⇒ T2 = (4π2 / GM) r3 ⇒ T ∝ r 3/2
11 D The force which the hinge exerts on the shelf must satisfy two conditions.
1. Provide an anti-clockwise moment about the CG of the shelf to
counterbalance the clockwise moment due to the tension in the chain about the same point.
2. Provide a horizontal force to the right to counterbalance the horizontal force to the left due to the tension in the chain.
Only force D satisfies the above two conditions.
12 A Resolving vertically, N cos θ = mg --------------------(1)
Resolving horizontally, N sin θ = mω2r ---------------(2)
(2)/(1) ⇒ tan θ = v2/gr All of the three equations are correct and possible.
13 A Triple point of water is precisely at 0.01oC and reproducible with simple lab apparatus easily.
14 C In one minute, energy supplied would raise temperature by 4 K In 40 minutes, energy supplied would raise temperature by 160 K That is, Q = m L = m c (160 K)
⇒ c / L = 1 / 160 K
15 D A Barton's Pendulums experiment demonstrates the physical phenomenon of resonance and the response of pendulums to vibration at, below and above their resonant frequencies.
16 A E = ½ m ω2 A2
In general, Ex = ½ m ω2 (A2 – x2)
At A/2, E1 = ½ m ω2 (A2 – [A/2]2) = ½ m ω2 (A2 – A2 /4) = ½ m ω2 (3A2/4) = 3E/4
17 B Diffraction is the "Bending” of wavefronts around obstacles. During diffraction, the wavelength should remain constant. We may use Huygens’ rule: “Each point on a wavefront acts as a source of secondary wavelets. The combination of these secondary wavelets produces the new wavefront in the direction of propagation”.
18 B Using Young Double Slit’s formula, x = Dλ/a
Increasing λ would increase x, the fringe spacing
19 A Time taken = 2 seconds to move one cm Speed = distance / time = 1 / 2 = 0.5 cm s-1
20 C At P, the intensity appears 4 times as bright as a single source.
The intensity alternates between bright and dark, or sinusoidally at high
CJC H2 Prelim Solutions 2007
3
frequency, although it is easily observable.
21 D C is incorrect because the arrows of field lines at the right-hand “charge” are pointing at two different directions. D is relatively correct although it only shows the upper half of the field lines between the two point charges.
22 D Using F = ma ⇒ eE = m a ⇒ a = e(V/d) / m = 5.3 X 1017 m s-2
23 D Total energy dissipated
= total energy generated by the battery = 2500 + 1500 = 4000 J E = W/ Q = 4000/ 2000 = 2.00 V
24 D A
ρlR = Since vol is constant, v1.1l
1.1
A=×
Resistance of new wire= 'Rρ
1.1
A
1.1l=× =
A
ρl21.1 = 1.2 R
25 B When the resistance of the top resistor is maximised to R, potential difference between X and Y = 3.0 V When the resistance of the top resistor is minimised to 0, potential difference between X and Y = 0 V
26 D Read the below from right to left.
Total resistance of circuit = 30 + 10 = 40 Ω I = V / R = 240 / 40 = 6.0 A
internal resistance 10 Ω
Series, total 100 Ω
Parallel with 300 Ω, total 75 Ω
Series, total 150 Ω
Parallel with 75 Ω, total 50 Ω
Series with 10 Ω, total 60 Ω
Parallel with 60 Ω, total 30 Ω
E.m.f. 240 V
CJC H2 Prelim Solutions 2007
4
27 A Using right hand grip rule, B field at P has a component acting upwards, using Fleming’s Left Hand rule, force is out of the page.
28 A Using Fleming’s Left Hand rule, force is east wards. Speed v of the proton can be calculated using ½ mv2 = 5 x16 x10-13. ½ x 1.6 x10-27 v2 = 5 x16 x10-13. solve for the value of v Sub into equation F = B q v = 7.4 x10 -12 N
29 C Faraday’s Law of EMI, induced e.m.f. equal to rate of change of flux linkage E= change of flux linkage/time =NB(A2-A1) cos 22.5/time =8x 50 x10-6 (0.124-0.120)cos 22.5 /1.59 =0.93 x10-6 V
30 C Using equation Q= change of flux linkage /Resistance =[NAB-(-NAB)]/R =2NAB/R =2 x50 x0.02x0.25/100 =0.005 V
31 A
oV
oI
oP = ,
22
oV
oI
oP
P =>=<
32 B R
rmsIP 2>=<
5
202 =rms
I
Arms
I 2=
33 C Of the three energy levels, Z is the lowest and Y is the lowest, hence produces
radiation of the shortest wavelength of 200nm.
34 B Energy of photon of blue light is higher than that of red, hence the number of photons of blue light must be lesser than that of red light, since the power of both beams are the same.
35 A E = qV = hc/λ ⇒ λ = hc/qV
36 D It could be debatable that Beta Decay should be explained by Electroweak theory. (Out of Syllabus) The important concept: A barrier, may be a form of energy state, analogous to a "hill" or incline in classical mechanics, which would suggest that passage through or over such a barrier would be impossible without sufficient energy. Hence, the term used is energy barrier or potential barrier rather than physical barrier.
37 C The closer the path to the nucleus, the greater the angle of scattering. It should occur when it is relatively nearer to the nucleus.
38 C By conservation of momentum, the Thorium, the heavier nucleus, will have a slower speed and smaller amount of kinetic energy. The alpha particle, will have a higher speed, and greater amount of kinetic energy.
CJC H2 Prelim Solutions 2007
5
39 D Energy of the pulse = 40 x 30 x 10-3 = 0.12 J Let n be the no. of photons emitted
E = n hλ
c
n = hc
λE= 3.74 x 1017
40 B Electrons and holes are urged towards the p-n junction causing the depletion
layer to narrow.
Paper 2
1(a)(i)
( ) t9.812
110t5
at2
1 ut s
m 5s ,ms 9.81a ,ms 10u
2
2
21
−=−
+=
−=−== −−
1
s 2.45
2(4.905)
5)4(4.905)(10)(10)(t
0510t4.905t
2
2
=
−−−±−−=
=−−
1
OR
( )
ms1.14v
sm 198.1
(-5)9.81-2102as u v
1-
2-2
222
−=
=
+=+=
s 2.45
81.9
101.14
a
u-vt
atuv
=−
−−==
+=
(ii) energy total initial
4
1mgh ×=
1
g8
uh
4
1h
mu2
1mgh
4
1
2
0
2
0
+=
+=
1
m 52.2
81.98
105
4
1 2
=×
+×=
1
OR Let un be the velocity before the nth impact vn be the velocity after the nth impact
CJC H2 Prelim Solutions 2007
6
KEn be the kinetic energy after the nth impact KE0 be the kinetic energy before the 1st impact KE1= ¼ KE0
( )
ms 7.04
14.1-2
1
u2
1v
mu2
1
4
1mv
2
1
1-
11
2
1
2
1
=
−=
−=
=
( ) m 52.2s
s9.81-204.70
2as u v2
22
=
+=
+=
(iii)
mu128
1
mu2
1
4
1mgh
2
2
1
3
=
×
=
1
m 0.158
81.9128
14.1
128g
u h
2
2
=×
=
=
1
(b)(i) 45cos15u Ox = ½
30cosRs O
x = ½
R0816.0
45cos15
30cosR
u
st
tus
O
O
x
x
xx
=
=
=
=
1
(ii) 45sin15u Oy = ½
30sinRs O
y = ½
CJC H2 Prelim Solutions 2007
7
2
2
2OO
2yyy
2
y
t81.9t2.21
t81.9t215R
t)81.9(2
1t45sin1530sinR
ta2
1tus
ms81.9a
−=
−=
−+=
+=
−= −
1
(iii) Sub t = 0.0816 R into the above equation
( )
m2.11Ror)NA(0R
0R732.0R0.0654
R0816.081.9)R0816.0(2.21R2
2
==
=−
−=
1
2(a)(i) Ayg W watereasρ=
1
(ii) Azg W waterreshfρ=
1
(iii)
y
z
AzgAyg
waterfresh
watereas
waterreshf watereas
=ρ
ρ
ρ=ρ
1
(iv) No. Since the ship is floating, the upthrust on the ship must always be equal to its weight.
1
(b)(i)
1
(ii)
O6.9
2.1
2.0sin
=θ
=θ
½
R = ke = 10 (0.2 + 0.3)
θ 1.2 m
0.2 m
Tension in string T
F Tension in spring R
Weight = 2g
θ
CJC H2 Prelim Solutions 2007
8
= 5 N
½
Resolving forces vertically, g2cosT =θ (1)
½
Resolving forces horizontally, RsinTF +θ= 5FRFsinT −=−=θ (2)
½
N 32.8
56.9tan81,92F
tang25F
g2
5Ftan
g2
5F
cosT
sinT ),1()2(
O
=
+××=
θ=−
−=θ
−=
θθ
÷
1
3(a) (i) Longitudinal 1 (ii) 1. wavelength = 0.5 m 1 (ii) 2. amplitude = 0.5 mm 1 (ii) 3. correct substitution into speed = frequency × wavelength;
to give v = 660 × 0.5 = 330 m s–1
2
(b)(i)
(b)(ii) A A
NA
N
pipe A pipe B
(i)correct wave shape for pipe A; correct wave shape for pipe B; (ii) correct marking of A and N for pipe A; correct marking of A and N for Pipe B;
B1
B1
(iii) for pipe A, λ = 2L, where L is length of the pipe;
c = f λ to give L = f
c
2;
substitute to get L = 0.317 m;
2
(iv) for 32 Hz, the open pipe will have a length of about 5 m; whereas the closed pipe will have half this length, so will not take up as much space as the open pipe
1
4a (i) Placed in vacuum so that the electrons are not scattered on its way to the anode, since mass of air molecule is so much larger than that of electron.
1
(ii) Using the equation hf = work function + eV h x4.5 x 1014 =Ф +1.6 x10-19 x0.25
1
CJC H2 Prelim Solutions 2007
9
h x6.5x 1014 =Ф +1.6 x10-19 x 1.00 Solve for values of h and work function h= 6.0 x 10-34 Js Ф =2.3 x10-19 J
1 1 1
5 (a) (b) (i)
(ii)
(c)
(d)
Using PV = nRT P(0.10) = (1)(8.31)(273.15 + 40) P1 = 2600 Pa Using PV = nRT P(0.40) = (1)(8.31)(273.15 + 40) P2 = 650 Pa Or use P1V1 = P2V2 P2 = P1/4 = 650 Pa
after process 1Change in internal energy = 0 J (No change in temperature) for one cycle Change in internal energy = 0 J (No change in temperature) Total work done = Area enclosed = 3610 – (0.40 – 0.10)(650) = 1660 J U = Q + W = 0 Q = – W = –(–1660) = +1660 J (Heat flows into the system)
3 1 1 2 3
6(a)
- Accuracy of Points plotted - Scale must be appropriate (Fit into space provided/ not too small) - Graph of R against T plotted - Labelled axis
1 1 ½ ½
2600
2700
2800
2900
3000
3100
3200
3300
3400
3500
3600
3700
3800
R/ Ω
1 2 3 4 5 6 7 8 9 10 T/ ºC
CJC H2 Prelim Solutions 2007
10
(b)(i) Best fit curve must be drawn. Curve must cover temp range from 0 ºC to 10 ºC.
1
(ii) From graph, Resistance at 0 ºC = 3820 Ω Resistance at 100 ºC = 2620 Ω
1 1
(c) Straight line drawn and the variation with temp between 0 ºC and 10 ºC shown. (Straight line can be line of best fit)
1
(d)(i) Temp at which resistance is 3060 Ω = 6.3 ºC
1
(ii) Difference in Temp = 6.3 – 5.2 = 1.1 ºC
21.1%
100%5.2
1.1
=
×
1 1 1
(e)(i) Refer to graph [Check for uncertainty in R and T]
2
(ii) The experimental uncertainties can be linear as the uncertainty at 9.6 ºC takes into consideration values of the linear line. (Answer depends on the straight line student has drawn)
2
Paper 3A
1(a)(i) 1. T + 300 – 70 g = 70 a T – 386.7 = 70 a (1)
1
2. T – 300 – 20 g = 20 a T – 496.2 = 20 a (2)
1
(ii) (1) – (2), 496.2 – 386.7 = 70 a – 20 a 50 a = 109.5 a = 2.19 ms-2
1
Sub a = 2.19 ms-2 into (1) T = 70 (2.19) + 386.7 = 540 N
1
(iii) R = 2T = 2 x 540 = 1080 N 1
(b)(i) By conservation of momentum, Momentum before collision = Momentum after collision mu = (M+m)v
1
u
mM
mv
+=
1
(ii) Initial KE = 2mu
2
1
Final KE = ( ) 2vmM2
1+
CJC H2 Prelim Solutions 2007
11
( )
)mM(2
um
umM
mmM
2
1
22
2
+=
++=
1
mM
m
mu2
1
)mM(2
um
KE total Initial
ncompressio maximum at KE Total
2
22
+=
+=
1
(iii) 1
KE total Initial
ncompressio maximum at KE Total< because some of the initial KE is now
converted into elastic potential energy stored in the spring,
½
some of it is lost as heat when compressing the spring, and the rest is manifested as the KE of the composite body.
½
2a (i) Work done on m converts to kinetic energy = F x s Kinetic energy = ma x s Since from kinematics v2 = u2 + 2as v2 - u2 = 2as Hence, kinetic energy = m (v2 - u2) /2
1 1
(b)(i) Using power = F x velocity 180= F x15 ms-1
F friction = 12N
1
(ii) On a slope F driving = F friction +mg sin 10 o
F driving = 12 + 50 x9.81 sin 10 o
Power = F driving x v =97.1 x15 =1460 W
1 1
3(i) Vo = (1/4πεo)(Q/ro)
1
(ii) ro = (1/4πεo)(Q/Vo) So ro = (9 x 109)(-2.2 x 10-11 / -10) = 0.0198 m = 1.98 cm
2
(iii) 1 mark for curve shape; 1 mark for horizontal line 2
CJC H2 Prelim Solutions 2007
12
4(a) 12
24
4 resistor 4 across Pd ×
+=Ω
1
= 8 V
1
(b) Let resistance of AB be R
V 8 AJand resistor 3 across Pd =Ω
1
820
2R3
R28.03=×
+++
1
8.33 R
20 R 2.4
40 R 8 R5.660
5)(R8R)0.28(320
Ω=
=
+=+
+=+
1
Q5 (i) - Correct value for max input power - Correct graph
1 1
(ii) 169.7
2
240
2
oV
rmsV === V
sV
169.7
1
50
sV
pV
sN
pN===
1 1
- 10
P/ W
t/ s
40
CJC H2 Prelim Solutions 2007
13
3.39sV = V
(iii) Heat loss due to the resistance in the wire of the transformer.
1
6(a) 1a. External energy source (eg. flashes of light or by high-voltage discharges) used to induce population inversion 1b. Majority of atoms are in the excited metastable state than in the grd state Population Inversion 2. The laser process begins when an excited atom spontaneously emits a photon parallel to the axis of the tube. 3. When this photon strikes an excited atom, it could stimulate the atom to emit another photon. [Explanation: These 2 photons, in turn, stimulate 2 more excited atoms, yielding 4 photons, the 4 photons yield 8, and so on. There is an exponential rate of increase.] 4. Mirrors (at both ends of the tube) repeatedly reflect the photons through the laser medium stimulation process multiply + maintain and enhance population inversion. 5. The partially reflective mirror at one end allows some of the photon to emerge from the tube to form the coherent laser beam.
½ ½ 1 1 1 1
b(i) n-type: Electrons p-type: Holes
(ii) For n- type doping, impurities with 5 valence electrons (eg. Arsenic) are added to an intrinsic semiconductor.
• Only 4 of the 5 valence electrons of each impurity atom will participate in the bonding.
• The impurities donate electrons to the semiconductor. • The energy level of the “extra” electrons is usually very close to the
base of the conduction band. • Therefore, they require very little energy to transit to the conduction
band. Hence n-type doping helps reduce the electrical resistivity of pure semiconductors and increase their conductivity. For p- type doping, impurities with 3 valence electrons (eg. Boron) are added to an intrinsic semiconductor.
• Only 3 of the 5 valence electrons of the impurity will participate in the bonding leaving a hole, associated with each impurity atom.
• The energy level of the “extra” holes is usually just above the top of the valence band.
• Hence it is easy for valence electrons to move into this hole as the impurities are seen to readily accept these electrons.
Therefore, p-type doping helps increase the electrical conductivity of pure semiconductors.
½
½
½
½
½
½
½
½
Paper 3B
7(a)(i) speed of object at Earth’s surface; so that it will escape from the gravitational field / travel to infinity;
2
(ii) gravitational potential energy at Earth’s surface = (–)
eR
GMn
This amount of energy is required for probe to escape from earth.
3
CJC H2 Prelim Solutions 2007
14
However, the kinetic energy provided is not sufficient.
(There is no need to calculate the escape velocity to obtain full marks;
however if students provide the working, marks will be awarded)
(b) (i) change = GMm
−
RR
11
e
; Accept GMm
−
e
11
RR
1
(ii) in orbit,
2
2
r
GMm
r
mv= ;
R
GMmmv
2
2
21 = ;
2
(c) idea of equating energies; Kinetic energy supplied on surface = ke in orbit + pe in orbit
R
GMm
R
GMm
R
GMm
R
GMm−+=
ee 24
3
R = 2Re;
height above surface = Re
3
(d) (i) probe collides with air molecules (inelastic); giving them kinetic energy and so probe loses energy;
2
(ii) Greater density;
Higher speed,
Acceptable: shape, sizes, surface area, direction of flight of probe
Not acceptable: mass (because question is on friction, so we are looking at factors affecting friction)
2
(iii) Height becomes less
Speed increases
2
(e) Spin the probe
Such that v = (√rg’) = 100 m/s
The centripetal force acts on the foot of the astronaut, an equal and opposite force acts on the walls by the foot of the astronaut (Newton’s 3rd Law)
3
8(a) Magnetic flux density B is defined as the force acting per unit current on a
unit length of conductor placed at right angles to the magnetic field.
2
(b) (i) Clock wise moment due to the mass m= anticlockwise moment due to the force acting on BC mg x AF = BIL x AB since AF = AB
1
CJC H2 Prelim Solutions 2007
15
Hence mg = BIL B=mg/(IL)
1
(ii) mg = BIL mg =(µnI) IL 0.1x10-3 x9.81 =4π x 10-7 x 1800 x I2 x 0.025 I = 4.17A
1 1
(iii) Acting down wards
1
(iv) Fleming Left Rule
1
(c) BIL sin 30o = mg BIL sin 30o = (A x L x density) 1.8 x10-3 x I x sin 30 =10-6 x 7.9x103. I = 86.1 A
1 1
(d)(i)
(d)(ii)
1 1 1
Flux linkage = NAB = 1 x 50 x10-4 x 2= 0.01T Time take t for the coil to enter the field = 0.1/1 = 0.1 s Induced e.m.f. = gradient of the graph = 0.01/0.1 =0.1 V
(d)(iii) Direction of induced current can be obtained using Fleming right hand rule When coil is entering current is flowing from S to Q anticlockwise When coil is leaving current is flowing from T to P current in clockwise direction.
1 1
(d) (iv) Force acting on the coil is F= BIL F =B L (E/R) = 2.0 x 0.05 (0.1/16) = 6.25 x10-4 N
1 1 1
Flux/Tesla
1.0x10-2
T
t/s 0.1 0.15 E/V
0.1
CJC H2 Prelim Solutions 2007
16
9(a)(i) (ii) (iii) (iv) (v) (b)(i) (ii) (iii) (c)(i) (ii) (iii) (iv) (v)
(53.7 ± 0.3) – 40 = 13.7 ± 0.3 as
(0.81 – 0.40) × 10-8 = 4.1 × 10-9 m
(0.81 – 0.40) × 10-8 / 13.7 × 10-18 = 2.9 × 108 m s-1
0.1 × 10-8 m / (35 – 8) × 10-18 = 3.7 × 107 m s-1
Initial momentum = 9.11 × 10-31 × 3.7 × 107 = 3.37 × 10-23 Ns
Change in momentum = 6.74 × 10-23 N s to the left 1. Electron in a potential well must be considered as a wave function 2. the square of the amplitude of the wave function gives the probability of finding the electron at a point 3. the electron as a wave has a small probability of existing outside the well 4. unlike a particle which could not escape without gaining energy.
Using ∆x ∆p ≥ h/4π or h /2π or h
E = (∆p)2/2m = 1.53 X 10-19 J, 6.12 X 10-19 J Gradient = 0.0857 year-1
λ = 2.718 X 10-9 s-1
e16.46 = 1.4 X 107
A = λ N ⇒ N = A / λ = 5.18 X 1015 The gradient is equal to the decay rate and is independent of the amount of radioactive isotopes. What Mr. bean has recorded should be readings obtained from the GM detector. This should be regarded as count rate rather than activity K corresponds to background radiation 0.83 implies that 83% of the radiation was recorded by the GM tubes. The GM tubes are relatively well arranged, surrounding the disposed waste.
1 1 1 1 2 3 1 2 2 2 1 1 2
Hwa Chong Institution (College) 2007 C2 Preliminary Examinations H2 9745 Physics Suggested Solutions Paper 1 1 D 11 A 21 C 31 B 2 D 12 D 22 D 32 B 3 C 13 A 23 B 33 D 4 A 14 A 24 D 34 C 5 B 15 A 25 D 35 B 6 D 16 C 26 C 36 C 7 B 17 C 27 C 37 C 8 A 18 B 28 A 38 A 9 D 19 B 29 D 39 C 10 B 20 C 30 B 40 A Paper 2 Q1 (a) Upthrust is the upward force exerted by a fluid on a body immersed in the fluid (due to
the pressure difference in the fluid). (The magnitude of upthrust is equal to the weight of the fluid displaced) [1]
(b) [1] UWF −= dgVmg fluid−=
[1] dgVmg obj−=
dgDmmg −=
)1(Ddmg −=
(c) (i) By principle of moments: Fstandardx=Fsamplex [1]
)1()1(tan
tansample
sampledards
dards Ddgm
Ddgm −=− [1]
)0.940
290.11()8493
290.11(17851.0 −=− samplem [1]
msample = 0.17873 kg [1]
(ii) % error = %123.0%10017873.0
17851.017873.0=×
− [1]
Q2 a) The Moon is moving in a direction that is tangent to a circle with its centre at the Earth.
The gravitational attraction provided the centripetal force for the Moon to move in a circle around the Earth. [2]
1
2007 HCI H2 Solutions
Comment: Without the gravitational attraction towards the Earth, the Moon will move in a straight line. The gravitational attraction pulls the Moon towards the Earth in such a way that it keeps falling but never reaches the Earth, causing the Moon to circle round the Earth.
b) Work Done = (GPE at 300 km above the Earth) - (GPE at surface of the Earth)
⎟⎠⎞
⎜⎝⎛−−
+−=
RGMm
hRGMmW [1]
where M is Mass of the Earth, R is radius of the Earth and h is height of satellite above the Earth
2RGMmmg =
2gRGM = Hence
( ) ⎟⎠⎞
⎜⎝⎛
+−=+
+−=
hRRmgRmgR
hRmgRW 1
2
[1]
( )( )( ) JW 866
66 1043.8
103.01038.61038.611038.681.9300 ×=⎟⎟
⎠
⎞⎜⎜⎝
⎛×+×
×−×= [1]
ci) In 12 hours, the satellite would have traversed an angle of π3 radians. [1]
Hence orbital period, hT 83
212=
×=
ππ
[1]
cii) Gravitational force = centripetal force
2
222
2
42T
mrrT
mrmr
GMm ππω =⎟⎠⎞
⎜⎝⎛== [1]
31
2
2231
2
2
44 ⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
ππTgRGMTr
( )( ) ( ) mr 73
1
2
226
1003.24
360081038.681.9×=⎟
⎟⎠
⎞⎜⎜⎝
⎛ ××=
π [1]
Q3
(a)A is the intercept on the y-axis consistent with line drawn (or by implication); = 12.6 × 102 = 1.26 × 103 Pa m3 [1]
B is the gradient;
= 56
2
100.110)180(
10)8.106.12( −×−=×−×−
[2]
(b) B = 0; [1]
2
2007 HCI H2 Solutions
(c) horizontal line at PV = 1.26 × 103 Pa m3 [1]
(di) substitute into PV = A + BP PV = 1.26 × 103 + (-1.0 × 10–5 ) (6.0 × 107); = 660 Pa m3; [2]
(dii) difference = 1260 – 660 = 600 Nm; [2] Q4 (ai) Electric field strength E ≅ ΔV/Δd = 5.0 x 106 / 500 = 1.0 x 104 V m-1 [1] (aii) The extent of the cloud (the linear dimension) is many times more than its distance
from the Earth’s surface, so that the electric field between the cloud & the Earth’s surface may be regarded as uniform. [1]
(bi) The charge carried by the drop must be opposite to the charge on the underside of
the cloud, ie. positive, so that an upward force due to the field could be present to counter the gravitational pull of the Earth. [1]
(bii) Net force on the raindrop = 0, weight of raindrop = electric force on the drop
qE = mg [1] mg = (4/3) π(d/2)3.ρ.g = (4/3) π(2x10-3/2)3.(1000)(9.81) q = mg/E = 4.11 x 10-12 C [1] (biii) The air must be calm (which is unlikely to be the case!). There are no other forces on
the drop. Upthrust, and air resistance from the air are neglected. [1] (c) For the same height, the charge on the cloud is proportional to the potential
difference between the cloud and the Earth. Q∝E ≅ ΔV/Δd, ∴Q∝ ΔV for the same ΔD.
At ΔV = 5.0 MV, Q = 475 mC, in magnitude
At the verge of breakdown ΔV’ = 1500 MV, Q’ = (1500/5)(475x10-3) = 142.5 C [1]
Mean discharge current = Q’/t = 142.5 /120 x 10-6 = 1.19 x 106 A [1] Q5 (a) Lenz’s law states that the polarity of the induced electromotive force (e.m.f.) is such
that it produces an effect to oppose the change in magnetic flux. [1] It is directly related to the law of conservation of energy because the induced current produces electrical energy due to an external force doing work against the opposing forces. By having to do work, energy is transferred from the external system. [1]
(bi) As the magnet falls through the tube, at any point in the tube, there is a change in the
magnetic flux produced by the magnet. Since aluminium is a conductor, according to Faraday’s law, this rate of change in the magnetic flux induces an e.m.f. [1] Since the tube is a closed circuit, the induced emf caused induced current to flow around the tube. [1] According to Lenz’s law, the induced current is in the direction that opposes the change in magnetic flux. [1] As the magnet falls vertically, the change in flux below the falling magnet produces a repulsive force acting upwards (opposite to gravity) and the change in flux above the falling magnet produces an attractive force acting upwards. Thus, these upward forces
3
2007 HCI H2 Solutions
slow down the fall of the magnet through the tube as there is a smaller net downward force as compared to the case of free fall magnet. [1] Alternative (using law of conservation of energy) As the magnet falls through the tube, at any point in the tube, there is a change in the magnetic flux produced by the magnet. Since aluminium is a conductor, according to Faraday’s law, this rate of change in the magnetic flux induces an e.m.f. [1] Since the tube is a closed circuit, the induced emf caused induced current to flow around the tube and electrical currents dissipate thermal energy as metal has resistance. [1] According to the law of conservation of energy, the kinetic energy of the magnet must have been transformed into the electrical energy of the currents, since energy can neither be created nor destroyed. [1] Thus, the magnet does not accelerate downwards but travels at constant speed as it does work to create the induced currents. [1]
(bii) Use a stronger magnet.
Use a lighter magnet. Use a tube of material with lower resistivity (e.g. copper, silver, gold). Use a tube with a thicker layer of aluminium. [Any two of the points - 2]
Q6 a) A laser beam’s power is incident on a smaller area and thus more intense. [1] A laser beam’s radiation energy is in the visible spectrum whereas a filament lamp
radiates energy in the invisible spectrum as well. [1] b)
Conduction band (almost empty)
Valence band (almost filled)
Small band gap electrons
holes
Clearly labelled valence, conduction bands and small band gap [1] Indicating bands are almost empty/filled OR carriers in the bands [1] c) A electron in the valence band can absorb energy of the photon and be excited into the
conduction band [1] This leads to more free electrons in the conduction band, and more holes in the valence
band, reducing the resistivity. [1] d) The addition of boron which has only 3 valence electrons introduces available vacant
energy levels very close to the top of the valence band. [1] Electrons in the valence band can easily be thermally agitated to fill the acceptor energy
level, creating holes in the valence band. [1]
4
2007 HCI H2 Solutions
Q7
a) tAAtAA
o
o
λλ−=−=
lnln)exp(
The gradient of the ln (A /s-1) against t graph will give the value for λ. ln (A /s-1) in 1990 = 16.80; ln (A/s-1) in 1997 = 16.20 [1]
360024365780.1620.16
×××−
−=λ [1]
191072.2 −−×= sλ [1]
(b) λ
2ln2
1 =T
yr
sT
1.8
1055.21072.22ln 8
921
=
×=×
=− [1]
(c) At year 1997, ln (A /s-1) = 16.20
A = 1.10 x 107 s-1. [1] The number of nuclei which ought to be present is
15
9
7
1
1005.410718.21010.1
×=××
== −λAN [1]
(d) The number of radioactive nuclei present at certain time is indicated by the activity, A (as A
= λN). To check the amount of radioactive material that has been removed, we have the check the activity of both lines at a single instance. This can be done by extrapolating any one of the two lines and checking the difference in the activity. Extrapolating the line after theft to 1997, the value of ln (A /s-1) after theft on 1997 is 16.00, corresponding to a value of A = 8.89 x 106 s-1. [1]
The number of nuclei left after the theft is 159
6
2 1027.310718.21089.8
×=××
== −λAN [1]
The number of nuclei stolen is given by
151515 1078.01027.31005.4 ×=×−× [1] (e) The gradient of the graph corresponds to the negative of the decay constant of the
radioactive isotope. The decay constant is a characteristic that depends only on the type of radioactive isotope and is independent of the number of radioactive nuclides present. Both sections of the graph are natural log of the activity of the same type of radioactive isotope. [1]
5
2007 HCI H2 Solutions
Paper 3 Section A Q1 (a) (i) Lost in GPE = Gain in KE [1] mgh = ½ mv2
v = 9.9 m s-1 [1] (ii) T – mg = mv2 / r [1] T = 80(9.81 + 9.92 / 5) [1] = 2350 N [1] (iii) At this lowest position, the tension needs to support the whole weight [1] of
the hero and at the same time provide the largest centripetal force [1] (v is he largest at the lowest point), hence it is most likely to break at this position.
(b) (i) By conservation of linear momentum mhero(9.9) = (mhero + mvillain)(vf) [1] vf = 5.28 m s-1 [1] (ii) In practice, friction might be present between the villain and the floor (external
force), hence momentum will not be conserved in (b) (i). [1] Others: chandelier may be swinging, villain may have rotational motion. Q2
Tsinθ
T
mg
Tcosθ
(a) (i) Horizontally, T sinθ = mv2/r [1] Vertically, T cosθ = mg
Dividing 2 equations yields tan θ = v2/rg [1]
and since v = rω [1] tan θ = rω2/g (ii) Since tan θ = r/l r/l = rω2/g hence l = g/ω2 [1]
6
2007 HCI H2 Solutions
(b) (i) If there is a power surge we can assume that the current delivered to the
motor increases and that the rotating shaft will spin faster.[1] Since ω increases from (a) (ii) we can see that l will decrease.[1] This means that the contact will slide up the variable resisitor thereby increasing the resistance in the circuit and reducing the current to the motor.[1] So the system has responded to the power surge by effectively “ chocking” the increase and hence regulating the speed of the motor.
(ii) Feasible – If car slows down in rough terrain or uphill, more power will be
delivered to the motor.[1] If car speeds up downhill, current drawn by motor will be reduced.[1]
Not feasible - The system would only work effectively in a vertical position.
- System could be quite cumbersome and take up a lot of space.[1]
Q3 (a) (i) The net displacement at a given place/position is the vector sum of the
individual displacements [1] which would have been produced by the individual waves.
(ii) 1. aDx λ
= [1]
Two conditions: (1) D>> a; (2) only for fringes must be close to the axis, or for light that are deviated at small angles from the
axis. [1]
2. 49
103.20080.0
00.410450 −−
×=××
=x m [1]
(b) (i) Since the wavelength of red light is larger than blue light [1], the fringe
separation will be larger, and the fringes will be further apart [1].
(ii) [background: White light is a continuous spectrum from about violet to red (400-700 nm). Each component wavelength forms its own fringe system. The fringe width increases with increasing wavelength, that is, red will have the widest fringe and blue the smallest fringe width.]
The central fringe is generally white. [1]. The zeroth order fringes of all the wavelengths coincide at the centre where the path difference from the 2 slits is zero for all wavelengths! The combined central fringes remains white![1],
The sides of the central fringes are likely to be more reddish. [1] This is because the wider red fringe extends beyond the narrower blue fringe. [1].
7
2007 HCI H2 Solutions
Q4 (a) (i) The electrostatic force is providing the centripetal force. CE FF =
r
mvr
e 2
20
2
4=
πε [1]
22
0
2
4vm
rme
=πε
( )24
100
21931
0
2
102.1105.14
)106.1(1011.94
−−
−−
×=×
×××==
πεπε rmep kg m s-1 [2]
(ii) If the momentum of the electron in the radial direction is fixed, then its
uncertainty is zero. Using the uncertainty principle, the uncertainty in displacement would be infinite. However the above model has a fixed radius, has it violates the uncertainty principle.
Or If the electron orbits with a fixed radius, then 0=Δx . Hence and
that violates the uncertainty principle which states that
0=ΔΔ px
π4hpx ≥ΔΔ . [2]
(b) (i) Since the total energy of the particle E is less than the potential barrier U0, the
KE of the particle in region B would be negative (E – U0) and hence the particle will not be able to penetrate the barrier and cannot be found at region C.
(ii) 1. 2)(xψ is called the probability distribution function. It describes how
the probability of finding a particle at different locations is distributed along a distance x or it gives the probability of finding a particle at a point. [1]
2. Since the wave function on the right of the barrier is non-zero [1], 2)(xψ
will be positive. The particle has a small probability of penetrating the barrier and be found at C.
Section B Q5 (a) (i) any 2 of the following: monochromatic, high intensity, coherent, minimal
divergence (ii) When an electron in an atom is in an excited state, it could de-excite to a
lower energy level as a result of the interaction with a photon of energy equal to the energy difference between the two levels. A photon is emitted as a result. This is called stimulated emission.
8
2007 HCI H2 Solutions
Key Point 1: excited electron interacts with photon and de-excites, emitting a photon. [1]
Key point(s) 2: [1] for any one of the following points - incoming photon energy is equal to the energy difference between the
excited and lower energy state - emitted photon has the same energy, phase and direction as the incident
photon (iii) Population inversion is a condition where there are more atoms in the excited
state as compared to ground state. [1] If there are more atoms in the higher metastable state than the ground state,
on the whole, the number of photons that cause stimulated absorption is clearly much less than the number of photons that produce stimulated emission. [1] (rate of stimulated emission is larger than rate of spontaneous absorption)
As such, a single stimulated emission event multiplies the number of photons by a factor of 2, and a series of stimulated emission events can cause a cascade or exponential increase in the number of photons emitted, resulting in amplification. [1]
(b) (i)
s.f.) 3 (to eV 96.1
[1] eV 101.60
10142.3J 10142.3
[1] J 106331000.31063.6
19-
19
19
9
834
=××
=
×=××
××=
=
−
−
−−
λchE photon
(b) (ii) Let n be the rate of emission of photons.
( )( )( )
[1] s 1019.3
[1] s 10142.3
1050.0
[1]
1-16
1-19
23410
1051004
3
×=×
×=
=
=
−
−×−
− π
photon
photon
EIAn
AnE
I
(b)(iii)
1-27
1-9
34
s m kg 1005.1
[1] s m kg 106331063.6
−
−
−
×=××
=
=λhp photon
(c)(i) change in momentum of atom = change in momentum of photon = (1.05 × 10-27 – 0 ) kg m s-1 = 1.05 × 10-27 kg m s-1 [1]
9
2007 HCI H2 Solutions
(c)(ii)
[1] s cm 13.3
[1] s m 1066.118.20
1005.1
1-
1-27
27
=××
×=
=
Δ=Δ
−
−
mp
mp
v
photon
atomatom
(c)(iii) Let N be the number of photons absorbed. If each absorption slows down the atom by 3.13 cm s-1 (student can estimate 1 cm s-1 if the answer to part(c)(i) was not correct or not worked out), then
310
[1]
5
≈
Δ=
vvN
If the atom is to be stopped in 1 ms, then there must be N absorptions in 1 ms, leading to a
rate of [1] s 103s 103
10s 10
1-71-3
51-
3 ×≈×
≈ −−
N
Alternatively: 1
10 5
≈Δ
=v
vN
⇒ rate -15 ms photons 10≈
Or rate 1-81-3
51-
3 s photons 10s photons 1010s photons
10≈=≈ −−
N
(c)(iv) Key point1: Stating whether it is feasible (from data) Since the rate of absorption is about 107 s-1 and the rate of photon emission from a laser is far greater at about 1016 s-1, from this data, it seems feasible [1] for atoms to absorb enough photons from a laser beam to be slowed significantly. It is also correct to say that it is not feasible if students cite limitations, e.g. those in key point2 below. Key point2: Evidence of further thought / deeper analysis (any one of the following) - Not more than approximately 109 atoms can be cooled in 1 s. - Assuming that the probability of an atom absorbing a photon is more than 1 in 109. - Since atoms have discrete energy levels, only atoms with energy difference equal to the photon energy can absorb the laser light.
10
2007 HCI H2 Solutions
- After absorption of a photon, the atom needs to de-excite and emit the photon. It will then recoil with the same momentum, but at a random direction. Hence, it will speed up again and will stay cooled. - There is a factor of 109 times more photons than required for stopping the atom so the atom may absorb more than the required number of photons and gain momentum in the opposite direction and not be cooled. - Atoms need to be moving in the opposite direction to the photon’s momentum; if they were moving in the same direction, the absorption of the photon will make the atom speed up. Q6 i) Irms = (48 x 10-3) / 2 = 0.034 A [1] ii) mean power dissipated in the coiled copper wire, Pm = Irms
2R [1] = (48 / 2 )2 (8) = 9.2 mW [1] iii) length of the copper wire in the magnetic field = N(2πr) = (250)(2π)(0.015) = 24 m
(shown) [2] iv) A current carrying wire placed perpendicular to a magnetic field would experience a
force. [1] The circular magnet produces a radial magnetic field. An alternating current (signal) through the coiled copper wire results in the coil experiencing an alternating force. [1] As the coiled copper wire is attached to the paper tube, the loudspeaker cone would thus vibrate.[1] This causes the air particles around the cone to vibrate, creating regions of compressions and rarefactions. Energy is thus transferred through the air and this would be heard as sound. [1]
v) Possible frequency of sound = 2000 / 2 = 1000 Hz [1] vi) Maximum magnitude of the force acting on the copper wire due to the magnetic field, Fmax = BImax l [1] = (3.6 x 10-2)(48 x 10-3)(24) = 0.041 N [1] vii) To create higher frequency waves, the loud speaker cone must vibrate more quickly.
[1] This is harder to do with a large cone because of the mass of the cone. [1]
viii) When a sound wave is produced by the speaker located at one end of the tube, it will propagate along the tube's volume until it hits the piston, where it is reflected back towards the speaker-end of the tube. [1] The piston can effectively alter the length of the tube, causing the incident and reflected waves within the tube to interfere in such a way as to maximize the amplitude of the sound waves. [1] When this occurs, the standing waves created within the tube are said to be in resonance and generate the loud sound. [1]
ix) When resonance first occur, L = λ / 4 [1]
Since v = f λ, L = v / (4 f) = (340) / (4000) [1]
= 0.085 m [1]
11
2007 HCI H2 Solutions
Q7 (a) (i) 1. The half life of a radioactive isotope is the average time taken for half the
number of atoms in the isotope to decay. [1] 2. The decay constant is defined as the fractional change in the number of
atoms of a radionuclide that occurs per unit time. Note that
⎟⎠⎞
⎜⎝⎛ −=−==
dtNdN
NdtdN
NA 1.1.λ . [1]
Alternate The decay constant is defined as the probability of decay per unit time of
atoms of a radionuclide. (a) (ii) The mass of β -particles are much smaller than those of α -particles, hence
the fractional loss of kinetic energy per collision for β -particles are smaller and therefore β -particles are more penetrating. [2]
Alternative Answer Alpha particles are doubly charged and suffer more loss due to ionising
radiation. It therefore has a shorter range. (b) (i)
112
21
1083.360602425.3655730
2ln2ln −−×=××××
== sT
λ
atomsAN 1012 1066.6
1083.3255.0
×=×
== −λ [1]
The total number of carbon in 1g = 2223
1002.512
1002.6×=
×
Fraction of carbon-14 atoms in atmospheric carbon = 1222
10
1033.11002.51066.6 −×=
××
[1]
(b) (ii)
toeAA λ−=
( )( 36002425.365
1083.3255.0
3600348lnln
12 ××÷ )×
÷
−=−= −λoA
A
t [1]
yearst 8030= [1] (c) (i) Mass difference in the beta decay,
( ) umd5100000.400055.090715.8990774.89 −×=−−= [1]
Assuming that the energy released are in the form of kinetic energy for the daughter particles, Conservation of energy:
12
2007 HCI H2 Solutions
222
21
21 cmvmvm dYY =+ ββ
Conservation of momentum ββ vmvm YY = where and are opposite in direction. [1] Yv βv
Solving these equations, we get
1882
2
1015.110146.12 −×≈×=
+
= smm
mm
cmv
Y
d
ββ
β [1]
1699 −== smvmm
vY
Y ββ [1]
(c) (i) The half life of strontium-90,
( ) yearsT 3.2836002425.3651075.72ln2ln
1021 =××÷
×== −λ
The half life of strontium-90 is very long. In order to have accurate measurements of its half life, one would need to take measurements for a very long period of time (a few years) or to use very large quantities of strontium-90. Using large quantities of a radioactive material would pose higher risk due to radiation hazards. [1]
The daughter particle, Yttrium-90, also undergoes beta decays. It is not possible to know whether a particular beta particle is due to the decay of strontium-90 or Yttrium-90. [1]
(ii) Need to account for background radiation [1] and beta decay due to the
daughter particle, Yttrium-90[1]. (iv) 1. Using Fleming’s left hand rule, the magnetic field is applied perpendicularly
into the plane of the page.
2. The estimated radius of curvature at A = 30 mm [1]
rvmveB
2
=
( )( )( ) 1631
194
1058.11011.9
030.01060.1100.3 −−
−−
×=×××
=∴ smv [1]
13
2007 HCI H2 Solutions
Answer ALL the questions in the spaces provided. 1 A space capsule of mass 260 kg enters the Earth’s atmosphere with a speed of
35 m s-1 and falls vertically downwards towards Earth. Due to the air resistance, the capsule experiences a deceleration. Assume the acceleration of free fall is constant at 9.81 m s-2. The air resistance acting on the capsule is given by:
R = kv2, where the value of k is 5.0 kg m-1.
Remarks
(a) Calculate the initial deceleration of the capsule at the instance that it enters Earth’s atmosphere.
[2]
Are you considering all the forces, which would give you resultant force? FR = R - mg = kv2 – mg = (5 x 352) – (260 x 9.81)= 3574.4 N FR = ma a = 3574.4 ÷ 260 = 13.7 m s-2
Deceleration =……………………..
C1 A1
R is mistaken to be the resultant force, without considering weight.
(b) Calculate the terminal velocity of the capsule. [2]
Terminal velocity is the maximum constant velocity that an object has moving with air resistance, i.e. when resultant force = 0 kv2 – mg = 0 0.5v2 – 260(9.81) = 0 vT = 22.6 m s-1 terminal velocity=…………………….
C1 A1
(c)
Calculate the rate at which the capsule is doing work against air resistance when it is traveling at terminal velocity.
[2]
W E mg hP mgv
t t t
∆ ∆= = = =
= 260 x 9.81 x 22.6 = 5.76 x 104 W OR: ONLY applicable for instantaneous power or power required for object moving at constant velocity) P = Fv = Rv = 5 x 22.63 = 5.77 x 104 W Rate of doing work=……………….
C1 A1
Kinetic energy is not changing, so no work is done on it, but rather GPE.
(d) State and explain whether the value obtained in (c) is an overestimation or underestimation.
[2]
As the capsule approaches Earth, the average g value will be less than the 9.81 used since the distance between the capsule and the centre of the Earth is larger than its radius. Since g is in general lower than 9.81, the power calculated in (c) will be an overestimation.
B1 B1
Look at the calculation in (c), what could be affecting any of the variable. Air resistance, which caused heat loss has already been included
related to the resultant force
2007 IJC H2 Solutions
2 A model aeroplane has a mass of 0.40 kg and has a control wire of length 5.0 m attached to it when it flies in a horizontal circle. Its wings are horizontal, creating a vertical upward lift on the aeroplane. The taut wire is then inclined 60° to the horizontal and fixed to a point O.
(a) If each revolution takes 3.5 s, show that its centripetal acceleration is 8.06 m s-2.
-12 2 1 80 rad s3 5 .T .π πω = = = [C1]
2 25.0cos60 1.795a rω= = °× [M1]
-28.06 m s= [A0]
[2]
(b) Calculate the tension T in the control wire.
Have you considered the centripetal force (resultant force) which can be related to acceleration computed in part (a)? How does tension relate to the acceleration? Draw a free-body diagram.
c cF ma=
cos60 0.40 8.057T ° = × [M1]
T = 6.45 N [A1]
[2] Tension is not the centripetal force, but its horizontal component is.
(c) Calculate the upward lift L on the model aeroplane due to the air. The aeroplane is moving in a horizontal circle, implying there is zero displacement and net force in the vertical direction.
0yF =∑ sin60L W T= + °
( )( )0.40 9.81 6.45sin60L = + ° [M1]
= 9.51 N [A1]
[2]
Not all the vertical forces are considered in the compution of
0yF =∑
60°
5.0 m
O
2007 IJC H2 Solutions
3 In an experiment to determine the characteristics of a diode, the results obtained were plotted and the graph was shown in Fig 3.1.
The diode was then connected in the circuit as given in Fig 3.2 and a current of 6.0 mA was found to flow through the circuit.
(a) Using the graph and the circuit above, calculate the resistance of the resistor R. [3] Either
When a current of 6.0 mA flows through diode, p.d. across diode = 7.5 V [C1] p.d. across 5R = 12 – 7.5 = 4.5 V p.d. across R = 0.90 V [C1]
Resistance of R = V/I = 0.90/ 6.0 × 10-3
= 150 Ω [A1]
Or
When a current of 6.0 mA flows through diode, p.d. across diode = 7.5 V [C1] Resistance of Diode
3
7.51250
6.0 10
dd
VR
I−
= = = Ω×
Total Resistance
3
122000
6.0 10T
ER
I−
= = = Ω×
[C1]
5R = 2000 – 1250 R = 150 Ω [A1]
(b) Explain, using the idea of depletion layer in a p-n junction, what happens when the diode is reversed.
[3]
When the diode is reversed, the diode is connected in the reverse-based mode. When the p-n junction is in reverse bias, the holes and electrons are pulled away from the junction. Since only very few new electron/hole pairs are created at the junction, the existing mobile carriers are swept away to leave a larger depletion zone and the mobile charge carriers cannot flow through the depletion region. Any 3 points
-2
0
2
4
6
8
10
12
-5 0 5 10 15
I/mA
V/ V
Fig 3.1
12 V
R
4R
Fig 3.2
The I-V graph is showing the variation of the diode’s current vs. potential difference.
The 12 V emf is connected in series to the R, 4R and the diode. Thus the current flowing through all components is 6.0 mA and the p.d. of the diode is not 12V.
2007 IJC H2 Solutions
4 A student sets up the apparatus in Fig.4.1 to determine a value for the specific latent heat of fusion of ice.
A heater is placed in the funnel, surrounded by pure melting ice. The student measures the mass of melted ice in the beaker at regular time intervals before and after switching on the heater. The variation with time t of the mass m of melted ice in the beaker is shown in Fig.4.2.
During the heating process, the current is adjusted so that the readings on the ammeter and voltmeter are constant.
(a) By reference to Fig.4.2,
(i) at what time was the heater switched on? Initially, the ice is melting at constant rate due to the heat gained from surrounding. Thus, once the gradient of the curve changes, it indicates that the heater is supplying heat to the ice, increasing the melting rate. Heater is on approximately 4.0 to 4.5 min
[1]
Fig.4.1
Fig.4.2
THINK:
• Why is the gradient of the graph changing as the time increases?
• Why is the ice melting even when the heater is not switched on yet?
the heater is switched after some time.
Gradient started to change.
2007 IJC H2 Solutions
(ii) determine the mass of ice melted in 1.0 minute
1. with the heater switched off,
−= =.mass per minute m . x kg. 30 011 2 75 104 0
mass per unit =……………..kg
[1]
2. with the heater switched on.
−−= =
−. .mass per minute m . x kg. . 20 100 0 038 1 34 1012 6 8 0
mass per unit =……………..kg
[2]
(b) The readings of the ammeter and the voltmeter are 5.2 A and 11.5 V respectively. Use your answers in (a) to calculate a value for the specific latent heat of fusion of ice. The mass of ice melted in 1.0 min in a(ii)2. consists of two portion, i.e. those that melted due to surrounding and due to heater. But the power computed in only meant for the mass of melted ice due to the heater alone. Thus there is a need to compute the change solely responsible by the heater.
( )
∆
∆
∆
− −
=
=
=
=−
=
f ff -1Q mLmVI LtVItL m . x . x. x . x. x J kg2 355 2 11 5 601 34 10 2 75 103 37 10
Specific latent heat of fusion of ice =……………………… J kg-1
[3]
Mistake of including mass of melted ice due to heat gained from surrounding
2007 IJC H2 Solutions
5 Water waves of wavelength 4.0 m are produced by two generators, S1 and S2 as shown in Fig.5. Each generator, when operated by itself, produces waves which have an amplitude of 20 mm. The frequency of the motor of both generators is 2.0 Hz and the generators are operating in phase. Assume the speed of water waves remain unchanged.
(a) What is resultant amplitude at point Q? To know the resultant amplitude, there is a need to know if the waves are interfering constructively, destructively or partially. (thus look at path difference)
Path difference, ∆x = 10 – 8.0 = 2.0 m = ½ λ , i.e. destructive interference. [M1] Resultant amplitude = 20 – 20 = 0 mm [A1] Amplitude = ………………………… mm
[2]
(b) (i) The frequency of the generator is now increased to 4.0 Hz. What is the new wavelength of the water waves? velocity of the waves is constant since it is traveling in the same medium.
f1 λ1 = f2 λ2 [C1]
(2.0)(4.0) = (4.0)(λ2) → (λ2) = 2.0 m [A1] Wavelength = ………………………… m
[1]
(ii) What is resultant amplitude at point Q using the new frequency of water wave which is 4.0 Hz? The path difference is still 2.0 m since the geometry of the setup is not changed. However since the wavelength is change to 2.0 m, thus the new path
difference in term of wavelength is 1.0 λ, i.e. constructive interference. [M1] Resultant amplitude = 20 + 20 = 40 mm [A1] Amplitude = ………………………… mm
[2]
Frequency of the wave does not affect its amplitude.
S1 S2 6.0 m
8.0 m
• Q
Fig.5
THINK:
• What happens when two waves interfere at a common point?
• What phenomenon is observed? A stationary wave or interference pattern?
10 m
2007 IJC H2 Solutions
6 Fig.6.1 shows the apparatus used in an experiment to investigate the photoelectric effect. Light falls on the photo-sensitive metal, the kinetic energy of the fastest moving emitted electrons can be measured by increasing the voltage provided by the battery until no current flow. The results obtained for monochromatic light of various frequencies are show in Fig.6.2.
(a) The graph indicates that there is a frequency below which no electrons are emitted. Explain why.
[1]
The product of the minimum frequency with the Planck’s constant is equal to the minimum amount of energy (work function of the metal) to liberate the electrons from the metal.
B1 NAQ: talk about threshold frequency
(b) Calculate the kinetic energy of the fastest moving electron emitted by light of
frequency 5.5 x 1014 Hz. [2]
From the graph, stopping potential Vs = 0.60 V Max KE of electrons = Max Electrical potential energy = eVs = 0.60 × 1.6 × 10-19 = 9.6 x 10-20 J (You have to use information from the graph)
B1 A1
Mistake max KE is not equal to energy of photon (hf)
(c) Use the graph to obtain a value for the Planck constant. [3] max photonE KEφ= +
ohf hf eV= +
o
h hV f f
e e= −
Gradient of a V-f graph is h/e
( )19 14
1.0 0
1.6 10 6.5 4.0 10
h−
−=
× − ×
h = 6.4 x 10-34 J s.
B1 C1 A1
Fig.6.1
Fig.6.2
2007 IJC H2 Solutions
(d) State the effect, if any, of increasing the intensity of the light on the 1. photo-current, and 2. on the graph
[2]
Current increases. (since increasing intensity will increase the rate of photon hitting on the surface, thus increasing the emission of photoelectrons) No change in the gradient and intercept (since intensity does not affect the work function of the method, i.e. threshold frequency).
B1 B1
(e) The metal is removed from the setup. It is then irradiated with photons of energy
3.2 eV. Loss of electrons causes the metal to acquire a positive potential. What is the potential at which no further loss of electrons from the surface occurs?
[2]
As the metal is losing electrons, the metal start to acquire a positive potential. Eventually, the positive potential becomes so high that any photoelectrons emitted will be pulled back to the metal piece. Work function of metal = hfo = (6.63 X10-34 x 4 x 1014) / 1.6 x 10-19 = 1.66 eV Energy of photon = hf = 3.2 eV Potential of surface (relative to a nearly earthed potential) = (3.2 – 1.7) = 1.5 V
B1 B1
2007 IJC H2 Solutions
7 The power of a car engine in watts is commonly measured in ‘horsepower’ (hp), an old way of measuring engine powers in comparison to the horse technology of those days. This power represents the maximum power that the engine can produce when the car is traveling at maximum speed on a level road. The ‘horsepower’ is defined as follows:
1 hp = 0.746 kW The engine produces power which is then transmitted to the wheels to drive the car, which then has to do work against frictional forces. There are two main frictional forces that the car has to go against - air resistance F and friction with the road R. Fig. 7.1 below shows the values of R and F at various speeds v for a particular model of a Toyota car.
v / m s-1 R / N F / N Total
Resistance FT/ N
Power Required / kW
9.0 228 52 280 2.52
18 223 206 429 7.72
27 219 470 689 18.6
36 212 834 1046 37.7
45 204 1300 1504 67.7
(a) Fill in the last 2 columns of the table in Fig. 7.1. [2] 1M for each column
(b) The most obvious point about the table is that, as the speed increases, air resistance F becomes the dominant energy loss but friction R is almost a constant (actually getting slightly less as speed increases). The effect of air resistance F depends not only on the speed but also on the shape and size of the car – in particular the cross-sectional area. This is usually expressed in a formula:
F = ½ CAρv2
where F is the air resistance for a vehicle traveling at speed v, A is the cross-sectional area, ρ is the density of the air and C is a ‘shape constant’ called the drag coefficient (a dimensionless constant). You may take ρ as 1.29 kg m-3.
(i) A Ferrari sports car with a surface area of 2.0 m2 experiences 1200 N of air resistance when it travels at a speed of 48 m s-1. Calculate the drag coefficient C for the sports car.
[2]
1200 = ½ C (2.0) (1.29) (48)2
C = 0.404
B1 B1
(ii) A lorry has a surface area of 4.0 m2 and it experiences 1200 N of air resistance when traveling at 30 m s-1. Calculate the drag coefficient C for the lorry.
[2]
1200 = ½ C (4.0 x 1.29 x 302) = 0.517
M1 A1
Fig. 7.1 Values of F and R for Toyota
2007 IJC H2 Solutions
(iii) Using the data provided in Fig. 7.1, estimate the average value of the product CA for the Toyota car.
[2]
From table, using columns 1 and 3, CA = 2F/(1.29 x v2) Obtain a new column for CA: v / m s-1 CA
9.0 0.9953 18 0.9857 27 0.9996 36 0.9977 45 0.9953
(CA)ave = 0.995 (Award 1 mark if candidate calculates a correct value of CA using 1 row of values from Fig. 7.1)
M1 A1
(iv) Is the Toyota car likely to be a large or small car? Explain your answer. [2] CA for Ferrari = 2 x 0.404 = 0.808
CA for lorry = 4 x 0.517 = 2.068 Since CA for Toyota car is slightly more than CA for Ferrari but much less than lorry, Toyota is a small car.
B1 B1
(c) A common way to analyse the resistive forces acting on a car is to assume that the
resistance R is constant at 220 N and that the total resistance FT follows the formula below-
FT = R + F = 220 + ½ CAρv2
(i) Use this equation and your answer in (b)(iii) to estimate the total resistance FT acting on the Toyota car when it is traveling at a speed of 36 m s-1.
[2]
FT = 220 + ½ (0.995) (1.29) (36)2 = 1050 N
C1 A1
(ii) Using the actual value of total resistance FT from the table in Fig. 7.1,
calculate the percentage error in this estimated value calculated in (c)(i).
[1]
% error = (1050 – 1046) / 1046 x 100% = 0.382 %
C1
(d) The Toyota car in Fig. 7.1 has an engine that can deliver a horsepower of 102 hp when it is traveling at its maximum speed of 45 m s-1.
(i) Calculate the power supplied by the engine in kilowatts. [1]
Power supplied by engine = 102 x 0.746 = 76.1 kW
(ii) The value calculated in (d)(i) is not the same as the required power stated in the table in Fig. 7.1. Account for the difference.
[2]
Power required < power supplied by engine due to energy loss Loss could be due to other energy uses in the car such as radiator, headlights, air-con, radio or other internal frictional forces in the car along the transmission line from engine to the wheels.
B1 B1
(iii) Calculate the efficiency of this engine.
[1]
Efficiency = 67.7 / 76.1 x 100 = 89.0 % C1
2007 IJC H2 Solutions
Innova JC 2007 9745/03
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Section A
Answer all the questions in the spaces provided. It is recommended that you spend about one hour on this section.
1 When a driver brings a car to a stop by braking as hard as possible, the stopping distance
can be regarded as the sum of a “thinking distance”, which is the distance travelled by the car during the driver’s reaction time, and ‘braking distance” which is the distance in which the car stops after the brakes have been applied. The following table gives typical values:
Initial Speed / m s-1
Thinking distance / m
Braking distance / m
Overall distance / m
10 7.5 5.0 12.5 20 15.0 20.0 35.0 30 22.5 45.0 67.5
(a) What is the reaction time of the driver?
Reaction time = Thinking distance / Initial speed = 0.75 s
A1
(b) What constant value of negative acceleration has been used to calculate the braking distance?
The deceleration = 102 / (2 x 5.0) = 10 m s-2
A1
(c) What is the car’s stopping distance if the initial speed is 25 m s-1?
Thinking distance = 25 x 0.75 = 18.75 m Braking distance = u2/2a = 252 / 20 = 31.25 m Overall distance = 50 m
C1 A1
(d) Draw a labelled distance time graph for the driver initially travelling at 25 m s-1. Indicate clearly on your graph the regions of ‘thinking distance’ and ‘braking distance’.
[2]
18.75
50.00
0.75 3.25
Distance / m
Time / s
B1: labeling of values
B1: Correct shape
2007 IJC H2 Solutions
Innova JC 2007 9745/03
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2 (a) A uniform wire AB of resistance 1.5 Ω is connected in series with a resistor of 5.0 Ω, a
cell of e.m.f. of 9.0 V and internal resistance 1.0 Ω, as shown in Fig. 2.1. Calculate the potential difference (p.d.) across the wire AB.
Total resistance = 1.5 + 5.0 + 1.0 = 7.5 Ω [C1] P.d. of AB = [(1.5) x (9.0)]/ (7.5) = 1.8 V [A1]
(b) A cell C of e.m.f. 1.5 V and internal resistance 0.80 Ω is connected to the circuit of Fig. 2.1, as shown in Fig. 2.2.
The moveable contact can be connected to any point along the wire AB. At point D, there is zero current in the galvanometer.
(i) Calculate the length of AD.
Where there is zero current in galvanometer, the p.d. of the length AD is equal to the e.m.f. of the cell C. [C1]
(Length of AB) 80.0 cm ∝ 1.8 V [C1]
AD ∝ 1.5 V
AD = (1.5 / 1.8) × 80.0 cm = 66.7 cm [A1]
80.0 cm
1.0 Ω 9.0 V
A B
5.0 Ω
1.0 Ω 9.0 V
A B
5.0 Ω
D
0.80 Ω C
1.5 V
G
Fig. 2.2
Fig. 2.1
2007 IJC H2 Solutions
Innova JC 2007 9745/03
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(ii) State and explain one advantage of using Fig. 2.2 to measure the e.m.f. of C compared to using a laboratory voltmeter.
Using Fig. 2.2 to measure e.m.f. of C is more accurate as it eliminates the systematic error of the internal resistance of C when the voltmeter is used. (1 mark per underline)
[2]
3 A pendulum bob in a particular clock oscillates so that its displacement from a fixed point varies with time t as shown in Fig. 3.1. The mass of the pendulum bob is 0.100 kg.
(a) (i) From Fig. 3.1, find the angular frequency of the oscillations. ω = 2π / T = 2π / 1.0 = 6.28 rad s-1. [A1]
(ii) Find the magnitude of the maximum velocity. For maximum velocity, x = 0 m.
From the graph, xo = 0.12 m [C1]
vmax = ω xo = (6.28) (0.12) = 0.754 m s-1. [A1]
(iii) Find the maximum kinetic energy of the system. KE = ½ m v2
= ½ (0.100) (0.754)2 = 28.4 mJ [A1]
(b) Sketch, with values on both axes, the graph of velocity against displacement.
[2]
Displacement / m
t / s
0.12
− 0.12
1.0 0.5 1.5 2.0
Fig. 3.1
0
v / m s-1
x / m
Fig. 3.2
− 0.12 0.12
0.754
−0.754
1 mark each for
• Correct x-values, y-
values
• Correct shape of graph
2007 IJC H2 Solutions
Innova JC 2007 9745/03
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(c) Sketch, with values on both axis, the graphs of kinetic energy and potential energy of system against displacement.
[3]
4 Using wave mechanics, we may consider electrons in a hydrogen atom as being trapped in
an infinite well of width, L. Given that its total energy is given by 2 2
2=
hE
mLn
n
8
where h is the Planck constant, m is the mass of electron and L is 120 pm.
(a) If an electron is in a state of n = 2, calculate its energy.
2 34 2
31 12 2
17
2 (6.63 10 )
8 9.11 10 (120 10 )
1.68 10
xE
x x x
x J
−
− −
−
=
=
C1 A1
(b) The lifetime of an electron in the state with n = 15 is about 10-8 s. What is the uncertainty in the energy of the n = 15 state?
3427
8
4
6.63 105.28 10
4 4 10
hE t
hE x J
t
π
π π
−−
−
∆ ∆ ≥
×∆ ≥ = =
∆ ×
C1 A1
(c) Calculate the momentum of the electron in (a).
2
31 17
24
2
2 2 9.11 10 1.68 10
5.52 10 N s
pE
m
p mE− −
−
=
= = × × × ×
= ×
C1 A1
E / mJ
x / m
Fig. 3.3
− 0.12 0.12
28.4
1 mark each for
• Correct x-values, y-
values
• Correct shape of KE and
PE graph respectively.
0
2007 IJC H2 Solutions
Innova JC 2007 9745/03
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4 (d) By making suitable approximation for the uncertainty in the momentum of the electron, calculate the uncertainty in its position.
24
12
4
1
4
1 4 5.52
9.56 10 m 10 pm
p p
hp x
hx
p
h
x
π
π
π −
−
∆ =
∆ ∆ ≥
∆ = ×
= ×
= ≈
allow ∆p to be 2p or ½ p
B1 C1 A1
5 (a) Distinguish between spontaneous emission and stimulated emission.
Spontaneous emission is the process by which an atom or molecule in an excited state drops to a lower state by emitting a photon without external provocation. Stimulated emission is the process by which, when perturbed by a photon, an atom in the excited state loses energy to return to the ground state, resulting in the creation of another photon.
B1 B1
(b) One of the conditions for lasing to occur is that population inversion must be achieved.
Explain what is meant by population inversion and state how it might be achieved.
Population inversion is the situation in which there are more atoms in the excited state than in the ground/lower state. Population inversion may be achieved by supplying energy to the system using a continuous light source to raise the atoms to a higher state, optical pumping or any other appropriate means.
B1 B1
(c) Pulsed dye lasers emit light of wavelength 585 nm in 0.45 ms pulses to remove skin
blemishes such as birthmarks. The beam is usually focused onto a circular spot of 5.0 mm in diameter. The output of one such laser is given to be 20.0 W.
(i) What is the energy of each photon?
Energy of each photons = hc/λ
34 8
19
9
6.63 10 3.0 103.40 10 J
585 10
−−
−
× × ×= = ×
×
C1 A1
(ii) How many photons per unit area are delivered to the blemish during each pulse?
Power of laser = rate of photons arrival x Energy = 20.0 W
3
32 19
21
20 0.45 10
5 10( ) 3.41 10
2
1.34 10
N
xA
N
A
π
−
−−
× ×=
× × ×
= ×
B1 C1 A1
2007 IJC H2 Solutions
Innova JC 2007 9745/03
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Section B
Answer two questions for this section. Each question carries 20 marks. It is recommended that you spend about one hour on this section.
6 (a) Define the following terms: (i) gravitational field strength, (ii) electric field strength and [3] (iii) magnetic flux density.
i. Gravitational field strength (at a point) is the gravitational force acting on unit mass (placed at that point in the field)
ii. Electrical field strength (at a point) is the electric force acting on unit positive charge (placed at that point)
iii. Magnetic flux density is the magnetic force acting per unit current in a wire of unit length placed at right angles to the magnetic field.
(b) For an isolated spherical mass, the mass behaves as if it were concentrated at the centre for calculation of gravitational potential of points on the surface or outside the sphere. Measurements of the distance x from the centre of the sphere and the corresponding
values of gravitational potential φ are given in the table below:
x / m φ / 10-9 J kg-1
0.15 - 8.67 0.21 - 6.19 0.26 - 5.00 0.32 - 4.06
(i) Without drawing a graph, use the data from the table to verify that the
gravitational potential φ is inversely proportional to the distance x.
x / m φ / 10-9 J kg-1 φx / 10-9 J m kg-1
0.15 - 8.67 - 1.30 0.21 - 6.19 - 1.30 0.26 - 5.00 - 1.30 0.32 - 4.06 - 1.30
Since all the product of φx are of the same value (- 1.30 × 10-9 J m kg-1)
[B1], therefore φ is inversely proportional to the distance x.
[2]
(ii) The gravitational potential at the surface of the spherical mass is – 9.22 × 10-9 J kg-1. Calculate the radius of the sphere.
GMrφ = − 99 1 30 109 22 10 .. r −− − ×
− × = − [M1]
r = 0.141 m [A1]
(iii) Hence calculate the mass of the sphere. For surface of the sphere: 119 6 67 109 22 10 0 141. M. . −
− ×− × = − [M1]
M = 19.5 kg [M1]
Calculation of φx is done. [B1]
2007 IJC H2 Solutions
Innova JC 2007 9745/03
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(iv) The gravitational potentials due to the spherical mass are all negative values. However when the spherical mass is electrically charged, explain why the electrical potentials can be positive values with reference to the definition of potentials. [2]
Electrical potential is the external work done in bringing an unit positive charge from infinity to that point. The electric force on a charged sphere can be attractive or repulsive depending on the polarity of the charge [B1]. The external work done in bringing an unit positive charge from infinity to a positively charged sphere is positive since the external force (which is acting in opposite direction to the repulsive electric force) is acting in the same direction as the displacement [B1].
(c) (i)
An electron is placed stationary in a region of space. A field acting into the plane of paper is then introduced into this region. Discuss the effects of the field on the electron and its subsequent motion, if any, when the field in each case is
1. a gravitational field, 2. an electrical field, 3. a magnetic field [6]
1. A gravitational force is exerted on the electron due to its mass. The direction
of the gravitational force is in the same direction as the gravitational field [B1]. Thus the electron moves into the plane of paper [B1].
2. An electrical force is exerted on the electron due to its charge. However the direction of electrical force is in the opposite direction of the electrical field [B1], since the electrical field denotes the direction of the electrical force acting on a positive charge. Thus the electron moves out of the plane of paper [B1].
3. No magnetic force is exerted on the electron since magnetic force only acts on moving charge [B1]. Thus the electron will remain stationary [B1].
(ii) State and explain how your answer in (c)(i)3. would change, if at all, if the
electron is moving to the right when the magnetic field acting into the page is introduced. [3]
With the electron moving, a constant magnetic force will act on the electron [B1] in a direction that is always perpendicular to its velocity [B1] (when the electron is in the magnetic field). Thus the electron moves in a circular path in the plane of the paper [B1].
Region of space where the field is acting into plane of paper
electron
2007 IJC H2 Solutions
Innova JC 2007 9745/03
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7 (a)
(i) (1) Define the term magnetic flux and magnetic flux through an area A placed in a uniform magnetic field of flux density B, such that the direction of B makes an angle θ with
the normal of the area is defined as Φ θ= BAN cos
[1]
1
(2) the weber. 1 Weber is the magnetic flux through an area of 1 metre square when the magnetic flux density perpendicular to the area is 1 Tesla.
[1]
1
(ii) A coil of 300 turns with area 2.0 cm2 is placed with its plane at an angle of 30o to a uniform magnetic field of flux density 3.0 mT. Calculate the flux linkage through the coil.
Φ − −
−
=
=
o. x x . x x cos. x Wb4 352 0 10 3 0 10 300 609 0 10
[2]
1 1
(b) A straight conductor PQ of length L moves with speed v perpendicularly to a uniform magnetic field with flux density B as shown in Fig.4.1
(i) By considering a small displacement x of the conductor in a short time t,
1. write down an expression for the ‘flux cut’ by the conductor in a short
time t.
∆Φ = BLx
[1]
1 2. show that the magnitude of e.m.f. ξ induced across the ends of the
conductor is given by BLv. By Faraday’s law
Φξ = =
=
d BLxdt tBLv
[2]
1
1
(ii) State which end of the conductor has the higher potential. End P has higher potential.
[1]
1 (c)
The conductor PQ in (b) is a tiny part of a metallic string stretched between two points RS, such that the conductor is in the middle of the string, as shown in Fig. 4.2.
B pointing out of page v x
L
P Q
Fig. 4.1
v x
R S
Fig. 4.2
P Q
CRO
2007 IJC H2 Solutions
Innova JC 2007 9745/03
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7 (c) The string is plucked at the center such that it vibrates with fundamental frequency f, cutting the uniform magnetic field at right angles. The displacement x of the conductor PQ varies with time t and may be given by the expression:
ox x sin tω=
where ox is the amplitude and ω is the angular frequency of the vibration
respectively.
(i) Write an expression to show the variation of the velocity v of PQ with time t.
ω ω= ov x cos t
[1]
1
(ii) 1. Using your answer to (c)(i) and the expression in (b)(i)(2), write an expression for the magnitude of the induced e.m.f. across PQ.
ξ ω ω= oBLx cos t
[1]
1
2. What is the peak value of the induced e.m.f?
ξ ω=o oBLx
[1]
1
(iii) Point R and S are connected across the y-plates of a cathode-ray oscilloscope to monitor the e.m.f. induced across RS. The trace on the CRO screen is shown in Fig. 4.3, with the time-base set at 5.0 ms cm-1 and the y-sensitivity at 1.0 mV cm-1. Determine
1. The peak value of the induced e.m.f.
ξ = × =o . .-11 5cm 1 mV cm 1 5 mV
[1]
2. The frequency of vibration of the string.
−= =
=
f T x . xHz 31 14 5 0 1050
[2]
1 1
Fig. 4.3
1.0 cm
1.0 cm
2007 IJC H2 Solutions
Innova JC 2007 9745/03
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7 (c) (iii) 3. The magnetic flux density B, given the following information:
L=1.5 cm, Amplitude of vibration of the string xo= 3.0 cm.
ξ ω
ξω π
−
− −
−
=
×∴ = =
× × × ×
=
o oooBLx .B Lx . . x. x T 32 22 1 5 101 5 10 3 0 10 2 501 06 10
[2]
1
1
(d) A resistor of 30 Ω is now connected across RS in Fig 4.2. (i) Estimate the maximum power dissipated across the resistor.
( )ξ−
−
= =
= ×
oo . xP R. 232 81 5 10307 5 10 W
[1]
1
(ii) Describe and explain how the trace on the oscilloscope would be affected as a result of connecting the resistor across RS. As current flows in the wire RS, RS experiences a force opposite to its motion (using Fleming’s Left Hand Rule) as it cuts the magnetic field, i.e. oscillation is damped. The speed of cutting decrease, thus the amplitude of the induced e.m.f. decreases. Hence, the amplitude of the trace will decrease to zero after a few cycles.
[3]
1
1
1
8 Radiocarbon dating is possible because of the presence of radioactive carbon-14 (
14
6 C )
caused by the collision of neutrons with nitrogen-14 ( 14
7 N ) in the upper atmosphere. The
equation for the reaction is:
14 1 14
7 0 6N n C X+ → +
(a) (i) From the above equation, determine the numbers of protons and neutrons
present in particle X. 1 proton and no neutron.
A1
(ii) Identify the particle X. X is proton/ hydrogen
A1
(iii) Using the data below, determine the mass of the particle X in u, given that the amount of energy released in one such reaction is 0.7060 MeV.
The masses of the nuclei involved are listed below: Carbon-14 14.003242 u Nitrogen-14 14.003158 u Neutron 1.008665 u
Energy released = ∆mc2 = 0.7060 MeV = 1.1296 × 10-13 J
∆m = 1.12511 × 10-30 kg = 7.561 × 10-4 u
C1
MN + Mn – MC - MX = 7.561 × 10-4 u MX = 1.007825u or 1.008u
C1 A1
2007 IJC H2 Solutions
Innova JC 2007 9745/03
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8 (b) The mass of carbon-14 produced by this reaction in one year is 7.5 kg. 14 g of
carbon-14 contains 6.0 × 1023 atoms.
The half-life of carbon-14 is 5.7 × 103 years.
(i) Define half-life and decay constant.
[2]
Half-life is the average time taken for half the number of nuclei present in the radioactive sample to decay. Decay constant is the probability that a radioactive nucleus in a sample would decay per unit time.
B1
B1
(ii) Show that the number of carbon-14 atoms produced each year is
approximately 3.2 × 1026.
Number produced = 7500/14 × 6.02 × 1023 = 3.2 × 1026
M1
(iii) Calculate the decay constant of carbon-14 in year-1.
λ = ln2/t1/2 = 1.21 × 10-4 yr-1
A1
(iv) Assuming that the number of carbon-14 atoms in the Earth and its atmosphere is constant, then 3.2 × 1026 carbon-14 atoms must decay each year. Use this fact and your answer to (ii) to calculate the number of carbon-14 atoms in the Earth and its atmosphere.
Number undecayed = N0 - 3.2 × 1026
N = N0e-λt
N0 - 3.2 × 1026 = N0e-(0.000121 )(1)
N0 = 2.64 × 1030
C1
C1 A1
(c) The carbon-14 dating technique for dating archaeological materials depends on the
assumption that when living organisms assimilate carbon (12
6 C ) from the atmosphere
they also assimilate some radioactive carbon-14 (14
6 C ) atoms. The organism thus
becomes slightly radioactive, emitting β-particles at a rate of 0.80 Bq for each gram of total carbon content in the organism. When the organism dies no more carbon is taken up and the carbon-14 present decays. The radioactive decay is said to be a random and spontaneous process.
(i) Discuss the statement that radioactive decay is a random and spontaneous process.
Spontaneous means that the nucleus decays without the need for any external stimulus. Random means there is no way of knowing when a particular nucleus will decay.
B1
B1
In a dating experiment 2.5 g of carbon from a sample of ‘ancient’ wood gave a count rate of 0.75 Bq after allowing for background radiation.
(ii) What does the term “background radiation” mean?
Background radiation is radiation that occurs naturally, e.g. cosmic rays from outer space.
B1
2007 IJC H2 Solutions
Innova JC 2007 9745/03
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8 (c) (iii) Sketch, on axes in the figure below, a graph to show how the activity of carbon-
14 in one gram of sample having an initial activity of 0.80 Bq will vary with time over a period of three half-lives.
[2]
(iv) Use the graph to estimate the age of the wood. Activity of 1 g of radioactive wood = 0.75/2.5 = 0.30 Bq.
From graph, age of wood = 8000 years.
C1
A1
age of wood = years [2] (v) Suggest why an activity of 0.80 Bq would be hard to measure in a typical A
Level Physics laboratory.
The background radiation in the laboratory is quite significant and this will affect the count rate recorded by the Geiger Muller tube. The percentage uncertainty in the count rate will be high. or Sensitivity of GM tube
B1
One mark for
1. correct plot of 3 points
for half life, i.e. (0, 0,8),
(~5.5, 0.4), (~11.5, 0.2) &
(~17, 0.1)
2. correct shape of the
slope.
2007 IJC H2 Solutions
JJC H2 Physics P2 Soln 2
Prelim Exam 2007 [Turn Over
1 (a) The equation must be homogeneous. Hence the unit of kv2 = unit of p i.e [kv2] = [p] [k] = [p] / [v2] = kg ms-2 m-2 / m2 s-2 = kg m-3
[working 1 mark, ans 1 mark] (b) (i)
resultant 3.0 m s-1
θ 4.0 m s-1
[correctly labelled – 1 mark, correction directions – 1 mark] (ii)
Magnitude of resultant velocity = 22 4.0 3.0 + = 5.0 m s-1 –[1]
Direction : θ = tan-1 (3/4) = 370 to the bank –[1]
2 (a) The amplitude of a simple harmonic motion is its maximum distance (or magnitude of maximum displacement) from its equilibrium position. –[1]
(b) (i) It refers to an object oscillating without external periodic force OR oscillating without any driving force. –[1]
(ii) Damping removes energy from the oscillating system. –[1]
(c) (i) 330 = 550λ
λ = 0.60 m–[1]
(ii) Angular frequency ω = 2π(550) = 3460 rad s-1–[1]
(iii) Maximum acceleration = (3460)2(2.0××××10-3) = 2.39××××104 m s-2
[Substitution – 1 mark, Ans – 1 mark] (iv) Power required = (550)(0.20)(8.0××××10-3) = 0.88 W
[Substitution – 1 mark, Ans – 1 mark] (d) A longitudinal wave is one in which the direction of oscillation of the wave particles
is parallel to the direction of propagation of the wave. –[1]
3 (a) (i) Electric potential at a point in an electric field is defined as the work done per unit positive charge in bringing an object from infinity to that point. –[1]
(ii) It is independent of any charges that may be placed in the field, unlike electric potential energy. –[1] OR Electric potential is a property of a point while electric potential energy is possessed by a charge placed at that point. –[1]
(b) (i) Change in electric potential = 0.6 x 9x104 = 54000 V –[1]
2007 JJC H2 Solutions
JJC H2 Physics P2 Soln 3
Prelim Exam 2007 [Turn Over
(ii) Change in potential energy = q(∆V) = 1.6x10-19 x (-54000) = -8.64x10-15 J –[1]
(iii) Gain in Ek = Loss in Ep ½ mv2 - ½ mu2 = 8.64x10-15 ½ mv2 - ½ (1.67x10-27)(0.5X106)2 = 8.64x10-15 v = 3.255x106
= 3.26x106 m s-1. [1 mark for subst, 1 mark for Ans]
(iv)
[correct shape-1 mark] (v)
[correct shape-1 mark] 4 (a) (i) Centripetal force = magnetic force
2
em vBev
r= ---[1 mark for equation]
e
v Be
r m= ----------[1 mark for working]
e
Be
mω =
(ii)
e
v Be
r m=
momentum = em v Ber= ----[1 mark for working]
Particles with the same radius of circular motion have the same momentum. [1 mark for correct explanation]
A B
v / m s-1
Fig.3.3
A B
Ep /J
Position / m
Fig.3.2
Position / m
2007 JJC H2 Solutions
JJC H2 Physics P2 Soln 4
Prelim Exam 2007 [Turn Over
(b) (i) Average power = total energy dissipated over one period / time of one period
=
2 22 1(0.002) (0.002)
5 5
0.01
+ = 0.200 W
[1 mark for working, 1 mark for Ans] (ii) (Irms)
2(5) = 0.2 Irms = 0.200 A
[1 mark for working, 1 mark for Ans] (iii)
(Irms)2(5) =
22(0.002)
5
0.01 (forward bias) or
21(0.002)
5
0.01 (reverse bias)
Irms = 0.179 A or 0.0894 A [1 mark for working, 1 mark for Ans]
5 (a) (i) Transition between E3 and E2 –[1]
(ii) The photons emitted have the same frequency, are in phase, and travel in the same direction, as the incident photon. OR Coherent and same direction/highly directional [1 mark for 2 points, 2 marks for 3 points]
(iii) E4 −−−− E3, E4 −−−− E2, E4 −−−− E1, E3 −−−− E2, E3 −−−− E1 [any 2 points]
(b) Thermal excitations at ordinary temperatures create a small number of electron-hole pairs [1] in the semiconductor, while metals have a large number of mobile electrons [1] in the conduction band.
(c) Electrons diffuse from n-type to p-type [1] region while holes diffuse from p-type to n-type region. This sets up an electric field, pointing from n-type to p-type [1] region across the junction, which prevents electrons and holes from further diffusing into the depletion layer.
6 (a) For a gas of a fixed mass, the volume depends on temperature and pressure. [1] Hence the density of a gas varies with temperature and pressure. [1]
(b) Helium [1], because at −−−−218 °°°°C = 55 K it is the only substance in Figure 6 that exists in the gaseous phase.[1]
(c) Liquid nitrogen is readily transportable / available / economical/safe/non-toxic. [1] Liquid nitrogen is able to maintain temperatures far below the freezing point of water. OR Boiling point of liquid nitrogen (77 K) is much lower than melting point of water (273 K). [1]
(d) Heat gain by 1 g of ice-then-water = heat loss by 10 g of water
( )( ) ( ) ( ) ( )( )θθ −=++ 20186.410186.43345060.2 mmmm ---[working 1 mark]
9.492046.46
86.412.837186.43.344
=
−=+
θθθ mmmm
2007 JJC H2 Solutions
JJC H2 Physics P2 Soln 5
Prelim Exam 2007 [Turn Over
θθθθ = 10.7°C –[1 mark]
(e) ( ) 805.07.257.20 =⇒=⇒= NNFNVHe mmLmLm kg -[working 1 mark]
Remaining mass = 1 − 0.805 = 0.195 kg –[Ans 1 mark]
7 (a) Correct labeling of line – [1] Since S is stable, the number of its nuclei can only increase. [1]
(b) (i) From Fig.7, half-life of parent isotope = 3.0 years. [1]
(ii) From Fig. 7, max % of D = 68 %
Therefore, maximum A of D = λ (max N) = 0.693
15 years
(0.68 (1.2×1015)) [1]
= 3.77××××1013 year-1 [1]
(c) The half-life of D (= 15 yrs) is much longer than half-life of parent isotope (= 3 yrs). In the initial stage, the relatively large number and short half-life of the parent nuclei results in a net rate of formation of D. [1] At t = 9 yrs, the relatively large number of D and small number of radioactive parent nuclei [1] results in rate of formation of D being approx equal to its rate of decay. After 9 yrs, the number of radioactive parent nuclei is insufficient to decay and replenish the D nuclei that decay [1], so the number of D decreases.
2007 JJC H2 Solutions
JJC H2 Physics P3 Soln 2
Preliminary Examination 2007 [Turn Over
Suggested Soln for JJ Prelims 2007 Physics (H2, 9745) Paper 3
1 (a) Velocity is defined as the rate of change of displacement (with respect to time).
(b) (i) [v = u + at]
0 = 15 sin 60°°°° + (-9.81)t t = 1.32 s
[Subst 1 mark, ans 1 mark] (ii) [v2 = u2 + 2as]
0 = (15 sin 60°°°°)2 + 2(-9.81)s s = 8.60 m
[Subst 1 mark, ans 1 mark] (iii) Range = (15 cos 60°)(2)(1.32) = 19.8 m
[ans 1 mark]
2 (a) 31029.1
360035.1
22 −×=×
==ππ
ωT
rad s-1 [ans 1 mark]
(b) ( )( )( ) 4731029.160cos108750652332 =×°××== −ωmrFC N [Ans 1 mark]
(c) The resultant (centripetal) force is always perpendicular to the linear
velocity / direction of motion. [Ans 1 mark]
(d) ( )
1390108750
651046.21067.623
2511
2=
×
××××==
−
r
GMmW N [Ans 1 mark]
(e) °⋅⋅−+= 60cos2222
CC FWFWC -[Method 1 mark]
°⋅⋅⋅−+= 60cos473139024731390 222C
1220=C N – [Ans 1 mark]
(f) (i) Direction: no change; Magnitude: increases [1]
(ii) Direction: changes; Magnitude: increases [1]
3 (a) Incident wave (from magnetron) and reflected wave from opposite wall [1]
superpose [1] according to Principle of Superposition. Since the incident and reflected waves travel in opposite direction, with same speed, magnitude and frequency, stationary waves are formed.
(b) 3.00×108 = (2.45×109)λ λ = 0.122 m [1]
(c) 3 complete nodes – 1 mark Nodes and antinodes both correctly labelled – 1 mark
Node Antinode
2007 JJC H2 Solutions
JJC H2 Physics P3 Soln 3
Preliminary Examination 2007 [Turn Over
(d) Food placed at nodes will not be heated OR those placed at antinodes may be overcooked. [1]
4 (a) (i) 6 –[1]
(ii) (-3.4) – (-54.4) = 51.0 eV –[1]
(iii)
E =hc
λλλλ λ =
hc
E=
34 8
19
(6.63 10 )(3 10 )
(51.0)(1.6 10 )
−−−−
−−−−
× ×× ×× ×× ×××××
--[1 mark for subst]
= 2.44×10-8 m [1 mark for ans]
(b) (i) It means 8.5×10-19 J is the minimum energy required to release an electron from the metal surface. [1]
(ii) KEmax = hf – φ =
hc
λλλλ– φ
=34 8
9
(6.63 10 )(3 10 )
(163.6 10 )
−−−−
−−−−
× ×× ×× ×× ×××××
– 8.51×10-19 = 3.65××××10-19 J
[Equation with λλλλ -1 mark] , [Ans – 1 mark]
(iii)
λ =h
p=
2
h
mE
=34
31 19
6.63 10
2(9.11 10 )(3.65 10 )
−−−−
− −− −− −− −
××××
× ×× ×× ×× ×= 8.13××××10-10 m
[Working – 2 marks, Ans – 1 mark] (c)
min
hc
λλλλ= eV
λmin =hc
eV=
34 8
19
(6.63 10 )(3 10 )
(1.6 10 )(3000)
−−−−
−−−−
× ×× ×× ×× ×××××
= 4.14×10-10 m [Subst – 1 mark, Ans – 1 mark]
5 (a) (i) a = 1, b = 0. X is a neutron. [1]
(ii) Carbon nucleus/nuclide with 13 nucleons and 6 protons. [1]
**Carbon atom is not acceptable.
(b) (i) Loss of mass = (9.0150 + 4.0040) - 13.0075 = 0.0115 u or 1.91 × 10-29 kg [1]
(ii) E = (1.91×10-29) (3.00×108)2 [1]
= 1.72××××10-12 J [1] (iii) Not valid. The energy of the radiation emitted (1.72×10-12 J) is less than the
energy of the penetrating radiation (8.8×10-12 J). [2 or zero; practice ecf from b(ii)]
2007 JJC H2 Solutions
JJC H2 Physics P3 Soln 4
Preliminary Examination 2007 [Turn Over
Section B
6 (a) The rate of change of momentum of an object is directly proportional to the resultant force acing on it. –[1] The change in momentum takes place in the direction of that force. -[1]
(b) • The total momentum of the system before the combustion is zero.
• After the combustion, the emitted gas and the rocket will have a momentum that are equal and opposite such that the total momentum is zero.
(c) (i)
(ii) By N3L, Downward force = 7.22 N
(iii) 2( ) ( )d mv d Ax
F v Avdt dt
ρ= = = ρ
7.22 = 1.3 (0.4)(v2) v = 3.7254
≈ 3.73 m s-1
(iv) Thus the change of momentum by the ground on the toy will take place over a longer time, hence reducing the force on the toy. OR Longer distance travelled during the collision (which implied longer time traveled)
(d) (i) When it is equilibrium, Weight of the submarine = Upthrust
= ρ V g = (1025)(100)(9.81) = 1005525 ≈ 1.00x106 N
(ii) 1. Pressure at the water outlet = ρ g h + 1.00 x 105
= (1025)(9.81)(20) + 1.00x105 = 301105
Energy required = ρ ∆V = 301105x10 = 3011050
≈ 3.01x106 J
2. Resultant force = Upthrust - weight of the submarine = weight of seawater pumped out = 10 x 1025 x 9.81= 100552.5
Mass of submarine = (1005525 − 10552.5) / 9.81 = 92250 kg
ΣF = ma 100552.5 = (92250)a a = 1.09 ms-2
0.7g
F F - 0.7g = 0.7 (0.5) F = 0.7(9.81) + 0.7(0.5) = 7.217
≈ 7.22 N
a
2007 JJC H2 Solutions
JJC H2 Physics P3 Soln 5
Preliminary Examination 2007 [Turn Over
(iii) The weight of the submarine reduces and the upthrust on the submarine remain constant. There will be a resultant upward force. By N2L, it will accelerate upwards. OR Explain using the principle of N3L or law of conservation of momentum due to water on the submarine. OR Density of the submarine decreases.
7 (a) (i) It is the energy required to convert unit mass of the liquid to vapour without any change in temperature.
(ii) 1. the power supplied by the kettle = the rate of thermal energy gained by the water during the temperature rise
= mc∆θ / t
=2.0 x 4.2x103 x (100 − 20) / (8 x 60) = 1.4x103 W
2. Rate of energy absorbed by the water during vaporisation = Power
supplied by the kettle (0.7 x L ) / (20 x 60) = 1.4x103 L = 2.4x106 J kg-1
3. Lm = (2.4x106 / 1000) x 18 = 4.3x104 J mol-1
(b) (i) The internal energy of a system is the sum of the potential and random kinetic energies of all the molecules in the system.
(ii) Any 2 among: Random KE of the molecules OR Temperature, amount of substance ( or mass), Volume, Pressure, Density
(iii) The speed with which the entire system moves
(iv) 1. Raising the temperature of the gas increases the random kinetic energy of the molecules and hence increases the internal energy.
2. Since the temperature is constant, the random kinetic energy of the molecules of the ideal gas is constant and hence the internal energy of the ideal gas is constant during this change in volume.
(c) (i) The no. of moles of helium gas, n = pV / RT
= (1.0x105 x 1.0x10-3) /( 8.31 x 300) = 0.04 Hence the number of helium molecules = 0.04 x 6.02x1023 = 2.4x1022
2007 JJC H2 Solutions
JJC H2 Physics P3 Soln 6
Preliminary Examination 2007 [Turn Over
(ii)
1-3
3
600rms
300rms
600rms
rms
ms 10 x 1.98
)104.1(2 )(v
2 300
600
)(v
)(v
T v
=
=
==
∝
xK
K
K
8 (a) (i) 6.0 Ω // 9.0 Ω => R = 3.6 Ω
6.0 Ω // (3.6+2.4) Ω
Hence, Reff = 1
3.01 1
6 6
=+
Ω
(ii) 1. 15
115
VI
R= = = A
Vac = (1)(6) = Va − Vc
Hence, Vc = −9 V 2. Vd = −9 – (1)(3) = −12 V
By principle of potential divider,
Ved = 3.6 3.6
( )( ) ( )(3.0) 1.83.6 2.4 6.0
cdV = =
+ V
1.80.30
6.0
VI
R= = = A
(b)
(i)
8
8
(49 10 )(0.75)30.625 30.6
1.2 10
xR
A x
ρ −
−= = = =
lΩ
(ii)
1. Pd across XY = 30.6
( )(2) 1.8230.6 3
=+
V
2. 50.0
2.0 ( )(1.82)75.0
XPI V= =
0.607I = A
(c) (i) ( )
(0.607)( 2) - - - - - (1)
(0.382)( 4) - - - - - (2)
= +
= +
= +
E I R r
E r
E r
Solve equations (1) and (2):
Internal resistance r = 1.40 Ω e.m.f. E = 2.06 V
(ii) If ammeter has non-negligible resistance, current / potential difference decreases. Hence length XP decreases.
2007 JJC H2 Solutions
Meridian Junior College J2 H2 Physics Prelim Paper 2 Solution 2007
1 (a) Consider motion along plane
v2=u2+2as = 0+ 2(4.00)(50) v =20.0 ms-1
Using v =u +at 20.0 0
5.004.00
v ut s
a− −= = =
[M1] [A1] [A1]
(b) Sign convention: horizontally to the right positive
Vertically down positive uy=20.0sin 370 = 12.0 ms-1
ux=20.0cos 370= 16.0 ms-1= vx
vy2= uy
2+2gsy= 12.02+2(9.81)(30.0) [M1] vy=27.1 ms-1
2 2 2 216 27.1 31.5x yv v v= + = + = ms-1 [A1]
tan θ = 27.116.0 , θ is the angle below horizontal
θ =59.40 [A1]
(c) vy= uy + g t1, t1 = time to fall from B. t1 = (27.1 -12.0)/9.81 =1.54 s total time = t + t1= 5.0 +1.54 = 6.54 s
[M1] [A1]
]
2 (a) P = WD / t
= F d / t = F v
[B1]
(b) (i) v2 = u2 + 2as
a = - 0.97 m s-2 F = ma = 0.97 x 1500 = 1460 or 1500 N (negative sign acceptable)
[C1] [M1] [A1]
(ii) P = F v = 1460 x 12
= 17500 W [M1] [A1]
(iii) On level road, WD by engine converted to WD against resistive
2007 MJC H2 Solutions
forces. With the same engine force, the WD by engine is still converted to WD against resistive forces, since both forces remain the same. But since the truck gains GPE, it’s KE will decrease. OR (or gain in GPE = loss in KE because WD by engine = WD against resistive forces)
[B1] [B1]
3 (a) x = λD / a
= (589 × 10-9) (2.5) / (0.8 × 10-3) = 1.84 × 10-3 m = 1.84 mm
[M1] [A1]
(b) (i) λ increases (red light)
hence fringe separation x increases. [M1] [A1]
(ii) Diffraction grating has more slits, hence interference pattern
sharper (clearer, more distinct) and/or brighter
OR
Using d sinθ = m λ, for grating with lines 1000 per mm [B1], d is smaller compared to double slit of d = 0.8mm, hence θ larger for diffraction grating, and fringe separation if greater.[B1]
[B1] [B1]
(iii) x is proportional to D.
Hence fringe separation at lower part of screen is closer. fringe separation at upper part of screen is further.
[M1] [A1]
4 (a) The electromotive force of the cell is 1.6 V [B1] (b) Ir = E – V [M1] (0.50)r = 1.6 – 1.2
r = 0.80 [A1]
(c) (i) V= (0.30)(0.80) [M1] = 0.24 V [A1] (ii) V = 1.6 – 0.24 [M1] = 1.36 V [A1] (d) Bulb Y : P=(1.36)(0.30) = 0.408 W [B1] Bulb X : P = (1.2)(0.50) = 0.60 W Bulb Y : Power supplied by cell Ps = (1.6)(0.3) = 0.48 W [B1] Bulb X : Power supplied by cell Ps = (1.6)(0.5) = 0.80 W
2007 MJC H2 Solutions
Bulb Y : Efficiency = 85 % [B1] Bulb X : Efficiency = 75 % 5 (a) An atom (or electron) in an excited state can be perturbed by an incident
resonant photon to emit another photon of the same frequency, in phase and in the same direction of travel as the incident photon.
[B1] [B1]
(b) There is an overlap in the valence and conduction bands in a metal, thus
there are more free (mobile) electrons to conduct electricity in the conduction band as compared to semiconductors. OR Semiconductors have a gap between the valence and conduction band, as such, semiconductors have less mobile electrons in their conduction band to conduct electricity as compared to metals.
[B1] [B1] [B1] [B1]
(c) (i) n-type semiconductor has an excess of electrons in the
conduction band. These electrons are free to conduct electricity and thus increase the conductivity of the semiconductor OR p-type semiconductor has an excess of holes in the valence band. These holes are free to conduct electricity and thus increase the conductivity of the semiconductor
[B1] [B1] [B1] [B1]
(ii) S, Se or Te (They have one extra electron as compared to As)
[B1]
(d) Acceptable answers: laser printers, laser pointers Not acceptable: laser surgery, laser cutters (they can cut and melt so would melt a CD too)
[B1]
6 (a) There is no way to predict which nuclei will decay and when it will
decay. The decay is not affected by external physical conditions and chemical composition of the sample.
[B1] [B1]
2007 MJC H2 Solutions
(b) Temperature OR pressure. [B1] (ci) 1 (from Fig 6.1, it can be deduced T1/2 for Co is around 5 years
and for Ni is around 90 years.) In region PQ, from Fig 6.1; AC >> AN Observe that PQ is approximately a straight line. Total Activity Amix = AC + AN ≈ AC ≈ AoC e - λc t
hence Amix ≈ AC = AoC e - λc t Hence ln Amix against t graph should give a straight line , which is shown in the line PQ.
[B1] [B1]
2 In region RS, most of Co has decayed, so activity is mainly due
to Ni hence Amix ≈ AN = AoN e - λn t Hence ln Amix against t graph gives a straight line.
[B1] [B1]
3 In region QR, both Co and Ni contribute to the activity Activity Amix = AC + AN = AoC e - λc t + AoN e - λn t
Hence ln Amix against t graph gives a curve.
[B1] [B1]
(cii) 1 grad of PQ = (7.63 – 8.87) / 10
= - 0.124 yr-1
[M1] [A1]
2 grad of RS = (4.77 – 5.08) / 40 = - 0.00775 yr-1
[M1] [A1]
(ciii) A = Ao e - λ t
ln A = ln A0 - λ t
2007 MJC H2 Solutions
Hence λ = magnitude of grad of ln A vs t graph. decay constant of cobalt = 0.124 yr-1
decay constant of nickel = 0.00775 yr-1
[A1]
(civ) T1/2 = ln 2 / λ = ln 2 / 0.124 = 5.59 yr
[A1]
(d) Initially, both will be hazardous (as they are in large dosage)
(also acceptable: if students observe that Co has very high activity at the start – hence more hazardous) As half life of Co is short (5.59 years) – activity will be small after that number of years, hence less hazardous. However Ni has a long half-life (about 90 years), it remains hazardous for a long time.
[B1] [B1] [B1]
2007 MJC H2 Solutions
Meridian Junior College J2 H2 Physics Prelim Paper 3 Solution 2007
Section A : SSQ 1 (a) (i) φ = -2.2- ( -2.6) x 106 = 4.0 x 105 Jkg-1 C1
Work done = m∆φ = 2 x 4.0 x 105 = 8.0 x 105 J B1
(ii) φ = 0 Work done = 0 B1
(iii)
V = 1493 ms-1
B1 M1 A1 M1 A1
(b) (i) Escape velocity is the minimum velocity of a mass/object needs to be projected from the surface of moon (or other planet) in order to have sufficient kinetic energy to overcome the gravitational field it experiences and move to infinity.
B1
(ii) The average speed of the molecules is much larger than the escape velocity, hence most of the nitrogen molecules will not stay in orbit. They will more likely to escape from the moon’s gravitational field, hence no atmosphere can be formed.
B1
2 (a) The mass is undergoing simple harmonic motion where its acceleration a
is always directed towards a fixed point and is directly proportional to its displacement x from that fixed point.
2a xω= −
B1 B1
(b)
( )
( )
2 2
2
2cos 8.00 10 cos 8.00 10 cos
8.00 10 cos 4.47
o
kx x t t t
T m
x t
πω − −
−
= = − × = − ×
= − ×
(positive cos is also acceptable if student defines downwards positive) (sine is not acceptable since question stated it was released after being pulled down by 8 cm)
M1 A1
6
5
dg
drd 0.4x10
F mg m 2.0x 2.0Ndr 4.0x10
φ
φ
= −
= = = =
6
5
d [ 2.0 ( 2.4 )] x10 J
dr (7.1 3.1)x10 J
φ = − − −= −
2
2
6 5
mvF
r2.0v
2.01.74x10 4.9x10
=
=+
2007 MJC H2 Solutions
(c) 2 2
2 2
1
4.47 0.0800 0.0600
0.237
ov x x
v
v ms
ω
−
= −
= −=
M1 A1
(d) ( )
( )0.04 0.08cos 4.47
0.5 cos 4.47
4.47 2.094 4.189
t
t
t rad or rad
= −
− ==
( )( )
0.04 0.08cos 4.47
0.5 cos 4.47
4.47 1.047 5.236
t
t
t rad or rad
− = −
==
Shortest time is to go from 4.47t = 4.189 rad to 5.236 rad t = (5.236 – 4.189)/4.47 = 0.234 s Alternative answer: Due to the symmetry of the cos curve, time required to go from x = -4 cm to +4 cm is also acceptable.
M1 M1 A1
(e) Answer must show amplitude getting smaller as time proceeds.
(Curve must remain cosine like, sin curve accepted as ecf only if (b) was written with sine)
B1
3 (a) (i)
1 mark for number of forces. 1 mark for directions
M1 M1
FB W
N
FE
2007 MJC H2 Solutions
(ii)
( ) ( )
2
22.0 2.192.0 9.81 1.5 0.20 2.19
0.50
39.46 N39 N
B cN W F F
mvN mg Bqv
r
− − =
= + +
×= + × + × ×
=≈
M1 M1 A1
(b) (i) When magnet X is pushed down towards solenoid A, it will start to
oscillate in the vertical plane. This causes a rate of change of magnetic flux linkage in solenoid A. By Faraday’s law of magnetic flux linkage, an e.m.f is induced within solenoid A and a current flows. Since the direction and velocity of X is constantly changing, the rate of change of magnetic flux linkage is constantly varying, thus causing an alternating current within A and hence B.
B1 B1 B1
(ii) When an alternating current flows in B, it induces a magnetic pole at the ends of the solenoid. Since the current is alternating, the magnetic poles induced are constantly changing in polarity. Hence, magnet Y is varyingly attracted and repelled, causing it to oscillate.
B1 B1
4 (a) (i) By potential divider principle (6.0) x 2.0 = 4.0 V
Alternatively 800
6.0 4.0V400 800 = +
B1
(ii) Equivalent resistance of 800 and 400 in parallel
C1 B1
(b) (i) VPrms = 2.4 V from (a)(ii) VSrms = 2.4 x 50 = 120V VSmax = 120 x √2 = 170V
C1 B1
(ii) Pave = ¼ Ppeak
OR
C1 B1
1
T1 1
R 267800 400
− = + = Ω
PQ6
V x267 2.4V400 267
= =+
S S
P P rms
N V50
N V
= =
= = =×
2 2peak
ave
V 170P 154W
4R 4 47
= = =
2peak
2rms
ave
V2V
P 154WR 47
2007 MJC H2 Solutions
(iii) Pnew = 2 Pave = 307 W OR Substitue Vrms = Vpeak/ 2 into Pave formula
B1
(iv) D.C B1
(v) Transformers work based on E.M.I. which requires a change in magnetic flux which is provided by an a.c. not a d.c.
B1
Section B : LSQ 5 (a) (i) The moment of a force about a point is the product of the force and the
perpendicular distance between the point and the line of action of the force.
B2 (ii) Total clockwise moments about any axis of rotation is equal to the total
anticlockwise moments about the same axis. OR B2 The net torque about any axis of rotation is zero, (b) (i) Taking moments about point P,
F(2.5) =1800(0.35) F = 252 N
M1 A1
(ii) F + FR = 1800 M1 FR = 1548 N = 1550 N A1 (c) (i) COM : 1 1 1 2 2m v m v m v= + Eqn 1 B1 (ii) Head-on : 2 1v v v−= Eqn 2 B1 (iii) Multiply Eqn 2 by 1m : 1 1 2 1 1m v m v m v= − Eqn 3
Eqn 3 – Eqn 1 : 1 2
1 21
)(2
m mv vm−
= Eqn 4 M1
From Eqn 4 and Eqn 2 : 1 2
1 1 2
)(m mv vm m−
= + M1
2007 MJC H2 Solutions
Fraction of initial K.E transferred =
2 2
1 1 11 2
22 1 2
1
1 142 2
1 ( )2
m v m v m mm mm v
−=
+
A1
(iv)
2
1
mm = 1.
A1
1m will stop moving. B1
2m will move off with the speed of v. B1
(v) 1. 1 2 0 6.0u u v− = − = m s-1
A1 2. COM : 2 2 2 1 2 22 (0) 2m v m m v m v+ = + M1 Head-on : 2 1 6.0v v− =
1 2.0v = m s-1, 2 8.0v = m s-1 A1 (vi) The velocities of the bodies after collision will not be along the
same line of motion before collision. B1
6 (a) The first law of thermodynamics states that the increase in internal energy is
equal to the sum of heat transferred into the system and work done on the system
(b) ∆U is the increase in internal energy.
p∆V is the work done by system
(c) (i) Using PV = n R T, T = 273 + 41 = 314 K P = 780 mmHg = ghρ = 13.6 x 103 x 9.81 x78.0 x 10-2
= 1.04 x 105 Pa
n = 1.04 x 105 x 2.0 x 10-4/(8.31 x 314) = 0.00797 moles .
(ii) Using PV = n R T, Tf =273 + 517 = 790 K Pf (2.9 x 10-5) = 0.00797 x 8.31 x 790 Pf= 1.80 x 106 Pa
(iii) ∆U = Q + W = 0 + 91 = 91 J The change is an increase.
(iv) Internal energy is the total translational kinetic energy of the molecules of an ideal gas.
2007 MJC H2 Solutions
The work done in compressing the gas increases the internal energy of the gas, thus the kinetic energy of its molecules increases. Since the kinetic energy of the gas molecules is proportional to the thermodynamic temperature. An increase in KE is thus associated with a rise in temperature.
(v) At low pressure or high temperature For low pressure, molecules are very far apart OR For high temperatures molecules move with very high speeds Intermolecular forces/interaction is negligible
(d) Given rate of water supplied to town , V/t = 2.7 m3 s-1 Let Vv/t be the rate of canal water vaporized. Area exposed to sun = 51 km x 9.2 m = 51 x 103 x 9.2 so rate of heat absorbed by water = 900 Wm-2 x Area = 4.22 x 108 W
From conservation of energy, rate of heat absorbed from sun = rate of vaporization of water.
4.22 x 108 W = vv
Ml
t, lv = 2.26 x 106 J kg-1
Mv/t = 4.22 x 108/(2.26 x 106) = 187 kg s-1
Rate at which water needs to be
supplied from reservoir = ρV/t + ρVv/t = 2700 + 187 = 2887 = 2900 kg s-1 Or Rate at which water needs to be supplied from reservoir = V/t + Vv/t = 2.700 + 0.187 = 2.887 = 2.9 kg s-1
7 (a) (i) An electron with an initial kinetic energy Ek initial collides with a
target atom. Upon collision, the electron decelerates and loses an amount of energy ∆Ek which appears as the energy of an X-ray photon that is radiated. This scattered electron which now has energy less than Ek initial may have a subsequent collision with another target atom, generating a second x-ray photon whose energy, in general, is different from the energy of the x-ray photon produced in the first collision. Hence all the photons generated by these collisions between electrons and target atoms form the continuous x-ray spectrum.
B1 B1 B1
2007 MJC H2 Solutions
(ii) max
min
hf eVc
h eVλ
=
=
( ) ( ) ( ) ( )8
34 19
min
3.0 106.63 10 1.6 10 40000
λ− −
×× = ×
11min 3.11 10 mλ −= ×
M1 A1
(b) (i) 6 possible transitions
B1
(ii) c
E hλ
∆ =
( ) ( ) ( ) ( )8
19 34
min
3.0 103.28 23.4 1.6 10 6.63 10
λ− −
×− + × = ×
min 61.8 nmλ =
M1 A1
(iii)
[B1] for 3 lines and correct relative spacing
(iv) The atoms will not be excited. The photon energy does not match any of the differences in the energy levels, hence the photon energy will not be absorbed by the atom.
A1 M1
(c) (i) 1. For a given metal, no photoelectrons are emitted if the
frequency of the incident light is lower than a certain frequency.
2. Emission of photoelectrons takes place almost
instantaneously after the light shines on the metal, with no detectable time delay.
3. Photoelectrons emitted from a metal have a range of
velocities from zero up to a maximum vmax.
Fig. 7.3
increasing wavelength
2007 MJC H2 Solutions
[B2] for any two observations (ii) 1. The ammeter reading remains zero.
As the intensity is increased, the rate of emission of photoelectrons will increase, but the KEmax of the electrons remain the same. Hence the electrons will still not be able to overcome the stopping potential. The photocurrent remains zero.
A1 M1
2. The ammeter reading will increase to a non-zero value.
When the wavelength is decreased, the KEmax of the photoelectrons will increase. Some of the electrons can now overcome the potential and reach the collector. A photocurrent will be registered.
A1 M1
(iii) No. of photoelectrons emitted per second
315 -1
19
0.34 102.125 10 s
1.6 10q
−
−
×= = = ××
I
No. of incident photons per second
( )
315 -1
834
9
2.85 103.224 10 s
3.0 106.63 10
225 10
P Pchf hλ
−
−−
×= = = = × ×× ×
15
15
2.125 10Quantum yield 0.659
3.224 10×= =×
M1 M1 A1
2007 MJC H2 Solutions
2007 NJC H2 Physics Prelim Paper 2 Solution
1. (a) Units of P = kg m-1s-2 (b) magnitude Unit The Earth’s magnetic field 10-5 to 10-6 T The resistance of a domestic filament lamp 103 Ω Wavelength of visible light 10-7 m 2 (a) (i)
( ) Ns
ptF
176676060
10606010003
=××
+=
Δ=Δ
(a) (ii)
NF 9298219.0
17667==
Therefore the crash distance is 158114 −== NmlFk
(a) (iii) Conservation of momentum is not applicable to the system of the block and the car as an external force is acted on the block to fix it to the ground.
(b) (i)
125
)(2)80(30−=
=−+
kmhvvmmm
towards left (b) (ii)
2985343600
108021
36001030
21
2323
=⎟⎟⎠
⎞⎜⎜⎝
⎛ ×+⎟⎟
⎠
⎞⎜⎜⎝
⎛ ×= mmKEini
511193600
1025221
23
=⎟⎟⎠
⎞⎜⎜⎝
⎛ ×= mKE fin
Lost in KE = 247415 J
2007 NJC H2 Solutions
(b) (iii) At the instant of collision, both dummy drivers experienced the same amount of force. Due to Newton’s 3rd law, both cars will experience the same equal amount of forces (action and reaction pair) acting on different bodies in the opposite direction. [Although both cars have different velocities, the relative speeds of approach for both cars are the same.]
3. (a) FvdtdsF
dtdWP ===
(c) FD = Fappllied = 200 / 5 = 40 N (d) P = Fv = kv2
(e) P’ = 200 x (6 / 5)2 = 288 W
(f) P = power to overcome drag + power to climb vertically
= 200 + mgv sinθ = 200 + 100(9.81)(5) sin 1.9o = 363 W
4.(a) P = hρg 101 000 = h(1.3)(9.81)
h = 7.92 km
(b) (ρwood)(0.90)Vg = (1000)(2/3)Vg
ρwood = 741 kgm-3
(c) (i) ΣFx = 0:
T1 sin 36.9° = T2 sin 53.1° T1 = 1.33 T2 (ii) ΣFy = 0: T1 cos 36.9° + T2 cos 53.1° = mg 1.664 T2 = mg
2007 NJC H2 Solutions
Στ about left-hand end of rod: mgx = T2 cos 53.1° x 6.10 mgx = mg(2.20) x = 2.20 m
5. (a) Magnitude of net force on the body must be directly proportional to displacement from
equilibrium position. Net force must also be in the opposite direction to its displacement OR directed towards the equilibrium position.
(b)(i) The graph of force against extension yields a straight line through the origin, meaning
the net force is directly proportional to the distance from it. . (ii) F = - kx
ma = - kx a = - kx / m = - ω2x (for SHM) Hence, ω2 = k/m
(iii) k = 3 / 0.1 = 30 Nm-1 ω2 = 30 / 0.20
T = 2π/ ω = 0.51 s
(iv) k (equilibrium length) = mg equilibrium length = 0.20 x 9.81 / 30 = 0.065 m
6. (a) WAB = FlAB = 5 x 10-7 x 20 x 10-3 = 1.00 x 10-8 J
WAC = 0 J , because force and displacement in the direction of the force are perpendicular to each other. WAD = 1.00 x 10-8 J, because the displacement in the direction of the force for moving from A to D is similar to that moving from A to B.
(b) VAB = WAB / Q = 1 x 10-8 / 2.5 x 10-11 = 400 V VAB = VB – VA 400 = VB – 200 VB = 600 V VC = VA = 200 V VD = VD = 600 V
(c) Need to ensure correct position and also indicate which the plate is earthed or which is 1000 V.
2007 NJC H2 Solutions
(d) Parallel lines perpendicular to plates, equal spacing, horizontal direction to the right, symmetry in the distribution of the lines
E [1]
1000 V
Fig 6 7. (a)(i)
n Energy /eV
1 -13.6
2 -3.40
3 -1.51
4 -0.85
5 -0.54
6 -0.35
2007 NJC H2 Solutions
(a) (ii)
(b) Consider a transition from level 4 to level 3, Energy of photo emitted = -0.85 – (-1.51) eV
λhc = 0.66 x 1.6 x 10-19
λ = 1.88 x 10-6 m Or Consider a transition from level 5 to level 3, Energy of photo emitted = -0.54 – (-1.51) eV
λhc = 0.97 x 1.6 x 10-19
λ = 1.28 x 10-6 m Or Consider a transition from level 6to level 3, Energy of photo emitted = -0.35 – (-1.51) eV
λhc = 1.16 x 1.6 x 10-19
λ = 1.07 x 10-6 m Since 8.00 x 10-7 m < λ <1 x 10-4 m, the wavelength lies in the infra-red region (c)(i) The angular spacing for 2nd order spectrum is wider than 1st order spectrum, hence it will
have a smaller percentage uncertainty.
2007 NJC H2 Solutions
(c)(ii) dsinθ = 2λ θ = 162.60 - 126.40 = 36.2o d = 1/N
λ = )105.4(2
2.36sin5×
= 6.56 x 10-7 m
(c)(iii) E = λhc
= 7
834
1056.6)103)(1063.6(
−
−
×
××
= 3.03 x 10-19 J = 1.9 eV E3 – E2 = -1.51 – (-3.4) = 1.9eV X is indicated in the graph above.
2007 NJC H2 Solutions
NJC Prelim 2007 H2 Physics Paper 3
Solutions Section A Question 1: Superposition
1(a) (i)
31
3 6
13.5 10tan ( )5.0
(0.50 10 )sin 1.35 10x m
θ
θ
−−
− −
×=
Δ = × = ×
(ii) (2 ) 4.5xφ π πλΔ
Δ = = or 2π
(iii)
1
(iv) For maximum intensity, resultant amplitude = 2E0 At X, resultant amplitude = 1.4 E0
Therefore ( )( )20
20
0 2
4.1
E
EII=
I = 0.49 I0 or 0.50 I0 (b)
S2
E
t
E0
− E0
− 2E0
2E0
0
R
Angle, θ°
Intensity, I
I0
0 90 180 270 360
0.25I0
2007 NJC H2 Solutions
Section A Question 2: Current of Electricity
2(a)
2
212030
480
VPR
RR
=
=
= Ω
(b) 50000 (120) 119
50000 480V V= =
+
(c) Effective resistance of Ammeter and Lamp, 11 1 1.99
480 2effR−
⎛ ⎞= + =⎜ ⎟⎝ ⎠
Ω
100 /1.99 50.2I A= =
120 100 50.20.398
E V Irr
r
= += +
= Ω (d) Let the no. of batteries needed be N
480 (1.5) 120
480 (1.2)100
NN
N
× =+
=
Section A Question 3: Thermal Physics 3(a) (i) Heat gained by milk = Heat loss by steam 0.2 (4000)(∆θ) = 0.001(2.2x106) + 0.001(4200)(15) Rise in temperature of milk, ∆θ = 2.83 K (b) (i) pV = nRT
=TpV
constant
=A
AA
TVp
C
CC
TVp
=2500
)95.0(105 5x
CTx )95.0(101 5
TC = 500 K (ii) ∆U = Q + W , W = 0 Q = 3/2 (PCVC – PAVA) = 3/2 (5-1) x105 x 0.95 = 570 kJ
(iii) ∆U = Q + W 3/2 (0.95-3) x105 = Q + (3-0.95) x 105 Q = - 513 kJ (iv) ∆U = Q + W For a cyclic process, ∆U = 0 W = - Q
2
2007 NJC H2 Solutions
= - (570 + (-513)) = - 57.5 kJ (v) PAVA = PBVB
γ γ
(5 x 105)(0.95) = (1 x 105)(3) γ γ
3ln95.0ln5ln γγ =+
40.1=γ (v) If it is not carried out rapidly, heat supplied to the system will not be zero.
Section A Question 4: Nuclear Physics 4 (a) (i) Binding energy of a nucleus is the amount of work needed to take all its constituent
nucleons apart so that they are separated an infinite distance from one another. (ii) Conservation of mass and conservation of energy (iii) Energy released in one reaction = (mU + mn – (mXe + mSr + 2mn)) uc2 = 2.94 x 10-11 J
(iv) There are 2 methods in calculating the total number of reactions. Method 1 No of reactions that will take place in 2 kg Uranium = (2000/235) x 6.02 x 1023
= 5.12 x 1024 Method 2 No of reactions that will take place in 2 kg Uranium = 2 / (235 x 1.66 x 10-27)
= 5.12 x 1024 Pbulb(t) = 1.51 x 1014 Amount of time for a 50 W light bulb to last, t = (1.51 x 1014/(50x60x60)) = 8.38 x 108 hours (b) Total momentum must be conserved. Since initial momentum is zero, final momentum must also
be zero. This will only happen if the gamma photons move in opposite direction.
Section A Question 5: Laser and Semiconductor 5 (a) (i) A longer lifetime in the E3 state means that the helium atoms can stay excited long enough for energy to be transferred to the ground state Neon atoms through collision. This ensures a constant supply of neon atoms in the E2 state necessary for laser action. OR If the lifetime of the E3 state is short compared to the E2 state, there won’t be enough excited helium atoms present in the medium that will excite the neon atoms to the E2 states and this will affect the degree of population inversion in neon. 5 (a) (ii) • The E0 state of neon is stable and highly populated. Hence it is impossible to achieve
population inversion between E0 and any excited states of neon. • The E1 state has the shortest lifetime among the three states. Hence it has the lowest
population among the 3 states. • The E2 state has a longer lifetime than the E1 and hence it is more highly populated compared
to the E1 state. • Hence population inversion can exist between the E2 and E1 states.
3
2007 NJC H2 Solutions
5 (a) (iii) ( )
⎟⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛==
λhc
tN
thfNP ⇒
834
93
1031063.6108.632102.1
×××
×××==
−
−−
hcP
tN λ = 3.82 x 1015 s-1
5 (b) (i) • Electrons in the n-type semiconductor diffuse across the junction towards the p-type
semiconductor thereby combining with the holes directly across the junction. • This causes an electric field to exist across the junction that slowly opposes the movement of
electrons from the n-type to the p-type semiconductor. • At steady state, the electric field will be strong enough to prevent the net movement of
electrons across the junction. • A depletion region is formed on either side of the junction characterized by the lack of mobile
charge carriers within this region. 5 (b) (ii) The width of the depletion region will increase because electrons in the n-type semiconductor will migrate away from the n-type type depletion region under the action of the applied reverse bias potential, causing an increase in the overall potential difference. Section B Question 6: Gravitaional Field (a) Work done by external agent in bringing a mass from infinity to a point within a gravitational
field.
(b)(i) Using R
mvR
GMm 2
2=
KE = ½ mv2 = R
GMm2
Total energy of satellite = KE + GPE
= ⎟⎠
⎞⎜⎝
⎛ −+R
GMmR
GMm2
= R
GMm2
−
(b)(ii) 2
22
2⎟⎠
⎞⎜⎝
⎛==T
mRmRR
GMm πω
EGM
RT32
2 4π=
(b)(iii) T = 24 x 3600s
R3 = 2
2
4πTGME = 7.567 x 1022
R = 4.23 x 107 m Distance above surface = 3.6 x 107 = 36000 km (Shown)
4
2007 NJC H2 Solutions
(b)(iv) Total energy of satellite at geostationary orbit = R
GMm2
− = - 5.676 x 108 J
By COE; ½ (120) + 2vER
GMm− = - 5.676 x 108 J
v = 10 752 ms-1 = 10.8 x 103 ms-1
(c) Using EGM
RT32
2 4π= = ( )
2411
3562
100.61067.6103104.64
xxxxx
−
+×π = 5447s = 5.4 x 103 s
Using R
mvR
GMm 2
2= ,
)103104.6(1061067.6
56
2411
xxxxxv
+=
−
= 7729 = 7.8 x 103 ms-1
(d) 200
1=ΔT = 5 x 10-3 revolution
=Δ
=Δ3/5
3/2
3
)2(
T
TmGME Eπ
=−
3/5
33/2
)5400(3
)105)(120()2( xGMEπ 2225 J
Average rate of loss of energy in 1 revolution = WTE 412.0
54002225
==Δ
From , 3vAP ρ=3Av
P=ρ =
3)7800)(4.2(412.0 = 3.62 x 10-13 kgm-3
(e) As the satellite spirals toward the Earth’s surface, the KE increases (i.e. it spirals faster and
faster towards the Earth) and its GPE decreases. This can be shown in the graph of Energy vs orbital radius.
r/m
E/J
Ek
Ep
Note that the curvature of the 2 curves is different.
Section B Question 7: EM/EMI
(a) (i) lVEM
Δ= ⇒ potential difference, lEV M=Δ
(a) (ii) Maximum current, RVI M
Δ=
lVAρΔ
= llAEM
ρ=
5
2007 NJC H2 Solutions
ρ
AEM= (shown)
(b) Maximum magnetic force on a current carrying conductor,
90sinlIBF MMB = lAE
B MM ⎟⎟
⎠
⎞⎜⎜⎝
⎛=
ρ
ρ
AlEB MM= (shown)
By Fleming’s Left Hand rule, the direction of FB is upwards
(c) Upwards (d) (i) For vertical equilibrium: Mg = FB
( )ρ
BEAlgAld =
dgBE ρ= (shown)
(d) (ii) From dgBE ρ= ⇒ dgc
E ρ=2
81.98900107.1103 88 ×××××= −E = 667 NC-1
(d) (iii) P = I2R or P = V2/R
ρ
ρρ
AlEAlEA 22
=⎟⎠⎞
⎜⎝⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛= ( )
ρρlAE
lAEl
22 =⎟⎟
⎠
⎞⎜⎜⎝
⎛=
Since ⇒ dgcE ρ=2 ( )ρ
ρ AldgcP =
( ) cMggAlcd == (shown)
(d) (iv) θΔ= rodMcQ ⇒ MgcdtdMc
dtdQ
rod =⎟⎠⎞
⎜⎝⎛=
θ
390
10381.9 8××=
dtdθ = 7.55 x 106 Ks-1
(e) (i) Assuming that 100% of the power supplied by the transmitter goes into supporting the reflector, minimum power required
P = Mgc = 50 x 9.81 x 3 x 108 = 1.47 x 1011 W
(e) (ii) Any 1 reason:
• Not all the transmitted power will be intercepted by the reflector
6
2007 NJC H2 Solutions
• Some of the transmitted energy will be absorbed by the atmosphere • Need to take into account the power reflected back to receiving stations • Or any other suitable reason.
(f) - Shorter distance between transmitter and reflector
- Less power is needed to relay a signal compared to a geostationary satellite.
(g) Any one reason: • Too much power is needed to support the reflector hence not economically viable. • Current in the reflector (or power dissipated by reflector or rate of temperature rise of
reflector) is too large may result in reflector melting. • Difficult to keep the reflector in stable equilibrium because of atmospheric disturbances.
Section B Question 8: Quantum Physics Q8. (a) (i) From the graph, λmin = 1.2 x 10-11 m
Since λhcEK = ,
minmax, λ
hcEK =
=1.7 x 10-14 J (a) (ii) Using law of conservation of energy, the work done by the electric field is converted into kinetic energy of the electron Ks EeV =Hence, Vs = 104 kV (a) (iii) [draw graph showing that intensity at all wavelength drops. All peaks are still at the same position and can be seen.] (a) (iv) [minimum wavelength is 3 x 1.2 x 10-11 = 3.6 X 10-11m. The K peaks should be absent; overall intensity also drops] (b) The continuous spectrum is due to the energy lost due to rapid deceleration of energetic electrons as they collide with the atoms in the target. As the electron can some or all its kinetic energy, the energy released while braking does not have discrete values. For the line emission spectrum, the energetic incoming electrons collide and ejected an inner shell electrons. An electron from a higher energy level drops to the vacant state resulting in discrete package of energy being released. (c) (i)
λtnhc
tnEP ==
λtnEhc
=×
99100600
n /t = 3.6 x 1016 s-1 (c) (ii) The bombardment of energetic electrons at the target will result in production of X-ray as well as other forms of electromagnetic radiation such as infra-red radiation. This will increased the temperature of the target and may even cause it to melt and breakdown. Hence a cooling liquid is required to constantly remove heat from the target. 7
2007 NJC H2 Solutions
(d) Any 2 effects: • Water is ionised to form free radicals that are highly reactive. Free radicals can combine to
make hydrogen peroxide H2O2 which is a powerful oxidising agent and can damage the DNA of the chromosomes;
• At the molecular level, enzymes, RNA and DNA are damaged, and metabolic pathways are interfered with.
• At the sub-cellular level cell membranes are damaged, along with the nucleus, chromosomes, and mitochrondria.
• Cellular level, cell division is damaged. Cells can die, or be transformed to malignant growth. • Tissue and organ damage. There can be disruption to the central nervous system, death of
bone marrow and the lining to the gastro-intestinal system, leading to sickness and death. Cancers may arise.
• Whole animal can die; or life is shortened; (e) (i) XY = d sin θ Path difference between ray 1 and 2 = XY + YZ = 2d sinθ (e) (ii) For Bragg peak to occur, there must be constructive interference, i.e. path difference between rays = nλ Since the path difference between rays is 2dsinθ, λθ nd =sin2
8
2007 NJC H2 Solutions
1
–RESTRICTED–
Answers to PJC Preliminary Examination Paper 2 (H2 Physics) Suggested Solutions: No. Solutions Remarks 1(a) Base unit is a unit that can be accurately and easily
reproduced and is unchanging with time. It cannot be expressed as product or quotient of other base units.
[1]
1(b)
SI unit magnitude
resistance of a domestic filament lamp
Ω 103
Earth’s magnetic field strength
T 10−5
size of nucleus m 10−15
[1] [1] [1]
1(c)(i) SI base units of
( )mm
msmkg2
2
×
×=
−
E
SI base units of =E kgm−1 s−2
[1]
1(c)(ii)
Ae
FLE =
ed
FLE
2
2
=
π
3
23
2
1050.02
1040.0
100.400.20
−−
−
××
×
××=
π
E
111027.1 ×≈E kgm−1 s−2
e
e
d
d
L
L
F
F
E
E ∆∆∆∆∆+++= 2
50.0
01.0
40.0
01.02
0.40
1.0
0.20
1.0
1027.1 11+++=
×
E∆
1010987.0 ×≈E∆ kgm−1 s−2 Therefore, the value of the Young modulus is
( ) 1010113 ×± kgm−1 s−2.
[1] [1] [1]
2(a) The gravitational field strength at point X due to the Earth is acting in opposite direction to the gravitational field strength at point X due to the Moon. At a certain point X, these values have the same magnitude and hence they cancel out each other.
[1] for correct explanation with reference to highlighted key ideas
2007 PJC H2 Solutions
2
–RESTRICTED–
2(b)(i) At point X, EarthMoon gg = .
22 EX
GM
XM
GM EarthMoon =
( )282 104.3012.0 ××=XM 7107.3 ×≈XM m
[1] for correct use of formulae [1] for correct substitution [1] for correct answer for XM
2(b)(ii) Since XMEXEM += ,
( ) ( )78 107.3104.3 ×+×≈EM 8108.3 ×≈EM m
[1] for correct answer for EM
2(b)(iii) By Netwon’s second law of motion,
2
2ωrM
r
MGMMoon
MoonEarth =
32
2 4r
GMT
Earth
π=
( )38
2411
22 108.3
100.61067.6
4××
×××=
−
πT
6103.2 ×≈T s
[1] for correct equation [1] for correct substitution [1] for correct answer
3(a) - total energy, or maximum kinetic energy, or maximum potential energy, - maximum velocity, minimum velocity - amplitude, - period, or frequency, or angular frequency
[2] for any two quantities stated
3(b)(i) From the graph, maximum potential energy = maximum kinetic energy
= 3100.1 −× J
[1]
3(b)(ii) Since 0.1
2
1 2 =omv mJ,
32 100.10.12
1 −×=×× m
3100.2 −×=m kg
[1] [1]
3(b)(iii)
( )200
2
12
1
222
22
.
xxm
xm
o
=−ω
ω
( )222 1020.0 xx −=
1.4≈x mm
[1] [1] [1]
2007 PJC H2 Solutions
3
–RESTRICTED–
4(a) - The electric field between both plates is uniform.
- The electric field strength is 200100.40
802
=× −
Vm−1.
- The electric field is in the direction QP (from higher potential to lower potential).
[1] for any 1 statement
4(b)(i) Since qEF =
qEma =
=
d
V
m
qa
××
×=
−−
−
231
19
100.40
80
1011.9
1060.1a
13105126.3 ×=a ms−2
131051.3 ×≈a ms−2 towards the right
[1] for correct formulae [1] for substitution [1] for answer
4(b)(ii)1.
When the electron moves from point A to point B, the change in its potential energy is the product of its charge and the potential difference.
( )AB VVqU −=∆
( )[ ]40101060.1 19 −−−××−= −U∆ 18108.4 −×−=U∆ J
Note that the change is negative, thus indicating a loss of potential energy.
[1] for correct use of formulae [1] for answer No marks awarded for correct magnitude
4(b)(ii)2.
By conservation of energy, gain in kinetic energy = loss in potential energy
Vqmv ∆=2
2
1
18231 108.41011.92
1 −− ×=××× v
6102.3 ×≈v ms−1
[1] for substitution [1] for answer
5(a) The decay constant λ of a radioactive isotope is the
fraction of the total number of atoms that decay per unit time.
[2] for correct definition
5(b)(i) Using toeAA λ−= ,
3600487 106.3105.4 ××−××=× λe 4104.1 −×≈λ s–1
[1] for correct equation [1] for correct answer
5(b)(ii) Since NA λ= ,
N××=× −48 104.1106.3 12105.2 ×≈N nuclei
[1] for correct equation [1] for correct answer
2007 PJC H2 Solutions
4
–RESTRICTED–
5(b)(iii) Since
V
M=ρ ,
V
32712 101066.180105.253.2
×××××=
−
10103.1 −×≈V cm3
[1] for correct mass [1] for correct volume
6(a)(i) When an electron collides with one of the target atoms, it may lose an amount of energy that corresponds to the energy of an X-ray photon. The electron may continue to lose energy in a series of collisions with other atoms, thereby giving off X-ray photons of different energies. Since different X-ray photons correspond to different wavelengths, the continuous spectrum is thus formed.
[1] for clear explanation
6(a)(ii) An electron may lose all its energy in a single collision with a target atom, thereby emitting a photon of energy
.maxE that corresponds to a minimum wavelength oλ .
[1] for clear explanation
6(a)(iii) The βK -line arises when an electron from the M-shell
( 3=n ) fill the vacancy in the K-shell, and in the
process emits a photon whose wavelength
corresponds to the value of the βK -line.
Since the electrons in the L-shell are nearer to the K-shell, there is then a greater probability that the vacancy in the K-shell is filled by an electron from the L-shell than from the M-shell. Hence, the intensity of
the βK -line is smaller than that of the αK -line.
[1] for clear explanation [1] for clear explanation
6(a)(iv) The lines are called the characteristic X-ray spectrum because for different target element, the wavelengths of the K-series lines are different and thus are unique for different elements.
[1] for clear explanation
6(b)(i) For the energy of the photon for the αK -line,
12 EEE −=α
( ) ( )
−−−
−−=
2
2
2
2
1
1
2
1 ZkZkEα
( )4
132−
=Zk
hf
( )h
Zkf
4
132−
=
( )1−= ZCf
where h
kC
4
3= and is a constant.
[1] for recognizing that the energy difference
between 2E and 1E
gives rise to the energy
for the αK -line
[1] for correct manipulation of equation
2007 PJC H2 Solutions
5
–RESTRICTED–
6(b)(ii)
[1] for best-fit line drawn
6(b)(iii) Since ( )1−= ZCf ,
CCZf −=
Since the equation is of the form cmXY += , the
constant C can be obtained either from the gradient of the graph, or from the vertical intercept of the graph.
[2] for stating and explaining the correct methods using the equation
6(b)(iv) From the graph,
gradient 7100.5 ×≈ ( 7102.0 ×± )
Therefore, the constant C is 7100.5 × Hz 2
1
.
[1] for correct calculation for gradient or C [1] for correct units for C
6(b)(v) Since ( )1−= ZCf ,
( )129100.5 7 −×=f
9104.1 ×=f
Since f
c=λ ,
( )29
8
104.1
100.3
×
×=λ
10105.1 −×≈λ m
[1] for correct substitution [1] for correct calculated value
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
21 22 23 24 25 26 27 28 29 30 31 32 33
X
X
X
X
X
X
Z
f / 910 Hz 2
1
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6(c)(i) Using the Pythagoras’ theorem,
θsind=x Since the difference in the path between ray 1 and ray 2 is 2x, the path difference is θsin2d .
[1] for identifying path difference [1] for correct calculation of path difference
6(c)(ii) The first two peaks are the 1st order maxima and the next two peaks are the 2nd order maxima for the two wavelengths of the X-ray beam. Using λθ md =sin2 ,
19 18.0sin1094.02 λ×=°××× −
261 ≈λ pm
29 24.2sin1094.02 λ×=°××× −
392 ≈λ pm
[1] for correct calculation of the shorter wavelength [1] for correct calculation of the longer wavelength Subtract 1 mark for wrong conversion of m to pm.
6(c)(iii) Using the diffraction grating expression λθ nd =sin ,
and considering the 1st order maxima,
99 101.01sin103000 −− ××=×× θ °≈ 0019.0θ
The 1st order maxima is too close to the central bright fringe and so is unable to discriminate between different wavelengths in the X-ray region. A grating with spacing approximately equal to the wavelength will be desirable, but such gratings cannot be constructed mechanically.
[1] for correct calculation of angle of diffraction using appropriate order [1] for clear explanation with reference to the calculation made
θ θ
ray 1
ray 2
d θ
x
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Answers to PJC Preliminary Examination Paper 3 (H2 Physics)
Qn Solution Remarks
1(a)(i) Let m be the mass of the neutron and 12m the mass of 12C. speed of separation = speed of approach
vVu −=− 0
( ) Vvu =+ -------------------(1)
[1] for illustrating understanding
1(a)(ii) By conservation of linear momentum, mVmvmu 12+=
( ) Vvu 12=− ---------------(2)
Solving by (1)+(2): uV13
2= , uv
13
11−=
Ratio is 13
11.
[1] for stating equation in C.O.M. [1] for answer calculated
1(a)(iii) Fractional loss of energy of neutron is
( )2
21
222
1
mu
vum −
28.013
1111
22
=
−−=
−=u
v(shown)
[1] for correct method [1] for correct answer
1(b) Let us first consider the collision between A and B. By conservation of linear momentum,
( )ABBABBAA vmmumum +=+
( ) ( ) ABv21131 =+
2=ABv m s-1
Carts A and B will combine and move off to the right. Consider it as a single body colliding with cart C. By applying conservation of momentum,
( ) ( ) ABCCBAccBA vmmmumvmm ++=++
ABCv3)5.0(1)2(2 =−+
1.1=ABCv m s-1, towards the right.
[1] for calculating the common speed when AB sticks together [1] for correct calculation, having used the fact that momentum is conserved. [1] for correct answer
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2(a) Vector sum of net external force acting on the object is zero to ensure translational equilibrium. Net external moment about any point is zero to ensure rotational equilibrium.
[1] for mention of either condition
2(b)
[1] for correct forces [1] for correct lines of action of forces (ensure that the point of intersection of the forces lies to the right of weight of boom)
2(c) Taking moments about hinge, Sum of clockwise moments = sum of anticlockwise moments
w (2
Lx cosθ ) + W (L cos θ )= T (L sin (θ - φ ))
( 2000 + 2
400 LL
× )cos 45° = T L cos 75°
T = 3100.6 × N
[1] for application of moments [1] for substitution [1] for correct answer
2(d) Power =
30
)()( boomcrate mghmgh +
P 30
)2.0400()5.02000( ×+×=
P = 36 W
[1] for substitution [1] for answer
Weight of boom
Tension
Tension in cable
Reaction force due to ground on boom
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3 (a)(i) The differences between X-ray and sound waves:
X-ray is transverse while sound waves are longitudinal. X-ray can be polarized but sound cannot be polarized.
X-ray has a speed of 8103 × m s-1 in vacuum while the speed of sound in air is approximately 330 m s-1.
X-ray can travel through vacuum while sound cannot.
[1] for any one
3 (a)(ii) The similarities between X-ray and sound waves:
The equation v = fλ applies to both. Both waves can undergo diffraction and interference.
[1] for any one
3(b)(i) Taking the two end points on the line shown,
Gradient = )5.00.5(
10)0.20.20( 3
−
×− −
= 31000.4 −×
[1] for calculation
3(b)(ii) ,
a
Dx
λ= where x is the fringe separation,
D is the distance between slit and screen and a is the slit separation.
xf
Dca
∆
∆= , where
D
x
∆
∆is the gradient
004.0
1
100.5
100.314
8
××
×=a
4105.1 −×=a m
[1] for derivation of method so that gradient from (b)(i) can be used in calculation. [1] for conversion of frequency to wavelength [1] for answer
3(b)(iii)1. Since
fd
Dcx
∆∆ = , when f is halved, ∆ x is doubled.
This implies that the fringe separation in the interference pattern doubles.
[1]
3(b)(iii)2. fd
c
D
x=
∆
∆
Since the gradient is inversely proportional to the frequency of the light, when experiment is repeated with half the frequency, gradient of the graph doubles.
[1]
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4(a) When the bulbs (cold) are first connected, the temperature is low and the resistances are small. However, when electrical energy is absorbed by the filaments, the temperature and resistance increases, and the amount of energy emitted increases. Hence, the bulbs will take some time to achieve their full brightness.
[2] for applying I-V characteristics of a filament lamp
4(b) Let the currents through 1L and 2L be 1I and 2I respectively.
When the two lamps are operating at full brightness,
30
601 =I and
60
452 =I
021 .=I A 7502 .=I A
Since I
VR = ,
752
601
.R =
221 ≈R Ω
02
302
.R =
152 =R Ω
[1] for correct calculation of currents [2] for correct substitutions of
values into I
VR =
[1] for correct answers
4(c)(i) When the filament of 1L breaks, the current through 2L
increases. Hence, 2L lights up very brightly.
[1]
4(c)(ii) As the operating power increases, the temperature of the filament increases. However, the rate at which heat energy is transferred (radiation) away from the filament may be higher than before. Hence the net rate of heat energy gained by the filament is slower than what is expected. The melting point of the filament may be high, thus causing it to take some time before the filament breaks.
[1] for any reasonable answer
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5(a)(i) ...... smrsmr VIP =
...
...
smr
smrV
PI =
120
1200... =smrI
10... =smrI A
0I ...2 smrI=
140 =I A
[1] Calculation of r.m.s value of current [1] Relationship between r.m.s. and peak current
5(a)(ii) Since the current is alternating, one should expect either
)sin(0 tII ϖ= or )cos(0 tII ϖ=
However, the condition is that the power output is zero at the start where t = 0. Hence one should write
)sin(0 tII ϖ= or )sin(0 tII ϖ−=
)2sin(0 ftII π= =I − )2sin(0 ftπI
)100sin(14 tI π= =I − )100sin(14 tπ
Both positive and negative answers are acceptable but values must be correct.
[1] for recognizing sine formula [1] for correct substitution of values
5(a)(iii) Calculation shows that along y axis, 1.7 squares are taken up by peak value while along the x axis, 4 squares are occupied by each period.
[1] for correct form [1] for two cycles
5(b) By conservation of energy, Power supplied to primary coil = Power output from secondary coil + power loss.
lossSSPP
lossSp
PVIVI
PPP
+=
+=
( ) 600120102400 +×=PI
75.0=PI A
[1] for applying C.O.E
[1] for final answer
1.0
1.0
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6(a)(i) PV = nRT (200 x 103)(0.25 x 0.030) = n x 8.31 x (300 + 273) n = 0.315 mol
[1] [1]
6(a)(ii) Pressure due to piston and atmosphere
= 310100
0300.0
81.9100×+
×
= 132700 Pa The piston starts to move down when the pressure in the cylinder is equal to the pressure due to the piston and atmosphere. To find the corresponding temperature at which this occurs, apply the ideal gas equation. PV = nRT, 132700 x (0.25 x 0.030) = 0.315 x 8.31 x T T = 380 K (shown)
[1] [1]
6(a)(iii) The volume of the cylinder decreases with temperature under constant pressure.
3
3
2
2
T
V
T
V=
27330380
030.025.0 3
+=
× V
33 1098.5 −×=V m-3
[1] for substitution [1] for answer
6(a)(iv)
Since there is a drop in volume, positive work is done on gas. W = p |Vf – Vi|
( ) 33 1050.798.5107.132 −×−××=W
202=W J
[1] for substitution [1] for answer
6(a)(v)
By the First Law of Thermodynamics,
WQU ∆+∆=∆ where nRTU2
3=
202)30030)(31.8)(3145.0(2
3+∆=− Q
Q = 1.26 kJ
[1] [1] for substitution [1] for answer
6(b)(i) True. When a solid melts, the increase in potential energy is less than the increase that occurs when a liquid vaporises. Work is also done against external pressure when a liquid vaporises.
[1] for answer [1] for explanation
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6(b)(ii) False. Pressure of a real gas is less than that of an ideal gas for the same mass, volume and temperature. This is because attractive forces exist between the atoms/molecules in a real gas and not in an ideal gas.
[1] [1]
6(c)(i) Assuming no heat lost to surroundings, Heat supplied by water = heat gained by ice
θθ ∆+=∆ cmlmcm iceficewater
( ) θθ ××+××=−×× 42001001034.3100854200900 5
°= 5.68θ C
[1] for stating the assumption [1] for substitution [1] for answer
6(c)(ii) Consider the same mass of steam m and the same mass of
water m at 100 °C. When in contact, both steam and skin reaches a thermal equilibrium. The same happens for boiling water with skin. However, the difference is that for steam, the process of condensation occurs prior to the drop in temperature.
θmcmlQ vsteam ∆+=
θmcQhotwater ∆=
When steam condenses into water at 100 °C, there is a release of a large amount of thermal energy. Since specific latent heat of vaporisation is far greater than specific heat capacity, the thermal energy transferred to the skin due to steam condensing is greater than that due to boiling water.
[1] for stating the difference between the processes. [1]
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7(a)
Magnetic flux density of a magnetic field is defined as the force acting on a conductor per unit length per unit current flowing through it when the conductor is placed at right angle to the field. Its unit is tesla (T).
[1] mark for correct definition [1] mark for correct unit
7(b)(i) 1. Resistance of coil needed, R =
P
V 2
= 1000
50 2
= 2.5 Ω
No. of turns needed = 2100.5
5.2−×
= 50
[1] [1]
7(b)(i) 2. B =
r
NI
20
µ
= D
V
PN )(
0µ
= (4π×10-7×50×50
100.1 3×) / 2.0
= 6.3×10-4 T (2 s.f.)
[1] for idea of P = IV [1] for substitution [1] for answer
7(b)(ii) Total resistance of coil of N turns is given by N
A
DNR ∝=
πρ
Since R
E=I ,
NR
11∝∝I
As N increases, R increases by the same factor, and hence I
decreases by the same factor. So, B remains the same.
[1] [1] [1]
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7(b)(iii) 1. 2.
The graphs below show variation of flux and induced e.m.f. with time.
[2] for shape (either sine, -sine, cosine or –cosine) [2] for shape being the negative of gradient of flux-time graph. (Faraday’s Law obeyed)
7(c)(i)
As the train moves towards the tunnel, there is a rate of change of magnetic flux linkage experienced by the wire-coiled tunnel. By Faraday’s Law of Electromagnetic Induction, an e.m.f. is induced in the coil. By Lenz’s Law, the induced e.m.f. will produce effects to oppose the increase in magnetic flux linkage. Since the circuit is closed, the induced current will flow through the coil and produce an induced magnetic field to oppose the increase in flux. Hence a North pole will be induced on the left end of the coil. Since like poles repel, the train will experience a resistive force that causes it to slow down.
[1] Faraday’s Law [1] Lenz’s Law [1]
7(c)(ii) If the switch is opened, no induced current flows even though the same e.m.f. is induced since the diode is reversed biased. Hence, there is no repulsion and the train continues moving with constant speed. One possible application of such a system in real life is to control the braking of the train. By closing the switch, the train will be slowed down and by opening the switch, the train will be able continue its journey smoothly.
[2] [1] for any reasonable answer
t
t
φ
E
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8 (a)(i) 1.
Electrons emitted from the surface will be emitted with the maximum energies. Electrons emitted deeper from the surface will be emitted with energies less than the maximum due to collisions with atoms along the way to the surface, thus losing energy.
[1]
8 (a)(i)2. t
nhfP =
λt
nhcIS =⇒ where S is the surface area of incidence
834
94
100.31063.6
103001045.0180
×××
××××==
−
−−
hc
IS
t
n λ
1161022.1 −×= st
n
[1] for correct formulae [1] for substitution [1] for answer
8 (a)(ii) 1.
Φ = Ephoton - KEmax
= maxKEhc
−λ
= )1060.1)(9.0(10300
)100.31063.6( 19
9
834−
−
−
×−×
×××
= 5.19 x 10-19 J = 3.24 eV
[1] [1] for substitution [1] for answer
8 (a)(ii) 2. t
ne
t
QI ==
19
9
106.1
104−
−
×
×==
e
I
t
n
110105.2 −×= st
n
[1] for substitution [1] for answer
8(b)(i) Energy / eV -0.38 A -1.54 B -3.72 C -13.6 D
[1] for the first 3 arrows drawn correctly [1] for second set of 3 arrows drawn correctly
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8(b)(ii)
Shortest wavelength is obtained from the transition that involves the largest energy difference (Transition from A to D).
ED – EA = λhc
λ
83419 100.31063.6
1060.1)]6.13()38.0[(×××
=××−−−−
−
λ = 9.40 x 10-8 m
UV region (Correct region according to the answer found)
[1] for substitution [1] for answer [1]
8(c)
2
2 )(8
h
EEmk
pot −=
π
19
234
47
1067.6
)1063.6(
10956.1
−
−
−
×=
×
×=
mk
k
)10750)(1067.6(22 129 −××−− =≈ eeT kL T = 4.5 x 10-5
[1] for calculation of k [1] for substitution [1] for answer
8(d) px = mvx = (9.11 x 10-31)(2.05 x 106) px = 1.87 x 10-24 kg m s-1
27
24
1034.9
)1087.1(100
50.0
−
−
×=∆
×=∆
x
x
p
p
By the Heisenberg’s Uncertainty Principle,
2
h≈∆∆ xpx
xpx
∆≈∆
2
h
27
34
1034.9
1
2
1063.6
2
1−
−
××
××≈∆
πx
91065.5 −×=∆x m
7.5=∆x nm
[1] for calculating px [1] for calculating uncertainty in px [1] for applying H.U.P. and correct calculation.
Kg m s-1
2007 PJC H2 Solutions
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SUGGESTED SOLUTIONS TO H2 PHYSICS PRELIMINARY EXAMINATION 2007
2007 RJC Prelim PAPER 2 1 (a) S.I. unit for E = Pa = N m−2
∴ Base units for E = kg m s−2 . m−2 = kg m−1 s−2
Base units for 2
2 2
kg . m s . m
m . m
kmgL
d e
−
= = = kg m−1 s−2
Since the units for the l.h.s. and r.h.s. of equation (1) are the same, the equation is dimensionally consistent.
(b) ( )
11
2 23 3
1.273 2.00 9.81 1.8741.9736 10 Pa
0.58 10 0.705 10
kmgLE
d e − −
× × ×= = = ×
× × ×
(((( ))))
11 11 11
11
2
0.002 0.05 0.01 0.005 2
1.874 2.00 0.58 0.705
0.0676
0.0676 1.9736 10 0.13 10 0.1 10 Pa
2.0 0.1 10 Pa
E L m d e
E L m d e
E
E
= + + += + + += + + += + + +
= + + += + + += + + += + + +
====
= × × = × ×= × × = × ×= × × = × ×= × × = × ×
∴ = ± ×∴ = ± ×∴ = ± ×∴ = ± ×
∆ ∆ ∆ ∆ ∆
∆
2 (a) (i) Impulse, Ft = mv – mu
t = m (v – u) / F = 0.060 (50) / 70 = 0.0429 = 0.043 s
(ii) consider horizontal motion to the net: using sx = uxt
11.50.2323 0.232 s (to 3 s.f.)
50cos8.0x
x
st
u= = == = == = == = =
°°°°
consider vertical motion to the net: using sy = uyt + ½ at2
(((( )))) (((( )))) (((( )))) (((( ))))2150sin8.0 0.2323 9.81 0.2323 1.881m 1.88 m
2ys = ° + == ° + == ° + == ° + =
distance above the net = 3.2 – 1.88 – 1.0 = 0.32 m
(b) consider vertical motion from point of contact to point where the ball first touches the ground
side view of tennis racket θ = 8.0°
90°
u = 50 ms-1
2007 RJC H2 Solutions
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using sy = uyt + ½ at2 3.2 = (50sin8.0°)t + ½ (9.81)t2 ½ (9.81)t2 + (50sin8.0°)t – 3.2 = 0
(((( )))) (((( )))) (((( ))))2 9.8150sin8.0 50sin8.0 4 3.2
2
9.812
2
0.3656 1.784 (NA)
t
s or s
− ° ± ° − −− ° ± ° − −− ° ± ° − −− ° ± ° − −
====
= −= −= −= −
distance opponent would have run = (3.0)(0.3656 – 0.30) = 0.197 m
3 (a) (i) At 25 ºC, resistance of thermistor = 300 Ω,
Hence, total resistance = 300 + 200 = 500 Ω
I 12
24500
= = mA
(ii) pd across thermistor = 300
(12) 7.2300 200
= +
V
(b) As temperature increases, resistance of thermistor decreases. The total resistance decreases and current in the circuit increases causing the pd across the resistor to increase. Since the pd across both is still at 12 V, pd across thermistor decreases.
OR
As temperature increases, resistance of thermistor decreases
Voltage of thermistor, 12 thth
th
RV
R R= ×
+
Denominator decreases less slowly than numerator, therefore Vth decreases
(c)
(i)
2 2120.48 W
300
VP
R= = = at 25 ºC
2 212
2.62W55
VP
R= = = at 45 ºC
(ii) Energy
Powertime
= . Hence, 0.48 2.62
( )(10 60) 930 J2
E Pt+
= = × =
(iii) Since power does not vary linearly with resistance (which varies linearly with temperature and time), this method will not yield an accurate value of power dissipated when resistance change. Method assumes linearity between power and resistance.
4 (a) Period = 2 × 5 = 10 ms
⇒ Frequency 3
1 1100 Hz
10 10T −= = =
×
(b) The area under one cycle of the V2 vs t graph is
2007 RJC H2 Solutions
3
2 20.75 2.5 0.5 7.5 3.28× + × =
Hence, the mean-squared voltage is
23.280.328 V
10=
⇒ rms 0.328 0.573 VV = =
(c) Mean Power 2 2
4rms 0.5736.56 10 W
500
V
R
−= = = ×
5 (a) (i) The graph is a straight line that passes through the origin; this means that the
acceleration is proportional to the displacement. The negative gradient means a and x are always in opposite direction; this shows a is always directed
towards the equilibrium. Hence, the motion can be represented by a = − ω2x and is simple harmonic.
(ii)
2
2
2 22
2
2.2 2.2gradient 55
0.04 ( 0.04)
4 40.85 s (proven)
55
a x
TT
ω
ω
π πω
= −
− −= − = = −
− −
= ⇒ = =
(b) (i)
( ) ( ) ( )
2 2
2
1
2
10 25 55 0 04
2
0 011J
max oKE m x
. .
.
ω=
=
=
(ii)
6 (a) (i) The two energy levels are the conduction and valence energy bands of the GaAs
layer.
(ii)
g
g
hc hcE hf
Eλ
λ= = ⇒ =
(b) The forward-bias voltage ensures a constant source of free charge carriers – the
positive holes moving from the p-doped GaAs to the n-doped GaAs and electrons in
K.E. / J
0 T 2T t
0.011
2007 RJC H2 Solutions
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the opposite direction. In the neighborhood of the junction, there is thus an increased number of excited electrons in the conduction band which will release photons when they fill up the holes in the valence band.
(c) A significant increase in temperature implies a greater concentration of mobile charge carriers in the semiconductors which can absorb photons to transit between energy bands other than the photon-producing ones, reducing the laser efficiency.
(d) The minimal length corresponds to half the wavelength 71.15 10−× m.
7 (a)
The velocity v may be resolved into two components – one parallel to the magnetic
field B , which is //v , and the other perpendicular to B , which is v⊥ . The first
component results in a translational motion of the proton along B , while the latter results in a circular motion of the proton about B. The resultant motion is therefore a spiral about B.
(b)
(i) The directions of the currents passing through each coil are as shown above.
(ii) Using sinF Bqv θ= , we get
( ) ( )( )
4 19 5 0 18
18
3.8 10 1.6 10 1 10 sin55 4.98 10
5.0 10 2
F N
N to sf
and directed perpendicularly into the page
− − −
−
= × × × = ×
= ×
v
B
θ
proton
Fig. 7.2
mid-plane
central axis
coil v = 1.0 × 105 m s−1
P 55o
Fig. 7.3
v⊥
//v
2007 RJC H2 Solutions
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(iii) Using max//
min
1Bv
v B⊥
= −
and recognizing that
//
tanv
vθ ⊥= , we get
max
min
0 0
1 1tan
6.311
3.8
50.95 51
B
B
θ
θ
= =
−−
∴ = =
(iv) Yes, the proton will be trapped, because its velocity is at an angle of 550 to B ,
which is larger than the critical pitch angle.
(c) The mass change per reaction is = 2(3.016029) – 4.002603 – 2(1.007825) = 0.013805 u This is equivalent to an energy release of 0.013805 x 1.66x10-27 x (3x108 )2 J
But in 1 reaction, the mass of 3
2He fused is 2(3.016029) x 1.66x10-27 kg.
Therefore 1.00 kg of 3
2He when fused, should produce
[0.013805 x 1.66x10-27 x (3x108 )2 ] ÷ [2(3.016029)x1.66x10-27] J of energy That is 2.06x1014 J is released.
(d) (i) For excitation to be possible, the kinetic energy of the protons should be higher than the excitation energy of the oxygen atoms. Hence, calculating for KE of protons, we get
( ) ( )22 27 5
18
18
19
1 11.67 10 1.0 10
2 2
8.35 10 J
8.35 10eV
1.6 10
52.2
mv
eV
−
−
−
−
= × × ×
= ×
×=
×=
Clearly, excitation is possible.
(ii) Using
( )8 34
19
7
3 10 6.63 10,
2.23 1.6 10
5.57 10 m
ch chE we get
Eλ
λ
−
−
−
× ×= = =
× ×
= ×
colour: Green or greenish yellow
2007 RJC H2 Solutions
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PAPER 3
Section A 1 (a) Newton’s law of gravitation states that the force of attraction between two point
masses is directly proportional to the product of their masses and inversely proportional to the square of their distance apart.
(b) (i)
(((( ))))(((( )))) (((( ))))(((( ))))
2 23
11 24
26 3
1
350 10
6.67 10 6.0 10
6.4 10 350 10
8.78
E E
E
GM GMg
r R
Nkg
−−−−
−−−−
= == == == =+ ×+ ×+ ×+ ×
× ×× ×× ×× ×====
× + ×× + ×× + ×× + ×
====
It is along the line joining point P and the centre of the Earth and is pointing towards the Earth’s centre.
(ii)
(((( ))))
2
2
2
2 6 3
2
4
4 6.4 10 350 10
8.78
5510
GF m r
gm m rT
rT
g
s
ω
π
π
π
====
====
====
× + ×× + ×× + ×× + ×====
====
2 (a) The First Law of Thermodynamics states that the internal energy of a system depends
only on the thermodynamic state of the system; the increase in the internal energy of a system is equal to the sum of the heat supplied to the system and the work done on the system.
(b) (i)
4
3
Using
0.20 8.31 3009.97 10 Pa
5.0 10A
A
pV nRT
nRTp
V −
=
× ×= = = ×
×
(ii) ( )4 3Work done 9.97 10 5.0 1.0 10 399 Jp V −= ∆ = × − × =
399 J of work is done by the gas.
∴ Work done on the gas = − 399 J
(iii) ∆U = ∆Q + ∆W
0 = ∆Q + (800 − 399) ⇒ ∆Q = − 401 J
Heat supplied to the gas = −401 J
2007 RJC H2 Solutions
7
3 (a)
6
614 1
20 10
20 101.25 10 s
Q t
Q ne
t t
n
t e
Ι
−
−−
=
= × =
×= = ×
(b) (i) Gain the KE = Loss in PE
21 2
2
eVmv eV v
m= =
For sodium ion, 19
5 1
26
2 1.6 10 50002.05 10 m s
3.81 10sv
−−
−
× × ×= = ×
×
For potassium ion, 19
5 1
26
2 1.6 10 50001.57 10 m s
6.49 10pv
−−
−
× × ×= = ×
×
(ii) Since p.d. applied for a very short time and ion travels with constant speed throughout the 20 m,
time taken for sodium ion to travel 20 m is 5209.76 10 ss
s
tv
−= = ×
time taken for sodium ion to travel 20 m is 52012.7 10 sp
p
tv
−= = ×
time difference = s pt t− = 2.94 × 10−5 s
(c) This time difference can be measured by connecting the output of the detector to the
Y-plates of an oscilloscope. The time/cm control can be set to for example 10 µs/cm. When the two pulses are fired by the emitter, they will appear at a distance of 2.94 cm apart on the screen.
OR
Connect the output to a current detector which is linked to a data logger. Set the time based logging in orders of microseconds. Two pulses of high intensity current are expected if plotted on a time based scale. The time difference could be found.
4 (a) p.d. across the 400 Ω resistor = 400
4.0 1.6 V400 600
× =+
(b) Let Reff be the effective resistance of the voltmeter and the 400 Ω resistor.
4.0 1.4600
4 1.4 840
323.08 323
eff
eff
eff eff
eff
R
R
R R
R
× =+
= +
= ≈ Ω
Hence,
400323.08
400
1680
R
R
R
=+
= Ω
2007 RJC H2 Solutions
8
(c) The required p.d. has to be at least 0.99 × 1. 60 = 1.584 V. Hence,
4.0 1.584600
393.4
eff
eff
eff
R
R
R
× =+
= Ω
This gives
4
400393.37
400
23760 2.38 10
R
R
R
=+
= Ω = × Ω
5 (a) 1. Random – do not know when or which particular nucleus will decay.
2. Spontaneous – not affected by external factors.
(b) (i)
Alpha particles are highly ionizing and will not penetrate far in air.
(ii) True count = observed count – background count
t / hour True count-rate / min-1
0 6,409 6.0 801
0 0
1/2
1/2
' ' 801 1
6409 2
801 1lg lg 3.0
6409 2
6.0hrs 3.0 t
2.0hrs
nA C
A C
n n
t
= = =
= ⇒ =
⇒ =
∴ =
'50 1
801 2
50 1lg ' lg ' 4.0
801 2
4.0 2.0 6.0 14.0 hrs (Ans)
n
n n
t
=
= ⇒ =
⇒ = × + =
6 (a) When the electrostatic field between the nucleus and electron is weakened, the barrier
height is lowered towards the right in the above diagram. This implies a greater amplitude (in general) for the wave function on the right of the potential well. Thus a greater probability for the electron to tunnel away towards the right.
(b) (i)
Classical electrons are attracted and held tightly to the metallic tip of the STM by electrostatic forces. Quantum mechanically, they can tunnel through the potential barrier due to the metallic surface, across the vacuum and reach the sample surface. The probability of tunneling depends on the electron density and distance between the tip and sample surface, hence the variation of the tunneling current can be used to generate surface profiles
(ii) 2 31 1910
34 2
8 (9.11 10 )(4 1.60 10 )exp( 2 (1.5 10 )) 0.046
(6.63 10 )T
π − −−
−
× × ×= − × ≈
×
2007 RJC H2 Solutions
9
Section B 7 (a) (i) The rate of change of velocity with respect to time.
(ii) The rate of change of momentum of a body is proportional to the resultant
force acting on it and occurs in the direction of the force.
(b) (i)
(ii) Block A: T = mA a ––– (1)
Block B: WB – T = mB a ––– (2)
(1) + (2): WB = (mA + mB) a
0.50 × 9.81 = (0.30 + 0.50) a
a = 6.13 m s–2
(iii) From (1): T = 0.30 × 6.131 = 1.84 N
(iv) Using v2 = u2 + 2as, v2 = 0 + 2 (6.131) (0.50)
v = 2.48 m s–1
(v) Using v = u + at, 2.476 = 0 + 6.131 t t = 0.404 s
(vi) Resultant force = 0 N, acceleration is zero.
Block B: WB – T = 0,
Block A: T – Fspring = 0
0.50 (9.81) – 360 x = 0
x = 0.0136 m
(vii) By conservation of energy,
mBgh = 1
2k xmax
2
(0.50)(9.81)(0.50+xmax)=1
2(360) xmax
2
180 xmax2 – 4.905 xmax – 2.4525 = 0
xmax = 0.131 m
Alternatively,
mBgh +1
2(mA+mB)v2=
1
2k xmax
2
0.50(9.81)(xmax)+ 1
2(0.80)(2.48)2
= 1
2(360) xmax
2
spring
0.50 m
Fig. 7
frictionless pulley
A
B
WA
NA
T
T
WB
2007 RJC H2 Solutions
10
(viii) After the maximum compression of the spring is reached, Block A will move back to its initial position where it comes to a rest momentarily, before moving towards the spring again to start a new cycle.
8 (a)
(b) (i)
52 1
4
2.00 108.89 10 Ω m
2.25 10
LR
A
R
L
ρ
−− −
−
=
×= = ×
×
(ii) ( )2
3
8.8889 10 0.100 0.100
0.01778 Ω
1.0 10
0.01778
56 kA
R
V
RΙ
−= × × +
=
=
×=
=
(iii) 1. Into the page
2. To the right OR away from the power source.
(iv)
7 3
22
4 10 56 102
0.0402
2
1.12 T
Br
οµ Ιπ
π
π
−
= ×
× × × ×= ×
× ×
=
Current flowing into the paper
Magnetic field lines
2007 RJC H2 Solutions
11
(v)
31.12 56 10 0.040
2508.81254.4
2500 N
F B LΙ=
= × × ×
=
=
(vi) The answer can be either lower or higher. Marks are only awarded for correct
physics explanation.
Lower
- current would be smaller because the projectile has resistance.
- due to the sudden surge in current when the switch is on, according to
Faraday’s law of induction, there will be an induced emf and hence an induced current. This will reduce the net current in the circuit and hence the force will be lower.
OR
Higher
- B-field is minimum at the centre as B is proportional to 1/r.
(vii) To the right OR same direction as the previous case.
(viii) - Use a magnetized projectile aligning it such that its B-field is in the same direction as the B-field generated by current. OR place a magnet between the rails.
- Use stronger current/voltage.
- Decrease resistance of track (includes increasing the cross-sectional area of the track)
(ix) Power supply: Generating the power necessary to accelerate rail gun projectiles is a real challenge. Capacitors must store electric charge until a sufficiently large current can be accumulated. While capacitors can be small for some applications, the capacitors found in rail guns are many cubic meters in size.
Resistive heating: When an electric current passes through a conductor, it meets resistance in the conductive material -- in this case, the rails. The current excites the rail's molecules, causing them to heat. In rail guns, this effect results in intense heat. The high velocity of the armature and the heat caused by resistive heating damages the surface of the rails.
Repulsion: The current in each rail of a rail gun runs in opposite directions. This creates a repulsive force, proportional to the current, that attempts to push the rails apart. Because the currents in a rail gun are so large, the repulsion between the two rails is significant.
Wear and tear on rail guns is a serious problem. Many break after a few uses, and sometimes they can only be used once.
Taken from http://science.howstuffworks.com/rail-gun2.htm
2007 RJC H2 Solutions
12
9 (a) (i) Interference is the superposition of waves in the same region and time so as to form regions of maxima (bright) and minima (dark) due to waves meeting constructively and destructively respectively.
(ii) Red light from bulb is not coherent.
(iii) 1. Phase difference = 0 6 or π
Path difference = 3 633 1899 nm × =
2. D
xa
λ= since D and a are constant, x is directly proportional to λ,
1 2 2
1 2
2
2
633 450
fringe spacing 1.42 mm
x x x
x
λ λ= ⇒ =
= =
(iv) The contrast between bright and dark fringes will be less as the bright fringes
will become less bright and dark fringe will not be as dark. This is because the amplitude of one of the waves reduces when the intensity reduces, and hence complete cancellation does not take place.
(b) (i) The particulate nature of light.
(ii) Rate of electrons produced =
910
19
2 101.25 10
1.6 10
−
−
×= ×
×
Rate of incident photons = 10 10 11.25 10 3 3.75 10 s−× × = ×
(iii)
34 819
9
6.63 10 3 108.12 10 J
245 10
hcE
λ
−−
−
× × ×= = = ×
×
(iv) power = 19 10 88.12 10 3.75 10 3.05 10 W− −× × × = ×
(v) Intensity = 8
4 2
4
power 3.05 101.53 10 W m
area 2.0 10
−− −
−
×= = ×
×
(vi) 1.
( )19 19
19
8.12 10 4.3 1.6 100.775 V
1.6 10
s
s
hf eV
hfV
e
φ
φ− −
−
= +
× − ×−= = =
×
2. Increasing intensity increases the rate of photons incident on metal surface. However the energy of the photon remains the same as the frequency of UV light is constant and light is incident on the same zinc surface ( work function is also constant). The stopping potential, which is the potential to stop the most energetic electron (depends on kinetic energy), does not change.
2007 RJC H2 Solutions
1
SAJC 2007 Prelim/9745/02 Solutions
SAJC Prelims H2 Physics Paper 2 Solutions 1(a) The rate of change of displacement (with respect to time.) [1] Or: Change in displacement per unit time, but NOT rate of change of
displacement per unit time. (b)(i) sx = vx t
20 × 10-3 = 9.0 × 106 × t t = 2.22 × 10-9 s [1] (ii) Electric force = m a
e (V / d) = m a
a = 331
19
10201011.9
160106.1
md
eV−−
−
×××
××= [1]
= 1.41 ×××× 1015 m s-2 [1]
(iii) s = u t + ½ a t2 = 0 + ½ (1.41 × 1015) (2.22 × 10-9)2 [1]
= 3.47 ×××× 10-3 m [1] (iv)
2(a) Thrust – Weight = mass × acceleration [1]
28.6 × 106 - 2 × 106 × 9.81 = 2 × 106 a Acceleration, a = 4.49 m s-2 [1]
(b) Contact force on astronaut - weight of astronaut = mass × acceleration [1] N - 65 x 9.81 = 65 x 4.49 N = 929.5 N
Reading = 929.5 / 9.81 = 94.75 kg [1]
(c) T = v dt
dm
28.6 × 106 = v (120
1 × 2 × 106) [1]
v = 1716 m s-1 [1] (d) Relative velocity of approach = relative velocity of separation (v) – (-U) = (-U) – (-vf) [1] Vf = 2 U + v [1]
Time / s
Displacement / m
3.47 × 10-3
2.22 × 10-9
Labelled axis with units and values [1] Shape of graph [1]
2007 SAJC H2 Solutions
2
SAJC 2007 Prelim/9745/02 Solutions
3(a) Diffraction is the bending of waves when it passes around an obstacle (or through a small gap). [1]
Note: Waves do NOT “split”; and bending of a wave when it passes through
different media is refraction. (b)(i) Light waves are transverse waves. [1] (ii)
Using the above set up, measure θ1 and use the equation (1 / N) sin θ1 = 1 λ to obtain the value of N, the number of lines per metre for the diffraction grating. [1] Replace the laser source with sodium light (followed by a collimator or two aligned slits), repeat experiment and calculate the wavelength of yellow light
with the previously calculated value of N and the new value of θ. [1]
(iii) Applying the equation: d sin θ = n λ
If the second order is just formed, (1 / N) sin 90 = 2 × 633 × 10-9 [1]
N = 7.89 ×××× 105 lines / m [1] (Value should be rounded down) 4(a) The electromotive force is defined as the energy transferred by a source in
driving a unit charge round (NOT through) a complete circuit. [1] Or: The energy transferred per unit charge from other forms of energy into
electrical energy when charge is moved round a complete circuit. Note: EMF is “energy per unit charge”; NOT a “force”.
(b) R = A
lρ =
8
8
102.1
75.0104.6−
−
×
×× [1]
= 4.00 Ω [1]
(c)(i) VXP = 243
475
50
×+
× [1]
= 0.762 V [1]
(ii) VXP = V2Ω = 0.762 = ×+ 23
2EQ [1]
EQ = 1.90 V [1]
Well-labelled diagram [2]
Alt: yellow2
1 633
sin
sin
λ=
θ
θ
2007 SAJC H2 Solutions
3
SAJC 2007 Prelim/9745/02 Solutions
(iii) The resistance of a voltmeter could affect the total resistance of the circuit, and therefore affect the p.d. measured across the 2 Ω. [1]
Or: No finite resistance in potential divider; voltmeter draws current. Note: The resistance in a voltmeter should be as large as possible; The resistance in an ammeter should be as small as possible.
5(a) K0 – K1 = λc h
, where K1 = ½ K0 [1]
½ (e × 50 k) = λ
××× − 834 1031063.6 [1]
λ = 4.97 × 10-11 m [1]
(b)(i) k = 2
2
h
)EU(m8 −π
= 234
19312
)1063.6(
106.1)1.58.6(1011.98−
−−
×
××−××π [1]
= 6.67 × 109 m-1
T = e-2 k d = -129 10 750 10 6.67 -2e ×××× [1]
Probability = 4.54 x 10-5 (or 45 per million or 4.5 x 10-3 %) [1] (ii) The probability is much smaller. [1] This is because the proton is much more massive (or has a larger mass.) [1] Reject: “larger”, “bigger”, larger size, etc. 6(a) Cells could be killed, damaged, mutated or heated up. (Any two) [2] (b) Each beam of gamma ray would have a lower dose of radiation. [1] Thus, individually, they will not damage the cells. [1] Or: If a single dose of killing radiation is directed at the tumour, all the healthy
cells along its path would be killed. (c) Beta particles have lower penetrative power. [1] (d) Cancer cells divide very rapidly, so they are much more likely to be killed by
radiation treatment. [1] (e) 28 protons, 32 neutrons. [1]
(f) Rest mass = 59.933820 u - 0.00055 u- 931
31.0u -
931
17.1u -
931
33.1u [2]
= 59.93025 u [1]
2007 SAJC H2 Solutions
4
SAJC 2007 Prelim/9745/02 Solutions
(g) Binding energy per nucleon = [(27 × 1.007825 + 33× 1.008665 - 59.933820) × 931] / 60 [1] = 8.74 MeV [1]
Or = 619
28-27
10106.160
)10(310 1.66 59.933820) - 1.008665 33 1.007825 (27
×××
×××××+×−
= 8.77 Mev
(h)
(i) A = t
t
2ln
02/1eA
×−
12102.3 × = 3
27.5
2ln
0eA×−
[1]
Ao = 4.75 × 1012 Bq [1]
A = λ N
4.748 × 1012 = 60602436527.5
2ln
××××N
N = 1.138 × 1021 [1] Mass of nuclides = 1.138 × 1021 × 59.933820 × 1.66 × 10-27 = 1.13 × 10-4 kg [1] (j) Advantage: The activity of the source is fairly constant during treatment. [1] Or: The source can be used for a long period of time. Disadvantage: A long storage time is needed before disposing such nuclides
(so as to reduce the risk of contaminating the environment. [1]
Note: The (α, β or γ) radiation that enters the brain kill, mutate or damage cells, but they are NOT radioactive. They will NOT decay further in the brain.
Nucleon number
Binding energy per nucleon/ MeV
8.74
7.6
8.79
56 60 238
Shape, labelled axes with values [1] Position of Co-60 [1]
Co-60
2007 SAJC H2 Solutions
1
SAJC 2007 Prelim/9745/03 Solutions
SAJC Prelims H2 Physics Paper 3 Solutions 1(a) Base units are units by which all other units may be defined. [1]
(b) [weber] = [(F / I L) × A] [1]
= mA
s m kg -2
× m2 = kg m2 A-1 s-2 [1]
(c) Since weber is an SI unit, it can only be defined by SI units, (i.e. “unit area” must be
expressed as “one metre-squared.”) [1]
(d) B = φ / A = (50 × 10-6) / (4 × 10-4) = 0.125 T
A
A
B
B ∆+
φφ∆
=∆
(A
A
B
B ∆−
φ
φ∆=
∆ : WRONG!) [1]
4
01.0
50
5
125.0
B+=
∆ => ∆B = 0.01 (1 s.f.) [1]
B = (0.13 ±±±± 0.01) T (2 d.p) [1]
(e) The percentage (or fractional) uncertainty of φ is more significant than that of A. [1]
2(a) Net power = Area of the triangle × 30 [1]
= ½ (6 – 2) (3 - 1) × 103 × 30 = 120 kW [1] (b) Change in internal energy is 0 J. After one complete cycle, the gas returns to its original state. [1] As internal energy is dependent on its state, it returns to its original value. [1] (c) Net work is done by the gas. Work done is done by the gas when it expands. [1]
Work done (= ∫ p dV) equals to the area under the graph. [1] The area under the graph is larger when the gas is expanding [1] (d) It is suitable for use as an engine as work is done by the gas. [1] 3(a) The field is directed out of the paper. [1] (b) Applying the equation, B q v = m v2 / r [1]
3.48 × 10-3 (1.6 × 10-19) v = 1.67 × 10-27 v2 / 0.03 v = 10000 m s-1 [1] (c) The speed remains the same. Changing the magnetic field density changes the magnitude of the force. [1] Since the force is perpendicular to the velocity, [1] it does not affect the magnitude of the velocity. [1] (d) The time taken for the motion is very short [1] The vertical velocity (or displacement) caused by gravity is negligible. [1] Reject: Since the proton’s mass is small, the effects of gravity is negligible, or Gravitational acceleration / force is small compared to centripetal acceleration / force. Note: Regardless of (large) mass or gravitational force, a projectile shot horizontally at high
speed over a short distance can be considered to be moving horizontally because its gain in vertical velocity or displacement is negligible due to the short time of flight.
2007 SAJC H2 Solutions
2
SAJC 2007 Prelim/9745/03 Solutions
V / V
t / s
4(a) (b) According to Faraday’s law, the induced EMF is proportional to the rate of change of
magnetic flux linkage, which is determined by the speed of the magnet. [1] Since the speed of the magnet varies with time in a sinusoidal manner, the emf also
varies in a sinusoidal manner. [1] At the amplitude positions, the speed of the magnet is zero, causing the induced
EMF to be zero; at its equilibrium position, its speed is highest, thus induced EMF is largest. [1]
(Accept any clear explanation for the difference in phase.) (c) (d) When a resistor is connected, an induced current flows in the coil, generating an
opposing magnetic field to that of the magnet. [1] Or: When a resistor is connected, energy is dissipated in the resistor. Note: When terminals X and Y are connected to a CRO / voltmeter, there is an induced
EMF but no induced current, as the CRO / voltmeter has “infinite” resistance. Thus there is no opposing B-field and no damping.
5(a)(i) Volume occupied by 1 mole of carbon = 12 / 2.3 = 5.2 cm3
Separation between two atoms = 3 231002.6/2.5 × = 0.205 nm [2]
Or: If assumed spherical, separation = 24
3)1002.6/2.5(3 23 ×
π×× = 0.254 nm
(ii) Wavelength, λ = v m
h =
427
34
101064.6
1063.6
××
×−
−
= 9.98 pm [1]
(iii) It cannot be used. Because its wavelength is too small compared with the spacing
between atoms, it cannot be diffracted. [1]
(b)(i) h f = φ + e Vs => 6.63 × 10-34 × f = (4.3 + 4.5) × 1.6 × 10-19 [1]
f = 2.12 ×××× 1015 Hz [1] (b)(ii),(iii)
2 cosine curves with axes labels [2] (Accept ‘+’ cosine curve, with sympathies)
Correct curves not starting at the
origin along with axes labels [2]
Voltage / V
Photocurrent / mA Curve A [1] Curve B [1] B
A
4.5
x / m
0 t / s
2007 SAJC H2 Solutions
3
SAJC 2007 Prelim/9745/03 Solutions
6(a)(i) Change in PE = )R
1
hR
1(GMm
E
−+
− [1]
= - 6.67 × 10-11 × 5.98 × 1024 × 300 (76 1058.31037.6
1
×+×)
1037.6
16×
−
= - 2.84 × 109 +1.88 × 1010 [1]
= 1.60 ×××× 1010 J [1]
(ii) Work done to move it up, W = Final GPE = 1.6 × 1010 J [1]
KE in orbit = ½ ×300 × (3.07 × 103)2 = 0.14 × 1010 J [1]
Total work done = 1.6 × 1010 + 1.41 × 109 = 1.74 ×××× 1010 J [1] (iii) Answer in (a)(ii) would be smaller. (No credit for guessing.) [1] Satellite would have been launched with some initial kinetic energy. [1] Or: Satellite would have been launched with an initial speed (or momentum). (b)(i) Friction of the Earth’s atmosphere would heat up the projectile [1] which would then burn up before it can escape. [1] Note: Question asked “what would happen….” Not: “What needs to be done….”
Suggestions to supply more energy etc. are irrelevant to the question. (ii) The rocket’s speed should be lower where air is denser (at lower heights, and
increased where air is less dense.) [1] This is because air resistance (or drag) increases with speed. [1] (c) Similarity: Gravity and the electric force are both inversely proportional to the
distance squared. [1] Differences: (Reject: “Gravity is … while electric force is not.”) 1. Gravity is the force between masses while electric force is between charges. [1] 2. The gravitational force is always attractive while [1] the electric force can be either attractive or repulsive depending on the
charges of the objects. [1]
(d)(i) m g = 2
2
o r
e
4
1×
πξ [1]
= 215
2199
)100.2(
)1060.1(109
−
−
×
××× [1]
m = 5.9 kg [1] (ii) The electric force on the proton is greater than the gravitational force. [1]
2007 SAJC H2 Solutions
4
SAJC 2007 Prelim/9745/03 Solutions
7(a) [2] During spontaneous emission, electron returns randomly to lower energy states, with
emission of a photon of energy E2 - E1. [1] For stimulated emission, downward transition is stimulated by the presence of photon
of energy E2 - E1. (2 photons with energy E2 - E1 are available.) [1] (b) Stimulated emissions are coherent, (parallel or in phase). [1] (c) Meta-stable states are excited states which have relatively longer lifetime compared
to normal excited states, (e.g. millisecond compared to nanoseconds) [1] Population inversion: situation where more atoms are in the excited states that
ground states. [1] (d)(i) Energy = 20 x 1.6 x 10-19 = 3.2 x 10-18 J [1] (ii) 15-eV, 5-eV and 10-eV [1], [1], [1] (iii) Nothing happens. [1]
(iv) Applying the equation: ∆E = h c / λ
3 x 1.6 x 10-19 = 6.63 x 10-34x 3 x 108 / λ [1]
λ = 4.125 x 10-7 m [1]
Applying the equation: d sin θ = n λ
6000
01.0 × sin θ = 4.125 x 10-7 [1]
θ = 14.3o [1] (v) A coloured spectrum is observed [1] with some dark lines [1] (e) At low temperature, the valence band is completely full and the conduction band
empty. The semiconductor behaves like an insulator because there is no electron in the conduction band to conduct electricity. [1]
Since the forbidden gap is narrow, at higher temperatures, electrons in the
valence band gain energy to move up to the conduction band, creating holes in the valence band. Both electrons and holes conduct electricity. [1]
2007 SAJC H2 Solutions
5
SAJC 2007 Prelim/9745/03 Solutions
8(a) A longitudinal wave is one whereby the oscillations are parallel to the direction of propagation and a progressive wave is one in which energy travels from source to surroundings. [1]
(b)(i) Period = 1 / f = 1 / 2500 = 4 x 10-4 s [1] (ii),(iv),(v)
(iii) a = ω2xo = (2 π 2500)2 x 1 x10-7 [1] = 24.7 m s-2 [1]
KE = ½ m vmax2 = ½ x [(2 π 2500) 1 x10-7]2 = 1.23 ×××× 10-6 x J [1]
(vi) Resonance is the tendency of a system to oscillate at maximum amplitude when
the driving frequency matches the natural frequency of the system. [1] Due to resonance, the amplitude of the vibrating ear drum (NOT sound) reaches a
maximum when sounds of about 3000 Hz reach it as it has a natural frequency of 3000 Hz. Hence, the ear picks up quieter (softer) sounds around that range. [1]
Note: Amplitude of sound is NOT amplified. But detection by ear is more sensitive. (c)(i) Similarity: wavelength, frequency, period or constant phase difference [1] Difference: amplitude or phase. (But NOT phase difference) [1] Note: “Different phase difference” means “different difference” which is nonsensical. (ii) Coherent, because phase difference is constant. [1]
(iii) Since xA2 ∝ I, 32 ∝ I. Thus, resultant intensity ∝ (3-2)2 = 0.11 I [1]
Resultant displacement = xA - xB = -2.6 x 10-4 +1.7 × 10-4 = - 0.9 x 10-4 cm [1] (d)(i) As M moves from X to Y, it will pass through equally spaced maxima and minima
(loud and soft). [1] This is in line with the Principle of Superposition. When the two waves superimpose,
the resultant is the algebraic sum of the two displacements. [1] (d)(ii) Doubling the frequency halves the wavelength and according to the young’s double
slit formula, ∆x = λD/a, [1] the distance between two adjacent maxima (loud) will be halved. [1]
0.1
0.4
a
k
Sine curve passing through 0.2 and 0.4 [2] Point A [1] Point k [1]
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2007 SRJC Paper 2 Solutions
1. (a) The compression of spring = (6.00 − 3.00) ×10−2 = 3.00 ×10−2 m
The energy stored in the spring
= J 1000.9)1000.3)(0.20(2
1kx
2
1 3222 −− ×=×= [1]
Assume that all the energy stored in the spring is transferred to the toy car as kinetic energy.
(Assume no energy loss due to friction in rotating parts of the car, air resistance and sound.) [1]
By conservation of mechanical energy, Loss in KE = Gain in GPE
J 1000.9 3−× = mgh (where m is the mass of the toy car and h is the height attained by the car.)
h = )81.9)(200.0(
1000.9 3−×
h = m 10587.4 3−× [1]
Actual distance travelled along the slope
= m 1017.9sin30
104.587
sin30
h 3-3
−×=°
×=
° [1]
(b) In order to have the toy moving at constant speed, net force on object must be zero.
A force F of N 981.0)5.0)(81.9(200.0mgsin30 ==° must be acting on
the object, along the slope, to keep it moving at constant velocity. [1] Power required to keep the object moving = Fv = (0.981)(5.00) = 4.91 W [1]
mg
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2. (a) As the ball is not moving in the vertical direction, Resultant force in the vertical direction = 0 N [1] Taking force exerted by man as T, mgT =°30cos
1130.30cos
)81.9(0.10=
°=T N [1] for substitution, [1] for correct answer.
(b) Centripetal force = horizontal component of T. [1]
°= 0.30sin2
Tr
mv
°=°
30sin11330sin20.1
0.10 2v
Solving, v = 1.84 m s-1 [1] 3. (a) Assume that the orbit around of ISS about the Earth is a uniform circular
one. [1/2] Assume that the only force acting on the ISS is the Earth’s gravitational
pull on the ISS. [1/2] Gravitational Force = Resultant Force
)+(Rm)+(R
GMm 2
2h
hω=
2
2
2 T
)+(R4
)+(R
GM h
h
π=
GM
)+(R4T
32hπ
= [1]
(b) If the definition of weightlessness is the fact that the force on the astronaut
by the Earth is zero, the astronaut does not experience weightlessness as the force of the Earth’s pull on him in the space station is still present. [1]
T
mg
300
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If the weightlessness refers to the feeling of the astronaut when he does not feel a reaction force from the ground of the space station on him, then, the astronaut does not feel weightlessness. [1]
As the ISS itself is experiencing drag forces and orbit corrections with thrusters, the ISS is not free falling towards the Earth. The astronaut in the ISS experiences a force from the ground on him. Thus the astronaut does not experience weightlessness. [1]
(c) The golf ball initially loses KE due to atmospheric friction. [0.5] The centripetal force required decreases. Gravitational force pulls golf ball nearer to Earth. [0.5] Loses PE and gains KE. [0.5] Spirals down to Earth with increasing speed. [0.5] As atmosphere is thin, the decrease in radius of the orbit is gradual.
Or
Assuming the ball travels in circular motion, the reason is because the gravitational pull of the Earth on the golf ball acts as a centripetal force. This force is perpendicular to the direction of motion thus only causing a change in direction and no change in magnitude of velocity. [2]
(Lack of consideration of the resistive forces stated in the question means that the above alternative answer does not warrant the full 3 marks.)
4. (a) Diffraction occurs at the single slit which produces a coherent wave [1] which then passes through the double slits where further diffraction occurs. [1/2]
Beyond the double slits, superposition occurs. [1/2]
If the path difference to a point P on the screen equals to wavelength, then, constructive interference produces a maxima. If path difference
equals to (n+1/2)λ (can be expressed as p.d. = (n+1/2)λ. Or
The two slits will produce two overlapping cylindrical wave trains that have a constant phase to each other over time. [1]
The phase difference between both wave trains depends on the position and distance to the two slits. [1] Additional learning points:
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Along the centre line between the two slits, the phase difference is zero because both trains travel equal distance. Superposition produces a resultant twice the displacement of each trains. Intensity is four times that of each train alone. Therefore a bright band is seen at the centre. On both sides of the centre line are positions where the path difference is an integer number of wavelengths, so that the two waves meets in phase and also give bright bands at these positions. In between these places where crest meet trough, net displacement and intensity are zero. [1]
(b) For Young’s Double-Slit experiment,
D
xa
a
Dx =⇒= λ
λ [1]
Hence,
700)00.3(
)1000.1](6/)106.12[( 33
=××
=−−
λ nm [1]
(c) For diffraction grating,
λθ nd n =sin [1]
°=×
×==
−−− 1.57
)1000.4/(1
)10700(3sin
3sin
5
911
3d
λθ
°=×
×==
−−− 1.34
)1000.4/(1
)10700(2sin
2sin
5
911
2d
λθ [1]
°=°−°=−=∆ 0.231.341.5723 θθθ [1]
5 (a) It means that the electron, a particle, can exhibit wave-like properties. [1]
A wave equation, Schrodinger equation, can be used to explain its behaviour. The square of the amplitude of this wave function gives the probability of finding that electron at a particular point. [1]
(b) The electron may be considered to be in a well of depth equivalent to the
maximum potential. [1] Classically, this electron does not possess sufficient energy to overcome this potential and hence is unable to leave the well. [1] By considering the electron as a wave, given by a suitable wave equation, there exists a finite probability that the electron may be found outside this
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well. As the electron does not possess enough energy to clear the well, it must have ‘tunneled’ its way through the wall of the well, the barrier. [1]
(c) The STM has a stylus, whose end is as sharp as a single atom. This stylus moves over the surface of the material to be scanned, and a voltage is applied between the probe and the sample surface. [1] Depending on the voltage applied, electrons will tunnel through the potential barrier between the surface and probe, resulting in a weak electric current. [1] This weak electric current is detected and maintained at a constant value. As this current is exponentially dependent on the distance between probe and the surface, current is kept constant by adjusting the distance between the tip and the surface. [1] This adjustment is extremely fine and corresponds to the contours of the surface of the material. [1] In this manner, the images of surfaces on the atomic scale may be obtained.
sketch is not necessary – will be considered if it aids in explanation
6 (a) Electrons are excited by an external source. Some materials have energy level that electrons are able to stay there slightly longer than usual, this level is called the metastable level. [2]
This enables more electrons to be excited and accumulated in that level for that duration of time.
Now there will be more electrons in the metastable state than in the ground state. [1]
This is the meaning of population inversion. Or Population inversion means there are more atoms in the excited state than
in the ground state. [2] This is possible in some materials where there exists a metastable state
where electrons stay there slightly longer than usual. [1] (b) When population inversion occurs, all that is needed now is for an electron
to undergo spontaneous emission to produce a photon. [1] The photon that is being produced will then stimulate other electrons to de-
excite and produce some more photons. [1] Together with a reflecting mirror, these photons could in turn stimulate others to de-excite producing a very intense beam. [1]
(c) Laser is highly directional, the intensity is high, and it is coherent. Laser is
produced by stimulated emission. [any 2 of the 3] Note: Both are monochromatic and the production of light are fairly similar,
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7. (a) (i) The data shows that more energetic alpha particles are emitted
from small t1/2 nuclides. [1]
(ii) The more energetic alpha particles have a higher probability of overcoming the nuclear potential barrier, have a higher decay rate and hence shorter half-life. [1]
(iii) The alpha particle escapes via the process of quantum tunneling.
[1]
(b)
NUCLIDE Ke α / MeV lg(Ke α / MeV) t1/2 / s lg( t1/2 /s) 232 Th 4.0 0.602 4.5 × 1017 17.7 228 Th 5.4 0.732 6.0 × 107 7.78 224 Ra 5.7 0.755 3.1 × 105 5.49 220 Rn 6.3 0.799 52 1.72 216 Po 6.8 0.833 0.16 -0.796 212 Po 8.8 0.944 3.0 × 10-7 -6.52
Potential well
Probability density
[1]
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(c)
lg( keα/MeV)
lg( t1/2 /s)
(d) lg 3700 = 3.56 from the graph, the lg(Keα /MeV) = 0.78 [1]
⇒ Keα = 6.0 MeV [1]
(e) (i) λ = ln 2/ t½ = 0.693/(90 x 365 x 24 x 60 x 60) [1]
= 2.44 x 10−10 s−1 OR
λ = ln 2/ t½ = 0.693/90
= 7.7 x 10−3 yr−1
(ii) For minimum number of atoms to be present, we assume that one
decay yields one alpha particle. [1] Initial activity = Decay constant x initial number of atoms
0.600
0.900
0.800
0.700
0 5.0 -10.0 10.0 20.0 15.0 -5.0
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Ao = λNo
No = Ao /λ
= [10 x 10−3/ (5.1 x 106 x 1.60 x 10−19)] / 2.44 x 10−10 [1] = 5.02 x 1019 atoms. [1]
(iii) Alpha particle radiation is not penetrating, [1] hence less harmful to the human body. [1] OR Having long half-life, it will have a slow decay rate. [1]
It will then most probably be able to maintain a constant supply of power within the lifetime of the patient and need not be replaced. [1]
END OF PAPER
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2007 SRJC Paper 3 Solutions
SECTION A
1 (a) 2.5 λ = 0.80
λ = 0.32 m [1]
v = fλ = 780 x 0.32 = 250 ms-1 [1]
(b) As the wave must have an anti-node at the end,
L = 2.75 λ = 2.75 x 0.32 = 0.88 m [1]
(c) Since the length of the tube and the wavelength are still the same, the standing wave might not be formed. For two end open tube, the
ends must be anti-nodes [1], and the length of the tube L = 2
λn. [1]
But,
2.5λ = 0.80 m
3λ = 0.96 m [1]
which is not equal to the length of the tube, thus no standing wave will be formed. Therefore, the powder will be seen moving in random motion about the tube. [1]
2. (a) Electric field strength at a point is defined as the electrostatic force acting per unit positive charge placed at that point. [1]
(b) Consider a unit positive charge brought from point 1 to point 2 in an
uniform field as follows:
1 2 F
d
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Work done on the charge = F x d = E Q x d = E d (Q=1) [1] By conservation of energy, the work done on the charge = gain in electric pe of the charge = V2Q – V1Q = V2 – V1 (Q=1) = V [1] (where V is p.d. bet 1 & 2) Hence V = Ed Or E = V/d = potential gradient [1]
(c) (i) For equilibrium, net force = 0 mg = Eq [1] where q is negative q = mg/E
= mgd
V
= 61.20 10 9.81 0.040
4.0
−× × × [1]
= 1.18 x 10-7 C hence g = - 1.18 x 10-7 C [1]
(ii) Upthrust on oil drop is negligible. [1]
3 (a) One volt is the potential difference between two points in a circuit in which one joule of electrical energy is converted into other forms of energy when one coulomb of charge passes from one point to the other. [1]
(b) The circuit is shown below.
0.50 Ω 0.50 Ω
2.00 Ω
1.5 V 1.5 V
3.00 Ω
3.0 V
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(i) Current flowing in the circuit = (E1+E2)/(R+2r) = 3.0/3.00 = 1.0 A
[1]
(ii) Power dissipated in 2.00 Ω = I2R = (1.0)2(2.00) = 2.0 W [1] (iii) Total power provided by two batteries
= (E1+E2)I = (3.0)(1.0)= 3.0 W [1] Efficiency of set up =power dissipated in load/Power provided by two batteries x100 Efficiency = 2/3*100= 66 % [1]
(c) The potential difference measured across the terminals of a battery
is normally lower than the battery’s e.m.f because there is internal resistance in the battery. [1] Electrical energy is dissipated as heat in the internal resistance of the battery when it supplies current. [1] The potential difference will be equal to the e.m.f of the battery when no current is flowing through the battery/ internal resistance is negligible or when there is no internal resistance. (Note: It is actually not possible for a real battery to have no internal resistance.) [1]
4. (a) For a fixed intensity of incident radiation, a fixed number of photons
‘strike’ the target metal per unit time. In doing so, a fixed number of delocalized electrons are emitted per unit time from that metal. [1] By increasing the potential difference between the plates, the emitted electrons are attracted to the collecting plate. The maximum number of electrons that will reach the plate per unit time is the number of electrons emitted per unit time. This value will remain the same even if the potential difference is further increased. [1]
Hence the current, which measures the rate of electrons reaching
the collecting plate reaches a maximum too.
(b) (i)
Photocurrent
Fig. 4
Potential difference between emitter and collector
½ mark for depicting larger current
½ mark for depicting larger stopping potential
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(ii) Increase in intensity of incident radiation means that more
photons are incident on target. This in turn will cause an increase in number of delocalized electrons emitted from target. [1]
Larger frequency implies that each photon has a larger quantum of energy. [1]
Work function Φ is constant hence the kinetic energy of the most energetic delocalized electron also increases. To stop this more energetic electron, a higher stopping potential is required. [1]
(c) p
h
f
c
p
h=⇒=λ [1]
27
34
12
8
1089.2
1063.6
101308
1000.3
−
−
=
=
xp
p
x
x
x
Initial momentum, pi = 2.89x10-27 Ns Final momentum, pf = - 2.89x10-27 Ns Change in momentum, ∆p = pf - pi.=-5.78x10 -27 Ns. [1]
5 (a)
Si Si Si
Si As Si
Si Si Si
Outer most electrons
atom [1]
[1] to draw the ‘normal’ atom
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As: +5 Si: +4
(b) Due to high concentration of electrons, the electrons in the n-type semiconductor start to diffuse into the p-type semiconductor and the holes start to diffuse into the n-type semiconductor. [1] The electrons which have diffused into the p-type semiconductor will then combine with the hole there while the holes which have been diffused into the n-type semiconductor will combine with the electrons. [1] This results in the formation of a depletion zone. [1]
(c)
When the positive end of the source is connected to the n-type semiconductor, electrons are attracted to it and move towards it. Similarly, the holes in the p-type will also starts to move towards the source. [1] When this happened, it widens the depletion layer, thus the potential difference increase. [1]
p-type n-type
Depletion layer
+
+
+
_
_
_
p-type n-type
Depletion layer
+
+
+
_
_
_
_
_
_
+
+
+
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Migration of the electrons/holes will stop once the potential difference of the depletion layer is the same as that of the source [1]. [1 mark for diagram]
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SECTION B
Answer two out of three questions in this section.
1 (a) (i) Upthrust is a result of the pressure difference between top and bottom surfaces of the body, resulting in a net upward force being exerted on the body by the fluid medium in which the body is located. [1]
According to Archimede’s principle, the upthrust is equal in
magnitude to the weight of the fluid displaced by the body. [1]
(ii) According to Archimede’s Principle,
Upthrust NgV air 0439.081.928.1103500 6 =×××== −ρ [1]
(iii) Resultant force acting on one balloon = Upthrust – weight of balloon and helium
N0279.0
)81.9](1000.1)103500180.0[(0439.0 36
=
×+××−= −−
[1]
For lifting of the banner, Resultant force on n balloons ≥ weight of banner [1]
8.175
81.9105000279.0 3
≥
××≥ −
n
n
The minimum number of balloons is 176. [1]
1. Net force acting on body is zero. [0.5]
Net moment about any point is zero. [0.5] 2. Taking Pivot about Point B By Principle of moments, [1] Sum of anticlockwise moments = sum of clockwise moments
°=
+=+
5.46
)00.200.4(2.24)00.2)(81.9)(500.0()4)(sin7.46(
θ
θ
[1] 3 As the setup is symmetrical, the tension within string
5 should be the same as the tension in string 4. [1]
The angle of T5 should be the same as the angle of T4 as well. [1]
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(b) The total momentum of a closed system is constant, provided no
external resultant force acts on it. [1] (c) (i) Initial speed of particle 1
15
26
15
1045.21064.6
)102(22 −−
−
×=×
×== sm
m
KE [1]
Initial speed of particle 2
15
26
15
1074.21064.6
)105.2(22 −−
−
×=×
×== sm
m
KE [1]
(ii) Applying Conservation of Momentum, [1] Total initial momentum = Total final momentum
15
1
5
1
55
1045.2
)1074.2()1074.2()1045.2(
−×=
×−=×−×
smv
mmvmm
Total initial kinetic energy =
1122525 1075.6)1074.2(2
1)1045.2(
2
1 −×=×+× smmm
Total final kinetic energy
= 1122525 1075.6)1074.2(2
1)1045.2(
2
1 −×=×+× smmm
Total initial kinetic energy = total final kinetic energy Hence, collision is elastic. [1]
OR prove that relative speed of approach = relative speed of separation..
2.74x105 m s-1 2.45x105 m s-1
2.74x105 m s-1
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(iii) Total momentum is still conserved in this case. [1] The mass of the helium particle is much smaller than the mass
of the balloon. Hence, although there is conservation of momentum, the change in velocity of the balloon is almost imperceptible. [1]
During the impact, the force of the molecule on the wall is equal
to the force of the wall on the molecules hence net force is still zero.
(iv) For the case of the helium particle not bouncing of the wall of
the balloon, for the case of it not bouncing off, we can take the final velocity of the particle to be zero. This means that the change in momentum will not be as big as the case when the particle does bounce off. [1]
As the duration of impact for both cases can be assumed to be the same, the rate of change of momentum for the case whereby it does not bounce off the wall is smaller. This implies a smaller force which the particle acts on the balloon wall, which leads to a lower pressure as pressure is the force exerted on a unit area. [1]
2 (a) Specific heat capacity of a body is the thermal energy required to
raise a unit mass of that body by unit temperature. [1]
(b) Specific latent heat of fusion of a body is the thermal energy required to convert a unit mass of that body from its solid state to its liquid state without any change in its temperature. [1]
(c)
= [1] Let m be the mass of ice needed. (mc∆T)ice +mlf + mc∆T = mc∆T. m(2.11 x 103)(0-(-5))+ m(3.34 x 105)+ m(4.20 x 103)(20-0) [1] = 1.00(2.40 x 103)(25-20)
energy absorbed by ice to change from – 5.00 °C to 0 C
heat absorbed by ice to become water
heat absorbed by melted ice at 0 C to 20.0 °C
heat energy lost by cola from room temperature to 20.0 °C
+ +
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m = 2.80 x 10 -2 kg. [1]
(d) (i) The system expands considerably. Work is done by the system. [1]
(ii) While the temperature remains constant, the potential
energy of the particles increase as the system changes its state. [1]
Hence the internal energy increases. [1]
(e) The substance increases in volume considerably during boiling. In
doing so, it has to do work against the surroundings (atmosphere). [1]
The change in volume when the substance melts is tiny in comparison. [1] Hence a larger amount of energy per unit mass must be supplied to boil a substance than melting it. [1] Thus, its specific latent heat of vaporization is higher than its latent heat of fusion.
(f) When a fluid evaporates, the particles of the fluid lose kinetic
energy. [1]
As a result, the temperature drops and it is no longer in thermal equilibrium with the surrounding. [1]
Hence, heat is transferred from the surrounding to the fluid.. When this heat is taken from the skin, the temperature of that skin drops quickly. This gives rise to the cooling sensation. [1]
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(g)
Set up apparatus as shown in diagram with the solid of known mass m1 but unknown specific heat capacity c1 thermally insulated with a felt jacket to reduce heat lost to the surrounding. Initial temperature θ1 is recorded. Switch is closed and a steady current is passed in circuit. Voltmeter reading V1 and ammeter reading I1 are recorded. After a time t, measured using a stopwatch, the switch is opened. Final temperature θ2 is recorded. By considering the principle of conservation of energy, Electrical energy supplied = heat received by solid + heat lost to surroundings
I1V1t = m1c1(θ2 – θ1) + h (eqn1) To eliminate h, experiment is repeated with another similar block of known mass m2 and known specific heat capacity c2. The circuit is adjusted so that in the same time t the temperature of this known mass also increases from θ1 to θ2.
A
V
ammeter
voltmeter
switch D.C. source
rheostat
thermometer
Felt jacket heater
Block of known mass but unknown specific heat capacity
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If the ammeter reading in this case is I2 while the voltmeter reading is V2, then
I2V2t = m2c2(θ2 – θ1) + h (eqn 2)
Eliminating h by simultaneous equation and manipulating the resulting equation the specific heat capacity of the unknown solid may be found as
)(
)()(
121
122222111 θθ
θθ−
−+−=
m
cmtVIVIc
2 marks for fully, well labeled, sketch 1 mark for taking into consideration heat lost to surroundings 1 mark for correct mathematical expression for “c” - the specific heat capacity. 2 marks for aptly describing the experiment.
3 (a) θsinIL
FB = [1]
Direction of force: into plane of paper. [1]
(b) From B = θsinBIL
F where θ = 90°
Force on opposite sides of the coil perpendicular to the field, F = BIL x N [1]
For equilibrium: Taking moments about the axis of rotation Sum of clockwise moment = sum of anti-clockwise moment [1]
θ
I
L
Mg NBIL
NBIL
L
x
F
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2 (NBIL x L/2) = Mg x [0.5]
thererfore 2
MgxB
IL N= [0.5]
(i) from 2
MgxB
IL N= (with left hand side being a constant)
with N doubled, - it means the length of wire making the coil has doubled [0.5]
- resistance of the coil is also doubled (since A
lR ρ= ). [0.5]
- since emf is constant, current flowing is halved. [0.5] - no net effect on x. [0.5]
(ii) from 2
MgxB
IL N= (with left hand side being a constant)
with sides of L/2,
- the perimeter of the coil is halved, [0.5] - the length of wire making the coil is halved, [0.5] - the resistance is halved, [0.5] - therefore current is doubled (since emf is constant) [0.5] - L2 is now ¼ of its original value [0.5] - Net effect on x is halved, i.e. x / 2 [0.5]
(c) (i) The alternating current in the power cable causes a continuous
changing flux linkage through the windings. [0.5] This causes an e.m.f. to be induced across the windings. [1]
By calibrating the e.m.f. induced with various known alternating currents in other power cables, the r.m.s. current of the unknown cable can be determined. [0.5]
(ii) A changing flux through the iron core will induce em.f. and
consequently induced current in the core, which results in heating effect. [1]
To minimize heating effect, have the core made of sheets of soft iron with insulated lamination in between each sheet. [1]
(iii) tB 314sin0.25= [1]
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(iv) Peak magnetic flux linkage through the windings
= NBAcos0o [1] = 100 x 25.0 x (13 x 7 x 10-6) x 1 = 0.228 Wb [1]
(v) flux linkage, t314sin228.0=φ [1]
e.m.f. is rate of change of flux linkage = dt
dφ = 314(0.228)cos314t [1]
peak e.m.f. = 314 x 0.228
since this is sinusoidal a.c., r.m.s. e.m.f. = (314 x 0.228)/ 2 [0.5] = 50.6 V [0.5]
END OF PAPER
2007 SRJC H2 Solutions
Temasek Junior College (Suggested Solution for Prelim 2007)
1
Paper 2 (SSQ) 1 a Taking upwards as positive: v = u + at = 2.5 + (- 9.81)(2.0) = - 17.1 m s-1 b c Taking helicopter as the reference, the distance between the helicopter and the parcel
after 2.0 s is the area bounded by the helicopter and the parcel.
Distance = ½ × 2.0 × [2.5 – (– 17.1)] = 19.6 m
OR
Displacement 2
2
1atuts +=
For helicopter, m 0.50.25.2 =×=Hs
For parcel, ( )( ) m 62.140.281.92
10.25.2
2 −=−+×=Ps
Distance apart = 5.0 – (– 14.62) = 19.6 m 2 a Having a gravitational potential energy of – 2.5 × 107 J means that the work done by an
external force in bringing the body from infinity to that point is – 2.5 × 107 J. The negative sign denotes that the external force and the displacement from infinity to
that point is in the opposite direction OR the gravitational potential energy is decreasing from infinity to distance less than infinity (taking GPE at infinity to be zero).
b i. 1. s 617014
606024=
××=T
2. 2ωrg =
2
2ωr
r
GM=
2
2
2411
6171
21097.51067.6
=××× − π
rr
r = 7.27 × 106 m
0 1.0 2.0 t /s
- 17.1
2.5 helicopter
parcel
v/m s-1
The helicopter is ascending at constant velocity of 2.5 m s
-1.
But the parcel is accelerating downwards at 9.81 m s
-2.
Distance apart
The parcel was travelling with the same velocity of the helicopter when it was released.
2007 TJC H2 Solutions
Temasek Junior College (Suggested Solution for Prelim 2007)
2
ii. Having a vertical launch would mean that the rocket will take a shorter time to travel
out into outer space, less time in the atmosphere. Thus there is less work done against air resistance, i.e. less energy loss due to air resistance.
3 a Amplitude = (0.80)4
1= 0.20 m
Frequency Hz 1.330.75
11===
Tf
b Since the object is pulled down to its amplitude position (0.20 m), its displacement from
equilibrium is described by tT
xtxx 2
cos cos 00
==π
ω
×= 12.075.0
2cos20.0
πx
= 0.1071 m Distance moved = 0.20 – 0.1071 = 0.0928 m c 4 a Electric field strength at a point is the electric force acting per unit positive charge
(placed at that point) b i. The magnitude of electric field due to +Q at B is E. The magnitude of electric field due to – 2Q at B is 2E (since the charge is twice).
Net electric field due to both charge, ( ) EEEE 5222 =+=
Magnitude of electric force at B, F = Eq = EQ5=
ii.
frequency of the driver (vibrator) /Hz 1.33
Amplitude of the driven system (brick) /m
0 3.0
+Q
+Q
-2 Q
F
The resultant force, F is the vector addition of the two forces due to the attraction of – 2Q and the repulsion of + Q. The direction of F is in general not related to the geometrical setup of the 3 charges.
2007 TJC H2 Solutions
Temasek Junior College (Suggested Solution for Prelim 2007)
3
iii. 5 a The rating of 80 W is the average power given out when a d.c. current is passed through
it, thus it is equivalent to the average power when an a.c. current passes through it.
rmsrmsave VIP =
2
11080 rmsI=
Irms = 1.03 A
b 80220 ×== avePP
= 160 W c
6 a HeUPu 42
23492
23894 +→
b i. The kinetic energy is produced from the decay of the 238 Pu into 234U and 4He.
2
mcKE ∆=
( )28196109979.2106022.110649.5 ×∆=××× −
m
∆m = 1.007057 × 10-29 kg = 0.0060644 u ∆m = mPu – mU - mα
0.0060644 = 238.0496 – mU – 4.0026 mU = 234.0409 u ii. The decay of 238Pu into 234U and 4He is an inelastic ‘collision’, similar to a bullet flying
out from a rifle.
B
C A
+Q
+Q
-2 Q
The number of electric field lines going into the – 2Q is higher than that going into + Q.
160
80
0.020 0.040
time / s
Power / W π
πω 100
2==
T
T = 0.020 s Note: the graph is a sin
2 curve.
2007 TJC H2 Solutions
Temasek Junior College (Suggested Solution for Prelim 2007)
4
Using Principle of conservation of momentum, mPuuPu = muvu + mαvα Since Pu is initially at rest, uPu = 0, 0 = muvu + mαvα
α
α
m
m
v
v U
U
−=
Ratio of kinetic energy: α
ααα
m
m
vm
vm
T
T U
UUU
==2
2
2
12
1
(since muvu = - mαvα)
5.580026.4
0409.234==
UT
Tα
iii. Total KE = Tα + TU = 5.649
649.547.58
11 =
+αT
Tα = 5.55 MeV
7 a speed = 3600
100054 × = 15 m s-1
b i. v2 = u2 + 2as 0 = 152 + 2a(1.25) a = - 90 m s-2 ii. v = u + at 0 = 15 – 90t t = 0.167 s
c i. F = kx = 30 × 3.6 × 10-2
= 1.08 N
ii. F = ma
120
081
.
.a = = 9.0 m s-2
d Assuming there is no leakage, the amount of gas in the airbag remains constant. At constant temperature, p1V1 = p2V2
250 × 103 × 0.060 = p2 × 3.0 × 10-4
p2 = 5.0 × 107 Pa e The drop in the pressure is due to the reduced in the no of moles of gas in the cylinder. (∆p)V = (∆n)RT
0.20 × 5.0 × 107 × 3.0 × 10-4 = ∆n × 8.31 × (17 + 273) ∆n = 1.245
2007 TJC H2 Solutions
Temasek Junior College (Suggested Solution for Prelim 2007)
5
Change in number of molecules, ∆N = ∆n × NA = 1.245 × 6.02 × 1023 = 7.494 × 1023
Average number of gas molecules leaving per sec = 36002474
10494.7 23
×××
×= 3.1 × 1017 s-1
f i. 8
26
10752
10221051−
−−
×
×××==
.
..
AR
lρ = 1.2 Ω
ii. E = Pt = tR
VPtE )(
2
==
21
12960
2
.
t. =
t = 8.0 × 10-3 s Paper 3 Section A 1 a Scale reading = 8.900 mm b Every reading will have the same zero error (of the same magnitude and sign). Thus the
omission introduces a systematic error. c Micrometer screw gauge has a small uncertainty, i.e. the readings will differ slightly from
each other. Thus the readings are precise. The mean value is not close to the true value due to the presence of zero error. Thus the
readings are not accurate. 2 a A field of force is a region where a body placed in the field will experience a force.
b Similarity: Both are inverse square law fields. / Both are action-at-a-distance type of fields./ Both are conservative fields.
Difference: A mass experiences an attractive gravitational force in a gravitational field where as a charge experiences an attractive/repulsive electric force in an electric field.
c A uniform electric field that is directed downwards.
(This electric field provides an upward electric force on the negative charged oil that is acting in the opposite direction to its weight)
d Release the oil drop and apply a uniform electric field downwards until the oil drop falls at a constant speed. Then apply a uniform magnetic field (to provide a constant magnetic force) horizontally (so that the magnetic force is always perpendicular to the velocity). The magnetic force will provide the centripetal force for the drop to rotate in a vertical plane.
3 a Since it is in a horizontal circular motion, sy = 0 Fy = 0
81.9500.030cos ×==° mgT
2007 TJC H2 Solutions
Temasek Junior College (Suggested Solution for Prelim 2007)
6
T = 5.66 N
r
mvT
2
30sin =°
°
=°30sin500.0
500.030sin66.5
2v
v = 1.19 m s-1
b ( )22230sin30sin fmlmrT πω °==°
For greatest number of revolution, T has to be the greatest as well.
( )2
max230sin500.0500.030sin0.15 fπ°×=°
fmax = 1.23 Hz 4 a i. Two bodies having the same temperature would mean that they are in thermal
equilibrium, i.e. no net heat flow between the two bodies when they are in thermal contact.
ii. Body H having a higher temperature than body C would mean that there is a net heat
flow from body H to body C when they are in thermal contact. b Two gases having the same internal energies would mean that the sum of random
distribution of potential and kinetic energies of the molecules are the same. c i. Using equation of state, pV = nRT
2
2
1
1
T
V
T
V= for constant pressure expansion
2
00550.0
300
00322.0
T=
T2 = 512 K = 239 °C ii. The statement is incorrect since degree Celsius is not the SI unit for the
thermodynamic temperature. Thus 40 °C (313 K) is not twice as hot as 20 °C (293 K). 5 a i. The e.m.f. of the cell is 2.0 V (when there is no current flowing in the circuit, i.e.
terminal potential difference is equivalent to the e.m.f.)
ii. Ω=−−
= 67.100.084.0
00.260.0Gradient
E = I(R+r) E = V + Ir V = E – Ir with y-axis as V and x-axis as I. The gradient will be – r. The significance of this gradient is the internal resistance of the cell.
b When R is very small, V ≈ 0,
hence E = V + Ir ≈ Ir
2007 TJC H2 Solutions
Temasek Junior College (Suggested Solution for Prelim 2007)
7
2.0 = I (1.67) I = 1.2 A Or from the graph when V = 0, I = 1.2 A.
c rR
R
r)(RI
RI
P
P2
2
cell
R
+=
+=
6 a i.
ii. Magnetic field at point P due to current I1=d2π
µo 1I
Magnetic field at point P due to current I2=d)-(2π
µo
l
I 2
Resultant magnetic field at point P: B =
−
d)-(d2π
µ 21o
l
II = 0
=> lII
I
2 1
1
+=d
b i. As the conductor moves through the magnetic field, there is a change in magnetic
flux linkage or cutting of flux. According to Faraday’s law, an induced e.m.f. is set-up in the conductor. Since the circuit is not closed, there is no induced current.
ii. Section B 7 a i. Momentum is the product of the object’s mass and its velocity.
Force is equal to the rate of change of momentum and the change in momentum takes place in the direction of the force.
I1 I2
(out of paper) (out of paper) P
B
The magnetic field due to I1 is stronger than that of I2. Thus the resultant magnetic field is in the direction of magnetic field due to I1.
v
× × × ×
×
×
× ×
×
×
× ×
× ×
×
×
× ×
×
× B
+
2007 TJC H2 Solutions
Temasek Junior College (Suggested Solution for Prelim 2007)
8
ii. 1. According to Newton’s 3rd law, both the car and the truck experience the same magnitude of force during the collision. However, having a smaller mass, the car would experience a greater acceleration, thus the driver is more likely to suffer more injury.
2. The seatbelt helps to prolong the time of contact. For a particular change of
momentum during car collision, the impact force will be decreased, reducing the injury on the driver.
3. The principle of conservation of energy is not violated as some kinetic energy lost
is converted into work done in deforming the car, heat and sound during collision.
4. The helium nucleus do not physically hit the gold nucleus due to the electrostatic repulsion whereas the car and the truck physically hit each other. OR Collision is elastic between helium and gold nucleus while collision is inelastic between the car and the truck.
b i. By conservation of momentum,
(3m + 2m)Vo = 3m (- 3
1Vo) + 2mvA
5mVo = - mVo + 2mvA
vA = 3 Vo ii. Having the second stage of the launching outside the earth’s atmosphere helps to
reduce the effects of air resistance on its velocity. The discarding of component B helps to reduce the total mass of the rocket, increasing the acceleration with the same thrust.
c i. As the object is falling through liquid, it experiences a drag force that retards its
motion downward. Thus the resultant force acting on the object = weight – upthrust – viscous force. With the speed of the object increases due to a resultant force, the viscous force increases too (viscous force is proportional to the square of the velocity). Thus the resultant force would eventually become zero, resulting in zero acceleration or constant velocity.
ii. When the object is falling through a column of liquid, it loses gravitational potential
energy. However this energy loss is not transformed into gain in kinetic energy, but work done against the viscous force.
iii. v ∝ a2
1
2
v
v =
21
22
a
a
0.5
0.25 =
2
22
2.0
a
a2 = 1.4 mm 8 a i. coherent refers to constant phase difference between two waves
2007 TJC H2 Solutions
Temasek Junior College (Suggested Solution for Prelim 2007)
9
monochromatic refers to a single wavelength plane-polarised refers to one plane of vibration diffraction refers to the spreading of waves around an obstacle or aperture ii. Allow the beam to pass through a polariser. Then rotate the polariser in a plane
perpendicular to the beam, the intensity of the beam should vary from maximum to a zero if it is plane polarised.
iii. arc length = radius × angle subtended = 6.0 × 1.2 × 10-3 = 7.2 × 10-3 m
b i. 1. Point R is the second bright fringe, thus the path difference is 2λ.
path difference = 2λ = 2 × 633 × 10-9 = 1.27 × 10-6 m 2. Point S is the second dark fringe, thus the path difference is 1.5 λ.
path difference = 1.5λ = 1.5 × 633 × 10-9 = 9.50 × 10-7 m ii. Assuming that it is a coherent source, the central fringe appear white since all the
wavelength will interfere constructively (i.e. path difference is zero). As for higher orders, coloured fringes are seen. The fringe separation for each colour depends on its wavelength. Larger wavelength will diffract more, leaving red fringes further apart as compared to the violet fringes. As such, there will be an overlapping of fringes.
c i. wavelength = f
v =
3
8
10 × 341
10 × 00.3 = 880 m
ii. 1. Since M is equidistant from X and Y, the signals arrive there in phase. So the signals interfere constructively.
2. Number of wavelengths between M and N = 880
1100 = 1.25
Signal from X will travel 1.25λ less as signal from Y will travel 1.25 λ more to reach point N. Thus the path difference is 2.5λ. So the signals arrive at N in antiphase and hence they interfere destructively.
iii.
N M
0 220 440 660 880 1100 distance/m
amplitude
2007 TJC H2 Solutions
Temasek Junior College (Suggested Solution for Prelim 2007)
10
9 a i. Ionization energy = 13.6 eV (from n = 1 to n = ∞)
ii. 9
834
10660
103.00106.63
λ
hcE
−
−
×
×××==
J103.0 19−×=
eV1.9=
iii & iv.
b i. m
2Evmv
2
1E 2 =⇒=
m103.3
101.6013.6109.112
106.63
2mE
h
mv
hλ
10
1931
34
−
−−
−
×=
×××××
×=
==
ii. 4π
h∆E∆t ≥
Electron in the ground state is stable and stays there for a long time. Hence the uncertainty in the time interval ∆t is large and from the expression above, ∆E is small. However, in the excited state, electron is unstable and stays there for a short time.
Hence ∆t will be small and ∆E will be large. Hence the statement is consistent with the uncertainty principle.
c i. Conduction band is an empty or partially filled energy band (higher than the valence band) in which electrons can move freely, producing net transport of charge while valence band is the highest electronic energy band in a semiconductor or insulator which can be filled with electrons.
ii. The conduction band and the valence band of a metal overlap so that the conduction
band is partially filled. When a small voltage is applied, electrons can accelerate and gain energy easily due to the higher unoccupied energy levels within the conduction band for the electrons to jump to. However for intrinsic semiconductors, there is a small energy gap between the conduction band and valence band. At higher temperatures, there is chance for the
2007 TJC H2 Solutions
Temasek Junior College (Suggested Solution for Prelim 2007)
11
electrons to gain sufficient energy to jump from the valence band into the conduction band. The migrated electron leaves behind a ‘hole’ in the valence band. Both the electrons in the conduction band and the ‘holes’ in the valence band act as charge carriers for the semiconductor.
2007 TJC H2 Solutions
1
2007 TPJC Prelim P2 Solutions 1 (a) Radius = (10.0 ± 0.1) mm
Volume of coin, V = π r2 h = π (10)2 (1.5) = 471.2 mm3 --- (M1)
∆V / V = 2∆r / r + ∆h / h --- (M1) ∆V = (2∆r / r + ∆h / h) V
= [2(0.1/10) + (0.1/1.5)](471.2) = (0.02 + 0.067) (471.2) = 40 mm3 (1 s.f)
Therefore V = (470 ± 40) mm3 --- (A1) (b) (i) Linear graph, maximum s = 25 m, max time = 2 s. Gradient of graph is
the horizontal velocity. (ii) Horizontal graph, v = 12.5 m s-1 until time = 2 s (iii) See iv)
(iv) Maximum height can be deduced from graph of vertical velocity with time = 5 m
(v)
Maximum height lower, horizontal range shorter, non parabolic path
2 (a) distance = (4.225 - 0.638) x 107 m width of area = 3.587 x 107 x sin 0.5 1 = 3.13 x 105 m . area = (3.13 x 10 5)2 = 9.80 x 1010 m2 1
(b) Intensity = 8000 W / 9.80 x 10 10 m2 = 8.16 x 10-8 W m-2 1
(c) distance travelled = 2 x 3.587 x 107 = 7.17(4) x 107 m 1 time = 7.17(4) x 107 / 3.00 x 108 = 0.239 s 1
3 (a)(i) The same amount of energy is distributed over a larger circumference [1] so the
energy per unit length decreases.
v / m s-1
t / s
2 1
10
-10
Maximum height = 5m
2007 TPJC H2 Solutions
2
(ii) Q = Rate of energy arrival at r/circumference [1]
= P / 2πr [1] (b)(i) This is because of inteference between the ripples from the dipper and the ripples
that are reflected off the slab [1]. The reflected ripples seem to originate from a point on the right hand side of the slab and equidistant from the surface of reflection.
(ii) Decrease the frequency of the dipper (so λ bigger and so x = λD/a is bigger) [1] Move the dipper closer to the slab (decrease a) [1] Move the dipper further away from AB (increase D) [1] [any two methods]
4 4(a) State the first law of thermodynamics, indicating clearly all the terms being used.
The first law of thermodynamics states that the internal energy of a closed
system depends only on its state. The increase in the internal energy of a
system is the sum of the heat supplied TO the system and work done ON the
system.
b (i) find the work done by the gas in pushing back the piston in stage 1
At const pressure, vol is directly proportional to T.
2 2
1 1
47.5 10
2
410
work done by gas = P
2.5
25
V T
V T
V X
XV
V J
=
−=
−=
=
(ii) find the total heat input in stage 1
2 methods:
dipper
a rectangular slab
screen
ripple tank
water surface a
D
2007 TPJC H2 Solutions
3
method 1
consider the whole process (stage 1 and stage 2)
1 2 1 2
1
1
0 ( 63) 25 0
88
U q q w w
q
q J
∆ = + + +
= + − − +
=
(where q1 and w1 are heat supplied to gas and work done on gas during stage 1
respectively)
method 2
consider only stage 2:
2 2U q w∆ = + = -63+ 0 = - 63 J
so for stage 1: 1
1
63 ( 25)
88 J
U J q
q
∆ = = + −
=
(iii) find the ratio of the specific heat capacity when the gas is at constant pressure to
the specific heat capacity when it is at constant volume.
1 1
2 2
1 1
2 2
since m and are const,
88
63
q mc
q mc
c q
c q
θ
θ
θ
=
=
= =
ratio = 88: 66 or 88/63
5 (a) 3.6 J of energy has been converted to electrical energy for every 1 coulomb of
charge passing through the battery. [1] (b)(i) correct parallel connection [1] (ii) Resistance of wires & internal resistance of the battery – [1]
(c) using P = R
V 2
, resistance of the LCD = 1.44 Ω
resistance of the GSM = 6.48 Ω [1/2] each for each correct resistance
LCD GSM
2007 TPJC H2 Solutions
4
Total power output = power of LCD + Power of GSM at 3.2 V
= 48.6
2.3
44.1
2.3 22
+ [1]
= 8.69 W [1]
(d)(i) using s
p
s
p
V
V
N
N=
6.3
220=
s
p
N
N
61.1 : 1 (ii) At 3.7 V rms, the voltage will sometimes be > 3.6 V and sometimes < 3.6 V.
It may even be -ve at times (if not rectified). For a significant portion of the time, the output of the charger is expected to be
less than 3.6 V [1]. Whenever the charger’s output voltage is < 3.6 V, the battery would not be charged. In fact it will be discharging [1].
In contrast, the steady 3.7 V of the DC supply will be charging the battery all the time. Hence the DC will charge faster [1].
6
(i) Peak magnetic flux through 1000 turns Φ0 = BAN = (6.0 x 10-5)(4.0 x 10-4)1000 = 2.4 x 10–5 Wb [1]
Ave induced emf = Ave change of magnetic flux per second
= (Φ0 – 0)/t [1] = 2.4 x 10–5 / 5.0 x 10-3 = 4.8 x 10-3 V [1]
(ii) Average induced current = emf / resistance [1] = 4.8 x 10-3 / 23 = 2.1 x 10–4 A [1] (iii) Same magnitude but opposite direction. [1]
7 (a) (i)
[B1] A chain reaction refers to a series of reactions in which one causes another to happen, and this goes on continuously. [B1]
(ii) Electromagnetic radiation (e.g. gamma radiation) [B1] (b) (i)8.6 x 107 Bq kg-1 [A1]
neutron U
235
92
fission product
fission product
further reactions
2007 TPJC H2 Solutions
5
(ii)
t
OeAAλ−= [M1]
For iron 55, 17000102400)100(
7.2
2ln
12 =×=−
eA Bq [A1]
For nickel 59, 12
)100(80000
2ln
12 102.2102.2 ×=×=−
eA Bq [A1] Nickel 59 is a bigger problem than iron 55 because of its much longer half life, causing it to decay much more slowly. [B1]
(iii) Longer half life means a lower probability of any particular nucleus decaying per unit
time (λ), hence it is possible for activity to be lower. [B1]
(iv) t
OeAAλ−=
t
e 12
2ln
1212 101101043.0−
×=× [M1] t = 96 years [A1]
(c)
(i) No. of atoms = atom 1 of mass
mass
= 2710661.1963999.39
100
0117.0
100
3.01
−××
×× [M1]
= 5.3 x 1018 [A1]
(ii) NA λ=
= 18
9103.5
3600243651028.1
2ln××
×××× [M1]
= 91.0 Bq [A1]
(iii) Carry out the measurement in a container of lead or concrete which is able to shield off any background radiation. OR Measure background count first (without source), then subtract from the final count with source. OR Any other logical method.
2007 TPJC H2 Solutions
1
2007 TPJC Prelim P3 Solutions 1
(a) The rate of change of momentum of a body is proportional to the resultant force acting on it [B1] and the change in momentum takes place in the direction of the resultant force. [B1]
(b) (i) F = Ma
50 = (2.0 + 4.0) a [M1] a = 8.3 ms-2 [A1]
(ii) compression [B1]
(iii) Consider m1, F – FS = m1a FS = 50 – 2.0 (8.33) [M1] FS = 33.33 N FS = kx [M1] 33.33 = 12 x x = 2.8 cm [A1]
2
(a) total energy PE KE∆ = ∆ + ∆ = (500)(9.81)(2) + ½ (500)(5)2 [M1]
= 16060 J =16.1 KJ
(b) Possible answers: Consider the girder to be a system on its own. Its total energy is not conserved due to the net work done by the crane [1] motor on the girder. The total energy of a system is conserved only if there is no external net work done/energy transfer on the system.[1]
Consider the entire crane as the system. Even though the girder’s total energy is not conserved, because there is no external net work done/energy transfer on the crane-system, [1] the total energy of the crane-system is conserved. [1] Note: there is no fixed answer to this part. As long as the student shows an adequate understanding of the identification of a system and the conditions in which the total energy of a system is conserved, full credit should be given. c) Here the efficiency of the crane-system is considered. [B1] ~ 1 mark given simply for identifying the system considered by the question although the student did not note this down. Useful output power of the crane = 16060/ 5 = 3212 W = 3.21 KW [M1] Efficiency of the crane during first 5 seconds = 3212/4000 x 100% = 80.3 % [A1]
2007 TPJC H2 Solutions
2
3(a) An astronaut in an orbiting satellite is falling freely towards the Earth. He experiences a sensation of weightlessness because all parts of his body are falling freely. [1] Normal reaction due to satellite platform on astronaut is zero. He does not exert
any force on platforms. [1] (b(i) Gravitational potential at a point is the work done per unit mass in bringing the
mass from infinity to the point. (ii) The potential has a negative value because
the value of potential is taken as zero at infinity & the work done by external agent in moving the mass from infinity towards an attracting body is negative.
(c) Let h = 0. 12x 10 6 m above the Earth's surface Gain in potential from launch to height h
11 24
6 6
6
( ) ( ) [1]
1 1( )
1 1(6.67 10 )(5.987 10 )(3000)( ) )
(6.38 0.12) 10 6.38 10
3.47 10
m
GM GMm
R h R
GMmR h R
x xx x
x J
φ
−
= ∆
= − − −+
= − ++
= − ++
=
Substitution –[1] Ans – [1]
2
2
11 24
6
10
( )
1. ( ) [1]
2
(6.67 10 )(5.987 10 )(3000) 1( )
2 (6.38 0.12) 10
9.22 10
mv GMm
R h R h
GMmK E
R h
x x
x
x J
−
=+ +
=+
=+
=
Ans – [1] 4 (a) 10-10 m [1] (b) Crystals were used as diffraction grating as
they are made up of a regular arrangement of atoms atomic spacing is comparable to the de Broglie wavelength of the
electrons
[1]
[1]
2007 TPJC H2 Solutions
3
Hence, diffraction effects are observable. (c) Wave nature of matter. [1] (d) Although the number of electrons arriving per unit time is doubled, the
momentum of each electron is still the same. Therefore, the de Broglie wavelength of the electron is the same. Hence,
no effect on the spacing between diffraction fringes.
[1]
[1]
(e) Each neutron is about 2000 times more massive than an electron
De Broglie wavelength for neutron is thus much shorter than atomic spacing. Consequently, diffraction effects are much more difficult to observe.
[1]
[1]
5 (a) Excited atom is perturbed by an incoming photon with photon energy hf
corresponding to a difference in energy levels (hf = E2 – E1) Releases a second photon with the same frequency. Accounts for why
laser light is monochromatic. Emitted photon is in phase with the original photon. Accounts for why
laser light is coherent. Emitted photon travels in the same direction as the original photon.
Accounts for why laser light is collimated.
[1]
[1]
[1]
[1]
(b) Population inversion refers to having a greater number of atoms in an
excited state than in the ground state. This is to ensure that stimulated emission is more favorable than
stimulated absorption. (or, allow more photons to be emitted by stimulated emission so that an intense beam can be formed).
[1]
[1]
(c) Intense pulse of light from the flash lamp excite the atoms in the ruby to
create a population inversion. (This is known as “optical pumping” as compared to other mechanisms)
[1]
6
(a)(i) S1P - S2P = nλ where n is an integer [1]
(ii) The principle states that the resultant displacement at a point where two or more
waves of the same type meet [1] is given by the vector addition of the individual
displacements [1].
(b)(i) (Here, the waves from speaker S travel by 2 different paths to reach the
microphones M1 and M2. Then the electrical signals travel to the CRO and get
added.) When the speaker is moved from A to B, the path difference is varied [1],
leading to alternate constructive and destructive inteference of the signals[1].
(ii) In this case, the path difference is constant, hence there is no variation.
2007 TPJC H2 Solutions
4
(c)(i) λ = 10 cm = 0.10 m [1]
v = fλ
v = 3000(0.1) = 300 m s-1 [1]
(ii) Intensity of waves drop when they are further away from the source [1] due to
loss(attenuation) and the fact that the wave energy is spread over a larger area.
The waves from S1 and S2 travel different distances to reach M [1], hence they
don’t have same amplitudes and hence do not cancel completely at minima
points [1].
(d)(i) Because the presence of resistive forces would cause the oscillations to die out.
[1]
(ii)
(iii) a0 = ω2x0 [1] for formula
= (2π/T)2x0
= [2π/(2.5 × 10-3)]2 (0.00012) [1] for ω formula
= 760 m s-2 [1] for answer
(iv) [1] for shape of curve
[1] for labelled axes if curve correct
2 The presence of a resonant frequency in the operating range would mean
that for different signal frequencies, the amplitudes of oscillation are different
[1] leading to different loudness for different frequencies given the same
electrical signal amplitude [1]. Or membrane will rupture [1]
t / ms
Energy
0.25
Total energy [1]
PE [1]
KE [1]
Amplitude
Driving
frequency Resonant
frequency
2007 TPJC H2 Solutions
5
-------------------------------------[1]
-------------------------------------[1]
-------------------------------------[1]
-------------------------------------[1]
-------------------------------------[1]
----------------[1]
-------------------------------------[1]
-------------------------------------[1]
-------------------------------------[1]
-------------------------------------[1]
-------------------------------------[2]
-------------------------------------[2]
7 Parabolic path
3 1
3
19 314 2
31
39
7
14 9 6 1
7
2 2
802 10
40 10
1.6 10 2 103.513 10
9.11 10
:
80 105.33 10
1.5 10
0 3.513 10 (5.33 10 ) 1.87 10
1.5 10
1.51
E
net E
y y
x
y x
VE x Vm
d x
F eE
F F ma
eE x x xa x ms
m x
Direction
xt x s
x
V u at x x x ms
V x
Find v V V x
−−
−−
−
−−
− −
= = =
=
= =
=> = = =
↑
= =
↑ = + = + =
=
= + = 7 1
60
7
10
1.87 10tan 7.1
1.5 10
y
x
ms
V x
V xθ θ
−
= = => =
(d) Horizontal component – unchanged
Vertical component – increased. [1] When the separation decreases, the electric field strength between plates is increased and since the electric force is acting vertically upwards, there is not change in the magnitude of the horizontal component but it results in larger vertical acceleration. [1]
(i) The velocity is comprised to vsinθ and vcosθ
By Flemming LHR, there is a magnetic force perpendicular to vsinθ and this
makes the electron go in circular motion. [1]
The horizontal component of the velocity vcosθ is unaffected by the magnetic field and is unchanged. The combined effect results in helical motion. [1]
2
31 7 0
4 19
7
7 0
( sin )sin
sin 9.11 10 (1.51 10 )sin 7.10.0656
1.62 10 (1.6 10 )
2 2 (0.0656)2.2 10
sin (1.51 10 )sin 7.1
B
m vF Bqv
R
mv x xR m
Bq x x
RT x s
v x
θθ
θ
π πθ
−
− −
−
= =
= = =
= = =
A circular patch of light is seen on the screen. [1] 8 (a) (i) Transition from Level 1 to 2, or 1 to 3. [1]
2007 TPJC H2 Solutions
6
(ii) For 1 to 2, KE left is 2.1 eV.
For 1 to 3, KE left is 0.3 eV.
[1] [1]
(iii) No transitions will be made.
The incoming photon must be absorbed entirely or not at all. However, no energy transition is exactly 7 eV.
[1]
[1]
(b) (i) When an electron (having sufficient energy) collides with an atom
in the metal target, the impact could remove an electron from the inner shells of the atom.
This vacancy will be quickly filled by an electron from the higher energy levels. An X-ray photon is emitted, whose energy is equal to the difference in energy between the two levels.
As there are fixed energy transitions that are allowed in the atoms of the metal target, a line spectrum (rather than band spectrum) is formed, that is characteristic of the metal target.
[1]
[1]
[1] (ii) Electrons in the L-shell (compared to the M shell) are nearer to the
K-shell, hence There is a greater probability that the vacancy in the K-shell is
filled by an electron from the L-shell than from the M-shell. (Or: At any instant, the number of transitions of electrons from the
L-shell to the K-shell is more than from the M-shell.)
[1]
[1]
(iii) With a higher potential difference, accelerated electrons gain more
K.E. When the electrons with higher K.E lose ALL their K.E upon impact
to emit an X-ray photon, the photon emitted will be of a higher maximum frequency, which corresponds to a lower minimum wavelength
[1]
[1]
(c) (i) Classically, the particle whose energy is less than the peak potential energy will not be able to escape without gaining energy and so will be stuck in the potential well on the left.
However, an electron with wave properties obeys the Schrodinger wave equation. The solution of the equation is a wave function that is non-zero inside the barrier and beyond.
Recall that square of the amplitude of the wave function gives the probability of finding the electron at a point
That means that the electron has a non-zero probability of existing inside the barrier and beyond, thus it has a chance of tunnelling through the barrier.
[1]
[1]
[1]
2007 TPJC H2 Solutions
7
(ii)
(d) U – E = (10 – 4.0) eV = 6 eV = 6 × 1.6 × 10-19 J = 9.6 × 10-19 J
For d = 0.20 nm = 0.20 × 10-9 m,
2
2 )(8
h
EUmk
−=
π=
234
19312
)1063.6(
)106.9)(1011.9(8−
−−
×
××π = 1.25 × 1010 m-1
2kd = )1020.0()1025.1(2 910 −×××× = 5.01
Hence, tunnelling probability T = kd
e2−
= 01.5−e = 0.0066
and reflection probability R = 1 – T = 1 – 0.0066 = 0.99
[1]
[1]
[1]
Material
Probe Empty Space
U
ψ
x
Fig. 8.3
2007 TPJC H2 Solutions
m VICTORIA JUNIOR COLLEGE SUGGESTED SOLUTIONS TO 2007 PHYSICS PRELIM EXAMS PAPER 2 H2 Q1(a) Point A has the highest potential. [1] The farther a point is away from the Earth, the higher the potential. [1] (b)(i) Work done by gravitational field in bringing mass from A to C = loss in GPE as mass goes from A to C = UA - UC
= mφA - mφC [1] = 5000 (-4.0 x 107 – [-5.0 x 107]) J = 5.0 x 1010 J [1] (b)(ii) No work is done by gravitational field as C and D are at the same potential. [1] (c) The gravitational field strength is not constant but decreases as one goes away from the Earth. [1]
(d) B
B rGM
−=φ and C
C rGM
−=φ [1]
Br)10x0.6)(x1067.6(10x 5.4
24-117 −=−
Hence rB ≈ 8.89 x 106 m Similarly, rC ≈ 8.0 x 106 m The distance BC = 8.9 x 105 m [1] Q2(a). Consider rope, bucket and water as a single system and let the tension in the rope be T. For circular motion of bucket of water, net force towards the centre of motion = centripetal force
v T
r
At top of circle, r
mvmgT2
=+ [1]
)81.9)(25.3(950.0
)23.3(25.3 2
2
−=
−=⇒ mgr
mvT
[1] ≈ 3.81 N [1] (b) If the speed of rotation is high enough, the weight of the water alone in the bucket is not enough to provide for the centripetal force required. A force will be exerted by the bottom of the bucket on the water for the water in the bucket to continue to move in a circle. By N3L, the water pushes against the bottom of the bucket and thus stays in the bucket.
Alternative: By N1L, if the speed of rotation is fast enough, the water in the bucket tends to move strongly away from the center of the circle, resulting in it pushing against the bottom of the bucket. Thus, the water will not fall out. [2]
1
2007 VJC H2 Solutions
(c) Consider the circular motion of the water only.
rvm
gmR ww
2
=+ [1]
)81.9)(25.2(950.0
)23.3(25.2 2
2
−=
−=⇒ gmrvmR w
w
≈ 2.64 N [1] R is towards the center of the circle. [1]
Q3(a) Initial electric PE = r
oπε421 [1]
= )100.1)(1085.8(4
)106.12)(106.1(912
1919
−−
−−
××××+×+
π
= 4.6 × 10-19 J [1] (b) By the law of conservation of momentum, Total initial momentum = total final momentum ∴ 0 = mpvp - mαvα [1] ∴ 0 = (1.67 × 10-27)(2.0 × 104) - (4 × 1.67 × 10-27)vα
∴ speed of α-particle = vα = 5.0 × 103 m s-1 [1]
(c) Total KE = 22
21
21
ααvmvm pp +
[1]
2327
2427
)100.5)(1067.14(21
)100.2)(1067.1(21
×××+
××=
−
−
= 4.2 × 10-19 J [1] R (d) Initial total energy = final total energy [1]
mwg
∴ KE1 + PE1 = KE2 + PE2
∴ 0 + 4.6 × 10-19 = 4.2 × 10-19 + r
oπε421
r)1085.8(4)106.12)(106.1(10 x 4.0 12
191919-
−
−−
×××+×+
=∴π
∴ separation of particles at this point = r = 1.1 × 10-8 m [1] Q4(a) The net emf of the wire structure between points P and Q is given by
vLBvLBvLBEnet )4
()2
()4
( ++= [1]
Hence E net= BLv [1] By Fleming’s Right Hand Rule, end Q is at the higher potential. [1] (b)(i) As the loop is crossing the boundary of the two magnetic fields, the magnetic flux through the loop changes with respect to time. According to Faraday’s Law, an emf that is generated drives a current around the loop. [1] The induced current flows counter-clockwise around the loop. [1] (b)(ii) X X X X X
X X X X
X X X X X X X X X
v
X X X X
E1 P Q
E2 S T
2
2007 VJC H2 Solutions
The rods QS and PT behave as sources of emf of magnitude B’L2v and BL2v respectively. The net emf is thus (B’-B)L2v. [1] The induced current is then
RvB)L(B'I 2−
= [1]
(iii) The flux through the coil remains constant and hence no net emf is induced (Faraday’s Law). [1] Q5(a). When a nucleus is separated into its individual components, the sum of the masses of its individual constituents is greater than the total mass of the nucleus. [1] This difference in mass is called the mass excess. [1] 5(b)(i) [1] PtME =0
where E0 = energy produced per kilogram of coal
0EP
tM
=
Mass of coal burnt per second = 40 000 / 20 = 2 000 kg s-1 [1]
(b)(ii) m = E / c2 or 2cPtm = [1]
= 40 000 x 106 (1.0)/ (3.0 x 108)2
= 4.44 x 10-7 kg [1] (b)(iii) Power Input = (40 000 x 106) / 0.20 = 2.0 x 1011 W [1] Mass of coal burnt per second = 2.0 x 1011 / 20 x 106
= 10 000 kg s-1 Extra coal that needs to be burnt per second = 10 000 – 2000 = 8000 kg s-1
[1]
Q6(a)(i) E = I(r+RL) [1]
rIERL −= [1]
RL / Ω I /A 1/1 −A
I
2.00 0.590 1.69 3.00 0.420 2.38 4.00 0.330 3.03 5.00 0.270 3.70 6.00 0.230 4.35 [1 mark for correct headings, units and computations]
L
[1 mark for correct plottings and axis labeling] The gradient of the graph is E = 1.51 V (accept 1.36 V ≤ E ≤ 1.66 V) [1] The negative intercept of the graph is –I = -0.563 Ω ⇒ r ≈ 0.563 Ω [1] (accept 0.507 Ω ≤ r ≤ 0.619 Ω) (a)(ii) When RL is dissipating maximum power, it resistance is equal to r = 0.563 Ω (maximum power transfer theorem). [1] Current flowing in the circuit is
rREI
L += [1]
Graph of R L vs 1/I
R L = 1.51(1/I ) - 0.563
0
1
2
3
4
5
6
7
0 1 2 3 4 5
(1/I)/A-1
R/ohm
3
2007 VJC H2 Solutions
= ≈+ 563.0563.051.1 1.34 A [1]
(accept 1.21 ≤ I ≤ 1.47 A) (b)(i) When the temperature of a metal or alloy specimen increases, its atomic kernels vibrate with bigger amplitudes. [1] Charge carriers, namely electrons, will experience an increased incidence of scattering as they drift through the specimens under the action of an external electric field. [1] As a result, less current flows for a fixed potential difference. This translates to an increase in electrical resistance with rise in temperature. [1]
(b)(ii)1. Given θ
α θ
oRRR 0−
=
Looking at the graph, when θ = 20.0 0C, R20 = 30.0 Ω . When θ = 60.0 0C, R60 = 34.5 Ω
)1(0 += αθθ RR Hence 30.0 = R0 (α20+1) …(1) [1] 34.5 = R0(α60+1) …(2) [1]
Taking )1()2( gives
120160
0.305.34
++
=αα
690α+34.5 = 1800α +30.0 ⇒ α ≈ 4.05 x 10-3 (0 C)-1 [1] (b)(ii)2. The graph will cut the vertical axis at a value of Rθ = R0 .
[1] In equation (1), 30.0 = R0 (20 x 4.05 x 10-3 + 1)
0R∴ = 27.8 Ω [1] (b)(ii)3. A negative value for α indicates that the semiconductor resistance decreases with rise in temperature. [1]
(c) Below the temperature TC , the specimen has no resistance and is basically superconducting. [1] When an induced current flows in the ring, practically no energy is dissipated to the surroundings via Joule heating. In consequence, the current flows in the ring for months on end. [1]
********* END *******
4
2007 VJC H2 Solutions
VICTORIA JUNIOR COLLEGE SUGGESTED SOLUTIONS TO 2007 PHYSICS PRELIM EXAMS P3 H2 Section A Q1(a) Using ‘s = ut + (1/2)at2’ For vertical motion and taking downwards as positive, 20 = 0 + (1/2)(9.81)t2 [1] Thus, t = 2.02 s [1] (b) Horizontal distance traveled by package backwards = (14-8.0)(2.02) = 12.1 m Horizontal distance traveled by helicopter forwards = (8.0)(2.02) = 16.2 m [1] Thus, distance between them = 12.1 + 16.2 = 28.3m [1] (c) Horizontal velocity of package just before hitting the ground is 14-8.0 = 6.0 m s-1 backwards. [1] For vertical velocity, Using ‘vy = uy + ay t’ ⇒ vy = 0 + (9.81)(2.02) = 19.8 m s-1
[1]
Thus, magnitude of the velocity, 22 )0.6()8.19( +=⇒ v = 20.7 m s-1
[1]
at angle ⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛= −−
0.68.19tantan 11
x
y
vv
θ
≈ 73.10 with respect to horizontal. [1]
Q2(a) The amplitude is x0 = 0.15 m [1] (b) One-quarter of a period later, the diver would be halfway between the highest and lowest points, where his speed is at the maximum.
410075
kmω = = = 7.39 rad s-1 [1]
ω0max xv = [1]
= 4100(0.15)( )75 = 1.1 m s-1 [1]
(c) Since 2Tπω = , 2T π
ω= [1]
27.39π= = 0.85 s [1]
(d) When the diver just ceases to remain in contact with the platform, the force the board exerts on the diver, R = 0. By N2L,
maRmg =− or a = g Numerically, gx =2ω
Or xgf
π21
=
Hence 02
1xgflowest π
= [1]
15.081.9
21π
=lowetsf ≈ 1.28 Hz [1]
vx Q3(a) At B, RTpV µ= [1]
T××=×× − 31.80.3)102.1)(103.4( 36 ∴ T ≈ 207 K [1] (b) At C, RTpV µ=
T××=×× − 31.80.3)108.3)(102.2( 36 ∴ T = 335 K For monatomic ideal gas, Internal energy, U = total KE of all molecules, since PE = 0
∴ RTU µ23
=
θ
v vy
1
2007 VJC H2 Solutions
∴ change in internal energy,
TRU ∆=∆ µ23 [1]
∴ from B to C,
)207335(31.80.323
−×××=∆U ≈ 4790 J
[1] (c) Work done from B to C = ∆W = area under graph from B to C [1]
= 36 10)2.18.3(10)2.23.4(21 −×−××+
= 8450 J [1] (d) By 1st Law of Thermodynamics, for process BC, ∆U = ∆Q + ∆W [1] ∴ 4790 = ∆Q + (−8450) ∴ heat absorbed by gas in process BC, ∆Q = 13240 J ≈ 1.32 x 104 J [1] Q4(a)(i) When forward biased at V = 3.0 V, the current flow is I = 55 mA.
[1] Thus, the resistance, R = V/I = 3.0/(0.055) = 54.5 Ω [1] (a)(ii) At V = -1.5 V, the current increases to a very large value. The resistance of device T has possibly dropped to a zero value. [1] (b)(i). From the I-V characteristic curve, the current flow through T at 3.0 V is 0.055 A. The current through the 40 Ω resistor is 3.0/40 or 0.075 A. [1] Hence the total current drawn from the d.c. source is 0.130 A. [1] (b)(ii) If the polarity of the power supply were to be reversed, a very large current will be drawn from the d.c. source due to the shorting of T.The d.c. source will likely be damaged.
[1]
(b)(iii). The component T can be used for rectification of an alternating voltage.
[1] Rectification is, however, limited to peak value of less than 1.5 V in the reverse direction. Device T will be shorted and damaged for a negative voltage whose magnitude is 1.5 V or larger. [1] Q5(a) Applying N= No te λ−
⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛−=⇒ )(
57002lnexp0925.0 tNN oo
(1 mark for oNN 0925.0= and 1 mark for
57002ln
=λ )
410 x 1.96=⇒ t years [1]
(b) 4x60x605700x365x2
2ln=λ [1]
Probability of decay per sec is given by the decay constant = 3.86 x 10-12 s-1 [1] (c) A charged particle moving in a B-field perpendicular to it will have a circular path. For circular motion,
rmvBqv
2
= [1]
Bqmvr =⇒
Taking v ≈ 1.0 x 108 m s-1 , [1]
)1.6x10)(20.0()10()(9.11x10
19
831
−
−
=⇒ r
r = 2.85 x 10-3 m [1] Section B 1(a) The Law of Conservation of Momentum states that the total momentum of a system of interacting bodies is
2
2007 VJC H2 Solutions
constant provided no external forces act on the system. [2] 1(b)(i)1. An elastic collision is one in which both total momentum and total kinetic energy of the system of interacting bodies are conserved. [1] (b)(i)2 An inelastic collision is one in which total momentum of interacting bodies is conserved but not the total kinetic energy of the system. [1] (b)(ii) An inelastic collision involves loss of KE from colliding bodies. The bodies do not necessarily stick together. [1] This is different from a perfectly inelastic collision in which the bodies coalesce and stick together as a result of the collision.
[1] (c)(i) [1] α4222226 +→ RnRa (c)(ii) From conservation of momentum,
[1] RnRnv vmm −= αα0
uu
mm
vv Rn
Rn 4222
==∴α
α = 55.5
[1]
(c)(ii) 2
2
2121
RnRnRn vm
vm
KK αα
α = [1]
22
5.552224
⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛=
uu
vv
mm
RnRn
αα = 55.5
[1]
(d) (i) 2
21
ααα vmK =
[1]
27-
-13
10 x 64.6) x102.9(2
=αv ≈ 1.66 x 107 m s-1
[1] (d)(ii) [1 mark for each curve] (d)(iii)
[1] Consider motion from A to B: Loss in KE of particle = gain in EPE
00
2
40
21
rqqvm Au
A πεα
α =− [1]
Ra pi = 0
Rn vα vRn
α
After decay
+ve
energy
Electric PE
KE of α-particle
Au
Point of closest approach
α
Au
Point of closest approach
α
B A
r0
3
2007 VJC H2 Solutions
20
0 2)79)(2(
Avmeer
απε=
= 2727-12-
2-19
)x1066.1)(x1.66x104)(x1085.8(2)x106.1)(79(2
π
≈ 3.97 x 10-14 m [1] (d)(iv) It would still be possible for the α-particle to pass through the fields undeflected. [1] The first situation would be when the particle passed through crossed magnetic and electric fields in such a way that the magnetic force Bqv exactly equalled the electric force Eq in magnitude but opposite in direction. The second situation would be if both fields were aligned parallel or anti-parallel to each other, and the particle passed through both fields along the field lines. [1] [Accept either situation] Q2(a) The Principle of Superposition states that when two or more waves of the same kind meet at a point in space, the resultant wave displacement is given by the vectorial sum of the individual wave displacements at the point. [2] (b)(i) The wavelength of the water waves set up = 0.20/40.0 = 0.0050 m. [1] The path difference of the waves from dippers 1 and 2 to D = 0.130 - 0.120 = 0.010 m. Since p.d. = 2 λ and the 2 sources are 180 o out of phase, the interference at D will be destructive. [1] Hence, resultant wave at D will have minimum amplitude. [1] (b)(ii) In increasing the frequency to 80.0 Hz, the wavelength would decrease from 0.0050 m to 0.0025m. [1]
At a wavelength of 0.0025m, the path difference = 4λ, hence a destructive interference would be produced at D.
[1] A minimum would also be observed when the path difference was 3λ. [1] Maxima would be observed when the path difference was 2.5λ and 3.5λ. [1] (c)(i) Upon striking the barrier, the plane waves are reflected and will interfere with the incident waves. This leads to the formation of stationary waves. [1] There will be regions of nodes and antinodes. [1] The fine sand will settle along the nodal regions resulting in the formation of regularly spaced ridges. [1] (c)(ii) The successive ridge separation corresponds to nodal separation which is = ½ λ = 0.018m. Thus, wavelength λ = 0.036 m. [1] The wave velocity is now v = fλ = 12 x 0.036 = 0.432 m s-1. [1] Hence, time taken to travel a distance of 0.300 m = 0.300/0.432 = 0.694 s. [1] (d)(i) The first order maximum is diffracted at and angle θ = tan-1 (0.46/2.00) = 12.95 o. [1] From ‘d sinθ = nλ’ or (sin θ)/N = nλ,
[1] where N = 300 x 103 m-1 and n = 1:
λx110x30095.12sin
3
0
=
⇒ λ = 7.47 x 10-7 m. [1] (d)(ii) Using the second-order diffracted light to measure the wavelength is more accurate. [1] This is because the larger angle of diffraction can be measured experimentally with a lower percentage error for a given precision of the measuring instrument used. [1]
4
2007 VJC H2 Solutions
Q3(a)(i) E = h c / λ [1] = (6.63 x 10-34 x 3.0 x 108) / (254 x 10-9) = 7.83 x 10-19 J [1] (a)(ii) Ptotal = 210 x 12 x 10-6 = 2.52 x 10-3 W [1] Rate of incidence = Ptotal / E [1] = 2.52 x 10-3 / 7.83 x 10-19 = 3.22 x 1015 per second [1]
a(iii) Current = etN⎟⎠⎞
⎜⎝⎛ [1]
=2.7 x 1013 x 1.6 x 10-19 = 4.32 x 10-6 A [1] (b)(i) Stopping potential refers to the minimum reverse potential difference that must be applied across the electrodes so that the most energetic photoelectron emitted will just fail to reach the collector electrode [1] (b)(ii) The energy of photons hitting the silver surface would decrease with increase in wavelength. [1] The energy of the electrons emitted would also decrease, [1] hence the stopping potential would decrease. [1] (b)(iii) [1 mark for each graph] (c)(i) It is the minimum frequency of radiation for a given metal below which no photoelectrons will be emitted from the metal surface.
[1] (c)(ii) Since the frequency of radiation = 3.0 x 108 / 254 x 10-9 = 1.18 x 1015 Hz, it was higher than the threshold frequency of the new metal surface used. [1] Hence, there would still be photoelectric emission. [1] (c)(iii) h f = h f0 + Kmax Kmax= 6.63 x 10-34 (1.18 x 1015 – 1.0 x 1015) [1] = 1.19 x 10-19 J [1] (c)(iv) maxKeVs = [1]
≈= 19-
-19
10x 6.110 x .191
sV 0.75 V [1]
********* END ********
i / A
for λ = 254 nm
0 V/ V
for λ > 254 nm
5
2007 VJC H2 Solutions
Yishun Junior College 2007 Preliminary Examinations H2 Physics Paper 9745/2 Suggested Solutions
1
1 (a) Horizontally, 10 = u cos30°°°° t
Vertically, 5 = u sin30°°°° t + 0.5 (-9.81) t2 Working
Solving, u = 29.1 m s−−−−1
(b)
Energy needed prior to transfer to trebuchet projectile = 0.5*2*29.12/0.6 = 1409.17J Energy at cliff A top = Energy just before landing on trebuchet 0.5*40*u2 + 40*9.81*[3.5] = 1409.17J
u = 1.34 m s−−−−1
(c)
71 u + 0 = 71*0.371 u + 40*1.34
u =1.20 m s−−−−1 Kinetic energy is NOT conserved for this question.
2
(a) (i) P.d across AB = 150 x 0.01 =1.5 V Current = 1.5/50 = 0.03 A R = 0.5/ 0.03 (p.d across R is 0.5V since r = 0)
R = 17 ΩΩΩΩ (ii) Power = IV = 0.03(2) = 60 mW
(iii) Power dissipated = I2R = V2/R = 1.52/50 = 45 mW (b) (i) Yes.
When the potentiometer is balanced, the resistance seen by the driver cell is due to the potentiometer wire and the variable resistor only. However, when the potentiometer is not balanced, R1 and the internal resistance of the cell Y is in parallel to the resistance of the length AC. The effective resistance seen by the driver cell varies with the position of C. Thus the power delivered by the driver cell varies.
(ii) At balance point, no current flows through R1.
Thus, its resistance does not affect the balance point.
2007 YJC H2 Solutions
Yishun Junior College 2007 Preliminary Examinations H2 Physics Paper 9745/2 Suggested Solutions
2
3
(b) Change in current in the coil ⇒⇒⇒⇒ Change in magnetic flux linkage in brain
⇒⇒⇒⇒ e.m.f induced in the brain according to Faraday’s law
⇒⇒⇒⇒ induced current flows due to presence of conducting ions in the brain
(c) ( )( )( ) V ..
t
BNA
dt
)NBA(d33910003101311 42 =××=== −ε
(d) Increase number of turns in the coil; Increase the current in the coil; Increase the
frequency of the a.c. in the coil; Increase the area of the coil (Any two of the above statements).
4 (a) Description and explanation of any one of the following observations:
- no photoelectrons emitted if frequency is below a certain minimum - maximum kinetic energy of photoelectrons is independent of the intensity of the
incident light - graph of KEmax / stopping potential versus frequency of incident light is a
straight line - photoelectric emission is almost instantaneous, even if intensity of light is very
low
(b) (i)
19
19
10
100.3106.1
108.4 −−
−
×=×
×=⇒= si current,
t
N
t
Ne
(ii)
115
9
834
6
102.3
10254
1031063.6
1012210 −
−−
−
×=
×
×××
××=⇒
=⇒=
s
II
n
AnhfAP
(iii)
7
15
9
10491023
1003 −×=×
×= .
.
.ratio
Wire Coil
Scalp
Fig. 3.1
I
(a)
2007 YJC H2 Solutions
Yishun Junior College 2007 Preliminary Examinations H2 Physics Paper 9745/2 Suggested Solutions
3
(c)
By de-Broglie’s relation, the momentum of the scattered photon λ'
h is less than the
momentum of the incident λ
h, since λλλλ’ is longer than λλλλ.
This is consistent with the conservation of linear momentum if we treat the photon as a particle in a collision – the incident photon loses some of its momentum as a result of collision with the electron.
5 (a) (i)
∆∆∆∆E∆∆∆∆t ≥≥≥≥ π
h
4
∆∆∆∆t ≥≥≥≥
π
h
4÷÷÷÷( 8.0 ×××× 10−−−−14) = 6.59 ×××× 10−−−−22 s = 7 ×××× 10−−−−22 s
(b) A voltage is applied between the surface and the conducting tip of the STM probe so
as to enable electrons to tunnel between surface and tip. The magnitude of the current detected is exponentially dependent on the distance
between the tip and the surface (actually kdeT
2−= ). The current is monitored and kept constant by adjusting the distance between the tip and the surface as it sweeps across the surface. By measuring the relative distances as the tip moves across the surface, the surface structure of the material can be reconstructed.
(c) The system must undergo population inversion.
The excited state of the system must be in a metastable state. Emitted photons are confined between two reflecting surfaces long enough to allow them to stimulate further emission from other excited atoms.
6 (a) (i)
n =
filmf
f
λλ
⇒⇒⇒⇒ λfilm = n
λ
(ii) Path difference = 2t
For constructive interference,
2t = ( m + ½ ) λλλλfilm (phase change of ππππ radians)
2t = ( m + ½ ) n
λ
2nt = ( m + ½ ) λλλλ
2007 YJC H2 Solutions
Yishun Junior College 2007 Preliminary Examinations H2 Physics Paper 9745/2 Suggested Solutions
4
(iii)
0.80
1.00
1.10
0.90
480 500 520 540 560 580 600 620
thickness/10 m-7
λ m/10−9
For minimum value, m = 0
2nt = ½ λλλλ
t = 1
4nλλλλ
1
4n = gradient =
9
7
10)480600(
10)80.000.1(−
−
×−
×−
n = 1.50 (b) (i) Since film is thin, t is small ⇒⇒⇒⇒ m is small
Hence, at any one viewing angle, only one or two values of λλλλ will satisfy the equation in (a)(ii)
(ii) At different viewing angles, the wavelength(s) that reinforce will be different
Thus, a spectrum will be observed
2007 YJC H2 Solutions
Yishun Junior College 2007 Preliminary Examinations H2 Physics Paper 9745/3 Suggested Solutions
1
1 a) (i)scalar (ii) vector (iii) vector (iv) scalar b) P = (297 + 209) x 2 = 1012 mm
∆P = 1 x 4 = 4 mm P = (1012 + 4) mm A = (297 x 209) = 62073 mm2
500
209
1
297
1
=∆
+=∆
+∆
=∆
A
b
b
l
l
A
A
A = (62100 ± 500) mm2 2a) b) .c) d)
22.8 – 1 = 0.5*9.81*t2 Time of drop, t = 2.108 s 5 = ux (2.108)
Horizontal speed = 2.4 m s−1.
k = 50/0.001 = 50000 N m−1 Total Energy (initial) = Total Energy (final) PEg + PEe + KE = PEg + PEe + KE
mg(21.8) + 0 + 0 = mg (−x) + 0.5 k x2. x = 0.8158 m = 0.82 m
Energy initial = Energy final
PEg + PEe + KE = PEg + PEe + KE
75*9.81*(21.8) + 0 + 0.5*75*2.42 = 75*9.81 (-xsin φ) + 0.5 k x
2.
k = 50000 N m−1
.
x = 0.8128 m
= 0.81 m
φ = tan −1
(2.4/5) = 25.641°
Let reference level be the
level of un-depressed net,
Advanced solution
Force exerted by net ranges from zero to maximum
value at max depression.
Average force exerted by wire = (0 + k∆x)/2
= 20000 N (2 sf)
Elasticity of material
Cannot extend too much
Safety limit of weight of person and the height from
which he jumps must also be considered.
Impact force on victim
Tensile strength of material
Speed which person lands on net
(any 2 of the above) 2007 YJC H2 Solutions
Yishun Junior College 2007 Preliminary Examinations H2 Physics Paper 9745/3 Suggested Solutions
2
3a) b) ci) cii) d) e) 4ai) wavelength = 4 m ii) speed = 10 m/s
iii) v = fλ 10 = f (4) f = 2.5 Hz bi) Constant phase relationship/difference (Also accept same frequency) bii) 1. 2.
An object remains at rest or continues along a straight line motion at constant speed if there is no net external force acting on it.
∑F = 0 ⇒ F – FR = 0
⇒ F = 45(3)2 = 405 N
COM: 100 (∆v)= 0.06*0.75*1050*(2.24)
∆v =1.06 m s−1. (We are not told that the mechanical energy is conserved in this question nor can be assumed since final velocity is given)
Strokes needed = 3/∆v Min Rate = 2.8 per sec
Drag from water does not act on the boat when paddle is transferring momentum to water Or resultant force acting on system is zero, etc.
To balance the turning moment on the canoe such that the canoe travels straight.
I ∝ a2
a1/a2 = (√0.36)/ (√0.81) a1 = 0.667 a2 aR = a1
+ a2 = 1.667 a2 IR = I2 × aR
2/a22 = 0.81 × 1.6672
= 2.25 W/m2
aR = a2 − a1 = 0.333 a2
IR = I2 × aR2/a2
2 = 0.81 × 0.3332
= 0.0898 W/m2
2007 YJC H2 Solutions
Yishun Junior College 2007 Preliminary Examinations H2 Physics Paper 9745/3 Suggested Solutions
3
3.
5a) S
p
S
P
V
V
N
N=
0500.020
1
240
12===
S
P
N
N
bi) Power supplied by power station, P = IV
Hence increasing the voltage will reduce the current since power supplied is a constant. Because of the long distance, the resistance of the cable transferring energy from the power station to our home, R is significant.
Power loss, PL = I2R ⇒ Higher current will mean higher power loss.
ii) Alternating voltages give rise to change in magnetic flux which allows transformer to stepped-up or stepped-down the voltage.
c) P = IV
700 = 240 I I = 2.92 A
d) Vo = V x √2 =339 V
P/W
t/s
Fig 5.1
0.02
1400
P = V2/R which is the square of a V-t graph
I = 0.36 + 0.81 = 1.17 W/m2
[1]
e)
2007 YJC H2 Solutions
Yishun Junior College 2007 Preliminary Examinations H2 Physics Paper 9745/3 Suggested Solutions
4
6ai) ( ) 2
2
2
2
42 R
GM
R
GMF ==
6aii) 24R
GM
M
Fa ==
6aiii) 3
2
4R
GMRa =⇒= ωω
6aiv) R
GMv
R
va
4
2
=⇒=
6bi) Gravitational potential energy of satellite, Ep = –GMEm/r … eqn (1) 6bii) For a satellite in circular orbit, GMEm/r2 = mv2/r … eqn (2) Sub. Ek = ½mv2 (kinetic energy of the satellite) into (2): GMEm/r = 2Ek, therefore Ek = GMEm/2r… eqn (3) 6biii) Comparing eqn (1) & (2): Ek = –½Ep … eqn(4) Also, ET = Ek + Ep = –½Ep + Ep = ½Ep eqn (5) 6biv)
6bv) 1. correctly indicating ∆ET (it is a decrease) correctly labelling r2 to be smaller than r1.
2. ∆EP is a decrease (twice as large as ∆ET)
∆EK is an increase (equal in magnitude to ∆ET)
3. Satellite’s speed increases.
2007 YJC H2 Solutions
Yishun Junior College 2007 Preliminary Examinations H2 Physics Paper 9745/3 Suggested Solutions
5
7a) The depletion layer does not have any mobile charge carriers (or consists of fixed ion cores), unlike the p-type and n-type semiconductors with holes and electrons respectively as majority mobile charge carriers Or Depletion layer has an internal electric field
bi) The mobile holes are attracted/pulled towards the negative junction X while the
mobile electrons are attracted/pulled towards the positive junction Y, The depletion layer is enlarged.
ii) A single p-n junction or diode can be used to convert an a.c to d.c.
During one half-cycle, diode is functioning in the forward-biased state where X is positive w.r.t Y. Current flows through as resistance is low since depletion layer has diminished/collapsed. For next half-cycle, diode is functioning in the reverse-biased state where X is negative w.r.t Y. Current does not flow through because resistance is high since depletion layer is enlarged.
ci) Ep = k(0.3 x 10-6)/ 0.5 2 – k(0.9 x 10-6)/1.22
= 5170 V m-1
ii) φ p = k(0.3 x 10-6)/0.5 – k(0.9 x 10-6)/1.2
=−1350 V
d) E = −dr
dV
ei) The proton initially moves in a straight line towards the right. ii) Initially, as its velocity increases (because it accelerates, its associated kinetic energy
increases. Its electric potential energy hence decreases (by conservation of energy).
8ai) α is most ionising and γ the least.
X Y
a.c input
d.c output
–3Q
+Q
Fig. 7.2
f)
2007 YJC H2 Solutions
Yishun Junior College 2007 Preliminary Examinations H2 Physics Paper 9745/3 Suggested Solutions
6
γ is the most penetrating and α the least.
ii) γ is the least massive Thus it interacts the least with matter and thus loses the least amount of energy
during its travel As a result, it is least ionising and hence most penetrating.
b) High energy → barrier width shorter
→ transmission coefficient greater (as T = exp(-2kd))
→ shorter average waiting time before tunnelling
→ shorter half-life
ci) e CaK 040
20 1
40
19 −+→
ii) 1. Change in mass = 1.00728 + 0.00055 − 1.00867
= − 0.00084 u
2. Energy released = 23
13
2827
1002.6106.1
)103(1066.100084.0××
×
××××−
−
= 4.72 × 1023 MeV d) (Any two of the following three – for short-term, damage is immediate rather than
delayed, symptoms are burns and sickness rather than cancer and hereditary defects, and cause is by large dose as opposed to small dose.) Suggested answer: Short-term: Ionising radiation can cause immediate damage to human tissue and could be accompanied by radiation burns, radiation sickness, etc. Short-term effects usually occur when there’s a large amount of exposure to radiation.
Long-term: On the other hand, delayed effects such as cataracts, hair loss, leukaemia, cancer, etc. could appear years after the exposure. Hereditary defects may also occur in succeeding generations as a result of gene damage. These effects take longer to become apparent and can be caused by much lower levels of radiation.
e) i) State: The majority of the mass of the atom is concentrated in a very small positively
charged region (nucleus).
α-particle
target nucleus
Fig. 8.2
(e)(ii)1.
2007 YJC H2 Solutions
Yishun Junior College 2007 Preliminary Examinations H2 Physics Paper 9745/3 Suggested Solutions
7
Explain: As shown by its path, the α-particle experiences a strong electrostatic repulsive force at a certain distance from the core which causes it to deviate significantly from its original direction.
ii)2. 193 106.11050 −×××=
minλ
hc
λmin = 2.49 × 10−11 m
2007 YJC H2 Solutions