Paper 1 1. Ans: A Reasoning: By definition, velocity is the displacement traveled per unit time. 2. Ans: D Reasoning: Option A – for a bus traveling at 70 km h-1 and KE 30 000 J, will result in a bus of mass of only 55.6 kg. Option B – A power of 300 W is too powerful for a domestic light. Option C – 300 K is equivalent to 27 oC. A hot oven clearly has a temperature much higher than this. 3. Ans: D Reasoning: accuracy of 1% = 0.04 and uncertainty 0.01 reading = (4.07 ± 0.05) V 4. Ans: C Reasoning:
v y = u y + ay t
When ball in air, before it hits the sand -
= 0 + ( −9.81)(1) = −9.81 m s-1
v y = uy + 2ay s
When ball in sand - 0 = ( −9.81)2 + 2a(8 × 10 −3 ) a = 6.0 × 103 m s-2
5. Ans: D Reasoning: The gradient of s-t graph gives the speed. Initially, the cyclist accelerates, hence his speed will increase, the gradient should also increase. When he travels at constant speed, the gradient should be a straight line. Finally he decelerates, hence his speed will decrease, the gradient should then decrease. 6. Ans: C Reasoning: Total momentum before the collision = 6(5) + 10(-3) = 0 kg m s-1 Consider 6.0 kg trolley - m v = t = 6(5) / 0.20 = 150 N
7. Ans: C Reasoning:
f
C N
C
N
W f
When the ladder is in equilibrium, all the line of action of the forces will intersect at one point. Contact force (C) is the resultant force of the normal contact force (N) and frictional force (f). 8. Ans: B Reasoning:
mw (0.9 ) = mr (1.6 ) mr 0.9 = mw 1.6
ρr mr /( A ) 4mr 0.9 = = =4 = 2.25 ρw mw /(4 A ) mw 1.6
9. Ans: D Reasoning: Work done by car = Gain in KE = ½m(v2 – u2) 10. Ans: B Reasoning: ε =B v ε B v I=
R
=
R B v R W ∝v
F = BI = B ∴F ∝ v
=
B 2 2v R
11. Ans: B Reasoning: ac = r ω 2 2
2π T
=r
= 8.0 × 10−2 = 7900 m s
2π 60 / 3000
2
-2
12. Ans: C Reasoning: mv 2 r mv 2 T = + mg r 1.27(0.575)2 = + (1.27 × 9.81) 0.600 = 13.2 N
T − mg =
13. Ans: D Reasoning: GMms r ∆U = Ufinal − U initial U =−
=−
GMm GMm − − r2 r1
= GMm
1 1 − r1 r2
14. Ans: B Reasoning: Gms me 2π = me r ω 2 = me r 2 r T ms =
r 3 (2π )2 GT 2
(1.50 × 10 ) 11
=
2
3
(2π )2
6.67 × 10−11(365 × 24 × 60 × 60)2
= 2.01× 1030 kg
T v =0.575 m s mg
-1
15. Ans: A Reasoning:
Amplitude = 5.0 cm Gradient of graph = − ω 2 =
10 − −10 = −200 −0.05 − 0.05
2π = 200 T T = 0.44 s
ω=
16. Ans: D Reasoning: During damping, the amplitude of the oscillation will decrease. 17. Ans: C Reasoning: PtQ = mLv PtS = mLf From graph, since tQ > tQ and P, m are constants, ∴ Lv > Lf
18. Ans: D Reasoning: pV = nRT 1 V = p nRT
gradient of 1/p vs V graph, m =
1 nRT
now if n and T doubles, new gradient of 1/p vs V graph, m ' =
19. Ans: B Reasoning:
Pt = mc ∆T mc ∆T t= P 500 × 10 −3 =
(
= 81 s
)( 4.2 × 10 ) (100 − 15 ) 3
2.2 × 103
1 1 = (m ) 4nRT 4
20. Ans: C Reasoning:
let the width of the area be a since I = → I1 =
∴
P A
P P ; I2 = 2a 16(8a )
I1 1 2a = I2 1 128a I2 =
I1 64
since I ∝ A2 A1 I = A2 I 64 A2 =
A1 A = 8 8
21. Ans: D Reasoning: E = h c/ λ hence K = hc Since E is in eV and λ in nm, need to express hc in eV nm h c = (6.63x 10-34 J s) (3 x 108 m s-1) = (6.63x 10-34 / 1.6x10-19 eV) (3 x 108 x 109 nm) = 1.24 x 103 22. Ans: A Reasoning: L=¾λ And at the node, pressure changes from low (first diagram) to high (second diagram) 23. Ans: D Reasoning: d sin θ = m λ sin θ = m λ / d < 1 m < d / λ = 5.5 where d = 1/ 3x105 hence total number of lines = 5 + 1 + 5 = 11 24. Ans: A Reasoning: (choosing the best option) 25. Ans: C
Reasoning: Consider P2 at higher potential than P1. WD will therefore be negative when –q is moved from P1 to P2 (the charge naturally attracted to P2, and force in opp dir is needed to prevent it from accelerating) Option C will give a negative value. 26. Ans: B Reasoning: R=ρl/A L = 2π (0.75/100)(30) 27. Ans: C Reasoning: P = I2R = (6 / (2.9+0.4)) 2 (2.9) 28. Ans: C Reasoning: From graph, at 7:30, R of thermistor is 1.5 k Using potential divider principle, R + 1.5 / (1.5+1.5+1.5+R) x 9 = 6 R = 4.5 29. Ans: B Reasoning: High resistance voltmeter - assume ideal. When RT = R, the voltmeter reading is 0 V. (same pd at both points, symmetrical) When RT = 5R, the voltmeter reading is 2V (can be calculated by earthing pt between the two resistor R. That point will be 0V. At the top of the resistor R will therefore be 1 V (as emf is 6 V) At the bottom of the other resistor R is 3 V. Hence the pd across the voltmeter is 3 – 1 = 2 V) 30. Ans: B Reasoning: Since path is not circular, it should be an electric field acting on the charged particle. 31. Ans: B Reasoning: By Fleming’s LHR, the magnetic force is directed out of the paper. The electric force deflects the protons into the paper. As a result, the protons’ beam will pass through undeflected.
32. Ans: B Reasoning: The component of the magnetic flux density that will induce an e.m.f. across the coil when the coil is rotated is Bsin . The change in magnetic flux linkage is then given by N(Bsin )A. 33. Ans: C Reasoning: The negative of the gradient of the variation of magnetic flux with time graph gives the e.m.f induced in the coil. The e.m.f. is given by = -d /dt. 34. Ans: D Reasoning: Irms =
I 2T Area under I2 -t graph = 0 = I0 T T
35. Ans: B Reasoning: Pattern of concentric circles on the fluorescent screen exhibits the wave characteristic of electrons when they are diffracted by the atoms of the carbon film. Diameters of the circles decrease when the velocity of the electrons is increased. This is due to increase in momentum of the electrons which in turn cause a decrease in the wavelength.
36. Ans: B Reasoning: The square of the wave function of the electron gives the probability of finding the electron at any position. 37. Ans: D Reasoning: An applied voltage with the indicated polarity further impedes the flow of electrons across the junction. For conduction in the device, electrons from the n-type region must move to the junction and combine with holes in the p-type region. A reverse voltage drives the electrons away from the junction, preventing conduction. 38. Ans: D Reasoning: After population inversion buils up, the metastable states of neon de-excite by stimulated emission of red light into the next state below the metastable state which is not efficiently populated in the discharge and because it undergoes very rapid spontaneous emission to the ground state. As the neon atoms de-excite through spontaneous emission to the state below the metastable state, the emitted photons
also trigger stimulated emissions to occur in other metastable states excited neon atoms, resulting in the lasing process. After photon emission, the neon atoms will be at the state below the metastable state. These neon atoms at this state then return to ground state via fast radiative decays (infra red emission). 39. Ans: C Reasoning: After two half lives of the source, the count rate is reduced by 4 times which gives 127 counts per minute. The final reading now is 127 + 24 = 151. 40. Ans: C Reasoning: The mass number decrease by 12 and the atomic number decrease by 6 after three alpha particle emissions. The number of neutrons remains unchanged. The difference in mass number and atomic number of the final daughter nucleus gives the number of neutrons in the parent nucleus.
Paper 2 Q1(a) (i) v 2 = 2gh = 2(9.81)(32) v = 25 m s
-1
C1 A1
-1 if g =10 used (penalized once only on the paper) if t is calculated using s = 1/2at2, t =2.55 s, this scores the first mark. (ii)
25 34 or 25 tan θ = 23 θ = 47o sin θ =
(b)
F=
∆p ∆t
or
C1 A1
m(v − u ) t 0.13 × 32 = 0.95 = 4.4 N
F=
C1
A1
(c) Splash mean KE of stone transferred to KE/PE of water. Stone has less KE hence less energy to be lost to heat (through friction of water)
Q2(a) For SHM, either a is proportional to x or a = Negative sign needed. Graph is a straight line through origin With negative gradient (b)(i)
(ii)
2
x where
B1 B1
is a constant
B1 B1 B1
(acceleration of sand is) g / 9.81 m s-2 / acceleration of free fall Plate must be moving downwards from maximum displacement Contact force is zero
B1 B1 B1
3.8 mm (allow ± 0.1 mm)
A1
Q3a
(i)
Centripetal force is provided by the gravitation force. mv 2 GMm = − − − − − − − B1 r R2 GMm Ek = − − − − − − − − B1 2R Ep = −
(ii)
Ep Ek
(b)
(i)
GMm negative sign essential---------- B1 R
= −2.0 − − − − − − − − A1
smooth curve in positive region and R ≥ R p ---------B1 curve passing through correct points from R = 1.5 R p and 4 R p -----------B1
(ii) either at R = 4 R p , Ek = 1.25x109 J , at R = 2 R p , Ek = 2.5x109 J --------C1, C1
1 (1600)v 2 = 1.25x109 , 2 v = 1250 m s -1 --------C1
1 (1600)v 2 = 1.25x109 2 v = 1770 m s -1 --------------C1
so change in speed = 1770 − 1250 = 520 m s -1 --------------A1
Q4 (a) (i) from graph, when P = max = 5.62 W, V = 4.5V hence I = p / V = 1.24 A (ii) R = V / I = 4.5 / 1.24 = 3.6 (shown) (b) V = R I = 2.03 x 1.6 = 3.248 V using ε - Ir = V we have ε - (1.6) r = 3.248 ε - (1.24) r = 4.5
(1) (2)
solving (1) – (2) r = 3.5 (note: this value of r is approx that of R : maximum power theorem)
Q5 For semiconductors, with a small energy gap (about 1 eV) at low temp, electrons cannot jump the energy gap. When temp rises, electrons gain sufficient energy to move from valence band to conduction band. A hole will be created in the valence band. When an external electric field is applied, both conduction electrons and holes contribute to current flow. Since the electrons and holes have opposite signs, both of these motions correspond to a current in the same direction. Since number of charge carriers increases, implying better conduction, or lower resistance.
Q6 (a)
Radioactive refers to the nuclide exhibiting radioactivity, where the nucleus of an atom spontaneously disintegrates by emitting by emitting an alpha, a beta particle or a gamma photon to become more stable.
(b) (i) (ii)
The decay constant of a nucleus is defined as its probability of decay per unit time.
A = λN 1.26 × 105 = λ
2.40 × 10 −8 × 6.02 × 1023 90
λ = 7.85 × 10 −10 s-1
(c)
Q7 (a)
This is because nuclides having small decay constants do not decay as fast, hence their mass and activity remains approximately constant over a short time interval.
(i)
Initially the rate of increase of the fireball radius is very high (high gradient), but the rate decreases with time.
(ii)
As the initial rate of increase of the fireball radius is very high, and given that the dimension of the room is at most 12 m, the fireball will spread throughout the entire room is a very short time (about 25 ms), posing a danger to any occupants in the room as the occupants will not enough time to escape.
(b) (i)
1. When t = 40 ms, R = 14.5 m. lg t = 1.60, lg R = 1.16 2.
×
(ii)
Gradient =
(iii)
R n = kt m
1.22 − 0.925 = 0.393 1.75 − 1.00
n lg R = m lg t + lg k m 1 lg R = lg t + lg k n m m Gradient = = 0.393 ≈ 0.40 n Since n and m are integers, suggested values could be m = 2, n = 5.
(c)
R 5 = cV c=
R5 V
(14.5 )
5
−3
For V = 12.5 × 10 , c =
12.5 × 10 −3
(13.9 )
= 5.13 × 107 m2
5
−3
For V = 10.0 × 10 , c = −3
For V = 7.5 × 10 , c = −3
For V = 5.0 × 10 , c = −3
10.0 × 10
(13.1)
= 5.19 × 107 m2
5
7.5 × 10
(12.1)
−3
−3
= 5.14 × 107 m2
5
5.0 × 10
(10.5 )
−3
= 5.19 × 107 m2
5
= 5.11× 107 m2 2.5 × 10 −3 Since the value of c remains approximately constant, the relationship is valid. For V = 2.5 × 10 , c =
(d) (i) (ii)
Calorific value of the liquefied gas. Explosion of fuel tank in an aircraft.
Paper 3 Q1
(a)(i)
(ii) The graph of P against 1/V showed that their product is a constant i.e. P = k/V where k is a constant. (b)
y1 − y 2 x1 − x 2 0.7 − ( −0.25) = = -1.98 6.78 − 7.26 1 (c)(ii) g is proportional to 2 since lg g = lg k − 2 lg r hints that the original r −2 relationship is g = kr
(c)(i) gradient =
Q2
(a)(i) This is in accordance to Faraday’s Law. An e.m.f. will be generated when there is a cutting of magnetic flux by a conductor. According to Faraday’s Law, dΦ E = −N . dt (ii) 1. Area enclosed by the coil 2. Strength of magnetic field
(iii) The magnetic flux Φ normal to turning coil is sinusoidal with respect (with dΦ respect to time). Since E = − N , the e.m.f. generated will be sinusoidal as dt well. (b)(i) peak input e.m.f. = 72 2 =102 V Vs Ns = (ii) Vp Np Vs 600 = 72 30 Vs = 1440 V 1440 (iii) r.m.s. current = = 9A 160 (iv) Power = VsIs = 1440 (9) = 13000 W (v) VpIp = VsIs 72(Ip) = 1440(9) Ip = 180 A Q3 a) Stationary waves are waves whose waveform does not advance and there is no translation of energy. The amplitude of the waves varies according to position from zero at the nodes (permanently at rest) to a maximum at the antinodes. b) Diffraction refers to the bending or spreading out of waves when they travel through a small opening or when they pass round a small obstacle. c) Coherent waves have the same frequency and a constant phase difference. d) Polarization is said to occur when a wave vibrates in a single plane. It is a phenomenon associated with transverse waves only. Q4 a)(i) The internal energy, U, is the sum of the total microscopic kinetic and potential energies of the particles (atoms and/or molecules) that make up the system. a)(ii) The first law of thermodynamics states that internal energy is a function of state. An increase in the internal energy of a system is sum of the heat supplied to the system and the work done on the system. b)(i) work done=1×105 × b)(ii) Section of cycle A B C
B C A
( 20-5 )
100×100×100
=1.5 J
Heating supplied to Work done on gas/ J gas/J Zero 4.2 -8.5 Zero 5.8 - 1.5
Increase in internal energy of gas / J 4.2 -8.5 4.3
5
(a)
(b)
(i)
Electric field strength is defined as the electric force experienced by a unit charge in an electric field.
(ii)
F F = Eq q W = F × d = Eqd
(iii)
∆V V = d d V W = Eqd = × q × d = qV d
(i)
Q = It = Ne
E=
E is uniform, therefore F is a constant
E=
N I 8.6 × 10 −3 = = = 5.4 × 1016 electrons −19 t e 1.6 × 10
(ii)
W = qV 1 mv 2 = qV 2 v=
(iii)
v=
2qV m
2 × 1.60 × 10 −19 × 60 × 103 = 2.1× 108 m s-1 9.11× 10 −31
Power supplied to anode = KE of all electrons hitting anode in 1 second N N × W = × qV t t 16 = 5.4 × 10 × 1.6 × 10−19 × 60 × 103 = 520 W
P=
Q = mc ∆T Q m = × c × ∆T t t m 520 = × 3500 × 30 t
(c)
(d)
(i)
(ii)
m = 0.0050 kg s-1 or 5.0 g s-1 t
The electric field lines should start at the charge position. The electric field lines all meet at the centre so it appears like all the charge is concentrated in the center. E=
Q 4πε o r 2
=
0.06 × 10 −6 4 × π × ( 0.1)
2
= 5.4 × 104 N C-1
(iii)
V=
k r
VAB = VA − VB = k
1 1 1 1 − = VBC = VB − VC = k − r A rB r B rC
1 1 1 1 − = − r A rB r B rC 1 1 1 1 − = − 0.40 0.50 0.50 rC 0.5 = 2 − 1 = 1.5 rC
1 rC rC = 0.67 m
Q6 a)
Frequency, f, refers to the number of oscillations of object completed in one second, while angular frequency, 2 f, is the angle (measured in radians) traveled by an object in one period. The unit of frequency is Hz or s-1 while the unit for angular frequency is rad s-1.
b)
(i)
Loss in g.p.e. = mg( h) = (0.400 kg)(9.81 m s-2)(0.200 m) = 0.785 J
(ii)
Assume Hooke’s Law is still valid, therefore, gain in e.p.e of spring = ½ kx2 = ½ (mg)(x) = ½ (0.400 kg) (9.81 m s-2)(0.200 m) = 0.3924 J ~ 0.392 J (3 s.f.)
c)
Some of the energy loss in g.p.e. transformed to e.p.e. of spring. The remaining amount of energy loss in g.p.e is used to overcome the negative work done by the hand in supporting the mass when it is moving downwards gently or to suppress the gain in kinetic energy when the mass is dropped freely without the support of the hand. Energy is not conserved here. (Note: If the hand is not there, the mass will be oscillating when it is dropped gain in k.e., but according to the question it is lowered gently until equilibrium is reached.)
d)
(i)
At the lowest point: Resultant force = Fspring – mg = k(0.200 m + 0.200 m) – mg = k(0.200 m) + k(0.200 m) – mg = mg + mg – mg = mg
= (0.400 kg)(9.81 m s-2) ~ 3.92 N (3 s.f.) (ii)
For S.H.M.: a = – 2x Take upwards as +ve. Resultant force (upwards) = mg = ma Therefore, g = a = – 2x0 = – 2(– 0.200 m) So,
(iii)
{from 6d(i)}
= (9.81/0.200)1/2 = 7.004 rad s-1 ~ 7.00 rad s-1 (3 s.f.)
Maximum speed, vmax = x0 = (7.00 rad s-1)(0.200 m) = 1.40 m s-1 (3 s.f.)
e)
lowest point equilibrium position highest point
gravitational potential energy /J 0
elastic potential energy / J 1.57
kinetic energy /J
total energy /J
0
1.57
0.785
0.392
0.392
1.57
1.57
0
0
1.57
Note: At the highest point, if the spring does not return to its original (unstretched) length, then EPE should not be zero. f)
Using values from table in 6(e):
energy / J
1.57
0.785
0.392
lowest point
eqm point
highest point
Legend: Total energy G.P.E (linear – mg h) E.P.E (½ kx2 – symmetric about highest point) K.E. (½m 2(x02 – x2) – symmetric about equilibrium point)
7
(a)
Detector Vacuum
Source Gold foil
Experimental arrangement used to study the scattering of -particles by thin metal foils. The detector can be rotated to various scattering angles .
The Rutherford’s scattering experiment involves injecting a beam of alpha particles into a gold foil. The experimental observations were used to construct a picture of the nucleus. These were as follows i)
A large number of the alpha particles passed through the gold foil with very small deflections. This indicates that most of the atom must be empty space.
ii)
A small number of alpha particles are deflected by very large angles, some even deflected by 180o. This indicates that the positively charged alpha particles have interacted with a region containing a very large positive charge.
The nucleus therefore contains all the positive charge and it occupies a very small region in the atom.
(b)
(i)
By the conservation of Mass Number
235 + 1 = x + 98 x = 138 By the conservation of Atomic Number 92 + 0 = 52 + y y = 40
(ii)
Energy released in the fission reaction E can be written as 235 1 138 98 92 U+ 0 n→ 52Te + 40 Zr
+ E released
Using Einstein’s Mass-Energy Equivalence, E released = M reac tan ts − M products c 2
[
[ (
E released = m
]
) + m ( n) − m(
235 92 U
1 0
138 52 Te
) − m(
98 40 Zr
)]c
2
E released = [235.0439u + 1.0087u − 137.9603u − 97.9197u] c 2 E released = [0.1726u] c 2
[
]
E released = 0.1726 × 1.66 × 10 -27 c 2 E released = 2.58 × 10
(iii)
(iv)
−11
J
The remaining energy is released as heat dissipated into the surroundings
1 The system initially at rest (assume speed of incident neutron is negligible so we can consider neutron to be essentially at rest). The conservation of momentum follows p initial = pfinal 0 = pTe + p Zr mTe v Te = m Zr v Zr pTe = p Zr .......................................................(eqn 1) v Te = speed of Te nucleus v Zr = speed of Zr nucleus speed of Zr nucleus mTe 137.9063 = = = 1.408 speed of Te Nucleus m Zr 97.9197
2 The ratio of kinetic energies can be determined from the ratio of mass & speed
1 2 m Zr (v Zr ) m Zr v kinetic energy of Zr nucleus 2 = = × Zr kinetic energy of Te Nucleus 1 mTe v Te 2 mTe (v Te ) 2 kinetic energy of Zr nucleus 97.9197 2 = × (1.408 ) = 1.408 kinetic energy of Te Nucleus 137.9603
2
ALTERNATIVE METHOD kinetic energy =
p2 2m 2
p Zr 2m Zr
mTe kinetic energy of Zr nucleus = = 2 kinetic energy of Te Nucleus m Zr pTe 2mTe
.......using eqn 1
kinetic energy of Zr nucleus 137.9603 = = 1.408 kinetic energy of Te Nucleus 97.9197
(v)
Total kinetic energy of the fragments = 2.3 x 10-11 J KETe + KE Zr = 2.3 × 10 −11
1 (KE Zr ) + KE Zr = 2.3 × 10 −11 1.408 1 + 1 KE Zr = 2.3 × 10 −11 1.408 KE Zr = v Zr =
(vi)
2.3 × 10 −11 = 1.345 × 10 −11 1.71 2 × 1.345 × 10 −11 97.9197 × 1.66 × 10 −27
= 1.27 × 10 7 m s -1
Assumption 1: The speed of the slow neutron is neglected so that the initial momentum of the system can be taken as zero. Assumption 2: The speed of the ejected particles << c. This will allow for the use of the classical (non-relativistic) equation for momentum.
