1
2012 A Levels
H2 Physics
Suggested Solutions
Anderson Junior College
1 1 1
 
f u v
Using extreme values of u and v,
When smallest values of u and v are used,
1
f
 37.87 mm
1
1

47 195
When largest values of u and v are used,
1
f
 42.11 mm
1
1

53 205
f  ( 42.112  37.87 ) / 2  2.1 mm
Paper 1 (40 marks)
1
B
6
C
11
A
16
A
21
A
26
B
31
D
36
A
2
A
7
C
12
B
17
D
22
B
27
A
32
A
37
C
3
D
8
B
13
A
18
C
23
D
28
C
33
B
38
C
4
D
9
B
14
D
19
B
24
D
29
A
34
D
39
D
5
D
10
B
15
C
20
B
25
A
30
A
35
C
40
C
1
B
% uncertainty = 0.01/2.16 x 100 = 0.46%
→ Precise.
Zero error: 0.08/2.16 x 100 = 3.7%
→ not accurate.
2
A
1 1 1
 
f u v
1
1
1
let x  , y  , z  ,
u
v
f
 z  x  y  z  x  y
x u

x
u
u
3  1 
 x 
(x) 
   0.0012 mm
u
50  50 
y v

y
v
v
5  1 
(y ) 

  0.000125 mm
v
200  200 
z  x  y  0.0012  0.000125
 0.001325 mm
z f

z
f
z
0.001325
40   2.1mm
 f 
(f ) 
z
1 / 40
OR
 y 
3
D
s1 = ½ a1t2
s2 = ½ at22
s2 – s1 = ½ t2 (a2-a1)
t = 3 s,
12 = ½ (32) (a2-a1)
(a2-a1) = 2.67 ms-1
t = 6 s,
s2 – s1 = ½ (6)2 (2.67) = 48 m
4
D
Resolving forces along horizontal dir,
Initial Fnet = 50 – 2 (40 cos 600) = 10 N
→ Accelerate in dir XG at first.
Final Fnet = 10 – 10 = 0N
→ Moves at constant speed in dir XG
5
D
p  mv, E k  21 mv 2


p  m 2v 2  2 21 mv 2 m  2E k m
6
C
Since the forces that the spheres act on
each other are equal in magnitude and
opposite in direction (N3L), they undergo
the same rate of change of momentum
(N2L).
As they have different magnitudes of initial
momentum, they will not reach zero speed
at the same time.
H2 Physics/9646/01/ALevelsSolutions
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2
7
C
Variation in T = Tbottom – Ttop = 2mg
U
W
13
Fv
a
Constant v, so Fnet = 0
U = W + Fv
Fv = U – W
= weight of fluid displaced – weight of
air bubble = ρwVg – ρaVg
= Vg (ρw – ρw) = 2.322 x 10-4 N
8
B
At equilibrium, lines of action of all the
forces should pass through the same
point.
9
B
P = FDv, where is driving force.
At max speed, FD = resistive force FR.
FR = FD = P/v = (2.0 x 106)/ 40
= 5.0 x 104 N
At v = 10 ms-1,
Fnet = FD – FR
= P/v – FR = (2.0 x 106)/ 10 – 5.0x104
= 1.5 x 105 N
10
11
12
A
B
Taking the top position as reference point,
using COE,
Gain in EP = mg∆h = 4.2 x 9.81 x (-0.29)
= -12 J
Ep at top = Es + Ek + Ep at eqbm for 1st time
0 = Es + Ek + (-12)
12 = Es + Ek
Since Es = ½ kx2 > 0, Ek > 0
A
ω = 2π/T = 2π / (365 x 24 x 60 x 60)
= 1.99 x 10-7 rad s-1
v = r ω = 1.50 x 1011 x 1.99 x 10-7
= 2.99 x 104 ms-1
B
Since ω is constant, Fnet = mrω2 is
constant.
When stone is at the top, T is directed
downwards.
Fnet = Ttop + mg
Ttop = Fnet – mg
When stone is at bottom, T is directed
upwards.
Fnet = Tbottom – mg
Tbottom = Fnet + mg
GM
r2
GM
R2
GM
GM 1
at r  3R, g 's 

 gs
2
(3R )
9R 2 9
at r  R, a  g s 
14
D
(a) is incorrect: g field of Sun is not
dependent on presence of Earth.
(b) is incorrect: g force of Earth on the
satellite (call it S) is less than g force of
Sun on S, so the two forces do not
balance each other.
(c) does not explain: T 2 
4 2 3
r ,
GM sun
T is independent of mass of S.
(d) is correct.
Since S and Earth has same period, using
T = 2π/ω, so they have the same ω.
→ ωS = ωEarth
Centripetal force Fc = mrω2
→ Fnet on S = ms(0.99R)ωEarth2
Consider another satellite A of same mass
as S, at distance 0.99R from the Sun, but
far
away from Earth.
Fsun on S = Fsun on A = Fnet on A = ms(0.99R)ωA2
Since T2 ∝ r3, and T = 2π/ω,
rEarth > rA, so TEarth > TA, so ω Earth < ω A
Hence, Fnet on S < Fsun on S. A lower Fnet will
result in a lower ω.
H2 Physics/9646/01/ALevelsSolutions
3
15
C
22
U  Ufinal  U initial
B
v = fλ, f = v/λ
L
GMm  GMm 


  
rfinal
 rinitial 
 1
1 

 GMm 

 rinitial rfinal 
 6.67 x 10 x 5.98 x 10
-11
  4Lf 
24
1
1


x 810 x 

3
3 
6462 x 10 
 6370 x 10
8
 7.22 x 10 J
16
A
17
D
First law of thermodynamics:
∆U = Q + W → d(∆U)/dt = dQ/dt + dW/dt
Constant temperature, so
∆U = d(∆U)/dt = 0
23
D
24
D
 V  T 
P1V1 T2
x
 P1  1  2 
T1 T1
 V2  T1 
1 273  25
x
2 273  50
 4.6 x 10 4 Pa
20
26
B
Using E ∝ 1/r2 and V ∝ 1/r
27
A
I = Q/t = ne/t =(n/V)Ve/t
= 8.5 x 1028 x 3.2 x 10-7 x 0.0047
x 1.60 x 10-19 / 60
= 0.34 A
28
C
Let R be resistance of variable resistor.
Terminal pd V = E – Ir ,
I = E / (R+r)
When R↑, I ↓, so V ↑.
PV  NkT
21
B
Sound wave is a longitudinal wave, hence
it oscillates horizontally.
At displacement antinodes, there is no
variation in pressure (pressure nodes).
1
λD
x
5
a
5 λ D 5 x 600 x 10- 9 x 2.5
x

a
0.40 x 10- 3
 1.9 x 10- 2 m
A
∆V = W/q = KE/q = (½ mv2)/q
∆V ∝ v2 → v ∝ √(∆V)
B
PV
9.80 x 10 4 x 8.70
T 

Nk 1.44 x 1026 x 1.38 x 10-23
 429 K
4L
5v
f 
5
4L
25
 1.0 x 105 x
19

 nc 
2  2 sin1 

 df 


3 x 3 x 10 8
  740
 2 sin1 1 x 10-2
14


 4x103 x 6.0 x 10 
C
P2 
4L
3v
f 
3
4L
nc
f
For angle between two third order
diffraction maxima,
find 2 when n  3.
Heat is lost by the filament.
dQ/dt = d(∆U)/dt – dW/dt = 0 – 20 = –20 W
P1V1 P2V2

T1
T2

d sin  n 
Work is done by electricity when energy is
converted from electrical form to other
forms in the filament.
Rate of work done by electricity on
filament, dW/dt = 20 W
18
v
4L
By maximum power theorem,
P across R will increase and reach
maximum when R = r, then decreases.
A
H2 Physics/9646/01/ALevelsSolutions
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4
29
A
R = ρl/A
E = Pt = V2t/R = V2tA/ρl
30
A
Circuit can be redrawn into
35
Assume t = 0, and θX = 00 when coils are
at their positions as shown in the diagram.
For both options B and C, emf for X is a
positive sine function.
B
12 V
A
V
0V
C
D
Hence, emf for X is positive from t = 0 to
t = ½ T, or from θX = 00 to θX = 1800
pd across AB = pd across AD.
Hence, pd across BC = 0
31
32
33
34
Hence, emf for X is positive when θX
=1200, which is the position of Z at t = 0.
D
The three resistors on the right are
arranged in parallel. The less the number
of resistors in the circuit, the higher the
effective resistance.
A
Resistor and thermistor arranged in series,
so they have the same current passing
through.
Sum of V across each component = 3.0 V
This is fulfiled when I = 0.10 A
B
No deflection, so FB in opposite direction
as FE.
FE directed upwards, so FB directed
downwards.
Using FLHR, B directed into the paper.
D
F x 0.23 = 7.4 x 10-3 Nm
F = 3.22 x 10-2 N
F = BIL, I = F/(BL)
I = 3.22 x 10-2 / (3.6 x 10-2 x 0.093) = 9.6 A
C
emf against time should be a sinosoidal
function, so options A and D are
eliminated.
It follows that emf of Z is positive at t = 0.
Only Option C fulfils this condition.
Alternatively, since Z is ahead of X by
1200, the emf of Z is also ahead of emf of
X by 1200.
36
A
I = I0 sin (ωt)
Irms = I0/√2 = 13.4 / √2 = 9.5 A
ω = 2πf, f = ω/(2π) = 380/(2π) = 60 Hz
37
C
38
C
According to band theory, conductivity
increases with temperature because some
valence electrons acquire thermal energy
greater than EG and hence move into the
conduction band to become free electrons,
leaving behind holes in the valence band.
Both free electrons and holes are the
charge carriers of electricity.
H2 Physics/9646/01/ALevelsSolutions
5
39
D
A = A0 exp(-λt)
For X :
9
 e ( 4  X )
10
 9 
ln   4 X
 10 
9
For Y :
 e ( 2Y )
10
 9 
ln   2Y
 10 
A e ( 8  X )
A
At t  8h, X  0 ( 8 Y )
AY
A0 e

40
[e ( 4  X ) ]2

[e ( 2 Y ) ] 4
109 2
109 4

100
81
C
If both decays are alpha emissions,
Initial neutron number = 135 + 4 = 139
Initial proton number = 93 + 4 = 97
Hence, S could be the initial isotope.
If both decays are beta emissions,
Initial neutron number = 135 + 2 = 137
Initial proton number = 93 – 2 = 91
Hence, P could be the initial isotope.
If there is one alpha and one beta
emission,
Initial neutron number = 135 + 2 + 1 = 138
Initial proton number = 93 + 2 – 1 = 94
Hence, R could be the initial isotope.
H2 Physics/9646/01/ALevelsSolutions
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H2 Physics/9646/01/ALevelsSolutions