Solutions to 2010 Prelim H2 9646/9745 Physics Paper 3
Section A
1 (a)
head-on : collision takes place along the line joining the centres of mass of the colliding
bodies.
[1]
elastic collision: both the momentum and the kinetic energy are conserved.
[1]
(b)
v2
u1
v1
Before collision
After collision
speed of separation = speed of approach
v 2  v 1  u1  0
or u1  v 1  v 2 - - - (1)
[1]
(c)
By conservation of linear momentum,
mu1  mv1  12mv 2
 u1  v 1  12v 2 - - - (2)
(1)  (2) 2u1  13v 2  v 2 
from (1) v 1  v 2 - u1 
2
u1
13
2
11
u1  u1   u1
13
13
v 1 11

.
u1 13
1
2
2
2
m u1  v 1
v 
2
(d) required fraction 
 1   1 
1
2
 u1 
mu1
2
COM
[1]
Working [1]
Answer [1]
Thus ratio


2
 11 
 1     0.28
 13 
Equation [1]
Working [1]
Answer [1]
(e) This is a head-on elastic collision between two equal masses. After collision, the
incident neutron will stop while the target neutron moves off with the velocity of the
incident neutron.
[1]
The ratio of the final to the initial speed of the incident neutron would be zero.
[1]
The fraction of the kinetic energy of the incident neutron that is transferred to the target
neutron would be 1 or 100 %.
[1]
2
2(a)
(i)
Curve concave downwards, crossing all field lines at right angle (by visual)
(ii)
(b)
(i)
Electrons are stripped off the molecule.
[1]
Electrons and positive ions move off in opposite direction, giving rise to an electric
current.
[1]
2

12

3
2
7
1. Q  4 r E  4  8.85  10  (2.00  10 )  1.13  10
[1]
o
-9
= 5.03 x 10 C
[1]
2.
(c)
(i)
(ii)
[1]
Q
= 1.13 x 107
4o r 2
1 Q
V
 E  r = 1.13 x 107 x (2.0 x 10-3 )
4o r
= 2.26 x 104 V
E
1
1
(5.03  10 9 ) 2
4 o r 2
4  8.85  10 12 ( 4.0  10 2 ) 2
= 1.42 x 10-4 N
This Coulomb’s repulsion results in greater apparent weight of sphere.
N = mg + F
N mg  F
F
1.42  10 4

 m   8.205 
 1000
 m’ =
g
g
g
9.81
= 8.219 g
F
1
Q1Q2
[1]
[1]

[1]
[1]
[1]
Due to repulsion, more charges accumulate on the outer sides of the 2 spheres.
Thus the distance between centre of charge of the spheres is increased. [1]
The electrostatic force F will be lesser. The reading m’ will be lesser. Thus
answer in (b)(ii)1 is an over-estimate.
[1]
3
3.(a) The magnetic flux linkage through the coil is the product of the number of turns
of the coil and the magnetic flux density that is perpendicular to the coil and
the area of the coil.
[1]
(b) Field is in XY direction
[1]
(c) When the current increases, the magnetic flux density of the solenoid XY increases.
[1]
Flux linkage with the small coil increases.
[1]
According to Faraday’s law, an emf is induced in the coil.
[1]
(d)
e.m.f.
time
Fig 4.3
-cosine graph
[1]
4
4 (a)
(i)
The binding energy of a nucleus is the energy released when a nucleus is
formed from its constituent protons and neutrons.
[1]
(ii) 1.
BE per
nucleon/MeV
8.8
correct shape [1]
0
56
nucleon number A
(iii) 2 In nuclear fission, a heavy nucleus splits to give two daughter nuclei of greater
binding energy per nucleon.
[1]
The difference in the total BE results in a large amount energy release during the
process. This is a potential source of energy.
[1]
(b)(i) Energy released in one reaction = (mu+ mn – (mxe + msr + 2mn)) uc2
[1]
= [235.043929 + 1.008665 – (138.918793 + 94.919359 + 2(1.008665)) x 1.66 x 10-27 x
(3.00 x 108)2
[1]
= 2.94 x 10-11 J
[1]
(ii)
235 u of Uranium produces 2.94 x 10-11 J in each fission reaction.
2000 kg of Uranium produces, by proportion,
2000
2.94 x 10-11 x
= 1.51 x 1017 J
[1]
 27
235 x1.66 x10
Useful Energy from 2000 kg of Uranium = 0.080 x 1.51 x 1017 = 1.21 x 1016 J [1]
Amount of time to last, t = (1.21 x 1016/1.15 x1017) = 0.105 years
= 38 days
[1]
5
Section B
5
(a) In a uniform circular motion, although the linear speed remains unchanged, its
direction changes with time, hence the velocity is always changing. So the motion
is accelerated. By Newton’s second law, a resultant force must be acting on it.
Since the change in velocity is directed towards the centre of the circle, the
acceleration, and hence the resultant force is directed towards the centre of the
circle.
[2]
(b) (i)
T = Mg = 1.0 x 9.81 = 9.81 N
[1]
(ii)
Resultant force = T 2   mg 
=
 9.81
2
2
  0.10×9.81
2
[1]
[1]
=9.76 N
(iii)
2
Resultant force  mr
9.76  mr  2 f
r

2
9.76
  180  
0.10  2 

  60  
2
 0.275 m
[2]
(iii) It would not be possible for the string to be horizontal because the string needs
to make an angle with the horizontal such that there is a vertical component of
the tension to balance the weight the weight of the mass.
[2]
F G
(c)
m1m2
r2
[2]
Where
F is the gravitation force of attraction between the two point masses m1, m2;
r is the distance between the two point masses, and
G is the universal gravitational constant.
(d)
(i)
6
GMm
 2 
 mr 2  mr 

2
r
 T 
4 2 3
T2 
r
GM
3
4 2
7 2
 3.45 10   GM  2.00 1011 
M  3.98  1030 kg
2
(ii)
Concept [1]
Working [1]
Answer [1]
Since both the star and the planet have a tangential velocity, the
gravitational force of attraction provides the centripetal force for both the
star and the planet to orbit about their centre of mass.
[1]
(iii)
1
 GMm 
mv 2 +  =0+0
2
r 

v=
v=
2GM
r
2  6.67×10 -11 1.20×10 24 
Concept [1]
Working [1]
Answer [1]
 7.5×10 6 


2


=6.53×103 m s-1
(iv)
Since average speed of the nitrogen gas is higher than the escape on the
planet, most gases would have escaped and hence the planet does not
have an atmosphere.
[1]
7
6 (a)
It is a phenomenon in which waves from 2 or more coherent sources superpose
with one another producing a resultant wave.
[1]
(b)(i) Sound from the 2 sources undergo interference. As the source S1 moves, the path
difference changes.
[1]
When the path difference between S1P and S2P is an integral number of the
wavelength, constructive interference occurs at P and a maxima is detected. [1]
When the path difference between S1P and S2P is an odd integral number of half
wavelength, destructive interference occurs at P and a minima is detected. [1]
6 (b)(ii)
Path difference =


= 0.082 m
2
 = 0.164 m
S1 X = 1 = 0.082 m
(b)(iii)
v  f  4100  0.082
= 340 m s-1
(c)
[1]
[1]
[1]
[1]
[1]
(i) Wave travel down the tube and gets reflected.
[1]
The incident and the reflected waves, both having the same amplitude, frequency
and speed travelling in opposite directions superpose to form standing wave. [1]
(c) (ii)
Air column in tube has natural frequency of vibration.
[1]
When fork frequency is equal to natural frequency, resonance occurs; there is
maximum energy transfer and maximum amplitude of vibrations occurs, leading
to maximum loudness.
[1]
When fork frequency is not equal to natural frequency, no resonance occurs
and loudness drop.
[1]
(c) (iii) Sketch: Antinode at top, node at surface of water, 1 antinode and 1 node in
between
[1]
(c) (iv)

 L1  c -------------(1)
4
3
 L2  c --------------(2)
4
1
  L2  L1  32.4 cm
2
  = 64.8 cm
v  f  512  0.648  332 m s-1
(2) – (1) 
[1]
[1]
[1]
1
1
[1]
  L1  c  15.7  c  c=  64.8  15.7 = 0.50 cm
4
4
Therefore, antinode is 0.50 cm above the top of the tube OR antinode is 16.2 cm above
water surface.
[1]
(c)(v)
8
7
(a)
1. Emission line spectrum is observed is at A. It is due to the emissions of photons of
specific energy when the excited He atoms return to ground states. The emission
occurs in all directions and hence is detected at A.
[2]
2. Absorption line spectrum observed is at B. It is due to the absorption of photons of
specific energy from the white light by cool He atoms to jump to excited states.
These photons are re-emitted in all directions and hence missing from the
transmitted beam at point B.
[2]
(b) (i) The transitions are from n=3 to n=1, n=3 to n=2 and n=2 to n=1.
[2]
(ii) The longest wavelength comes form transition n=3 to n =2.
E = hc/
[-6.04-(-13.6)] x 1.60 x 10-19 = 6.63 x 10-34 x 3.00 x108/,
[1]
 = 1.64 x 10-7 m
[1]
(iii) 1. A metastable state is the state in which the electrons remain longer than
usual so that the transition to the lower state occurs more likely by stimulated
emission than by spontaneously emission.
[2]
2. Population inversion is a condition of having more atoms in an excited state
than in the lower energy state.
[1]
(c) (i) P = (N/t) hf
N/t = P/hf
= 3.8 x 10-3/4.7 x 1.60x10-19
= 5.1 x 1015 s-1
(ii) Current I = = N’e/t,
from graph, max current = 0.80x10-8 A
N’/t = 0.80x10-8 /1.60x10-19 = 5.0 x 1010 s-1
[1]
[1]
[1]
[1]
(iii) Rate of incident photons is very much larger than the rate of electrons emitted.
Most of the photons rebound/reflect from the metal surface/electrons deep below
surface requires more energy/presence of impurities on surface
[2]
(iii)
Stopping potential Vs = -0.50 V.
Maximum KE of electrons = eVs = 8.0 x 10-20 J
(v) Work function  = hf – KEmax = 4.7 x 1.60 x10-19 - 8.0 x 10-20 = 6.7 x 10-19 J
[1]
[1]
[1]