Solution to 2008 TJC Physics Prelims H2 Paper 3
1
(a)
(i)
strain energy 

1 2 1
kx  800  3.5  10 2
2
2

[1]
2
 0.49 J
[1]
By conservation of momentum: 0.80 x v1 = 2.40 x v2
[1]
(ii)
v1
3
v2
By conservation of energy: 0.49 
Sub (1) to (2):
0.49 
(b)
2
(a)
(b)
(i)
(ii)
1
0.80v12  1 2.40v 22 --------(2)
2
2
1
0.80 v 12  1 2.40  1 v 1 
2
2
3 
v 1  0.96 m s
----------(1)
[1]
2
[1]
[1]
1
- Since initial momentum of magnets is zero, when released magnets
move off in opposite directions with equal momentum such that final
momentum of the system is zero.
[1]
- Potential energy is stored in the magnetic field when the magnets are
pushed
together, and is converted into kinetic energies of the magnets when the
magnets are released.
[1]
The idea that more energy has to be supplied to separate liquid molecules
into gaseous state than to break solid bonds, hence there is a greater
change in PE during change from liquid water to gaseous than from solid
ice to water.
[1]
and during vaporisation work is done against atmospheric pressure
[1]
in Experiment 1, volume constant so no work is done
all energy absorbed raises temperature of the gas or becomes internal
energy i.e. U = Q
[1]
in experiment 2, work done in expanding the gas or work done by gas =
pV
less energy to produce temperature rise or less energy becomes internal
energy or internal energy change is lower.
1 mark for each correct answer with relevant workings shown.
Expt 1 : W = 0, so U = Q = 150 J
Expt 2: U  T => U = 18/29 x 150 = 93 J
W = 93 – 150 = -57 J
Experiment 1
U/J
+150
W/J
0
Q/J
+150
Experiment 2
93
-57
+150
[1]
[4]
2
3
(a)
The 2 conductors A and B are both negatively charged.
[1]
This is because the total electric potential between the line joining their centres
are negative for all points between them.
[1]
(b)
V/V
A
B
x/ m
-5.0
VB
-45.0
VA
-80.0
The separate potentials VA and VB due to charged conductors A and B are
as shown above.
The total potential at a point between them is the scalar sum of the
individual potentials.
Maximum potential occurs at P, where the scalar sum of the potentials is a
maximum. (P does not correspond to the point where the two potentials
intersect.)
[3]
(c)
Let x be the distance between point P and the centre of conductor A, and
y be the distance between point P and the centre of conductor B.
QA
4o x 2
1 QB
Electric field strength at point P due to B alone , E B 
4o y 2
Electric field strength at point P due to A alone , E A 
At point P, resultant E field = 0

E A  EB

1
[1]
[1]
Q A QB
 2
x2
y
Since x < y ,  QA < QB
[1]
3
4
(a)
For the 4 lamps in parallel, Rt = 30/4 = 7.5 .
(b)
Placed parallel to the four lamps so as to reduce the effective [1]
resistance of the circuit
(c)
For maximum power, Req = 0.50 = R(7.5) / R + 7.5
R = 0.54 
[1]
[1]
(d)
With R in parallel, Req = 0.50 , hence Vt = 6.0 V
[1]
New total power Pt = Vt 2 / Rt = 6.02 / 7.5 = 4.8 W
[1]
He did not achieved his aim because the new total power of the
lamps is less than before(16.9 W)
[1]
Most of the alpha particles passed straight through or have small
deflection.
so the atom must have mostly empty space and the nucleus is small target
[1]
about 1 in 8000 were deflected at large angles. Deflection too large to be
gravitational
so must be electrostatic repulsion i.e. nucleus is positively charged.
[1]
(d)
5
(a)
(i)
(ii)
(b)
(i)
(ii)
[1]
[1]
[1]
There is no external forces present, so momentum is conserved in this
system. Since initial total momentum of the positron and electron is zero,
the final total momentum of the 2 gamma photons must be zero, that is,
they have equal and opposite momentum and hence should move in
opposite directions with equal speeds and kinetic energies
[1]
2 hf = 2 me c2
f = me c2/h = 9.11 x 10-31 x (3.00 x 108)2/6.63x10-34
= 1.24 x 1020 s-1
[1]
[1]
[1]
[1]
4
Section B
6
(a)
(b)
(i)
(ii)
(iii)
Gravitational potential at a point is the work done per unit mass by an
external agent in bringing a body from infinity to that point (without any
change in its kinetic energy.
Potential at infinity is assigned value of zero
work done to bring a mass from infinity to surface is negative.
dV
g
dr
 4.0 x10 5  ( 5.0 x10 5 )
g
0.6 x10 3
[2]
[1]
[1]
[1]
[1]
[1]
= -167 N kg-1
Using surface and first line,
GM
R
GM
 4.0  10 5  
R  600
 5.0  10 5  
[1]
[1]
working
Solving, R = 2400 m
Sub R into either eqn, M = 1.8 x 1019 kg
(iv)
COE:
(ii)
Take
1
GMm
mv 2  
2
R
2GM
v
R
 0
[1]
[1]
[1]
[1]
v = 1000 ms-1
ω = 1.0 rev min-1
= 2π/ 60
= 0.1047 rad s-1
[1]
[1]
m r ω2 = m g
r = 9.81 / 0.10472
= 8.95 x 102 m
[1]
[1]
5
6
(c)
(i)
[1]
(ii)
As the 'tube' rotates, the outer 'tube' wall constantly pushes him inwards
with normal contact force N towards the centre of rotation (centre of
colony), and this force acts as the centripetal force for its circular motion.
[1]
According to Newton's 3rd law, the man must exert an equal and
opposite force on the outer 'tube' wall, this appears as the 'weight' of the
man under simulated gravity.
[1]
Take
ω = 1.0 rev min-1
= 2π/ 60
= 0.1047 rad s-1
[1]
m r ω2 = m g
r = 9.81 / 0.10472
= 8.95 x 102 m
[1]
[1]
6
7
(a)
(i)
Waves that have a constant phase difference between them.
[1]
(ii)
The same wavefront reaches slit S1 and S2 at the same time.
[1]
(iii)
Path difference=
(iv)
[1]
3

2
3
x  12mm
2
[1]
 x = 8.0 mm
x 
D
a
[1]

 3.0  108 

  1.5
4.7  1014 

a
 1.20  10 4 m
3
8.0  10
[1]
(v) Each wavetrain is of intensity I  A2 where A = amplitude
At O, constructive interference occurs  resultant amplitude = 2A
Hence, resultant intensity at O Ir  (2A)
[1]
2
 Ir = 4 I
At the dark fringe, the intensity is zero since the resultant amplitude
cancel out due to destructive interference.
[1]
On the whole, energy is distributed over the dark and bright fringes and the
average intensity is
½ (4 I + 0) = 2 I
[1]
[1]
which is equal to the sum of intensities from the two slits
(vi)
I+I= 2I
Law of conservation of energy is not violated!
A continuous coloured spectrum will be observed for each order, with the
violet end of each spectrum nearer to the central maximum and the red
end furthest away, since the different wavelengths of white light is
diffracted at different angles.
The zeroth order will appear as a narrow white fringe for all colours are
undeviated.
(b)
(i)
[1]
[1]
1
p2
eV  mv 2 
2
2m
1.6 10
19
p2
 500 
2  9.1110 31
P = 1.21 x 10-23 kg m s-1
h 6.63  10 34

p 1.21 10 23
= 5.48 x 10-11 m

[1]
[1]
[1]
7
(ii)
For 1st order,
1 = 0.00578
th
d sin 1  
2.50  10 2
sin 1  5.48  10 11
4.60  10 4
[1]
o
For 10 order,
2.50  10 2
sin 1  10  5.48  10 11
4
4.60  10
10 = 0.0578o
(iii)
[1]
Average angular separation between adjacent orders =
[1]
10  1
10
 0.00520
[1]
Since the angular separation between adjacent orders is very small, the various
orders cannot be resolved by naked eyes and the effects are not likely to be
[1]
observed in practice.
8
(a)
(i)
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 lose 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 eject inner shell electrons.
(ii)
(b)
(i)
(ii)
(iii)
An electron from a higher energy level drops to the lower vacant state [4]
resulting in a discrete package of energy being released.
[1]
nE neV
P

t
t
[1]
600  100 n( 1.6  10 19 )10000

99
t
[1]
n /t = 3.8 x 1017 s-1
Work function is the minimum amount of energy required for an
electron to escape from the surface of a metal.
Voltmeter reading increases since resistance across AS increases.
Thus, the potential of plate C is increased so that the
photoelectrons must do more work to reach the anode.
The milliammeter reading will decrease.
KE = hv - 
3  10 8
 3  10 19
= 6.6  10 34 
7
4.0  10
= 1.95 x 10-19 J = 1.22 eV
[1]
[1]
[1]
[1]
[1]
[1]
[1]
8
Using x p ≥h/4
1.0 × 10-10)p ≥6.63 × 10-34/4
p ≥5.3× 10-25 Ns
p2
( 5.3 x10 25 ) 2
=
2m
2x 9.1x10 31
[1]
=1.53 x 10-19 J = 0.95 eV
[1]
Energy E =
(ii)
[1]
The sharp tip of the STM is placed at approximately one nanometer
from the sample with a potential difference between them.
[1]
The tunneling current due to the electrons tunneling is kept constant
by electronically controlling the tip, so that the tip-surface distance is
kept constant.
[1]
The movement of the tip is recorded and can be displayed as an
image of the surface topography.
[1]