Answers to VJC Prelim 2010 - H2 Physics Paper 2
1(a)
(b)
Electric potential energy of a charge at a point is defined as the work done in
bringing the charge from infinity to that point.
(i)
(ii)
Electric potential energy
I assumed that the uranium nucleus remains at rest.
By conservation of energy,
k.e. of proton  electric potential energy
qq
1 2
mv  1 2
2
40 r




1
1.60  10 19 92  1.60  10 19
1.67  10 27 v 2 
2
4 8.85  10 12 1.5  10 14





v  4.1 10 7 m s 1
Initial speed of proton of proton = 4.110 7 m s 1
(iii)
2 (a)
The alternative particle would be the neutron.
The neutron has no charge, thus it can penetrate the nucleus easily without
being repelled.
Faradays’s law of electromagnetic induction states that the magnitude of
induced emf E is proportional to the rate of change of magnetic flux
linkage.
(b)
Given that B = 2.0 × 10-3 T
(i)
1.
From P to Q, there is no change in flux linkage of the coil,
average induced e.m.f. = 0
2.
Time for coil to move out of the magnetic field,
diameter of coil 5.0  102
t

 3.3  103 s
v
15.0
1
Average induced e.m. f . E 
 NBA  0

t
t
  5.0 10  2  2 
 
50 2.0 10 T  
2
 
 

3
3.3 10
 0.059 V

-3

(ii)
1
Work done on coil from P to Q, W  (average power dissipated ) x time
E2
(t )
R
 0 (since induced e.m.f. is zero)

2
(c)(i)
(ii)
E2
(t )
R
0.059 2

(3.3  10 3 )
5.0
2.3  10 -6 J
Work done on coil from Q to R, W 
E = Blv
= (2.0 × 10-3)(5.0 × 10-2)(15.0)
= 1.5 × 10-3 V
N is at a higher potential.
2
3(a)
Ns
(Vrms)p
Np
1
 240
=
20
= 12 V
(Vms)s =
V 
2
(b)
Maximum power dissipated =

=

=
peak
R

2
2 Vrms
R
2  12
6.0
= 48 W

2
(c)
A
B
(d)
V
0
(e)
t
T
Half of the power supplied is not transferred to the resistor.
(f)
+
A
B
-
3
4(a) Kinetic energy K of the electron can be expressed as: K 
p2
2m
So, its momentum, p  2mK  2(9.11  10 31 )(5.00)(1.60  10 19 )
p  1.207  10 24 kg m s 1
Uncertainty in p, p = 1.207  10 28 kg m s 1
According to the uncertainty principle, xp 
h
4
So, the minimum uncertainty in the electron’s position is
(x)min 
h
6.63  1034

 4.4  10 7  4  107 m
 28
4p 4 (1.207  10 )
(b)(i)
Tunneling probability, T = exp(2kd)
8 2m(U  E )
8 2 (9.11  1031)(10.0  5.0)(1.60  1019 )
where k 

h2
(6.63  1034 ) 2
k  1.144 1010 m 1
So, T = exp[2(1.144  1010)(0.200 109)] = 0.0103
(b)(ii) Since R + T = 1, the reflection probability is R = 1  0.0103 = 0.99
(c) If the barrier height is reduced, the tunneling probability T will increase.
The reflection probability R is reduced. Since R is proportional to the square of the
amplitude of the reflected wavefunction, the amplitude will decrease.
4
5(a)




Atoms in a laser medium are excited to a metastable state, which is an excited
state with a longer than usual lifetime.
This continues until the number of excited atoms in the metastable state is more
than that in the lower energy state. Thus population inversion is achieved.
Eventually, one of the atoms in the metastable state de-excites to the lower energy
level, producing a photon, which triggers other atoms in the metastable state to
de-excite to the same energy level, producing photons which travel in the same
direction and are in the same phase as the triggering photon. This is called
stimulated emission.
All the photons produced in this way form the laser light.
(b)(i) Laser light is coherent.
(ii)
The photons which are produced by stimulated emission have the same phase and
frequency as the triggering photon.
(c)(i) Energy of 1 photon, E = 3.98 x 10-19 J
(ii)
Power = (Energy of 1 photon) x (no. of photons per second)
= 3.98 x 10-19 x 2.51 x 1015
= 9.99 x 10-4 W
(iii)
Intensity =
(e)
This is because the cross sectional area of the beam is very narrow.
6(a)
R1 =
power
cross sectional area of beam
9.99  10 4
=
7.07  10 6
= 1.41 x 102 W m-2
1
0.10
= 10 
1
0.01
= 100 
R2 =
10
0.001
= 10 000 
R4 =
5
100
0.001
= 100 000 
R5 =
(b)
47 
2000 
(d) Region of safe use
(d)
(c)
Potential difference / V
1000
100
10
1
Maximum current / A
0.001
0.01
0.1
1
(d)
The lower half of the graph is the region of safe use, as the values of V x I are
below 1 W.
(e)
In real life, as the current increases, the heating effect on the resistor increases.
Thus the resistance will increase, and the graph will curve downwards.
6
7.
Fixed distance
Lamp
Variable
power
supply
Solar
panel
Light meter
Colour filter
V
Solar Panel
A
Resistor
Switch
Procedure:
(a) Set up as shown in the diagram.
(b) The lamp and filter is placed at a fixed distance measured using a ruler from the
solar panel.
(c) Place a light meter next to the solar panel.
(d) With the room lights switched off, switch on the power supply to the lamp and
adjust so that the light meter registers a reading say x. (this should be adjusted to
the same value x for the other filters)
(e) With the circuit connected to the solar panel, record the current I and the p.d. V.
(f) Calculate the power generated by the solar panel, P = IV.
(g) The frequency f of the light through each filter is determined by using the
spectrometer and diffraction grating by finding the angle of diffraction  and its
wavelength. (or any other method). By taking note of the spacing d of the lines on
the grating and the order n of the diffracted light being observed, f can be
nc
calculated using dsin =
, where c is the speed of light.
f
(h) Replace the filter with another colour and repeat steps (d) to (f).
(i) Plot power generated P against the frequency f of light used.
7
Control of variables:
1) Intensity of light of different colours (frequencies) is kept fixed by
adjusting the power supply to the lamp and that the light meter registering
reading x throughout.
2) Solar panel is placed perpendicular to axis of lamp so that incident angle
of light does not affect intensity.
3) Ensure that there is no surrounding light (room light), so that all the light
falling on the solar panel is from the lamp.
Accuracy / Safety
(See marking scheme below)
Diagram shows
Setup for lamp to shine light through filter onto the solar panel
Circuit connected with ammeter in series and voltmeter in parallel
across the solar panel, with resistor.
Basic procedure
Direct light through filters at solar panel
Repeat for different coloured filters to change the frequency of light
Measurements
Measure the current and potential difference across the solar panel
Frequency of light through each filter is determined by using the
spectrometer and diffraction grating. (Measure angle of diffraction)
Control of variables
Intensity of light on solar panel constant by ensuring that the light meter
registers a constant reading OR adjusting the power supply to the lamp
OR Keeping distance between lamp and solar panel constant
OR Keeping orientation of solar panel constant (perpendicular to axis of
lamp).
Room lights switched off (Dark Room)
Switching off lamp between readings to prevent solar panel from
heating to maintain constant conductivity.
Any further detail
Calculate the power generated by the solar panel, P = IV.
Equation to determine frequency from dsinθ= nc/f
Room lights switched off (Dark Room)
Light meter placed next to solar panel.
Adjust position of lamp to get maximum intensity on solar panel
Ensure that all light on solar panel is through the coloured filters
Safety (1) concerning electrical power supply to lamp – handle with care
if voltage is high.
Safety (2) concerning lamp – handle with care if lamp gets too hot.
Safety (3) concerning light intensity – looking directly at lamp over a
long period of time.
8
D1
D2
[2]
P1
P2
[2]
M1
M2
[2]
C1
C2
C3
[max 2]
A1
A2
A3
A4
A5
A6
A7
A8
[max 4]