Solutions to 2010 Prelim H2 9646/9745 Physics Paper 2
1
(a)
Distance = Area under the graph =
1
(6  12)0.6
2
[1]
= 5.4 m
[1]
(b)
1
Max height =Area under the graph =  1.0  10  5.0 m
2
(c)
a = Gradient=
(d)
From graph, time taken to reach max height on rebound = time taken to next reach
ground = 1.0 s.
Thus, time taken to next reach ground = 1.0 + 1.6 = 2.6 s
[1]
[1]
12  6
 10 m s 2
0.6
[1]
(e)
[2]
s/m
0.6
2.6
0
t/s
0.4
1 m for correct shape,
1 m for correct values
5.4
2) (a) Simple harmonic motion is an oscillatory motion of a particle whose acceleration is
directly proportional to its displacement from that equilibrium point and the
acceleration is always directed towards that position.
[1]
(b) (i)
Ep vs s
1 mark for the correct shape (max EP = 0.20 J at s=  0.16 m and min 0 at center)
1 mark to show the crossing of curves at E = 0.10 J

2
(b) (ii)
From Fig. 2.2, Max E K  0.20 J =
0.40
 =
f=
(b) (iii)
3 (a)
1
2
m 2 x 0
2
m x0
2
= 5.1 rad s -1
[1]
[1]
 5 .1

= 0.81 Hz
2 2
Fmax = mamax = m2xo
= 0.50x5.12x0.16
= 2.5 N
[1]
[1]
Magnetic flux density at a point is defined as the force acting per unit current per unit
length of conductor when the conductor is placed at right angles to the field.
[2]
3 (b) (i)
FB = FC
Bqv  m
v2
r

[1]


Bqr
6.5  10 4 1.6  10 19 0.083 
v 
m
9.11 10 31
v  9.5  10 6 m s 1
[1]
[1]
3 (b) (ii)
Correct trace [1]
Diameter of circle
roughly 166 mm
[1]
magnetic field
down into paper
-200
-100
0
100
200 mm
3
3 (b)(iii) Component of velocity along the field caused zero force so electron moved constant
velocity along the field.
[1]
Component of velocity perpendicular to the field caused centripetal force so electron
moved in a circular direction.
[1]
4 (a)
(i)
(b)
(i)
The increase in internal energy of a gas is equal to the sum of the heat absorbed
by the gas minus the work done by the gas.
P=Patm +hρg
=1.0×105 + 14 1000  9.81
[1]
=2.4×105Pa
(ii)
PsVs = PV
[1]
4  0.030 
1.0×105 Vs  2.4×105 × π 

3  2 
 Vs=3.4×10-5 m3
(iii)
[1]
[1]
[1]
P/Pa
2.4 x 105
1 m correct line and arrow
1 m for correct labeling
T = 298K
1.0 x 105
-5
1.4 x 10
(iv)
3
-5
V/m3
3.4 x 10
At constant temperature, T = 0. The air bubble is assumed to be an ideal gas,
so U = 0. Since volume of air bubble increases, positive work is done by the air
bubble. From the relation stated in (i), U = Q - wb => Q = wb is positive. Hence
heat must be added.
[2]
4
5 (a)
Conduction Band
1 - 2 eV
Valence Band
Diagram [1]
In semiconductors, at low temperature, there is small energy gap (about 1 to 2 eV)
between the valence band and the conduction band.
[1]
As temperature increases, electrons in the valence band acquire sufficient energy to be
promoted across the smaller "energy gap" into the conduction band, which leads to
conduction of electricity.
[1]
(b)
Due to concentration gradient, the free electrons in the n-type will diffuse across the p-type
and the holes from p-type diffuse across the n-type. This creates an internal electric field
near the junction that will stop further diffusion of electrons across the junction. This region
at the junction is known as the depletion region.
[1]
In forward bias condition, when emf of source is greater than the junction voltage,
electrons from n type are able to cross over to the p-type and vice versa for holes in p-type
resulting in a narrow depletion region and conduction of current.
[1]
In reversed biased condition, the electrons and holes moves away from the junction to
produce a wider depletion region and prevents conduction of current.
[1]
6
(a)
Activity is the rate of disintegration of a radioactive nuclide.
2.0 × 10
M
L =
0.239
Mm
(b)
N=
(c)
A = N
 =
(d)
(i)
 6.02  1023 = 5.0  1018
[1]
[1]
4.4 × 10
6
5.0 × 1018
Nu =
N
No
[1]
-6
=
= 8.8  10-13 s-1
[1]
6.0 × 10 -6
 6.02  1023 = 1.5  1019
0.235
5.0 × 1018
5.0 × 1018 + 1.5 × 1019
[1]
[1]
= 0.25 (shown)
(ii)
N = No e-t  t = -
1
N
ln
λ
No
1
=-
8.8 × 10 -13
[1]
ln 0.25
= 1.6  1012 s
= 5.1 x 104 years
[1]
[1]
5
7(a) (i)
(ii)
I=
P 20

 1.7 A
V 12
R=
[2]
V 2 12 2

 7.2 
P
20
[1]
7 (b)
A
correct placement of voltmeter
and ammeter
[1]
switch connected
in series
[1]
rheostat used as potentiodivider
[1]
V
7 (c) (i) The potentiometer in the lamp circuit in Fig. 7.1 is set to its minimum resistance so that
the lamp is dimmest. The resistance of the LDR is thus greatest and the current in the
LDR circuit is at its minimum.
[1]
The resistance of the potentiometer is then slowly increased so that the illumination of
the lamp is increased, the resistance of the LDR decreases and the current in the LDR in
Fig. 7.2 is increased to 10 mA.
[1]
(ii)
7 (c) (iii)
(iii)
minimum resistance of the LDR =
emf
1.50

 150 
current 10  10 3
lg R = lg (150) = 2.18
From graph, this correspond to lg L = 3.0
Thus L = 1000 W m-2
when L = 10 W m-2,
lg (10) = 1
this
corresponds to lg R = 4
[2]
[1]
[1]
(from graph)
R  10000 
emf 1.50
 4  0.15 mA
Current in the circuit =
R
10
No. Cannot detect as milliammeter has a precision of up to  0.5 mA.
[1]
[1]
[1]
6
8. Planning Question Mark Scheme
Defining the problem
Independent variable is p.d. VAB
Dependent variable is count rate C
Control of variables : keep the following constant
distance from the radioactive source to the mica window
activity of the source
Method of data collection
1
1
diagram showing voltmeter and ratemeter/scalar/datalogger connected across
AB + correct circuit diagram with variable d.c. supply
1
radium or cobalt source used + reason of choice – must be related to long half
life(this mark is not given if choice of source is wrong)
1
Method of removing  or  radiation (depending on source used)
Appropriate absorber is expected. Accept ‘aluminium’ or thin lead
Could be shown on the diagram. Allow electric or magnetic deflection
1
Measure Count Rate using ratemeter/scalar/datalogger connected across AB
Method of analysis
1
Plot appropriate graph (C vs VAB)
Safety considerations (any 2)
1
2
wear gloves/use tongs to handle radioactive source
store source in lead boxes when not in use
do not point source at people/do not look directly at source
Do not allow ‘protective clothing’, ‘lead suits’, ‘lead gloves’, ‘goggles’, etc.
Additional details (any 2)
2
Repeat readings (to allow for randomness of activity) or long time
Sensible value of p.d. applied to GM tube (i.e. 50 V to 1000 V)
Subtract count rate due to background radiation
Aluminium sheets must be mm or cm thickness
Diagram:
1