2013 Physics H2 Mock P3 solution
Section A
1(a)
Random errors are errors with different magnitudes and signs in repeated measurements, and
can be reduced by taking the average of many readings.
Systematic errors are errors of measurements that occur according to some 'fixed rule or
pattern' such that they yield a consistent over-estimation or under-estimation of the true value.
(b)
k

F
mg

e (x2  x1 )
0.020 x 9.81
(36.30 32.00)
 4.563 N m1
k

k

m

e
m
e
1 0.1


 0.073
20 4.3
k  0.073 x 4.563 = 0.3 N m-1 (to 1 sf)
Hence k = (4.6  0.3) N m-1

(c)
2 (a)
Advantages of many sets of readings and drawing graph: (state any 3)
1.
By drawing best fit line, random errors are reduced.
2.
Systematic error can be spotted eg. if the F-e graph does not pass
through origin.
3.
Can spot erroneous points that are out of the trend.
4.
Can check if the proportionality limit is exceeded eg. if the F-e
graph turns into a curve instead of a straight line.
(i)
Incident waves travel along the wire to the ends are reflected.
The incident and reflected waves travelling in opposite directions have same frequency
and amplitude. They superimpose to form standing waves.
(ii)
v = fλ = 50 (0.80)= 40 m s-1
1
(iii)
0.40 m
v = fλ
40 = f (0.40)
f = 100 Hz
0.40 m
v = fλ
40 = f (2/3 x 0.40)
f = 150 Hz
(b)
(i)
When the support is shifted, the natural frequency of the wire is changed and no longer
matches that of the driving frequency of the periodic force produced by the alternating
current in the wire and magnetic field.
The system no longer resonates and hence its amplitude decreases.
The frequency of the a.c. source must therefore be the natural frequency of the wire
which is 50 Hz.
(ii)
The length l can be reduced to 0.20 m to double the fundamental frequency.
The weights attached to the wire can be increased to increase the speed of the waves to
80 m s-1.
3 (a)
In a system of interacting bodies, the total momentum of the
system remains constant provided no external force acts on the system.
(b)
An elastic collision is a collision in which the kinetic energy of the system
is conserved.
(c)
(i) Strain energy stored = .


1 2 1
kx  2.0  10 3 2.0  10  2
2
2

2
 0.40 J
(ii) By Conservation of Energy,
KE of block after collision = Strain energy stored in spring
2
1
1
 mv 2  0.40  (1.0)v 2  0.40
2
2
1
v  0.894 m s
(iii) By conservation of momentum, m1u1  m2u2  m1v1  m2 v2
(0.2)u1  0  (0.2)v1  (1.0)v2
For elastic collision, u1  u2  v2  v1
u1  0  v2  v1
 v1  v2  u1
 0.2u1  0.2(v2  u1 )  v2
0.4u1  1.2v2  u1 
1.2
(0.894)  2.68 m s 1
0.4
Speed of ball = 2.68 m s-1
(iv) By Conservation of Energy,
Gain in KE of ball = Loss in GPE of ball
1
 mv 2  0  mg (l  l cos )
2
v2
2.68 2
 (1  cos ) 

2 gl 2(9.81)(1.0)
  50.7
4
(a)
The resistance of a conductor is the potential difference across the
conductor per unit current flowing through it. ( R 
V
)
I
The ohm is defined as the resistance of a conductor such that it experiences a voltage
per unit current of one volt per ampere.
3
(b)
Using R 
l
for same density of material,
 6 - - - - - - - - - - - - - (1)
A
 (3l )
R' 
- - - - - - - - - - - - - (2)
A
3
'
R
(2)
9
(1) R
R '  9 x 6.0  54 
 
(c)
original area,
A
m
l
new area,
A' 
m
A

 (3l ) 3
C
X
160
B
1000
1000
Y
(i)
A
If the safe power rating is 0.40 W,
Using P 
V2
, the maximum safe voltage for
R
V2
1000
V  20 V
1000  :
0.40 
160  :
V2
0.40 
160
V  8V
4
1000
V XY  I (
 160) ,
2
I 
V XY
500  160
V AB  IR AB
 V XY

20  
  500 ,
 500  160 
V XY  26.4 V
if VAB  20 V ,
 V XY

8
 160 ,
 500  160 
V XY  33 V
if VBC  8V ,
 Max safe voltage would be 26.4V
Or
By Potential divider principle
2 V
XY
500  160
500
20 
 V XY
500  160
VXY  26.4 V
V AB 
if VAB  20 V ,
1000
(ii)
if VAB  8V ,
8
VXY
(ii)
160
 V XY
500  160
 33V
If this potential difference were exceeded, one of the 1000 resistor would most likely
fail. This is because when VXY exceeds 26.4 V , the max safe power for the 1000  would
be exceeded first.
Section B
5(a)(i) Magnetic flux density is defined as the force per unit current per unit length acting on a straight
conductor that is placed perpendicularly to the field.
SI units: tesla (T)
5
(ii)
(b)(i)
F
IL
N
Units of B =
Am
kg m s -2
=
= kg s -2 A -1
Am
B
Gain in KE = Loss in EPE
1
2
mv 2  qV
v
2qV
(shown)
m
(ii)
The ion will experience a force that acts perpendicularly to its velocity.
This force will cause the velocity to change its direction but not its speed.
Consequently, the ion moves in a circular path.
(iii)
The direction of the B-field is out of the plane of the paper.
(iv)
Magnetic force on ion provides the centripetal force
mv 2
r
mv
r=
(shown)
Bq
Bqv 

(v)
Since x is the diameter, ie. x =2r, then from (iv)
x = 2r =
2mv
Bq
From b(i), substitute for v we get
x=
=
c(i)
2m 2qV
Bq
m
8mV
B2q
Since 14C is more massive than 12C, then from the equation in b(v) 14C would have a larger
diameter in path than 12C. Thus it would hit the photographic plate at two spots.
6
8 x 14 x 1.66 x 10 -27 x 4000
0.50 2 x 1.6 x 10 -19
= 0.136 m
[1]
(ii)
For 14C , x1 =
8 x 12 x 1.66 x 10 -27 x 4000
For C , x 2 =
0.50 2 x 1.6 x 10 -19
= 0.126 m
12
Hence, x = 0.136 – 0.126 = 0.010 m
(d)
If an electron were to be introduced, it would deflect to the left because electron is negative
charge.
The diameter of its path would also be much smaller since the mass of electron is only
9.11 x 10-31 kg.
6 (a)(i)
Binding energy is the energy released in forming the nucleus from its
component nucleons.
(ii)
Mass defect, m = 8(1.007276) + 8(1.008665) – 15.990527
= 0.137001 u


Binding energy, mc 2  0.137001  1.66  10 27 3.00  108

2
 2.04  10 11 J
 127.6 MeV
 128 MeV
(iii)
126.43
 7.43 MeV
17
127.6
 7.98 MeV
Binding energy per nucleon for 168O 
16
Binding energy per nucleon for
17
8
O
Since 168O has a higher binding energy per nucleon, it would be more
stable.
(b)(i)
40
19
40
K 18
Ar  10
7
(ii) The probability that a particular potassium-40 nucleus will decay
within a year is 5.3 x 10-10.
A
(iii)
Ao
Ao/2
0
(iv)
t1/2
t
M  M 0 e  t
1 0
0.1  (0.1  0.84)e  ( 5.310
)t
 t  4.23 10 9 years
(v) All the argon-40 is formed from the decay of potassium-40.
(vi) Activity,
A  N 
 0.110 6

5.3 10 10

 6.02 10 23 
365  24  60  60  40

 0.025 Bq
Since the activity is quite low, simple safety precautions would suffice.
7
(a)
hf is the energy of the incident photon
 is the work function, which is the minimum energy required to cause photoelectric
emission
1 2
mv is the maximum kinetic energy of the emitted electrons
2
(b)
Work function,  
hc
th
, where th is the threshold wavelength.
hc (6.63  1034 )(3.00  108 )
So, th 

 3.68  107 m = 368 nm

3.38(1.60  1019 )
Since 540 nm is above th, the incident photons do not possess sufficient energy to
cause photoelectric emission.
8
(c)
(i)
Momentum of an incident photon,
p
(ii)
h


6.63  1034
 1.23  1027 kg m s1
540  10 9
Incident power on the metal surface, P = Intensity  Area
= (1.2 mW m2)(1.4  104 m2)
= 1.68  107 W
Energy of a single incident photon, E 
hc


(6.63  1034 )(3.00  108 )
540  10 9
= 3.683  1019 J
No. of photons incident per unit time, n =
(iii)
P
1.68  107

 4.56  1011 s1
E 3.683  1019
Magnitude of the change in momentum of each photon as it is absorbed,
p = | pf  pi | = | 0  1.23  1027 | = 1.23  1027 kg m s1
 p 
 = n  p
 t 
Force exerted on the surface, F = nt 
= (4.56  1011)(1.23  1027)
= 5.61  1016 N
(iv)
Suppose the piece of foil has a mass of 1 g, the acceleration resulting from F
would be
F 5.61  1016
a 
 5.61  1013 m s2
m
1  10 3
This is extremely small, so the effect on the foil would be negligible. Hence, the
force would not be a practicable means to move the piece of foil.
(d)
(i)
Momentum of the electron, p 
h


6.63  1034
 1.326  1025 kg m s1
5.0  10 9


2
p2
1.326  10  25

 9.65  1021 J
Kinetic energy of the electron, K 
2m 2(9.11  10  31 )
9
(ii)
The de Broglie wavelength of a particle may be expressed as:  
h
h

p
2mK
If we increase the kinetic energy K of the particle, the above relation tells us that 
would be reduced.
So, the electrons can be accelerated through a larger potential difference to achieve a
greater K, so that  is smaller.
If we use a more massive particle (larger m), then  would also be reduced.
So, instead of electrons, we can accelerate more massive particles like protons to
achieve a shorter .
10