VICTORIA JUNIOR COLLEGE
SUGGESTED SOLUTIONS TO 2013
PHYSICS H2P3 PRELIM EXAMS
1(a)
2(b) By conservation of momentum
mau = mbvb + mava where ma and mb are the
mass of the bullet and block respectively.
10(100) = 9mb + 10(73)
mb = 30 g = 0.030 kg
t
t
2(c) Loss in KE = Initial KE of bullet–
Final KE of bullet and block
= (1/2)(0.010)(100)2 – (1/2)(0.030)(9.0)2 –
(1/2)(0.010)(73)2
= 22 J
d
D
D  d 0.365  0.240

2
2
= 0.0625 mm
t
Dd
1
 t  D  d 
2
2
t = 0.001 mm
t
Percentage uncertainty =
t
 100%
t
 0.001 

  100%
 0.0625 
3(a) From conservation of energy, the
elastic energy stored in the spring will be
transformed into gravitational potential
energy. Hence:
½ kx2 = mgH
 H = ½ kx2/(mg)
= ½ (500)(0.0202)/(0.005 x 9.81)
= 2.04 m = 204 cm
3(b) (i) As the spring recovers from its
2 cm depression, the stored elastic
potential energy will be transformed to
kinetic energy as well as gravitational
potential energy.
=1.6 %
1(b) t = ( 0.063 0.001) mm
= (6.3 0.1) x10-5 m
The ball bearing’s kinetic energy will
finally be all transformed to gravitational
potential energy when the ball bearing
reaches its maximum height H.
2(a)(i) Using v = u +at
va = 100 – 900(0.030) (for bullet)
= 73 m s-1
2(a)(ii) vb = 300(0.030) (for block)
= 9.0 m s-1
1
Energy /J
0.10
about the centre of mass is unbalanced,
and will cause him to topple.
epe
gpe
5(a)(i)
ke
p/ Pa
A
height
risen /m
p1
B
p2
C
0.020
2.04
V1
V2
V/ cm3
Lowest point
4.(a) R can be resolved into its vertical and
horizontal components Ry and Rx
respectively. Ry = W = mg for vertical
equilibrium.
5(a)(ii) The First Law of Thermodynamics
states that the increase in internal energy
of the system is equal to the sum of the
heat supplied to the system and the work
done on it i.e. U  Q  W .
For an isothermal expansion, ΔU = 0 since
the temperature is kept constant.
Rx provides the centripetal force and is
equal to mr(2/T)2
tan = Rx/Ry = mr(2/T)2/mg
= (83)(2/20.0)2/9.81
 = 39.9 0
4(b)
f
wall
W
For an adiabatic expansion, ΔQ = 0 and
ΔW is negative. Hence, there is a decrease
in internal energy of the gas.
As the internal energy of the gas decreases,
temperature also decreases.
Using pV = nRT, at the same volume, since
Tc < TB, PC < PB.
N
f is the frictional force, N is the normal
force of wall on motorcycle, W is the
weight of the motorcycle and man.
Q6(a) R 
L
A
1.25  10-7  0.80

 4.0 Ω
2.5 108
If the motorcyclist is oriented
perpendicular to the wall surface, the
anticlockwise moment due to friction
2
6(b)(i) P.D. across XY =
I
4.0
 5.0 V = 3.33 V
4.0  2.0
Balance length =
3.0
 0.80 = 0.721 m
3.33
6(b)(ii)V = IR, or I 
P
4r 2
4.18 x1013 x 2.50x1.6x10 13

4 (1.50) 2
= 0.591 W m-2
3.0
= 1.5 A
2.0
Section B
6(b)(iii) P.D across the XP is now
1.5 x 1.0 Ω = 1.5 V
New balance length =
1.5
 80  36.0
3.33
8(a) The speed of the wave is given by

v  f 
T

cm
= 0.360 m
Q7(a) The possible reactions include:
0.4
 0.32 m s -1
1.25
8(b) The displacement time relation for
2π
point A is y A  0.020sin(
t)
1.25
Or y A  0.020sin(5.03t )
60
Cu01n27
Co24He
63
29
60
Ni  12H  27
Co 24He
62
28
60
Co 01n 27
Co
59
27
Q7(b) The activity of the 1.00 g sample,
A  N
ln 2
1

x x 6.02x10 23 s -1
(5.27 x365x 24x3600) 60
8(c) A and B are separated by distance =
2.25 . The phase difference is equivalent
x
to a separation of 0.25.   
x 2


0.25

x 2 
π
rad  1.57 rad
2
Power of radiation from source is P = AE
= 4.18 x 1013 (1.17+1.33)(1.6x10-13) W
8(d) As the wave spreads out from a point
source, the total energy of the wave will be
distributed to and shared by the spreading
ripples.
Hence the intensity of radiation at 1.50 m
is
This results in a reduced amplitude in the
wave away from the source.
 4.18x1013 s -1
3
8(e)(i) The frequency of the oscillation of
1
0.10
the seed =
2 2.5x1 0 3
= 1.0 Hz

1
2 2
(3.95x10 3 )(
) (1.8x10  2 ) 2
2
1.25
 1.62x105 J
9(a)
8(e)(ii) The frequency of the ripples is f
=1/T = 1/1.25 = 0.80 Hz. The seed will
be set into forced oscillation at the
frequency of the ripples, which is 0.80 Hz.
As the natural frequency of the seed is
higher at 1.00 Hz, its amplitude of
oscillation is small.
9(b) Induced e.m.f. in vertical side is
E = B0lvsin
This is because there is no resonance.
8(e)(iii)1. The mass of the seed will
increase causing its natural oscillation
frequency to decrease.
At a certain critical mass when its natural
frequency has dropped to close to the
ripple frequency of 0.80 Hz, it will be
driven into resonance with increased
amplitude of oscillation.
But as its mass continues to increase with
more water taken in, its forced oscillatory
amplitude will decrease due to its much
smaller natural frequency.
8(e)(iii)2. When the seed is oscillating
with maximum amplitude, its frequency of
oscillation will be 0.80 Hz (resonance).
Its soaked mass is now given by:
1 0.10
0.80 
 m  3.95x10 3 kg .
2
m
The total energy of the seed’s oscillatory
motion,
1
E  m 2 A 2
2
(
)
9(c) Vertically downwards.
9(d) Q will be at a higher potential than P.
9(e) When t = 0,  = 0o
When t = T / 4,  = 90o
This is the e.m.f. induced in each of the
two vertical sides.
Hence, the total e.m.f. induced in the
structure would be equal to
Hence, the induced e.m.f. at time t is given
by:
E
𝐵 𝑙 𝜔
4
0
T
t
with time. Consequently, the induced
e.m.f. will also be reduced with time.
Induced e.m.f., E
t
0
9(k) The rotational mechanical energy of
the ring is converted to electrical energy
()
v
9(g)
I
A
and finally into heat energy which is
dissipated to the surroundings.
B
𝑙

F
F
𝑙
I
D
v
9(h) Torque on ring
(
)
=
10(a)(ii) n
(
)
(
)
(
10(a)(i) The square of the amplitude of the
wave-function is a measure of the
probability of locating a particle in a small
region of space. As there is no probability
of locating an electron beyond the
infinitely high walls of the box, the
amplitude of the wave-function must
vanish at the walls.
)
λ
 L ….(1)
2
10(a)(iii) The kinetic energy of the
p2
electron is given by E 
2m
E
h2
h
…..(2) (since p  )
2

2m
2L
. Substituting into (2), we
n
h2 n2
n2 h2
have E 

2m( 4 L2 ) 8mL2
In (1),  
9(i) The torque has the effect of retarding
the rotational motion of the ring.
9(j) Due to the retarding torque, the
rotational speed of the ring will slow down
5
10(a)(iv) The longest wavelength of a
photon that can be absorbed corresponds to
a transition from n = 1 to n =2 (smallest
energy difference)
h2
hc
(2 2  12 ) 
2
longest
8mL
L
3hlongest
8mc
3(6.67 x10 34 )(1.0x10 7 )

8(9.11x10 31 )(3.0x10 8 )
 3.0 x 10-10 m
10(b)(i) When an electron comes close to a
positive nucleus, it will veer in its path and
lose a certain amount of kinetic energy.
This loss of KE manifests itself as a
photon of X-ray that is emitted.
Large numbers of bombarding electrons
lose different amounts of KE in their
interactions with nuclei. This leads to a
large number of wavelengths being
emitted, resulting in the continuous
spectrum.
10(b)(ii) From the Duane-Hunt rule, the
entire KE of the bombarding electron gets
converted into the energy of an X-ray
photon.
p2
h2
hc
K


2
2m 2m
min
10(b)(iii) The first group of sharp peaks
corresponds to the K-series lines.
The energy differences for the formation
of K lines are bigger than those for L-lines,
resulting in shorter wavelengths for the
former.
10(b)(iv) One possibility is the electron
gets ejected from the atom, changing it
into an ion.
If the outer electron is excited but not
ejected from the atom, it may de-excite but
produce a photon of longer wavelength
compared to the K or L series lines.
10(b)(v) The minimum X-ray wavelength
originates from an incoming electron
losing all its energy in its interaction with
a target nucleus. The wavelength thus
depends only its initial KE and therefore
only the tube voltage and is independent of
the nature of the target nucleus.
Characteristic X-rays are formed from
electron transitions within atoms. Different
characteristic wavelengths are due to
different energy levels within atoms.
Hence characteristic wavelengths depend
on the nature of the target atom.
******** END ********
where  is the de Broglie wavelength.
2mc2
min 
h
2(9.11x10 31 )(3.0x10 8 )(8.0x10 12 ) 2

6.63x10 34
-11
5.27 x 10 m
6