1
FOUNDATION STUDIES
EXAMINATIONS
September 2007
PHYSICS
Third Paper
February Program
Time allowed 1 hour for writing10 minutes for reading
This paper consists of 3 questions printed on 6 pages.PLEASE CHECK BEFORE COMMENCING.
Candidates should submit answers to ALL QUESTIONS.
Marks on this paper total 50 marks, and count as 10% of the subject.
Start each question at the top of a new page.
2
INFORMATION
a · b = ab cos θ
a× b = ab sin θ c =
∣∣∣∣∣∣i j kax ay az
bx by bz
∣∣∣∣∣∣v ≡ dr
dta ≡ dv
dt
v =∫
a dt r =∫
v dt
v = u + atx = ut + 1
2at2
v2 = u2 + 2ax
a = −gjv = u− gtjr = ut− 1
2gt2j
s = rθ v = rω a = ω2r = v2
r
p = mv
N1 : if∑
F = 0 then δp = 0N2 :
∑F = ma
N3 : FAB = −FBA
W = mg Fr = µR
g =acceleration due to gravity=10m s−2
τ ≡ r× F∑Fx = 0
∑Fy = 0
∑τP = 0
W ≡∫ r2r1
F dr W = F · s
KE = 12mv2 PE = mgh
P ≡ dWdt
= F · v
F = kx PE = 12kx2
dvve
= −dmm
vf − vi = ve ln( mi
mf)
F = |vedmdt|
F = k q1q2
r2 k = 14πε0
≈ 9× 109 Nm2C−2
ε0 = 8.854× 10−12 N−1m−2C 2
E ≡limδq→0
(δFδq
)E = k q
r2 r
V ≡ Wq
E = −dVdx
V = k qr
Φ =∮
E · dA =∑
qε0
C ≡ qV
C = Aεd
E = 12
q2
C= 1
2qV = 1
2CV 2
C = C1 + C2 + C3
1C
= 1C1
+ 1C2
+ 1C3
R = R1 + R2 + R3
1R
= 1R1
+ 1R2
+ 1R3
V = IR V = E − IR
P = V I = V 2
R= I2R
K1 :∑
In = 0K2 :
∑(IR′s) =
∑(EMF ′s)
F = q v ×B dF = i dl×B
F = i l×B τ = niA×B
v = EB
r = mq
EBB0
r = mvqB
T = 2πmBq
KEmax = R2B2q2
2m
dB = µ0
4πidl×r
r2∮B · ds = µ0
∑I µ0 = 4π×10−7 NA−2
φ =∫
areaB · dA φ = B ·A
ε = −N dφdt
ε = NABω sin(ωt)
3
f = 1T
k ≡ 2πλ
ω ≡ 2πf v = fλ
y = f(x∓ vt)
y = a sin k(x− vt) = a sin(kx− ωt)= a sin 2π(x
λ− t
T)
P = 12µvω2a2 v =
√Fµ
s = sm sin(kx− ωt)
∆p = ∆pm cos(kx− ωt)
I = 12ρvω2s2
m
n(db′s) ≡ 10 log I1I2
= 10 log II0
where I0 = 10−12 W m−2
fr = fs
(v±vr
v∓vs
)where v ≡ speed of sound = 340 m s−1
y = y1 + y2
y = [2a sin(kx)] cos(ωt)
(m = 0, 1, 2, 3, 4, ....)
N : x = m(λ2) AN : x = (m + 1
2)(λ
2)
y = [2a cos(ω1−ω2
2)t] sin(ω1+ω2
2)t
fB = |f1 − f2|
y = [2a cos(k∆2
)] sin(kx− ωt + k∆2
)
∆ = d sin θ
Max : ∆ = mλ Min : ∆ = (m + 12)λ
I = I0 cos2(k∆2
)
E = hf c = fλ
KEmax = eV0 = hf − φ
E2 = p2c2 + (m0c2)2
E = m0c2 E = pc
λ = hp
(p = m0v (nonrelativistic))
∆x∆px ≥ hπ
∆E∆t ≥ hπ
dNdt
= −λN N = N0 e−λt
R ≡ |dNdt| T 1
2= ln 2
λ= 0.693
λ
MATH:
ax2 + bx + c = 0 → x = −b±√
b2−4ac2a
y dy/dx∫
ydx
xn nx(n−1) 1n+1
xn+1
ekx kekx 1kekx
sin(kx) k cos(kx) − 1k
cos kxcos(kx) −k sin(kx) 1
ksin kx
where k = constant
Sphere: A = 4πr2 V = 43πr3
CONSTANTS:
1u = 1.660× 10−27 kg = 931.50 MeV1eV = 1.602× 10−19 Jc = 3.00× 108m s−1
h = 6.626× 10−34 Jse ≡ electron charge = 1.602× 10−19 C
particle mass(u) mass(kg)
e 5.485 799 031× 10−4 9.109 390× 10−31
p 1.007 276 470 1.672 623× 10−27
n 1.008 664 904 1.674 928× 10−27
PHYSICS: Third Paper. February Program 2007 4
x-axis
y-axis
3 kg
x-axis
y-axis
2 kg
1 kg 3 kg
3 m/s
v
4 m/s
Before After
1 kg
2 kg
Figure 1:
Question 1 ( 17 marks):
Three stationary bodies, of masses 1, 2, and 3 kg, are in contact, on a horizontal
frictionless surface, near the origin of x- and y-axes, as shown in Figure 1 (Before). An
explosion occurs between the bodies, which blows them apart, on the surface. The 1 kg
body moves away in the negative direction of the x-axis, with a velocity of 4 m/s, while
the 2 kg body moves away in the negative direction of the y-axis, with a velocity of
3 m/s, as shown in Figure 1 (After).
Use momentum principles, to find the magnitude and direction of the velocity, v, of the
third (3 kg) body, after the explosion.
PHYSICS: Third Paper. February Program 2007 5
2 Ω
5 Ω
3 Ω
4 V 8 V
6 V
i I
Figure 2:
Question 2 ( 12 + 4 = 16 marks):
The circuit of Figure 2 shows three emf’s with their respective internal resistances,
connected in a two loop network.
(i) Write down two Kirchoff loop equations for this circuit, in terms of unknown
currents, i and I.
(ii) Hence determine the numerical values of the two currents, i and I, as labeled.
Draw a copy of the circuit, and on it label the physical magnitudes, and directions of
these two currents.
PHYSICS: Third Paper. February Program 2007 6
L
+Qa
b
r
P
Figure 3:
Question 3 ( 3 + 14 = 17 marks):
Figure 3 shows a hollow conducting pipe, of length, L, internal radius, a, and externalradius, b (where b L). This pipe carries a total positive charge, Q, which is distributeduniformly along the length of the pipe.
(i) Draw a diagram of the pipe, showing the electric field about it.
(ii) Use Gauss’ law to derive an expression, in terms of the parameters labeled onFigure 3, for the magnitude, E, of the electric field, at a point P, which is in the middleregion of the pipe, at a radial distance of r (r > b), from its central axis. Clearly describethe Gaussian surface that you use. Be careful to justify each step of your derivation, asmarks will be given for this.
END OF EXAM
ANSWERS:
Q1. 2.40 m/s in a direction 56.3 deg above the x-axis.
Q2. (i) 12 = 2i− 3I, −14 = 5i + 8I; (ii) i : 1.74 A ↑, I : 2.84 A ↓.
Q3. (ii) E = Q2πε0Lr
.
1
FOUNDATION STUDIES
EXAMINATIONS
November 2007
PHYSICS
Final Paper
February Program
Time allowed 3 hours for writing10 minutes for reading
This paper consists of 6 questions printed on 13 pages.PLEASE CHECK BEFORE COMMENCING.
Candidates should submit answers to ALL QUESTIONS.
Marks on this paper total 120 marks, and count as 45% of the subject.
Start each question at the top of a new page.
2
INFORMATION
a · b = ab cos θ
a× b = ab sin θ c =
∣∣∣∣∣∣i j kax ay az
bx by bz
∣∣∣∣∣∣v ≡ dr
dta ≡ dv
dtv =
∫a dt r =
∫v dt
v = u + at a = −gjx = ut + 1
2at2 v = u− gtj
v2 = u2 + 2ax r = ut− 12gt2j
s = rθ v = rω a = ω2r = v2
r
p ≡ mv
N1 : if∑
F = 0 then δp = 0N2 :
∑F = ma
N3 : FAB = −FBA
W = mg Fr = µR
g =acceleration due to gravity=10m s−2
τ ≡ r× F∑Fx = 0
∑Fy = 0
∑τP = 0
W ≡∫ r2r1
F dr W = F · s
KE = 12mv2 PE = mgh
P ≡ dWdt
= F · v
F = kx PE = 12kx2
dvve
= −dmm
vf − vi = ve ln( mi
mf)
F = |vedmdt|
F = k q1q2
r2 k = 14πε0≈ 9× 109 Nm2C−2
ε0 = 8.854× 10−12 N−1m−2C 2
E ≡limδq→0
(δFδq
)E = k q
r2 r
V ≡ Wq
E = −dVdx
V = k qr
Φ =∮
E · dA =∑
qε0
C ≡ qV
C = Aεd
E = 12
q2
C= 1
2qV = 1
2CV 2
C = C1 + C21C
= 1C1
+ 1C2
R = R1 + R21R
= 1R1
+ 1R2
V = IR V = E − IR
P = V I = V 2
R= I2R
K1 :∑
In = 0K2 :
∑(IR′s) =
∑(EMF ′s)
F = q v ×B dF = i dl×B
F = i l×B τ = niA×B
v = EB
r = mq
EBB0
r = mvqB
T = 2πmBq
KEmax = R2B2q2
2m
dB = µ0
4πidl×r
r2∮B · ds = µ0
∑I µ0 = 4π×10−7 NA−2
φ =∫
areaB · dA φ = B ·A
ε = −N dφdt
ε = NABω sin(ωt)
f = 1T
k ≡ 2πλ
ω ≡ 2πf v = fλ
y = f(x∓ vt)
y = a sin k(x− vt) = a sin(kx− ωt)= a sin 2π(x
λ− t
T)
P = 12µvω2a2 v =
√Fµ
s = sm sin(kx− ωt)
∆p = ∆pm cos(kx− ωt)
3
I = 12ρvω2s2
m
n(db′s) ≡ 10 log I1I2
= 10 log II0
where I0 = 10−12 W m−2
fr = fs
(v±vr
v∓vs
)where v ≡ speed of sound = 340 m s−1
y = y1 + y2
y = [2a sin(kx)] cos(ωt)
N : x = m(λ2) AN : x = (m + 1
2)(λ
2)
(m = 0, 1, 2, 3, 4, ....)
y = [2a cos(ω1−ω2
2)t] sin(ω1+ω2
2)t
fB = |f1 − f2|
y = [2a cos(k∆2
)] sin(kx− ωt + k∆2
)
∆ = d sin θ
Max : ∆ = mλ Min : ∆ = (m + 12)λ
I = I0 cos2(k∆2
)
E = hf c = fλ
KEmax = eV0 = hf − φ
L ≡ r× p = r×mv
L = rmv = n( h2π
)
δE = hf = Ei − Ef
rn = n2( h2
4π2mke2 ) = n2a0
En = −ke2
2a0( 1
n2 ) = −13.6n2 eV
1λ
= ke2
2a0( 1
n2f− 1
n2i) = RH( 1
n2f− 1
n2i)
(a0 = Bohr radius = 0.0529 nm)
(RH = 1.09737× 107 m−1)
(n = 1, 2, 3....) (k ≡ 14πε0
)
E2 = p2c2 + (m0c2)2
E = m0c2 E = pc
λ = hp
(p = m0v (nonrelativistic))
∆x∆px ≥ hπ
∆E∆t ≥ hπ
dNdt
= −λN N = N0 e−λt
R ≡ |dNdt| T 1
2= ln 2
λ= 0.693
λ
MATH:
ax2 + bx + c = 0 → x = −b±√
b2−4ac2a
y dy/dx∫
ydx
xn nx(n−1) 1n+1
xn+1
ekx kekx 1kekx
sin(kx) k cos(kx) − 1k
cos kxcos(kx) −k sin(kx) 1
ksin kx
where k = constant
Sphere: A = 4πr2 V = 43πr3
CONSTANTS:
1u = 1.660× 10−27 kg = 931.50 MeV1eV = 1.602× 10−19 Jc = 3.00× 108m s−1
h = 6.626× 10−34 Jse ≡ electron charge = 1.602× 10−19 C
particle mass(u) mass(kg)
e 5.485 799 031× 10−4 9.109 390× 10−31
p 1.007 276 470 1.672 623× 10−27
n 1.008 664 904 1.674 928× 10−27
PHYSICS: Final Paper. February Program 2007 8
4V3V
7V
6Ω 5Ω 2Ω
i I
Figure 5:
Question 3 ( (1 + 9) + (3 + 7) = 20 marks):
Part (a):
(i) Sketch the circuit of Figure 5 in your answer book, and label on it, the current
through the 5 Ω resistor, in terms of currents, i and I.
(ii) Use the Kirchoff rules, to find the numerical values for the currents, i and I.
Hence determine the numerical value of the current through the 5 Ω resistor. Label the
physical magnitudes and directions of all three currents, on your circuit sketch.
PHYSICS: Final Paper. February Program 2007 9
3 m
4 m
P
A B(−q) (+Q)
Figure 6:
Part (b):
Figure 6 shows a negative charge, (−q), and a positive charge, (+Q), at corners, A and
B, of a rectangle. Dimensions of the figure are labeled.
(i) Derive an expression for the potential, V , at the corner, P, of the rectangle, in
terms of q, Q, and the Coulomb constant, k. (This is the potential relative to V = 0
at ∞).
(ii) Derive an expression for the electric field vector, E, at the corner, P, of the
rectangle, in terms of q, Q, Coulomb constant, k, and unit vectors i and j.
PHYSICS: Final Paper. February Program 2007 10
irP
Figure 7:
Question 4 ( (2 + 8) + (3 + 6 + 1) = 20 marks):
Part (a):
Figure 7 shows a long straight conducting wire, that carries a constant current of i.
Point P is a radial distance of r, from the wire, as shown, where r (length of the wire).
(i) Draw a clear diagram to illustrate the magnetic field near P. In particular, show
the direction of the magnetic field vector, B, at P.
(ii) Use Ampere’s law to derive an expression for the magnitude, B, of the magnetic
field vector, at point P, in terms of i, r, and the permeability constant, µ0. Clearly
describe the Amperian path that you use. Be careful to justify each, and every, step of
your reasoning, as marks will be given for this.
PHYSICS: Final Paper. February Program 2007 11
t = 0
v
b
a
t = 0
B
b
a
(a)t
v
(b)
Figure 8:
Part (b):
Figure 8 shows a flat rectangular loop, of length, a, and width, b, and total resistance,
R, being withdrawn, at a constant velocity, v, from a region of uniform magnetic field,
with field vector, B. The field is directed perpendicularly downward into the plane of
the loop. Figure 8(a) shows the loop, at time t = 0, as it just begins to emerge from the
field; Figure 8(b) shows the loop, at a later time t.
(i) Derive an expression for the total magnetic flux, φ, that cuts through the loop, at
time, t, during the loop’s emergence from the field, in terms of the parameters labeled
on Figure 8.
(ii) Use Faraday’s law to derive an expression for the current, i, induced in the loop,
at a time t, as it is being withdrawn from the field, in terms of the parameters labeled
on Figure 8.
(iii) In what direction will the induced current, i, flow in the loop?
PHYSICS: Final Paper. February Program 2007 12
M
SvP
Figure 9:
Question 5 ( (3 + 4 + 3) + (4 + 6) = 20 marks):
Part (a):
A string, of total length, l = 2.0 m, is measured to have a total mass of m = 40 gram.
One end of this string is connected to a wave source, S. The string then passes over a
frictionless pulley, and is securred, at its other end, to a block, of mass M = 0.8 kg, as
depicted in Figure 9. The wave source, S, transmits a wave along the string. Take the
acceleration of gravity, g = 10 ms−2.
(i) Calculate the velocity, v, with which the wave passes along the string.
(ii) What output power, P , would the wave source, S, need, in order to send a wave,
with frequency, f = 100 Hz, and amplitude, a = 1 mm, along the string?
(iii) Using your answers above, write down a numerical wave function for the wave
transmitted along the string from S, in Figure 9.
Part (b):
Nitrogen isotope 137 N has a half-life of 10.0 minutes. A given sample of this isotope is
measured to have an activity of 10 MBq.
(i) How many atoms of 137 N are in this sample?
(ii) How long will it take for the activity of this sample to fall to 1.0 MBq ?
PHYSICS: Final Paper. February Program 2007 13
Question 6 ( (3 + 4 + 3) + (10) = 20 marks):
The following data may be useful in answering this question -
Planck const. ≡ h = 6.626× 10−34 Js
Coulomb const. ≡ k = 8.988× 109 Nm2C−2
Rydberg const. ≡ RH = 1.0974× 107 m−1
Bohr radius ≡ a0 = 0.0529 nm
speed of light ≡ c = 3.00× 108 ms−1
electron mass ≡ m = 9.109× 10−31 kg
electron charge ≡ e = 1.602× 10−19 C
Part (a):
(i) Starting from the expression for the de Broglie wavelength, λ, of an electron, in
terms of its momentum, p, derive the quantisation rule for the angular momentum, L,
of the electron in the Bohr model of the hydrogen atom.
(ii) Calculate the orbital radii, and energies, for an electron in its lowest, and second
lowest, energy states, in the Bohr model of the hydrogen atom.
(iii) Calculate the wavelength of the spectral line produced by an electron transition
from the second lowest to the lowest energy level, in the hydrogen atom.
Part (b):
When a photoelectric cell with a copper cathode, is illuminated by light of wavelength,
λ = 245 nm, the reverse potential, that just stops the photo current, is 0.181 V . What
is the photoelectric threshold wavelength for copper?
END OF EXAM
PHYSICS: Final Paper. February Program 2007 14
ANSWERS:
Q1.(a).(i) AB−→ = (−4i + 3j) m, AC−→ = (−4i + 2k) m; (ii) A = (3i + 4j + 6k) m2.
Q1.(b). (i) u = (3i + 16j) m s−1 ; (iii) 9.60 m.
Q2.(a) (ii) a = g(M
m+ 4
5− 3µ
5)
(Mm
+1).
Q2.(b) (i) u = 2v; (ii) u =√
8gr (M−2m)(M+4m)
.
Q3.(a) (ii) i : (0.25 A ←), I : (0.25 A →); through 5 ohm : ( 0.50 A ↑).
Q3.(b) (i) V = k( |Q|3− |q|
5); (ii) E = k(−4|q|
125i + ( |Q|
9− 3|q|
125) j).
Q4.(a) (ii) B = µ0i2πr
.
Q4(b). (i) φ = bB(a− vt); (ii) i = bBvr
; (iii) clockwise.
Q5.(a) (i) 20 m/s; (ii) 79 mW ; (ii) y = 10−3 sin 10π(x− 20t).
Q5.(b) (i) 8.66× 109 atoms; (ii) 33.2 min.
Q6.(a) (i) L = n h2π
; (ii) 0.0529 nm, 0.2116 nm, −13.6 eV , −3.4 eV ; (iii) 121.5 nm.
Q6. (b) 254.1 nm.
