1 FOUNDATION STUDIES EXAMINATIONS March 2009 PHYSICS First Paper February Program Time allowed 1 hour for writing 10 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.
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1
FOUNDATION STUDIES
EXAMINATIONS
March 2009
PHYSICS
First 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 ✓
a⇥ b = ab sin ✓ c =
������i j kax ay az
bx by bz
������v ⌘ dr
dta ⌘ dv
dtv =
Ra dt r =
Rv dt
v = u + at a = �gjx = ut + 1
2at
2 v = u� gtjv
2 = u
2 + 2ax r = ut� 12gt
2j
s = r✓ v = r! a = !
2r = v2
r
p ⌘ mv
N1 : ifP
F = 0 then �p = 0N2 :
PF = ma
N3 : FAB
= �FBA
W = mg Fr = µR
g =acceleration due to gravity=10m s�2
⌧ ⌘ r⇥ FPF
x
= 0P
Fy
= 0P
⌧P = 0
W ⌘R
r2
r1F dr W = F · s
KE = 12mv
2PE = mgh
P ⌘ dWdt
= F · v
F = kx PE = 12kx
2
dvve
= �dmm
vf � vi = ve ln( mimf
)
F = |vedmdt
|
F = k
q1q2
r2 k = 14⇡✏0
⇡ 9⇥ 109 Nm2C�2
✏0 = 8.854⇥ 10�12 N�1m�2C 2
E ⌘lim�q!0
⇣�F�q
⌘E = k
qr2 r
V ⌘ Wq
E = �dVdx
V = k
qr
� =H
E · dA =P
q✏0
C ⌘ qV
C = A✏d
E = 12
q2
C= 1
2qV = 12CV
2
C = C1 + C21C
= 1C1
+ 1C2
R = R1 + R21R
= 1R1
+ 1R2
V = IR V = E � IR
P = V I = V 2
R= I
2R
K1 :P
In = 0K2 :
P(IR
0s) =
P(EMF
0s)
F = q v ⇥B dF = i dl⇥B
F = i l⇥B ⌧ = niA⇥B
v = EB
r = mq
EBB0
r = mvqB
T = 2⇡mBq
KEmax = R2B2q2
2m
dB = µ0
4⇡i
dl⇥r
r2HB · ds = µ0
PI µ0 = 4⇡⇥10�7 NA�2
� =R
areaB · dA � = B · A
✏ = �N
d�dt
✏ = NAB! sin(!t)
f = 1T
k ⌘ 2⇡�
! ⌘ 2⇡f v = f�
y = f(x⌥ vt)
y = a sin k(x� vt) = a sin(kx� !t)= a sin 2⇡(x
�� t
T)
P = 12µv!
2a
2v =
qFµ
s = sm sin(kx� !t)
�p = �pm cos(kx� !t)
3
I = 12⇢v!
2s
2m
n(db0s) ⌘ 10 log I1I2
= 10 log II0
where I0 = 10�12 W m�2
fr = fs
⇣v±vrv⌥vs
⌘where v ⌘ speed of sound = 340 m s�1
y = y1 + y2
y = [2a sin(kx)] cos(!t)
N : x = m(�2 ) AN : x = (m + 1
2)(�2 )
(m = 0, 1, 2, 3, 4, ....)
y = [2a cos(!1�!22 )t] sin(!1+!2
2 )t
fB = |f1 � f2|
y = [2a cos(k�2 )] sin(kx� !t + k�
2 )
� = d sin ✓
Max : � = m� Min : � = (m + 12)�
I = I0 cos2(k�2 )
E = hf c = f�
KEmax = eV0 = hf � �
L ⌘ r⇥ p = r⇥mv
L = rmv = n( h2⇡
)
�E = hf = Ei � Ef
rn = n
2( h2
4⇡2mke2 ) = n
2a0
En = �ke2
2a0( 1
n2 ) = �13.6n2 eV
1�
= ke2
2a0hc( 1
n2f� 1
n2i) = RH( 1
n2f� 1
n2i)
(a0 = Bohr radius = 0.0529 nm)
(RH = 1.09737⇥ 107m
�1)
(n = 1, 2, 3....) (k ⌘ 14⇡"0
)
E
2 = p
2c
2 + (m0c2)2
E = m0c2
E = pc
� = hp
(p = m0v (nonrelativistic))
�x�px � h⇡
�E�t � h⇡
dNdt
= ��N N = N0 e��t
R ⌘ |dNdt
| T
12
= ln 2�
= 0.693�
MATH:
ax
2 + bx + c = 0 ! x = �b±p
b2�4ac2a
y dy/dx
Rydx
x
nnx
(n�1) 1n+1x
n+1
e
kxke
kx 1ke
kx
sin(kx) k cos(kx) � 1k
cos kx
cos(kx) �k sin(kx) 1k
sin kx
where k = constant
Sphere: A = 4⇡r
2V = 4
3⇡r
3
CONSTANTS:
1u = 1.660⇥ 10�27kg = 931.50 MeV
1eV = 1.602⇥ 10�19J
c = 3.00⇥ 108m s
�1
h = 6.626⇥ 10�34Js
e ⌘ electron charge = 1.602⇥ 10�19C
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: First Paper. February Program 2009 4
P
Q
R
S
x
y
z
4 m
2 m
3 m
✓
Figure 1:
Question 1 ( 6 + 11 = 17 marks):
Figure 1 shows a rectangular box, aligned with the x-, y-, and z-axes. The dimensions
of the box are labeled.
(i) Write down expressions for the diagonal vectors PQ�! and RS�!, in terms of unit
vectors ijk.
(i) Using the dot product (PQ�!•RS�!) determine the angle ✓, between PQ�! and RS�!.
PHYSICS: First Paper. February Program 2009 5
m
g
k
Figure 2:
Question 2 ( 16 marks):
Figure 2 shows a spring, of spring constant, k (N/m), with one end attached to a fixed
beam. A mass, m (kg), hangs from the other end of the spring. The mass is pulled
down and released, so that it oscillates vertically up and down, at the end of the spring,
with a period of P (s). Assume that P could depend on k, m, and the acceleration due
to gravity, g (m s
�2).
Note : The spring constant, k, is the force, in Newton, required per metre of extension,
to extend the spring.
Note : N = Newton = kg m s
�2.
Use dimensional analysis to derive an equation, giving P in terms of some, or all, of k,
m, and g.
PHYSICS: First Paper. February Program 2009 6
x
y
R
M
vR
✓
vM
✓ = 30 deg
vR = 20 m/s
vM = 40 m/s
Figure 3:
Question 3 ( 17 marks):
A rocket, R, moves vertically upward with a velocity of 20 m/s, while a meteorite, M ,falls downward, with a velocity of 40 m/s, at an angle of 30 deg, to the horizontal.These measurements are relative to the xy-axes, and are illustrated in Figure 3.
Calculate the relative velocity of the meteorite, M , as seen by the rocket, R. Givemagnitude and direction.
END OF EXAM
ANSWERS:
Q1. (i) PQ�! = �4i + 3j� 2k m, RS�! = +4i + 3j� 2k m ; (ii) 95.9 deg.
Q2. P = C
pmk, where C is a dimensionless constant.
Q3. 52.9 m/s at an angle of 49.1 deg below the (-x)-axis.
1
FOUNDATION STUDIES
EXAMINATIONS
June 2009
PHYSICS
Second 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 ✓
a⇥ b = ab sin ✓ c =
������i j kax ay az
bx by bz
������v ⌘ dr
dta ⌘ dv
dtv =
Ra dt r =
Rv dt
v = u + at a = �gjx = ut + 1
2at
2 v = u� gtjv
2 = u
2 + 2ax r = ut� 12gt
2j
s = r✓ v = r! a = !
2r = v2
r
p ⌘ mv
N1 : ifP
F = 0 then �p = 0N2 :
PF = ma
N3 : FAB
= �FBA
W = mg Fr = µR
g =acceleration due to gravity=10m s�2
⌧ ⌘ r⇥ FPF
x
= 0P
Fy
= 0P
⌧P = 0
W ⌘R
r2
r1F dr W = F · s
KE = 12mv
2PE = mgh
P ⌘ dWdt
= F · v
F = kx PE = 12kx
2
dvve
= �dmm
vf � vi = ve ln( mimf
)
F = |vedmdt
|
F = k
q1q2
r2 k = 14⇡✏0
⇡ 9⇥ 109 Nm2C�2
✏0 = 8.854⇥ 10�12 N�1m�2C 2
E ⌘lim�q!0
⇣�F�q
⌘E = k
qr2 r
V ⌘ Wq
E = �dVdx
V = k
qr
� =H
E · dA =P
q✏0
C ⌘ qV
C = A✏d
E = 12
q2
C= 1
2qV = 12CV
2
C = C1 + C21C
= 1C1
+ 1C2
R = R1 + R21R
= 1R1
+ 1R2
V = IR V = E � IR
P = V I = V 2
R= I
2R
K1 :P
In = 0K2 :
P(IR
0s) =
P(EMF
0s)
F = q v ⇥B dF = i dl⇥B
F = i l⇥B ⌧ = niA⇥B
v = EB
r = mq
EBB0
r = mvqB
T = 2⇡mBq
KEmax = R2B2q2
2m
dB = µ0
4⇡i
dl⇥r
r2HB · ds = µ0
PI µ0 = 4⇡⇥10�7 NA�2
� =R
areaB · dA � = B · A
✏ = �N
d�dt
✏ = NAB! sin(!t)
f = 1T
k ⌘ 2⇡�
! ⌘ 2⇡f v = f�
y = f(x⌥ vt)
y = a sin k(x� vt) = a sin(kx� !t)= a sin 2⇡(x
�� t
T)
P = 12µv!
2a
2v =
qFµ
s = sm sin(kx� !t)
�p = �pm cos(kx� !t)
3
I = 12⇢v!
2s
2m
n(db0s) ⌘ 10 log I1I2
= 10 log II0
where I0 = 10�12 W m�2
fr = fs
⇣v±vrv⌥vs
⌘where v ⌘ speed of sound = 340 m s�1
y = y1 + y2
y = [2a sin(kx)] cos(!t)
N : x = m(�2 ) AN : x = (m + 1
2)(�2 )
(m = 0, 1, 2, 3, 4, ....)
y = [2a cos(!1�!22 )t] sin(!1+!2
2 )t
fB = |f1 � f2|
y = [2a cos(k�2 )] sin(kx� !t + k�
2 )
� = d sin ✓
Max : � = m� Min : � = (m + 12)�
I = I0 cos2(k�2 )
E = hf c = f�
KEmax = eV0 = hf � �
L ⌘ r⇥ p = r⇥mv
L = rmv = n( h2⇡
)
�E = hf = Ei � Ef
rn = n
2( h2
4⇡2mke2 ) = n
2a0
En = �ke2
2a0( 1
n2 ) = �13.6n2 eV
1�
= ke2
2a0hc( 1
n2f� 1
n2i) = RH( 1
n2f� 1
n2i)
(a0 = Bohr radius = 0.0529 nm)
(RH = 1.09737⇥ 107m
�1)
(n = 1, 2, 3....) (k ⌘ 14⇡"0
)
E
2 = p
2c
2 + (m0c2)2
E = m0c2
E = pc
� = hp
(p = m0v (nonrelativistic))
�x�px � h⇡
�E�t � h⇡
dNdt
= ��N N = N0 e��t
R ⌘ |dNdt
| T
12
= ln 2�
= 0.693�
MATH:
ax
2 + bx + c = 0 ! x = �b±p
b2�4ac2a
y dy/dx
Rydx
x
nnx
(n�1) 1n+1x
n+1
e
kxke
kx 1ke
kx
sin(kx) k cos(kx) � 1k
cos kx
cos(kx) �k sin(kx) 1k
sin kx
where k = constant
Sphere: A = 4⇡r
2V = 4
3⇡r
3
CONSTANTS:
1u = 1.660⇥ 10�27kg = 931.50 MeV
1eV = 1.602⇥ 10�19J
c = 3.00⇥ 108m s
�1
h = 6.626⇥ 10�34Js
e ⌘ electron charge = 1.602⇥ 10�19C
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: Second Paper. February Program 2009 4
4
3 5
6 kg
2 kg
1 kg
✓
µ
a
T
Figure 1:
Question 1 ( 4 + 13 = 17 marks):
Figure 1 shows three blocks, of masses, 1 kg, 2 kg, and 6 kg, connected by a massless
string, that passes over two massless, frictionless pulleys. The dimensions of the slope,
upon which the 2 kg block slides, are labeled, in metres, on the figure. There is a
coe�cient of friction, µ = 0.2, between the 2 kg block and the slope. The system is
released from rest. Take the acceleration due to gravity, g = 10 ms
�2.
(i) Draw a separate diagram of each of the three blocks, and on each, label the
particular forces that act on that block.
(ii) Use Newton’s laws of motion to find the value of the acceleration, a, of the 6 kg
block.
PHYSICS: Second Paper. February Program 2009 5
M
M
k2 m
3 m 3 m
v
A B A
B
2 m
C C
frame frame
rest
(a)(b)
hinge
Figure 2:
Question 2 ( 2 + 15 = 17 marks):
Figure 2 (a) shows a horizontal bar, AB, of negligible mass, and total length 6 m hinged
at end A, to a rigid, right-angled frame. A vertical spring, of spring constant, k, and
unstretched length, 2 m is attached between point C of the frame, and end B, of the
bar. A mass M = 16 kg is attached at the mid point of the bar.
The bar is released from rest, and rotates downward, about the hinge (point A), stretch-
ing the spring. Figure 2 (b) shows the bar when it has reached the vertical position. At
this moment, the mass, M, is moving with a horizontal velocity of v = 2 m/s. Take the
acceleration due to gravity g = 10 ms
�2.
(i) What is the length of the spring in Figure 2 (b), and hence, by what length has
the spring been extended?
(ii) Use energy principles to determine the value of the spring constant, k.
PHYSICS: Second Paper. February Program 2009 6
M
A
B
8 m
6 m
AB = 10 m
µ
T
About to slip
✓
Figure 3:
Question 3 ( 3 + 2 + 11 = 16 marks):
Figure 3 shows a ladder, AB, of length 10 m, and mass, M . The coe�cient of frictionbetween end A of the ladder, and the horizontal floor on which it stands, is µ. Ahorizontal rope attached to end B, keeps the ladder inclined at the angle illustrated. Atthis angle, the ladder is just about to slip. Other dimensions of the figure are labeled.
(i) Draw your own sketch of Figure 3 and on it label all of the forces that act on theladder.
(ii) State the conditions for equilibrium of the ladder.
(iii) Use the conditions for the equilibrium of the ladder to determine the coe�cientof friction, µ, between the ladder and the floor.
END OF EXAM
ANSWERS:
Q1. a = 6.53 ms
�2.
Q2. (i) length = 10 m, therefore extension = 10� 2 = 8 m; (ii) k = 14 N/m.
Q3. (iii) µ = 38 .
1
FOUNDATION STUDIES
EXAMINATIONS
November 2009
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.