foundation studies examinations march 2009 …flai/theory/exams/feb09_1.pdf1 foundation studies...
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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.
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
2at2 v = u� gtjv2 = u2 + 2ax r = ut� 1
2gt2j
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 = 12mv2 PE = mgh
P ⌘ dWdt
= F · v
F = kx PE = 12kx2
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 q
r2 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= I2R
K1 :P
In = 0K2 :
P(IR0s) =
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⇡idl⇥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!2a2 v =
qFµ
s = sm sin(kx� !t)
�p = �pm cos(kx� !t)
3
I = 12⇢v!2s2
m
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 = n2( h2
4⇡2mke2 ) = n2a0
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⇥ 107 m�1)
(n = 1, 2, 3....) (k ⌘ 14⇡"0
)
E2 = p2c2 + (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:
ax2 + bx + c = 0 ! x = �b±p
b2�4ac2a
y dy/dxR
ydx
xn nx(n�1) 1n+1x
n+1
ekx kekx 1kekx
sin(kx) k cos(kx) � 1k
cos kxcos(kx) �k sin(kx) 1
ksin kx
where k = constant
Sphere: A = 4⇡r2 V = 43⇡r3
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 2009 4
8 m
6 m
10 m
µ
M1310 m
A
B
wheel
ladder
Figure 1:
Question 1 ( (3 + 7) + (10) = 20 marks):
Part (a):
Figure 1 shows a ladder, of length, 10 m, but negligible mass, leaning between a
horizontal floor, and a vertical wall. End B of the ladder has a wheel, so there is zero
friction between the ladder and the wall, at that end. End A of the ladder leans directly
on the floor, and there is a coe�cient of friction of µ with the floor, at end A. A climber,
of mass, M , commences climbing the ladder, starting from the floor. When the climber
has climbed a distance that is 13 of the length of the ladder, the ladder slips. Dimensions
of the figure are labeled.
(i) Draw a labeled diagram of the ladder, showing all forces that act upon it, just
before it slips.
(ii) Use the conditions for the equilibrium of the ladder to determine the coe�cient
of static friction, µ, between the ladder and the floor.
PHYSICS: Final Paper. February Program 2009 5
x
y
5 m/s
10 m/s
3M M
x
y
M
before(b)
✓
✓tan = 43
(a)after
rest
Figure 2:
Part (b):
Figure 2(a) shows a stationary ball, of mass 3M , on a flat horizontal, frictionless surface,
at the origin of the x- and y-axes. An explosion occurs inside the ball, blowing it apart
into three fragments, each of mass M . Figure 2(b) shows two of the fragments after the
explosion; the third is not shown. All three fragments remain in the xy-plane.
Using momentum principles, determine the magnitude and direction of the velocity of
the third fragment of the ball, after the explosion.
PHYSICS: Final Paper. February Program 2009 6
!
y
disk
4 m
3 m5 m
M
✓
µ
wedge
viewfromtop
d
M
!
d
Figure 3:
Question 2 ( (4 + 6) + (10) = 20 marks):
Part (a):
Figure 3 shows a horizontal disk that rotates about the vertical y-axis as its spin axis.
A wedge, whose dimensions of 3 m, 4 m, and 5 m, are labeled, is firmly attached to the
surface of this disk, as illustrated.
The block of mass, M , lies a distance, of d, along the inclined face of the wedge, from its
base, and has a coe�cient of friction, µ, with the wedge surface. The angular velocity,
!, of the disc is slowly increased. When ! exceeds a value !s, the block slips.
(i) Draw a labeled diagram of the block, of Figure 3, showing all forces that act on
it, just before it slips. Label also, the acceleration of the block, just before the block slips.
(ii) Use Newton’s laws of motion, to derive an expression for !s, in terms of the
parameters labeled in Figure 3, and the acceleration due to gravity, g.
PHYSICS: Final Paper. February Program 2009 7
M
6 m
8 m
10 m
A
B
5 m
k
�
rest
rest
µ
Figure 4:
Part (b):
Figure 4 shows a block, of mass, M , at a point A, on a slope. The coe�cient of friction
between the slope and the block is µ. The block is released from rest, slides a distance
of 10 m down the slope, and then compresses a spring, of spring constant, k, near the
bottom of the slope, a distance of �. The block is then projected back up the slope, by
the spring. It comes momentarily to rest again at a position B, which is a distance of
5 m, below its original release point, A.
Using energy principles, derive an expression for the distance, �, that the spring was
compressed, by the block, in terms of µ. Hence calculate � if µ = 0.4.
PHYSICS: Final Paper. February Program 2009 12
S DL
m FP, f
Figure 9:
Question 5 ( (3 + 3 + 4) + (5 + 5) = 20 marks):
Part (a):
A wave source, S, is connected to a wave detector, D, by means of a string, of total
length, L = 4.00 m, and mass, m = 0.16 gram, which is stretched to a tension,
F = 100 N , as shown in Figure 9. S is set to a power of P = 500 W , and generates a
continuous wave of frequency, f = 1000Hz.
(i) Calculate the speed, v, of the wave from S to D.
(ii) Calculate the amplitude, a, of the wave along the string.
(iii) Write down a possible wave function for the wave along the string from S to D.
Part (b):
A photoelectric cell has an aluminium electrode. The work function for aluminium is
4.08 eV .
(i) Calculate the threshold wavelength for this photocell.
(ii) Calculate the stopping potential for this photocell, when illuminated with light
of wavelength, � = 200 nm?