FOUNDATION STUDIES EXAMINATIONS March 2009 …flai/Theory/exams/Feb09_1.pdf1 FOUNDATION STUDIES EXAMINATIONS March 2009 PHYSICS First Paper February Program Time allowed 1 hour for

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�!.


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.


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





2

INFORMATION

a · b = ab cos ✓


��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


⌧ ⌘ 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


)

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)



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


✏ = �N

d�dt

✏ = NAB! sin(!t)

f = 1T

k ⌘ 2⇡�

! ⌘ 2⇡f v = f�

y = f(x⌥ vt)


�� t

T)

P = 12µv!

2a

2v =

qFµ



3

I = 12⇢v!

2s

2m


= 10 log II0

where I0 = 10�12 W m�2

fr = fs

⇣v±vrv⌥vs


y = y1 + y2


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�



L = rmv = n( h2⇡

)


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)


(RH = 1.09737⇥ 107m

�1)

(n = 1, 2, 3....) (k ⌘ 14⇡"0

)

E

2 = p

2c

2 + (m0c2)2

E = m0c2

E = pc

� = hp


�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


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


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:


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.


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:


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.


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





2

INFORMATION

a · b = ab cos ✓


��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


⌧ ⌘ 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


)

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)



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


✏ = �N d�dt

✏ = NAB! sin(!t)

f = 1T

k ⌘ 2⇡�

! ⌘ 2⇡f v = f�

y = f(x⌥ vt)


�� t

T)

P = 12µv!2a2 v =

qFµ



3

I = 12⇢v!2s2

m


= 10 log II0

where I0 = 10�12 W m�2

fr = fs

⇣v±vrv⌥vs


y = y1 + y2


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�



L = rmv = n( h2⇡

)


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)


(RH = 1.09737⇥ 107 m�1)

(n = 1, 2, 3....) (k ⌘ 14⇡"0

)

E2 = p2c2 + (m0c2)2

E = m0c2 E = pc

� = hp


�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


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


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.


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.


!

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.


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.


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?

Documents

FOUNDATION STUDIES EXAMINATIONS March 2009 …flai/Theory/exams/Feb09_1.pdf1 FOUNDATION STUDIES EXAMINATIONS March 2009 PHYSICS First Paper February Program Time allowed 1 hour for