FOUNDATION STUDIES EXAMINATIONS March 2008 PHYSICS …flai/Theory/exams/Feb08_1.pdf · =dm m vf vi...

1

FOUNDATION STUDIES

EXAMINATIONS

March 2008

PHYSICS

First Paper

February Program

Time allowed 1 hour for writing10 minutes for reading

This paper consists of 2 questions printed on 5 pages.PLEASE CHECK BEFORE COMMENCING.

Candidates should submit answers to ALL QUESTIONS.

Marks on this paper total 20 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 2008 4

x

y

10N

8N

9N

✓

4m

3mframe

A B

CD

Figure 1:

Question 1 ( (3 + 4 + 3) = 10 marks):

Figure 1 shows a frame, ABCD, with sides of lengths 4 m and 3 m, aligned with the x-

and y-axes. Forces of 10 N , 8 N , and 9 N , act at D, in the directions illustrated.

(i) Express each of the forces in terms of the ijk unit vectors.

(ii) Hence find the vector sum, F, of these three forces, in terms of the ijk unit vectors.

(iii) Find the magnitude, and direction, of F.

PHYSICS: First Paper. February Program 2008 5

Question 2 ( (5 + 5) = 10 marks):

A force, F = 3i + 2j + 5k Newton, drags a block through a displacement ofr = 6i� 4j + k metre.

(i) Given, that the work, W , done by a force, F, when it drags a body through adisplacement. r, is given by the dot product -

W = F • r

find the work, W , done by the force, F, above.

(ii) Use the dot product, to find the angle between F and r.

END OF EXAM

ANSWERS:

Q1.(i) 8�! = 8i N , 9�! = �9j N , 10�! = 8i + 6j N ; (ii) F = 16i� 3j N ;(iii) 16.3 N 10.6 deg below +x-axis

Q2. (i) 15 J ; (ii) 70.5 deg

1

FOUNDATION STUDIES

EXAMINATIONS

June 2008

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

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: Second Paper. February Program 2008 4

2m

3m

2m

H E

F

80 kg

T

20 kg

Figure 1:

Question 1 ( (3 + 13) = 16 marks):

Beam HE, of mass 20 kg and length 4 m, is hinged to a vertical post at H. This beam

is held horizontally by a cable EF, attached to its end E. The other end of the cable is

secured to the post at F, which is 3 m above H. A mass of 80 kg hangs from the centre

point of the beam. Take the acceleration of gravity g = 10 m s�2.

(i) Draw a diagram of the beam, showing all the forces that act upon it.

(ii) Using the conditions for the equilibrium of the beam, find the tension, T , in the

cable, and the vertical and horizontal components of the reaction of the hinge on the

beam, at H. Draw a second diagram of the beam, HE, and on it label these forces.


m

3m

2m

µ

µ = 0

a

P

Figure 2:

Question 2 ( (5 + 10 + 2) = 17 marks):

Figure 2 shows a wheeled trolley, of mass 2m, connected to a block, of mass 3m, by

means of a string that passes over a pulley. A second block, of mass m, rests on the

top surface of the trolley, but is restrained by a second string, that is attached to a

fixed post at P. All strings are massless, and all wheels and pulleys are massless and

frictionless. The system is released from rest. As block 3m falls, it pulls the trolley

along the horizontal surface. Because of the string attached to P, block m is held at

rest, and so slides along the top surface of the trolley. There is negligible friction in the

wheels of the trolley. The coe�cient of friction between the bottom surface of block m

and the top surface of the trolley is µ.

(i) Draw three diagrams - one for the trolley, one for the block of mass, 3m, and one

for the block of mass m. Label each of these diagrams with the particular forces that

act on each body. Label also the acceleration of each body.

(ii) Write equations of motion (Newton’s second law) for the trolley, and each of the

two blocks, in vertical and horizontal directions, as appropriate (five equations).

(iii) Using the above equations, derive an expression for the acceleration, a, of the

3m block, in terms of µ, and the acceleration of gravity g.


M

m

C

M

m

C

v

v(a) (b)

rest

rest

L

Figure 3:

Question 3 ( (14 + 3) = 17 marks):

Two balls, of masses, m and M (M > m), are secured to the ends of a rod of negligible

mass. The distance between the balls is L. Initially, this rod is in the vertical position,

as illustrated in Figure 3 (a). When released from rest, the rod rotates clockwise about

a pivot at its centre point, C. When the rod is next vertical, both balls are traveling

with a speed, v, as shown in Figure 3 (b) . There is negligible friction at the pivot.

(i) Using energy principles, derive an expression for the speed, v, of the balls in

Figure 3 (b), in terms of l, m, M , and the acceleration of gravity, g.

(ii) Hence, write down an expression for the angular velocity, !, of the system of rod

and balls, in Figure 3 (b).

END OF EXAM


ANSWERS:

Q1.(ii) T = 8.33 kN , Rx = +6.66 kN , Ry = �4.00 kN .

Q2. (ii) 3mg � T = 3ma, r �mg = 0, +µr � t = 0, R � r � 2mg = 0, +T � µr = 2ma;(iii) a = (3�µ)

5 g.

Q3. (i) v =q

2Lg (M�m)(M+m) ; (ii) ! =

q8gL

(M�m)(M+m) .

1

FOUNDATION STUDIES

EXAMINATIONS

November 2008

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 2008 4

P

Q

x

y

z7

6

5

(dimensions in m )

B

Figure 1:

Question 1 ( (4 + 6) + (2 + 8) = 20 marks):

Part (a):

Figure 1 shows a rectangular box, with a corner at the origin, 0, and its sides aligned

along the x-, y-, and z-axes. Dimensions of the box are labeled in metre. A wire, PQ, is

stretched between corners P and Q of the box. A constant force, F, moves a bead, B,

along the wire, from P to Q. This force is given by -

F = 2i + 3j + 4k N

(i) Derive an expression for the displacement vector, PQ, in terms of unit vectors ijk.

(ii) Calculate the work, W , that is done on the bead by the force, F, between P and Q.


5 m/sB = 3.0 kgC = 4.0 kg

A = 2.0 kg

A

BC

x

y

B

Cx

y

Before

After

6 m/s

(a)

(b)

Figure 2:

Part (b):

Figure 2(a) shows three stationary balls, A, B, and C, with masses as labeled, on a

flat horizontal, frictionless surface, near the origin of the x- and y-axes. An explosion

occurs between the balls, blowing them apart. Figure 2(b) shows balls B and C after the

explosion; ball A is not shown. Ball B moves in the +y direction, while ball C moves in

the +x direction, with the velocities as labeled.

(i) Redraw Figure 2(b), showing the general direction in which you would expect ball

A to move, after the explosion.

(ii) Using momentum principles, determine the magnitude and direction of the

velocity of ball A, after the explosion.


y

mM

d

e!y

mM

d

c(a) (b)

L✓

Figure 3:

Question 2 ( (3 + 1 + 6) + (10) = 20 marks):

Part (a):

A mad inventor devises an instrument for the measurement of the acceleration ofgravity, g. This invention is illustrated in Figure 3, and consists of a ball of mass m, anda block of mass M , connected together, over a massless, frictionless pulley, of diameter,d, by a massless string. The length of the string from the pulley, to the ball m, is L.The y-axis in the figure passes symmetrically through block M , and is aligned vertically.The instrument is spun, with a slowly increasing angular velocity about the y-axis. Asthe angular velocity increases, the angle of the string to the vertical, at ball m, increases,from zero. Figure 3 (a) shows the system before it begins to rotate. Figure 3 (b) showsthe system as block M is just about to lift o↵ the shelf on which it stands. At this stage,the angular velocity of ball m is !, and the angle of the string to the vertical is ✓, aslabeled. ! and ✓ are then recorded, and the acceleration of gravity, g, calculated.

(i) Draw a diagram of each of masses m and M , as in Figure 3 (b). Label on eachdiagram all forces that act on each of the two masses. Label also, the acceleration ofeach mass.

(ii) What is the radius of ball m’s circular path in Figure 3 (b)?

(iii) Use Newton’s laws of motion, to derive an expression for the acceleration ofgravity, g, in terms of !, ✓, L, and d, as in Figure 3 (b).


M

✓

µ

k

d

D

�

Figure 4:

Part (b):

Figure 4 shows a block, of mass, M , 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, D, down

the slope, compresses the spring, of spring constant, k, at the bottom of the slope, and

is projected back up the slope. It comes momentarily to rest again at a distance of d,

up the slope, beyond the end of the uncompressed spring. D and d are labeled on Figure 4.

Using energy principles, derive an expression for the distance, �, that the spring was

compressed, by the block, in terms of the parameters labeled in Figure 4.


G

M

µ

µ = 0.01 kg/m

H

Figure 9:

Question 5 ( (4 + 3 + 3) + (5 + 5) = 20 marks):

Part (a):

In order to measure the mass, M , of a block, a mad invertor designs the apparatus,shown in Figure 9. A string, with mass per unit length, µ, is stretched over a pulley, H,between a wave generator, G, at one end, and the block, of mass M , at the other. Thevalue of µ is labeled on the figure. Generator G transmits an harmonic wave of fixedfrequency and amplitude, from G to H, which is measured using a wave analyser.

In a particular measurement, the wave function was found to be -

y = 10�3 sin(100x� 600t) (SI units)

Take the acceleration of gravity, g = 10 m s�2.

(i) Find the amplitude, frequency, wavelength, and velocity of this harmonic wave.

(ii) Find the power that generator G must output to continuously transmit thisharmonic wave along the string.

(iii) Find the mass, M , of the block, for this particular measurement.

Part (b):

A photoelectric cell has a silver electrode. The work function for silver is 4.73 eV .

(i) Calculate the threshold wavelength for this photocell.

(ii) Calculate the stopping potential for this photocell, when illuminated with lightof wavelength, � = 200 nm?

FOUNDATION STUDIES EXAMINATIONS March 2008 PHYSICS …flai/Theory/exams/Feb08_1.pdf · =dm m vf vi...

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