foundation studies examinations april 2007 physics first ...flai/theory/exams/feb07_1.pdf · =dm m...

1

FOUNDATION STUDIES

EXAMINATIONS

April 2007

PHYSICS

First Paper

February Program 2007

Time allowed 1 hour for writing10 minutes for reading

This paper consists of 3 questions printed on 5 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 ka

x

a

y

a

z

b

x

b

y

b

z

��v ⌘ dr

dt

a ⌘ dvdt

v =R

a dt r =R

v dt

v = u + at

x = ut + 12at

2

v

2 = u

2 + 2ax

a = �gjv = u� gtjr = ut� 1

2gt

2j

s = r✓ v = r! a = !

2r = v

2

r

p = mv

N1 : ifP

F = 0 then �p = 0N2 :

PF = ma

N3 : FAB

= �FBA

W = mg F

r

= 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 ⌘ dW

dt

= F · v

F = kx PE = 12kx

2

dv

ve= �dm

m

v

f

� v

i

= v

e

ln( mimf

)

F = |ve

dm

dt

|

F = k

q1q2

r

2 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

r

2 r

V ⌘ W

q

E = �dV

dx

V = k

q

r

� =H

E · dA =P

q

✏0

C ⌘ q

V

C = A✏

d

E = 12

q

2

C

= 12qV = 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

= I

2R

K1 :P

I

n

= 0K2 :

P(IR

0s) =

P(EMF

0s)

F = q v ⇥B dF = i dl⇥B

F = i l⇥B ⌧ = niA⇥B

v = E

B

r = m

q

E

BB0r = mv

qB

T = 2⇡m

Bq

KE

max

= R

2B

2q

2

2m

dB = µ0

4⇡

i

dl⇥r

r

2HB · ds = µ0

PI µ0 = 4⇡⇥10�7 NA�2

� =R

area

B · dA � = B · A

✏ = �N

d�

dt

✏ = NAB! sin(!t)

3

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 =

qF

µ

s = s

m

sin(kx� !t)

�p = �p

m

cos(kx� !t)

I = 12⇢v!

2s

2m

n(db0s) ⌘ 10 log I1I2

= 10 log I

I0

where I0 = 10�12 W m�2

f

r

= f

s

⇣v±vrv⌥vs

⌘where v ⌘ speed of sound = 340 m s�1

y = [2A sin(kx)] cos(!t)

y(x=0) =h2A cos (!1�!2)

2 t

isin (!1+!2)

2 t

y =h2A cos (k�+�)

2

isin(kx� !t + k�+�

2 )

f

beat

= f1 � f2

� = n� � = (n + 12)�

d sin ✓ = n�

E = hf c = f�

KE

max

= eV0 = hf � �

E

2 = p

2c

2 + (m0c2)2

E = m0c2

E = pc

� = h

p

(p = m0v (nonrelativistic))

�x�p

x

� h

⇡

�E�t � h

⇡

dN

dt

= ��N N = N0 e��t

R ⌘ |dN

dt

| T

12

= ln 2�

= 0.693�

MATH:

If ax

2 + bx + c = 0

then x = �b±p

b

2�4ac

2a

y dy/dx

Rydx

x

n

nx

(n�1) 1n+1x

n+1

e

kx

ke

kx

1k

e

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�27 kg = 931.50 MeV

1eV = 1.602⇥ 10�19 J

c = 3.00⇥ 108 ms�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. 2007 4

Question 1 ( (9 + 8) = 17 marks):

Figure 1:

A rectangular box aligned along thex-, y-, and z-axes, with sides of 4 m,

4 m, and 3 m, as labeled, is illustratedin Figure 1. Three forces act at

corner, P, of the box. 2 N acts up inthe direction of the y-axis; 4 N actsalong the diagonal, PQ, of the top

face; 6 N acts along the diagonal, PR,of the right face.

(i) Express each of the three forcesin terms of the ijk unit vectors.

(ii) Hence find the vector sum ofthe three forces. Again express youranswer in terms of ijk unit vectors.

Question 2 ( 16 marks):

The power, P , generated by a wind turbine depends on the total area, A, swept out by

its blades, the velocity, v, of the air that passes through the blades, and the density

(mass per unit volume), ⇢, of that air.

Use dimensions to find an expression for P in terms of ⇢, A and v.

Hint: dimensions of P are - [P ] = ML

2T

�3

PHYSICS: First Paper. February Program. 2007 5

Question 3 ( 4 + 6 + 7 = 17 marks):

Figure 2:

A rectangular box aligned along thex-, y-, and z-axes, with sides of 2 m,

3 m, and 4 m, as labeled, is illustratedin Figure 2. A triangle, PQR, is

drawn between corners of the box asillustrated. Displacement

�!PQ ⌘ a and

displacement�!PR ⌘ b.

(i) Express vectors a and b interms of the ijk unit vectors.

(ii) Use the dot product (a · b) tocalculate the angle ✓, between a and b.

(iii) Find an expression for vectorarea, A, of triangle PQR. Give youranswer in terms of ijk unit vectors.

You are given that the vector area, A,of PQR is given by the cross product

A = 12(a⇥ b)

END OF EXAM

ANSWERS:

Q1.(i)2 = (2j) N , 4 = 15(�16i + 12k) N , 6 = 1

5(�24j + 18k) N .(ii) F = 1

5(�16i� 14j + 2k) N .

Q2. P = k⇢Av

3.

Q3. (i) a = (+3i � 2k) m, b = (�4i + 3j � 2k) m. (ii) ✓ = 47.9 deg.(iii) A = (+4j + 6k) m

2.

1

FOUNDATION STUDIES

EXAMINATIONS

June 2007

PHYSICS

Second Paper

February Program

Time allowed 1 hour for writing10 minutes for reading





2

INFORMATION

a · b = ab cos ✓


��i j ka

x

a

y

a

z

b

x

b

y

b

z

��v ⌘ dr

dt

a ⌘ dvdt

v =R

a dt r =R

v dt

v = u + at

x = ut + 12at

2

v

2 = u

2 + 2ax

a = �gjv = u� gtjr = ut� 1

2gt

2j

s = r✓ v = r! a = !

2r = v

2

r

p = mv

N1 : ifP

F = 0 then �p = 0N2 :

PF = ma

N3 : FAB

= �FBA

W = mg F

r

= 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 ⌘ dW

dt

= F · v

F = kx PE = 12kx

2

dv

ve= �dm

m

v

f

� v

i

= v

e

ln( mimf

)

F = |ve

dm

dt

|

F = k

q1q2

r

2 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

r

2 r

V ⌘ W

q

E = �dV

dx

V = k

q

r

� =H

E · dA =P

q

✏0

C ⌘ q

V

C = A✏

d

E = 12

q

2

C

= 12qV = 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

= I

2R

K1 :P

I

n

= 0K2 :

P(IR

0s) =

P(EMF

0s)



v = E

B

r = m

q

E

BB0r = mv

qB

T = 2⇡m

Bq

KE

max

= R

2B

2q

2

2m

dB = µ0

4⇡

i

dl⇥r

r

2HB · ds = µ0

PI µ0 = 4⇡⇥10�7 NA�2

� =R

area

B · dA � = B · A

✏ = �N

d�

dt

✏ = NAB! sin(!t)

3

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 =

qF

µ

s = s

m

sin(kx� !t)

�p = �p

m

cos(kx� !t)

I = 12⇢v!

2s

2m


= 10 log I

I0

where I0 = 10�12 W m�2

f

r

= f

s

⇣v±vrv⌥vs


y = y1 + y2

y = [2a sin(kx)] cos(!t)

(m = 0, 1, 2, 3, 4, ....)

N : x = m(�

2 ) AN : x = (m + 12)(

�

2 )

y = [2a cos(!1�!22 )t] sin(!1+!2

2 )t

f

B

= |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�

KE

max

= eV0 = hf � �

E

2 = p

2c

2 + (m0c2)2

E = m0c2

E = pc

� = h

p


�x�p

x

� h

⇡

�E�t � h

⇡

dN

dt

= ��N N = N0 e��t

R ⌘ |dN

dt

| T

12

= ln 2�

= 0.693�

MATH:

ax

2 + bx + c = 0 ! x = �b±p

b

2�4ac

2a

y dy/dx

Rydx

x

n

nx

(n�1) 1n+1x

n+1

e

kx

ke

kx

1k

e

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 2007 4

2M

M

µ

3 m

4 m

a

5 m

Figure 1:

Question 1 ( 2 + 15 = 17 marks):

Figure 1 shows blocks, of masses 2M and M , connected by a string of negligible mass,

via two pulleys of negligible mass and friction. The system is released from rest. The

coe�cient of friction between the block of mass M , and the inclined surface, on which it

slides, is µ. Dimensions of the incline are illustrated.

(i) Draw a labeled diagram to show all the forces that act on each block.

(ii) User Newton’s laws of motion to find an expression, in terms of µ, and the

acceleration of gravity, g, for the acceleration, a, of the block of mass 2M .

(iii) Find a similar expression for the tension, T , in the string.


M

CB = 8 m

A

B

C

AC = 6 m

L

AB =10 m

2M

3M

horizontal

Figure 2:

Question 2 ( 3 + 10 + 3 = 16 marks):

A beam, AB, of mass M , and length 10 m, is hinged at B, as illustrated in Figure

2. The other end, A, is attached, by means of a light, horizontal rope, over a fric-

tionless pulley, to a block of mass, 3M . In order to lift the block o↵ the ground, a

worker, of mass 2M , walks up the beam from B. The dimensions of the system are labeled.

(i) Draw a diagram showing all the forces that are acting on the beam AB, just

before the block lifts.

(ii) Using the conditions for equilibrium, find the distance, L, along the beam AB,

that the worker, will need to walk, to just lift the block.

(iii) Given that M = 30 kg, find the reaction of the hinge on the beam, as the block

is just lifted. Take acceleration of gravity g = 10 m s

�2.


2M

Mµ

L

k

unstretched

rest

rest

Figure 3:


Two blocks, of masses M and 2M , are connected together, and to an unstretched spring,of spring constant, k, by means of two lengths of massless string, that pass over twomassless and frictionless pulleys, as depicted in Figure 3. The block, of mass 2M , isreleased from rest, and falls a vertical distance of L, before coming momentarily to restagain. The other block, which has a coe�cient of friction, µ, with the horizontal surface,is simultaneously pulled along the horizontal surface, as the spring is stretched.

Using energy principles, derive an expression for the distance, L, that the 2M blockfalls, in terms of M , µ, k, and the acceleration of gravity, g.

END OF EXAM

ANSWERS:

Q1.(ii) a = (2�µ

5 )g ; (iii) T = 2Mg

5 (3 + µ).Q2. (ii) L = 8.75 m ; (iii) R

x

= 900 N , R

y

= 900 N .

Q3. L = 2Mg

k

(2� µ).

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:


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.

1

FOUNDATION STUDIES

EXAMINATIONS

November 2007

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

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)



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


= 10 log II0

where I0 = 10�12 W m�2

fr = fs

⇣v±vrv⌥vs


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

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 ⌘ 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 2007 4

O

OC = 2 mB

C

A

z

y

x

OB = 3 mOA = 4 m

4

2

3

Figure 1:

Question 1 ( (2 + 8) + (5 + 5) = 20 marks):

Part (a):

Figure 1 shows a surface, ABC, whose corners are on the x-, y-, and z-axes, as illustrated.

(i) Express each of the two vectors, AB�! and AC�!, in terms of the ijk unit vectors.

(ii) Hence express the vector area, A�! of the surface, in terms of ijk unit vectors. Youare given the following, cross-product expression -

A�! = 12(AB�!⇥ AC�!)


O

B

u

6m

12m

A C

2s

Figure 2:

Part (b):

Figure 2 shows the trajectory of a ball which is projected from a point, O, at ground

level, which is 6 m from the base of a 12 m high vertical pole, AB. The ball passes

through a loop at the top, B, of the pole, precisely 2 s. later, then continues until it

strikes the ground at C. Assume friction e↵ects of the air to be negligible.

(i) Determine the initial velocity, u, with which the ball was projected from point O.

You may express your answer in ijk form.

(ii) Find the total horizontal distance, OC, between the point of projection, O, and

the point C, where the ball hits the ground.


M

m

3 m

4 m

✓

µ

a

Figure 3:

Question 2 ( (2 + 8) + (1 + 9) = 20 marks):

Part (a):

Figure 3 shows Two blocks of masses, m and M , connected by a massless string, over

a massless, frictionless pulley. The dimensions of the system are labeled (tan ✓ = 43).

There is a coe�cient of friction, µ, between the block of mass, m, and the inclined

surface on which it slides. The system is released from rest.

(i) Draw a labeled diagram, showing all the forces that act on each block.

(ii) Use Newton’s laws of motion, to derive an expression for the acceleration, a, of

the block of mass, M , vertically downward, in terms of M , m, µ, and the acceleration

of gravity, g.


mM

m

M

rest rest

r 2r

PP

u

v

(a)

(b)

S

bar

Figure 4:

Part (b):

Figure 4(a) shows a catapult, of length 3r, consisting of a straight bar, which has a

block, of mass, M , fixed at one end, and a basket, containing a ball, of mass, m, at the

other. The bar is initially horizontal (Figure 4(a)), but can rotate about a fixed pivot,

P. The bar, and the basket have negligible mass, and there is negligible friction, at the

pivot. Relevant dimensions of the figure are labeled.

The catapult is released, and because M � m, it rotates anticlockwise, until the

bar becomes vertical. It then hits stopper S, and the ball is ejected from the basket.

Figure 4(b) shows the bar in the vertical position, just before it hits stop, S. At this

stage, the block, M , has a horizontal velocity, v, and the ball, m, has a horizontal

velocity, u, as labeled.

(i) Express the velocity of the block, v, in terms of the velocity of the ball, u, as

illustrated in Figure 4(b).

(ii) Using energy principles, derive an expression for the velocity, u, of the ball, in

Figure 4(b), in terms of M , m, r, and the acceleration of gravity, g.


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 ?

foundation studies examinations april 2007 physics first ...flai/theory/exams/feb07_1.pdf · =dm m...

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