FOUNDATION STUDIES EXAMINATIONS …flai/Theory/exams/FT08_1.pdfFOUNDATION STUDIES EXAMINATIONS September 2008 PHYSICS ... Hence ﬁnd the maximum angular velocity, ω m, ... Express

1

FOUNDATION STUDIES

EXAMINATIONS

September 2008

PHYSICS

First Paper

July Fast Track

Time allowed 1.5 hour for writing10 minutes for reading

This paper consists of 4 questions printed on 6 pages.PLEASE CHECK BEFORE COMMENCING.

Candidates should submit answers to ALL QUESTIONS.

Marks on this paper total 60 Marks, and count as 15% of the subject.

Start each question at the top of a new page.

2

INFORMATION

a · b = ab cos θ

a× b = ab sin θ c =

∣∣∣∣∣∣i j kax ay az

bx by bz

∣∣∣∣∣∣v ≡ dr

dta ≡ dv

dtv =

∫a dt r =

∫v dt

v = u + at a = −gjx = ut + 1

2at2 v = u− gtj

v2 = u2 + 2ax r = ut− 12gt2j

s = rθ v = rω a = ω2r = v2

r

p ≡ mv

N1 : if∑

F = 0 then δp = 0N2 :

∑F = ma

N3 : FAB = −FBA

W = mg Fr = µR

g =acceleration due to gravity=10m s−2

τ ≡ r× F∑Fx = 0

∑Fy = 0

∑τP = 0

W ≡∫ r2r1

F dr W = F · s

KE = 12mv2 PE = mgh

P ≡ dWdt

= F · v

F = kx PE = 12kx2

dvve

= −dmm

vf − vi = ve ln( mi

mf)

F = |vedmdt|

F = k q1q2

r2 k = 14πε0

≈ 9× 109 Nm2C−2

ε0 = 8.854× 10−12 N−1m−2C 2

E ≡limδq→0

(δFδq

)E = k q

r2 r

V ≡ Wq

E = −dVdx

V = k qr

Φ =∮

E · dA =∑

qε0

C ≡ qV

C = Aεd

E = 12

q2

C= 1

2qV = 1

2CV 2

C = C1 + C21C

= 1C1

+ 1C2

R = R1 + R21R

= 1R1

+ 1R2

V = IR V = E − IR

P = V I = V 2

R= I2R

K1 :∑

In = 0K2 :

∑(IR′s) =

∑(EMF ′s)

F = q v ×B dF = i dl×B

F = i l×B τ = niA×B

v = EB

r = mq

EBB0

r = mvqB

T = 2πmBq

KEmax = R2B2q2

2m

dB = µ0

4πidl×r

r2∮B · ds = µ0

∑I µ0 = 4π×10−7 NA−2

φ =∫

areaB · dA φ = B ·A

ε = −N dφdt

ε = NABω sin(ωt)

f = 1T

k ≡ 2πλ

ω ≡ 2πf v = fλ

y = f(x∓ vt)

y = a sin k(x− vt) = a sin(kx− ωt)= a sin 2π(x

λ− t

T)

P = 12µvω2a2 v =

√Fµ

s = sm sin(kx− ωt)

∆p = ∆pm cos(kx− ωt)

3

I = 12ρvω2s2

m

n(db′s) ≡ 10 log I1I2

= 10 log II0

where I0 = 10−12 W m−2

fr = fs

(v±vr

v∓vs

)where v ≡ speed of sound = 340 m s−1

y = y1 + y2

y = [2a sin(kx)] cos(ωt)

N : x = m(λ2) AN : x = (m + 1

2)(λ

2)

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

y = [2a cos(ω1−ω2

2)t] sin(ω1+ω2

2)t

fB = |f1 − f2|

y = [2a cos(k∆2

)] sin(kx− ωt + k∆2

)

∆ = d sin θ

Max : ∆ = mλ Min : ∆ = (m + 12)λ

I = I0 cos2(k∆2

)

E = hf c = fλ

KEmax = eV0 = hf − φ

L ≡ r× p = r×mv

L = rmv = n( h2π

)

δE = hf = Ei − Ef

rn = n2( h2

4π2mke2 ) = n2a0

En = −ke2

2a0( 1

n2 ) = −13.6n2 eV

1λ

= 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 ≡ 14πε0

)

E2 = p2c2 + (m0c2)2

E = m0c2 E = pc

λ = hp

(p = m0v (nonrelativistic))

∆x∆px ≥ hπ

∆E∆t ≥ hπ

dNdt

= −λN N = N0 e−λt

R ≡ |dNdt| T 1

2= ln 2

λ= 0.693

λ

MATH:

ax2 + bx + c = 0 → x = −b±√

b2−4ac2a

y dy/dx∫

ydx

xn nx(n−1) 1n+1

xn+1

ekx kekx 1kekx

sin(kx) k cos(kx) − 1k

cos kxcos(kx) −k sin(kx) 1

ksin kx

where k = constant

Sphere: A = 4πr2 V = 43πr3

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: First Paper. July Fast Track 2008 4

y

disc

ω

M

m

µ

r

Figure 1:

Question 1 ( 15 marks):

When a meteor falls through the atmosphere, the friction force, f (N) , of the atmo-sphere on the meteor, depends on the density of the air, ρ (kg m−3) , the cross-sectionarea, A (m2), of the meteor, and the velocity, v (m s−1), of the meteor, relative to the air.

Use dimensions to derive an expression for f , in terms of ρ, A, and v.

Question 2 ( (6 + 6 + 3) = 15 marks):

Figure 1 shows two blocks, of masses M and m, connected by a string, of negligiblemass, over a pulley, of negligible mass and friction. Block m rests on the horizontalsurface of a disc, at a distance r from its centre. Block M rests on a horizontal surface,at the bottom of a hole in the centre of the disc. The coefficient of static friction betweenblock m, and the surface of the disc is µ. The disc is rotated with a slowly increasingangular velocity, ω, with the vertical y-axis as its spin axis. If ω continues to increase,block m will eventually slip, on the surface of the disc.

(i) Draw a diagram of each block, labeling all particular forces that act on it, justbefore block m slips. Label also, any acceleration of each block.

(ii) Write down Newton’s equation of motion for each block, in both the vertical, andhorizontal directions, just before block m slips (three equations).

(iii) Hence find the maximum angular velocity, ωm, at which the disc can spin,before block m slips on the disc surface. Express ωm in terms of M , m, µ, r, and theacceleration of gravity, g.


6 m

8 m

A

B

10 m

M = 100 kg

M

C

Figure 2:

Question 3 ( (4 + 8 + 3) = 15 marks):

Figure 2 shows a beam, AB, of length 10 m, and mass 100 kg. End A of the beam is

hinged to a horizontal floor. End B of the beam is held in position by a rope BC, which

is perpendicular to the beam. End B of the beam is 6 m above the floor. Dimensions of

the system are labeled. Take the acceleration of gravity, g = 10 m s−2.

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

(ii) Write down the equations for equilibrium of the beam (three equations).

(iii) Hence find the tension force in the rope, and the reaction force of the floor on

the beam, at end A. Give vertical and horizontal components of the reaction.


d

m m

Mµ µ

k k restrest

M

(a) (b)

Figure 3:


Figure 3 (a) shows two blocks, of masses, m and M , connected by a string, over africtionless pulley. Mass, m, is in turn, connected to a fixed pivot by means of anunstretched spring, of spring constant, k. The string, pulley and spring, all havenegligible mass. The system is released from rest. M falls vertically, pulling m along thehorizontal surface, and stretching the spring. The coefficient of friction between m andthe horizontal surface is µ. Figure 3 (b) shows the system as it just comes momentarilyto rest, after the spring has been stretched.

Use energy principles to derive an expression for the vertical distance, d, that the blockof mass, M , falls, before coming momentarily to rest. Express d in terms of m, M , k, µ,and the acceleration of gravity, g.

END OF EXAM

ANSWERS:

Q1. f = kρAv2, where k ≡ dimensionless const.

Q2. (ii) T −Mg = 0 , R−mg = 0 , T + µR = mω2r ; (iii) ωm =√

gr(M

m+ µ) .

Q3. (ii) Rx−T sin θ = 0 , Ry −Mg + T cos θ = 0 , −(Mg)4 + (T sin θ)6 + (T cos θ)8 = 0;(iii) T = 400 N , Rx = 240 N , Ry = 680 N .

Q4. d = 2gk(M − µm).

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FOUNDATION STUDIES

EXAMINATIONS

December 2008

PHYSICS

Second Paper

July Fast Track

Time allowed 1.5 hour for writing10 minutes for reading



Marks on this paper total 60 Marks, and count as 15% of the subject.


PHYSICS: Second Paper. July Fast Track 2008 4

x

y

x

AB

C D

AB

C

θ

1 m/s

θ1 m/s

2 m/s

tanθ = 43

(a) (b)

y

Figure 1:


Figure 1(a) shows Four balls, A, B, C, and D, of equal mass, m, at rest near the

origin of the xy axes. An explosion between the balls causes them to move apart in the

frictionless xy plane. Three of the balls are shown in Figure 1(b). A moves at 1 m/s,

at angle θ to the y-axis; B moves at 1 m/s, in the negative direction of the x-axis;

C moves at 2 m/s, at angle θ to the negative x-axis. The value of θ is labeled on the figure.

Use momentum principles to calculate the velocity of ball D, after the explosion. You

can give your answer in terms of the ijk unit vectors.

1

FOUNDATION STUDIES

EXAMINATIONS

January 2009

PHYSICS

Final Paper

July Fast Track

Time allowed 3 hours for writing10 minutes for reading



Marks on this paper total 120 marks, and count as 45% of the subject.


PHYSICS: Final Paper. July Fast Track 2008 4

x

y

z

P

3 m

4 m

2 m

r

F = 10N

Q

Figure 1:

Question 1 ( (2 + 3 + 5) + (10) = 20 marks):

Part (a):

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

aligned along the x-,y- and z-axes. Dimensions of the box are labeled. A force of 6 N

acts at corner, P, of the box, along diagonal, PQ.

(i) Write down an expression for the position vector, r, of point P , in terms of unit

vectors ijk.

(ii) Express force, F, in terms of unit vectors ijk.

(iii) Find the torque, τ , of force, F, on the box, about the origin, O. Give your answer

in terms of ijk unit vectors, noting that -

τ = r × F (cross product)


A

B

m 5m

u = 6 ms−1

restu

Figure 2:

Part (b):

Figure 2 shows two balls, of masses, m, and 5m, hanging together by means of strings.

The ball of mass, m, is drawn aside and then released, so that it swings down and

collides elastically with the other stationary ball, with a velocity of 6 m/s.

Use conservation principles to determine the magnitude and direction of the velocity of

each ball, immediately after the collision.


6M

Mµ

PT

2Mµ

a

rest

Figure 3:

Question 2 ( (3 + 7) + (10) = 20 marks):

Part (a):

Figure 3 shows a block, of mass 2M , on top of another block of mass, M , on a fixed

horizontal surface. The M block is connected, via a string-and-pulley, to a third block,

of mass 6M . The 2M block is tethered to a fixed post, at P. When the system is

released from rest, block M accelerates, with acceleration, a, as labeled, while block

2M remains at rest, on top of M . There is a coefficient of friction, µ, between M , and

the surfaces above and below it, past which it slides. The tension in the string con-

nected to M is T , as labeled. The strings and the pulley have negligible mass and friction.

(i) Draw a separate diagram of each of blocks M , 2M and 6M . Label on each of these

diagrams, all the forces that act on the particular block. Label also, the acceleration of

each block.

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

block M , in terms of µ, and the acceleration due to gravity, g.


P

Q

u 3 mµ

4 m

v

4 m

5 m

M

horizontal

Figure 4:

Part (b):

Figure 4 shows a block, of mass, M , positioned at point P, at the bottom of an elevated

slope. The coefficient of friction between the slope and the block is µ. The block is given

an initial velocity of u, at P, slides up the slope, and is projected into space, landing at

a lower point, Q. Dimensions of the figure are labeled.

Use energy principles to derive an expression for the speed, v, of the block at point Q,

just before impact with the horizontal surface. Express v in terms of the parameters

labeled on the figure, and the acceleration due to gravity, g.


G

M

µ

µ = 0.01 kg/m

H

M = 2.0 kg

Figure 9:

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

Part (a):

Figure 9 shows a string, with mass per unit length, µ, which is stretched over a pulley,H, between a wave generator, G, at one end, and the block, of mass M , at the other.The values of µ and M are labeled on the figure. Generator G transmits an harmonicwave of frequency f = 500 Hz, and amplitude a = 2 mm, from G to H. Take theacceleration due to gravity, g = 10 m s−2.

(i) Find the velocity of the harmonic wave along the string.

(ii) Hence, write down a possible wave function for this harmonic wave.

(iii) Find the power that generator G must output to produce this harmonic wave.

Part (b):

The Balmer series of spectral lines for atomic hydrogen, are formed by electron transi-tions terminating on the n = 2 energy level. Use the Bohr theory for the hydrogen atomto answer the following questions.

(i) Calculate the two longest wavelengths, and the shortest wavelength, in thisspectral series.

(ii) What is the total energy of an electron in the n = 2 level?

(iii) What is the radius of the orbit of an electron in the n = 2 level?

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FOUNDATION STUDIES EXAMINATIONS …flai/Theory/exams/FT08_1.pdfFOUNDATION STUDIES EXAMINATIONS September 2008 PHYSICS ... Hence ﬁnd the maximum angular velocity, ω m, ... Express