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Queen’s University at Kingston

Faculty of Arts and Science

Department of Physics

PHYSICS 106

Final Examination

April 16th, 2009

Professor: A. B. McLean

Time allowed: 3 HOURS

Instructions This examination is three hours in length.

Part I: Put your student number on the scantron sheet in two places. At-tempt all ten multiple choice questions and record your answers on thescantron sheet. Each question is worth one mark. The boxes should bepenciled so that the letter is not visible. Select test form ”A”.

Part II: Put your student number on the front of the worksheets. Answertwo out of three worksheet questions. Indicate clearly which questions youwant marked or the first two answers in the worksheets will be marked. Youmay do rough working in the answer booklet. Each question is worth fivemarks. When finished put your scantron sheet and worksheet inside yourrough working booklet and write your student number on the front.

Please note: “Proctors are unable to respond to queries about the inter-pretation of exam questions. Do your best to answer exam questions aswritten.”

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Part I: Attempt all 10 questions Page 2 of 5

Qu 1 Seven identical charges of magnitude Q = +1.0 nC are located at thevertices of a cube of side a = 1.0 cm. The potential at point P is?A 2400 V B 3200 V C 4600 V D 5100 V E 6200 V

Pa

Q

Qu 2 While spinning down from 500.0 r.p.m. to rest, a solid uniform flywheeldoes 5.1 kJ of work. If the radius of the disk is 1.2 m, what is its mass? Youmay use the fact that the moment of inertia of a uniform disk about thecenter is Ic = 1

2mr2.

A 4.1 kg B 5.2 kg C 4.4 kg D 6.0 kg E 6.8 kg Answer: B

Qu 3 Three blocks of length L are arranged as shown. The block on top hasmass 2m. The block in the middle has mass m and the block on the bottomhas mass 3m. What is the maximum value for x for which the blocks arestable and do not collapse?A L/4 B L/3 C L/5 D 2L/3 E 5L/6

x

L

2m

m

3m

2

Part I: Attempt all 10 questions Page 3 of 5

Qu 4

4

22 V

2 V

Ω

Ω

Ω

I4 V

1

I2

I1+I2

2

4 V

Which of the following is correct:A I1 = 3/5 AB I1 = 7/5 AC I2 = 6/5 AD I1 + I2 = 4/5 AE I1 + I2 = −7/5 AAnswer: D

Qu 5 An insulating solid sphere, with radius a = 1.0 m, has a charge ofQ = +1.0×10−9 C distributed uniformly throughout its volume. Use Gauss’slaw to find the magnitude of the electric field a distance of 0.5 m from thecenter of the sphere.A 2.3 N/CB 4.5 N/CC 7.2 N/CD 7.2 N/CE 9.0 N/C Answer: B

3

Part I: Attempt all 10 questions Page 4 of 5

Qu 6 Three charges, Q1 = Q2 = 20 × 10−6 C and Q3 = 40 × 10−6 C, arelocated in an equilateral triangle of side a = 1.0 m. The net force (Fx, Fy)acting on Q1 is?

Q

1

2

a

a

a Qx

yQ3

A (3.6, -1.4) N B (7.2, -2.4) N C (7.2, -3.2) ND (9.4, -1.8) N E (9.4, -2.4) N

Qu 7 Positive charge Q is placed on a conducting spherical shell with innerradius a1 and outer radius a2. A point charge q is placed at the center ofthe cavity. The magnitude of the electric field at a point outside the shell, adistance r from the center, is:A. κ(q +Q)/r2

B. κQ/(a21 − r2)

C. κq/r2

D. κQ/a21

E. κ(q +Q)/(a21 − r2)

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Part I: Attempt all 10 questions Page 5 of 5

Qu 8 The figure shows a current entering a truncated solid cone made of aconducting metal. The electron drift speed at the 3.0 mm diameter end ofthe cone is 4.0 × 10−4 m/s. What is the electron drift speed at the 1.0 mmdiameter end of the wire?A 3.6 ×10−3 m/s B 1.2 ×10−3 m/s C 1.3 ×10−4 m/s D 4.4 ×10−5

m/s E 5.2 ×10−5 m/s

3mm1mm

I = 1.0 A

Answer: A

Qu 9 A projectile is launched over horizontal ground as shown. At which ofthe points A-E is ~v perpendicular to ~a?

0

y

xA

B

C

D

E45o

vi

Qu 10 Two boys, with masses of 40 kg and 60 kg, respectively, stand on ahorizontal frictionless surface holding the ends of a light 10 m long rod. Theboys pull themselves together along the rod. What distance will the 40 kgboy have moved when they meet?A 4 m B 5 m C 6 m D 10 m E need to know the forces they exertAnswer: C

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Part II: Long Question Worksheets (attempt 2 out of 3)Student Number:

Worksheet One

(a) A marble rolls down a track and around a loop-the-loop of radius Rwithout slipping. The marble has mass m and radius r. Calculate a formulafor the minimum height h the track must have for the marble to make itaround the loop-the-loop without falling off. For the marble Ic = 2

5mr2.

hR

2r

continued over . . .

1

(b) A cube slides down a frictionless track and around a loop-the-loop ofradius R. The cube has mass m and the length of each side is 2r. Calculatea formula for the minimum height h the track must have for the cube tomake it around the loop-the-loop without falling off.

hR

2r

continued over . . .

2

(c) If the answers to parts (a) and (b) are the same, explain why they arethe same. If the answers to parts (a) and (b) are different, explain why theyare different.

3

Part II: Long Question Worksheets (attempt 2 out of 3)Student Number:

Worksheet Two

a The figure shows a thin rod with charge Q that has been bent into asemi-circle of radius R. Find an expression for the electric potential at thecenter.

center

Q R

b A disk with a hole has inner radius a and outer radius b. The disk isuniformly charged with total charge Q and it lies in the xy plane. Show thatthe on-axis electric potential at a distance z from the center of the disk is

V (z) = κ2πσ[√b2 + z2 −

√a2 + z2

]You may use: ∫ b

a

xdx√x2 + z2

=[√x2 + z2

]b

a

continued over . . .

4

Put your answer to part b on this page.

continued over . . .

5

c Using the result from part b, and Ez = −dV/dz, calculate the electric fieldat distance z from the center of the disk.

d Use the result from part c to calculate the electric field above an infiniteuniformly charged plane.

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Part II: Long Question Worksheets (attempt 2 out of 3)Student Number:

Worksheet Three

A rectangular current loop, with sides of length a and b, is placed in amagnetic field that is uniform and parallel to the x-axis. The loop is pivotedabout points P1 and P2, and its axis of rotation is the line that joins thesetwo points. Conventional current flows in the direction shown. θ is the anglebetween the normal to the current loop (n) and the ~B field.

a Draw the forces on each side of the loop. Indicate in what direction theloop will rotate?

n

bB

x

z

0

2

1

θ

a4

3

y

I

Ipivot

pivot

nbθ

Bx

z

0

I

pivot

perspective view side view

P1

P2

b Add a vector to both diagrams, showing the magnetic moment ~µ.

c How is the magnitude of the magnetic moment related to the size of theloop and the current I?

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d Explicitly calculate the forces on each side of the loop (~F1 → ~F4) andderive a formula for the torque on the current loop.

e Graph the magnitude of the torque for angles in the range −π ≤ θ ≤ π.

continued over . . .

8

f Can you make a motor using the current loop shown above? If not, whatcould be changed to cause the loop to rotate continuously in one direction?

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PHYS106 EQUATION SHEET Page 1 of 6

Vectors

• a = axx+ ayy + az z and b = bxx+ byy + bz z

• a · b = ab cos θ = axbx + ayby + azbz cos θ = (a · b)/ab

• a× b = ab sin θe = x(aybz − azby) + y(azbx − axbz) + z(axby − aybx)

• |a| =√a2

x + a2y + a2

z a = a/a

Math

f xn ex 1x

lnx sinx cosx tanx f(g(x))

dfdx

nxn−1 ex − 1x2

1x

cosx − sinx sec2 x dfdg

dgdx

Roots of quadratic ax2 + bx+ c = 0 are x = 12a

(−b±√b2 − 4ac).

Area of circle = πr2, circumference of circle = 2πr, area of sphere = 4πr2,volume of sphere = 4πr3/3

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PHYS106 EQUATION SHEET Page 2 of 6

Kinematics

r = xx+ yy + zz v =dr

dta =

dv

dt

Constant acceleration t ≡ t12 Constant angular acceleration t ≡ t12

• x12 = v1t+ 12at2 θ12 = ω1t+ 1

2αt2

• v12 = at ω12 = αt

• v22 = v2

1 + 2ax12 ω22 = ω2

1 + 2αθ12

Projectiles: R = v2o sin 2θ/g h = v2

yo/2g

Uniform circular motion (ω = dθ/dt = 2πf = 2π/T )

v = ωr v = ωrθ θ is tangential

ar = ω2r = v2

rar = −ω2rr = −v2

rr r is radial

DynamicsFnet =

∑i

Fi = ma

Gravity

F = −GMm

r2r F = mg U = mgh

G = 6.67× 10−11 Nm2kg−2 g = 9.81 ms−2

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PHYS106 EQUATION SHEET Page 3 of 6

Momentum, impulse and torque

p = mv F =dp

dtl = r× p ~τ =

dl

dt

Impulse p12 = p2 − p1 = J =∫ t2

t1F(t)dt

~τ = r× F τ = rF sin θ = r⊥F = rF⊥

v1 =m1 −m2

m1 +m2

u1 +2m2

m1 +m2

u2 v2 =2m1

m1 +m2

u1 +m2 −m1

m1 +m2

u2

Constant Value Unitse 1.6 ×10−19 Cεo 8.854 ×10−12 N−1m−2C2

κ = 1/4πεo 9 ×109 Nm2C−2

me 9.1 ×10−31 kgmn 1.7 ×10−27 kgmp 1.7 ×10−27 kg

κm = µ/4π 10−7 TmA−1

µo 4π × 10−7 TmA−1

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PHYS106 EQUATION SHEET Page 4 of 6

Systems

∑i

Fexti = P If

∑i

Fexti = 0 then P = P′

Elastic collision : K1 = K2 Inelastic collision : K1 6= K2

Center of mass : xc =1

M

∑i

mixi yc =1

M

∑i

miyi zc =1

M

∑i

mizi

I =∑

i

mir2i Ip = Ic +Md2

P = Mvc F =dP

dt= Mac L = I~ω

∑i

~τi =dL

dt= I~α

For rolling motion with no slipping, xc = θr, vc = ωr, ac = αr

K =1

2Icω

2 +1

2Mv2

c K =1

2Icω

2 (fixed axis)

W12 =∫ 2

1F · dl W = F · d P =

dW

dt= F · v W12 =

∫ 2

1τdθ P = τω

F = −kx U =1

2kx2 W12 = K12 =

1

2mv2

2−1

2mv2

1 W12+U12 = 0 (conservative)

4

PHYS106 EQUATION SHEET Page 5 of 6

Statics ∑i

Fi = 0∑

i

~τi = 0

Electrostatics

F =κQqtr

r2E =

F

qtE =

κQr

r2U =

κQqtr

V =U

qtV =

κQ

r

E =∑

i

Ei E = κ∫ dQr

r2dQ = λdl = σdA = ρdν

ΦE =∮

AE · dA =

Qenc

εoE =

Qenc

A‖ε

p = QD (pointing from −Q to +Q) ~τ = p×E U = −p·E Vp =κp cos θ

r2

V12 = −∫ 2

1E · dl U12 = −

∫ 2

1F · dl E = −dV

dl⊥

Magnetostatics

F = qv ×B dF = Idl×B rc =mv⊥qB

T =2πm

qB

~τ = ~µ×B ~µ = I ~A U = −~µ ·B dB = kmIdl× rr2

∮LB · dl = µIenc

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PHYS106 EQUATION SHEET Page 6 of 6

Circuits

Kirchhoff’s Laws: The Loop law:∑

i Vi = 0. The Node law:∑

i Ii = 0.

Equivalent resistances: Series req =∑

i ri. Parallel r−1eq =

∑i r−1i

V = IR P = IV = I2R =V 2

R

R = ρL

Aρ =

1

σJ = nevd J =

I

A

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