Physics 111

Optional Extra Credit Assignment

Assigned Wednesday, March 9, 2011 @ 9:00am

Due on or before Friday, March 11, 2011 @ 8:00am

Name___________________________________

Problem #1 /20 Problem #2 /20 Problem #3 /20 Problem #4 /20 Problem #5 /20

Total /100 Extra Points on

Final Total

/5

RulesofEngagement:

1. This optional extra credit assignment is assigned on Wednesday, March 9, 2011. 2. This optional extra credit assignment is due on or before Friday, March 11, 2011

at 8am. 3. Late extra credit assignments will not be looked at or graded and no extra credit

points will be assigned. 4. This optional extra credit assignment will award you up to 5 additional points on

top of your final class score. You are under no obligation to do any of this assignment.

5. You may do some or all of the assignment. Any amount you do will earn you some number of extra points on your final class score.

6. I will grade this optional assignment and the percent you earn on this assignment will be multiplied by 5 to determine the number of extra credit points you will earn on your final score.

7. I will determine grades for the class as if this assignment was not present and after assigning grade ranges will then add your extra credit points onto your final score. This has the potential to raise your grade by one letter grade. For example, if your final grade were at the class average of a B-, these extra points could potentially bring your grade up to a B.

8. Since this assignment is optional, you do not have to do it and your final grade will be whatever you earn.

9. This optional assignment can only help you, not hurt you. 10. You must work independently and consult no one on the preparation of this

optional extra credit assignment. 11. You may consult only your textbook, class notes, or class homework assignment

solutions for guidance. 12. You may not consult tutors of any kind, other faculty, other students, or the

Internet for solutions. 13. I will treat this optional assignment as if it were an exam. I will answer questions,

but will not help you with the solutions. 14. This optional assignment is also good preparation for the final exam, as many of

these same topics will be covered. FreeResponseProblems:Thefiveproblemsbelowareworth100pointstotalandeachseparateproblemisworth20points.Pleaseshowallworkinordertoreceivepartialcredit.Ifyoursolutionsareillegibleorillogicalnocreditwillbegiven.Anumberwithnoworkshown(evenifcorrect)willbegivennocredit.Pleaseusethebackofthepageifnecessary,butnumbertheproblemyouareworkingon.

1. A particular kind of mass spectrometer is shown below. Carbon from a sample is ionized in the ion source at the left. The resulting singly ionized

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612C+and

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614C+ ions

have negligibly small initial velocities and can be considered at rest. They are accelerated through the potential difference ΔV1. They then enter a region where the magnetic field has a fixed magnitude of B = 0.2T. The ions pass through electric deflection plates that are 1cm apart and have a potential difference ΔV2 that is adjusted so that the electric deflection and magnetic deflection cancel each other for a particular isotope: one isotope goes straight through, and the other isotope is deflected and misses the entrance to the next section of the spectrometer. The distance from the entrance to the fixed ion detector is a distance of 20cm.

There are controls that let you vary the accelerating potential ΔV1 and the deflection

potential ΔV2 in order that only

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612C+ or

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614C+ ions go all the way through the system

and reach the detector. You count each kind of ion for fixed times and thus determine relative abundances. The various deflections insure that you count only the desired type of ion for a particular setting of the two voltages.

1a. Which accelerating plate is positive (left or right), which deflection plate is positive (upper or lower) and what is the direction of the magnetic field? Be sure to explain your reasoning for the choices you make.

Accelerating plates: Left is + and right is – so that the charges are accelerated across the gap.

Deflection plates: Upper is + and lower is – so that the electric field (and the electric force on a positively charged particle) points vertically down and points opposite to the magnetic force. The positively charged particle bends upwards in the magnetic field, by the RHR the magnetic field has to point into the page.

1b. Derive expressions for ΔV1 and ΔV2 in terms of the given quantities and the

masses and charges of the ions.

To determine ΔV1, work is done accelerating charge q across the accelerating

region.

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qΔV1 = 12mv

2 →ΔV1 =mv 2

2q where the velocity is determined from the

bending of the charged particle in the far right external magnetic field. The

velocity is thus

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FB = qvB = m v 2

R→ v =

qRBm

. Thus we get

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ΔV1 =mv 2

2q=m2q

qRBm

2

=qR2B2

2m.

To determine ΔV2 we equate the electric and magnetic forces in the deflection region. We have

€

FB − FE = 0→ FB = FE → qvB = qE = qΔV2l

→ΔV2 = Blv = Bl qRBm

=

qRlB2

m

1c. What are the appropriate values for ΔV1 and ΔV2 for

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612C+ and

€

614C+ respectively?

(The spectroscopic masses for

€

612C+ and

€

614C+ are 12.00000u and 14.003242u

respectively.)

For 12C

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ΔV1 =qR2B2

2m=1.6 ×10−19C × 0.1m( )2 × 0.2T( )2

2 12.00000u × 1.66 ×10−27kg

1u

=1606V =1.6kV

ΔV2 =qRlB2

m=1.6 ×10−19C × 0.1m × 0.01m × 0.2T( )2

12.00000u × 1.66 ×10−27kg

1u

= 321.3V = 0.32kV

For 14C

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ΔV1 =qR2B2

2m=1.6 ×10−19C × 0.1m( )2 × 0.2T( )2

2 14.003242u × 1.66 ×10−27kg

1u

=1377V =1.4kV

ΔV2 =qRlB2

m=1.6 ×10−19C × 0.1m × 0.01m × 0.2T( )2

14.003242u × 1.66 ×10−27kg

1u

= 275.3V = 0.28kV

1d. Suppose that your sample is actually a piece of cloth from an old artifact. If is found to contain only 6.0% of

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614 C that a sample of newly made cloth contains,

how old is your sample?

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N t( ) = 0.06No = Noe−λt → t = −

1λln 0.06( ) = −

5730yr0.693

ln 0.06( ) = 23,262yrs ~ 23000yrs

2. The Bell helicopter shown below has two blades extending out from a hub at the center of the aircraft. Each blade has a length from the hub to the tip of the rotor of 6.7m. The hub and blade assembly rotates at 6 rev/sec.

2a. What is the angular velocity of one of the blades and what is the translational velocity of the tip of the blade?

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ω = 2π radrev × 6 rev

s = 37.7 rads

vtip = Rω = 6.7m × 37.7 rads = 252.6 m

s

2b. What is the induced potential difference between the blade tip and the center hub

if the Earth has a component of its magnetic field 50µT and is pointing vertically?

There are at least two ways that you can do this problem. One way: Since all parts of the blade do not have the same translational velocity, we need to determine the average velocity to use in the equation for the motional

emf

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ε = Blvaverage = Blvtip + vhub

2

= Bl

vtip2

= 50 ×10−6T × 6.7m ×

252.6 ms

2= 0.042V

Second way: The blade sweeps out some fraction of a full circle in a time Δt. The fraction of the circle that is swept out in this time Δt is given as

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θswept out

2π=ωΔt2π

. Using Faraday’s law we have

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ε =Δ BA( )Δt

= BΔAΔt

= B

ωΔt2π

πl2

Δt

=

ε =12Bl2ω =

12× 50 ×10−6T × 6.7m( )2 × 37.7 rad

s( ) = 0.042V

2c. What is the magnitude of the electric field that is induced over the blade?

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E =ΔVΔx

=0.042V6.7m

= 0.0063 Vm

hub

blade tip

3. The Physics of Rainbows: Suppose that you are standing outside with your back to the sun just after a rainstorm and you are looking at the rain clouds off in the distance. Consider white light (white light is composed of the colors red, orange, yellow, green, blue, and violet) incident on a spherical raindrop located in the storm clouds in front of you and that the light is incident on the raindrop at an angle of 30o as shown in the figure below.

Each of the colors has its own frequency of oscillation and the index of refraction of a

material varies slightly with frequency. This is known as dispersion. When white light passes from air into a material with a higher refractive index, the material of higher refractive index disperses the white light into its component colors. Thus the different colors of light get bent by varying amounts in the material and we have a rainbow of color produced, in this case in the raindrop, and this color separation is maintained as the colors exit the raindrop and we see a rainbow of color in the sky.

Assume that the indices of refraction for water for red and blue light are nred = 1.331

and nblue = 1.343 respectively and that a typical light ray (orange) is also shown below.

3a. Suppose that red and blue light enter the drop at the 30o angle shown. Determine the angles of refraction for red and blue light as well as which color (red or blue) gets bent more. On the raindrop above draw and label the rays corresponding to red and blue light in relation to the orange ray shown.

Sincetheseanglesaremeasuredwithrespecttothenormal,bluelightbendsmorethanred.

3b. What are the critical angles (θc,r and θc,b) for red and blue light striking the back surface of the raindrop? Draw and label on the raindrop above the rays corresponding to red and blue light in relation to the orange ray shown reflecting off of the back surface. (Note: A rainbow of light is formed from many such drops with the red light emerging at the same angle from each drop and adding together to form a bright red band, with similar effects for the other colors.)

3c. Draw the distribution of colors when the light leaves the bottom of the water

droplet? Try to be as accurate as possible with your sketch and make sure you show the ordering of the colors and explain your choice for that ordering. On the drawing.

5Ω

5Ω

6Ω 7Ω

4Ω

12V

3Ω

4. Given the circuit shown below answer the following questions.

4a. What is the equivalent resistance of the circuit?

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Rright branch = 5Ω+ 5Ω+1

6Ω+

17Ω

−1

=13.2Ω

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Rleft branch = 3Ω+ 4Ω = 7Ω

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Req =1

Rright branch

+1

Rleft branch

−1

=1

13.2Ω+

17Ω

−1

= 4.6Ω

4b. What is the total current supplied by the battery?

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Itotal =VReq

=124.6Ω

= 2.6A

4c. What are the potential drop across and the current through the 6Ω resistor?

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Iright branch =Vright

Rright branch

=12V

13.2Ω= 0.91A

V6 =V7 =12V − 2V5Ω =12V − 2 Iright branchR5Ω( ) =12V − 2 0.91A × 5Ω( ) = 2.9V

V6 = I6R6 → I6 =V6

R6

=2.9V6Ω

= 0.48A

4d. What is the power dissipated across the 3Ω resistor?

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Ileft branch = Itotal − Iright branch =Vleft branch

Rleft branch

=12V7Ω

=1.7A

P3 = I32R3Ω = (1.7A)2 3Ω( ) =

V32

R3Ω

=5.1V( )2

3Ω= I3V3 =1.7A × 5.1V = 8.7W

4e. Suppose that we represent your body as a resistive circuit and that currents

between 0.1A and 1.0A can cause ventricular fibrillation (uncontrolled beating of the heart which can lead to death.) Suppose that the resistance of a single foot on wet soil is 200Ω and the resistance of the body is 1400Ω, and you are standing on wet ground, what is the current that flows through the body if the potential difference between your feet was 1500V? Is this enough of a potential difference to cause your heart to stop?

Yourfeetareinserieswithyourbodysotheequivalentresistanceisthesumofthebody’sresistanceandtwicetheresistanceofafoot,or1800Ω.Thus

thecurrentthatflowsisgivenbyOhm’sLaw

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V = IR→ I =VR

=1500V1800Ω

= 0.83A ,

andthisisenoughtocauseaheartattack.

5. Age of the Iceman: Early one afternoon in the fall of 1991 a German couple were hiking in the Italian Alps and noticed something brown sticking out of the ice 8 – 10m ahead. At first they took the object to be a doll or some rubbish, but upon closer inspection discovered it to be the body of a person trapped in the ice, with only part of the body exposed above the snow. Subsequent investigation revealed the remarkably well-preserved body of a stone-aged man, shown below, who had died in the mountains and had become entombed in the ice. When 14C-dating methods were applied to the remains of the iceman it was found that the 14C activity was 0.121Bq per gram of carbon.

5a. What is the decay constant that characterizes the decay of 14C given the half-life of 14C is 5730 yrs?

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λ =0.693t 12

=0.6935730yr

=1.21×10−4 yr−1 × 1yr31.5 ×106s

= 3.8 ×10−12 s−1

5b. What was the initial activity of the sample when the person died?

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Ao = λNo = 3.8 ×10−12 s−1( ) × 1g612C×1mol12g

×6.02 ×1023atoms

1mol×1.3×10−12

= 0.248Bq

5c. How long ago did the iceman die?

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A t( ) = Aoe−λt → t = −

1λln

A t( )A0

= −

1yr1.21×10−4

ln 0.1210.248

= 5930yr

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F = qvB sinθF = IlB sinθτ = NIAB sinθ = µB sinθPE = −µB cosθ

B =µ 0I2πr

ε induced = − N ΔφBΔt

= −NΔ BA cosθ( )

Δt

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c = fλ =1εoµo

S t( ) =energy

time × area= cεoE

2 t( ) = cB2 t( )

µ0

I = Savg = 12 cεoEmax

2 = c Bmax2

2µ0

P =Sc

=ForceArea

S = So cos2θ

v =1εµ

=cn

θinc = θrefln1 sinθ1 = n2 sinθ21f

=1do

+1di

M =hiho

= −dido

Mtotal = Mii=1

N

∏

d sinθm = d tanθm = d ymD

= mλ or m + 12( )λ

asinφm ' = atanφm' = aym 'D

= m'λ

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F = k Q1Q2

r2ˆ r

E = F q

E Q = k Q

r2ˆ r

PE = k Q1Q2

r

V r( ) = k Qr

Ex = −ΔVB ,A

ΔxWA ,B = qΔVA ,B

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g = 9.8 ms2

G = 6.67 ×10−11 Nm 2

kg 2

1e =1.6 ×10−19C

k =14πεo

= 9 ×109 C 2

Nm 2

εo = 8.85 ×10−12 Nm 2

C 2

1eV =1.6 ×10−19Jµo = 4π ×10−7 TmAc = 3×108 m

s

h = 6.63×10−34 Js

me = 9.11×10−31kg =0.511MeV

c 2

mp =1.67 ×10−27kg =937.1MeV

c 2

mn =1.69 ×10−27kg =948.3MeV

c 2

1amu =1.66 ×10−27kg =931.5MeV

c 2

NA = 6.02 ×1023

Ax 2 + Bx + C = 0→ x =−B ± B2 − 4AC

2A

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F = Δ

p Δt

=Δ mv( )Δt

= m a F = −k y F C = m v 2

Rˆ r

W = ΔKE = 12 m v f

2 − vi2( ) = −ΔPE

PEgravity = mgy

PEspring = 12 ky 2

x f = xi + vixt + 12 axt

2

v fx = vix + axt

vvx2 = vix

2 + 2axΔx

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Circles : C = 2πr = πD A = πr2

Triangles : A = 12 bh

Spheres : A = 4πr2 V = 43 πr

3

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