Phyricr l21B-RosenbergAutumn 2013

Pritrted NaEe

Fioal ErsmDecember ll,2013

Sert Numberlqst lirtt

I certify thrt thc work I shrll submit ic my own creriion, not copled froD any sourc, rrdthrt I lhall abide by the ex{mlnrtio! procedures oullincd below.

Sigtrrtur Studcnt ID NuDberf,rrn! wtll bc rv.llrbl. for pick up In Rm. C-136 (wlth photo lD) rtler Dc.cmber l6th.

READ THIS ENTIRE PAGE NOW, BEFORE TEE HALF.HOI'R BELL.Do gl! open the exrm before the half-hour bell,

You will hsve 110 mitrutes .ftcr the bell to complete the ersminrtion.Erqm prpeB will !o longer be sccepted rfter l15 mlnutes have elapsed.

P CELL PHONES, TEXT MSG, etc. ALLOWED AT AIYY TIME

Bfore the xam beginsi. Write your name, student ID number and exam version on your bubble sheet, and

fill in the concsponding "bubbles" using dark pencil marks.

Durhg ahc eram:. Inportrrt lirlt itep: Print your name and student ID at the top of each page.o lf you are confused about a question, raise your hand and ask for ao explanalion.. Ifyou cannot do one part ofa problem, move on to the next part.. This is a closed book examination. You may not bring any information o! an

equation sheet to this exam.. You may use a calculator, but you may Doa use text storage capabilities, graphics

capabilities, intcmet connectioDs, phones, nor any programmable device,o You may use scratch paper. Do not tum in scratch paper, it will not be gaded.. You may not communicate with any person.

For multiple-choice problcms (those ou white psper):r Fill iq bubble sheets carefully and darkly, Make no stsy marks, Erase carefully.. AIso circle your choices directly on the exam paper for later reference.

For haod graded problcm! (those otr colored prper):. lf you need more space than is available to answer any palt of a problem, use thc

b|ck lide of the lsme pag to complete your answer. Clearly indicate to thegrader that you used the back side. Do 49! use scratch papr,

r Show your work in enough detail so that the grader can follow your teasoning andyour method ofsolution. Circle your answe$, and state units ifappropriate. Fornumerical answers significant figures should match the number of significantfigures in the numerical values given in the goblem (usually 2 or 3).

Auuldr Qulncl 2013l2lB (horors) Finrl ExaD EqultioDs

= CIDdt

i -nf; g-e,8N/kg-dr-ctvy-- a--dt dt_ I r - _ _\rcu - fi\\rr+ nzrz+...+ mNrN )D - |,fiE - K,+ K,+ K,+...+v(ar)+r(ar)+v(r.)+.^

..tD2Kt't -:mvz -:- ' K- -!Ia222m'2dK-F.aP-tF./t-wl,,l!yl

' I ldt I ldt ledUf. r --; i! drFcrioo r

Y"(z)-nv

v (r) - rlfz; t - 8.99 x tqeJ. rnlct; lrl - t.ooz rro-'tc

r(r)--cWz;at \ ^ ntnt- ^r l r l - t r - r;/ \ , I \i l r l - - r - l f - r" l' \ " / spdng

Lr14-;k,1,-n)_ spnnt

, ' -k//mx(t)- Acns( +e) or r(r) - lcos(ot) + Bsin(@o

c - 6.67 x 10-rrJ.m/kg2

d2xft\-if --''xO.i - t1 ,v, + v ,y, +u "t/, - w cosonag 0 xl -W sn?u " I - ;(u,rr, - u,y y) + i (u,y, - u,r,), t (u "v, - u,r,,)L-lxF -mrza (circular motion) -16 (azimuthally symnetric)-AL-=

d,t- t JItt l- -; - ld9l^ Y - 2,v(t) - l - - : - l ( , l ( ))a- I --ldt l r a)

.t R: rrt ilirl

tht l

ln ,ws.r,*',^, 'n"'^*t ,- ':l-r *r,,

i,llx:\(tlid

j R: slt''"'

t - znllv_

p

U^' -1Nk.T; f" - l.38xl0-'J/K2-dUw = M cdT: qg;nct*

dtltu -mL; trrat iar io vrporiz.c *ltRocket! whh no exteanal forces. .d l -= dMM--r ._

dt-dr-c ley3- a--dt dt

u'a ' -uaiform circular motioo: R nonuaifonn circular motion:

- dv^ v2^dtR

v,(t)-a,t+vo, xg1-)a72 +vo,t + xo

&ot"'t'.rt"'rr*-)cen]t'(t) - t(t) - F(t) caua,o valocity rr$formltiotrdO-aO-f,O cr or! rccolqtioo irodomrtionnd--nl+F,+F-+,, , - . . .tlcuuouS tonocs

dtLKoplcr's rccood law:

Kptci'sthird lEw: "

-*ot "t* v - 1ff

f>fnlli(rr o{ E .rd L u .2Gt L;- t - , dno -rr"ant6 = ioLl r',tLnr .rt8/L -- ry.tat| .nauhr rtloas't8G q tnvltdiird (urr.nl

r = dbLrr. frqrr Dilrn.tyr - ..ldllb !p..d !t th.t pdntt = rrh b.trt !tr| t .nd t, - rir |n.F. olb for .lltFe,.'ulogour tor htFttol.r, = dbLnc. ftq|r Ftr|'ry !o

dcr pd dr lh..ibll4E dLt|tGfrnr Fltrl ry to

,.dlE*poinl ar d|eorbltI= ortit t pcrtdo = rnsL ol dtnpaoa..

(:rkuhdur.lf ccttn!.lcily ,-, ,i"b,(*)'(!i)hin Cl|clkrl C..

Co.tltc.tltr| hll[!cr| n .nd E 2E CM 2E ,CM

\+rt =Alrr -.0 +.)()rhr t&f'd rEhtlirb e-{""

(&?hrt drird Lw)

trnd = .,/FJr /1\ ,28= er (;, v;

K(YNaEa

laaFrStud.lt lD

01 Decr> ol ioaScorc

rtrstI' spring In an experimcnt, an cnd ofa sprhg on a frictionress horizoqtal surface is fixed inplace. The spring is found to be stretched g.O cm fiom its equilibrium posirion when a force of3.0 N acts on the other end. In another experimenl a 0.7 kg mass attacied to the some springrcplaces the 3.0 N force. When the mass is pulled +l0.0 cm and released at 14 sec, the massundergoes simple harmonic motion.

l. (3 prs) Vr'hich answer is closest to the sp.ing-constant?

i:liil,,l * 'h, = z?,e ",/rv2. (3 pts) Which answer is closest to the force exened by fte spring on the 0.7 kg massjustbefore it's released?

F= 37,5"o, 1 .3,75NA, IN.8.2N.

,llijls) Which answer is closesr ro the oscillation period ofthe mass?\9-:-.r-T?-r1" T - zn\ fn/x = 2r(o,z/z l . f | . o,36 sC.3s. ' v ' t 'D.4s.8.5s.

4. (3 prs) Which answer is closest ro the amplitude ofthe harrnonic motion?A, 0.01 cm.B.0. lcm. O, l na

5. (3 pts) Which answer is closest to the maximum speed ofthe 0.1 kg mass?f : f j [ ' , /z kxz=r/rmvZ

E. 100. rr/s. V=lqlY =i" ,?,r7;7o,t - o,?3)/s

c.3N.

B. 0.1 cm.

Phylics l2l, Autunn t3 Final .xsm, pagc I

Ntna Srrd.rt lD Scorafirsl

6. (4 pts) f4fch answer is closst to the maximum acceleration ofthe 0.2 kg mass?i *l'-$' 4 . c, ?k =u/. x

E. 500. m,/s'.=Ho,r =S, jc^t /z

7. (4 ps) Which answer is closest to the s?ecd ofthe 0.7 kg mass wten it is halfway betwecn itsequilibrium positioD and its maximum cxtension?

40..06 rtls.@!-M)A,0.006rn/s. r , ._ 2.!-.0.06 m/s. -t K 4. . - -

= +kX,; + +h4vzD. 6.0 ltrls.E. 60. m/s. v =tffi = o,s3)/ tt. (4 pts) Which answsr is closst to the magnitude ofthe accelention of the 0.7 kg mass whenit is halfi^,ay-bctween its equilibrium position and its maximum extension?

f : ; . i ' " f f \G )-kryq,(o!t- i ; 4k)-- azx4r- (oect4y", hJLXn^^ ' /z

= 5,7. : o, t + j1?. t - 7.(7 , /s.

c.D.E.

E. 300. m/s2.

9. (4 prs) Which answer is closest trio t -7st to the kinetic cnergy ofthe 0.7 kg mass when it is halfiray

librium position and its maximum extension?k= t \vL

= + o, ' t ( t , t3) ' . o, l t r t10. Joules.l0O. Joules.I 000. Joules.

10. (4 pts) Which answer is closst to the tottl cnergy ofthe oscillating systm? (Assume rhe,fotEotialtle$y is zero at the equilibrium position)./A. 0.1 JoulcJ

C. 10. Joules.D. 100. Joules.E. 1000. Joules.

24oq'

Physics | 2I, Autunn l3 Final cxrm, page 2

Nrm Studnt IDJirst

(4 pts) Which expression is closest to the value ofx for which the potential is a minimum?hl4o/ . - , , ^ J l t t t t ' . t /\3 0 =?xU(x), O' ' t /4/xtz+ bb/x7

zqA=at'2a./,

/D=i

11. (3 pts) Which expression is closest to the displacement of the patlftom its equilibrium

B' xC) = 0.1m sin(2rr 7/s t)

.-6:##d#.-#i x(+) = o,tvn .ae l'/ *\E. x(t) - 0.lm sin(;7/s t)

II. PoteDliel between two rtoms. The poiential energy between two atoms in a diatomicmolecule can be expressed approximately as uAt=$-$ with a and b positive constants and xthe distance between the centers of the two aloms.

12. (4 pts) Which value(s) ofx are closest to those for which,the potential is equal to zero?A'o r , -a/ ' /B.oando O' ' / * ' . - o/X t ' /b --x

6

i:#:l: x - 6,P6 -L @

13.A.

\WE,14. (3 pts) Which ofthe following statements is most co.rect rcgarding the atom-atom force forseparations precisly at the potential minimum?A. Conservation ofenergy requires the position ofthe potential minimum be at zero energy.B. The position ofthe potential minimum depends on the convention for the potential at @.C. The atoms are in unstable equilibrium at the potential minimumD. The$rce-begeelthe atoms is unbalanced, so the molecule will oscillate.

C-T:fhe fo'ce betoeen rhfiiiiiG@q;>\\----=.

Physics 12l, Aut'rnn l3 Finsl exarn, page 3

A. The force between rheii6friffiulsive for separations slightly less than that at thercly morethanthatatthe potential n

Saudent lI) Score

15. (4 pts) Which ofthe follo\ving statements is most correct rcgarding the atom_abm fbrce forminimum?

rrlllllmutnC. For positions greater than the porentiat minimum the atomic forc"i;;;lr;r".D. Since all we know about the constants a and b is that they are positive,'we donl knowwhether for_ positioas less than the potential minimum the force is attractive or repulsive.E. Since all we know about the constants a and b is rhat they are positive, we don,t knowwhether for positions greater than the potential minimum rhe force is atuactive or repulsive..16' (4 pts) which ofthe following statements is most conect regarding rhe oscillation ofrhemolecule for separ"ations neat the potential minimum?A. The molecule will not oscillate.B. The molecule is unstable equilibrium; fo owing any displacement, the atoms either cometogether or lly apart.C. In general the oscillation is that of simple hamonic motion.

aE, Since the potential energ5r becomcs infinitely large at close distances, it will take an infiniteamount ofenergy to separate the atoms.

1t. (3 pts) suppose rhe molecule consists oflwo cqual-mass atoms. Also suppos the moleculerotates at-an average angular frequency o about its centcr-of_mass in addition to its oscilaforymotion. What's the total energy ofthe molecular system?4,0

Ll1\ lB' i lh* + K'/w a v.! F-zTar rap4t

c.q krN5r,c NT).6Y4^h?

D. I

E. The molecute witt oscillate, bu iG motion rvitt nor t" aGiEA-Tiffi-n"r-onie_soli

E. It can't be derermin

Physics l2l , Aulumn t3 Finalcxam, pagc 4

Name Student ID Score

III. Lab question. Consider the system shown. The radius of the pulley #l on which the string iswound is 2.0 cm. Assume the string and the pulleys are massless and thin. Assume there's nofiiction. The hanging weight has a mass of0.50 .r 0.01 kg. The rotating system is initially held atrest and it is released at time F 0 s. At F0.5 s the hanging weight has dropped by 0.2 m.

19. (3 pts) Wllat's a good unit for to.que?A.N

c.wD. rad./secE. Nsec

20. (3 pts) In units you established in the p.evious problem, what's nearcst to the torquenecessary to hold the roror in place (at time t

Strd.lt lllNrnc

D. 0.4 JouleE. -0.4 Joule

t"ti------ 1"" -

22. (4 pts) Which value bcst describes the rotational kinetic energy ofthe rolating syslem at hme

flt1,it"*-rq"' ' M1 11 -Ke'1--- O,5 1,YO,Z-O,6 to ' lLe

O'?2 ral t23. (4 pts) Whafs ihe kiuetic energy ofhanging mass/at r=0 5 s? ., * ,/l . l9^r : : \ x. 'hqt ' i 4 =zxAa' v =4t ' - /+ffi,**r = Jr-vL . .zrq. (*Al.;: ; :d;f f i _L 0.5 (o,z7o,-, )z = o. t{ d-d,rL?2,1. (4 pts) what's thc angulsr velocity of the rorating weight at r{'5 s?

i f f i ht=v1. '4 f.tor@ t rn-g: -- za:l L -

t1o, 1se-A7 ' . i

25. (4 pts) Whst's the angular velocrty. (D' Jf i# 'otuting -*t "r Fl '0 s rlative to to d H) 5 s?

s)s,s,s,

1.0nt1.01.0

(FG(F(F

(0d|

oo

B.cD.E.

z(D1.41.00.5

Nk=o,5 t )q [ t=t .o e )

{ffirrr (r=0.5 s)ro (r=0.5 s)

-a--q^(t

O,, 5t t o t- -o, f s)au(t=l .o s) ' 7

E. l0 rad/s

PhFics l2l, Autumn ll Fin l cxsrn, Prgc 6

Nsme StudeDt ID Scorclast rtrsl

lV Il5 points totall For each of the following problems, a point particle is part of a system. The potentialenergy U of the system is shown as a function of the position .r of the particle. Assume that the only formsof energy are potmtial energy of the system atrd translatiotral kinetic energy of the particle. The only forceacting on the particle is that associated with the potential energy.

Use the diagram ar right for the lollowing two pmblems. Thehorizontal line Ed represents the total energy of the system.26. [5 ptsl When the panicle is at .r = 6 cm, which of thefollowing best describes how the spd of the p|rticle iscbatrging?

A. Speeding upB. Slowins down

Lj-z/ uttrer sPemlng uP of slowlnS oownD. The particle is not movingE, None of the above

4J3J2t

OJIJ2l

27. [5 ptsl Consider lhe positions .r that satisfy the relationship U(,r) < 0 on the diagr-am above.Which of the followiDg is a troe statement about these po6itions?A. These positions describe the cntire allowed region.B. These positions descdbe the enlire forbidden region.C. The force on the particle at these positions is in a single direction.D. The padcle stops and changes dircction at one or more ofthese positions.

,'6}oo. of rh" "bou".

2t. 15 ptsl Consider rhe gr-aph of potential energy vr. positionshown at.ight.Suppo6e that the panicle were rcleased from rcst at x = Icm.What is the mrxihom kiDctic eners/ obtained by thepanicle?A. IJ

( B. NJc. 3JD.4JE. 5J

4l]J2JIJOJ

- lJ2I

Posilion (cn)

Posirion (cm)

Physics l2lB, Autum 2013

Name Studenr IDfirst

V U 5 points totall Spools A and B are attached to a ceiling bymassless strings. The strings are wrapped many times arcundeach spool such that the spools may unwind. The spools areheld at the same heighl lhen released from resl al time tr. asshown near right. At time rr. spool B has fallen a greaterdistance than spool A.

The spools have the same mass and mdius, but the momentof inertia of spool A is greater than that of spool B.(/o>1u)

29. [5-pts] How does the magnitude ofthe netforceon spoolA,l4'1,compare to thaton spool B,

A. rFY i> rA" l(9rrF." ' r l ,(" '- l

@rl-r

t121H Fina.l Exam SolutionsVI. Peadulum. As ahcn'n, a pendulum of st ns-length Z rnd mass m is ob6ervd t harc s sped h

wheu the cord srakes atr atrgle do (0 < ,o < r/2) with th vEtticd.

r. (7 pts) In terms of the gilleD quantities rnd 9, vhat's th total ede!6/ ofth systm? Tble the potdialonersr of the podulum bob st it"e lo*E6t position to be zso.

Sitrcc iEk dclnins h : o at the lln.rt pint o! tl' pzn&thnn ati,,s, tl],- h'l.,1!n ol th. pe'!h&n ;tat h = t. Ihe Inb ir lsBo btow that, Md ro tlE hc4ti ol t u to, nt lr : l(1 - ccd). So, tlE tntt!ercW o, thc tgtten, ushg tle tuitbl Lwlucs we'ft gii'en, is

E=KE+PE

t'l:,vi +nst(r ,rxqi)

Cs

2. (? pt) Whai's the lped of the pndulum bob at its lc'wst psitiotr?

ttncr ,r : 0 at the bb's kruest p!;tiorl, and tttll ctuw is 6 craen, dI of tlE ii^itial cnsglt mxnruu b it tJe lorm ol Eiftti, erwtU.

i''/,2-*: ir'r.ui +msr(\- (Iueo\i,3+29r(r-co6ro)

3. (8 pt5) WLar's the llst tdu the spd Uo could hav so the strins i.s hodzontal at some poht dunnsits motion?

To fta.l a horiza d piit (h - 2), tln ssstan nulrt haoc ot bast rnsl ol energs. Thir mea'], thcminl,'}d q ir stuen W

t'llu)A+mst(r-crxeo)=msli'I'.ii : nsr (,xeo

t-- : lth:1\/2ezsxeol

4. (8 pt) What's the le,st v8lue th spd rb cordd have so the !'elrdulum will lot 6cill"st, but rathrwill continue to mtst around in a rutical circl?

To m&e it oll thc un! ar?,!wl tlu ciir.cle, thc bob m'r'tt h goiry latt enowh aa the top ol the ciftle to'tot "toll in', i-e., the loiz of snl,tts drn te no

greatr thsd tlr .al@al lonE nerded to rtry on tiecit nlor

',''},. fr just banly md*e it atu'n4 ue need

So, thc tan ti erErss at tlrc top muct tE d hort imst. A&iw to tM tl' 2msl ol pt 'ttid e erysot the to? ol the ciftle, atuI setttns thot eqtd to otlr capft.'.irn lor total eftrw fiwn poblen (1),

r''x?*--7= : ^g+nu|.e: ir'l4t.

lnst + 2mst : tmd + mst(l- cr'9s)\sl : tt + 2st(r - s' eo)uA= sl(3 +2(Xseo)