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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.
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Durhg ahc eram:. Inportrrt lirlt itep: Print your name and
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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
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For multiple-choice problcms (those ou white psper):r Fill iq
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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).
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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
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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
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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;
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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
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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. . -
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= +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$y is zero at the
equilibrium position)./A. 0.1 JoulcJ
C. 10. Joules.D. 100. Joules.E. 1000. Joules.
24oq'
Physics | 2I, Autunn l3 Final cxrm, page 2
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Nrm Studnt IDJirst
(4 pts) Which expression is closest to the value ofx for which
the potential is a minimum?hl4o/ . - , , ^ J l t t t t ' . t /\3 0
=?xU(x), O' ' t /4/xtz+ bb/x7
zqA=at'2a./,
/D=i
11. (3 pts) Which expression is closest to the displacement of
the patlftom its equilibrium
B' xC) = 0.1m sin(2rr 7/s t)
.-6:##d#.-#i x(+) = o,tvn .ae l'/ *\E. x(t) - 0.lm sin(;7/s
t)
II. PoteDliel between two rtoms. The poiential energy between
two atoms in a diatomicmolecule can be expressed approximately as
uAt=$-$ 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
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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
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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
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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
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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
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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
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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)
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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)