2. v = 3 4 c. 3. v = 4c. 5. v = 2 3 c. 6. v = 9. v = 4 3 c

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1 Vesion 074 Exam Final Daf swinney (55185) 1 This pin-ou should have 30 quesions. Muliple-choice quesions may coninue on he nex column o page find all choices befoe answeing poins AballofmassM ishownowadhegound fom a heigh H above Eah wih a speed (gh) 1/. A ficionalfoce of consan magniude (3/4)M g acs opposie o he diecion of moion. How long does i ake he ball o each he gound? H 1. g 1 H. g 3. 4( H 1) g coec H 4. 4 g H 5. 4 g ( ) H g H 7. g 8. ( H ) g Apply he kinemaic equaion of moion in he y diecion y = y 0 +v 0y + 1 a y o he ball. Fis calculae he acceleaion fom he momenum pinciple: F ne,y = = 1 4 = ma y. Hence, a y = (1/4)g. Now leing T be he ime of fligh and using y f = 0,y 0 = H,v 0y = (gh) 1/,a y = (1/4)g we have he equaion: 0 = H (gh) 1/ T 1 8 gt which can be solved using he quadaic equaion. Hence gh ± gh T = g/4 and aking T > 0 gives H T = 4 g ( 1) poins An elecon of mass m has a speed of v. A wha speed would he aio of he magniudes of he elaivisic and non-elaivisic momena be equal o? 3 1. v = 4 c coec. v = 3 4 c 3. v = 4c 4 4. v = 3 c 5. v = 3 c 6. v = 3 c 1 7. v = 4 c 5 8. v = 4 c 9. v = 4 3 c v = c The magniude of he elaivisic momenum is p = γmv = mv 1 v c.

2 Vesion 074 Exam Final Daf swinney (55185) and he non-elaivisic momenum so he aio is p p n = = p n = mv, 1 1 v c so by squaing boh sides we obain 4 = and wih a lile algeba 1 1 v c v c = 3 4 Now we muliply each side by c and ake he squae oo of boh sides o ge 3 v = 4 c poins A paicle of mass m moves along he x axis as descibed by x() = A 3 B, due o an applied foce F x (). Wha powe P() does he foce delive o he paicle? 1.P() = m(6a B)(3A B)coec. Zeo, because he objec s velociy is insananeously consan 3. P() = m(6a+b)(a 3 B ) 4. P() = m(6a B)(3A B) 5.P() = m(3a B)(3A B) 6. P() = m(6a B)(3A +B) 7. P() = m(3a B)(3A B) 8. P() = m(6a+b)(3a B).. 9. P() = m(6a B)(A 3 +B ) 10. P() = m(6a B)(A 3 B ) Thus x = A 3 A v x = dx d = 3A B and a x = d x = (6A B), so d F x = ma x = m(6a B). P = F v = F x v x = m(6a B)(3A B) poins When you ae moving up a consan speed in an elevao, hee ae wo foces acing on you: he floo pushing up on you (F 1 ) and gaviy pulling down (F ). Wha is he elaionship beween he magniudeoff 1 andf andhephysical pinciple ha explains his elaionship? 1. F 1 = F fom he momenum pinciple. coec. I depends on which diecion he elevao is moving. 3. F 1 = F fom he pinciple of ecipociy. 4. F 1 > F fom he pinciple of ecipociy. 5. F 1 < F fom he pinciple of ecipociy. 6. F 1 < F fom he momenum pinciple. 7. F 1 > F fom he momenum pinciple.

3 Vesion 074 Exam Final Daf swinney (55185) 3 Since he speed is consan, hee is no change in momenum. The momenum pinciple saes ha he ne foce mus heefoe be zeo, which equies F 1 = F poins A plane of mass m is obiing a sa of mass M inaciculapahofadiusr. Foasysem consising of only he plane and he sun, whaisheoalenegyofhesyseminems ofm, M, R,andGheunivesalgaviaional consan? Assume he sun is moionless and ha he gaviaional poenial enegy goes o zeo a infinie sepaaion. 1. GM m R. GM m 4R 3. 4GM m R 4. GM m 4R 5. GM m R coec 6. 0 Theefoe he oal enegy is given by K +U = GM m R poins A ubbe ball is dopped fom es ono he floo, andbounces back upohesameheigh fom which i saed. Ignoe he foce of ficion due o he ai. Which of he following ses of plos mos accuaely depic his moion? (The foce plos depic he foce on he ball by he envionmen.) y 1. y. v y v y F y F y 7. GM m R 8. GM m R 9. 4GM m R Since he plane is moving on a cicula obi, mv R = GM m R, which implies mv = GM m R. Theefoe, K = 1 mv = GM n R. The gaviaional poenial enegy is given by U = GM m R. y 3. coec y 4. y 5. v y v y v y F y F y F y

4 Vesion 074 Exam Final Daf swinney (55185) 4 Since hee is a consan foce, gaviy, downwad, he plo of he y-posiion should be paabolic wih negaive slope whose magniude is inceasing ove ime, unil i his he floo. A his poin he slope should swich sign and poin upwad and decease in magniude ove ime. The velociy should be iniially 0 m/s since he ball saed fom es, and decease linealy unil i his he floo, a which poin i is vey quickly given a boos o a posiive value, and againconinues o decease linealy. The foce should be always a negaive consan value, excep fo when he ball his he floo, a which poin i should be a naow posiive peak o indicae he bief upwad foce exeed on he ball by he floo poins A mass aached o a sping is given an iniial displacemen x 0 and an iniial velociy v 0. If he posiion of he mass is given by x() = Asin(ω+α), wha is he phase angle α? ( ) 1. α = co 1 x0 ωv ( 0 ). α = an 1 ωx0 coec v ( 0 3. α = co 1 ωx ) 0 v ( 0 4. α = co 1 x ) 0 ωv ( 0 5. α = an 1 ωx ) 0 v ( 0 ) 6. α = an 1 x0 ωv ( 0 ) 7. α = co 1 ωx0 v ( 0 8. α = an 1 x ) 0 ωv ( ) α = cos 1 v α = sin 1( x ) 0 ω Hee, use he iniial posiion and iniial velociy o solve fo α. The iniial posiion is x( = 0) = Asin(α) = x 0 and he iniial velociy is v( = 0) = ωacos(α) = v 0. Now dividing x 0 by v 0 eliminaes A o give: ( ) α = an 1 x0 ωv poins If a apeze ais oaes once each second while sailing hough he ai, and conacs o educe he oaional ineia o one hid of wha i was, how many oaions pe second will esul? 1. 1/9 oaions pe second.. 1/3 oaions pe second oaion pe second oaions pe second oaions pe second. coec The ais will oae 3 imes pe second. By he consevaion of angula momenum, he ais will incease oaion ae by 3. Tha is, I 1 ω 1 = I ω I 1 ω 1 = I 1 3 3ω poins Two wies wih equal lenghs ae made of pue coppe. The diamee of wie A is wice he diamee of wie B. You make caeful measuemens and compue Young s modulus fo boh wies. Wha do you find? 1. Y A = Y B coec. Y A > Y B 3. Y A < Y B Since boh wies ae made of coppe, Y A = Y B poins

5 Vesion 074 Exam Final Daf swinney (55185) 5 The moe massive Eah and he less massive moon ae aaced o each ohe by he gaviaional foce. Is he acceleaion expeienced by he Eah due o his gaviaional ineacion wih he moon is geae han, less han, o he same as he acceleaion expeienced by he moon? 1. Unable o deemine. less han coec 3. geae han 4. he same as The gaviaional foces expeienced by he Eah and moon ae he same, bu he Eah is much moe massive so he acceleaion i expeiences is much less han ha expeienced by he moon poins Conside a hin unifom od of mass M and lengh L. L Wha is he oaional ineia of he od abou a poin a disance L fom he neaes end of he od, along he line of he od? 1. Sill he same, since empy space has no mass and hus no oaional ineia M L 3. 1 M L 7. M (L) = 4M L M L coec M L M L Accoding o he paallel axis heoem, I p = I cm + M h, whee h is he disance fom he CM o he new paallel axis pivo poin p. In his case, he oaional ineia of he od abou is cene of mass is 1 1 M L. (We don cae abou he oaional ineia abou is end; hee is no heoem elaing ha oaional ineia o any abiay axis oaional ineia.) The addiional disance h fomheceneofmassohenewpivopoin is L+ L = 3L, so he new oaional ineia is ( ) M L = M L poins A ball moves in he diecion of he aow labeleddinhefollowingdiagam. Theballis suck by a sick ha biefly exes a foce on he ball in he diecion of he aow labeled f. Which aowdescibeshediecionof p, he change in he ball s momenum? g h f a e b d c M L M L 6. Zeo. 1. e. h 3. c

6 Vesion 074 Exam Final Daf swinney (55185) 6 4. d 5. a 6. g 7. f coec 8. b Recall he definiion of impulse: Impulse = F ne = p. Theefoe, whaeve diecion he ne foce poins in is he same diecion as he change in he ball s momenum. We ae oldha he foceisinhe diecionlabeledf so pisalso in he diecion labeled f poins A flywheel slows fom an iniial angula velociy ω 0 o es in ime T. If he angula acceleaion α is consan, wha is he angula displacemen θ? 1. ω 0 T. ω 0T 3. ω 0 T 4. ω 0 T 5. 1 ω 0T 6. ω 0 T 7. 1 ω 0 T 8. ω 0 T 9. 1 ω 0T ω 0T coec Foconsanangulaacceleaion,ω = ω 0 + αt, bu since he flywheel comes o a sop, ω = 0. Hence α = ω 0 /T. The angula displacemen θ fo consan α and θ 0 = 0isa ime T θ = ω 0 T + 1 αt Subsiuing α = ω 0 /T, we have θ = ω 0 T + 1 ( ω 0/T)T = 1 ω 0T poins Given: The baleship and enemy ships 1 and lie along a saigh line. Neglec ai ficion. baleship 1 Conside he moion of he wo pojeciles fied a = 0. Thei iniial speeds ae boh v 0 bu hey ae fied wih diffeen iniial angles θ 1 and θ. Wha is he aio of he maximum heighs, h 1 of he pojecile dieced a ship 1 and h dieced a ship, ha he shells each? 1. h 1 h = sinθ 1 sinθ. h 1 h = cos θ 1 cos θ 3. h 1 h = sin θ 1 sin θ coec 4. h 1 h = sinθ sinθ 1 5. h 1 h = sin θ sin θ 1 6. h 1 h = anθ anθ 1 7. h 1 h = 1

7 Vesion 074 Exam Final Daf swinney (55185) 7 8. h 1 h = an θ 1 an θ 9. h 1 h = cos θ cos θ h 1 h = cosθ 1 cosθ A is maximum heigh, he y componen of is velociy veco is 0, so 0 = v y = v y gh h = v y g = v sin θ g Theefoe, he aio of he heighs is h 1 h = v 0 sin θ 1 g v 0 sin θ g = sin θ 1 sin θ poins The figue below shows wo negaively chaged objecs and one posiively chaged objec. Wha is he diecion of he ne elecic foce on he posiively chaged objec? Choose you answe fom he diagam labeled I-IX VII 4. II 5. VIII 6. VI 7. V 8. IV 9. IX We need o keep in mind he expession fo he magniude of elecic foce: F = 1 q 1 q 4πǫ 0. The foce is dependen on disance, wih less disance meaning geae foce. The posiively chaged objec is aaced elecomagneically o boh of he negaively chaged objecs, bu since i is close o he one onhe igh, he aacive foce beween i and he negaive chage on he igh will be songe. Theefoe he ne elecic foce poins owad he igh poins A paicle moves in he x-diecion as indicaed in he posiion vs. ime plo below. I VIII II VII III VI IV V IX zeo ne foce 1. III coec. I A poin P, he mass has 1. negaive velociy and negaive acceleaion.. negaive velociy and posiive acceleaion. 3. posiive velociy and negaive acceleaion. coec

8 Vesion 074 Exam Final Daf swinney (55185) 8 4. posiive velociy and zeo acceleaion. 5. negaive velociy and zeo acceleaion. 6. zeo velociy bu is acceleaing (posiively o negaively). 7. zeo velociy and zeo acceleaion. 8. posiive velociy and posiive acceleaion. The velociy is posiive because he slope of he cuve a P is posiive. The acceleaion is negaive because he cuve is concave down a P poins Conside he following diagam. A B If angle beween B and C is a igh angle, which of he following ae accuae veco elaions? I. A B = C II. B C = A III. B B + C C = A A IV. B B C C = A A 1. IV. III, IV 3. II, III coec 4. I, II 5. I, IV 6. III C 7. I 8. II, IV 9. II 10. I, III Adding C o B gives A hence B C = A. Also, aking he do poduc of each side of his elaion wih iself gives bu B C = 0 so ( B C) ( B C) = A A B B + C C = A A whichisjushepyhagoeantheoem: B + C = A poins A od has a pivo a one end and is fee o oae wihou ficion a he ohe end, as shown. A foce F is applied o he fee end a an angle θ o he od ceaing a oque τ abou he pivo. If insead he same foce is applied pependiculaoheod,awhadisancedfomhe pivo should i be applied in ode o poduce he same ne oque τ abou he pivo? 1. d = L/cosθ. d = L 3. d = L 4. d = L/sinθ 5. d = L/ anθ L m θ F

9 Vesion 074 Exam Final Daf swinney (55185) 9 6. d = L cosθ 7. d = L/ 8. d = L sinθ coec 9. d = L anθ 10. d = 5 L The foce geneaes a oque of τ = F Lsinθ, so he disance is L sinθ poins How fas mus a olle coase ca go hough a cicula dip fo you o feel hee imes as heavy as usual, due o he upwad foce of he sea on you boom being hee imes as lage as usual? The cene of he ca moves alongaciculaacofadiusrasinhefigue below. R v 9. v = gr gr 10. v = A he boom of a hill (o a dip ), he ne foce on he ide is upwad. The foce by he sea on he ide is also upwad, and in his case has a magniude of 3. The ne foce on he ide is F ne = F sea +F gav 0, m v,0 = 0,3,0 + 0,,0 R m v R = v = gr poins The enegy levels of a paicula quanum objec ae 8. ev, 6.4 ev, and 1.8 ev. If a collecion of hese objecs is bombaded by an elecon beam so ha hee ae some objecs in each excied sae, wha ae he enegies of he phoons ha will be emied? ev, 1.8 ev, 6.4 ev. 1.8 ev, 14.6 ev, 4.6 ev 1. v = 3 gr. v = gr 3. v = gr 4. v = gr 5. v = gr coec 6. v = 1 gr 7. v = 3gR 8. v = 3gR ev, 6.4 ev, 1.8 ev ev, 6.4 ev, 1.8 ev ev, 6.4 ev, 4.6 ev coec ev, 6.4 ev, 4.6 ev ev, 4.6 ev, 10 ev ev, 6.4 ev, 1.8 ev ev, 10 ev, 8. ev ev, 8. ev, 14.6 ev

10 Vesion 074 Exam Final Daf swinney (55185) ev, 6.4 ev, 4.6 ev is he coec answe. The 3 emied phoons coespond o 3 ansiions beween he saes. Label hem as E 1 = 8. ev E = 6.4 ev E 3 = 1.8 ev Then he possible ansiions ae given by E phoon = E 1, = E 1 E = 8. ev+6.4 ev = 1.8 ev E phoon = E 1,3 = E 1 E 3 = 8. ev+1.8 ev = 6.4 ev E phoon = E,3 = E E 3 = 6.4 ev+1.8 ev = 4.6 ev poins Wha is he appoximae diamee of a coppe aom in mees? m m m m m coec Anaomisabou 0.1nm indiamee which is m poins A ypicalhuman die is000 foodcaloiespe day. Each food caloie is he enegy equivalen of J. Wha ishe powe consumpion fo a ypical human die? W coec W W 4. 1W W W 7. 10W W W Each day a peson consumes 000 caloies. Muliplying 000 caloies by J pe caloiegiveshedailyenegyinakeinj.then muliplying 4 hous pe day imes 60 minues pe hou imes 60 sec pe min gives he numbeofsecondsinaday. Dividingheoal enegy inakeinjoules by he imeove which he enegy is consumed in seconds gives he powe consumpion in a die, i.e., m m = m m m poins The following gaph epesens a hypoheical poenial enegy cuve fo a paicle of mass m.

11 Vesion 074 Exam Final Daf swinney (55185) 11 U() 9U 0 6U 0 3U 0 O 0 0 If he paicle is eleased fom es a posiion 0, wha will be is speed v a posiion 0? 1. v = U0 m U0. v = 4m 1U0 3. v = m coec 8U0 4. v = m 6U0 5. v = m 4U0 6. v = m 10U0 7. v = m U0 8. v = m U0 9. v = m The oal enegy of he paicle is conseved. So he change of he poenial enegy is conveed ino he kineic enegy of he paicle, which gives 1 mv = 9U 0 3U 0 1U0 v = m poins TwoobjecsshaeaoalenegyE = E 1 +E. Thee ae 5 ways o aange an amoun of enegy E 1 in he fis objec and 10 ways o aangeanamoun ofenegy E inhesecond objec. How many diffeen ways ae hee o aange he oal enegy E = E 1 +E so ha hee is E 1 in he fis objec and E in he ohe? 1. 14! 5!9! 15!. 5!10! ! 4!10! coec The oal numbe of micosaes o numbe of ways of aanging enegy in he sysem is he poduc of he numbe of ways of aanging he enegy in especive objecs, i.e. Ω oal = Ω 1 Ω poins A ball of mass M collides head-on and inelasically wih a ball of mass 3M. Befoe he collision, he ball of mass M is moving o he igh a speed v 0 and he ball of mass 3M is moving o he lef a he same speed v 0. M v 0 v 0 3M Afe he collision, he ball of mass 3M is moving o he lef a speed v 0. Wha is he aio of he final kineic enegy o he iniial kineic enegy?

12 Vesion 074 Exam Final Daf swinney (55185) 1 v v 0 The aio of he final o iniial kineic enegy is jus M 3M Wha facion of he iniial kineic enegy is los in he collision? coec K f K i = (1/)M v 0 M v 0 = poins A block of mass m is on a ficionless plane inclined a θ wih espec o he hoizonal and is pushed a a consan velociy up he plane by a foce F ha is inclined a α wih espec o he hoizonal. Gaviy acs downwad wih a magniude F α m θ Take componens of p i = p f along he line of he collision o obain ( v0 ) M v 0 (3M)v 0 = M v (3M) v 0 = v + 3v 0 v = v 0. Thus K i = 1 [ M v 0 +(3M)v0 ] = M v 0, K f = 1 [ ( v0 ) ( v0 ) ] M +(3M) = 1 ( 1 M v ) = 1 4 M v 0, and Wha is he magniude of he applied foce F applied o he block in ems of,θ, and α? 1. F = cosα(anα+anθ). F = anα(sinα+cosθ) 3. F = cosα(coα+coθ) 4. F = cosα(anα+coθ) coec 5. F = sinα(anα+anθ) 6. F = coα(sinα+cosθ) 7. F = coα(cosα+sinθ) 8. F = anα(cosα+sinθ) 9. F = sinα(coα+coθ) 10. F = sinα(anα+coθ)

13 Vesion 074 Exam Final Daf swinney (55185) 13 We apply he Momenum Pinciple F ne = m a. Le he x-diecion o be o he igh and hey-diecionobeup. Thenefoceineach diecion of his coodinae sysem is and F x,ne = F cosα N sinθ F y,ne = F sinα+n cosθ We apply he Momenum Pinciple and since he velociy is consan, he acceleaion is a = 0. Thus, he equaion in he x-diecion gives N = F cosα sinθ. Plugging his ino he y-equaion gives F sinα+f cosα cosθ = sinθ o eaanging his gives F cosα(anα+coθ) = finally solving fo F gives F = cosα(anα+coθ) poins A block of mass m slides up a amp wih ficion ha makes an angle θ wih he hoizonal. The block has an iniial velociy v 0 as i sas up he amp and hen avels a disance L along he amp befoe coming momenaily o es. L v 0 How much wok did kineic ficion do on he block beween is saing poin and he poin i came o momenay es? ( ) 1. m gl sinθ v 0 coec. L ( ) 3. m gl sinθ + v 0 4. L sinθ 5. L cosθ ( ) 6. m gl cosθ v 0 7. Zeo 8. m( v 0 9. m( v mv 0 ) +gl cosθ ) gl sinθ θ L L sinθ If we ake he gaviaional poenial enegy o be zeo a he poin whee he block has is iniial speed, hen fom E = K +U g E i = 1 mv 0 +0 and E f = 0+L sinθ. The wok done by ficion, a nonconsevaive foce, is hus ( ) W f = E f E i = m gl sinθ v poins Which of he following gaphs coesponds o a sysem of one elecon and one poon ha sa ou fa apa, moving owad each ohe (ha is, hei iniial velociies ae

14 Vesion 074 Exam Final Daf swinney (55185) 14 nonzeo and hey ae heading saigh a each ohe)? Noe ha he hoizonal and veical axes in each plo ae he sepaaion beween he paicles and enegy, especively. 1. None of hese gaphs could epesen he enegies of he sysem. 6. U K K +U K +U K. U 7. K K +U U K K +U 3. U 4. K U K +U coec When he elecon and poon ae vey fa away hei poenial enegy poenial enegy ends o zeo, and since hey have nonzeo iniial velociies, his means ha hey have a posiive kineic enegy and hence a posiive oal enegy. As he elecons ge close, due o hei Coulomb aacion, i.e., U 1/, hei kineic enegies inceases as he poenial enegy deceases. Thus he coec answe K K +U is. 5. K +U K U U poins A small sphee of mass m is conneced o he end of a cod of lengh and oaes in a veical cicle abou a fixed poin O. The ensionfoceexeedbyhecodonhesphee is denoed by T.

15 Vesion 074 Exam Final Daf swinney (55185) 15 diecion, so θ O F c = mv = T cosθ. Wha is he equaion fo he ne foce in he adial diecion when he cod makes an angle θ wih he veical? 1. None of hese keywods: poins Jane swings fom a vine, acing ou an ac as in he following figue.. T sinθ = + mv 3. T sinθ = + mv cosθ anθ 4. T sinθ = + mv 5. T +sinθ = + mv When she is a he boom of he swing, which aow bes descibes he diecion of he ae of change of Jane s momenum? 6. T +cosθ = + mv 7. T cosθ = + mv coec VII VIII I II III 8. T sinθ = mv 9. T sinθ = mv anθ 1. VII VI V IV O. II T θ 3. V θ The cenipeal foce is F c = mv This cenipeal foce is povided by he ension foce and he adial componen of he weigh. In his case, hey ae in opposie. 4. IV 5. I coec 6. VIII 7. III 8. VI

16 Vesion 074 Exam Final Daf swinney (55185) 16 A he lowes poin Jane s momenum is hoizonal. A sho ime befoe Jane eaches he lowes poin, he momenum veco has a downwad componen, while a ime afe eaching his lowes poin, his momenum veco has an upwad componen; hese momena ae equal in magniude. Theefoe, he change in momenum, p, is upwad, in diecion I.

PHYS PRACTICE EXAM 2

PHYS PRACTICE EXAM 2 PHYS 1800 PRACTICE EXAM Pa I Muliple Choice Quesions [ ps each] Diecions: Cicle he one alenaive ha bes complees he saemen o answes he quesion. Unless ohewise saed, assume ideal condiions (no ai esisance,