Solution: b All the terms must have the dimension of acceleration. We see that, indeed, each term has the units of acceleration

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PHYS 54 Tes Pracice Soluions Spring 8 Q: [4] Knowing ha in he ne epression a is acceleraion, v is speed, is posiion and is ime, from a dimensional v poin of view, he equaion a is a) incorrec b) correc c) The dimensional validiy of he epression depends on he unspecified sysem of coordinaes. d) The dimensional validiy of he saemen depends on he unspecified sandard of unis. Soluion: b ll he erms mus have he dimension of acceleraion. We see ha, indeed, each erm has he unis of acceleraion a v v m s m s so he relaion is dimensionally correc. Q: [4] Which of he following is a rue saemen? a) The componens of a vecor do no depend on he sysem of coordinaes. b) vecor s magniude is negaive if he force acs agains he direcion of moion. c) vecor componen of a vecor can be equal o he vecor. d) The angle giving he direcion of a vecor canno be negaive. e) None of he above. Soluion: c vecor aligned wih an ais is equal wih is componen along he respecive ais (he oher componen will be zero hen). Q3: [4] When vecors and are added ogeher hey form a vecor and heir magniudes saisfy he relaionship + =. Which of he following saemens is rue? a) They mus be parallel o each oher. b) They mus be perpendicular on each oher. c) They mus oppose each oher. d) They mus be a 45º wih respec o each oher. e) They mus have he same magniudes. Soluion: b Only when he vecors are perpendicular heir resulan will have a magniude given by he relaionship above. This is because he vecors and will be equal o he componens of : = and y =. If he vecors and are no perpendicular, a leas one of he componens of will depend boh on and. Q4: [4] Wha is he magniude of he resulan R = 3 of he coplanar vecors = (3,) and = (3,)? a) Incomplee informaion b) c) 3 d) 4 e) 5 Soluion: e To find he resulan, one has o find is componens: R 3 3 and R 3 4 y y y Then, he magniude of he resulan can be calculaed from R R R 3 4 5. y

PHYS 54 Tes Pracice Soluions Spring 8 Q5: [4] The adjacen moion diagram shows he posiions a equal ime inervals of four idenical paricles moving horizonally wih zero or consan acceleraion. Which of he four paricles has an acceleraion larger in magniude? a) b) c) d) e) oh and have he same acceleraion magniude. Soluion: e acceleraion will correspond o varying disances raveled successively in equal inervals of ime, so we see immediaely ha paricles and do no accelerae. On he oher hand, if he acceleraion a is consan, he segmens raveled successively in equal ime inervals will vary by he same amoun. Inspecing he diagram, we noice ha he paricle increases he disance raveled successively by one square, whereas paricle decreases is disance by one square. So, while he acceleraions of he wo paricles are opposie ( speeds up while slows down), heir magniude is he same. Q6: [4] car ravels due Souh slowing down. If he Norh direcion is aken o be posiive, and we denoe Δ, v and a va,, is displacemen, velociy and acceleraion respecively, hen a), v, a N+ b), v, a c), v, a d), v, a e) None of he above. S- Soluion: c The displacemen and velociy have he same direcion: due Souh. Since Norh is declared as posiive direcion, Souh is negaive such ha he respecive vecors, v. Since he car is slowing down, he acceleraion mus oppose velociy, such ha i mus be posiive, a. Q7: [4] Which of he adjacen posiion vs. ime graphs mos likely represen he moion of he car described in he previous quesion? a) b) c) d) e) None of hem. Soluion: c The car ravels due Souh, so in he negaive direcion. This eliminaes graph where he objec advances in he posiive direcion. Then noe ha he car slows down, so he slope of he - curve (represening velociy) mus decrease. In case he slope increases, while in case i says consan. So we remain wih graph represening an objec raveling in he negaive direcion slowing down. Q8: [4] suden drops a rubber ball verically on a fla surface so he ball bounces back ino her hand. Which of he following is rue abou he average acceleraion of he ball during he shor insan when i is in conac wih he floor? a) I is zero. b) I is non-zero, verically upwards c) I is non-zero, verically downwards d) I is he graviaional acceleraion. e) ny of he answers above may be possible, depending on he velociy of he ball before i his he floor. Soluion: b efore i his he floor, he velociy of he ball is verically downwards, whereas afer i is verically upwards. To flip he velociy like his, he ball mus be pushed (by he floor) upwards, so he acceleraion is ha way.

PHYS 54 Tes Pracice Soluions Spring 8 Q9: [4] Wha can you say abou he following saemen? The graviaional acceleraion is always negaive. a) True, since i poins downwards. b) True, because i is opposie o he moion of an ascending objec. c) False by popular voe, since we all know ha g = 9.8 m/s d) False, because i is always in he direcion of moion of a freely falling objec. e) epends on he sysem of coordinaes Soluion: e The graviaional acceleraion is negaive only when he sysem of coordinaes is chosen wih an ais (such as y-ais) poining verically upwards. Nohing prevens you o pick a frame wih an ais poining in he direcion of graviy, in which case he graviaional acceleraion should be considered as posiive. Q: [4] On he Moon, an objec is hrown verically upward wih speed v =.5 m/s. I reaches a maimum heigh h =.9 m. Therefore, he graviaional acceleraion on he Moon is: a) 9.8 m/s b) m/s c).6 m/s d) e) Insufficien informaion Soluion: c We can employ he kinemaics wih consan acceleraion on he moon. Since he speed is zero a maimum heigh v v g h g v h.6 m s. M M Q: [4] uddhis monk mediaes a he edge of a cliff, holding one ball in each hand (since we all know ha surreal posures deepen he insigh). He simulaneously osses one ball sraigh up wih speed v and he oher sraigh down, also wih speed v. Which ball has he greaer average speed during he -s inerval afer release? a) The ball hrown upward b) The ball hrown downward c) oh have he same average speed d) Insufficien informaion e) oesn maer: on he pah o Nirvana, up or down are irrelevan mundane illusions. Soluion: b The average speed is defined as he disance raveled per inerval of ime. We see ha he disances raveled by he balls raveling upward and downward, are respecively y v g y v g v g y Therefore, he average speed of he ball moving downward is larger. (Moreover, illusion or no illusion, he order of realiy mus be assumed as dicaed by he mos commonly acceped percepion, oherwise he uddhis monk won reach Nirvana for he s no humble enough.) Q: [4] If he shadow of a ball moving in a circle of radius R wih consan speed v is projeced perpendicular uno a wall (say along a y-ais), i will oscillae wih a velociy v y depending on ime as given by v y = vcos(ω), where ω is some consan. Seeing how a ime = he shadow moved wih maimum speed, which of he following gives he ime dependency of he y-posiion of he shadow relaive o he origin? a) y = R + v ω sin(ω) b) y = R + v ω cos(ω) c) y = v ω sin(ω) d) y = v ω cos(ω) e) None of he above. all R v y Shadow v y 3

PHYS 54 Tes Pracice Soluions Spring 8 Soluion: c y definiion, he posiion is given by y y v d, y where y is he posiion a ime =. Noe ha he shadow of he ball will reach is maimum speed when going hrough he origin of he y-ais aligned wih he cener of he circle. Therefore y = and v y v cos d sin 4

PHYS 54 Tes Pracice Soluions Spring 8 P: The posiion-versus-ime ( ) - graph below represens he one-dimensional moion of a paricle. (N: On he represenaion, he graph in he inervals - and - is parabolic and beween - is a sraigh line. ssume zero slopes - in poins and.) (m) 5..5..5 5. v (m/s) 5..5..5 5. 3 4 5 6 3 4 5 6 a) [3] alculae he acceleraion a of he paricle beween poins and. a v ms. a.5 m s b) [3] alculae he velociy v of he paricle before poin. The velociy in he inerval - is consan: 5. v v m s 5. m s.. c) [3] alculae he acceleraion a of he paricle beween poins and. You may consider he speed equal on he wo sides of poin. Noice ha, albei equal in magniude, he velociy a he beginning of he - inerval is opposie in direcion o he velociy a he end of -, so i is posiive. Therefore v v a v v a 6. 3. 5. m s.7 m s d) [8] Skech he paricle s moion graphs in he velociy vs. ime (v-) and hen in he acceleraion vs. ime (a-) represenaions. Noice ha in poin he velociy swiches direcion and his corresponds o a raher large acceleraion since he vecor flip happens in a very shor ime inerval. This acceleraion was ignored on he graph a-. a (m/s )... a. a 3 4 5 6 e) [3] ircle he inerval(s) where he paricle moves wih negaive velociy: [-] [-] [-] ircle he poin(s) where he paricle sopped. ircle he inerval(s) where he paricle moves wih posiive velociy. [-] [-] [-] ircle he inerval(s) where he paricle deceleraes [-] [-] [-] ircle he inerval(s) where he acceleraion of he paricle is zero [-] [-] [-] 3. 5

PHYS 54 Tes Pracice Soluions Spring 8 (m) P: The velociy-versus-ime (v-) graph below represens he moion of a oy car saring a iniial posiion = 8. m. 5. 5. 3 4 5 6 a) [4] alculae he acceleraions a, a, a of he car hrough he respecive inervals, and. In he inerval he velociy is consan, so a. On he oher hand, in he ne wo inervals, v v 6. a ms 4.. a v v 3. ms 6. 4.. m s.5 m s v (m/s) 5. b) [5] alculae he posiions of he car in poins, and. v a. m v a 7. m v a m. 5. 3 4 5 6 a (m/s ) 3.. a c) [8] Skech he paricle s moion graphs in he posiion vs. ime (-) and hen in he acceleraion vs. ime (a-) represenaions. d) [3] Each do on he moion diagrams below marks he locaion of a moving paricle a equal ime inervals. Which moion diagram is he closes represenaion for he moion of he car in our problem?.. a 3 4 5 6.. 3. 4.. 5. None of he above, since our car doesn move in only one direcion. 6

PHYS 54 Tes Pracice Soluions Spring 8 P3: rocke rises verically, from res. Is engine works for a ime = 5. s hen i runs ou of fuel. The ime dependen velociy of he rocke while he engine works is given by v() = + 3, where =.3 m/s and =.46 m/s 4. fer he engine sops, he rocke moves freely. a) [5] Find ou symbolical epressions (in erms of, and ime ) for he verical posiion and acceleraion as funcions of ime unil he rocke runs ou of fuel. Wih he origin of he y-ais on he ground, we can assign symbols o he kinemaic quaniies in he significan posiions. Since y = and we know he ime dependency of he velociy, we ge 3 4 4, y y v d d dv d a d d 3 3. Useful derivaive and aniderivaive (n is a consan): y, v, g b) [5] alculae numerically he rocke s aliude y, velociy v and acceleraion a a he momen when he engine runs ou of fuel (ha is, a he end of he acceleraion inerval). he insan = 5. s when he engine runs ou of fuel 4 y y m 3 3 5. s 4 5. s v v 69 m s 5. s 5. s a a 5. s 5. s 36 m s. fuel ou y, v, a, c) [5] fer i runs ou of fuel, he rocke is under he sole effec of is weigh. alculae he maimum aliude y wih respec o he ground subsequenly reached by he rocke, and he oal ime i akes o reach i, relaive o he sarup ime. fer he rocke runs ou of fuel, he acceleraion is eclusively due o graviy. he maimum aliude y, he velociy v =. Therefore, considering again he moion from he insan he fuel was ehaused, we have v v g y y y v g y 34 m. v v g v g s. a d) [5] alculae he ime i akes he rocke o fall back o he ground (from he maimum aliude). If we denoe he reurn ime wih r, a he momen when he rocke reurns o he ground we have y v g y g 8.4 s r r r ground y, v, a, y (m) e) [5] The adjacen graph represens he posiion of he rocke before he fuel runs ou. omplee he graph by skeching he curve represening he posiion of he rocke beween he momen when is fuel runs ou and when i reurns back o he ground. (Noice ha he graph shows you wha you have o obain eplicily in par (b) for he aliude a ime ). 4 y 3 5 5 5 7