m m m m m correct

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1 Version 055 Miderm 1 OConnor (05141) 1 This prin-ou should have 36 quesions. Muliple-choice quesions ma coninue on he ne column or pae find all choices before answerin. V1:1, V:1, V3:3, V4:, V5: (par 1 of 1) 10 poins The raph below shows he veloci v as a funcion of ime for an objec movin in a sraih line. v 0 Q R S P Which of he followin raphs shows he correspondin displacemen as a funcion of ime for he same ime inerval? 1. 0 Q R S P correc. 0 Q R S P Q R S P 0 Q R S P 0 Q R S P 6. None of hese raphs are correc Q R S P Q R S P 0 Q R S P The displacemen is he ineral of he veloci wih respec o ime = v d. Because he veloci increases linearl from zero a firs, hen remains consan, hen decreases linearl o zero, he displacemen will increase a firs proporional o ime squared, hen increase linearl, and hen increase proporional o neaive ime squared. From hese facs, we can obain he correc answer. 0 Q R S P 00 (par 1 of 1) 10 poins A man jos a a speed of 1.5 m/s. His do wais 1.6 s and hen akes off runnin a a speed of 4 m/s o cach he man. How far will he have each raveled when he do caches up wih he man? m m m m m correc

2 Version 055 Miderm 1 OConnor (05141) m m m m m We calculae firs he disance ha separaes he man and he do when he do akes off. I is d 0 = v man wai =.4 m. Once he do akes off, he posiion of he do and he man a an ime is man = d 0 + v man do = v do. The mee when man = do, so d 0 +v man = v do and he ime is = The disance is d 0 v do v man = 0.96 s. d = v do = 3.84 m. 003 (par 1 of 1) 10 poins An objec was suspended in a fied place ( = 0) and hen allowed o drop in a free fall. Takin up as he posiive verical direcion, which of he followin raphs correcl represens is verical moion as displacemen vs ime? correc.

3 Version 055 Miderm 1 OConnor (05141) The objec is underoin a consan downward raviaional acceleraion. The slope of a posiion vs ime curve represens he veloci. Given v0 hma 9.8 m/s Since 0 m/s a < 0 m/s a = m/s v 0 = 0 m/s 0 = 0 m. = 0 + v a, Wha is is maimum heih, h ma erms of he iniial speed v 0 )? 3 v 1. h ma = 0. h ma = 5 v h ma = v 0 4. h ma = 3 v 0 4 (in herefore 5. h ma = 5 v 0 = h ma = 3 v 0 This is a parabolic shaped curve sarin a = 0 wih a coninuousl decreasin slope as ime increases. Onl he fiure below is correc. 004 (par 1 of 1) 10 poins A ball is hrown upward. Is iniial verical speed v 0, acceleraion of ravi, and maimum heih h ma are shown in he fiure below. Nelec: Air resisance. The acceleraion of ravi is 9.8 m/s. 7. h ma = v h ma = v 0 correc For consan accelera- 9. h ma = v 0 Basic Concep: ion, we have v = v 0 + a ( 0 ). (1) Soluion: The veloci a he op is zero. Since we know velociies and acceleraion, Eq. 1 conainin v, a, 0, and (no ime) is he easies one o use. Choose he posiive direcion o be up; hen a = and 0 = v 0 + ( ) (h ma 0)

4 Version 055 Miderm 1 OConnor (05141) 4 or h ma = v 0. Basic Concep: For consan acceleraion, we have 005 (par 1 of 1) 10 poins A ball is hrown upward. Is iniial verical speed v 0, acceleraion of ravi, and maimum heih h ma are shown in he fiure below. Given: = 9.8 m/s. Nelec: Air resisance. v0 hma 9.8 m/s v = v 0 + a. (1) Soluion: The veloci a he op is zero. Since we know velociies and acceleraion, Eq. 1 conainin v, a, and. Choose he posiive direcion o be up; hen a = and or 0 = v 0 + ( ) up up = v (par 1 of 1) 10 poins A ball is hrown upward. Is iniial verical speed v 0, acceleraion of ravi, and maimum heih h ma are shown in he fiure below. Nelec: Air resisance. The acceleraion of ravi is 9.8 m/s. Whais is ime inerval, up (in erms of he iniial speed v 0 ), beween he release of he ball and he ime i reaches is maimum heih? 1. up = v 0. up = 4 v 0 v0 3. up = 3 v0 4. up = 5. up = v 0 6. up = v 0 v0 7. up = 3 8. up = v 0 correc 9. up = v 0 4 v0 hma 9.8 m/s Wha is is ime inerval, up (in erms of he maimum heih h ma ), beween he release of he ball and he ime i reaches is maimum heih? 1. up = h ma 4. up = h ma 3. up = h ma

5 Version 055 Miderm 1 OConnor (05141) 5 4 h ma 4. up = 5. up = h ma h ma 6. up = correc h ma 7. up = 8. up = h ma 9. up = h ma h ma 10. up = Basic Concep: For consan acceleraion, we have = 0 + v a. (1) while i is in Michael s hand and afer he les i o, assumin i has no e hi he round? 1. correc. 3. a a a a Soluion: The veloci a he op is zero. Since we know velociies and acceleraion, Eq. 1 conainin h ma = 0, a, and. Choose he posiive direcion o be up, hen a = and h ma = v a = 1 = 1, where v 0 =. Therefore up = ma 6. a 007 (par 1 of 1) 10 poins Michael sands moionless holdin a baseball in his hand. Afer a while he osses i upwards, and i ravels up for a while before urnin abou and headin owards he round. Define upwards o be posiive. Which of he followin diarams can describe he verical acceleraion of he ball,. The ball firs eperiences a period of zero acceleraion when Michael is jus holdin he ball. Then, as he ball is hrown upward, i feels an upward acceleraion. Quickl he ball is le loose. Once he ball leaves Michael s hand i is in free-fall. In free-fall all objecs feel a downward (here neaive) consan raviaional acceleraion. This scenario is shown below.

6 Version 055 Miderm 1 OConnor (05141) 6 a 008 (par 1 of 3) 10 poins Given: The acceleraion of ravi on Earh is 9.8 m/s. Consider a ball hrown up from he round (he poin O). I passes a window (he semen AB) in he ime inerval s (see he fiure). The poins in he fiure represen he sequenial order, and are no drawn o scale. The disance AB = 1.4 m. C B 1.4 m A O Find he averae speed as he ball passes he window m/s m/s m/s m/s m/s m/s m/s correc m/s m/s m/s Basic Conceps: a = v For consan acceleraion v = s = v i + v f v = v 0 + a. For he acceleraion of ravi (a = ) v = v 0. Soluion: The averae veloci is iven b v = s = AB = 1.4 m s = 4.59 m/s. Alernaive Par 1: v A = d = B A = v A 1 d m + 1 = (9.8 m/s ) (0.314 s) (0.314 s) = m/s v B = v A = m/s (9.8 m/s ) (0.314 s) = m/s v = v A + v B m/s m/s = = 4.59 m/s. 009 (par of 3) 10 poins Wha is he maniude of he decrease of he veloci from A o B? m/s

7 Version 055 Miderm 1 OConnor (05141) m/s m/s m/s m/s m/s m/s m/s correc m/s m/s For a consan acceleraion a = v, = v = a =. Thus he decrease in veloci is v = = (9.8 m/s )(0.314 s) = m/s. Alernaive Par : v = v A v B = m/s m/s = m/s. 010 (par 3 of 3) 10 poins If he ball coninues is pah upward wihou obsrucion, find he ravel ime beween B and C, where poin C is a he ball s maimum heih s s s s s s correc s s s s The veloci chane is v A v B = v = v A = v B + v and he averae veloci is Thus v = v A + v B v B = v v = 4.59 m/s = m/s. = v B + v m/s The ball reaches is maimum heih when is veloci is zero. Based on v = v 0 + a v C = 0 = v B = v B m/s = 9.8 m/s = s. 011 (par 1 of 3) 10 poins The posiion of a sofball ossed vericall upward is described b he equaion = c 1 c,

8 Version 055 Miderm 1 OConnor (05141) 8 where is in meers, in seconds, c 1 = 8.59 m/s, and c = 7.4 m/s. Find he ball s iniial speed v 0 a 0 = 0 s m/s m/s m/s m/s m/s m/s m/s m/s correc m/s m/s Basic Conceps: v = d d a = dv d = d d Soluion: The veloci is simpl he derivaive of wih respec o : v = d d = 8.59 m/s (7.4 m/s ), which a = 0 is v 0 = 8.59 m/s. 01 (par of 3) 10 poins Find is veloci a = 1.08 s m/s m/s m/s correc m/s m/s m/s m/s m/s m/s m/s Subsiuin = 1.08 s ino he above formula for v, we obain v = 8.59 m/s (7.4 m/s )(1.08 s) = m/s. 013 (par 3 of 3) 10 poins Find is acceleraion a = 1.08 s m/s m/s correc m/s m/s m/s m/s m/s m/s m/s m/s The acceleraion is he derivaive of veloci wih respec o ime: a = dv d = (7.4 m/s ) = 14.8 m/s. 014 (par 1 of ) 10 poins

9 Version 055 Miderm 1 OConnor (05141) 9 A commuer airplane sars from an airpor and akes he roue shown in he fiure. I firs flies o ci A locaed a 198 km in a direcion 9 norh of eas. Ne, i flies 135 km 16 wes of norh o ci B. Finall, i flies 180 km due wes o ci C. (km) N C R km 198 km km B W 16 A S (km) How far awa from he sarin poin is ci C? km km km correc km km km km km km km E a α The -componen of he resulan is r = a + b + c = a cos α b sin β c = (198 km) cos 9 (135 km) sin km = km. The -componen of he resulan is r = a + b + c = a sin α + b cos β + 0 = (198 km) sin 9 + (135 km) cos 16 = km. and he resulan is R = r + r = ( km) + (5.763 km) = km. 015 (par of ) 10 poins Wha is he direcion of he final posiion vecor r, measured from Norh? Use counerclockwise as he posiive anular direcion, beween he limis of 180 and correc b β Given : a = 198 km, α = 9, b = 135 km, β = 16, and c = 180 km

10 Version 055 Miderm 1 OConnor (05141) r r γ r Since γ is he anle beween r and he - ais, for r is in he second quadran Thus an γ = r r. ( ) γ = an 1 r r = an 1 ( km km = (par 1 of 3) 10 poins Denoe he iniial speed of a cannon ball fired from a baleship as v 0. When he iniial projecile anle is 45 wih respec o he horizonal, i ives a maimum rane R. v 0 45 The ime of flih ma of he cannonball for his maimum rane R is iven b 1. ma = 1 4 v 0 R. ma = 3 v 0 3. ma = v 0 correc 4. ma = 4 v 0 ) v 0 5. ma = 3 6. ma = 1 v 0 7. ma = 3 v 0 8. ma = v 0 9. ma = v ma = 1 v 0 3 The cannonball s ime of flih is = v 0 = v 0 sin 45 = v (par of 3) 10 poins The maimum heih h ma of he cannonball is iven b 1. h ma = 4 v 0. h ma = 1 v0 3. h ma = v h ma = 1 v0 5. h ma = v 0 6. h ma = 3 v 0 7. h ma = 1 v0 4 correc 8. h ma = v 0 9. h ma = v h ma = 1 3 v 0

11 Version 055 Miderm 1 OConnor (05141) 11 Use he equaion v = v 0 h. Book A he op of is rajecor v = 0. Solvin for h ields h = v 0 = v 0 sin 45 = 1 v0 4. Which free bod diaram bes describes his ssem? 018 (par 3 of 3) 10 poins The speed v hma of he cannonball a is maimum heih is iven b 1. fricion normal raviaional 1. v hma = 1 v 0. v hma = v 0 3. v hma = 3 v 0. normal fricion correc 4. v hma = v 0 raviaional 5. v hma = 4 v 0 6. v hma = v 0 3. normal 7. v hma = 1 v 0 correc raviaional 8. v hma = 3 v 0 9. v hma = 1 3 v v hma = 1 4 v 0 4. normal fricion A he op of he cannonball s rajecor, v = 0. Hence he speed is equal o v. raviaional v = v = v 0 cos 45 = 1 v normal 019 (par 1 of 3) 10 poins Consider a book ha remains a res on an incline. raviaional

12 Version 055 Miderm 1 OConnor (05141) 1 01 (par 3 of 3) 10 poins Are he opposie in direcion? 6. fricion raviaional 1. Yes, he normal force acs opposie he weih force because he book is saionar.. Yes, he normal force alwas acs opposie he weih force. Oherwise, he book would fall hrouh he inclined plane. 7. fricion normal raviaional 3. No, he normal force acs perpendicular o he surface of he inclined plane. correc 4. No, he normal force acs up he incline so ha he book does no slide down. 8. fricion raviaional The fricional force keeps he book from slidin down. The normal force acs perpendicular o he inclined surface. The raviaional force acs down. 00 (par of 3) 10 poins Compare he normal force eered on he book b he inclined plane and he weih force eered on he book b he earh. Are he equal in maniude? The normal force acs perpendicular o he surface of he inclined plane. Because he plane is inclined, his is no opposie in direcion o he weih force. 0 (par 1 of 1) 10 poins A man sands in an elevaor in he universi s adminisraion buildin and is accelerain upwards. (Durin peak hours, his does no happen ver ofen.) Elevaor Cable 1. Yes. No correc 3. Their maniudes canno be deermined since he forces are no in he same direcion. The are no equal in maniude. The normal force iven b N = m cos θ, wih θ he anle he incline makes wih he round. Since cos θ is less han 1 as lon as he incline is no horizonal, he maniude of he normal force, N, will be less han he maniude of he weih, m. Choose he correc free bod diaram for he man, where F i,j is he force on he objec i, from he objec j. 1. F man, elevaor F elevaor, cable

13 Version 055 Miderm 1 OConnor (05141) 13. F F elevaor, cable man, earh shown below. The lef-hand cable had ension T and makes an anle of 44 wih he ceilin. The rih-hand cable had ension T 1 and makes an anle of 48 wih he ceilin. 3. T T 1 F man, cable 679 N 4. a) Wha is he ension in he cable labeled T 1 slaned a an anle of 48? N F elevaor, cable F man, floor N N 5. F F man, earh man, floor N N correc 6. F man, earh F man, acceleraion N N N N N correc Observe he free-bod diaram below. Onl he forces acin direcl on he man are o be in he free bod diaram. Therefore, he force from he cable should be omied, while hose from ravi and from he floor s normal force should be included. θ F F 1 W θ 1 03 (par 1 of ) 10 poins Consider he 679 N weih held b wo cables

14 Version 055 Miderm 1 OConnor (05141) 14 Noe: The sum of he - and -componens of F 1, F, and W are equal o zero. Given : W = 679 N, θ 1 = 48, and θ = 44. Basic Concep: Vericall and Horizonall, we have F ne = F 1 F = 0 = F 1 cos θ 1 F cos θ = 0 (1) F ne = F 1 + F W = 0 = F 1 sin θ 1 + F sin θ W = 0 () Soluion: Usin Eq. 1, we have F = F 1 cos θ 1 cos θ. Subsiuin F from Eq. ino Eq. 1, we have N N N N Usin Eq. 1, we have F = F 1 cos θ 1 cos θ = ( N) = N. cos 48 cos (par 1 of 3) 10 poins Consider a man sandin on a scale which is placed in an elevaor. When he elevaor is saionar, he scale readin is S s. F 1 sin θ 1 + F sin θ = W cos θ 1 F 1 sin θ 1 + F 1 sin θ = W cos θ F 1 sin θ 1 + F 1 cos θ 1 an θ = W W F 1 = sin θ 1 + cos θ 1 an θ 679 N F 1 = sin 48 + cos 48 an 44 = N 04 (par of ) 10 poins a) Wha is he ension in he cable labeled T slaned a an anle of 44? N N N N N N correc Scale Find S up, he scale readin when he elevaor is movin upward wih acceleraion a = 1 5 ĵ, in erms of S s. 1. S up = 7 6 S s. S up = 3 S s 3. S up = 4 5 S s 4. S up = 0 m/s

15 Version 055 Miderm 1 OConnor (05141) 15 Equaion 1 in he upward direcion reads 5. S up = 3 5 S s 6. S up = 6 5 S s correc 7. S up = 3 S s 8. S up = 5 4 S s 9. S up = 7 5 S s 10. S up = 5 6 S s Basic Conceps: Newon s nd law F = m a. (1) Soluion We consider he forces acin on he man. Takin up (ĵ) as posiive, we know ha m acs on he man in he downward ( ĵ) direcion. The onl oher force acin on he man is he normal force S s from he scale. B he law of acion and reacion, he force on he scale eered b he man (i.e., he scale readin) is equal in maniude bu opposie in direcion o he S s vecor. Iniiall, he elevaor is movin upward wih consan speed (no acceleraion) so S s m = 0, or S s = m as we would epec. Call he scale readin for his par S up (in unis of Newons). Consider he free bod diaram for each he case where he elevaor is accelerain down (lef) and up (rih). The man is represened as a sphere and he scale readin is represened as S. a m S down S up a m S up m = m a, and since a = 1 a his paricular insan, 5 S up m = 1 5 m S up = m m = 6 5 m = 6 5 S s. Thus he scale ives a hiher readin when he elevaor is accelerain upwards. 06 (par of 3) 10 poins Find S down, he scale readin when he elevaor is movin downward wih consan veloci, in erms of S s. 1. S down = 7 5 S s. S down = 5 6 S s 3. S down = 4 5 S s 4. S down = 5 4 S s 5. S down = 3 S s 6. S down = 3 5 S s 7. S down = 6 5 S s 8. S down = S s correc 9. S down = 3 S s 10. S down = 7 6 S s A consan veloci he acceleraion is zero. Equaion (1) in he upward direcion reads S down m = 0,

16 Version 055 Miderm 1 OConnor (05141) 16 since a = 0 a his paricular insan S down m = 0 S down = m 0 = m = S s. Thus he scale ives he same readin as when saionar if he elevaor is movin downward a consan veloci. 08 (par 1 of ) 10 poins Hin: sin θ + cos θ = 1. Consider he 674 N weih held b wo cables shown below. The lef-hand cable had ension 40 N and makes an anle of θ wih he ceilin. The rih-hand cable had ension 500 N and makes an anle of θ 1 wih he ceilin. 07 (par 3 of 3) 10 poins Suppose all of a sudden he rope breaks when he elevaor is movin upward. Wha is he readin of he scale S b afer he rope is broken? 40 N θ θ N 1. S b = 3 S s. S b = 3 S s 3. S b = 7 6 S s 4. S b = 4 5 S s 5. S b = 5 6 S s 6. S b = 7 5 S s 7. S b = 3 5 S s 8. S b = 5 4 S s 9. S b = 6 5 S s 10. S b = 0 N correc When he rope breaks, he whole ssem moves under he influence of ravi wih he acceleraion a = ĵ, so S b m = m S b = 0 N. Inuiivel his implies ha here is no force beween he person and he scale. Noe: There are four differen presenaions of his problem. 674 N a) Wha is he anle θ 1 which he rihhand cable makes wih respec o he ceilin? correc Observe he free-bod diaram below.

17 θ F F 1 Version 055 Miderm 1 OConnor (05141) 17 (40 N) ] (500 N) (674 N) = θ 1 W Noe: The sum of he - and -componens of F 1, F, and W are equal o zero. 09 (par of ) 10 poins b) Wha is he anle θ which he lef-hand cable makes wih respec o he ceilin? correc Given : W = 674 N, F 1 = 500 N, F = 40 N. Basic Conceps: F = 0 and and F1 = F F 1 cos θ 1 = F cos θ (1) F1 cos θ 1 = F cos θ () F = 0 F 1 + F + F 3 = 0 F 1 sin θ 1 + F sin θ F 3 = 0 F 1 sin θ 1 = F sin θ + F 3 F1 sin θ 1 = F sin θ F F 3 sin θ + F3, since (3) F 3 sin θ 3 = F 3 sin 70 = F 3, and F 3 cos θ 3 = F 3 cos 70 = 0. Soluion: Since sin θ + cos θ = 1 and addin Eqs. and 3, we have F = F 1 F 1 F 3 sin θ 1 + F 3 sin θ 1 = F 1 + F 3 F F 1 F ( 3 F θ 1 = arcsin 3 + F1 F ) F 1 F 3 [ (674 N) + (500 N) = arcsin (500 N) (674 N) Usin Eq. 1, we have cos θ = F 1 cos θ 1 F ( ) F1 θ = arccos cos θ 1 F ( ) 500 N = arccos cos N = (par 1 of 3) 10 poins A 3.14 k block slides down a smooh, fricionless plane havin an inclinaion of 3. The acceleraion of ravi is 9.8 m/s.

18 Version 055 Miderm 1 OConnor (05141) 18 W = m sin θ = m a 3.85 m 3.14 k µ = 0 3 Find he acceleraion of he block m/s m/s m/s Moion has a consan acceleraion. Recall he kinemaics of moion wih consan acceleraion. Soluion: Because he block slides down alon he plane of he ramp, i seems loical o choose he -ais in his direcion. Then he -ais mus emere perpendicular o he ramp, as shown. Le us now eamine he forces in he - direcion. Onl he weih has a componen alon ha ais. So, b Newon s second law, F = m a = m sin θ m/s m/s m/s m/s m/s correc m/s m/s Given : m = 3.14 k, θ = 3, and µ s = 0. Consider he free bod diaram for he block Thus a = sin θ Wih our paricular value of θ, a = (9.8 m/s ) sin 3 = m/s 031 (par of 3) 10 poins Wha is he block s speed when, sarin from res, i has raveled a disance of 3.85 m alon he incline m/s m/s m/s m/s m/s m sin θ N Basic Conceps: µ N N = m cos θ W = m F,ne = F cos θ W = m/s m/s m/s m/s correc m/s

19 Version 055 Miderm 1 OConnor (05141) 19 Since v 0 = 0, a = m/s and L = 3.85 m, v f = v 0 + a ( 0 ) = ( m/s ) (3.85 m) v f = m/s. I is ver imporan o noe a his poin ha neiher of hese values depended on he mass of he block. This ma seem odd a firs, bu recall wha Galileo discovered 300 ears ao objecs of differin mass fall a he same rae. 03 (par 3 of 3) 10 poins Wha is he maniude of he perpendicular force ha he block eers on he surface of he plane a a disance of 3.85 m down he incline? N N N N N N correc N N N N B eaminin he free bod diaram aain, we see ha he force in he direcion is iven b F = m cos θ = FN. Two blocks on a fricionless horizonal surface are conneced b a lih srin. The acceleraion of ravi is 9.8 m/s. T 1.4 k 0.7 k 49.8 N Find he acceleraion of he ssem m/s µ m/s correc m/s m/s m/s m/s m/s m/s m/s m/s Le : m 1 = 1.4 k, m = 0.7 k, F = 49.8 N, and µ = 0, Pars 1 and µ = 0.109, Pars 3 and 4. Appl Newon s second law o he each block F 1ne = m 1 a = T, (1) = F N = m cos θ = 3.14 k (9.8 m/s ) cos 3 = N. 033 (par 1 of 4) 10 poins F ne = m a = F T. () Addin hese equaions ives us (m 1 + m ) a = F a = F m 1 + m.

20 Version 055 Miderm 1 OConnor (05141) (par of 4) 10 poins Wha is he ension in he srin beween he blocks? N N N N N N N N correc N N From equaion 1, T = m 1 a. 035 (par 3 of 4) 10 poins If he surface were fricional, and he coefficien of kineic fricion beween each block and he surface is 0.109, wha would be he new acceleraion? m/s correc m/s m/s m/s m/s m/s m/s m/s m/s The fricion force on each block is F fr = µ m. so appl Newon s second law o each block F 1ne = m 1 a 1 = T 1 µ m 1 (3) F ne = m a 1 = F T 1 µ m (4) Addin hese equaions ives (m 1 + m ) a 1 = F µ m 1 µ m a 1 = F µ (m 1 + m ) m 1 + m F a 1 = µ. m 1 + m 036 (par 4 of 4) 10 poins Wha would be he new ension in he srin beween he blocks? N N N N N N N N correc N N From equaion 3, T 1 = µ m 1 + m 1 a m/s