finish line dr 1 L 1 v b v r finish line.

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1 Answer, Key { Homewor { Rubn H Landau 1 Ths prn-ou should have 1 quesons. Chec ha s complee before leavng he prner. Also, mulple-choce quesons may connue on he nex column or page: nd all choces before mang your selecon. Noe ha only a few (usually 4) of he problems wll have her scores ep for a grade. You may mae mulple res o ge a problem rgh, alhough 's worh less each me. Wored soluons o a numner of hese problems (even some of he scored ones) may be found on he Ph11 home page. Dsplacemen Curve 0:01, rgonomery,mulple choce, > 1 mn. 001 Consder a movng objec whose poson x s ploed as a funcon of he me on he followng gure: x 1 O I 1 II III Clearly, he objec moved n deren ways durng he me nervals denoed I, II and III on he gure. Durng hese hree nervals, when was he objec's speed hghes? Cauon: Do no confuse he speed wh he velocy. 1. Durng nerval I.. Durng nerval II.. Durng nerval III. correc 4. Durng nervals II and III (same speed durng hose wo nervals).. Same speed durng each of he hree nervals. The velocy v s he slope of he x() curve; he magnude v jvj of hs slope s he speed. Loong a he pcure we see ha he curve s seepes (n absolue magnude) durng he nerval III and ha s when he objec had he hghes speed. 00 Durng whch nerval(s) dd he objec have no change n dsplacemen? 1. Durng each of he hree nervals.. Durng nerval II only.. Durng nerval III only. 4. Durng nerval I only. correc. Durng none of he hree nervals. 6. Durng nervals II and III. The gure shows no change n dsplacemen durng he nerval I bu here are changes n dsplacemens durng he oher nervals II and III. 00 Durng whch nerval(s) dd he objec's velocy reman consan? 1. Durng each of he hree nervals. correc. Durng nerval II only.. Durng nerval III only. 4. Durng nerval I only.. Durng none of he hree nervals. For each of he hree nervals I, II or III, he x() curve s lnear, so s slope he velocy v s consan durng each nerval. Beween he nervals, he velocy dd change n a raher abrup manner bu dd reman consan durng each nerval. 004

2 Answer, Key { Homewor { Rubn H Landau Durng whch nerval(s) does he objec have non-zero, posve acceleraon? 1. Durng nerval I only.. Durng nerval II only.. Durng nerval III only. 4. Durng none of he hree nervals. correc. Durng each of he hree nervals. Durng each nerval he velocy remaned consan (see prevous explanaon), so he acceleraon was zero. Beween he nervals, he velocy changed abruply, whch means he acceleraons were undened bu he queson s abou he nervals, no he ranson nsans. Brd and Runner 0:01, rgonomery, numerc, > 1 mn. 00 A runner s joggng a a seady v r 8: mhr. When he runner s L :7 m from he nsh lne a brd begns yng from he runner o he nsh lne a v b : mhr ( 4 mes as fas as he runner). When he brd reaches he nsh lne, urns around and es bac o he runner. Even hough he brd s a dodo, we wll assume ha occupes only one pon n space,.e.,a zero lengh brd. v b v r L fnsh lne How far does he brd ravel? Correc answer: 4: m. Le, dodo brds y, and d r be he dsance he runner ravels. d b be he dsance he brd ravels. v r be he speed of he runner. v b be he speed of he brd. L d r be he orgnal dsance o he nsh lne. L 1 be he dsance o he nsh lne afer he rs encouner.. L be he dsance o he nsh lne afer he h encouner. dr 1 db1 L 1 fnsh lne Snce he brd ravels 4 mes as fas as he runner a he rs meeng beween he brd and runner, d b1 4d r1 : (1) The sum of he brd's and runner's dsances s 4 mes L. d b1 + d r1 L: () Therefore, subsung for d b1 from Eq. (1) d r1 +4d r1 L d r1 L (:7 m)1:08 m : () Thus he dsance he brd es s d b1 4d r1 8 L 8 (:7 m)4: m ; (4) and he dsance for he runner o ravel afer hs rs encouner s L 1 L (:7 m)1:6 m : 006 Afer hs rs encouner, he brd hen urns around and es from he runner bac o he nsh lne, urns around agan and es bac o he runner. The brd repeas he bac and forh rps unl he runner reaches he nsh lne. How far does he brd ravel from he begnnng? (.e. nclude he dsance raveled o

3 he rs encouner) Correc answer: 10:8 m. Repeang hs scenaro a second me he dsance for he runner o ravel afer he second encouner s L L 1 and he hrd me and he h me L L L L,1 Answer, Key { Homewor { Rubn H Landau L; L; L: () Noe: The dsance he brd ravels beween he (, 1) h and h me s [see Eq. (4)] d b 8 L (6) and summng over all erms d b " d b d b 8 # L (7) " 8 L (8) # Or, by facorng 1 from he second erm on ( " d b 8 L #) 1+ + (9) By comparng Eq. (8) wh (9), and generalzng ( 1, and ), he nne seres 1+ (10) hen solvng Eq. (10) for, ;, : Therefore [from Eq. (7)] d b 8 X 1 L 8 L 4L 4(:7 m)10:8 m: (11) Elegan Alernave Soluon: The brd wll ravel 4 mes as far as he runner n he same me. Snce he brd and jogger ravel for he same lengh of me, he brd wll ravel d b 4 L 4(:7 m)10:8 m Algorhm 4 (1) 6 +1:0 () h4 +1:0 h h + h:h j :0 () :0 h4 8 h h h, :0 (4) h,:0 h h, h:h v r 8: mhr () 9 v b v r (6) h4h8: : mhr hmhr h hmhr d r :7 m 9 (7)

4 Answer, Key { Homewor { Rubn H Landau 4 L 1 d r hh:7 h 1:6 m h hm hm h (8) How many g's would you pull? (Wha facor mes g 9:80 m/s s hs?) Correc answer: 4:060. a ng,so n a g d r1 :0 d r :0 h:7 h 1:08 m h hm hm h d b1 jd r h8h:7 h 4: m h hm hm h (9) (10) d b d r (11) h4h:7 10:8 m hm h hm Acceleraon of a Sled 0:0, rgonomery, numerc, > 1 mn. 009 A roce-drven sled runnng on a sragh, level rac has been used o sudy he physologcal eecs of large acceleraons on asronaus. One such sled can aan a speed of 448 ms n1:9 s sarng from res. Wha s he acceleraon of he sled, assumng s consan? Correc answer: :789 ms. v v 0 + a;so a v, v 0 (448 ms), (0 ms) 1:9 s :789 ms 010 :789 ms 9:8 ms 4:060 Acceleraon of a Sled 0:0, rgonomery, numerc, > 1 mn. 011 How far does he sled ravel n 1:9 s, sarng from res? Correc answer: 4:6 m. Assumng an nal poson and velocy of zero, x v a Algorhm 0+ 1 (:789 ms )(1:9 s) 4:6 m g 9:8 ms (1) v 0 0:0 ms () v 448 ms 0 40 () 1:9 s 1: (4) a v h448 h1:9 :789 ms hms hms hs n a g h:789 h9:8 4:060 h hms hms () (6)

5 Answer, Key { Homewor { Rubn H Landau x a:0 :0 h:789h1:9:0 :0 4:6 m hm hms hs :0 h (7) 01 A ball s dropped from res a pon O. Afer fallng for some me, passes by a wndow of hegh b 4 m and does so durng me AB 0:46 s. O A B The ball acceleraes all he way down (g 9:8 ms ); le v A be s speed as passes he wndow's op A and v B s speed as passes he wndow's boom B. Howmuch dd he ball speed up as passed he wndow? Calculae v down v B, v A. Correc answer: 4:08 ms. Assume down s posve. The ball falls under a consan acceleraon g, so g v b v B, v A and he change of s velocy durng me AB s smply v v B, v A g AB, assumng he downward drecon o be posve. v down (9:8 ms )(0:46 s) 4:08 ms: 014 Calculae he speed v A a whch he ball passes he wndow's op. Correc answer: 6:4416 ms. Down s posve. Gven he unform downward acceleraon g, wehave y() y( A )+v A (, A )+ 1 g(, A) and hence b y( B ), y( A ) v A ( B, A )+ 1 g( B, A ) v A AB + 1 g AB : Solvng hs equaon for he velocy v A, we oban v A b, g AB AB 4m 0:46 s, (9:8 ms )(0:46 s) Thus he speed v A jv A j 6:4416 ms. 6:4416 ms: Ball Dropped From Res 0:0, rgonomery, numerc, > 1 mn. 01 Now consder a new suaon: The ball s hrown upward from he ground wh an nal velocy ha aes exacly he same me BA AB 0:46 s o pass by he wndow, wh he ball movng up raher han down. Consder he ball's slowdown durng hs me: Le v 0 B be he ball's speed (do no confuse he speed wh he velocy) as passes he wndow's boom on he way up and le v 0 A be s speed as passes he wndow's op, also n s way up. How does he ball's slowdown v up v 0 B, v0 A compare o s speedup v down on he way down? 1. v up > v down :. v up v down : correc. v up < v down : 4. v up > v down f he mass of he ball s less han 0:1 g and v up < v down f he mass of he ball s greaer han 0:1 g. v up < v down f he mass of he ball

6 Answer, Key { Homewor { Rubn H Landau 6 s less han 0:1 g and v up > v down f he mass of he ball s greaer han 0:1 g v up v 0 A, v0 B The free-fall acceleraon always pons downward, regardless of wheher he velocy s downward or upward, so he velocy change durng he same me perod s always he same, Algorhm x 0 17:9 m (1) v 4: ms 18:8 () g 9:8 ms () p v 0 v :0, :0 gx 0 (4) q h4: :0, :0 h9:8h17:9 1: ms q hms hms :0,hhms hm v 0 A, v 0 B g BA g AB v down : Algorhm g 9:8 ms (1) b 4m 4 () 0: AB 0:46 s () 0: dv g AB (4) h9:8h0:46 4:08 ms hms hms hs v A b, 0: g:0 AB () AB h4,0: h9:8h0:46:0 h0:46 6:4416 ms hms hm, h hms hs :0 hs h v:0 :0 g h4::0 :0 h9:8 9:8796 m hm hms:0 h hms () Sone Thrown Up 0:0, arhmec, numerc, > 1 mn. 017 Wha s he maxmum dsance from he ground below (no from he cl) durng s gh? Correc answer: 9:8796 m. When he sone s a s maxmum hegh h, s velocy s zero: so h v gh ) h v g (4: ms) (9:8 ms ) 9:8796 m