Kinematics in 1D From Problems and Solutions in Introductory Mechanics (Draft version, August 2014) David Morin,


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1 Chaper 2 Kinemaics in 1D From Problems and Soluions in Inroducory Mechanics (Draf ersion, Augus 2014) Daid Morin, As menioned in he preface, his book should no be hough of as a exbook. The inroducion o each chaper is brief and is herefore no subsiue for an acual exbook. You will mos likely wan o hae a exbook on hand when reading he inroducions. 2.1 Inroducion In his chaper and he nex, we won be concerned wih he forces ha cause an objec o moe in he paricular way i is moing. We will simply ake he moion as gien, and our goal will be o relae posiions, elociies, and acceleraions as funcions of ime. Our objecs can be reaed like poin paricles; we will no be concerned wih wha hey are acually made of. This is he sudy of kinemaics. In Chaper 4 we will moe on o dynamics, where we will deal wih mass, force, energy, momenum, ec. Velociy and acceleraion In one dimension, he aerage elociy and acceleraion oer a ime ineral are gien by ag = x and a ag =. (2.1) The insananeous elociy and acceleraion a a paricular ime are obained by leing he ineral become infiniesimally small. In his case we wrie he as a d, and he insananeous and a are gien by = dx d and a = d d. (2.2) In calculus erms, is he deriaie of x, and a is he deriaie of. Equialenly, is he slope of he x s. cure, and a is he slope of he s. cure. In he case of he elociy, you can see how his slope arises by aking he limi of = x/, as becomes ery small; see Fig The smaller is, he beer he slope x/ approximaes he acual slope of he angen line a he gien poin P. In 2D and 3D, he elociy and acceleraion are ecors. Tha is, we hae a separae pair of equaions of he form in Eq. (2.2) for each dimension; he x componens are gien by x = dx/d and a x = d x /d, and likewise for he y and z componens. The elociy and acceleraion are also ecors in 1D, alhough in 1D a ecor can be iewed simply as a number (which may be posiie or negaie). In any dimension, he speed is he magniude of he elociy, which means he absolue alue of in 1D and he lengh of he ecor in 2D and 3D. So he speed is a posiie number by definiion. The unis of elociy and speed are m/s, and he unis of acceleraion are m/s 2. x acual slope a poin P P Figure 2.1 x 27
2 28 CHAPTER 2. KINEMATICS IN 1D area = x Figure 2.2 Figure 2.3 () Displacemen as an area If an objec moes wih consan elociy, hen he displacemen x during a ime is x =. In oher words, he displacemen is he area of he region (which is jus a recangle) under he s. cure in Fig Noe ha he displacemen (which is x by definiion), can be posiie or negaie. The disance raeled, on he oher hand, is defined o be a posiie number. In he case where he displacemen is negaie, he s. line in Fig. 2.2 lies below he axis, so he (signed) area is negaie. If he elociy aries wih ime, as shown in Fig. 2.3, hen we can diide ime ino a large number of shor inerals, wih he elociy being essenially consan oer each ineral. The displacemen during each ineral is essenially he area of each of he narrow recangles shown. In he limi of a ery large number of ery shor inerals, adding up he areas of all he hin recangles gies exacly he oal area under he cure; he areas of he iny riangular regions a he ops of he recangles become negligible in his limi. So he general resul is: The displacemen (ha is, he change in x) equals he area under he s. cure. Said in a more mahemaical way, he displacemen equals he ime inegral of he elociy. This saemen is equialen (by he fundamenal heorem of calculus) o he fac ha is he ime deriaie of x. All of he relaions ha hold beween x and also hold beween and a. In paricular, he change in equals he area under he a s. cure. And conersely, a is he ime deriaie of. This is summarized in he following diagram: slope (deriaie) slope (deriaie) x a area (inegral) area (inegral) Moion wih consan acceleraion For moion wih consan acceleraion a, we hae a() = a, () = 0 + a, x() = x a2, (2.3) where x 0 and 0 are he iniial posiion and elociy a = 0. The aboe expressions for () and x() are correc, because () is indeed he deriaie of x(), and a() is indeed he deriaie of (). If you wan o derie he expression for x() in a graphical manner, see Problem 2.1. The aboe expressions are echnically all you need for any seup inoling consan acceleraion, bu one addiional formula migh make hings easier now and hen. If an objec has a displacemen d wih consan acceleraion a, hen he iniial and final elociies saisfy 2 f 2 i = 2ad. (2.4) See Problem 2.2 for a proof. If you know hree ou of he four quaniies f, i, a, and d, hen his formula quickly gies he fourh. In he special case where he objec sars a res (so i = 0), we hae he simple resul, f = 2ad. Falling bodies Perhaps he mos common example of consan acceleraion is an objec falling under he influence of only graiy (ha is, we ll ignore air resisance) near he surface of he earh. The
3 2.2. MULTIPLECHOICE QUESTIONS 29 consan naure of he graiaional acceleraion was famously demonsraed by Galileo. (He mainly rolled balls down ramps insead of dropping hem, bu i s he same idea.) If we ake he posiie y axis o poin upward, hen he acceleraion due o graiy is g, where g = 9.8 m/s 2. Afer eery second, he elociy becomes more negaie by 9.8 m/s; ha is, he downward speed increases by 9.8 m/s. If we subsiue g for a in Eq. (2.3) and replace x wih y, he expressions become a() = g, () = 0 g, y() = y g2, (2.5) For an objec dropped from res a a poin we choose o label as y = 0, Eq. (2.5) gies y() = g 2 /2. In some cases i is adanageous o choose he posiie y axis o poin downward, in which case he acceleraion due o graiy is g (wih no minus sign). In any case, i is always a good idea o ake g o be he posiie quaniy 9.8 m/s 2, and hen hrow in a minus sign by hand if needed, because working wih quaniies wih minus signs embedded in hem can lead o confusion. The expressions in Eq. (2.5) hold only in he approximaion where we neglec air resisance. This is generally a good approximaion, as long as he falling objec isn oo ligh or moing oo quickly. Throughou his book, we will ignore air resisance unless saed oherwise. 2.2 Muliplechoice quesions 2.1. If an objec has negaie elociy and negaie acceleraion, is i slowing down or speeding up? (a) slowing down (b) speeding up 2.2. The firs figure below shows he a s. plo for a cerain seup. The second figure shows he s. plo for a differen seup. The hird figure shows he x s. plo for a ye anoher seup. Which of he wele labeled poins correspond(s) o zero acceleraion? Circle all ha apply. (To repea, he hree seups hae nohing o do wih each oher. Tha is, he plo is no he elociy cure associaed wih he posiion in he x plo. ec.) a x A B C D E F G H I J K L 2.3. If he acceleraion as a funcion of ime is gien by a() = A, and if x = = 0 a = 0, wha is x()? (a) A2 2 (b) A2 6 (c) A 3 (d) A3 2 (e) A Under wha condiion is he aerage elociy (which is defined o be he oal displacemen diided by he ime) equal o he aerage of he iniial and final elociies, ( i + f )/2? (a) The acceleraion mus be consan. (b) I is rue for oher moions besides consan acceleraion, bu no for all possible moions. (c) I is rue for all possible moions.
4 30 CHAPTER 2. KINEMATICS IN 1D 2.5. Two cars, wih iniial speeds of 2 and, lock heir brakes and skid o a sop. Assume ha he deceleraion while skidding is independen of he speed. The raio of he disances raeled is (a) 1 (b) 2 (c) 4 (d) 8 (e) You sar from res and accelerae wih a gien consan acceleraion for a gien disance. If you repea he process wih wice he acceleraion, hen he ime required o rael he same disance (a) remains he same (b) is doubled (c) is haled (d) increases by a facor of 2 (e) decreases by a facor of A car raels wih consan speed 0 on a highway. A he insan i passes a saionary police moorcycle, he moorcycle acceleraes wih consan acceleraion and gies chase. Wha is he speed of he moorcycle when i caches up o he car (in an adjacen lane on he highway)? Hin: Draw he s. plos on op of each oher. (a) 0 (b) 3 0 /2 (c) 2 0 (d) 3 0 (e) You sar from res and accelerae o a gien final speed 0 afer a ime T. Your acceleraion need no be consan, bu assume ha i is always posiie or zero. If d is he oal disance you rael, hen he range of possible d alues is (a) d = 0 T/2 (b) 0 < d < 0 T/2 (c) 0 T/2 < d < 0 T (d) 0 < d < 0 T (e) 0 < d < 2.9. You are driing a car ha has a maximum acceleraion of a. The magniude of he maximum deceleraion is also a. Wha is he maximum disance ha you can rael in ime T, assuming ha you begin and end a res? (a) 2aT 2 (b) at 2 (c) at 2 /2 (d) at 2 /4 (e) at 2 / A golf club srikes a ball and sends i sailing hrough he air. Which of he following choices bes describes he sizes of he posiion, speed, and acceleraion of he ball a a momen in he middle of he srike? ( Medium means a noniny and nonhuge quaniy, on an eeryday scale.) (a) x is iny, is medium, a is medium (b) x is iny, is medium, a is huge (c) x is iny, is huge, a is huge (d) x is medium, is medium, a is medium (e) x is medium, is medium, a is huge Which of he following answers is he bes esimae for he ime i akes an objec dropped from res o fall a erical mile (abou 1600 m)? Ignore air resisance, as usual. (a) 5 s (b) 10 s (c) 20 s (d) 1 min (e) 5 min
5 2.3. PROBLEMS You hrow a ball upward. Afer half of he ime o he highes poin, he ball has coered (a) half he disance o he op (b) more han half he disance (c) less han half he disance (d) I depends on how fas you hrow he ball A ball is dropped, and hen anoher ball is dropped from he same spo one second laer. As ime goes on while he balls are falling, he disance beween hem (ignoring air resisance, as usual) (a) decreases (b) remains he same (c) increases and approaches a limiing alue (d) increases seadily You hrow a ball sraigh upward wih iniial speed 0. How long does i ake o reurn o your hand? (a) 2 0 /2g (b) 2 0 /g (c) 0/2g (d) 0 /g (e) 2 0 /g Ball 1 has mass m and is fired direcly upward wih speed. Ball 2 has mass 2m and is fired direcly upward wih speed 2. The raio of he maximum heigh of Ball 2 o he maximum heigh of Ball 1 is (a) 1 (b) 2 (c) 2 (d) 4 (e) Problems The firs hree problems are foundaional problems Area under he cure A = 0 an objec sars wih posiion x 0 and elociy 0 and moes wih consan acceleraion a. Derie he x() = x a 2 /2 resul by finding he area under he s. cure (wihou using calculus) A kinemaic relaion Use he relaions in Eq. (2.3) o show ha if an objec moes hrough a displacemen d wih consan acceleraion a, hen he iniial and final elociies saisfy 2 f 2 i = 2ad Maximum heigh If you hrow a ball sraigh upward wih iniial speed 0, i reaches a maximum heigh of 0 2 /2g. How many deriaions of his resul can you hink of? 2.4. Aerage speeds (a) If you ride a bike up a hill a 10 mph, and hen down he hill a 20 mph, wha is your aerage speed? (b) If you go on a bike ride and ride for half he ime a 10 mph, and half he ime a 20 mph, wha is your aerage speed? 2.5. Colliding rains Two rains, A and B, rael in he same direcion on he same se of racks. A sars a res a posiion d, and B sars wih elociy 0 a he origin. A acceleraes wih acceleraion a, and B deceleraes wih acceleraion a. Wha is he maximum alue of 0 (in erms of d and a) for which he rains don collide? Make a rough skech of x s. for boh rains in he case where hey barely collide.
6 32 CHAPTER 2. KINEMATICS IN 1D 2.6. Raio of disances Two cars, A and B, sar a he same posiion wih he same speed 0. Car A raels a consan speed, and car B deceleraes wih consan acceleraion a. A he insan when B reaches a speed of zero, wha is he raio of he disances raeled by A and B? Draw a reasonably accurae plo of x s. for boh cars. You should find ha your answer for he raio of he disances is a nice simple number, independen of any of he gien quaniies. Gie an argumen ha explains why his is he case How far apar? An objec sars from res a he origin a ime = T and acceleraes wih consan acceleraion a. A second objec sars from res a he origin a ime = 0 and acceleraes wih he same a. How far apar are hey a ime? Explain he meaning of he wo erms in your answer, firs in words, and hen also wih regard o he s. plos Raio of odd numbers An objec is dropped from res. Show ha he disances fallen during he firs second, he second second, he hird second, ec., are in he raio of 1 : 3 : 5 : Dropped and hrown balls A ball is dropped from res a heigh h. Direcly below on he ground, a second ball is simulaneously hrown upward wih speed 0. If he wo balls collide a he momen he second ball is insananeously a res, wha is he heigh of he collision? Wha is he relaie speed of he balls when hey collide? Draw he s. plos for boh balls Hiing a he same ime A ball is dropped from res a heigh h. Anoher ball is simulaneously hrown downward wih speed from heigh 2h. Wha should be so ha he wo balls hi he ground a he same ime? Two dropped balls A ball is dropped from res a heigh 4h. Afer i has fallen a disance d, a second ball is dropped from res a heigh h. Wha should d be (in erms of h) so ha he balls hi he ground a he same ime? 2.4 Muliplechoice answers 2.1. b The objec is speeding up. Tha is, he magniude of he elociy is increasing. This is rue because he negaie acceleraion means ha he change in elociy is negaie. And we are old ha he elociy is negaie o sar wih. So i migh go from, say, 20 m/s o 21 m/s a momen laer. I is herefore speeding up. Remarks: If we had said ha he objec had negaie elociy and posiie acceleraion, hen i would be slowing down. Basically, if he sign of he acceleraion is he same as (or he opposie of) he sign of he elociy, hen he objec is speeding up (or slowing down). A commen on erminology: The word decelerae means o slow down. The word accelerae means in a colloquial sense o speed up, bu as a physics erm i means (in 1D) o eiher speed up or slow down, because acceleraion can be posiie or negaie. More generally, in 2D or 3D i means o change he elociy in any general manner (magniude and/or direcion) B,E,H,K Poin B is where a equals zero in he firs figure. Poins E and H are where he slope (he deriaie) of he s. plo is zero; and he slope of is a. Poin K is where he slope of he x s. plo is maximum. In oher words, i is where is maximum. Bu he slope of a funcion is zero a a maximum, so he slope of (which is a) is zero a K.
7 2.4. MULTIPLECHOICE ANSWERS 33 Remark: In calculus erms, K is an inflecion poin of he x s. cure. I is a poin where he slope is maximum. Equialenly, he deriaie of he slope is zero. Equialenly again, he second deriaie is zero. In he presen case, he angen line goes from lying below he x s. cure o lying aboe i; he slope goes from increasing o decreasing as i passes hrough is maximum alue e Since he second deriaie of x() equals a(), we mus find a funcion whose second deriaie is A. Choice (e) is saisfies his requiremen; he firs deriaie equals A 2 /2, and hen he second deriaie equals A, as desired. The sandard A 2 /2 resul is alid only for a consan acceleraion a. Noe ha all of he choices saisfy x = = 0 a = 0. Remark: If we add on a consan C o x(), so ha we now hae A 3 /6 + C, hen he x = 0 iniial condiion isn saisfied, een hough a() is sill equal o A. Similarly, if we add on a linear erm B, hen he = 0 iniial condiion isn saisfied, een hough a() is again sill equal o A. If we add on quadraic erm D 2, hen alhough he x = = 0 iniial condiions are saisfied, he second deriaie is now no equal o A. Likewise for any power of ha is 4 or higher. So no only is he A 3 /6 choice he only correc answer among he fie gien choices, i is he only correc answer, period. Formally, he inegral of a (which is ) mus ake he form of A 2 /2 + B, where B is a consan of inegraion. And he inegral of (which is x) mus hen ake he form of A 3 /6 + B + C, where C is a consan of inegraion. The iniial condiions x = = 0 hen quickly ell us ha C = B = b The saemen is a leas rue in he case of consan acceleraion, as seen by looking a he s. plo in Fig. 2.4(a). The area under he s. cure is he disance raeled, and he area of he rapezoid (which corresponds o consan acceleraion) is he same as he area of he recangle (which corresponds o consan elociy ( i + f )/2). Equialenly, he areas of he riangles aboe and below he ( i + f )/2 line are equal. If you wan o work hings ou algebraically, he displacemen is d = i a2 = 1 2 (2 i + a) = 1 2 ( i + ( i + a) ) = 1 2 ( i + f ). (2.6) The aerage elociy d/ is herefore equal o ( i + f )/2, as desired. The saemen is cerainly no rue in all cases; a counerexample is shown in Fig. 2.4(b). The disance raeled (he area under he cure) is essenially zero, so he aerage elociy is essenially zero and hence no equal o ( i + f )/2. Howeer, he saemen can be rue for moions wihou consan acceleraion, as long as he area under he s. cure is he same as he area of he recangle associaed wih elociy ( i + f )/2, as shown in Fig. 2.4(c). For he cure shown, his requiremen is he same as saying ha he areas of he wo shaded regions are equal. (a) (b) (c) f f f ( i + f )/2 ( i + f )/2 i i i Figure c The final speed is zero in each case, so he f 2 2 i = 2ad relaion in Eq. (2.4) gies 0 i 2 = 2( a)d, where a is he magniude of he (negaie) acceleraion. So d = i 2/2a. Since his is proporional o i 2, he car wih wice he iniial speed has four imes he sopping disance. Alernaiely, he disance raeled is d = i a 2 /2, where again a is he magniude of he acceleraion. Since he car ends up a res, he () = i a expression for he elociy
8 34 CHAPTER 2. KINEMATICS IN 1D ells us ha = 0 when = i /a. So ( i ) d = i a 1 ( 2 a i ) 2 = i 2 a 2a, (2.7) in agreemen wih he relaion obained ia Eq. (2.4). 2 Figure Alernaiely again, we could imagine reersing ime and acceleraing he cars from res. Using he fac ha one ime is wice he oher (since = i /a), he relaion d = a 2 /2 immediaely ells us ha wice he ime implies four imes he disance. Alernaiely ye again, he facor of 4 quickly follows from he s. plo shown in Fig The area under he diagonal line is he disance raeled, and he area of he large riangle is four imes he area of he small lowerlef riangle, because all four of he small riangles hae he same area. Remark: When raeling in a car, he safe disance (according o many sources) o keep beween your car and he car in fron of you is dicaed by he hreesecond rule (in good weaher). Tha is, your car should pass, say, a gien ree a leas hree seconds afer he car in fron of you passes i. This rule inoles ime, bu i immediaely implies ha he minimum following disance is proporional o your speed. I herefore can sricly be correc, because we found aboe ha he sopping disance is proporional o he square of your speed. This square behaior means ha he hreesecond rule is inadequae for sufficienly high speeds. There are of course many oher facors inoled (reacion ime, he naure of he road hazard, he fricion beween he ires and he ground, ec.), so he exac formula is probably oo complicaed o be of much use. Bu if you ake a few minues o obsere some cars and make some rough esimaes of how driers ou here are behaing, you ll find ha many of hem are following a asonishingly unsafe disances, by any measure e The disance raeled is gien by d = a 2 /2, so = 2d/a. Therefore, if a is doubled hen decreases by a facor of 2. 2a 2a a S 2 S 1 Figure Remark: Since = a for consan acceleraion, he speeds in he wo gien scenarios (label hem S 1 and S 2 ) differ by a facor of 2 a any gien ime. So if a all imes he speed in S 2 is wice he speed in S 1, shouldn he ime simply be haled, insead of decreased by he facor of 2 ha we jus found? No, because alhough he S 1 disance is only d/2 when he S 2 disance reaches he final alue of d, i akes S 1 less ime o rael he remaining d/2 disance, because is speed increases as ime goes on. This is shown in Fig The area under each s. cure equals he disance raeled. Compared wih S 1, S 2 s final speed is 2 imes larger, bu is ime is 1/ 2 imes smaller. So he areas of he wo riangles are he same c The area under a s. cure is he disance raeled. The car s cure is he horizonal line shown in Fig. 2.7, and he moorcycle s cure is he iled line. The wo ehicles will hae raeled he same disance when he area of he car s recangle equals he area of he moorcycle s riangle. This occurs when he riangle has wice he heigh of he recangle, as shown. (The area of a riangle is half he base imes he heigh.) So he final speed of he moorcycle is 2 0. Noe ha his resul is independen of he moorcycle s (consan) acceleraion. If he acceleraion is small, hen he process will ake a long ime, bu he speed of he moorcycle when i caches up o he car will sill be 2 0. Alernaiely, he posiion of he car a ime is 0, and he posiion of he moorcycle is a 2 /2. These wo posiions are equal when 0 = a 2 /2 = a = 2 0. Bu he moorcycle s speed is a, which herefore equals 2 0 when he moorcycle caches up o he car. Figure d A disance of essenially zero can be obained by siing a res for nearly all of he ime T, and hen suddenly acceleraing wih a huge acceleraion o speed 0. Approximaely zero disance is raeled during his acceleraion phase. This is rue because Eq. (2.4) gies d = 2 0 /2a, where 0 is a gien quaniy and a is huge. Conersely, a disance of essenially 0 T can be obained by suddenly acceleraing wih a huge acceleraion o speed 0, and hen coasing along a speed 0 for nearly all of he ime T.
9 2.4. MULTIPLECHOICE ANSWERS 35 These wo cases are shown in he s. plos in Fig The area under he cure (which is he disance raeled) for he lef cure is approximaely zero, and he area under he righ cure is approximaely he area of he whole recangle, which is 0 T. This is he maximum possible disance, because an area larger han he 0 T recangle would require ha he s. plo exend higher han 0, which would hen require a negaie acceleraion (conrary o he saed assumpion) o bring he final speed back down o T T Figure d The maximum disance is obained by haing acceleraion a for a ime T/2 and hen deceleraion a for a ime T/2. The s. plo is shown in Fig The disance raeled during he firs T/2 is a(t/2) 2 /2 = at 2 /8. Likewise for he second T/2, because he wo riangles hae he same area, and he area under a s. cure is he disance raeled. So he oal disance is at 2 /4. Alernaiely, we see from he riangular plo ha he aerage speed is half of he maximum, which gies ag = (at/2)/2 = at/4. So he oal disance raeled is ag T = (at/4)t = at 2 /4. Noe ha he riangle in Fig. 2.9 does indeed yield he maximum area under he cure (ha is, he maximum disance raeled) subjec o he gien condiions, because (1) he riangle is indeed a possible s. plo, and (2) elociies aboe he riangle aren allowed, because he gien maximum a implies ha i would eiher be impossible o accelerae from zero iniial speed o such a, or impossible o decelerae o zero final speed from such a. a(t/2) T/2 Figure 2.9 T b The disance x is cerainly iny, because he ball is sill in conac wih he club during he (quick) srike. The speed is medium, because i is somewhere beween he iniial speed of zero and he final speed (on he order of 100 mph); i would be exacly half he final speed if he acceleraion during he srike were consan. The acceleraion is huge, because (assuming consan acceleraion o ge a rough idea) i is gien by /, where is medium and is iny (he srike is ery quick). Remark: In shor, he ball experiences a ery large a for a ery small. The largeness and smallness of hese quaniies cancel each oher and yield a medium resul for he elociy = a (again, assuming consan a). Bu in he posiion x = a 2 /2, he wo facors of win ou oer he one facor of a, and he resul is iny. These resuls (iny x, medium, and huge a) are consisen wih Eq. (2.4), which for he presen scenario says ha 2 = 2ax c From d = a 2 /2 we obain (using g = 10 m/s 2 ) 1600 m = 1 2 (10 m/s2 ) 2 = 2 = 320 s 2 = 18 s. (2.8) So 20 s is he bes answer. The speed a his ime is g = 200 m/s, which is abou 450 mph (see MulipleChoice Quesion 1.4). In realiy, air resisance is imporan, and a erminal elociy is reached. For a skydier in a spreadeagle posiion, he erminal elociy is around 50 m/s b Le he ime o he op be. Since he ball deceleraes on is way up, i moes faser in he firs /2 ime span han in he second /2. So i coers more han half he disance in he firs /2.
10 36 CHAPTER 2. KINEMATICS IN 1D Remark: If you wan o find he exac raio of he disances raeled in he wo /2 ime spans, i is easies o imagine dropping he ball insead of firing i upward; he answer is he same. In he upper /2 of he moion, he ball falls g(/2) 2 /2, whereas in he oal ime he ball falls g 2 /2. The raio of hese disances is 1 o 4, so he disance in he upper /2 is 1/4 of he oal, which means ha he disance in he lower /2 is 3/4 of he oal. The raio of he disances raeled in he wo /2 ime spans is herefore 3 o 1. This also quickly follows from drawing a s. plo like he one in Fig d If T is he ime beween he dropping of each ball (which is one second here), hen he firs ball has a speed of gt when he second ball is dropped. A a ime laer han his, he speeds of he wo balls are g( + T) and g. So he difference in speeds is always gt. Tha is, he second ball always sees he firs ball pulling away wih a relaie speed of gt. The separaion herefore increases seadily a a rae gt. This resul ignores air resisance. In realiy, he objecs will reach he same erminal elociy (barring any influence of he firs ball on he second), so he disance beween hem will approach a consan alue. The reallife answer is herefore choice (c) e The elociy as a funcion of ime is gien by () = 0 g. Since he elociy is insananeously zero a he highes poin, he ime o reach he op is = 0 /g. The downward moion akes he same ime as he upward moion (alhough i wouldn if we included air resisance), so he oal ime is 2 0 /g. Noe ha choices (a) and (b) don hae he correc unis; choice (a) is he maximum heigh d From general kinemaics (see Problem 2.3), or from conseraion of energy (he subjec of Chaper 5), or from dimensional analysis, he maximum heigh is proporional o 2 /g (i equals 2 /2g). The 2 dependence implies ha he desired raio is 2 2 = 4. The difference in he masses is irrelean. 2.5 Problem soluions Alhough his was menioned many imes in he preface and in Chaper 1, i is worh belaboring he poin: Don look a he soluion o a problem (or a muliplechoice quesion) oo soon. If you do need o look a i, read i line by line unil you ge a hin o ge going again. If you read hrough a soluion wihou firs soling he problem, you will gain essenially nohing from i! 0 Figure 2.10 a 2.1. Area under he cure The s. cure, which is simply a iled line in he case of consan acceleraion, is shown in Fig The slope of he line equals he acceleraion a, which implies ha he heigh of he riangular region is a, as shown. The area under he s. cure is he disance raeled. This area consiss of he recangle wih area 0 and he riangle wih area (1/2) a. So he oal area is 0 + a 2 /2. To find he presen posiion x(), we mus add he iniial posiion, x 0, o he disance raeled. The presen posiion is herefore x() = x a 2 /2, as desired A kinemaic relaion Firs soluion: Our sraegy will be o eliminae from he equaions in Eq. (2.3) by soling for in he second equaion and plugging he resul ino he hird. This gies ( 0 ) x = x a + 1 ( 2 a 0 ) 2 a = x a ( ) + 1 a = x a (2 0 2 ). (2.9)
11 2.5. PROBLEM SOLUTIONS 37 Hence 2a(x x 0 ) = Bu x x 0 is he displacemen d. Changing he noaion, f and 0 i, gies he desired resul, 2ad = f 2 2 i. A quick corollary is ha if d and a hae he same (or opposie) sign, hen f is larger (or smaller) han i. You should conince yourself ha his makes sense inuiiely. Second soluion: A quicker deriaion is he following. The displacemen equals he aerage elociy imes he ime, by definiion. The ime is = ( 0 )/a, and he aerage elociy is ag = ( + 0 )/2, where his second expression relies on he fac ha he acceleraion is consan. (The firs expression does oo, because oherwise we wouldn hae a unique a in he denominaor.) So we hae ( + 0 ) ( 0 ) d = ag = 2 a Muliplying by 2a gies he desired resul Maximum heigh = a. (2.10) The soluions I can hink of are lised below. Mos of hem use he fac ha he ime o reach he maximum heigh is = 0 /g, which follows from he elociy () = 0 g being zero a he op of he moion. The fac ha he acceleraion is consan also plays a criical role in all of he soluions. 1. Since he acceleraion is consan, he aerage speed during he upward moion equals he aerage of he iniial and final speeds. So ag = ( 0 + 0)/2 = 0 /2. The disance equals he aerage speed imes he ime, so d = ag = ( 0 /2)( 0 /g) = 2 0 /2g. 2. Using he sandard expression for he disance raeled, d = 0 g 2 /2, we hae ( ) 0 d = 0 g g 2 ( ) 2 0 = 2 0 g 2g. (2.11) 3. If we imagine reersing ime (or jus looking a he downward moion, which akes he same ime), hen he ball sars a res and acceleraes downward a g. So we can use he simpler expression d = g 2 /2, which quickly gies d = g( 0 /g) 2 /2 = 2 0 /2g. 4. The kinemaic relaion f 2 2 i = 2ad from Eq. (2.4) gies = 2( g)d = d = 0 2 /2g. We hae been careful wih he signs here; if we define posiie d as upward, hen he acceleraion is negaie. 5. The firs hree of he aboe soluions hae graphical inerpreaions (alhough perhaps hese shouldn coun as separae soluions). The s. plos associaed wih hese hree soluions are shown in Fig The area under each cure, which is he disance raeled, equals 2 0 /2g. (a) (b) (c) /2 0 /g 0 /g 0 /g Figure We can also use conseraion of energy o sole his problem. Een hough we won discuss energy unil Chaper 5, he soluion is quick enough o sae here. The
12 38 CHAPTER 2. KINEMATICS IN 1D iniial kineic energy m0 2 /2 ges compleely conered ino he graiaional poenial energy mgd a he op of he moion (because he ball is insananeously a res a he op). So m0 2/2 = mgd = d = 2 0 /2g Aerage speeds (a) Le he lengh of he hill be l, and define 10 mph. Then he ime up he hill is l/, and he ime down is l/2. Your aerage speed is herefore ag = d oal 2l = oal l/ + l/2 = 2 3/2 = 4 = 13.3 mph. (2.12) 3 (b) Le 2 be he oal ime of he ride, and again define 10 mph. Then during he firs half of he ride, you rael a disance. And during he second half, you rael a disance (2). Your aerage speed is herefore ag = d oal oal = = 3 2 = 15 mph. (2.13) 2.5. Colliding rains Remark: This resul of 15 mph is simply he aerage of he wo speeds, because you spend he same amoun of ime raeling a each speed. This is no he case in he scenario in par (a), because you spend longer (wice as long) raeling uphill a he slower speed. So ha speed maers more when aking he aerage. In he exreme case where he wo speeds differ grealy (in a muliplicaie sense), he aerage speed in he scenario in par (a) is ery close o wice he smaller speed (because he downhill ime can be approximaed as zero), whereas he aerage speed in he scenario in par (b) always equals he aerage of he wo speeds. For example, if he wo speeds are 1 and 100 (ignoring he unis), hen he answers o pars (a) and (b) are, respeciely, The posiions of he wo rains are gien by ag (a) 2l = l/1 + l/100 = = 1.98, ag (b) = = 101 = (2.14) 2 2 These are equal when x A = d a2 and x B = a2. (2.15) d x x A = d + a 2 /2 x B = 0 a 2 /2 d a2 = a2 = a d = 0 0 ± 0 2 4ad = =. (2.16) 2a The rains do collide if here is a real soluion for, ha is, if 2 0 > 4ad = 0 > 2 ad. The relean soluion is he roo. The + roo corresponds o he case where he rains pass hrough each oher and hen mee up again a second ime. The rains don collide if he roos are imaginary, ha is, if 0 < 2 ad. So he maximum alue of 0 ha aoids a collision is 2 ad. In he cuoff case where 0 = 2 ad, he rains barely ouch, so i s semanics as o wheher you call ha a collision. Noe ha ad correcly has he unis of elociy. And in he limi of large a or d, he cuoff speed 2 ad is large, which makes inuiie sense. 0 /2a A skech of he x s. cures for he 0 = 2 ad case is shown in Fig If 0 is smaller Figure 2.12 han 2 ad, hen he boom cure says lower (because is iniial slope a he origin is 0 ),
13 2.5. PROBLEM SOLUTIONS 39 so he cures don inersec. If 0 is larger han 2 ad, hen he boom cure exends higher, so he cures inersec wice. Remarks: As an exercise, you can show ha he locaion where he rains barely collide in he 0 = 2 ad case is x = 3d/2. And he maximum alue of x B is 2d. If B has normal fricion brakes, hen i will of course simply sop a his maximum alue and no moe backward as shown in he figure. Bu in he hypoheical case of a je engine wih reerse hrus, B would head backward as he cure indicaes. In he a 0 limi, B moes wih essenially consan speed 0 oward A, which is essenially a res, iniially a disance d away. So he ime is simply = d/ 0. As an exercise, you can apply a Taylor series o Eq. (2.16) o produce his = d/ 0 resul. A Taylor series is required because if you simply se a = 0 in Eq. (2.16), you will obain he unhelpful resul of = 0/ Raio of disances The posiions of he wo cars are gien by x A = 0 and x B = a2. (2.17) B s elociy is 0 a, and his equals zero when = 0 /a. The posiions a his ime are ( 0 ) x A = 0 = 2 0 a a and ( 0 ) x B = 0 a 1 ( 2 a 0 ) 2 = 0 2 a 2a. (2.18) The desired raio is herefore x A /x B = 2. The plos are shown in Fig Boh disances are proporional o 0 2/a, so large 0 implies large disances, and large a implies small disances. These make inuiie sense. The only quaniies ha he raio of he disances can depend on are 0 and a. Bu he raio of wo disances is a dimensionless quaniy, and here is no nonriial combinaion of 0 and a ha gies a dimensionless resul. Therefore, he raio mus simply be a number, independen of boh 0 and a. Noe ha i is easy o see from a s. graph why he raio is 2. The area under A s elociy cure (he recangle) in Fig is wice he area under B s elociy cure (he riangle). And hese areas are he disances raeled. x x A = 0 x B = 0  a 2 /2 0 /a Figure How far apar? A = 0 A ime, he firs objec has been moing for a ime +T, so is posiion is x 1 = a(+t) 2 /2. The second objec has been moing for a ime, so is posiion is x 2 = a 2 /2. The difference is x 1 x 2 = at at 2. (2.19) B = 0 a 0 /a The second erm here is he disance he firs objec has already raeled when he second objec sars moing. The firs erm is he relaie speed, at, imes he ime. The relaie speed is always at because his is he speed he firs objec has when he second objec sars moing. And from ha ime onward, boh speeds increase a he same rae (namely a), so he objecs always hae he same relaie speed. In summary, from he second objec s poin of iew, he firs objec has a head sar of at 2 /2 and hen seadily pulls away wih relaie speed at. Figure = a(+t ) 2 = at The s. plos are shown in Fig The area under a s. cure is he disance raeled, so he difference in he disances is he area of he shaded region. The riangular region on he lef has an area equal o half he base imes he heigh, which gies T (at)/2 = at 2 /2. And he parallelogram region has an area equal o he horizonal widh imes he heigh, which gies (at) = at. These erms agree wih Eq. (2.19). T at Figure 2.15
14 40 CHAPTER 2. KINEMATICS IN 1D 2.8. Raio of odd numbers This general resul doesn depend on he 1second alue of he ime ineral, so le s replace 1 second wih a general ime. The oal disances fallen afer imes of 0,, 2, 3, 4, ec., are 0, 1 2 g2, 1 2 g(2)2, 1 2 g(3)2, 1 2 g(4)2, ec. (2.20) The disances fallen during each ineral of ime are he differences beween he aboe disances, which yield 1 2 g2, g2, g2, g2, ec. (2.21) These are in he desired raio of 1 : 3 : 5 : Algebraically, he difference beween (n) 2 and ( (n + 1) )2 equals (2n + 1) 2, and he 2n + 1 facor here generaes he odd numbers. Geomerically, he s. plo is shown in Fig The area under he cure (a iled line in his case) is he disance raeled, and by looking a he number of (idenical) riangles in each ineral of ime, we quickly see ha he raio of he disances raeled in each ineral is 1 : 3 : 5 : Figure /g Figure = 0 g 1 = g 2.9. Dropped and hrown balls The posiions of he wo balls are gien by y 1 () = h 1 2 g2 and y 2 () = g2. (2.22) These are equal (ha is, he balls collide) when h = 0 = = h/ 0. The heigh of he collision is hen found from eiher of he y expressions o be y c = h gh 2 /20 2. This holds in any case, bu we are gien he furher informaion ha he second ball is insananeously a res when he collision occurs. Is speed is 0 g, so he collision mus occur a = 0 /g. Equaing his wih he aboe = h/ 0 resul ells us ha 0 mus be gien by 0 2 = gh. Plugging his ino y c = h gh 2 /20 2 gies y c = h/2. The wo elociies are gien by 1 () = g, and 2 () = 0 g. The difference of hese is 0. This holds for all ime, no jus a he momen when he balls collide. This is due o he fac ha boh balls are affeced by graiy in exacly he same way, so he iniial relaie speed (which is 0 ) equals he relaie speed a any oher ime. This is eiden from he s. plos in Fig The upper line is 0 aboe he lower line for all alues of Hiing a he same ime The ime i akes he firs ball o hi he ground is gien by g = h = 1 = 2h g. (2.23) The ime i akes he second ball o hi he ground is gien by 2 + g2 2 /2 = 2h. We could sole his quadraic equaion for 2 and hen se he resul equal o 1. Bu a much quicker sraegy is o noe ha since we wan 2 o equal 1, we can jus subsiue 1 for 2 in he quadraic equaion. This gies 2h g + g 2 ( ) 2h = 2h = g 2h gh g = h = = 2. (2.24) In he limi of small g, he process will ake a long ime, so i makes sense ha should be small. Noe ha wihou doing any calculaions, he consideraion of unis ells us ha he answer mus be proporional o gh.
15 2.5. PROBLEM SOLUTIONS 41 Remark: The inuiie inerpreaion of he aboe soluion is he following. If he second ball were dropped from res, i would be a heigh h when he firs ball his he ground a ime 2h/g (afer similarly falling a disance h). The second ball herefore needs o be gien an iniial downward speed ha causes i o rael an exra disance of h during his ime. Bu his is jus wha he middle equaion in Eq. (2.24) says Two dropped balls The oal ime i akes he firs ball o fall a heigh 4h is gien by g 2 /2 = 4h = = 2 2h/g. This ime may be diided ino he ime i akes o fall a disance d (which is 2d/g), plus he remaining ime i akes o hi he ground, which we are old is he same as he ime i akes he second ball o fall a heigh h (which is 2h/g). Therefore, 2 2h g = 2d g + 2h g = 2 h = d + h = d = h. (2.25) Remark: Graphically, he process is shown in he s. plo in Fig. 2.18(a). The area of he large riangle is he disance 4h he firs ball falls. The righ small riangle is he disance h he second ball falls, and he lef small riangle is he disance d he firs ball falls by he ime he second ball is released. If, on he oher hand, he second ball is released oo soon, afer he firs ball has raeled a disance d ha is less han h, hen we hae he siuaion shown in Fig. 2.18(b). The second ball raels a disance ha is larger han h (assuming i can fall ino a hole in he ground) by he ime he firs ball raels 4h and his he ground. In oher words (assuming here is no hole), he second ball his he ground firs. Conersely, if he second ball is released oo lae, hen i raels a disance ha is smaller han h by he ime he firs ball his he ground. This problem basically boils down o he fac ha freefall disances fallen are proporional o 2 (or equialenly, o he areas of riangles), so wice he ime means four imes he disance. (a) (b) 4h oal d = h 4h oal d < h Figure 2.18 h second ball released > h second ball released
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