Physics 18 Exam 1 wih Soluions Fall 1, Secions 51-54 Fill ou he informaion below bu o no open he exam unil insruce o o so! Name Signaure Suen ID E-mail Secion # ules of he exam: 1. You have he full class perio o complee he exam.. Formulae are provie on he las page. You may NOT use any oher formula shee. 3. When calculaing numerical values, be sure o keep rack of unis. 4. You may use his exam or come up fron for scrach paper. 5. Be sure o pu a box aroun your final answers an clearly inicae your work o your graer. 6. Clearly erase any unwane marks. No crei will be given if we can figure ou which answer you are choosing, or which answer you wan us o consier. 7. Parial crei can be given only if your work is clearly explaine an labele. 8. All work mus be shown o ge crei for he answer marke. If he answer marke oes no obviously follow from he shown work, even if he answer is correc, you will no ge crei for he answer. Pu your iniials here afer reaing he above insrucions: Table o be fille by graers only! Exam Par Toal Par 1 (15) Par () Par 3 () Par 4 () Bonus (5) Score 1
Par 1: Basic ieas of unis, conversions, an vecors. Problem 1.1: (1p) Wha sysem of unis is use in his course? Wha are he basic unis of mass, lengh, an ime of ha sysem? Inernaional Sysem (SI), Kilogram, meer, secons. Problem 1.: Joule, erg an ev are unis of energy efine as: 1 J (Joule) = 1 Kg m /s 1 erg (erg) = 1 gram cm /s 1 ev (elecron- Vol) = 1.6 1-1 erg Quesion 1..1: (p) Express 1 erg in unis of Joules. 1erg =1 gcm =1 Kg m s s 1 1 1 Kgm = 1 *1 *1 s =1 7 J Quesion 1..: (p) The LHC acceleraor in Swizerlan acceleraes proons o he worl s larges energy of 3.5x1 1 ev. Express ha energy in Joules. 3.5 1 1 ev = 3.5 1 1 1.6 1-1 erg = 3.5 1.6 erg = 3.5 1.6 1-7 J = 5.6 1-7 J Problem 1.3: The following plo shows he posiion x as a funcion of ime Quesion 1.3.1: (5p) For each ime range A,B,C I, fill he able below wriing in each cell wheher he velociy an acceleraion are <, >, or =. A B C D E F G H I egion Velociy Acceleraion A = = B > > C > = D > < x[cm] E < < F < = G < > H = = [s] I > > Quesion 1.3.: (p) Is he magniue of he velociy greaer in region C han i is in F? Why? The magniue of he velociy a a given ime is he magniue of he slope of he angen line in he above graph a ha given ime. The slope a ime range C is abou + squares/ squares, wih a magniue of +1. The slope a ime range F is abou -4 squares/ squares wih a magniue of -. Hence, he answer is NO; he magniue of he velociy a region C is smaller han ha a region F.
Par : A car epars from res uner a consan acceleraion of 1 m/s moving in a sraigh line. Afer ravelling some isance i passes firs a ancing club an secons laer a gas saion. The isance beween he ancing club an he gas saion is 4meers. Quesion.1.1: (6p) In he space below raw a schemaic iagram of he problem an wrie any associae imes. In aiion choose an raw a coorinae sysem an clearly inicae is origin. 4 m = C C+ s Origin Club Gas saion Quesion.1.: (4p) Wrie he equaions of moion of he acceleraing car accoring o your coorinae sysem. () = 1m s Quesion.1.3: (6p) Fin he ime i ook he car o ravel from he original poin of eparure o he club. (Hin: use he fac ha you know he isance an he ime beween he Club an he gas saion) 1m s ( c + s) 1m s C = 1m s 4s + 1m s C s = 4m C = 4m C = ms m =1s ( c + s) X( C ) = 4m m + m s Quesion.1.4: (4p) Fin he isance beween he ancing club an he original poin from where he car epare. ( c ) = 1m s C = 1m s 1s = 5m 3
Par 3: Acceleraion in boh componens. Problem 3.1: A car is fie wih a rocke propulsion engine ha provies he car wih a consan acceleraion in he horizonal irecion. As epice below he car mus jump of a 1m high brige an lan on a flabe ruck moving wih a consan velociy of m/s. A he momen he car leaves he brige he ruck is a a isance of 15m from he brige an he car has an iniial velociy of m/s. Ignore he heigh of he flabe, air resisance an any mass loss ue o he rocke. The following quesions mus be answere in he form of a number wih proper unis. Y 1m X 15m Quesion 3.1.1: (3p) Choose an raw your coorinae sysem on he figure above an associae imes o he ifferen evens. Quesion 3.1.: (5p) Wrie he posiion of he car an he ruck as a funcion of ime For he ruck : () =15m + m s Y T () = For he car : X C () = m s + a x Y C () =1m g Quesion 3.1.3: (5p) Fin he ime a which he car lans on he ruck. Y C ( L ) =1m g L = L = m g =1.43 s Quesion 3.1.4: (7p) Fin he minimum horizonal acceleraion ha he rocke propulsion engine in he car nees o give he car so i can successfully lan on he ruck. X C ( L ) = ( L ) replacing we ge m s L + a x L =15m + m s L a x = 3m L = g 3m m = 14.7 m s 4
Par 4: A more complex problem. Problem 4.1: A ball is ie up o a ro of raius connece o a moor ha makes i spin in he verical plane wih a uniform moion once every TA secons. A secon similar evice is locae a a isance an roaing in opposie irecion wih perio TB as shown in he picure below. Graviy is presen an he cener of boh evices is locae a isance h wih respec o he groun. All answers mus be expresse in erms of known parameers. TA Quesion 4.1.1: (p) Fin he raio of he spees of he balls in heir movemen aroun heir respecive circles. Quesion 4.1.: (p) Fin he raio of he magniue of he acceleraion of he balls in heir movemen aroun heir respecive circles. In general, wha is he irecion of he acceleraion? a A a B h Y = VA = v A v B = T B The acceleraion vecor of he balls poin owars he cener of heir respecive circles. X g VB = h Quesion 4.1.3: (7p) When boh balls are simulaneously a heir maximum heighs he balls break free of heir respecive ros an sar moving agains each oher. Fin he ime i akes he balls o collie assuming he heigh h is big enough. TB v A v B = " " T B = T B. X A ( c ) = X B ( c ) v A c = v B c c = = v A + v B π 1 + 1 T B T c = A T B π T B + ( ) Quesion 4.1.4: (4p) In your coorinae sysem fin he horizonal posiion a which boh balls collie. X A ( c ) = π c = π T B ( ) = T B π T B + ( + T B ) Quesion 4.1.5: (5p) Fin he minimum verical isance h necessary for he balls o collie in he air. The verical posiion where he paricles collie is given by YA(c) Since I pu my coorinae sysem a he cener of he circle which is from where h is measure i follows h=- YA(c) h = Y A ( c ) = + g C = + g T B π T B + ( ) 5
Formula shee: Vecors: A = A xˆ i + A yˆ j + Azˆ k A = A x + A y + A z an( ϕ) = A y,where ϕ = angle beween x ˆ axis an projecion of vecor A o he (ˆ x y ˆ ) plane. A x In - D, ϕ = angle beween x ˆ axis an vecor A. Mahemaical Formulae: a + b + c = = -b ± b 4ac a If x = a n x = na n -1. a If x = a n x( ) = n +1 n +1 n +1 ( 1 ) The following equaions are always rue: v = r a = v r ( ) = r + v ( ) = v + v x = x a x = v x v ( ) x a ( ) v x ( ) = x + v x ( ) v av -x = (x - x ) 1 ( - 1 ), a av -x = (v - v ) x 1x ( - 1 ) ( ) = v x + a x ( ) The following apply for consan acceleraion: r = r + v o + 1 v = v + a v x v x ( ) = v x + a x Δx ( ) Δx = v + v x x Oher Equaions: a ra = v = 4π T v P/A = v P/B + v B/A a x = x + v x + 1 a x ( ) = v x + a x 6