A 1.3 m 2.5 m 2.8 m. x = m m = 8400 m. y = 4900 m 3200 m = 1700 m

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1 PHYS : Soluions o Chper 3 Home Work. SSM REASONING The displcemen is ecor drwn from he iniil posiion o he finl posiion. The mgniude of he displcemen is he shores disnce beween he posiions. Noe h i is onl he iniil nd finl posiions h deermine he displcemen. The fc h he squirrel jumps o n inermedie posiion before reching his finl posiion is no imporn. The rees re perfecl srigh nd boh growing perpendiculr o he fl horizonl ground beneh hem. Thus, he disnce beween he rees nd he lengh of he runk of he second ree below he squirrel s finl lnding spo form he wo perpendiculr sides of righ ringle, s he drwing shows. To his ringle, we cn ppl he Phgoren heorem nd deermine he mgniude A of he displcemen ecor A. SOLUTION According o he Phgoren heorem, we he A.3 m.5 m.8 m A.3 m.5 m 6. REASONING AND SOLUTION The horizonl displcemen is x = 9 6 m m = 84 m The ericl displcemen is = 49 m 3 m = 7 m The mgniude of he displcemen is herefore, r x 84 m 7 m 86 m. REASONING AND SOLUTION The ericl moion consiss of he bll rising for ime sopping nd reurning o he ground in noher ime. For he upwrd porion Noe: = m/s since he bll sops he op. Now g g V = sin = (5. m/s) sin 6. =.7 m/s = (.7 m/s)/(9.8 m/s ) =. s The required "hng ime" is = 4.4 s. 5. REASONING Trigonomer indices h he x nd componens of he dolphin s eloci re reled o he lunch ngle ccording o n = / x. SOLUTION Using rigonomer, we find h he componen of he dolphin s eloci is n n m/ s n m/ s x x b g

2 5. REASONING The mgniude nd direcion of he iniil eloci cn be obined using he Phgoren heorem nd rigonomer, once he x nd componens of he iniil eloci x nd re known. These componens cn be clculed using Equions 3.3 nd 3.3b. SOLUTION Using Equions 3.3 nd 3.3b, we obin he following resuls for he eloci componens: 3775 m / s 5. m / s 565 s m / s x x x 486 m / s 7.3 m / s 565 s m / s Using he Phgoren heorem nd rigonomer, we find c hb g c hb g m / s m / s 3 m / s x n F HG x I b g b g m / s 37 7 KJ F H G I n K J.. m / s 7. SSM REASONING AND SOLUTION The wer exhibis projecile moion. The x componen of he moion hs zero ccelerion while he componen is subjec o he ccelerion due o gri. In order o rech he highes possible fire, he displcemen of he hose from he building is x, where, ccording o Equion 3.5 (wih x = m/s ), x ( cos ) x wih equl o he ime required for he wer he rech is mximum ericl displcemen. The ime cn be found b considering he ericl moion. From Equion 3.3b, When he wer hs reched is mximum ericl displcemen, = m/s. Tking up nd o he righ s he posiie direcions, we find h nd Therefore, we he sin sin x ( cos ) cos sin (5. m/s) cos 35. sin 35. x 9.8 m/s 3. m 3. REASONING The d for he problem re summrized below. In he bles, we use he smbol o denoe he speed wih which he bll is hrown nd choose upwrd nd o he righ s he posiie direcions. x-direcion D 3. x x x x 83 m m/s cos 3.? m x PHYS : Soluions o Chper 3 Home Work pg. /6

3 -Direcion D m 9.8 m/s sin 3. Sme s for x direcion Noe h x = m/s, becuse ir resisnce is being ignored. In ddiion, noe h = m, becuse he foobll rises nd hen reurns o he sme ericl leel from which i ws lunched. Finll, we he used rigonomer o express he componens x nd of he iniil eloci in erms of he speed nd he 3. lunch ngle. The ke here is o remember h he horizonl nd ericl prs of he foobll s moion cn be reed seprel, he ime for he moion being he sme for ech. Since ir resisnce is being ignored, we cn ppl he equions of kinemics seprel o he moions in he x nd direcions. SOLUTION Since here is no ccelerion in he horizonl direcion, moion in h direcion is consn-eloci moion, nd he horizonl displcemen x is simpl he iniil eloci componen x imes he ime : x x cos3. An expression for cn be obined b considering he moion in he ericl direcion. Thus, we use Equion 3.5b from he equions of kinemics nd recognize h he displcemen is zero nd = sin 3.: sin 3. or m sin3. or Subsiuing his resul for he ime ino he expression for x gies sin 3. cos 3. sin 3. x cos 3. cos m9.8 m/s x cos3.sin 3. cos3.sin m/s 37. SSM REASONING. The drwing shows he iniil eloci of he pckge when i is relesed. The iniil speed of he pckge is 97.5 m/s. The componen of is displcemen long he ground is lbeled s x. The d for he x direcion re indiced in he d ble below x x x-direcion D x x x x? m/s +(97.5 m/s) cos 5. = +6.7 m/s PHYS : Soluions o Chper 3 Home Work pg. 3/6

4 Since onl wo ribles re known, i is no possible o deermine x from he d in his ble. A lue for hird rible is needed. We know h he ime of fligh is he sme for boh he x nd moions, so le s now look he d in he direcion. -Direcion D 73 m 9.8 m/s +(97.5 m/s) sin 5. = m/s? Noe h he displcemen of he pckge poins from is iniil posiion owrd he ground, so is lue is negie, i.e., = 73 m. The d in his ble, long wih he pproprie equion of kinemics, cn be used o find he ime of fligh. This lue for cn, in urn, be used in conjuncion wih he x-direcion d o deermine x. b. The drwing he righ shows he eloci of he pckge jus before impc. The ngle h he eloci mkes wih respec o he ground cn be found from he inerse ngen funcion s n / x. Once he ime hs been found in pr (), he lues of nd x cn be deermined from he d in he bles nd he pproprie equions of kinemics. SOLUTION. To deermine he ime h he pckge is in he ir, we will use Equion 3.5b nd he d in he -direcion d ble. Soling his qudric equion for he ime ields 4 + x +x 74.7 m/s 74.7 m/s m/s 73 m 9.8 m/s 6.78 s nd. s We discrd he firs soluion, since i is negie lue nd, hence, unrelisic. The displcemen x cn be found using =. s, he d in he x-direcion d ble, nd Equion 3.5: x x 6.7 m/s. s m/s. s 38 m x PHYS : Soluions o Chper 3 Home Work pg. 4/6

5 b. The ngle h he eloci of he pckge mkes wih respec o he ground is gien b n / x. Since here is no ccelerion in he x direcion ( x = m/s ), x is he sme s x, so h x = x = +6.7 m/s. Equion 3.3b cn be emploed wih he -direcion d o find : 74.7 m/s m/s. s 4 m/s Therefore, 4 m/s n n m/s x where he minus sign indices h he ngle is 66. below he horizonl. 43. SSM WWW REASONING The horizonl disnce coered b sone is equl o he disnce coered b sone fer i psses poin P in he following digrm. Thus, he disnce x beween he poins where he sones srike he ground is equl o x, he horizonl disnce coered b sone when i reches P. In he digrm, we ssume up nd o he righ re posiie. SOLUTION If P is he ime required for sone o rech P, hen x x ( cos ) x P P For he ericl moion of sone, for gies sin. Soling P sin When sone reches P, o rech P is Then, sin, so he ime required P sin sin x xp ( cos ) x x 9.8 m/s sin cos (3. m/s) sin 3. cos m PHYS : Soluions o Chper 3 Home Work pg. 5/6

6 49. SSM REASONING Since he horizonl moion is no ccelered, we know h he x componen of he eloci remins consn 34 m/s. Thus, we cn use Equion 3.5 (wih x m/s ) o deermine he ime h he bulle spends in he building before i is embedded in he wll. Since we know he ericl displcemen of he bulle fer i eners he building, we cn use he fligh ime in he building nd Equion 3.5b o find he componen of he eloci of he bulle s i eners he window. Then, Equion 3.6b cn be used (wih m/s) o deermine he ericl displcemen of he bulle s i psses beween he buildings. We cn deermine he disnce H b dding he mgniude of o he ericl disnce of.5 m wihin he building. Once we know he ericl displcemen of he bulle s i psses beween he buildings, we cn deermine he ime required for he bulle o rech he window using Equion 3.4b. Since he moion in he x direcion is no ccelered, he disnce D cn hen be found from D x. SOLUTION Assuming h he direcion o he righ is posiie, we find h he ime h he bulle spends in he building is (ccording o Equion 3.5) x 6.9 m.3 s 34 m/s x The ericl displcemen of he bulle fer i eners he building is, king down s he negie direcion, equl o.5 m. Therefore, he ericl componen of he eloci of he bulle s i psses hrough he window is, from Equion 3.5b,.5 m (window).3 s ( 9.8 m/s )(.3 s) = 4.5 m/s The ericl displcemen of he bulle s i rels beween he buildings is (ccording o Equion 3.6b wih m/s) Therefore, he disnce H is ( 4.5 m/s) 3.6 m ( 9.8 m/s ) H 3.6 m.5 m 3 m The ime for he bulle o rech he window, ccording o Equion 3.4b, is Hence, he disnce D is gien b ( 3.6 m).5 s ( 4.5 m/s) D x (34 m/s)(.5 s) 85 m PHYS : Soluions o Chper 3 Home Work pg. 6/6

Physics 2A HW #3 Solutions

Physics 2A HW #3 Solutions Chper 3 Focus on Conceps: 3, 4, 6, 9 Problems: 9, 9, 3, 41, 66, 7, 75, 77 Phsics A HW #3 Soluions Focus On Conceps 3-3 (c) The ccelerion due o grvi is he sme for boh blls, despie he fc h he hve differen