Physics 2A HW #3 Solutions


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1 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 33 (c) The ccelerion due o grvi is he sme for boh blls, despie he fc h he hve differen velociies. Focus On Conceps 34 (d) The ccelerion of projecile is he sme ll poins on he rjecor. I poins downwrd, owrd he erh, nd hs mgniude of 9.8 m/s. Focus On Conceps 36 (c) The ime for projecile o rech he ground depends onl on he componen (or vericl componen) of is vribles, i.e.,, v, nd. These vribles re he sme for boh blls. The fc h Bll 1 is moving horizonll he op of is rjecor does no pl role in he ime i kes for i o rech he ground. Focus On Conceps 39 () The ime projecile is in he ir is equl o wice he ime i kes o fll from is mimum heigh. Projecile 1 reches he greer heigh, so i spends he greer moun of ime in he ir Problem 39 REASONING. We designe he direcion down nd prllel o he rmp s he + direcion, nd he ble shows he vribles h re known. Since hree of he five kinemic vribles hve vlues, one of he equions of kinemics cn be emploed o find he ccelerion. Direcion D v v +1. m? +7.7 m/s m/s b. The ccelerion vecor poins down nd prllel o he rmp, nd he ngle of he rmp is 5. relive o he ground (see he drwing). Therefore, rigonomer cn be used o deermine he componen prllel of he ccelerion h is prllel o he ground. 5. prllel
2 SOLUTION. Equion 3.6 ( v v ) = + cn be used o find he ccelerion in erms of he hree known vribles. Solving his equion for gives ( + ) ( ) ( + ) v v 7.7 m/s m/s = = =.47 m/s 1. m b. The drwing shows h he ccelerion vecor is oriened 5. relive o he ground. The componen prllel of he ccelerion h is prllel o he ground is prllel ( ) = cos 5. =.47 m/s cos 5. =.4 m/s Problem 39 REASONING The vericl displcemen of he bll depends on he ime h i is in he ir before being cugh. These vribles depend on he direcion d, s indiced in he ble, where he + direcion is "up." Direcion D v v? 9.8 m/s m/s? Since onl wo vribles in he direcion re known, we cnno deermine his poin. Therefore, we emine he d in he direcion, where + is ken o be he direcion from he picher o he ccher. Direcion D v v +17. m m/s +41. m/s? Since his ble conins hree known vribles, he ime cn be evlued b using n equion of kinemics. Once he ime is known, i cn hen be used wih he direcion d, long wih he pproprie equion of kinemics, o find he vericl displcemen. SOLUTION Using he direcion d, Equion 3.5 cn be emploed o find he ime h he bsebll is in he ir: Solving for gives ( ) = v + = v since = m/s m = = =.415 s v +41. m/s
3 The displcemen in he direcion cn now be evlued b using he direcion d ble nd he vlue of =.415 s. Using Equion 3.5b, we hve ( )( ) ( )( ) = v + = + = 1 1 m/s.415 s 9.8 m/s.415 s.844 m The disnce h he bll drops is given b he mgniude of his resul, so Disnce =.844 m. Problem 33 REASONING Since we know he lunch ngle θ = 15., he lunch speed v cn be obined using rigonomer, which gives he componen of he lunch veloci s v = v sin θ. Solving his equion for v requires vlue for v, which we cn obin from he vericl heigh of = 13.5 m b using Equion 3.6b from he equions of kinemics. SOLUTION From Equion 3.6b we hve Using rigonomer, we find Problem 341 REASONING The speed v of he soccer bll jus before he golie cches i is given b v v v = +, where v nd v re he nd componens of he finl veloci of he bll. The d for his problem re (he + direcion is from he kicker o he golie, nd he + direcion is he up direcion): Direcion D v v m m/s? +(16. m/s) cos 8. = m/s Direcion D v v 9.8 m/s? +(16. m/s) sin 8. = m/s
4 Since here is no ccelerion in he direcion ( = m/s ), v remins he sme s v, so v = v = m/s. The ime h he soccer bll is in he ir cn be found from he direcion d, since hree of he vribles re known. Wih his vlue for he ime nd he direcion d, he componen of he finl veloci cn be deermined. SOLUTION Since = m/s, he ime cn be clculed from Equion 3.5 s m = = = 1.19 s. The vlue for v v m/s cn now be found b using Equion 3.3b wih his vlue of he ime nd he direcion d: ( )( ) v = v + = m/s m/s 1.19 s = 4.15 m/s The speed of he bll jus s i reches he golie is ( ) ( ) v = v + v = m/s m/s = 14.7 m/s Problem 366 REASONING The mgniude nd direcion of he iniil veloci v cn be obined using he Phgoren heorem nd rigonomer, once he nd componens of he iniil veloci v nd v 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 veloci componens: Using he Phgoren heorem nd rigonomer, we find
5 Problem 37 REASONING When he bll is hrown srigh up wih n iniil speed v, he mimum heigh h i reches cn be found b using wih he relion v = v + (Equion 3.6b). Since he bll is hrown srigh up, v = v, where v is he iniil speed of he bll. Also, he speed of he bll is momenril zero is mimum heigh, so v = m/s h poin. The ccelerion is h due o grvi, so he onl unknown besides is he iniil speed v of he bll. To deermine v we will emplo Equion 3.6b second ime, bu now i will be pplied o he cse where he bll is hrown upwrd n ngle of 5º bove he horizonl. In his cse he mimum heigh reched b he bll is 1 = 7.5 m, he iniil speed in he direcion is v = v sin 5º, nd he componen of he speed he mimum heigh is v = m/s. SOLUTION We will sr wih he relion v = v + (Equion 3.6b) o find he mimum heigh h he bll ins when i is hrown srigh up. Solving his equion for, nd subsiuing in v = v nd v = m/s gives v = (1) To deermine v, we now ppl he equion v = v + o he siuion where he bll is hrown upwrd n ngle of 5º relive o he horizonl. In his cse we noe h v = v sin 5º, v = m/s, nd = 1 (he mimum heigh of 7.5 m reched b he bll). Solving for v, we find ( ) 1 sin 5 1 or v = v + v = sin 5 Subsiuing his epression for v ino Equion (1) gives 1 v sin m = = = = = sin 5 sin 5 1 m Problem 375 SSM REASONING The horizonl disnce covered b sone 1 is equl o he disnce covered b sone fer i psses poin P in he following digrm. Thus, he disnce beween he poins where he sones srike he ground is equl o, he horizonl disnce covered b sone when i reches P. In he digrm, we ssume up nd o he righ re posiive.
6 θ P θ θ 1 SOLUTION If P is he ime required for sone o rech P, hen = v = ( v cos θ ) P P For he vericl moion of sone, v = v sin θ +. Solving for gives v = v sin θ When sone reches P, Then, v = v sinθ, so he ime required o rech P is P v sinθ = v sinθ vp ( vcos θ ) = = Problem 377 REASONING Using he d given in he problem, we cn find he mimum fligh ime 1 of he bll using Equion 3.5b ( = v + ). Once he fligh ime is known, we cn use he definiion of verge veloci o find he minimum speed required o cover he disnce in h ime. SOLUTION Equion 3.5b is qudric in nd cn be solved for using he qudric formul. According o Equion 3.5b, he mimum fligh ime is (wih upwrd ken s he posiive direcion)
7 1 ( ) ( ) 1 ( ) ± ± + v v 4 v v = = ( ) ( ) 15. m/s sin 5. ± 15. m/s sin 5. +( 9.8 m/s ) (.1 m) = 9.8 m/s =. s nd.145 s where he firs roo corresponds o he ime required for he bll o rech vericl displcemen of = +.1 m s i rvels upwrd, nd he second roo corresponds o he ime required for he bll o hve vericl displcemen of = +.1 m s he bll rvels upwrd nd hen downwrd. The desired fligh ime is.145 s. During he.145 s, he horizonl disnce rveled b he bll is [ ] = v= ( v cos θ ) = (15. m/s) cos 5. (.145 s) =.68 m Thus, he opponen mus move.68 m 1. m = 1.68 m in.145 s.3 s = s. The opponen mus, herefore, move wih minimum verge speed of v min 1.68 m = = s 5.79 m / s
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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