WEEK3 Recitation PHYS 131. of the projectile s velocity remains constant throughout the motion, since the acceleration a x


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1 WEEK3 Reciaion PHYS 131 Ch. 3: FOC 1, 3, 4, 6, 14. Problems 9, 37, 41 & 71 and Ch. 4: FOC 1, 3, 5, 8. Problems 3, 5 & 16. Feb 8, 018 Ch. 3: FOC 1, 3, 4, 6, (a) The horizonal componen of he projecile s velociy remains consan hroughou he moion, since he acceleraion a x in he horizonal direcion is zero (a x = 0 m/s ). The verical componen v y, however, changes as he projecile moves. This componen is greaes a poin 1, decreases o zero a poin a he op of he rajecory, and hen increases o a magniude less han ha a poin 1 as he projecile approaches poin (c) The acceleraion due o graviy is he same for boh balls, despie he fac ha hey have differen velociies. 4. (d) The acceleraion of a projecile is he same a all poins on he rajecory. I poins downward, oward he earh, and has a magniude of 9.80 m/s. 6. (c) The ime for a projecile o reach he ground depends only on he y componen (or verical componen) of is variables, i.e., y, v 0y, and a y. These variables are he same for boh balls. The fac ha Ball 1 is moving horizonally a he op of is rajecory does no play a role in he ime i akes for i o reach he ground. 14. (c) The velociy v PC of he passenger relaive o he car is given by v PC = v PB + v BC, according o he subscriping mehod discussed in Secion 3.4. However, he las erm on he righ of his equaion is given by v BC = v BG + v GC. So, v PC = v PB + v BG + v GC = + m/s + 16 m/s + ( 1 m/s) = +6 m/s.
2 Ch. 3: Problems 9, 37, 41 & REASONING The verical displacemen y of he ball depends on he ime ha i is in he air before being caugh. These variables depend on he ydirecion daa, as indicaed in he able, where he +y direcion is "up." ydirecion Daa y a y v y v 0y? 9.80 m/s 0 m/s? Since only wo variables in he y direcion are known, we canno deermine y a his poin. Therefore, we examine he daa in he x direcion, where +x is aken o be he direcion from he picher o he cacher. xdirecion Daa x a x v 0x m 0 m/s m/s? Since his able conains hree known variables, he ime can be evaluaed by using an equaion of kinemaics. Once he ime is known, i can hen be used wih he ydirecion daa, along wih he appropriae equaion of kinemaics, o find he verical displacemen y. SOLUTION Using he xdirecion daa, Equaion 3.5a can be employed o find he ime ha he baseball is in he air: Solving for gives ( ) x= v + a = v since a = 0 m/s x x x x x m = = = s v m/s 0x The displacemen in he y direcion can now be evaluaed by using he ydirecion daa able and he value of = s. Using Equaion 3.5b, we have y 1 1 = v + a = ( 0 m/s)( s) + ( 9.80 m/s )( s) = m 0 y y The disance ha he ball drops is given by he magniude of his resul, so Disance = m.
3 37. SSM REASONING a. The drawing shows he iniial velociy v 0 of he package when i is released. The iniial speed of he package is 97.5 m/s. The componen of is displacemen along he ground is labeled as x. The daa for he x direcion are indicaed in he daa able below. v y +y +x xdirecion Daa x a x v 0x? 0 m/s +(97.5 m/s) cos 50.0 = +6.7 m/s Since only wo variables are known, i is no possible o deermine x from he daa in his able. A value for a hird variable is needed. We know ha he ime of fligh is he same for boh he x and y moions, so le s now look a he daa in he y direcion. ydirecion Daa x y a y v y v 0y 73 m 9.80 m/s +(97.5 m/s) sin 50.0 = m/s? Noe ha he displacemen y of he package poins from is iniial posiion oward he ground, so is value is negaive, i.e., y = 73 m. The daa in his able, along wih he appropriae equaion of kinemaics, can be used o find he ime of fligh. This value for can, in urn, be used in conjuncion wih he xdirecion daa o deermine x. b. The drawing a he righ shows he velociy of he package jus before impac. The angle ha he velociy makes wih respec o he ground can be found from he inverse 1 angen funcion as θ = an ( v / v y x). Once he ime has been found in par (a), he values of v y and can be deermined from he daa in he ables and he appropriae equaions of kinemaics. v y +y θ v +x
4 SOLUTION a. To deermine he ime ha he package is in he air, we will use Equaion 3.5b 1 ( y v a 0 y y ) = + and he daa in he ydirecion daa able. Solving his quadraic equaion for he ime yields 1 ( ay ) ( )( ) 1 v ± v 4 a y 0y 0y y = ( 74.7 m/s) ( 74.7 m/s) 4( 1 )( 9.80 m/s )( 73 m) ± = = 6.78 s and.0 s 9.80 m/s 1 ( )( ) We discard he firs soluion, since i is a negaive value and, hence, unrealisic. The displacemen x can be found using =.0 s, he daa in he xdirecion daa able, and Equaion 3.5a: b. The angle θ ha he velociy of he package makes wih respec o he ground is given by 1 θ = an v / v. Since here is no acceleraion in he x direcion (a x = 0 m/s ), is he ( y x) same as v 0x, so ha = v 0x = +6.7 m/s. Equaion 3.3b can be employed wih he ydirecion daa o find v y : Therefore, ( )( ) v = v + a= m/s m/s.0 s = 141 m/s y 0 y y v 1 y m/s = an = an = 66.0 v m/s x θ where he minus sign indicaes ha he angle is 66.0 below he horizonal.
5 41. REASONING The speed v of he soccer ball jus before he goalie caches i is given by v = v + v, where and are he x and y componens of he final velociy of he ball. x y The daa for his problem are (he +x direcion is from he kicker o he goalie, and he +y direcion is he up direcion): xdirecion Daa x a x v 0x m 0 m/s? +(16.0 m/s) cos 8.0 = m/s ydirecion Daa y a y v y v 0y 9.80 m/s? +(16.0 m/s) sin 8.0 = m/s Since here is no acceleraion in he x direcion (a x = 0 m/s ), remains he same as v 0x, so = v 0x = m/s. The ime ha he soccer ball is in he air can be found from he xdirecion daa, since hree of he variables are known. Wih his value for he ime and he ydirecion daa, he y componen of he final velociy can be deermined. SOLUTION Since a x = 0 m/s, he ime can be calculaed from Equaion 3.5a as x m = = = 1.19 s. The value for v v m/s y can now be found by using Equaion 3.3b 0x wih his value of he ime and he ydirecion daa: ( )( ) v = v + a= m/s m/s 1.19 s = 4.15 m/s y 0 y y The speed of he ball jus as i reaches he goalie is x y ( ) ( ) v = v + v = m/s m/s = 14.7 m/s
6 71. SSM REASONING Once he diver is airborne, he moves in he x direcion wih consan velociy while his moion in he y direcion is acceleraed (a he acceleraion due o graviy). Therefore, he magniude of he x componen of his velociy remains consan a 1.0 m/s for all imes. The magniude of he y componen of he diver's velociy afer he has fallen hrough a verical displacemen y can be deermined from Equaion 3.6b: v = v + a y. y 0 y y Since he diver runs off he plaform horizonally, v 0 y = 0 m/s. Once he x and y componens of he velociy are known for a paricular verical displacemen y, he speed of he diver can be obained from v = v + v. x y SOLUTION For convenience, we will ake downward as he posiive y direcion. Afer he diver has fallen 10.0 m, he y componen of his velociy is, from Equaion 3.6b, v = v + a y = 0 + (9.80 m / s )( m) = m / s y 0 y y Therefore, v = v + v = (1.0 m / s) + ( 14.0 m / s) = 14.1 m / s x y Ch. 4: FOC 1, 3, 5, (b) If only one force acs on he objec, i is he ne force; hus, he ne force mus be nonzero. Consequenly, he velociy would change, according o Newon s firs law, and could no be consan. 3. (e) Newon s firs law saes ha an objec coninues in a sae of res or in a sae of moion a a consan speed along a sraigh line, unless compelled o change ha sae by a ne force. All hree saemens are consisen wih he firs law. 5. (c) Newon s second law gives he answer direcly, provided he ne force is calculaed by vecor addiion of he wo given forces. The direcion of he ne force gives he direcion of he acceleraion. 8. (b) Newon s hird law indicaes ha Paul and Tom apply forces of equal magniude o each oher. According o Newon s second law, he magniude of each of hese forces is he mass imes he magniude of he acceleraion. Thus, we have m Paul a Paul = m Tom a Tom, or m Paul /m Tom = a Tom /a Paul.
7 Ch. 4: Problems 3, 5 & REASONING In each case, we will apply Newon s second law. Remember ha i is he ne force ha appears in he second law. The ne force is he vecor sum of boh forces. SOLUTION a. We will use Newon s second law, ΣF x, o find he force F. Taking he posiive x direcion o be o he righ, we have so F F 1 F = (3.0 kg)(+5.0 m/s ) (+9.0 N) = + 6N b. Applying Newon s second law again gives F F 1 = (3.0 kg)( 5.0 m/s ) (+9.0 N) = 4 N c. An applicaion of Newon s second law gives F F 1 = (3.0 kg)(0 m/s ) (+9.0 N) = 9.0 N 5. SSM REASONING The magniude ΣF of he ne force acing on he kayak is given by Newon s second law as Σ F = ma (Equaion 4.1), where m is he combined mass of he person and kayak, and a is heir acceleraion. Since he iniial and final velociies, v 0 and v, and he displacemen x are known, we can employ one of he equaions of kinemaics from Chaper o find he acceleraion. SOLUTION Solving Equaion.9 ( v v ax) = 0 + from he equaions of kinemaics for he acceleraion, we have v v0 a = x Subsiuing his resul ino Newon s second law gives ( ) ( ) ( ) v v m/s 0 m/s Σ F = ma = m = 73 kg = 3 N x ( 0.41 m)
8 16. REASONING Since here is only one force acing on he man in he horizonal direcion, i is he ne force. According o Newon s second law, Equaion 4.1, he man mus accelerae under he acion of his force. The facors ha deermine his acceleraion are (1) he magniude and () he direcion of he force exered on he man, and (3) he mass of he man. When he woman exers a force on he man, he man exers a force of equal magniude, bu opposie direcion, on he woman (Newon s hird law). I is he only force acing on he woman in he horizonal direcion, so, as is he case wih he man, she mus accelerae. The facors ha deermine her acceleraion are (1) he magniude and () he direcion of he force exered on her, and (3) he her mass. SOLUTION a. The acceleraion of he man is, according o Equaion 4.1, equal o he ne force acing on him divided by his mass. a man ΣF 45 N = = = m 8 kg 0.55 m / s (due eas) b. The acceleraion of he woman is equal o he ne force acing on her divided by her mass. a woman ΣF 45 N = = = m 48 kg 0.94 m / s (due wes)
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