{"id":1395,"date":"2015-09-29T20:28:03","date_gmt":"2015-09-29T20:28:03","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1395"},"modified":"2024-08-23T12:57:39","modified_gmt":"2024-08-23T12:57:39","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-conservacao-da-quantidade-de-movimento","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/dinamica\/conservacao-da-quantidade-de-movimento\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-conservacao-da-quantidade-de-movimento\/","title":{"rendered":"Conserva\u00e7\u00e3o da Quantidade de Movimento – Resolu\u00e7\u00e3o"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre<\/b><\/span><\/span><\/span><\/p>\n

Conserva\u00e7\u00e3o da Quantidade de Movimento<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

\u00a0<\/span>01-<\/b><\/span><\/span><\/span>\u00a0R- C\u00a0 (veja teoria)<\/b><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=mV=(250 ).10\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=2.500kg.m\/s<\/b><\/span><\/span>\u00a0 —\u00a0 Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0+ m<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>=50.(-10) + 200.(+V<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>= – 500 + 200.V<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 2.500= – 500 +200.V<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 V<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>=15m\/s R- E<\/b><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0(01) verdadeira (veja teoria)\u00a0 —\u00a0 (02) falsa \u2013 sua massa \u00e9 diferente da massa do tanque \u00a0—\u00a0 (04) verdadeira –\u00a0 (ambos estavam parados)\u00a0 —\u00a0 (08) falsa \u2013 ambos possuem velocidades de sentidos contr\u00e1rios\u00a0 —\u00a0 (16) antes do choque do tanque com o astronauta B \u2013 Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>t<\/b><\/span><\/span><\/sub>\u00a0+ m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 10.5 + 90X0\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=50kg.m\/s\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 ap\u00f3s o choque, astronauta + tanque (m=90 + 10=100kg) tem velocidade comum V\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=100.V\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 50=100V\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V=0,5m\/s<\/b><\/span><\/span>(verdadeira)<\/b><\/span><\/span><\/p>\n

(01+04+08+16)=29<\/b><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0antes\u00a0 —\u00a0\u00a0 Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u00a0+ m<\/b><\/span><\/span>SH.<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>SH<\/b><\/span><\/span><\/sub>\u00a0=10.000X100 + 50.(-V<\/b><\/span><\/span>SH<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>\u00a0\u2013 50V<\/b><\/span><\/span>HS<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 depois\u00a0 —\u00a0 Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>SH.<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>SH<\/b><\/span><\/span><\/sub>\u00a0+ m<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=0 + 0=0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub>\u00a0—\u00a0 Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 100.10<\/b><\/span><\/span>4\u00a0<\/b><\/span><\/span><\/sup>– 50.V<\/b><\/span><\/span>SH<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>SH<\/b><\/span><\/span><\/sub>=2.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>km\/h R- A<\/b><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0Sendo a velocidade uma grandeza vetorial, vamos orient\u00e1-la como, por exemplo, para a direita positiva e para a esquerda, negativa e vamos supor que, ap\u00f3s o choque, ambos os fragmentos se movam para a direita\u00a0 —\u00a0 Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u00a0(antes da explos\u00e3o)\u00a0 —\u00a0 Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=mV=m.10\u00a0 —\u00a0<\/b><\/span><\/span>Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=10m\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>\u00a0(depois da explos\u00e3o)\u00a0 —\u00a0 Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=m\/2V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ m\/2V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span>Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=(mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ mV<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>)\/2\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u00a0= Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 10m=(mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0+ mV<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>)\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>+ V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=20\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>com a trajet\u00f3ria orientada para a direita, a soma alg\u00e9brica das duas velocidades deve ser 20.(01) +15 -5=10m\/s (falsa)\u00a0 —\u00a0 (02) +20 + 0=20m\/s (verdadeira)\u00a0 —\u00a0 (04) +30 -10=20m\/s (verdadeira)\u00a0 —\u00a0 (08) +25 + 0=25m\/s (falsa)\u00a0 —\u00a0 (16) +25 \u2013 5=20m\/s (verdadeira)\u00a0 —\u00a0 (32) +10 + 0=10m\/s (falsa)\u00a0 —\u00a0 (64) +50 \u2013 30=20m\/s (verdadeira)\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

Soma (02 + 04 + 16 + 64) = 86<\/b><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=0<\/b><\/span><\/span>\u00a0 —\u00a0 Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>= mv +M(-V)=60.\u2206s\/\u2206t \u2013 80.\u2206S\/\u2206t= 60.\u2206s\/\u2206t \u2013 80.12\/\u2206t\u00a0\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=\u00a0 60.\u2206s\/\u2206t \u2013 960\/\u2206t<\/b><\/span><\/span>\u00a0\u00a0 —\u00a0 Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 0=60.\u2206s\/\u2206t \u2013 960\/\u2206t\u00a0 —\u00a0\u00a0\u2206t \u00e9 o mesmo e se cancela \u00a0\u00a0—\u00a0 60.\u2206s=960\u00a0 —\u00a0\u00a0\u2206s=16m\u00a0 —\u00a0 a dist\u00e2ncia entre eles \u00e9 d=12m + 16m\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 d=28m<\/b><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span> Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=0<\/b><\/span><\/span>\u00a0 —\u00a0 Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=(45 + 1).(-V<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>) + (m + 1).V<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>= – 46V\u00ad<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>\u00a0+ (m + 1)V<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>= – 46.10\/\u2206t + (m + 1).30\/\u2206t\u00a0 —\u00a0 Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=( – 460 +<\/b><\/span><\/span><\/p>\n

 <\/p>\n

\"\"\"\"<\/p>\n

30m + 30)\/\u00a0\u2206t\u00a0 —Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 0==( – 460 + 30m + 30)\/\u00a0\u2206t\u00a0 —\u00a0 30m \u2013 430=0\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>m=14,33kg R- D<\/b><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0a) Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=M<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ M<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=234.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>.1,0.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 4.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=234.10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\u00a0\u2013 4.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 0=234.10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\u00a0\u2013 4.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=5,85.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>m\/s<\/b><\/span><\/span><\/p>\n

b) a varia\u00e7\u00e3o da energia cin\u00e9tica da part\u00edcula fornece o trabalho necess\u00e1rio para par\u00e1-la\u00a0 —\u00a0 E<\/b><\/span><\/span>ca<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 E<\/b><\/span><\/span>cd<\/b><\/span><\/span><\/sub>=234.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>(1,0.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 +4.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>.(5,85.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 — W=E<\/b><\/span><\/span>cd<\/b><\/span><\/span><\/sub>\u00a0\u2013 E<\/b><\/span><\/span>ca<\/b><\/span><\/span><\/sub>= – 6,8.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>J<\/b><\/span><\/span>\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

c) W=F.d.cos0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 6,8.10<\/b><\/span><\/span>4<\/b><\/span><\/span><\/sup>=F.10\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 F=6,8.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>N<\/b><\/span><\/span><\/p>\n

09<\/b><\/span><\/span><\/span>–<\/b><\/span><\/span>\u00a0a) c\u00e1lculo da componente vertical (V<\/b><\/span><\/span>oy<\/b><\/span><\/span><\/sub>) da velocidade inicial V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\u00a0—\u00a0 na altura m\u00e1xima (h=45m) \u2013 V<\/b><\/span><\/span>y<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 V\u00ady<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= V<\/b><\/span><\/span>oy<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.(-g).h\u00a0 —\u00a0 0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=V<\/b><\/span><\/span>oy<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 2.10.45\u00a0 —\u00a0 V<\/b><\/span><\/span>oy<\/b><\/span><\/span><\/sub>=30m\/s\u00a0 —\u00a0 tempo para atingir a altura m\u00e1xima\u00a0 —\u00a0 V<\/b><\/span><\/span>y<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>oy<\/b><\/span><\/span><\/sub>\u00a0\u2013gt\u00a0 —\u00a0 0=30 \u2013 10.t\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 t=T<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=3s<\/b><\/span><\/span><\/p>\n

b) para atingir a altura m\u00e1xima ele percorreu uma dist\u00e2ncia horizontal x=60m em t=3s\u00a0 —\u00a0 x=x<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ V<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>.t\u00a0 —\u00a0 60=0 + V<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>.3\u00a0 —\u00a0 <\/b><\/span><\/span>V<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>=20m\/s<\/b><\/span><\/span><\/p>\n

c) no ponto mais alto da trajet\u00f3ria o roj\u00e3o, antes de explodir, tem apenas velocidade horizontal V<\/b><\/span><\/span>x<\/b><\/span><\/span><\/sub>=20m\/s e sua quantidade de movimento vale Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=mV=0,5.20\u00a0 —\u00a0 Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=10kg.m\/s (horizontal e para a direita)\u00a0 —\u00a0 ap\u00f3s a explos\u00e3o, as componentes horizontal e vertical do fragmento A s\u00e3o nulas, sobrando apenas a componente horizontal do fragmento B, que tamb\u00e9m deve ser horizontal e para a direita, pois, Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=0,25.V\u00a0 —\u00a0 10=0,25.V\u00a0 —\u00a0 V=40m\/s (componente horizontal da velocidade do fragmento B)\u00a0 —\u00a0 a energia fornecida pelo combust\u00edvel corresponde \u00e0 varia\u00e7\u00e3o de energia cin\u00e9tica antes e depois da explos\u00e3o, tomando este ponto como n\u00edvel zero de altura (E<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=0)\u00a0\u00a0 —\u00a0 E<\/b><\/span><\/span>ca<\/b><\/span><\/span><\/sub>=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 =0,5.400\/2\u00a0\u00a0 —\u00a0 E<\/b><\/span><\/span>ca<\/b><\/span><\/span><\/sub>=100J —\u00a0 E<\/b><\/span><\/span>cd<\/b><\/span><\/span><\/sub>=0,25.1600\/2=200J\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>DE<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=200 \u2013 100=100J<\/b><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0Sendo a massa do maior o dobro da massa do menor, a velocidade do menor (V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>) dever\u00e1 ser o dobro da velocidade do maior (V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>), pelo princ\u00edpio da conserva\u00e7\u00e3o da quantidade de movimento\u00a0 —\u00a0 V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=2V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0\u00a0\u2206S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/\u2206t=2\u2206S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/\u2206t\u00a0 —\u00a0 sendo\u00a0\u2206t o mesmo\u00a0 —\u00a0\u2206S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=2\u2206S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\u00a0<\/b><\/span><\/span>—\u00a0\u00a0\u2206S<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/\u2206S<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=2 R- C<\/b><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span>\u00a0O pescador deve correr para B, a fim de que a for\u00e7a que seus p\u00e9s exercem sobre a jangada a<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

acelere no sentido B para A, fazendo com que ela se afaste da trajet\u00f3ria do torpedo.<\/b><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span>\u00a0a)<\/b><\/span><\/span><\/span> Pela 3.a lei de Newton, ou princ\u00edpio fundamental da a\u00e7\u00e3o e rea\u00e7\u00e3o o barco ir\u00e1 para tr\u00e1s.<\/b><\/span><\/span><\/p>\n

b) F.\u2206t=m\u2206V\u00a0 —\u00a0 250.2=(180 + 70).(V \u2013 0)\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 V=2,0m\/s<\/b><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>(<\/b><\/span><\/span>01) Falsa \u2013 massas diferentes, velocidades diferentes.<\/b><\/span><\/span><\/p>\n

02-Verdadeira \u2013 verdadeira, pois a massa de A \u00e9 o dobro da massa de B<\/b><\/span><\/span><\/p>\n

(04) Falsa<\/b><\/span><\/span><\/p>\n

(08) Correta \u2013 a quantidade de movimento do sistema antes de puxar a corda \u00e9 igual \u00e0 quantidade de movimento do sistema ap\u00f3s o pux\u00e3o.<\/b><\/span><\/span><\/p>\n

(16) Falsa \u2013 vide teoria<\/b><\/span><\/span><\/p>\n

Soma \u2013 (02 + 08) = 10<\/b><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0O sistema dever\u00e1 conservar a quantidade de movimento horizontal inicial. Desta forma como a velocidade foi reduzida \u00e0 metade, a massa do sistema dever\u00e1 dobrar, passando de 30 g para 60 g. A diferen\u00e7a de 30 g corresponde a massa da bolinha de isopor.\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span>\u00a0a) W=F.d.cosa=25.4.cos0<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>W=100J<\/b><\/span><\/span><\/p>\n

b) velocidade com que a poltrona chega em B\u00a0 —\u00a0 F=ma\u00a0 —\u00a0 25=10.a\u00a0 —\u00a0 a=2,5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0 V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.a.DS\u00a0 —\u00a0 V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>=0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.2,5.4\u00a0 —\u00a0 V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=\u221a20m\/s\u00a0 —\u00a0 velocidade com que a poltrona chega \u00e0 Maria (V<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>) \u2013 conserva\u00e7\u00e3o da energia mec\u00e2nica \u2013 E<\/b><\/span><\/span>mB\u00a0<\/b><\/span><\/span><\/sub>=m.V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\/2 + mgh=10.(\u221a20)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 + 10.10.0,8\u00a0 —\u00a0 E<\/b><\/span><\/span>mB =<\/b><\/span><\/span><\/sub>180J\u00a0 —\u00a0\u00a0 E<\/b><\/span><\/span>mM<\/b><\/span><\/span><\/sub>=m.V<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\/2 + m.g.h=10.V<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\/2 + 0\u00a0 —\u00a0 E<\/b><\/span><\/span>mM<\/b><\/span><\/span><\/sub>=5V<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 180=5V<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>=6m\/s<\/b><\/span><\/span><\/p>\n

c) conserva\u00e7\u00e3o da quantidade de movimento \u2013 antes de Maria sentar na poltrona \u2013 Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=m.V=10.6\u00a0 —\u00a0 Q<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=60kg.m\/s\u00a0 —\u00a0 depois que Maria sentou, o sistema Maria + poltrona movem-se juntos com velocidade V\u00a0 —\u00a0 Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=60.V\u00a0 —\u00a0 Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=Q<\/b><\/span><\/span>d\u00a0\u00a0<\/b><\/span><\/span><\/sub>—\u00a0 60=60V\u00a0 —\u00a0 <\/b><\/span><\/span>V=1m\/s<\/b><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=50.(+2) + 10.(-2)=80kg.m\/s\u00a0 —\u00a0 Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=50.(V) + 10.(-0,5)=50V \u2013 5\u00a0 —\u00a0 Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 80=50V \u2013 5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V=1,7m\/s R- D<\/b><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span>\u00a0R- A (veja teoria)<\/b><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span>\u00a0Velocidade de recuo do rifle\u00a0 —\u00a0 Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=15.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>.(+3.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>) + 5.(-V)\u00a0 —\u00a0 Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=45.10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\u00a0\u2013 5V\u00a0 —\u00a0 Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 0=45.10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\u00a0\u2013 5V —\u00a0\u00a0<\/b><\/span><\/span>V=0,9m\/s\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 velocidade de recuo do atirador\u00a0 —\u00a0 Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=15.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>.(+3.10<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>) + (95 +5).(-V)=45.10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\u00a0\u2013 100V\u00a0 —\u00a0 Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 0=45.10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>\/100V\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V=4,5.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>m\/s<\/b><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span>\u00a0a) Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=M<\/b><\/span><\/span>c.<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=10.5=50kg.m\/s\u00a0 —\u00a0 Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=(90 + 10).V=100V\u00a0 —\u00a0 Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 50=100V\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V=0,5m\/s<\/b><\/span><\/span><\/p>\n

b) E<\/b><\/span><\/span>ca<\/b><\/span><\/span><\/sub>=10.25\/2=125J\u00a0 —\u00a0 E<\/b><\/span><\/span>cd<\/b><\/span><\/span><\/sub>=100.0,25\/2=12,5J\u00a0 —\u00a0\u00a0\u2206E<\/b><\/span><\/span>c=<\/b><\/span><\/span><\/sub>E<\/b><\/span><\/span>cd<\/b><\/span><\/span><\/sub>\u00a0\u2013 E<\/b><\/span><\/span>ca<\/b><\/span><\/span><\/sub>= 12,5 \u2013 125\u00a0 —<\/b><\/span><\/span>\u00a0 <\/b><\/span><\/span><\/sub>\u2206E<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=-125J<\/b><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span>(<\/b><\/span><\/span>1) Verdadeira \u2013 desprezando-se os atritos ele est\u00e1 em queda livre sob a\u00e7\u00e3o exclusiva da for\u00e7a peso.<\/b><\/span><\/span><\/p>\n

(2) Velocidade com que ele chega ao solo \u2013 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2g.h\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>= 0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.10.0,5\u00a0 —\u00a0 V=\u221a10m\/s\u00a0 —\u00a0 m\u00f3dulo da acelera\u00e7\u00e3o durante o choque\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.a.\u00a0DS\u00a0 —\u00a0 0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=(\u221a10)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0-2.a.0,03\u00a0 —\u00a0 a=333m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 F=ma=70.333,3\u00a0 —\u00a0 F=23.310N (for\u00e7a que ele troca com o solo) que \u00e9 maior que o peso de 1.000kg (1.000 X 10=10.000N) \u2013 Verdadeira<\/b><\/span><\/span><\/p>\n

(3) Verdadeira \u2013 princ\u00edpio da a\u00e7\u00e3o e rea\u00e7\u00e3o<\/b><\/span><\/span><\/p>\n

(04) Verdadeira \u2013 (veja teoria).<\/b><\/span><\/span><\/p>\n

Todas verdadeiras\u00a0<\/b><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span><\/span>\u00a0a) Objeto A \u2013 queda livre com V<\/b><\/span><\/span>oA<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 a colis\u00e3o ocorre na altura de h=(45 \u2013 25)=20m\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=Vo<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.g.h\u00a0 —\u00a0 V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>= 0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.10.20\u00a0 —\u00a0 V=20m\/s\u00a0 —\u00a0 No encontro V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=20m\/s \u2013 antes da colis\u00e3o Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=M(+V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>) + M.(-V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>)=M.(+20) + M.(-20)\u00a0 —\u00a0 Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=2M.V<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>\u00a0—\u00a0 2M.V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 V<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>=0\u00a0 —\u00a0 a energia mec\u00e2nica dissipada na colis\u00e3o corresponde a E<\/b><\/span><\/span>mdepois<\/b><\/span><\/span><\/sub>\u00a0\u2013 E<\/b><\/span><\/span>mantes<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0DE<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=2M.V<\/b><\/span><\/span>AB<\/b><\/span><\/span><\/sub>\u00a0\u2013 (M.V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\/2 + M.V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\/2)= 2.4.0 \u2013(4.20<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 + 4.(-20)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2) \u00a0—\u00a0 —\u00a0\u00a0DE<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>= 0 \u2013(800 + 800)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>DE<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>= – 1600J<\/b><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

b) varia\u00e7\u00e3o de energia mec\u00e2nica\u00a0 —\u00a0 E<\/b><\/span><\/span>mA<\/b><\/span><\/span><\/sub>= E<\/b><\/span><\/span>pA\u00ad<\/b><\/span><\/span><\/sub>\u00a0+ E<\/b><\/span><\/span>cA<\/b><\/span><\/span><\/sub>\u00a0= mgh + 0= 8.10.25\u00a0 —\u00a0 E<\/b><\/span><\/span>mA<\/b><\/span><\/span><\/sub>=2.000J\u00a0 —\u00a0 E<\/b><\/span><\/span>md<\/b><\/span><\/span><\/sub>=E<\/b><\/span><\/span>cd<\/b><\/span><\/span><\/sub>\u00a0+ E<\/b><\/span><\/span>pd<\/b><\/span><\/span><\/sub>=E<\/b><\/span><\/span>cB\u00a0<\/b><\/span><\/span><\/sub>+ 0\u00a0 — E<\/b><\/span><\/span>mA<\/b><\/span><\/span><\/sub>=E<\/b><\/span><\/span>md\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/sub>E<\/b><\/span><\/span>cB<\/b><\/span><\/span><\/sub>=2.000J<\/b><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span>Orientando a trajet\u00f3ria como positiva no sentido do movimento da sonda (positivo para a esquerda)\u00a0 —\u00a0 antes \u2013 bola chegando \u00e0 sonda\u00a0 com velocidade de 2.000m\/s e a sonda inicialmente parada\u00a0 —\u00a0 Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>b\u00a0<\/b><\/span><\/span><\/sub>+ m<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>=10.2.000 + 1.000 X 0\u00a0 —\u00a0 Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=20.000kg.m\/s\u00a0 —\u00a0 depois \u2013 bola saindo com velocidade V<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>=- 400m\/s e a sonda se movendo para a esquerda com velocidade +V\u00a0 —\u00a0 Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>b<\/b><\/span><\/span><\/sub>\u00a0+ m<\/b><\/span><\/span>s<\/b><\/span><\/span><\/sub>.(+V)=10.( – 400) + 1.000V\u00a0 —\u00a0 Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>= – 4.000 + 1.000V\u00a0 —\u00a0 Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 20.000= – 4.000 + 1.000V\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

V=24m\/s\u00a0<\/b><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span>\u00a0Como a dire\u00e7\u00e3o e o sentido da velocidade coincidem com a dire\u00e7\u00e3o e sentido da quantidade de movimento, e como a quantidade de movimento do sistema antes da explos\u00e3o \u00e9 nula, depois da explos\u00e3o tamb\u00e9m dever\u00e1 ser nula. Assim, a \u00fanica alternativa em que todos os vetores se anulam \u00e9 a\u00a0<\/b><\/span><\/span>D.<\/b><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span>\u00a0Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 — \u00a0m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0e V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=Q, que somados vetorialmente fornecem\u00a0 Q<\/b><\/span><\/span>12<\/b><\/span><\/span><\/sub><\/p>\n

\"\"<\/p>\n

Como a quantidade de movimento antes da explos\u00e3o \u00e9 nula, a mesma, depois da explos\u00e3o tamb\u00e9m dever\u00e1 ser nula, o que s\u00f3 poder\u00e1 ocorrer se\u00a0\"\"\u00a0\u00a0anular\u00a0\"\"\u00a0\u00a0\u00a0<\/b><\/span><\/span>R- III<\/b><\/span><\/span><\/p>\n

25-<\/b><\/span><\/span><\/span>\u00a0Multiplicando as velocidades pelas respectivas massas obtemos as quantidades de movimento de cada elemento do sistema antes e depois da fiss\u00e3o<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe na figura acima que\u00a0\"\"\u00a0e\u00a0\"\"\u00a0se anulam, sobrando:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Em m\u00f3dulo\u00a0 —\u00a0 Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Q<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>=Q<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 m.v<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>=m\/3.v<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>v<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>=v\/3\u00a0 R- E<\/b><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span><\/span>\u00a0Sendo a quantidade de movimento da granada antes da explos\u00e3o vertical e para cima, sua quantidade de movimento depois da explos\u00e3o tamb\u00e9m dever\u00e1 ser vertical e para cima e na horizontal deve se anular.\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

27-<\/b><\/span><\/span><\/span>\u00a0C\u00e1lculo do m\u00f3dulo de Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0e de Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0imediatamente depois da explos\u00e3o\u00a0 —\u00a0\u00a0 Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>= 20.20=400kg.m\/s\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>= Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=400kg.m\/s<\/b><\/span><\/span>\u00a0 —\u00a0 somando vetorialmente\u00a0\"\"\u00a0e\u00a0\"\"\u00a0obtemos\u00a0\"\"\u00a0cujo\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

m\u00f3dulo \u00e9 calculado pela lei dos cossenos\u00a0 —\u00a0 Q<\/b><\/span><\/span>12<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0= Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0+ 2Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>cos120<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 Q<\/b><\/span><\/span>12<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>= 400<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup>\u00a0+ 400<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 2.400.400.(-1\/2)\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 Q<\/b><\/span><\/span>12<\/b><\/span><\/span><\/sub>=400kg.m\/s<\/b><\/span><\/span>\u00a0 —\u00a0 como a quantidade de movimento antes da explos\u00e3o \u00e9 Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=0, a quantidade de movimento depois da explos\u00e3o ser\u00e1 tamb\u00e9m nula (Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=0) e, para que isso ocorra\u00a0\"\"\u00a0deve anular\u00a0\"\"\u00a0, ou seja, deve ter a mesma intensidade, mesma dire\u00e7\u00e3o e sentido contr\u00e1rio (figura abaixo<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

R- D<\/b><\/span><\/span><\/p>\n

28-<\/b><\/span><\/span><\/span>\u00a0a) Quantidade de movimento do n\u00facleo de tr\u00edtio antes de se desintegrar\u00a0 —\u00a0 Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=0 (repouso)\u00a0 —\u00a0 ap\u00f3s a desintegra\u00e7\u00e3o \u2013 n\u00facleo do h\u00e9lio \u2013 Q<\/b><\/span><\/span>He<\/b><\/span><\/span><\/sub>3<\/b><\/span><\/span><\/sup>=12.10<\/b><\/span><\/span>-24<\/b><\/span><\/span><\/sup>kg.m\/s\u00a0 —\u00a0 eletron \u2013 Q<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>–\u00a0<\/b><\/span><\/span><\/sup>=6,0.10<\/b><\/span><\/span>-24<\/b><\/span><\/span><\/sup>kg.m\/s\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Sen60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sup>= Q<\/b><\/span><\/span><\/span>v\u00a0<\/b><\/span><\/span><\/span><\/sub>\/ Q<\/b><\/span><\/span><\/span>He<\/b><\/span><\/span><\/span><\/sub>3<\/b><\/span><\/span><\/span><\/sup>\u00a0 — \u00a0<\/b><\/span><\/span><\/span>\u221a<\/b><\/span><\/span>3\/2=Q<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/span><\/sub>\/12.10<\/b><\/span><\/span><\/span>-24<\/b><\/span><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/span><\/sub>=6,0.\u221a3.10<\/b><\/span><\/span><\/span>-24<\/b><\/span><\/span><\/span><\/sup>kg.m\/s<\/b><\/span><\/span><\/span><\/p>\n

b) Q<\/b><\/span><\/span><\/span>He<\/b><\/span><\/span><\/span><\/sub>3<\/b><\/span><\/span><\/span><\/sup>=m.V\u00a0 —\u00a0 12.10<\/b><\/span><\/span><\/span>-24<\/b><\/span><\/span><\/span><\/sup>=5,0.10<\/b><\/span><\/span><\/span>-27<\/b><\/span><\/span><\/span><\/sup>.V\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V=2,4.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup>m\/s<\/b><\/span><\/span><\/span><\/p>\n

29-<\/b><\/span><\/span><\/span>\u00a0\u00a0Antes\u00a0 —\u00a0 Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=m.v<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0<\/b><\/span><\/span><\/p>\n

Depois \u2013 Q<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=m\/3.3v<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=m.v<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\u2013 Q<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=m\/6.v<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0\u2013 Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=m\/2.v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/p>\n

\"\"<\/p>\n

a) Como a quantidade de movimento (\"\") antes da explos\u00e3o \u00e9 horizontal e para a direita, a quantidade de movimento de movimento (\"\") depois da explos\u00e3o tamb\u00e9m dever\u00e1 ser horizontal e para a direita. Assim,\u00a0\"\"\u00a0e\u00a0\"\"\u00a0devem se anular, ou seja\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

Q<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=Q<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 m.v<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=m\/2.v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=2.v<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/p>\n

b) Como os vetores verticais se anulam,\u00a0\"\"\u00a0deve ser igual a\u00a0\"\", ou seja,\u00a0 —\u00a0 m.v<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=m\/6.v<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>v<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=6.v<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 \u00a0<\/b><\/span><\/span><\/p>\n

c) Considerando a energia mec\u00e2nica como sendo apenas a cin\u00e9tica\u00a0 —\u00a0 antes \u2013<\/b><\/span><\/span>E<\/b><\/span><\/span>ma<\/b><\/span><\/span><\/sub>=m.v<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\/2\u00a0<\/b><\/span><\/span>\u00a0— depois\u00a0 —\u00a0 E<\/b><\/span><\/span>md<\/b><\/span><\/span><\/sub>=m\/3.(3v<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 + m\/6.(6v<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 + m\/2.(2v<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0<\/b><\/span><\/span>E<\/b><\/span><\/span>md<\/b><\/span><\/span><\/sub>=11m.v<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\/2<\/b><\/span><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>E<\/b><\/span><\/span>md<\/b><\/span><\/span><\/sub>>E<\/b><\/span><\/span>ma<\/b><\/span><\/span><\/sub>\u00a0– aumenta<\/b><\/span><\/span><\/p>\n

30-<\/b><\/span><\/span>\u00a0a) eixo X\u00a0 —\u00a0 antes\u00a0 —\u00a0 Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=m.V=6.V\u00a0 —\u00a0 depois\u00a0 —\u00a0 Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=Q<\/b><\/span><\/span>AX<\/b><\/span><\/span><\/sub>\u00a0+ Q<\/b><\/span><\/span>BX<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>+ m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.V\/2 + m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.2V<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Q<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=Q<\/b><\/span><\/span>d<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 6.V=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.V\/2 + m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.2V\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>12=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0+ 4m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 I<\/b><\/span><\/span>\u00a0 —\u00a0 m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0+ m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=6\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=6 \u2013 m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 II\u00a0\u00a0<\/b><\/span><\/span>—\u00a0 II em I\u00a0 —\u00a0 12=6 \u2013 m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0+ 4m<\/b><\/span><\/span>B\u00a0<\/b><\/span><\/span><\/sub>\u00a0—\u00a0<\/b><\/span><\/span><\/p>\n

m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=2kg<\/b><\/span><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=4kg<\/b><\/span><\/span><\/p>\n

b) na vertical a resultante \u00e9 nula\u00a0 —\u00a0 Q<\/b><\/span><\/span>AY<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>Y<\/b><\/span><\/span><\/sub>=2.V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Q<\/b><\/span><\/span>BY<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>Y<\/b><\/span><\/span><\/sub>=4.V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/p>\n

 <\/p>\n

\"\"<\/p>\n

Q<\/b><\/span><\/span>AY<\/b><\/span><\/span><\/sub>=Q<\/b><\/span><\/span>BY\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/sub>2V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>=4V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>A<\/b><\/span><\/span><\/sub>\/V<\/b><\/span><\/span>B<\/b><\/span><\/span><\/sub>=2<\/b><\/span><\/span><\/p>\n

31-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) Dados\u00a0 — \u00a0m<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u00a0= 60 kg\u00a0 —\u00a0 \u00a0m<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\u00a0= 80 kg\u00a0 —\u00a0 \u00a0v<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u00a0= 0,15 m\/s\u00a0 —\u00a0 como se trata de um sistema isolado, h\u00e1 conserva\u00e7\u00e3o do momento linear (quantidade de movimento) do sistema (Q)\u00a0 —\u00a0 Q<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>=Q<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Q<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>=0 (inicialmente em repouso)\u00a0 —\u00a0 Q<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u00a0+ m<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 Q<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>=Q<\/b><\/span><\/span>f<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 0= m<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>\u00a0+ m<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 60.V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=80.(0,15)\u00a0 —\u00a0 V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>= -0,2m\/s\u00a0 (sinal negativo significa que o astronauta recuou)\u00a0 —\u00a0<\/b><\/span><\/span>V=0,2m\/s<\/b><\/span><\/span><\/p>\n

|b) Ap\u00f3s o empurr\u00e3o, o momento linear do painel \u00e9\u00a0 —\u00a0 Q<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\u00a0= m<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\u00a0v<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\u00a0 = 80.(0,15) = 12 kg.m\/s\u00a0 —\u00a0 como a for\u00e7a aplicada pelo astronauta \u00e9 a respons\u00e1vel pela varia\u00e7\u00e3o da velocidade do painel, temos, pelo teorema do impulso\u00a0 —\u00a0 I<\/b><\/span><\/span>Fa<\/b><\/span><\/span><\/sub>=Q<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=12kg.m\/s\u00a0 —<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

O impulso \u00e9 numericamente igual \u00e0 \u00e1rea\u00a0 —\u00a0 I<\/b><\/span><\/span>Fa<\/b><\/span><\/span><\/sub>=\u00e1rea do trap\u00e9zio=(B + b).h\/2=(0,9 + 0,3).F<\/b><\/span><\/span>max<\/b><\/span><\/span><\/sub>\/2\u00a0 —\u00a0 I<\/b><\/span><\/span>Fa<\/b><\/span><\/span><\/sub>=0,6F<\/b><\/span><\/span>max<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 12=0,6F<\/b><\/span><\/span>max<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>F<\/b><\/span><\/span>max<\/b><\/span><\/span><\/sub>=20N<\/b><\/span><\/span><\/p>\n

32-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>01) Correta\u00a0 — trata-se de um sistema mecanicamente isolado (a resultante das for\u00e7as externas \u00e9 nula), portanto ocorre conserva\u00e7\u00e3o da quantidade de movimento ou momento linear\u00a0 —\u00a0 como a quantidade de movimento inicial do sistema \u00e9 nula, para ocorrer conserva\u00e7\u00e3o, o m\u00f3dulo da quantidade de movimento adquirida pelo homem deve ser igual ao m\u00f3dulo da quantidade de movimento adquirida pelo menino\u00a0 —\u00a0 M<\/b><\/span><\/span>H<\/b><\/span><\/span><\/sub>\u00a0v<\/b><\/span><\/span>H<\/b><\/span><\/span><\/sub>\u00a0= M<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>\u00a0v<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>\u00a0 — \u00a060.(0,3) = 30. v<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 v<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>\u00a0= 0,6 m\/s\u00a0 —\u00a0 a velocidade relativa de afastamento entre eles tem m\u00f3dulo\u00a0 —\u00a0 v<\/b><\/span><\/span>rel<\/b><\/span><\/span><\/sub>\u00a0= 0,6 + 0,3 = 0,9 m\/s\u00a0 —\u00a0 ent\u00e3o, 2 s ap\u00f3s o empurr\u00e3o, a dist\u00e2ncia (d) entre eles \u00e9:
\nd = v<\/b><\/span><\/span>rel.<\/b><\/span><\/span><\/sub>\u0394t = 0,9.(2)\u00a0 —\u00a0 d = 1,8 m.
\n02) Falsa\u00a0 —\u00a0 o sistema \u00e9 n\u00e3o-conservativo, pois homem e menino consomem energia de seus organismos, trocando for\u00e7as que realizam trabalho mec\u00e2nico, transferindo energia cin\u00e9tica ao sistema.
\n04) Falsa. For\u00e7as conservativas transformam energia potencial em cin\u00e9tica ou vice-versa\u00a0 —\u00a0 no caso h\u00e1 transforma\u00e7\u00e3o de energia qu\u00edmica dos organismos em energia cin\u00e9tica.
\n08) Correta\u00a0 —\u00a0 as for\u00e7as trocadas entre eles t\u00eam mesma intensidade, pois formam um par a\u00e7\u00e3o-rea\u00e7\u00e3o\u00a0 —\u00a0 como s\u00e3o for\u00e7as internas ao sistema, a resultante dessas for\u00e7as \u00e9 nula.<\/b><\/span><\/span><\/p>\n

16) Falsa\u00a0 —\u00a0 veja (01)<\/b><\/span><\/span><\/p>\n

32) Correta\u00a0 —\u00a0 apenas for\u00e7as externas alteram a quantidade de movimento do sistema, conforme afirma o teorema do impulso\u00a0 —\u00a0 \u00a0resultante das for\u00e7as externas \u00e9 igual \u00e0 varia\u00e7\u00e3o da quantidade de movimento do sistema.\u00a0<\/b><\/span><\/span><\/p>\n

R- (01 + 08 + 32) = 41<\/b><\/span><\/span><\/p>\n

33-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Trata-se de um sistema mecanicamente isolado, pois apenas for\u00e7as internas provocam varia\u00e7\u00f5es de velocidades\u00a0 —\u00a0 assim, ocorre conserva\u00e7\u00e3o da quantidade de movimento do sistema\u00a0 —\u00a0 como se trata de uma grandeza vetorial, as part\u00edculas\u00a0\"\"\u00a0e\u00a0\"\"devem ter velocidades de sentidos e de mesmo m\u00f3dulo, uma vez que as massas s\u00e3o iguais\u00a0 —\u00a0 essas velocidades tamb\u00e9m devem se anular, pois como a quantidade de movimento inicial \u00e9 nula, a final tamb\u00e9m dever\u00e1 ser nula\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

34-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>As varia\u00e7\u00f5es de velocidades na colis\u00e3o ocorrem somente pela intera\u00e7\u00e3o entre a massa m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0e a massa m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0formando, ent\u00e3o, um sistema mecanicamente isolado\u00a0 —\u00a0 assim, h\u00e1 conserva\u00e7\u00e3o da quantidade de movimento do sistema que engloba m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>, m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0e a mola\u00a0 —\u00a0 a compress\u00e3o m\u00e1xima da mola ocorre quando os dois corpos t\u00eam a mesma velocidade<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

35-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>a) Dados\u00a0 — \u00a0h<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0= 2 m\u00a0 —\u00a0 \u00a0H = 20 m\u00a0 — v<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0= 21 m\/s\u00a0 —\u00a0\u00a0 massa da flecha = m\u00a0 —\u00a0 \u00a0massa do alvo = 5 m\u00a0 —\u00a0 conserva\u00e7\u00e3o da energia mec\u00e2nica\u00a0 —\u00a0 c\u00e1lculo da velocidade com que a flecha atinge o alvo\u00a0 —\u00a0 E<\/b><\/span><\/span>mi<\/b><\/span><\/span><\/sub>=E<\/b><\/span><\/span>mf<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 mV<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\/2 + mgh=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 + mgH\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

multiplicando por 2 a equa\u00e7\u00e3o\u00a0\"\"\u00a0\u00a0—\u00a0 v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= 21<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 20(2 \u201320)\u00a0 —\u00a0 \u00a0v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>= 81\u00a0 — v = 9 m\/s\u00a0 —\u00a0 como a flecha atinge o alvo no ponto mais alto, no momento de impacto a velocidade do alvo \u00e9 nula\u00a0 —\u00a0 o sistema pode ser considerado mecanicamente isolado, e a colis\u00e3o \u00e9 inel\u00e1stica\u00a0 —\u00a0 assim, pela conserva\u00e7\u00e3o da quantidade de movimento (momento linear), sendo v\u2019 a velocidade final do sistema\u00a0 —\u00a0\u00a0\"\"\u00a0\u00a0—\u00a0\u00a0 mv = (m + 5m)v\u2019\u00a0 —\u00a0 \u00a0v = 6v\u2019\u00a0 —\u00a0 \u00a09 = 6v\u2019\u00a0 —\u00a0 \u00a0<\/b><\/span><\/span>v\u2019 = 1,5 m\/s.<\/b><\/span><\/span><\/p>\n

b) C\u00e1lculo do tempo de queda (t<\/b><\/span><\/span>q<\/b><\/span><\/span><\/sub>) do sistema\u00a0 —\u00a0 H=gt<\/b><\/span><\/span>q<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 t<\/b><\/span><\/span>q<\/b><\/span><\/span><\/sub>=\u221a2×20\/10\u00a0 —\u00a0 t<\/b><\/span><\/span>q<\/b><\/span><\/span><\/sub>=2s\u00a0 —\u00a0 na dire\u00e7\u00e3o horizontal voc\u00ea tem um movimento uniforme\u00a0 —\u00a0 X=V\u2019.t<\/b><\/span><\/span>q<\/b><\/span><\/span><\/sub>=1,5×2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>X=3m<\/b><\/span><\/span><\/p>\n

36-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Dados\u00a0 —\u00a0 \u00a0massa da mala \u2013 m<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>\u00a0= 20 kg\u00a0 —\u00a0 \u00a0velocidade da mala \u2013 v<\/b><\/span><\/span>l<\/b><\/span><\/span><\/sub>\u00a0= 4 m\/s\u00a0 —\u00a0 \u00a0massa do carrinho \u2013 m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0= 60 kg\u00a0 —\u00a0 \u00a0velocidade do carrinho \u2013 v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0= 0\u00a0 —\u00a0\u00a0 velocidade final do sistema (m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>) \u2013 V\u00a0 —\u00a0 observe que o sistema \u00e9 mecanicamente isolado apenas nadire\u00e7\u00e3o horizontal\u00a0 —\u00a0 assim, s\u00f3 h\u00e1 conserva\u00e7\u00e3o da Quantidade de Movimento (ou Momento Linear) apenas nessa dire\u00e7\u00e3o\u00a0 —\u00a0 Q<\/b><\/span><\/span>antes<\/b><\/span><\/span><\/sub>\u00a0= Q<\/b><\/span><\/span>depois<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0 m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>v<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0= (m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>)V\u00a0 —\u00a0 \u00a020(4) + 60(0) = (20 + 60)V\u00a0 —\u00a0 \u00a080 = 80V\u00a0 —\u00a0\u00a0 V = 1 m\/s\u00a0 —\u00a0 quanto a Energia Mec\u00e2nica, seria desnecess\u00e1rio c\u00e1lculo, pois pode-se analisar esse caso como uma colis\u00e3o inel\u00e1stica (os corpos seguem juntos), onde h\u00e1 dissipa\u00e7\u00e3o de Energia Mec\u00e2nica (a Energia Mec\u00e2nica s\u00f3 se conserva em choques perfeitamente el\u00e1sticos)\u00a0 — provando\u00a0 —\u00a0 como a altura n\u00e3o varia ocorre apenas varia\u00e7\u00e3o de energia cin\u00e9tica\u00a0 —\u00a0\u00a0 E<\/b><\/span><\/span>mi<\/b><\/span><\/span><\/sub>=m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>V<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\/2=20×4<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span>mi<\/b><\/span><\/span><\/sub>=160J\u00a0 —\u00a0 E<\/b><\/span><\/span>mf<\/b><\/span><\/span><\/sub>=(m<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0+ m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>)V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 =80×1<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=40J\u00a0 —\u00a0 a energia mec\u00e2nica passou de 120J para 40J, ou seja, diminuiu\u00a0 —<\/b><\/span><\/span>\u00a0 R- C<\/b><\/span><\/span><\/p>\n

37-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>Dados\u00a0 —\u00a0 M<\/b><\/span><\/span>G<\/b><\/span><\/span><\/sub>\u00a0= 300 g\u00a0 —\u00a0 M<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>\u00a0= 100 g\u00a0 —\u00a0 V<\/b><\/span><\/span>G<\/b><\/span><\/span><\/sub>\u00a0= 80 km\/h\u00a0 —\u00a0 V<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>\u00a0= 24 km\/h\u00a0 —\u00a0 antes da ca\u00e7a, os m\u00f3dulos das quantidades de movimento do gavi\u00e3o e do melro s\u00e3o, respectivamente\u00a0 —\u00a0 Q<\/b><\/span><\/span>G<\/b><\/span><\/span><\/sub>\u00a0= 300 (80) g.km\/h=24.000gkm\/h\u00a0 —\u00a0 Q<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>\u00a0= 100 (24) g. km\/h\u00a0 —Q<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>=2.400gkm\/h\u00a0 —\u00a0 como ocorre conserva\u00e7\u00e3o da quantidade de movimento no momento da ca\u00e7a, o vetor velocidade\u00a0\"\"\u00a0tem a<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

mesma dire\u00e7\u00e3o da quantidade de movimento do sistema gavi\u00e3o-melro\u00a0 —\u00a0 observe na figura que tg\u03b1=cateto oposto\/cateto adjacente\u00a0 —\u00a0 tg\u03b1=Q<\/b><\/span><\/span>G<\/b><\/span><\/span><\/sub>\/Q<\/b><\/span><\/span>M<\/b><\/span><\/span><\/sub>=24.000\/2.400\u00a0 —\u00a0 tg\u03b1=10\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- B<\/b><\/span><\/span><\/p>\n

 <\/p>\n

38- <\/b><\/span><\/span><\/span>\u00a0a) Se voc\u00ea considerar o sistema carro-\u00f4nibus como um sistema isolado, voc\u00ea pode utilizar o teorema da conserva\u00e7\u00e3o da<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0quantidade de movimento\u00a0 —\u00a0 Q<\/b><\/span><\/span>antes<\/b><\/span><\/span><\/sub>=M.V<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ m<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>.V<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>=9000.80 +1000.0\u00a0 —\u00a0 Q<\/b><\/span><\/span>antes<\/b><\/span><\/span><\/sub>=720000kg.m\/s\u00a0 —\u00a0 Q<\/b><\/span><\/span>depois<\/b><\/span><\/span><\/sub>=(M + m<\/b><\/span><\/span>a<\/b><\/span><\/span><\/sub>).V=10.000.V<\/b><\/span><\/span><\/p>\n

Q<\/b><\/span><\/span>depois<\/b><\/span><\/span><\/sub>=10000V\u00a0 —\u00a0 Q<\/b><\/span><\/span>antes<\/b><\/span><\/span><\/sub>\u00a0= Q<\/b><\/span><\/span>depois<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 720000=10000V\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V=72km\/h<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

b) Pela figura fornecida voc\u00ea pode determinar a intensidade da for\u00e7a lateral\u00a0\"\"\u00a0\u00a0—\u00a0 sen3<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=F<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>\/F<\/b><\/span><\/span>at<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 0,05=F<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>\/8000\u00a0 —\u00a0 F<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>=400N\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

A acelera\u00e7\u00e3o lateral\u00a0\"\"\u00a0do carro tem intensidade\u00a0 —\u00a0 F<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>=m.a<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 400=1600.a<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0a<\/b><\/span><\/span>L<\/b><\/span><\/span><\/sub>=0,25m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

 <\/p>\n

Voltar para os exerc\u00edcios<\/span><\/a><\/h3>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre Conserva\u00e7\u00e3o da Quantidade de Movimento \u00a0 \u00a001-\u00a0R- C\u00a0 (veja teoria) 02-\u00a0Qa=mV=(250 ).10\u00a0 —\u00a0\u00a0Qa=2.500kg.m\/s\u00a0 —\u00a0 Qd=mc.Vc\u00a0+ mt.Vt=50.(-10) + 200.(+Vt)\u00a0 —\u00a0\u00a0Qd= – 500 + 200.Vt\u00a0\u00a0—\u00a0 Qa=Qd\u00a0 —\u00a0 2.500= – 500 +200.Vt\u00a0\u00a0—\u00a0 Vt=15m\/s R- E 03-\u00a0(01) verdadeira (veja teoria)\u00a0 —\u00a0 (02) falsa \u2013 sua massa \u00e9 diferente da massa do tanque \u00a0—\u00a0 (04) verdadeira –\u00a0 (ambos<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1391,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"class_list":["post-1395","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1395","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/comments?post=1395"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1395\/revisions"}],"predecessor-version":[{"id":10816,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1395\/revisions\/10816"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1391"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1395"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}