{"id":2572,"date":"2016-10-26T03:52:49","date_gmt":"2016-10-26T03:52:49","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2572"},"modified":"2024-09-02T14:55:13","modified_gmt":"2024-09-02T14:55:13","slug":"dinamica-impulsiva-resolucao","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/raciocinio-logico\/dinamica-impulsiva\/dinamica-impulsiva-resolucao\/","title":{"rendered":"Din\u00e2mica Impulsiva – Resolu\u00e7\u00e3o"},"content":{"rendered":"

Resolu\u00e7\u00f5es comentadas dos exerc\u00edcios de vestibulares sobre <\/b><\/span><\/span><\/span><\/p>\n

Din\u00e2mica Impulsiva<\/b><\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Alguns autom\u00f3veis disp\u00f5e de um eficiente sistema\u00a0 de prote\u00e7\u00e3o para o motorista, que consiste de uma bolsa infl\u00e1vel de ar (air-bag). <\/b><\/span><\/span><\/span><\/p>\n

Essa bolsa \u00e9 automaticamente inflada quando o autom\u00f3vel sofre uma desacelera\u00e7\u00e3o s\u00fabita, de modo que a cabe\u00e7a e o t\u00f3rax do motorista, em vez de colidirem com o volante, colidem com a bolsa.<\/b><\/span><\/span><\/span>\"\"<\/p>\n

Veja pela express\u00e3o F.\u0394t = m.\u0394V, onde foram igualados o Impulso I=F.\u2206t com o momentum (quantidade de movimento), que\u00a0<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span>\u00a0e\u00a0<\/b><\/span><\/span><\/span>\u0394V\u00a0<\/b><\/span><\/span><\/span>s\u00e3o constantes (mesmo motorista e mesma varia\u00e7\u00e3o de velocidade) e assim, teremos:<\/b><\/span><\/span><\/span><\/p>\n

F.\u0394t=constante, portanto se voc\u00ea aumentar o intervalo de tempo do choque, estar\u00e1 diminuindo a intensidade da for\u00e7a m\u00e1xima que o motorista troca com o volante.<\/b><\/span><\/span><\/span><\/p>\n

Assim, a bolsa aumenta o intervalo de tempo da desacelera\u00e7\u00e3o da cabe\u00e7a do motorista, diminuindo, portanto, a for\u00e7a m\u00e1xima que atua sobre a cabe\u00e7a e o t\u00f3rax\u00a0\u00a0—\u00a0\u00a0<\/b><\/span><\/span><\/span>R- A.<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span> <\/b>\u201c O impulso da for\u00e7a resultante de um sistema de for\u00e7as que age sobre um corpo \u00e9 igual \u00e0 varia\u00e7\u00e3o da quantidade de movimento do corpo\u201d<\/b><\/span><\/span><\/span><\/p>\n

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

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

Unidades no SI:<\/b><\/span><\/span><\/span><\/p>\n

Impulso \u2013 newton vezes segundo (N.s)<\/b><\/span><\/span><\/span><\/p>\n

Quantidade de movimento \u2013 quilograma vezes metro por segundo (kg.m\/s)<\/b><\/span><\/span><\/span><\/p>\n

\"\"A rela\u00e7\u00e3o\u00a0\"\"\u00a0\u00e9 uma rela\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/span>vetorial.<\/b><\/span><\/span><\/span><\/p>\n

\"\"\u00a0A quantidade de movimento tamb\u00e9m \u00e9 chamada de momentum ou momento.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0F.\u0394t=m\u0394V\u00a0 —\u00a0 F.0,01=0,12.(6)\u00a0 —\u00a0 F=0,72\/0,01\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 F=72N\u00a0 R- D<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>P\u00e1ssaro- antes V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>=0 e depois ele adquire a mesma velocidade que o avi\u00e3o V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>=900\/3,6=250m\/s\u00a0 —\u00a0 F.\u0394t=m(V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0– V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>)\u00a0 — F.0,001=2.(250 \u2013 0)\u00a0 —\u00a0 F=500\/0,001\u00a0 —\u00a0<\/b><\/span><\/span><\/span>F=500.000N ou F=5,0.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>N<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe a figura abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

P=mg=10m\u00a0 —\u00a0 F=60.p=60.10m\u00a0 —\u00a0 F=600m\u00a0 —\u00a0 F.\u0394t=m(V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 V<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 600m.0,2=m(+3V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>) \u2013 (-V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 120=3V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=120\/4\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=30m\/s\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

05-<\/b><\/span><\/span><\/span>1- Verdadeira \u2013 quando ele deixa o degrau, est\u00e1 em queda livre e, nesse caso, a for\u00e7a resultante sobre ele \u00e9 seu peso e a varia\u00e7\u00e3o do seu momento linear (quantidade de movimento) \u00e9 I=P.\u0394t=m\u0394V=\u0394Q.<\/b><\/span><\/span><\/span><\/p>\n

2 Verdadeira \u2013 c\u00e1lculo da velocidade com que ele chega ao solo \u2013 queda livre com Vo=0 \u2013 Torricelli \u2013 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> =V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span> + 2.g.\u0394S\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> =0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> + 2.10.0,5\u00a0 —\u00a0 V=\"\"10 m\/s \u2013 velocidade com que chega ao solo\u00a0 —\u00a0 durante o choque com o solo sua velocidade varia de\u00a0\"\"10 m\/s a zero e o m\u00f3dulo dessa varia\u00e7\u00e3o \u00e9\u0394V=\"\"10 m\/s enquanto que sua altura diminui de\u00a0\u0394S=3cm=0,03m , numa desacelera\u00e7\u00e3o a de\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> =V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span> +2.(-a). \u0394S\u00a0 —\u00a0 02=(\"\"10)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span> -2.a 0,03\u00a0 —\u00a0 a=500\/3 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 for\u00e7a que ele troca com o solo e que produz essa desacelera\u00e7\u00e3o\u00a0 —\u00a0 F=m.a=70.500\/3\u00a0 —\u00a0 F\u00bb11.667N\u00a0 —\u00a0 peso correspondente \u00e0 massa de 1ton=1.000kg \u2013 P=1.000X10 \u2013 P=10.000N\u00a0 —\u00a0 portanto – F>P<\/b><\/span><\/span><\/span><\/p>\n

3- Verdadeira \u2013 \u00e9 a for\u00e7a de contato, que os p\u00e9s trocam com o solo.<\/b><\/span><\/span><\/span><\/p>\n

4- Verdadeira \u2013 (veja teoria)<\/b><\/span><\/span><\/span><\/p>\n

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

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados\u00a0 —\u00a0 \u00a0m = 0,8 kg\u00a0 — v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0= 93,6 km\/h = 26 m\/s\u00a0 —\u00a0 \u00a0v = 280,8 km\/h = 78 m\/s\u00a0 —\u00a0 o enunciado afirma que antes da colis\u00e3o a velocidade relativa entre a mola e o capacete \u00e9 V<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=78 \u2013 26=52m\/s, e no instante ap\u00f3s a colis\u00e3o a velocidade relativa \u00e9 nula\u00a0 —\u00a0 assim, a quantidade de movimento da mola ap\u00f3s a colis\u00e3o em rela\u00e7\u00e3o ao capacete \u00e9 nula\u00a0 —\u00a0 I=\u0394Q\u00a0 —\u00a0 F.\u0394t=m\u0394V\u00a0 —\u00a0 F.\u0394t=m.(V \u2013 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 F.0,026=0,8.(52 \u2013 0)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>F=1.600N<\/b><\/span><\/span><\/span><\/p>\n

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

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Sendo 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><\/span>a<\/b><\/span><\/span><\/sub><\/span>\u00a0(antes da explos\u00e3o)\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=mV=m.10\u00a0 —\u00a0<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=10m\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>\u00a0(depois da explos\u00e3o)\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=(m\/2)V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0+ (m\/2)V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=(mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0+ mV<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>)\/2\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>\u00a0= Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 10m=(mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0+ mV<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>)\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=20 m\/s\u00a0 —\u00a0\u00a0<\/b><\/span><\/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><\/span><\/p>\n

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

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O pescador deve correr para B, a fim de que a for\u00e7a que seus p\u00e9s exercem sobre a jangada a <\/b><\/span><\/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><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/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\u00a0devem 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><\/span>\u00a0R- A<\/b><\/span><\/span><\/span><\/p>\n

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

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

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

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

11-<\/b><\/span><\/span><\/span><\/p>\n

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

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

e=V<\/b><\/span><\/span><\/span>rdepois<\/b><\/span><\/span><\/sub><\/span>\/V<\/b><\/span><\/span><\/span>rantes<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 1=(V\u2019<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 10)\/(20 \u2013 10)\u00a0 —\u00a0 10=V\u2019<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 10\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V\u2019<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>=0<\/b><\/span><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Como a perda de energia cin\u00e9tica \u00e9 m\u00e1xima, o choque \u00e9 inel\u00e1stico\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=4.3 + 0\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=12kg.m\/s\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=(m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>2)<\/b><\/span><\/span><\/sub><\/span>.V\u2019=(4 + 2).V\u2019\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=6V\u2019\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 12=6V\u2019\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V\u2019=2m\/s<\/b><\/span><\/span><\/span><\/p>\n

b) E<\/b><\/span><\/span><\/span>ca<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=4.9\/2 + 0\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>ca<\/b><\/span><\/span><\/sub><\/span>=18J\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>cd<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>1V\u2019<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>V\u2019<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2=4.4\/2 + 2.4\/2\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>cd<\/b><\/span><\/span><\/sub><\/span>=12J\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u0394E<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=18 \u2013 12 = 6j<\/b><\/span><\/span><\/span><\/p>\n

Calcule a modifica\u00e7\u00e3o na velocidade do cometa e fa\u00e7a um coment\u00e1rio sobre a alega\u00e7\u00e3o da astr\u00f3loga russa.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

13- <\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>=1kg\/L=1kg\/10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>\u00e1gua=<\/b><\/span><\/span><\/sub><\/span>10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>kg\/m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=5.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.5.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>.10.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=25.1010m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>\u00e1gua<\/b><\/span><\/span><\/sub><\/span>=M<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\/V<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>=M<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\/25.10<\/b><\/span><\/span><\/span>10<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>—\u00a0 M<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=25.10<\/b><\/span><\/span><\/span>13<\/b><\/span><\/span><\/sup><\/span>kg<\/b><\/span><\/span><\/span>\u00a0\u00a0—\u00a0\u00a0<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sub><\/span>=100kg<\/b><\/span><\/span><\/span>\u00a0\u00a0 —\u00a0 antes – V<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=10km\/s\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=M<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0+ M<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sub><\/span>=25.10<\/b><\/span><\/span><\/span>13<\/b><\/span><\/span><\/sup><\/span>.10 + 0\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=25.10<\/b><\/span><\/span><\/span>14<\/b><\/span><\/span><\/sup><\/span>kg.m\/s\u00a0 choque inel\u00e1stico \u2013 movem se juntos com velocidade V\u2019\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=(M<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0+ M<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sub><\/span>)V\u2019=(25.10<\/b><\/span><\/span><\/span>13<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 100)V\u2019\u00a0 —\u00a0 como 100kg \u00e9 desprez\u00edvel em rela\u00e7\u00e3o a 25.10<\/b><\/span><\/span><\/span>13<\/b><\/span><\/span><\/sup><\/span>kg\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=25.10<\/b><\/span><\/span><\/span>13<\/b><\/span><\/span><\/sup><\/span>\u00a0V\u2019\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 25.10<\/b><\/span><\/span><\/span>14<\/b><\/span><\/span><\/sup><\/span>=25.10<\/b><\/span><\/span><\/span>13<\/b><\/span><\/span><\/sup><\/span>\u00a0V\u2019\u00a0 —\u00a0 V\u2019=10km\/s (observe que a velocidade do cometa, ap\u00f3s o choque com a nave praticamenten\u00e3o sofreu altera\u00e7\u00e3o e a astr\u00f3loga russa estava errada, pois essa altera\u00e7\u00e3o \u00e9 muito pequena e n\u00e3o afeta o equil\u00edbrio do Sistema Solar).<\/b><\/span><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/span>ca<\/b><\/span><\/span><\/sub><\/span>=3m<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 antes\u00a0 — Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>ca<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>ca<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>ca<\/b><\/span><\/span><\/sub><\/span>=3m<\/b><\/span><\/span><\/span>ca<\/b><\/span><\/span><\/sub><\/span>.10<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0 Q<\/b><\/span><\/span><\/span>ca<\/b><\/span><\/span><\/sub><\/span>=30m<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0\"\"\u00a0(veja figura)\u00a0 <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

—\u00a0 aplicando sen45<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=cateto oposto\/hipotenusa\u00a0 —\u00a0 sen45<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>=Q<\/b><\/span><\/span><\/span>ca<\/b><\/span><\/span><\/sub><\/span>\/Q<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 \u221a2\/2=30m<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\/m<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>c<\/b><\/span><\/span><\/sub><\/span>=60\/\u221a2 \u224860\/1,4\u224843m\/s X 3,6 \u2248145km\/h \u2013\u00a0<\/b><\/span><\/span><\/span>a declara\u00e7\u00e3o \u00e9 falsa, pois a velocidade do carro era aproximadamente de 145km\/h<\/b><\/span><\/span><\/span><\/p>\n

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

15-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Esquematizando a situa\u00e7\u00e3o e supondo que ap\u00f3s o choque, eles se movam para a direita. \"\"Aplicando o teorema da conserva\u00e7\u00e3o da quantidade de movimento, supondo velocidades positivas para a direita e negativas para a esquerda\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=Q<\/b><\/span><\/span><\/span>d\u00a0\u00a0<\/b><\/span><\/span><\/sub><\/span>—\u00a0 m<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>\u00a0= m<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>\u2019 + m<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>\u2019\u00a0 —\u00a0 13.(-3) + 15.(5) = 13.V<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>\u2019 + 15.V<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>\u2019\u00a0 —\u00a0 -39 +75 = 13.V<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>\u2019 + 15.V<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>\u2019\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>13.V<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>\u2019 + 15.V<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>\u2019=36 I<\/b><\/span><\/span><\/span>\u00a0 \u00a0—\u00a0 aplicando a express\u00e3o do coeficiente de restitui\u00e7\u00e3o \u2013 e=V<\/b><\/span><\/span><\/span>rdepois<\/b><\/span><\/span><\/sub><\/span>\/V<\/b><\/span><\/span><\/span>rantes\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0—\u00a0 3\/4 = (V<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>\u2019 \u2013 V<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>\u2019)\/(5 + 3)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>\u2019<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>\u2013 V<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>\u2019=6 II\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>resolvendo o sistema composto por I e II\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>\u2019=4,5m\/s (para a direita) e V<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>\u2019= -1,5m\/s (para a esquerda).<\/b><\/span><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Durante o choque eles trocam for\u00e7as que obedecem ao Princ\u00edpio da A\u00e7\u00e3o e Rea\u00e7\u00e3o, pois trocam for\u00e7as que tem a mesma intensidade, mesma dire\u00e7\u00e3o, mas sentidos contr\u00e1rios. Como essas for\u00e7as tem a mesma intensidade, F<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=F<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0e\u00a0<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/F<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=1<\/b><\/span><\/span><\/span><\/p>\n

b) Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.(10) + m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.(-6) = m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.(-3) + m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>(9)\u00a0 —\u00a0 13m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=15m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=15\/13<\/b><\/span><\/span><\/span><\/p>\n

c) e=V<\/b><\/span><\/span><\/span>rdepois<\/b><\/span><\/span><\/sub><\/span>\/V<\/b><\/span><\/span><\/span>rantes<\/b><\/span><\/span><\/sub><\/span>= (3 + 9)\/(10 + 6)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>e=3\/4<\/b><\/span><\/span><\/span><\/p>\n

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

17-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O choque que ocorre na posi\u00e7\u00e3o 1 \u00e9 parcialmente el\u00e1stico, pois n\u00e3o \u00e9 el\u00e1stico (do enunciado) e nem inel\u00e1stico (n\u00e3o se movem juntas ap\u00f3s o choque)\u00a0 —\u00a0 adotando o sentido hor\u00e1rio como positivo\u00a0 –<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

— Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0+ m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=3m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>Vo \u2013 m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=2m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=3m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>V + m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 2m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=3m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>V + m<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>V<\/b><\/span><\/span><\/span>o\u00a0\u00a0<\/b><\/span><\/span><\/sub><\/span>—\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=Vo\/3<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=3V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=3V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/sub><\/span>—\u00a0 equa\u00e7\u00e3o de cada bola\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.t=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.t\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.t=3V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.t<\/b><\/span><\/span><\/span><\/p>\n

Supondo que o encontro seja no instante t e no ponto P.<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Nesse instante, a bola A est\u00e1 na posi\u00e7\u00e3o S<\/b><\/span><\/span><\/span>A\u00a0<\/b><\/span><\/span><\/sub><\/span>e percorreu essa dist\u00e2ncia (S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>) enquanto que a bola B deu uma volta completa (2\"\"R)\u00a0 e percorreu mais (S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>) at\u00e9 o encontro\u00a0 —\u00a0 (S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>) + (2\"\"R) =S<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.t<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0+ 2\"\"R=3V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.t\u00a0 —\u00a0 t=\"\"R\/V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0I\u00a0 —\u00a0 substituindo I em\u00a0 S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.t=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.\"\"R\/V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>A\u00a0<\/b><\/span><\/span><\/sub><\/span>=\"\"R (percorreu meia circunfer\u00eancia a partir do ponto 1) ou S<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=3V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.t=3V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.\"\"R\/V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=3\"\"R (percorreu uma circunfer\u00eancia e meia a partir do ponto 1) \u2013\u00a0<\/b><\/span><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Como os caixotes tem a mesma massa, e os choques s\u00e3o perfeitamente el\u00e1sticos, em todos os choques eles trocam ou conservam suas velocidades<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe que ap\u00f3s o 3<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0choque, V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=0 e V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>o\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span>R- E<\/b><\/span><\/span><\/span><\/p>\n

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

19-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Colis\u00e3o perfeitamente el\u00e1stica de A com B e eles n\u00e3o trocam suas velocidades, pois suas massas s\u00e3o diferentes<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Q<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=Q<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 2.(10) + 1.(0) = 2V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0+\u00a0 1V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 2V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=20 I\u00a0 —\u00a0 e=V<\/b><\/span><\/span><\/span>rdepois<\/b><\/span><\/span><\/sub><\/span>\/V<\/b><\/span><\/span><\/span>rantes<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 1=(V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>-V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>)\/10\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=10 II\u00a0 —\u00a0 resolvendo I com II\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=10\/3m\/s e V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=40\/3m\/s<\/b><\/span><\/span><\/span><\/p>\n

B se desloca at\u00e9 a parede, onde chega com velocidade de 40\/3m\/s (n\u00e3o existe atrito) e retorna com a mesma velocidade (choque perfeitamente el\u00e1stico).<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Trata-se de um encontro de dois m\u00f3veis em MRU com o referencial (origem da trajet\u00f3ria em A) de equa\u00e7\u00f5es\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>A\u00a0<\/b><\/span><\/span><\/sub><\/span>= S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>t\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>A\u00a0<\/b><\/span><\/span><\/sub><\/span>= 0 + 10\/3.t\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>S<\/b><\/span><\/span><\/span>A\u00a0<\/b><\/span><\/span><\/sub><\/span>= 10\/3.t\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>S<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>= S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.t= 4 + (-40\/3).t\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>= 4 \u2013 40\/3.t\u00a0 —\u00a0 no encontro \u2013 AS=S<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 10\/3.t = 4 \u2013 40\/3.t\u00a0 —\u00a0 t=6,25s (tempo do encontro)\u00a0 —\u00a0 S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=10\/3.t=10\/3.6\/25\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/sub><\/span>S<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=0,8m (dist\u00e2ncia que A percorreu enquanto B voltava)\u00a0 —\u00a0 como B foi e voltou, A percorreu\u00a0 —\u00a0 \u0394S=0,8 (ida) + 0,8 (volta)\u00a0 —\u00a0 \u0394S=1,6m\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A previs\u00e3o de Pedro \u00e9 correta, pois a quantidade de movimento do sistema sempre se conserva, ou seja,\u00a0\"\".<\/b><\/span><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>O vetor quantidade de movimento antes do choque tem dire\u00e7\u00e3o horizontal e sentido para a direita. Como\u00a0\"\", e o choque \u00e9 el\u00e1stico, a \u00fanica alternativa que satisfaz \u00e9 a <\/b><\/span><\/span><\/span>C<\/b><\/span><\/span><\/span>, cuja soma vetorial tamb\u00e9m tem dire\u00e7\u00e3o horizontal e sentido para a direita e, como o choque \u00e9 el\u00e1stico e as bolas id\u00eanticas, a bola 1 tem que parar.<\/b><\/span><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span><\/p>\n

1. Sendo os atritos desprez\u00edveis e considerando o sistema conservativo voc\u00ea pode utilizar o teorema da conserva\u00e7\u00e3o da energia mec\u00e2nica.<\/b><\/span><\/span><\/span><\/p>\n

A. Energia mec\u00e2nica no ponto M onde V<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>=0 e h=1,35m\u00a0\u00a0\u00a0—\u00a0\u00a0E<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>cM<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>pM\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0\u00a0E<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m.g.h=m.0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m.10.1,25\u00a0\u00a0—\u00a0\u00a0E<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>=0 + 12,5m\u00a0\u00a0—\u00a0\u00a0E<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>=12,5m\u00a0\u00a0—\u00a0\u00a0energia mec\u00e2nica no ponto N onde h=0 e V<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>=?\u00a0\u00a0—\u00a0\u00a0E<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>cN<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>pN\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

–\u00a0\u00a0E<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m.g.h=m.V<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m.10.0\u00a0\u00a0—\u00a0\u00a0E<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 0\u00a0\u00a0—\u00a0\u00a0E<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0\u00a0—\u00a0\u00a0pelo teorema da conserva\u00e7\u00e3o da energia mec\u00e2nica\u00a0\u00a0—\u00a0E<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>N\u00a0\u00a0<\/b><\/span><\/span><\/sub><\/span>—\u00a0\u00a012,5m = mV<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0\u00a0—\u00a0\u00a0V<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>=\u221a25\u00a0\u00a0—\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>=5m\/s.<\/b><\/span><\/span><\/span>\"\"<\/p>\n

B. Energia mec\u00e2nica no ponto P onde V<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>=? e h=0\u00a0\u00a0\u00a0—\u00a0\u00a0E<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>cP<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>pP\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0\u00a0E<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m.g.h=m.V<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m.10.0\u00a0\u00a0—\u00a0\u00a0E<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0\u00a0—\u00a0energia mec\u00e2nica no ponto Q onde h=0,8m e V<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/sub><\/span>=0\u00a0\u00a0—\u00a0\u00a0E<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>cQ<\/b><\/span><\/span><\/sub><\/span>\u00a0+ E<\/b><\/span><\/span><\/span>pQ\u00a0<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0\u00a0E<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/sub><\/span>=mV<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m.g.h\u00a0\u00a0— E<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/sub><\/span>=m.0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + m.10.0,8\u00a0\u00a0—\u00a0\u00a0E<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/sub><\/span>= 0 + 8m\u00a0\u00a0—\u00a0\u00a0E<\/b><\/span><\/span><\/span>Q<\/b><\/span><\/span><\/sub><\/span>=8m\u00a0\u00a0—\u00a0\u00a0pelo teorema da conserva\u00e7\u00e3o da energia mec\u00e2nica\u00a0\u00a0—\u00a0\u00a0E<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>Q\u00a0\u00a0<\/b><\/span><\/span><\/sub><\/span>—<\/b><\/span><\/span><\/span><\/p>\n

mV<\/b><\/span><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/span><\/sup><\/span>\/2 = 8m\u00a0\u00a0—\u00a0\u00a0V<\/b><\/span><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/sub><\/span>=\u221a16\u00a0\u00a0—\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span><\/span>P<\/b><\/span><\/span><\/span><\/sub><\/span>=4m\/s.<\/b><\/span><\/span><\/span><\/span><\/p>\n

2. C\u00e1lculo do coeficiente de restitui\u00e7\u00e3o na primeira colis\u00e3o com o solo onde a velocidade antes da colis\u00e3o \u00e9 V<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=5m\/s e a velocidade depois da colis\u00e3o \u00e9 V<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=4m\/s\u00a0\u00a0—\u00a0\u00a0pelo enunciado e=V<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>\/V<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=4\/5\u00a0\u00a0—\u00a0\u00a0e=0,8\u00a0\u00a0—\u00a0\u00a0na segunda colis\u00e3o a velocidade com que ela sai do solo e atinge a altura de h=0,8m \u00e9 a mesma com que ela chega ao solo a partir dessa altura\u00a0\u00a0—\u00a0\u00a0V\u2019<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=4m\/s\u00a0\u00a0—\u00a0\u00a0ela abandona o solo com V\u2019<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>=?\u00a0\u00a0—\u00a0\u00a0usando novamente o coeficiente de restitui\u00e7\u00e3o e=0,8\u00a0\u00a0—<\/b><\/span><\/span><\/span>\"\"<\/p>\n

e=V\u2019<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>\/V\u2019<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0\u00a00,8=V\u2019<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>\/4\u00a0\u00a0—\u00a0<\/b><\/span><\/span><\/span>\u00a0V\u2019<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=3,2m\/s.<\/b><\/span><\/span><\/span><\/p>\n

3.\u00a0\u00a0C\u00e1lculo das quantidades de movimento da bolinha 1 antes e depois da primeira colis\u00e3o, orientando, a partir do solo a trajet\u00f3ria para cima\u00a0\u00a0—\u00a0Q<\/b><\/span><\/span><\/span>1antes<\/b><\/span><\/span><\/sub><\/span>=m.(-V<\/b><\/span><\/span><\/span>N<\/b><\/span><\/span><\/sub><\/span>)=m.(-5)= – 5m\u00a0\u00a0—<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

–\u00a0\u00a0Q<\/b><\/span><\/span><\/span>1depois<\/b><\/span><\/span><\/sub><\/span>=m.(+V<\/b><\/span><\/span><\/span>P<\/b><\/span><\/span><\/sub><\/span>)=m.4=4m\u00a0\u00a0—\u00a0\u00a0somando vetorialmente \"\"\u00a0—\u00a0\u00a0Q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>= 4m \u2013 (-5m)\u00a0\u00a0—\u00a0\u00a0Q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=9m=\u2206Q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0\u00a0analogamente,\u00a0calculando as quantidades de movimento da bolinha 2 antes e depois da colis\u00e3o e, como ela foi abandonada da mesma altura que a bolinha 1 ela chega ao solo com a mesma velocidade de V<\/b><\/span><\/span><\/span>2antes<\/b><\/span><\/span><\/sub><\/span>=5m\/s e p\u00e1ra V<\/b><\/span><\/span><\/span>2depois<\/b><\/span><\/span><\/sub><\/span>=0\u00a0\u00a0—\u00a0\u00a0orientando, a partir do solo a trajet\u00f3ria para cima\u00a0\u00a0—\u00a0\u00a0Q<\/b><\/span><\/span><\/span>2antes<\/b><\/span><\/span><\/sub><\/span>=m.(-V<\/b><\/span><\/span><\/span>2antes<\/b><\/span><\/span><\/sub><\/span>)=m.(-5)= – 5m\u00a0\u00a0—\u00a0\u00a0Q<\/b><\/span><\/span><\/span>2depois<\/b><\/span><\/span><\/sub><\/span>=m.0=0 —\u00a0\u00a0somando-os vetorialmente voc\u00ea obter\u00e1\u00a0\u00a0—\u00a0\u00a0Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=0 \u2013 (-5m)\u00a0\u00a0—\u00a0\u00a0Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=5m=\u2206Q<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0—\u00a0\u00a0pelo teorema do impulso\u00a0\u00a0—\u00a0\u00a0I=\u2206Q=F.\u2206t\u00a0\u00a0—\u00a0\u00a0F= \u2206Q\/\u2206t\u00a0\u00a0—\u00a0\u00a0F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=\u2206Q<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/\u2206t\u00a0\u00a0—\u00a0\u00a0F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=9m\/\/\u2206t\u00a0\u00a0—\u00a0\u00a0F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=5m\/\/\u2206t\u00a0\u00a0—\u00a0\u00a0sendo o intervalo de tempo das colis\u00f5es os mesmos\u00a0\u00a0—\u00a0<\/b><\/span><\/span><\/span>\u00a0F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0> F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Quantidade de movimento de cada grupo de 3 alunos de massa m com velocidade de 5m\/s\u00a0\u00a0—\u00a0\u00a0Q=Mv=3m5=15m\u00a0\u00a0— quantidade de movimento do quarto grupo com 6 alunos e com velocidade V\u00a0\u00a0—\u00a0\u00a0Q\u2019=6mV\u00a0\u00a0—\u00a0\u00a0a velocidade V, em m\u00f3dulo, desse grupo de 6 alunos para satisfazer a coreografia do professor, deve ser tal que a quantidade de movimento total (grandeza vetorial) deve ser nula\u00a0\u00a0—\u00a0\u00a0para que isso ocorra (veja figura) Q\u2019 = Q\u00a0\u00a0—\u00a0\u00a06mV = 15m\u00a0\u00a0—\u00a0\u00a0V=2,5m\/s\u00a0\u00a0—<\/b><\/span><\/span><\/span>\"\"<\/p>\n

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

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

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

 <\/p>\n

 <\/p>\n

 <\/p>\n

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

Resolu\u00e7\u00f5es comentadas dos exerc\u00edcios de vestibulares sobre Din\u00e2mica Impulsiva 01-\u00a0Alguns autom\u00f3veis disp\u00f5e de um eficiente sistema\u00a0 de prote\u00e7\u00e3o para o motorista, que consiste de uma bolsa infl\u00e1vel de ar (air-bag). Essa bolsa \u00e9 automaticamente inflada quando o autom\u00f3vel sofre uma desacelera\u00e7\u00e3o s\u00fabita, de modo que a cabe\u00e7a e o t\u00f3rax do motorista, em vez de colidirem com o volante, colidem<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2570,"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-2572","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2572","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=2572"}],"version-history":[{"count":4,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2572\/revisions"}],"predecessor-version":[{"id":10972,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2572\/revisions\/10972"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2570"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=2572"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}