{"id":4883,"date":"2020-01-10T03:13:28","date_gmt":"2020-01-10T03:13:28","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=4883"},"modified":"2024-09-04T14:38:22","modified_gmt":"2024-09-04T14:38:22","slug":"resolucao-comentada-dinamica-impulsiva-2019-2018","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/vestibulares-recentes\/mecanica\/dinamica-impulsiva-2018-2019\/resolucao-comentada-dinamica-impulsiva-2019-2018\/","title":{"rendered":"Resolu\u00e7\u00e3o Comentada \u2013 Din\u00e2mica Impulsiva \u2013 2019\/2018"},"content":{"rendered":"

Resolu\u00e7\u00e3o Comentada \u2013 Din\u00e2mica Impulsiva \u2013 2019\/2018<\/span><\/span><\/span><\/p>\n

01-<\/span><\/span><\/span><\/p>\n

Express\u00f5es matem\u00e1ticas<\/b><\/span><\/span>\u00a0dos\u00a0<\/b><\/span><\/span>dois tipos de energias envolvidos no fen\u00f4meno<\/b><\/span><\/span><\/span><\/span><\/p>\n

<\/p>\n

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

02-<\/span><\/span><\/span><\/p>\n


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

<\/p>\n

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

03-<\/span><\/span><\/span><\/p>\n

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04-<\/span><\/span><\/span><\/p>\n

A) Correta\u00a0<\/b><\/span><\/span>\u2013 A\u00a0<\/b><\/span><\/span>for\u00e7a resultante \u00e9 a for\u00e7a centr\u00edpeta (Fc)\u00a0<\/b><\/span><\/span>de\u00a0<\/b><\/span><\/span>dire\u00e7\u00e3o radial e sentido para o centro da<\/b><\/span><\/span><\/span><\/span><\/p>\n

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circunfer\u00eancia\u00a0<\/b><\/span><\/span>e, no caso,\u00a0<\/b><\/span><\/span>\u00e9 apenas a parcela horizontal de T (Tx) que n\u00e3o produz rota\u00e7\u00e3o,\u00a0<\/b><\/span><\/span>que \u00e9\u00a0<\/b><\/span><\/span>produzida apena pela parcela vertical de T (Ty).\u00a0<\/b><\/span><\/span>Assim,\u00a0<\/b><\/span><\/span>o momento gerado pela for\u00e7a resultante (for\u00e7a centr\u00edpeta) \u00e9 nulo.<\/b><\/span><\/span><\/span><\/span><\/p>\n

B) Falsa\u00a0<\/b><\/span><\/span>\u2013 a\u00a0<\/b><\/span><\/span>quantidade de movimento possui m\u00f3dulo Q = m.V constante.\u00a0<\/b><\/span><\/span>Mas sua\u00a0<\/b><\/span><\/span>dire\u00e7\u00e3o e sentido muda de acordo com a velocidade que \u00e9 tangente \u00e0 trajet\u00f3ria em cada ponto.<\/b><\/span><\/span><\/span><\/span><\/p>\n

C) Falsa\u00a0<\/b><\/span><\/span>\u2013 veja (A)<\/b><\/span><\/span><\/span><\/span><\/p>\n

D) Falsa\u00a0<\/b><\/span><\/span>\u2013 veja (A)<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

05-<\/span><\/span><\/span><\/p>\n

<\/p>\n


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

C\u00e1lculo da\u00a0<\/b><\/span><\/span><\/span>intensidade da velocidade inicial Vo<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>do\u00a0<\/b><\/span><\/span><\/span>sistema\u00a0<\/b><\/span><\/span><\/span>sabendo que ele est\u00e1\u00a0<\/b><\/span><\/span><\/span>freando\u00a0<\/b><\/span><\/span><\/span>com acelera\u00e7\u00e3o\u00a0<\/b><\/span><\/span><\/span>a = \u2013 5 m\/s2\u00a0at\u00e9 parar V = 0\u00a0<\/b><\/span><\/span><\/span>ap\u00f3s se\u00a0<\/b><\/span><\/span><\/span>deslocar de \u2206S = 10 m.<\/b><\/span><\/span><\/span><\/p>\n

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Equa\u00e7\u00e3o de Torricelli\u00a0\u00a0V2\u00a0= Vo2\u00a0+ 2.a. \u2206S\u00a0\u00a0<\/b><\/span><\/span><\/span>02\u00a0= Vo2\u00a0\u2013 2.5.10\u00a0Vo2\u00a0= 100\u00a0<\/b><\/span><\/span><\/span>\u00a0Vo\u00a0= 10 m\/s\u00a0<\/b><\/span><\/span><\/span>(velocidade dos dois ve\u00edculos juntos ap\u00f3s o impacto).<\/b><\/span><\/span><\/span><\/p>\n

Agora voc\u00ea deve\u00a0<\/b><\/span><\/span><\/span>aplicar<\/b><\/span><\/span><\/span>\u00a0o\u00a0<\/b><\/span><\/span><\/span>princ\u00edpio da conserva\u00e7\u00e3o da quantidade de moviment<\/b><\/span><\/span><\/span>o imediatamente\u00a0<\/b><\/span><\/span><\/span>antes<\/b><\/span><\/span><\/span>\u00a0e imediatamente\u00a0<\/b><\/span><\/span><\/span>ap\u00f3s o impacto<\/b><\/span><\/span><\/span>.<\/b><\/span><\/span><\/span><\/p>\n

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Antes<\/b><\/span><\/span><\/span>\u00a0\u00a0<\/b><\/span><\/span><\/span>Qsa\u00a0= mc.Vc\u00a0+ ms.Vs<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= 2000.Vc\u00a0+ 1000.0 = 2000Vc\u00a0+ 0\u00a0\u00a0<\/b><\/span><\/span><\/span>Qsa\u00a0= 2000Vc.<\/b><\/span><\/span><\/span><\/p>\n

Depois\u00a0\u00a0Qsd\u00a0= M.Vo<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= 3000.10\u00a0\u00a0<\/b><\/span><\/span><\/span>Qsd\u00a0= 30000 kgm\/s.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0R- E<\/b><\/span><\/span><\/span><\/p>\n

06-<\/span><\/span><\/span><\/p>\n

Considerando o\u00a0<\/b><\/span><\/span><\/span>sistema isolado\u00a0<\/b><\/span><\/span><\/span>de for\u00e7as externas voc\u00ea pode\u00a0<\/b><\/span><\/span><\/span>utilizar o princ\u00edpio da conserva\u00e7\u00e3o da quantidade de movimento do sistema imediatamente antes e imediatamente depois da explos\u00e3o.<\/b><\/span><\/span><\/span><\/p>\n

Lembre-se que neste ponto\u00a0<\/b><\/span><\/span><\/span>(altura m\u00e1xima) a velocidade VH\u00a0\u00e9 apenas a componente horizontal\u00a0<\/b><\/span><\/span><\/span>das velocidades vertical e horizontal do lan\u00e7amento obl\u00edquo. Assim ela tem\u00a0<\/b><\/span><\/span><\/span>m\u00f3dulo VH, dire\u00e7\u00e3o horizontal e sentido para a direita.<\/b><\/span><\/span><\/span><\/p>\n

Vamos\u00a0<\/b><\/span><\/span><\/span>supor que, ap\u00f3s a explos\u00e3o o fragmento A retorne com velocidade Va\u00a0<\/b><\/span><\/span><\/span>(veja figura).<\/b><\/span><\/span><\/span><\/p>\n

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Quantidade de movimento do sistema antes da explos\u00e3o\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Qsa\u00a0= (MA\u00a0+ MB).VH<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>= (2MB\u00a0+ MB).VH\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Qsa\u00a0= 3MB.VH.<\/b><\/span><\/span><\/span><\/p>\n

Quantidade de movimento do sistema depois da explos\u00e3o\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Qsd\u00a0= MA.(-VA) + MB.(VB)\u00a0<\/b><\/span><\/span><\/span>= \u2013 2MB.VA\u00a0+ MB.(5VH)\u00a0\u00a0<\/b><\/span><\/span><\/span>Qsd\u00a0= \u2013 2MB.VA\u00a0+ 5MB.VH.<\/b><\/span><\/span><\/span><\/p>\n

Qsa\u00a0= Qsd\u00a0\u00a03MB.VH\u00a0= \u2013 2MB.VA\u00a0+ 5MB.VH\u00a0<\/b><\/span><\/span><\/span>\u00a03VH\u00a0\u2013 5VH\u00a0= \u2013 2VA\u00a0\u00a0\u2013 2VH\u00a0= \u2013 2VA\u00a0\u00a0<\/b><\/span><\/span><\/span>VA\u00a0= VH.<\/b><\/span><\/span><\/span><\/p>\n

Ent\u00e3o o\u00a0<\/b><\/span><\/span><\/span>fragmento A retorna com velocidade VH<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>possuindo\u00a0<\/b><\/span><\/span><\/span>trajet\u00f3ria invertida do lan\u00e7amento obl\u00edquo,\u00a0<\/b><\/span><\/span><\/span>percorrendo o\u00a0<\/b><\/span><\/span><\/span>mesmo caminho na volta e atingindo o ponto II.<\/b><\/span><\/span><\/span><\/p>\n

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

07-<\/span><\/span><\/span><\/p>\n

<\/p>\n

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Pelo princ\u00edpio da conserva\u00e7\u00e3o da quantidade de movimento, em todo tipo de choques,\u00a0<\/b><\/span><\/span><\/span>a\u00a0<\/b><\/span><\/span><\/span>quantidade de movimento do sistema antes do choque Qsa<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00e9 igual \u00e0 quantidade de movimento do sistema depois do choque Qsd.<\/b><\/span><\/span><\/span><\/p>\n

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08-<\/span><\/span><\/span><\/p>\n

Sempre<\/b><\/span><\/span><\/span>\u00a0que a colis\u00e3o \u00e9\u00a0<\/b><\/span><\/span><\/span>perfeitamente el\u00e1stica\u00a0<\/b><\/span><\/span><\/span>e os\u00a0<\/b><\/span><\/span><\/span>corpos possuem a mesma massa,\u00a0<\/b><\/span><\/span><\/span>eles\u00a0<\/b><\/span><\/span><\/span>trocam<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

suas velocidades,\u00a0<\/b><\/span><\/span><\/span>ou seja, se o\u00a0<\/b><\/span><\/span><\/span>bloco A atinge o B (VB\u00a0= 0) com velocidade VA, imediatamente ap\u00f3s a colis\u00e3o, A para (VA) = 0 e B segue com velocidade VB\u00a0= VA.<\/b><\/span><\/span><\/span><\/p>\n

Desprezando os atritos e a resist\u00eancia do ar\u00a0<\/b><\/span><\/span><\/span>o\u00a0<\/b><\/span><\/span><\/span>sistema pode ser considerado conservativo\u00a0<\/b><\/span><\/span><\/span>(energia<\/b><\/span><\/span><\/span><\/p>\n

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mec\u00e2nica = energia cin\u00e9tica + energia potencial gravitacional) e assim o\u00a0<\/b><\/span><\/span><\/span>bloco B s\u00f3 pararia numa altura H. Como h < H, ele chega at\u00e9 o ponto P.<\/b><\/span><\/span><\/span><\/p>\n

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

09-<\/span><\/span><\/span><\/p>\n

Como elas\u00a0<\/b><\/span><\/span><\/span>ficam grudadas ap\u00f3s a colis\u00e3o<\/b><\/span><\/span><\/span>\u00a0o choque \u00e9\u00a0<\/b><\/span><\/span><\/span>inel\u00e1stico<\/b><\/span><\/span><\/span>\u00a0e elas se\u00a0<\/b><\/span><\/span><\/span>movem, ap\u00f3s o choque<\/b><\/span><\/span><\/span>, unidas com a\u00a0<\/b><\/span><\/span><\/span>mesma velocidade v\u2019\u00a0<\/b><\/span><\/span><\/span>(veja figura abaixo).<\/b><\/span><\/span><\/span><\/p>\n

<\/p>\n

<\/p>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o Comentada \u2013 Din\u00e2mica Impulsiva \u2013 2019\/2018 01- Express\u00f5es matem\u00e1ticas\u00a0dos\u00a0dois tipos de energias envolvidos no fen\u00f4meno R- B 02- R- B 03- 04- A) Correta\u00a0\u2013 A\u00a0for\u00e7a resultante \u00e9 a for\u00e7a centr\u00edpeta (Fc)\u00a0de\u00a0dire\u00e7\u00e3o radial e sentido para o centro da circunfer\u00eancia\u00a0e, no caso,\u00a0\u00e9 apenas a parcela horizontal de T (Tx) que n\u00e3o produz rota\u00e7\u00e3o,\u00a0que \u00e9\u00a0produzida apena pela parcela vertical de T<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":4881,"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-4883","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4883","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=4883"}],"version-history":[{"count":1,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4883\/revisions"}],"predecessor-version":[{"id":4884,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4883\/revisions\/4884"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/4881"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=4883"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}